text:
\section{Introduction}
In 1988, Cameron and Erd\H{o}{s} \cite{CameronEdros} conjectured that the maximum size of a sum-free subset $S\subseteq\left\{1,2,...,n\right\}$ is $O(2^{\frac{n}{2}})$, where ``sum-free'' refers to the condition that no two distinct elements in $S$ add to a third distinct element contained in $S$. This conjecture has since been upgraded to a theorem \cite{Green} \cite{Sapozhenko}, and recently a generalization of this sum-free property has arisen in the form of $(k,\ell)$-sumfree sets.

Calkin and Thomson first defined $(k,\ell)$-sumfree sets in \cite{CalkinThomson} as sets $S\subseteq\mathbb{N}$ such that the equation
$$x_1+x_2+\cdots+ x_k = y_1+y_2+\cdots+y_\ell$$
has no solutions with all $x_i,y_i\in S$.  Notice that when $k=2$ and $\ell=1$, the definition agrees with the classical definition of a sum-free set. These sets and their close variants have been studied by many authors; see e.g. \cite{BierChin} for an early study of $(k,\ell)$-sumfree sets in cyclic groups,
and \cite{menu-research-problems} for a detailed survey of recent directions.

We will consider the case of $(k,\ell)$-sumfree sets in $\mathbb{Z}/n\mathbb{Z}$ where $k=0$, which we call \textbf{$\bm{\ell}$-zero-sumfree sets} and denote by $\Delta_{n,\ell}$. Explicitly,
$$\Delta_{n,\ell} \coloneqq \Big\{\{s_1,\dots, s_k\}\subset{\mathbb{Z}/n\mathbb{Z}}:c_1s_1 + \cdots + c_ks_k\neq 0\text{ whenever } \sum_{i=1}^{k}c_i = \ell\Big\}.$$
This case appears to have qualitatively different behavior than when both $k$ and $\ell$ are strictly positive. For instance, although Bajnok and Matzke \cite{BajnokMatzke}, following on the work of \cite{HamidounePlagne}, recently determined an explicit formula for the maximum size of a $(k,\ell)$-sumfree set in $\mathbb{Z}/n\mathbb{Z}$ for any $k,\ell>0$, their methods do not directly extend to the $k=0$ case.

We are concerned with understanding global properties of $\Delta_{n,\ell}$. To this end, it is extremely useful to have a complete classification of all $\ell$-zero-sumfree sets. We are able to observe numerous properties of $\Delta_{n,\ell}$ for variations on parameters $n$ and $\ell$ by the use of an algorithm which we will describe in detail in the following section. In particular, we completely describe $\Delta_{n,\ell}$ for three families of parameters:

\begin{enumerate}
    \item a ``doubling'' class with parameters $n=2^{m+1}\rho$ and $\ell = 2^m\rho$, for odd $\rho\geq{3}$,
    \item a ``prime powers'' class with parameters $n = p^e$ and $\ell = p^e-1$ for prime $p\geq{2}$ and exponent $e\geq{1}$, and
    \item three ``arms and legs'' classes with parameters $n = 2p$, $\ell = 2p-s$ for prime $p\geq{3}$ and $s\in\left\{1,2,3\right\}$.
\end{enumerate}

One benefit of the complete descriptions of the families introduced in this paper is that we are also able to compute naturally related objects with relative ease. For instance, from each $\Delta_{n,\ell}$ we may construct an object called a \textbf{subspace arrangement}. In general, a subspace arrangement is simply a collection $\mathcal{A}$ of finitely many subspaces $S$ of some vector space. In particular, for each of the $\Delta_{n,\ell}$ above, we compute two common invariants of their associated subspace arrangements: the characteristic polynomial and the intersection poset.

\section{Simplicial Complexes}

Given a set $V$, a \textbf{simplicial complex} (or just a \textbf{complex}) on $V$ is a nonempty collection $\Gamma$ of subsets of $V$ which is closed under inclusion; that is, if $S\in\Gamma$ and $T\subseteq S$, then $T\in\Gamma$ as well. Elements of simplicial complexes are called \textbf{faces}, and faces which are maximal under inclusion are called \textbf{facets}. For any face $S\in\Gamma$, its \textbf{dimension} is defined to be $\dim(S)=|S|-1$, and then the \textbf{dimension} of the complex itself is defined to be $\dim(\Gamma) = \displaystyle\max_{S\in\Gamma} \dim(S)$.

Notice that simplicial complexes are entirely determined by their (nonempty) set of facets. Therefore, given any collection $\mathcal F$ of subsets of $V$, we write $\langle\mathcal F\rangle$ to mean $\left\{S\in 2^V: S\subseteq F \text{ for some } F\in \mathcal F\right\}$. If no two elements of $\mathcal F$ contain each other, then $\mathcal F$ is precisely the collection of facets in $\langle\mathcal F\rangle$.

We observe that for any $k\geq\ell\geq 0$, the collection of $(k,\ell)$-sumfree sets in any abelian group $G$ is a simplicial complex on $G$. This is because if there are no solutions to $x_1+\cdots+x_k=y_1+\cdots y_\ell$ with all $x_i,y_j$ in $S$, then the same is clearly true for any subset of $S$. The structure of these complexes are not well-understood; indeed, most prior research has been devoted merely to computing their dimensions.

In the case that $k=0$, this means that $\Delta_{n,\ell}$ is a simplicial complex for any $0<\ell<{n}$. In order to understand this class of complexes, we developed an algorithm for computing small examples. Key to this algorithm is the notion of the \textbf{Alexander dual}: $\Gamma^\vee = 2^V\setminus \left\{S\in 2^V: S\notin \Gamma \right\}$. Notice that $\Gamma^\vee$ is a complex if and only if $\Gamma$ is a complex, and $(\Gamma^\vee)^\vee=\Gamma$. We also recall that a \textbf{partition} $\lambda$ of $N$ into $\ell$ \textbf{parts} is a non-increasing ordered list of numbers $(\lambda_1,\dots, \lambda_\ell)$ which sum to $N$.

\begin{theorem
\label{algorithm}
The following procedure computes $\Delta_{n,\ell}$, the simplicial complex of of all $\ell$-zero-sumfree subsets of $\mathbb{Z}/n\mathbb{Z}$:
\begin{enumerate}
    \item For each $0\leq m\leq n-1$, generate the partitions of $mn$ into at most $\ell$ parts which also satisfy $\lambda_1\leq n-1$ and consider their underlying sets. That is, create the set (without multiplicity) $\sigma=\left\{\lambda_1,\dots,\lambda_h\right\}$ with $h\leq{\ell}$ and $\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_h$ from each such partition $\lambda$. Store these $\sigma$ into a set called NLC (short for ``$(n,
    \ell)$-congruent'' partitions).
    \item Take the complement of every element in $\langle\text{NLC}\rangle$ and denote the set of complements $\langle\text{NLC}\rangle_c$, which is a simplicial complex.
    \item Take the Alexander dual of $\langle\text{NLC}\rangle_c$ to obtain $\Delta_{n,\ell}$: $\langle\text{NLC}\rangle_c^{\lor} = \Delta_{n,\ell}$.
\end{enumerate}
\end{theorem}
\begin{proof}
The set $\langle\text{NLC}\rangle$ is the collection of all subsets of our vertex set $V$ such that the subset contains an $NLC$ partition; this is, by definition, the collection of all subsets of $V$ that are \textit{not} $\ell$-zero-sumfree. Thus, we can express this set as as
$$\langle\text{NLC}\rangle = 2^V\setminus{\Delta_{n,\ell}} = \Delta_{n,\ell}^c.$$
The set of all complements of elements in $\langle\text{NLC}\rangle$ is thus
$$\langle\text{NLC}\rangle_c^{\lor}= \left\{\sigma: (2^V\setminus\sigma)\in{\langle\text{NLC}\rangle}\right\}^{\lor} = (\Delta_{n,\ell}^{\lor})^{\lor} = \Delta_{n,\ell},$$
as desired.
\end{proof}

\subsection{Disjoint Unions of Simplices}

A \textbf{d-simplex} is the simplicial complex $\Delta=2^V$ on some set $V$ having $d+1$ elements. Note that the indexing is chosen this way so that a $d$-simplex has dimension $d$. Some of the $\Delta_{n,\ell}$ that we consider are simplices or \textbf{disjoint unions} of simplices: for any two simplicial complexes $\Gamma_1$ and $\Gamma_2$ on disjoint vertex sets $V_1$ and $V_2$, $\Gamma_1\cup\Gamma_2$ is a simplicial complex on $V_1\cup V_2$. Because we have not found this information elsewhere in the literature, we would like to state some elementary facts about such complexes.

For any simplicial complex $\Gamma$ with dimension $d$, its \textbf{f-vector} $f(\Gamma)$, or $f$ if there is no risk of confusion, is the list of numbers $(f_{-1}, f_0,\dots, f_d)$, where each $f_k$ is the number of faces in $\Gamma$ having dimension $k$. This information is also encoded in its \textbf{h-vector} $h(\Gamma)=h=(h_0,h_1,\dots, h_{d+1})$, where
$$h_k = \sum_{i=0}^k (-1)^{k-i}\binom{d+1-i}{d+1-k} f_{i-1}.$$

Notice that any simplicial complex contains $\varnothing$, the unique face of dimension $-1$, and so $f_{-1}=h_0=1$.

\begin{proposition} \label{f-vec-distinct-simplices}
If $\Gamma=\Delta_{1}\cup\ldots\cup\Delta_{\a}$ where each $\Delta_{i}$ is a $d_i$-simplex, then the $f$-vector of $\Gamma$ is given by $f_{-1}=1$ and $f_{k-1}=\sum_{i=1}^{\a}\binom{d_i+1}{k}$ for all $1\leq k\leq d$.
\end{proposition}

\begin{proof}
Let $\Gamma$ be a simplicial complex of dimension $d$ such that it is comprised of only $\a$ many $d_i$-simplices, not necessarily distinct. By definition of the binomial coefficient, a $d$-simplex has the $f$-vector given by $f_{k-1}=\binom{d+1}{k}$. One easily checks that the $f$-vector is nearly additive with respect to disjoint unions: $f_k(\Gamma_1\cup\Gamma_2) =f_k(\Gamma_1)+f_k(\Gamma_2)$ for all $k\geq 0$, from which the proposition follows.
\end{proof}

We now turn our attention to the $h$-vector. The non-additivity of $f_{-1}$ for disjoint unions causes more serious difficulties for the $h$-vector, since the defining sum for $h_k$ contains a term with $f_{-1}$ for any $k$. However, there is a surprisingly pleasing formula for the $h$-vector as well.

Given a partition $\l=(\l_1,\dots, \l_\a)$, we may draw its \textbf{Young diagram} (in French notation), a left-and-bottom-justified array of boxes with $\l_i$ boxes in row $i$.
For example, if $\l=(5,4,1,1)$, then its Young diagram is
\begin{center}
$\yng(1,1,4,5)$
\end{center}

Since $\l$ is a decreasing list, reflection across the line $y=x$ gives rise to another Young diagram, called the \textbf{conjugate partition} $\mu$ of $\lambda$. For instance, relfecting the $\lambda$ above yields $\mu=(4,2,2,2,1)$:

\begin{center}
\yng(1,2,2,2,4)
\end{center}

In the proof of the following proposition, it will be helpful to observe the formal definition of the conjugate partition $\mu=(\mu_1,\dots,\mu_\ell)$:
$$ \mu_m = \#\{j: \l_j\geq m \}, \qquad 1\leq m\leq\l_1=:\ell. $$
It will also be convenient to write $[N]$ as shorthand for the set $\{1,2,\dots, N\}$.

\begin{proposition}  \label{h-vec-distinct-simplices}
Let $\Gamma_\lambda$ be a disjoint union of simplices $\Delta_{1}\cup\cdots\cup\Delta_{\a}$ where $\l$ is a partition and each $\Delta_i$ is a $(\l_i-1)$-simplex. Then $h_0=1$, and for all $1\leq k\leq \l_1$
$$(-1)^{k-1} h_k = \sum_{m=1}^{\ell-k+1}\binom{\ell-m}{k-1}(\mu_m-1),$$
where $\mu$ is the conjugate partition of $\l.$
\end{proposition}

\begin{proof}
Beginning with the definition, we apply \textbf{Proposition \ref{f-vec-distinct-simplices}}, separating out the exceptional $f_{-1}$ term from the others and then swapping the order of summation.
\begin{align*}
h_k &= \sum_{i=0}^k (-1)^{k-i}\binom{\l_1-i}{k-i} f_{i-1} \\
 &= (-1)^k\binom{\l_1}{k} + \sum_{i=1}^k (-1)^{k-i}\binom{\l_1-i}{k-i} \sum_{j=1}^{\a} \binom{\l_j}{i} \\
 &= (-1)^k\binom{\l_1}{k} + \sum_{j=1}^\a  \sum_{i=1}^{k} (-1)^{k-i}\binom{\l_1-i}{k-i}\binom{\l_j}{i}.
\end{align*}
At this point, we observe the following combinatorial fact, which we will prove later.

\begin{lemma} \label{combinatorial-technicality}
Let $a, b,$ and $k$ be nonnegative integers. Then
$$ \sum_{i=0}^k (-1)^i \binom{a+b-i}{k-i}\binom{b}{i} = \binom{a}{k}. $$
\end{lemma}

We apply this fact with $a=\l_1-\l_j$ and $b=\l_j$. When plugging in to the formula for $h_k$ above, note that we need to be careful with the $i=0$ term:
$$ h_k = (-1)^k\binom{\l_1}{k} + \sum_{j=1}^r (-1)^k\left[\binom{\l_1-\l_j}{k}-\binom{\l_1}{k} \right]. $$
Observe that when $k>0$, the $j=1$ term contributes $\binom{\l_1-\l_1}{k}-\binom{\l_1}{k} = -\binom{\l_1}{k}$. Therefore, we can cancel it with the exceptional term and start the sum at $j=2$:
\begin{align*}
h_k &= (-1)^k\sum_{j=2}^r \left[\binom{\l_1-\l_j}{k}-\binom{\l_1}{k} \right].
\end{align*}
We divide both sides by $(-1)^{k+1}$ and interpret this sum as follows:
$$  (-1)^{k+1}h_k = \sum_{j=2}^r \left[ \sum_{\substack{ S\subseteq [\l_1] \\ |S|=k}} 1 - \sum_{\substack{ S\subseteq [\l_1]\smallsetminus[\l_j] \\ |S|=k}} 1 \right]. $$
In other words, each subset $S\subseteq[\l_1]$ contributes to the inner sum precisely if it is not a subset of $\{\l_j+1,\dots, \l_1\}$. This means that $S$ contributes to the $j$ term of the outer sum precisely when its smallest element is at least $\l_j$. In other words, by swapping the order of summation, we obtain
$$ (-1)^{k+1} h_k =  \sum_{\substack{ S\subset [\l_1] \\ |S|=k}} \#\{j\geq 2: \l_j\geq \min(S) \}. $$

Now split this sum into parts by tracking the minimum element of each set (which exists because $|S|=k>0$). Once the minimum element of a $k$-element set $S$ is known to be $m$, the other elements may form any $(k-1)$-element subset of $\{m+1,\dots, \l_1\}$. Therefore:
\begin{align*}
 (-1)^{k+1} h_k &= \sum_{m=1}^{\l_1}\left[\sum_{\substack{ S\subset \{m+1,\dots,\l_1\} \\ |S|=k}} \#\{j\geq 2: \l_j\geq m \}\right] \\
 &= \sum_{m=1}^{\l_1}\binom{\l_1-m}{k-1} \#\{j\geq 2: \l_j\geq m \}
\end{align*}
This is the desired identity, slightly disguised. First, the binomial coefficient ensures the terms vanish when $m\geq \l_1-k+1$, so we can match the upper limit of the sum. Finally, $\#\{j\geq 2: \l_j\geq m \}=\mu_m-1$, by definition of the conjugate partition and because $\l_1\geq m$ for all $m$ in the summation.
\end{proof}

For completeness, we now prove \textbf{Lemma \ref{combinatorial-technicality}}.

\begin{proof}[Proof (of Lemma \ref{combinatorial-technicality})]

The unsigned version of the left-hand side
$$ \sum_{i=0}^k  \binom{a+b-i}{k-i}\binom{b}{i}$$
counts the number of ways to choose $k$ elements from $[a+b]$ in two phrases: choosing first $i$ elements from $[a]$, and choosing second $k-i$ more (distinct) elements from $[a+b]$. There is a sign-reversing involution on such two-phase sets, given by swapping the phase in which the smallest element was chosen. This cancels the contribution of all subsets except for those with no elements in $[a]$, since it is impossible to choose any of their elements in the first phase. Therefore, each of these $\binom{a}{k}$ sets contributes exactly once to the sum, which gives the desired identity.
\end{proof}

\begin{corollary}  \label{h-vec-same-simplices}
If $\Gamma=\Delta_{1}\cup\ldots\cup\Delta_{\a}$ where each $\Delta_{i}$ is a $d$-simplex, then the $h$-vector of $\Gamma$ is given by $h_0=1$ and $h_k=(-1)^{k+1}(\a-1)\binom{d+1}{k}$ for all $1\leq k\leq d$.
\end{corollary}

We were led to \textbf{Proposition \ref{h-vec-distinct-simplices}} by observing two qualitative features of \textbf{Corollary \ref{h-vec-same-simplices}} which we tried to generalize. The first of these is that, $h_0$ notwithstanding, the signs in the $h$-vector alternate; we now have a satisfactory explanation. However, the second property remains mysterious even with the explicit formula. A sequence $(a_1,a_2,\dots)$ is called $\textbf{log-concave}$ if $a_k^2 \geq a_{k-1}a_{k+1}$ for all $k\geq 2$; this property means that the sequence increases ``smoothly'' until it hits a maximum, and then decreases smoothly afterward. Many combinatorial sequences, such as binomial coefficients and Eulerian numbers, are known to be log-concave; we conjecture that so too is the unsigned $h$-vector for general disjoint unions of simplices.

\begin{conjecture} \label{distinct-simplices-unimodal}
Let $\Gamma_\lambda$ be a disjoint union of simplices $\Delta_{1}\cup\cdots\cup\Delta_{\a}$ where $\l$ is a partition and each $\Delta_i$ is a $(\l_i-1)$-simplex. If the $d_i$ are all distinct, then the unsigned $h$-vector $( |h_1|, |h_2| ,\dots, |h_{\l_1}|)$ is log-concave.
\end{conjecture}

It is worth noting that the analogous conjecture is false for the $f$-vector; for instance, having a single high-dimensional simplex, together with extremely many low-dimensional ones, will cause the $f$-vector to spike in low dimension, which is not permitted in a log-concave sequence.

\subsection{Subspace Lattices and Intersection Posets}

The $f$-vector and $h$-vector are two combinatorial features of $\Gamma$, and we now wish to discuss a geometric one. Fix a field $\mathbb{K}$. For any simplicial complex $\Gamma$ on $V=\{v_1,\dots, v_n\}$, we define $\mathcal A_\Gamma$ (or $\mathcal A$ if there is no risk of confusion) to be the \textbf{associated subspace arrangement} in $\mathbb K^{V}$, which consists of all subspaces of the form
$$ S_F = \{(z_{v_1},\dots, z_{v_n}): z_{v}=0 \text{ for all } v\in V\setminus F \}, $$
where $F$ is an \emph{facet} of $\Gamma$.

This is a natural object to consider for the following reason. The \textbf{Stanley-Riesner ideal} $I_\Gamma$ of $\Gamma$, is the ideal in the polynomial ring $\mathbb K[v_1,\dots, v_n]$ which is generated by the monomials $\prod_{v\notin F} v$ for all faces $F\in\Gamma$. One studies the geometric properties of ideals $J$ in polynomial rings by considering their \textbf{variety} $V(J)\subseteq \mathbb K^n$, defined as $V(J) = \{z\in \mathbb K^n: f(z)=0 \text{ for all } f\in J\}.$ It is a straightforward but tedious exercise in element-chasing to show that $V(I_\Gamma)$ is the union of all the subspaces in $\mathcal A_\Gamma$.

We conclude this section by describing the features of subspace arrangements that are used in the statement of our results. Define the \textbf{intersection poset} of a subspace arrangement $\mathcal A$, denoted $\mathcal{L}(\mathcal{A})$, as the finite set containing all intersections of subspaces $S_F\in\mathcal{A}$, ordered by reverse-inclusion: $I\leq J$ if $J\subseteq I$. Notice that this set has a minimum element $\hat 0=\mathbb K^n$, the empty intersection, and a maximum element $\hat 1$, given by intersecting all of the subspaces in the arrangement.

The most well-studied subspace arrangements are those for which all subspaces have codimension one, known as \textbf{hyperplane arrangements}. Hyperplane arrangements have particularly well-behaved intersection posets.

For instance, say that a poset is \textbf{graded of rank $n$} if all maximal chains have length $n$. Note that a \textbf{maximal chain} is a chain that is not a proper subset of another chain. Hyperplane arrangements are graded of rank $n$, which can be seen by a closer examination of their structure: elements in a poset are called \textbf{atoms} if they \textbf{cover} the minimum element, i.e. such that there does not exist another element $x\in\mathcal{L}(\mathcal{A})$ with $\hat 0<x<a$. Atoms in the intersection poset of a hyperplane arrangement are precisely the subspaces in the arrangements. Since they all have codimension one, intersections having codimension two are precisely the intersections of two subspaces; similarly codimension three corresponds to the intersection of three subspaces, and so on.

For general subspace arrangements, the atoms of $\mathcal{L}(\mathcal{A})$ are still the subspaces in the arrangement. However, because the subspaces may have higher codimension, there is no longer a guarantee that $\mathcal{L}(\mathcal{A})$ is not graded at all. Despite this deficit, Athanasiadis \cite{Athanasiadis} showed that the so-called ``finite field method'', which counts the points in $\mathcal{L}(\mathcal{A})$ for $\mathbb K$ a sufficiently large finite field, can be extended to general subspace arrangements. To make this result work, the appropriate notion of a \textbf{characteristic polynomial} for a subspace arrangement $\mathcal{A}$ is $$\chi_\mathcal{A}(x)=\sum_{t\in\mathcal{L}(\mathcal{A})}\mu(\Hat{0},t)x^{\dim(t)},$$
where the \textbf{M{\"o}bius function} $\mu$ is defined recursively by $\mu(s,s)=1$ and
$$\mu(s,t) = -\sum_{s\leq z\leq t} \mu(s,z).$$

\begin{proposition}\label{max-chain-union}
 Let $\Gamma=\Delta_1\cup\ldots\cup\Delta_\alpha,$ be a simplicial complex on a vertex set $V$ where each $\Delta_i$ is a $d_i$-simplex. Let $\mathcal{L}(\mathcal{A})$ be the corresponding intersection lattice. Then every maximal chain in $\mathcal{L}(\mathcal{A})$ is of the form: $$C=\{\hat 0=a_0<a_1<a_2=\hat 1\}.$$
\end{proposition}

\begin{proof}
Let $C$ be a maximal chain in $\mathcal{L}(\mathcal{A}).$
We want to show that every maximal chain in $\mathcal{L}(\mathcal{A})$ is of the form: $C=\{\hat 0=a_0<a_1<a_2=\hat 1\}.$
Since $C$ is a maximal chain, it contains a minimal element $\hat 0$ and maximum element $\hat 1.$ Then clearly the minimal element is $\hat 0=a_0=\mathbb{K}^{|V|}.$ Since each subspace is an atom $a_1,$ then $\Hat{0}=a_0=\mathbb{K}^{|V|}<a_1.$

Let $F_1,F_2$ be facets in $\Gamma$ and let $S_1,S_2\in\mathcal{A}$ be the corresponding subspaces. Since $F_1,F_2\in\Gamma,$ then $F_1\cap F_2=\varnothing.$ Furthermore, since $F_1\cap F_2=\varnothing,$ then $(V\setminus F_1)\cup(V\setminus F_2)=V.$ Therefore, by definition of $S_1,S_2\in\mathcal{A},$ $S_1\cap S_2=\{0\}=a_2.$

Since any subspace in $\mathcal{A}$ covers $\mathbb{K}^{|V|}$ and the intersection of any subspaces in $\mathcal{A}$ is $\{0\},$ then $$C=\{\hat 0=a_0<a_1<a_2=\hat 1\},$$ is a maximal chain as desired.
\end{proof}

\begin{corollary}\label{lattice-rank2}
Let $\Gamma=\Delta_1\cup\ldots\cup\Delta_\alpha,$ be a simplicial complex on a vertex set $V$ where each $\Delta_i$ is a $d_i$-simplex. Let $\mathcal{L}(\mathcal{A})$ be the corresponding intersection lattice. Then $\mathcal{L}(\mathcal{A})$ is a graded poset of rank $2.$
\end{corollary}

\begin{proof}
Since $\Gamma$ is the disjoint union of $d_i$ simplices, then by Proposition \ref{max-chain-union}, any maximal chain $C\in\mathcal{L}(\mathcal{A})$ has the form $C=\{\hat 0=a_0<a_1<a_2=\hat 1\}.$ Since $C$ is arbitrary, then the length of all maximal chains is $2.$ Thus, $\mathcal{L}(\mathcal{A})$ admits grading, in particular, $\mathcal{L}(\mathcal{A})$ has rank $2.$
\end{proof}

\begin{proposition}\label{char-poly-union}
  Let $\Gamma=\Delta_1\cup\ldots\cup\Delta_\alpha,$ be a simplicial complex on a vertex set $V$, where each $\Delta_i$ is a $d_i$-simplex. Let $\mathcal{L}(\mathcal{A})$ be the corresponding intersection lattice. Then,
  $$\chi_\mathcal{A}(x)=x^{|V|}-\sum_{F\in\Gamma}x^{|F|}+\a-1.$$
\end{proposition}

\begin{proof}
Let $\Gamma=\Delta_1\cup\ldots\cup\Delta_\alpha,$ be a simplicial complex on a vertex set $V$ where each $\Delta_i$ is a $d_i$-simplex. By Proposition \ref{max-chain-union} we have that every maximal chain is of the form $C=\{\hat 0=a_0<a_1<a_2=\hat 1\}.$ Recall that by definition, $$\chi_\mathcal{A}=\sum_{t\in\mathcal{L}(\mathcal{A})}\mu(\hat{0},t)x^{\dim(t)}.$$

First, by Proposition \ref{max-chain-union} we have that $\hat{0}=a_0=\mathbb{K}^{|V|}.$ So, $\dim(z_0)=\dim(\mathbb{K}^{|V|})=|V|.$ Also,  $\mu(\hat{0},\hat{0})=1$.

Secondly, any atom $a_1$ is a subspace $S_F\in\mathcal{A}.$ Since $$S_F=\{(z_{v_1},\ldots,z_{v_n}):z_v=0,\forall v\in V\setminus F\},$$ then for $F\in\Gamma,$ $\dim(S_F)=\dim(a_1)=|F|$ and $\mu(\hat{0},a_1)=-1.$

Finally, by Proposition \ref{max-chain-union}, $\hat{1}=a_2=\{0\}.$ So $\dim(a_2)=0.$ Moreover, since the number of atoms is equal to the number of facets, which is equal to the number of $\Delta_i\in\Gamma,$ then $\mu(\hat{0},\hat{1})=-1(\mu(\hat{0},\hat{0})+\mu(\hat{0},a_1))=\alpha -1.$

Thus, $$\chi_\mathcal{A}(x)=x^{|V|}-\sum_{F\in\Gamma}x^{|F|}+\a-1,$$ as desired.
\end{proof}

\begin{corollary}\label{char-poly-union-eqdim}
If $\Gamma=\Delta_1 \cup \cdots\cup \Delta_\a$ where each $\Delta_i$ is a $\delta$-simplex, for some fixed dimension $\delta$, then
$$\chi_\mathcal{A}(x)=x^{\a(\delta+1)}-\a\cdot x^{\delta+1}+(\a-1).$$
\end{corollary}
\begin{proof}
By Theorem \ref{char-poly-union}, $$\chi_\mathcal{A}(x)=x^{|V|}-\sum_{F\in\Gamma}x^{|F|}+\a-1.$$
Since there are $\a$ many facets $F_i\in\Gamma,$
then $$-\sum_{F\in\Gamma}x^{|F|}=\sum_{i=0}^{\a}(-1)x^{|F_i|}=-\a\cdot x^{|F|}.$$ Moreover, each facet has $|F|=\delta+1$ vertices, for a total of $|V|=\a(\delta+1)$; this completes the proof.
\end{proof}

\section{Main Results and Conjectures}

Our main results are a complete description of $\Delta_{n,\ell}$ for certain families of $n$ and $\ell$, as well as some implications for their intersection lattices. In particular, it is interesting that these posets are graded for all of the families that we considered here, and we wish to stress that this is not true for every $\Delta_{n,\ell}$. In this section we only present the results, leaving the calculations themselves to Section \ref{all-the-proofs}.

\begin{restatable}{theorem}{doubling}

\label{total-doubling}
Consider $\Gamma=\Delta_{2\ell,\ell}$ for any integer $\ell$. Let $\ell=\rho\cdot 2^m$ such that $\rho$ is odd, and $\mathcal A$ be the subspace arrangement associated to $\Gamma$.
\begin{enumerate}[(a)]
    \item\label{doubling} $\Gamma = \Delta_0\cup \Delta_1\cup\cdots\cup \Delta_{2^{m-1}}$, where $\Delta_t$ is the $(\rho-1)$-simplex on the set $V_t$ of all $x\in \mathbb Z/2\ell\mathbb{Z}$ congruent to $2t + 1$ mod $2^{m+1}$.
    \item $\mathcal{L}(\mathcal{A})$ is a graded poset of rank $2$.
    \item The characteristic polynomial of $\mathcal A$ is $\chi_\mathcal{A}(x)=x^{\ell}-2^mx^{\rho}+(2^m-1).$
\end{enumerate}
\end{restatable}

\begin{example}
We draw some typical examples of the above theorem: $\Delta_{2^{m+1}\rho,2^m\rho}$ for $\rho=3$ and $m=1,2,3$.
\begin{center}
\begin{tikzpicture}[scale=0.35]
\pgfmathsetlengthmacro{\rad}{38pt};
\pgfmathsetlengthmacro{\vertsize}{\rad*0.1666};
\pgfmathsetlengthmacro{\distance}{1.5*\rad};
\pgfmathsetlengthmacro{\largedist}{8*\rad};

\pgfmathsetmacro{\triangleopacity}{0.4};
\definecolor{dimtwocolor}{RGB}{140,140,200}
\begin{scope}[shift={(0,0)}]
\begin{scope}[shift={(0,0)}]
	\coordinate (a1) at (90:\rad);
	\coordinate (b1) at (210:\rad);
	\coordinate (c1) at (330:\rad);
	\draw [ultra thick, draw=black, fill=dimtwocolor, fill opacity=\triangleopacity] (a1) -- (b1) -- (c1) -- cycle;
	\draw [black, fill=black] (a1) circle [radius=\vertsize];
	\draw [black, fill=black] (b1) circle [radius=\vertsize];
	\draw [black, fill=black] (c1) circle [radius=\vertsize];
	\node[yshift=2*\vertsize] at (a1) {1};
	\node[yshift=-2*\vertsize] at (b1) {3};
	\node[yshift=-2*\vertsize] at (c1) {5};

	\node[xshift=0*\distance,yshift=-.75*\distance] at (0,0) {\LARGE $\Delta_{6,\,3}$};
\end{scope}

\begin{scope}[shift={(6.5*\distance,0)}]
	\begin{scope}[shift={(-\distance,0)}]
		\coordinate (a1) at (90:\rad);
		\coordinate (b1) at (210:\rad);
		\coordinate (c1) at (330:\rad);
		\draw [ultra thick, draw=black, fill=dimtwocolor, fill opacity=\triangleopacity] (a1) -- (b1) -- (c1) -- cycle;
		\draw [black, fill=black] (a1) circle [radius=\vertsize];
		\draw [black, fill=black] (b1) circle [radius=\vertsize];
		\draw [black, fill=black] (c1) circle [radius=\vertsize];
		\node[yshift=2*\vertsize] at (a1) {1};
		\node[yshift=-2*\vertsize] at (b1) {5};
		\node[yshift=-2*\vertsize] at (c1) {9};
	\end{scope}

	\begin{scope}[shift={(\distance,0)}]
		\coordinate (a1) at (90:\rad);
		\coordinate (b1) at (210:\rad);
		\coordinate (c1) at (330:\rad);
		\draw [ultra thick, draw=black, fill=dimtwocolor, fill opacity=\triangleopacity] (a1) -- (b1) -- (c1) -- cycle;
		\draw [black, fill=black] (a1) circle [radius=\vertsize];
		\draw [black, fill=black] (b1) circle [radius=\vertsize];
		\draw [black, fill=black] (c1) circle [radius=\vertsize];
		\node[yshift=2*\vertsize] at (a1) {3};
		\node[yshift=-2*\vertsize] at (b1) {7};
		\node[yshift=-2*\vertsize] at (c1) {11};
	\end{scope}

	\node[xshift=0*\distance,yshift=-.75*\distance] at (0,0) {\LARGE $\Delta_{12,\,6}$};
\end{scope}

\begin{scope}[shift={(16*\distance,0)}]
	\begin{scope}[shift={(-3*\distance,0)}]
		\coordinate (a1) at (90:\rad);
		\coordinate (b1) at (210:\rad);
		\coordinate (c1) at (330:\rad);
		\draw [ultra thick, draw=black, fill=dimtwocolor, fill opacity=\triangleopacity] (a1) -- (b1) -- (c1) -- cycle;
		\draw [black, fill=black] (a1) circle [radius=\vertsize];
		\draw [black, fill=black] (b1) circle [radius=\vertsize];
		\draw [black, fill=black] (c1) circle [radius=\vertsize];
		\node[yshift=2*\vertsize] at (a1) {1};
		\node[yshift=-2*\vertsize] at (b1) {9};
		\node[yshift=-2*\vertsize] at (c1) {17};
	\end{scope}

	\begin{scope}[shift={(-\distance,0)}]
		\coordinate (a1) at (90:\rad);
		\coordinate (b1) at (210:\rad);
		\coordinate (c1) at (330:\rad);
		\draw [ultra thick, draw=black, fill=dimtwocolor, fill opacity=\triangleopacity] (a1) -- (b1) -- (c1) -- cycle;
		\draw [black, fill=black] (a1) circle [radius=\vertsize];
		\draw [black, fill=black] (b1) circle [radius=\vertsize];
		\draw [black, fill=black] (c1) circle [radius=\vertsize];
		\node[yshift=2*\vertsize] at (a1) {3};
		\node[yshift=-2*\vertsize] at (b1) {11};
		\node[yshift=-2*\vertsize] at (c1) {19};
	\end{scope}

	\begin{scope}[shift={(\distance,0)}]
		\coordinate (a1) at (90:\rad);
		\coordinate (b1) at (210:\rad);
		\coordinate (c1) at (330:\rad);
		\draw [ultra thick, draw=black, fill=dimtwocolor, fill opacity=\triangleopacity] (a1) -- (b1) -- (c1) -- cycle;
		\draw [black, fill=black] (a1) circle [radius=\vertsize];
		\draw [black, fill=black] (b1) circle [radius=\vertsize];
		\draw [black, fill=black] (c1) circle [radius=\vertsize];
		\node[yshift=2*\vertsize] at (a1) {5};
		\node[yshift=-2*\vertsize] at (b1) {13};
		\node[yshift=-2*\vertsize] at (c1) {21};
	\end{scope}

	\begin{scope}[shift={(3*\distance,0)}]
		\coordinate (a1) at (90:\rad);
		\coordinate (b1) at (210:\rad);
		\coordinate (c1) at (330:\rad);
		\draw [ultra thick, draw=black, fill=dimtwocolor, fill opacity=\triangleopacity] (a1) -- (b1) -- (c1) -- cycle;
		\draw [black, fill=black] (a1) circle [radius=\vertsize];
		\draw [black, fill=black] (b1) circle [radius=\vertsize];
		\draw [black, fill=black] (c1) circle [radius=\vertsize];
		\node[yshift=2*\vertsize] at (a1) {7};
		\node[yshift=-2*\vertsize] at (b1) {15};
		\node[yshift=-2*\vertsize] at (c1) {23};
	\end{scope}

	\node[xshift=0*\distance,yshift=-.75*\distance] at (0,0) {\LARGE $\Delta_{24,\,12}$};
\end{scope}
\end{scope}
\end{tikzpicture}
\end{center}
\qed
\end{example}

In the previous section we discussed properties of complexes which are disjoint unions of simplices. The complex $\Delta_{2\ell,\ell}$ is thus a ``naturally occurring'' instance of such an object in which all the simplices have the same dimension. One may wonder whether more general unions of simplices also arise from a $\Delta_{n,\ell}$, and the next result answers this in the affirmative.

\begin{restatable}{theorem}{primepowers}
\label{total-primepowers}
Consider $\Gamma=\Delta_{p^e, p^e-1}$ for prime $p$ and and positive integer $e$. Let $\mathcal A$ be the subspace arrangement associated to $\Gamma$.
\begin{enumerate}[(a)]
    \item $\Delta_{p^{e},p^{e}-1}$ is a disjoint union of $e(p-1)$ simplices, with $e-1$ many $\left(p^j-1\right)$-simplices for each $0\leq j\leq e-1$.
    \item $\mathcal{L}(\mathcal{A})$ is a graded poset of rank $2$.
    \item The characteristic polynomial of $\mathcal A$ is $\chi_\mathcal{A}(x)=x^{p^e-1}-(p-1)\sum^{e-1}_{j=0}x^{p^j}$.
\end{enumerate}
\end{restatable}

\begin{example}
We draw a typical example of the above theorem: $\Delta_{p^e,p^e-1}$ for $p=3,e=2$.
\begin{center}
\begin{tikzpicture}[scale=0.35]
\pgfmathsetlengthmacro{\rad}{38pt};
\pgfmathsetlengthmacro{\vertsize}{\rad*0.1666};
\pgfmathsetlengthmacro{\distance}{1.5*\rad};
\pgfmathsetlengthmacro{\largedist}{8*\rad};

\pgfmathsetmacro{\triangleopacity}{0.4};
\definecolor{dimtwocolor}{RGB}{140,140,200}

\begin{scope}[shift={(0,0)}]
    \begin{scope}[shift={(0*\vertsize,0)}]
	\coordinate (d1) at (210:\rad);
	\coordinate (e1) at (330:\rad);
	\draw [black, fill=black] (d1) circle [radius=\vertsize];
	\draw [black, fill=black] (e1) circle [radius=\vertsize];
	\node[yshift=-2*\vertsize] at (d1) {3};
	\node[yshift=-2*\vertsize] at (e1) {6};
    \end{scope}
	\begin{scope}[shift={(20*\vertsize,0)}]
		\coordinate (a1) at (90:\rad);
		\coordinate (b1) at (210:\rad);
		\coordinate (c1) at (330:\rad);
		\draw [ultra thick, draw=black, fill=dimtwocolor, fill opacity=\triangleopacity] (a1) -- (b1) -- (c1) -- cycle;
		\draw [black, fill=black] (a1) circle [radius=\vertsize];
		\draw [black, fill=black] (b1) circle [radius=\vertsize];
		\draw [black, fill=black] (c1) circle [radius=\vertsize];
		\node[yshift=2*\vertsize] at (a1) {1};
		\node[yshift=-2*\vertsize] at (b1) {4};
		\node[yshift=-2*\vertsize] at (c1) {7};
	\end{scope}

	\begin{scope}[shift={(40*\vertsize,0)}]
		\coordinate (a1) at (90:\rad);
		\coordinate (b1) at (210:\rad);
		\coordinate (c1) at (330:\rad);
		\draw [ultra thick, draw=black, fill=dimtwocolor, fill opacity=\triangleopacity] (a1) -- (b1) -- (c1) -- cycle;
		\draw [black, fill=black] (a1) circle [radius=\vertsize];
		\draw [black, fill=black] (b1) circle [radius=\vertsize];
		\draw [black, fill=black] (c1) circle [radius=\vertsize];
		\node[yshift=2*\vertsize] at (a1) {2};
		\node[yshift=-2*\vertsize] at (b1) {5};
		\node[yshift=-2*\vertsize] at (c1) {8};
	\end{scope}
	\node[xshift=-1*\distance,yshift=0*\distance] at (0,0) {\LARGE $\Delta_{9,\,8}$};
\end{scope}
\end{tikzpicture}
\end{center}
\qed
\end{example}

We note that these two families are not the only parameters $n$ and $\ell$ which yield a disjoint union of simplices; for instance, $\Delta_{12,9}$ is the disjoint union of a 5-simplex and a 2-simplex. It may be interesting to seek a complete classification of the parameters $n$ and $\ell$ which exhibit this phenomenon.

We also note that in the Theorem \ref{total-primepowers}, $n-\ell=1$. In all the explicit calculations we were able to carry out, we observed that $\Delta_{n,\ell}$ is somehow ``simpler'' when $n-\ell$ is small. In particular, the facets for such complexes seem to be determined by elementary number-theoretic conditions. The last theorem may be interpreted as some further evidence for this observation.

\begin{restatable}{theorem}{armslegs}
\label{total-armslegs}
Consider $\Gamma_s = \Delta_{2p,2p-s}$ for prime $p\neq 2$ and positive integer $s\leq 2p$. Let $\mathcal A_s$ be the subspace arrangement associated to $\Gamma_s$.
\begin{enumerate}[(a)]
    \item
    \begin{itemize}
        \item\label{2p,2p-2} $\Gamma_2$ is a disjoint union of the $p-1$ edges $\{i_1,i_2\}$ for which $i_1\equiv{i_2}\not\equiv{0}\bmod{p}$.
        \item\label{2p,2p-1} The facets of $\Gamma_1$ are the facets of $\Gamma_2$ together with $\{1,3,5,...,2p-1\}$.
        \item\label{2p,2p-3} If $p\geq 5$, the facets of $\Gamma_3$ are the facets of $\Gamma_1$ together with the $p-1$ edges $\{i_1,i_2\}$ for which $i_1$ is odd, $i_2\neq 0$ is even, and $i_2\equiv-2i_1\bmod{2p}$.
    \end{itemize}

    \item $\mathcal L(\mathcal A_s)$ is a graded poset. In particular, $\mathcal L(\mathcal A_2)$ is of rank $2$; and $\mathcal L(\mathcal A_1)$ and  $\mathcal L(\mathcal A_3)$ are each of rank $3$.

    \item The characteristic polynomial of $\mathcal A_s$ is
    $$ \chi_{\mathcal A_s}(x) =
        \left\{\begin{array}{lr}
        x^{2p-1}-x^{p}-(p-1)x^2+(p-1)x & : s=1\\
        x^{2p-2}-p\cdot x^{2}+p-1 & : s=2\\
        x^{2p-1}-x^p-2(p-1)x^2+2(p-1)x & : s=3
        \end{array}\right.
    $$
\end{enumerate}
\end{restatable}

\begin{example}

We draw some typical examples of the above theorem: $\Delta_{2p,2p-s}$ for $p=7$ and $s=1,2,3$.

\begin{center}
\begin{tikzpicture}[scale=0.25]
\pgfmathsetlengthmacro{\rad}{70pt};
\pgfmathsetlengthmacro{\vertsize}{\rad*0.1666};
\pgfmathsetlengthmacro{\distance}{1.5*\rad};
\pgfmathsetlengthmacro{\largedist}{5.5*\rad};

\pgfmathsetmacro{\p}{7};
\pgfmathsetmacro{\pp}{\p-1};
\pgfmathsetmacro{\angleSep}{360/\pp};
\pgfmathsetmacro{\angleOff}{\angleSep*0.5};

\pgfmathsetmacro{\triangleopacity}{0.4};
\pgfmathsetmacro{\subfaceopacity}{0.3};
\definecolor{dimtwocolor}{RGB}{140,140,200};
\definecolor{dimhighcolor}{RGB}{70,70,100};
\begin{scope}[shift={(0,0)}]
\begin{scope}[shift={(0*\distance,0)}]
	\coordinate (C) at (\vertsize,1.5*\vertsize);

	\coordinate (I1) at (\angleOff-0*\angleSep:\rad);
	\coordinate (I2) at (\angleOff-1*\angleSep:\rad);
	\coordinate (I3) at (\angleOff-2*\angleSep:\rad);
	\coordinate (I4) at (\angleOff-3*\angleSep:\rad);
	\coordinate (I5) at (\angleOff-4*\angleSep:\rad);
	\coordinate (I6) at (\angleOff-5*\angleSep:\rad);
	\coordinate (O1) at (\angleOff-0*\angleSep:2*\rad);
	\coordinate (O2) at (\angleOff-1*\angleSep:2*\rad);
	\coordinate (O3) at (\angleOff-2*\angleSep:2*\rad);
	\coordinate (O4) at (\angleOff-3*\angleSep:2*\rad);
	\coordinate (O5) at (\angleOff-4*\angleSep:2*\rad);
	\coordinate (O6) at (\angleOff-5*\angleSep:2*\rad);

	\draw [ultra thick, draw=black, fill=dimhighcolor, fill opacity=2*\triangleopacity] (I1) -- (I2) -- (I3) -- (I4) -- (I5) -- (I6)-- cycle;
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I1) -- (I3);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I1) -- (I4);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I1) -- (I5);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I1) -- (I6);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I2) -- (I4);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I2) -- (I5);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I2) -- (I6);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I3) -- (I5);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I3) -- (I6);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I4) -- (I6);
	\draw [ultra thick, draw=black, opacity=2*\subfaceopacity] (C) -- (I1);
	\draw [ultra thick, draw=black, opacity=2*\subfaceopacity] (C) -- (I2);
	\draw [ultra thick, draw=black, opacity=2*\subfaceopacity] (C) -- (I3);
	\draw [ultra thick, draw=black, opacity=2*\subfaceopacity] (C) -- (I4);
	\draw [ultra thick, draw=black, opacity=2*\subfaceopacity] (C) -- (I5);
	\draw [ultra thick, draw=black, opacity=2*\subfaceopacity] (C) -- (I6);

	\draw [ultra thick, draw=black] (I1) -- (O1);
	\draw [ultra thick, draw=black] (I2) -- (O2);
	\draw [ultra thick, draw=black] (I3) -- (O3);
	\draw [ultra thick, draw=black] (I4) -- (O4);
	\draw [ultra thick, draw=black] (I5) -- (O5);
	\draw [ultra thick, draw=black] (I6) -- (O6);

	\draw [black, fill=black] (C) circle [radius=\vertsize];
	\draw [black, fill=black] (I1) circle [radius=\vertsize];
	\draw [black, fill=black] (I2) circle [radius=\vertsize];
	\draw [black, fill=black] (I3) circle [radius=\vertsize];
	\draw [black, fill=black] (I4) circle [radius=\vertsize];
	\draw [black, fill=black] (I5) circle [radius=\vertsize];
	\draw [black, fill=black] (I6) circle [radius=\vertsize];
	\draw [black, fill=black] (O1) circle [radius=\vertsize];
	\draw [black, fill=black] (O2) circle [radius=\vertsize];
	\draw [black, fill=black] (O3) circle [radius=\vertsize];
	\draw [black, fill=black] (O4) circle [radius=\vertsize];
	\draw [black, fill=black] (O5) circle [radius=\vertsize];
	\draw [black, fill=black] (O6) circle [radius=\vertsize];

	\node[xshift=0.75*\vertsize, yshift=-0.5*\vertsize] at (I1) {3};
	\node[yshift=-0.95*\vertsize] at (I2) {9};
	\node[xshift=-0.90*\vertsize, yshift=-0.40*\vertsize] at (I3) {13};
	\node[xshift=-0.75*\vertsize, yshift=0.65*\vertsize] at (I4) {11};
	\node[yshift=1*\vertsize] at (I5) {5};
	\node[xshift=0.75*\vertsize, yshift=0.5*\vertsize] at (I6) {1};

	\node[xshift=1.25*\vertsize] at (O1) {10};
	\node[xshift=0.9*\vertsize, yshift=-0.9*\vertsize] at (O2) {2};
	\node[yshift=-1.25*\vertsize] at (O3) {6};
	\node[xshift=-1.25*\vertsize] at (O4) {4};
	\node[xshift=-0.9*\vertsize, yshift=0.9*\vertsize] at (O5) {12};
	\node[yshift=1.15*\vertsize] at (O6) {8};

	\node[xshift=0*\distance,yshift=-.75*\distance] {\LARGE $\Delta_{14,\,13}$};
\end{scope}

\begin{scope}[shift={(6*\distance,0)}]
	\coordinate (C) at (\vertsize,1.5*\vertsize);

	\coordinate (T1) at (-2.5*\rad,0.5*\rad);
	\coordinate (T2) at (-1.5*\rad,0.5*\rad);
	\coordinate (T3) at (-0.5*\rad,0.5*\rad);
	\coordinate (T4) at (2.5*\rad,0.5*\rad);
	\coordinate (T5) at (1.5*\rad,0.5*\rad);
	\coordinate (T6) at (0.5*\rad,0.5*\rad);
	\coordinate (B1) at (-2.5*\rad,-0.5*\rad);
	\coordinate (B2) at (-1.5*\rad,-0.5*\rad);
	\coordinate (B3) at (-0.5*\rad,-0.5*\rad);
	\coordinate (B4) at (2.5*\rad,-0.5*\rad);
	\coordinate (B5) at (1.5*\rad,-0.5*\rad);
	\coordinate (B6) at (0.5*\rad,-0.5*\rad);

	\coordinate (O1) at (-0*\angleSep:2*\rad);
	\coordinate (O2) at (-1*\angleSep:2*\rad);
	\coordinate (O3) at (-2*\angleSep:2*\rad);
	\coordinate (O4) at (-3*\angleSep:2*\rad);
	\coordinate (O5) at (-4*\angleSep:2*\rad);
	\coordinate (O6) at (-5*\angleSep:2*\rad);

	\draw [ultra thick, draw=black] (T1) -- (B1);
	\draw [ultra thick, draw=black] (T2) -- (B2);
	\draw [ultra thick, draw=black] (T3) -- (B3);
	\draw [ultra thick, draw=black] (T4) -- (B4);
	\draw [ultra thick, draw=black] (T5) -- (B5);
	\draw [ultra thick, draw=black] (T6) -- (B6);

	\draw [black, fill=black] (T1) circle [radius=\vertsize];
	\draw [black, fill=black] (T2) circle [radius=\vertsize];
	\draw [black, fill=black] (T3) circle [radius=\vertsize];
	\draw [black, fill=black] (T4) circle [radius=\vertsize];
	\draw [black, fill=black] (T5) circle [radius=\vertsize];
	\draw [black, fill=black] (T6) circle [radius=\vertsize];
	\draw [black, fill=black] (B1) circle [radius=\vertsize];
	\draw [black, fill=black] (B2) circle [radius=\vertsize];
	\draw [black, fill=black] (B3) circle [radius=\vertsize];
	\draw [black, fill=black] (B4) circle [radius=\vertsize];
	\draw [black, fill=black] (B5) circle [radius=\vertsize];
	\draw [black, fill=black] (B6) circle [radius=\vertsize];

	\node[yshift=1*\vertsize] at (T1) {1};
	\node[yshift=1*\vertsize] at (T2) {3};
	\node[yshift=1*\vertsize] at (T3) {5};
	\node[yshift=1*\vertsize] at (T4) {9};
	\node[yshift=1*\vertsize] at (T5) {11};
	\node[yshift=1*\vertsize] at (T6) {13};
	\node[yshift=-1*\vertsize] at (B1) {8};
	\node[yshift=-1*\vertsize] at (B2) {10};
	\node[yshift=-1*\vertsize] at (B3) {12};
	\node[yshift=-1*\vertsize] at (B4) {2};
	\node[yshift=-1*\vertsize] at (B5) {4};
	\node[yshift=-1*\vertsize] at (B6) {6};

	\node[xshift=0*\distance,yshift=-.75*\distance] {\LARGE $\Delta_{14,\,12}$};
\end{scope}

\begin{scope}[shift={(12*\distance,0)}]
	\coordinate (C) at (\vertsize,1.5*\vertsize);

	\coordinate (I1) at (\angleOff-0*\angleSep:\rad);
	\coordinate (I2) at (\angleOff-1*\angleSep:\rad);
	\coordinate (I3) at (\angleOff-2*\angleSep:\rad);
	\coordinate (I4) at (\angleOff-3*\angleSep:\rad);
	\coordinate (I5) at (\angleOff-4*\angleSep:\rad);
	\coordinate (I6) at (\angleOff-5*\angleSep:\rad);
	\coordinate (O1) at (-0*\angleSep:2*\rad);
	\coordinate (O2) at (-1*\angleSep:2*\rad);
	\coordinate (O3) at (-2*\angleSep:2*\rad);
	\coordinate (O4) at (-3*\angleSep:2*\rad);
	\coordinate (O5) at (-4*\angleSep:2*\rad);
	\coordinate (O6) at (-5*\angleSep:2*\rad);

	\draw [ultra thick, draw=black, fill=dimhighcolor, fill opacity=2*\triangleopacity] (I1) -- (I2) -- (I3) -- (I4) -- (I5) -- (I6)-- cycle;
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I1) -- (I3);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I1) -- (I4);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I1) -- (I5);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I1) -- (I6);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I2) -- (I4);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I2) -- (I5);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I2) -- (I6);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I3) -- (I5);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I3) -- (I6);
	\draw [ultra thick, draw=black, opacity=\subfaceopacity] (I4) -- (I6);
	\draw [ultra thick, draw=black, opacity=2*\subfaceopacity] (C) -- (I1);
	\draw [ultra thick, draw=black, opacity=2*\subfaceopacity] (C) -- (I2);
	\draw [ultra thick, draw=black, opacity=2*\subfaceopacity] (C) -- (I3);
	\draw [ultra thick, draw=black, opacity=2*\subfaceopacity] (C) -- (I4);
	\draw [ultra thick, draw=black, opacity=2*\subfaceopacity] (C) -- (I5);
	\draw [ultra thick, draw=black, opacity=2*\subfaceopacity] (C) -- (I6);

	\draw [ultra thick, draw=black] (I1) -- (O1);
	\draw [ultra thick, draw=black] (I2) -- (O2);
	\draw [ultra thick, draw=black] (I3) -- (O3);
	\draw [ultra thick, draw=black] (I4) -- (O4);
	\draw [ultra thick, draw=black] (I5) -- (O5);
	\draw [ultra thick, draw=black] (I6) -- (O6);
	\draw [ultra thick, draw=black] (I1) -- (O6);
	\draw [ultra thick, draw=black] (I2) -- (O1);
	\draw [ultra thick, draw=black] (I3) -- (O2);
	\draw [ultra thick, draw=black] (I4) -- (O3);
	\draw [ultra thick, draw=black] (I5) -- (O4);
	\draw [ultra thick, draw=black] (I6) -- (O5);

	\draw [black, fill=black] (C) circle [radius=\vertsize];
	\draw [black, fill=black] (I1) circle [radius=\vertsize];
	\draw [black, fill=black] (I2) circle [radius=\vertsize];
	\draw [black, fill=black] (I3) circle [radius=\vertsize];
	\draw [black, fill=black] (I4) circle [radius=\vertsize];
	\draw [black, fill=black] (I5) circle [radius=\vertsize];
	\draw [black, fill=black] (I6) circle [radius=\vertsize];
	\draw [black, fill=black] (O1) circle [radius=\vertsize];
	\draw [black, fill=black] (O2) circle [radius=\vertsize];
	\draw [black, fill=black] (O3) circle [radius=\vertsize];
	\draw [black, fill=black] (O4) circle [radius=\vertsize];
	\draw [black, fill=black] (O5) circle [radius=\vertsize];
	\draw [black, fill=black] (O6) circle [radius=\vertsize];

	\node[xshift=0.80*\vertsize, yshift=0.5*\vertsize] at (I1) {3};
	\node[xshift=0.8*\vertsize, yshift=-0.5*\vertsize] at (I2) {9};
	\node[yshift=-1.15*\vertsize] at (I3) {13};
	\node[xshift=-1*\vertsize, yshift=-0.5*\vertsize] at (I4) {11};
	\node[xshift=-0.9*\vertsize, yshift=0.6*\vertsize] at (I5) {5};
	\node[yshift=1*\vertsize] at (I6) {1};

	\node[xshift=1.25*\vertsize] at (O1) {10};
	\node[xshift=0.9*\vertsize, yshift=-0.9*\vertsize] at (O2) {2};
	\node[xshift=-0.9*\vertsize, yshift=-0.9*\vertsize] at (O3) {6};
	\node[xshift=-1.05*\vertsize] at (O4) {4};
	\node[xshift=-1*\vertsize, yshift=1*\vertsize] at (O5) {12};
	\node[xshift=0.9*\vertsize, yshift=0.9*\vertsize] at (O6) {8};

	\node[xshift=0*\distance,yshift=-.75*\distance] {\LARGE $\Delta_{14,\,11}$};
\end{scope}
\end{scope}
\end{tikzpicture}
\end{center}
\qed
\end{example}

We conclude this section with some avenues for further work. These conjectures were generated by explicitly computing $\Delta_{n,\ell}$ for all $n\leq 19$, using Theorem \ref{algorithm}. The following conjecture partially extends the investigation suggested by Theorem \ref{total-armslegs} to smaller $\ell$:

\begin{conjecture}
For any prime $p$ and even $\ell$ with $p>\ell\geq\frac{p-1}{2}$, the complexes $\Delta_{2p,\ell}$ have no isolated vertices; i.e., they have no facets of dimension 0.
\end{conjecture}

It is known, for instance by \cite[Theorem F.6]{menu-research-problems}, that all of these complexes are graphs. Therefore, this conjecture shows that all of their facets have dimension one. In general, a complex whose facets all have the same dimension is called \textbf{pure}.

\begin{conjecture}
For odd $n$ each of the complexes $\Delta_{n,\frac{n-1}{2}}$ and $\Delta_{n,\frac{n+1}{2}}$ is pure if and only if $n$ is prime.
\end{conjecture}

Surprisingly, the $h$-vector--- which in principle records nothing at all about facets--- seems to contain a sufficient condition for purity for $\Delta_{n,\ell}$:

\begin{conjecture}
The complex $\Delta_{n,\ell}$ is pure if $h_i\geq 0$ for all $i=1,\ldots ,d$.
\end{conjecture}

In a different direction, call a complex $\Gamma$ \textbf{connected} if for any two vertices $v$ and $w$, there is a sequence of vertices $(p_0=v,p_1,p_2,\dots p_{k-1}, w=p_{k})$ such that $\{p_i,p_{i+1}\}\in\Gamma$ for every $0\leq i\leq k-1$. We conjecture that a significant portion of all $\Delta_{n,\ell}$ are connected:

\begin{conjecture}
For any $n>2\ell$, the complex $\Delta_{n,\ell}$ is connected.
\end{conjecture}

Theorem \ref{total-doubling} shows that the inequality is sharp, in the sense that $\Delta_{2\ell,\ell}$ is connected if and only if $\ell$ is odd. However, if $n$ is odd, the data suggests that there may be some weaker bound on $\ell$; in particular, it appears that $\Delta_{n,\frac{n+1}{2}}$ is connected for $n\geq 7$.

The topologically sophisticated reader will be aware that a complex is connected if and only if its zeroth homology group vanishes. The small examples we computed suggest that higher homology groups also vanish when $n\gg h$, but the evidence is too weak to give a more quantitative estimate.

\section{Proofs of Theorems}
\label{all-the-proofs}

Recall that an integer $N$ is called even if there is another integer such that $N=2M$, and odd otherwise. In the proofs below, we will be performing modular arithmetic but it will be helpful to have these words available. So, we say that $N\in\mathbb{Z}/n\mathbb{Z}$ is \textbf{even} if $N=2M$ for some $M\in\mathbb{Z}/n\mathbb{Z}$, and \textbf{odd} otherwise. When $n$ is odd, there are no odd elements, but we will only be concerned with the case when $n$ is even, where these definitions are more intuitive.

\subsection*{Proof of Theorem \ref{total-doubling}}

\doubling*

\begin{enumerate}[(a)]
    \item The vertex set $V$ of $\Delta_{2\ell,\ell}$ contains precisely the odd elements of $\mathbb{Z}/2\ell\mathbb{Z}$; indeed, any $\ell$ even elements sum to zero.
    \newline\\
    All vertices in the same residue class modulo $2^{m+1}$ and thus contained in $V_t$ --- that is, vertices congruent to $2t+1\bmod{2^{m+1}}$ for a fixed $0\leq{t}\leq{2^m-1}$ --- are connected. If this were not the case, there would exist a set of elements that is not $\rho\cdot{2^m}$-zero-sumfree:
$$\left\{(2t + 1 + k_12^{m + 1}), (2t + 1 + k_22^{m + 1}),..., (2t + 1 + k_{\rho\cdot2^m}2^{m+1})\right\}$$
(where the $k_i$ for $0\leq{k_i}\leq{\rho-1}$ are not necessarily distinct), implying that
$$\sum_{i = 1}^{\rho\cdot{2^m}}(2t + 1 + 2^{m+1}k_i)\equiv{0}\bmod{\rho\cdot{2^{m+1}}}.$$
However, this is a contradiction since
\begin{align*}
\sum_{i = 1}^{\rho\cdot{2^m}}(2t + 1 + 2^{m+1}k_i) &= \sum_{i = 1}^{\rho\cdot{2^m}}2t + \sum_{i = 1}^{\rho\cdot{2^m}}1 + \sum_{i=1}^{\rho\cdot{}2^m}2^{m+1}k_i\\
&=(\rho\cdot{2^{m}})2t + (\rho\cdot{2^m})1 + (\rho\cdot{2^m})2\sum_{i=1}^{\rho\cdot{2^m}}2^mk_i\\
&= \rho\cdot{2^{m+1}}t + \rho\cdot{2^{m}} + \rho\cdot{2^{m+1}}\sum_{i=1}^{\rho\cdot{2^m}}2^mk_i\\
&\equiv{0 + \rho\cdot{2^m} + 0 \bmod{\rho\cdot{2^{m + 1}}}}\\
&\equiv{\rho\cdot{2^m}}\not\equiv{0\bmod{\rho\cdot{2^{m+1}}}}.
\end{align*}
Thus all vertices in the same residue class modulo $2^{m+1}$ are connected, forming a $\rho-1$-simplex.
\newline\\
No two vertices lying in different residue classes modulo $2^{m+1}$ are connected, demonstrated by the following proof: suppose we have any two vertices contained in different residue classes modulo $2^{m+1}$, $2t_1 + k_1\cdot{2^{m+1}}$ and $2t_2 + 1 k_2\cdot{2^{m+1}}$ where $t_1\neq{t_2}$. Then if it is possible for us to solve the congruence
$$r\cdot(2t_1 + 1 + k_1\cdot2^{m+1}) + (\rho\cdot{2^m}-r)\cdot(2t_2 + 1 + k_2\cdot{2^{m+1}}) \equiv 0 \bmod{\rho\cdot{2^{m+1}}}$$
for $r$, then $2t_1 + 1 + k_1\cdot2^{m+1}$ and $2t_2 + 1 + k_2\cdot{2^{m+1}}$ are not $\rho\cdot{2^{m}}$-zero-sumfree. Note that
\begin{align*}r\cdot(2t_1 + 1 + k_1\cdot2^{m+1}) &+ (\rho\cdot{2^m}-r)\cdot(2t_2 + 1 + k_2\cdot{2^{m+1}})\\
&\equiv 2r(t_1 - t_2 + 2^{m}(k_1-k_2)) + \rho\cdot{2^{m}}\bmod{\rho\cdot2^{m+1}}
\end{align*}
Thus we wish to find $r$ such that
\begin{align}r(t_1 - t_2 + 2^{m}(k_1-k_2))\equiv -\rho\cdot{2^{m-1}}\bmod{\rho\cdot2^{m}}.
\end{align}
We can express $t_1-t_2 = t_3\cdot2^e$ for some odd number $t_3$ and $0\leq{e}\leq{m-2}$. Then we have
\begin{align}
r(t_3 + 2^{m-e}(k_1-k_2))\equiv{-\rho\cdot{2^{m-1-e}}}\bmod{\rho\cdot2^{m-e}}.
\end{align}
Since $t_3 + 2^{m-e}(k_1-k_2)$ is odd, $\gcd(t_3 + 2^{m-e}(k_1-k_2), \rho\cdot2^{m-e}) = q$ if and only if $\gcd(t_3 + 2^{m-e}(k_1-k_2),\rho) = q$.
Suppose that $\gcd(t_3 + 2^{m-e}(k_1-k_2),\rho) = q$ for some $1\leq{q}\leq{\rho}$. Suppose also that
$$\frac{t_3 + 2^{m-e}(k_1-k_2)}{q} = s\text{ and } \frac{\rho}{q} = \rho'$$
and thus $gcd(s,\rho') = 1$, implying $s$ is invertible modulo $\rho'\cdot{2^{m-e}}$. Then we have from (2),
\begin{align*}
r(qs)&\equiv{-q\rho'\cdot{2^{m-1-e}}}\bmod{q\rho'\cdot{2^{m-e}}}\\
rs&\equiv-\rho'2^{m-1-e}\bmod{\rho'\cdot2^{m-e}}\\
r&\equiv \frac{-\rho'2^{m-1-e}}{s}\bmod{\rho'\cdot2^{m-e}}.
\end{align*}
and thus $2t_1 + k_1\cdot{2^{m+1}}$ and $2t_2 + k_2\cdot{2^{m+1}}$ are not $\rho\cdot{2^{m}}$-zero-sumfree. This implies that any two vertices contained in different residue classes modulo $2^{m+1}$ are disconnected in $\Delta_{\rho\cdot{2^m},\rho\cdot{2^{m+1}}}$. These facts show that $\Gamma = \Delta_0\cup\Delta_1\cup\cdots\cup\Delta_{2^m-1}$ is a collection of $2^m$ many $(\rho-1)$-simplices.

\item Let $\mathcal{L}(\mathcal{A})$ be the corresponding intersection lattice of subspaces. Since $\Gamma$ on a vertex set $V$ is a collection of $2^m$ many $(\rho-1)$-simplices, then by Proposition \ref{lattice-rank2}, $\mathcal{L}(\mathcal{A})$ is graded of rank $2$.
\item Now, notice that $|V|=\ell$ and for each facet $F\in\Gamma,$ $|F|=\rho.$ So, by Corollary \ref{char-poly-union-eqdim},  $$\chi_\mathcal{A}(x)=x^\ell-2^mx^\rho+(2^m-1).$$

\end{enumerate}

\begin{corollary}
If $n=\rho\cdot 2^{m+1}, \ell=\rho\cdot 2^{m}$ for all $m\geq{1}$ and for some odd number $\rho$, then $\Delta_{n,\ell}$ is a pure simplicial complex.
\end{corollary}

\begin{proof}
By \textbf{Theorem \ref{total-doubling}(\ref{doubling})}, $\Delta_{2^{m+1}\rho,2^m\rho}$ contains $2^m\rho$ disjoint $(\rho-1)$-simplices. Since every simplex is pure, $\Delta_{2^{m+1}\rho,2^m\rho}$ is pure.
\end{proof}

\subsection*{Proof of Theorem \ref{total-primepowers}}

\primepowers*

\begin{enumerate}[(a)]
    \item Partition the vertex set of $\Delta_{p^e,p^e-1}$ into $e\cdot(p-1)$ disjoint sets, each denoted $V_{i,j}$, where
    $$V_{i,j} = \left\{x\in{V}\;\ifnum\currentgrouptype=16 \middle\fi|\; x\equiv i\cdot{p^{j-1}}\bmod{p^{j}},1\leq{i}\leq{p-1}\right\}.$$
    One can comprehend the structure of $\Delta_{p^e,p^e-1}$ with greater ease by use of the following table:
        \begin{center}
        \begin{tabular}{c|c|c|c|c|c|c}
        $\bm{\bmod{\hspace{0.1cm}p^1}}$ & $p^{0}\cdot1$&$p^0\cdot{2}$&$p^0\cdot{3}$&\ldots&$p^0\cdot{(p-2)}$&$p^0\cdot{(p-1)}$\\\hline
        $\bm{\bmod{\hspace{0.1cm}p^2}}$& $p^1\cdot{1}$&$p^1\cdot{2}$&$p^1\cdot{3}$&\ldots&$p^1\cdot{(p-2)}$&$p^1\cdot{(p-1)}$\\\hline
        $\bm{\bmod{\hspace{0.1cm}p^3}}$& $p^2\cdot{1}$&$p^2\cdot{2}$&$p^2\cdot{3}$&\ldots&$p^2\cdot{(p-2)}$&$p^2\cdot{(p-1)}$\\\hline
        $\vdots$&$\vdots$&$\vdots$&$\vdots$&$\ddots$&$\vdots$&$\vdots$\\\hline
        $\bm{\bmod{\hspace{0.1cm}p^{e-1}}}$& $p^{e-2}\cdot{1}$&$p^{e-2}\cdot{2}$&$p^{e-2}\cdot{3}$&\ldots&$p^{e-2}\cdot{(p-2)}$&$p^{e-2}\cdot{(p-1)}$\\\hline
        $\bm{\bmod{\hspace{0.1cm}p^{e}}}$& $p^{e-1}\cdot{1}$&$p^{e-1}\cdot{2}$&$p^{e-1}\cdot{3}$&\ldots&$p^{e-1}\cdot{(p-2)}$&$p^{e-1}\cdot{(p-1)}$
        \end{tabular}
        \end{center}
        Note that each cell of this represents the set of all vertices in the vertex set of $\Delta_{p^e,p^e-1}$ that are equivalent to the quantity labelling the cell modulo the bolded value labelling the corresponding row. For example, in the first row, the set contained in the third cell labelled ``$p^0\cdot{2}$'' is the set $\left\{x\in{V}: x\equiv{p^0\cdot{2}}\bmod{p^1}\right\}$ where $V$ is the vertex set of $\Delta_{p^e,p^e-1}$.
    We will prove that (1) all vertices in $V_{i,j}$ form a $(p^e-1)$-zero-sumfree set, and (2) no vertex in $V_{i,j}$ is connected to any vertex in $V_{i',j'}$ unless both $i = i'$, and $j = j'$.
    \newline\\
    To prove $(1)$, suppose for the sake of contradiction that the vertices in $V_{i,j}$ for some $i$ and $j$ are not $(p^e-1)$-zero-sumfree. Then there must exist some subset of $p^e-1$ (not necessarily distinct) vertices from $V_{i,j}$, $\left\{i\cdot{p^{j-1}} + t_1p^j, i\cdot{p^{j-1}} + t_2p^{j},...,i\cdot{p^{j-1}} + t_{p^e-1}p^j\right\}$ such that
    $$\sum_{k = 0}^{p^{e}-1}\left(i\cdot{p^{j-1} + t_kp^j}\right)\equiv{0}\bmod{p^e}.$$
    However,
\begin{align*}
    \sum_{k=0}^{p^e-1}\left(p^{j-1}\cdot{i} + t_kp^j\right)
    &=\left(p^e-1\right)p^{j-1}\cdot{i} + p^j\sum_{k=0}^{p^e-1}t_k
    = p^{e+j-1}\cdot{i} - p^{j-1}\cdot{i} + p^j\sum_{k=0}^{p^e-1}t_k\\
    &\equiv -p^{j-1}\cdot{i} + p^j\sum_{k=0}^{p^e-1}t_k\bmod{p^e}
    \equiv p^{j-1}\left(p\sum_{k=0}^{p^e-1}t_k - i\right)\bmod{p^e}
\end{align*}
    For this to be equivalent to $0\bmod{p^e}$, it must be that
    $$p\sum_{k=0}^{p^e-1}t_k - i\equiv{0}\bmod{p^{e - j +1}}$$
    which is a contradiction since $1\leq{i}\leq{p-1}$, and thus $\gcd(i,p) = 1$. Thus the vertices in $V_{i,j}$ form a $(p^2-1)$-zero-sumfree set.
    \newline\\
    To prove $(2)$, consider vertices $(p^{j-1}\cdot{i} + t_1p^{j})\in{V_{i,j}}$ and $(p^{j'-1}\cdot{i'} + t_2p^{j'})\in V_{i',j'}$ such that either $i\neq{i'}$ or $j\neq{j'}$ (or both). These two vertices are $(p^{e}-1)$-zero-sumfree if and only if there exists no $r$ such that the following critical equivalence holds:
    $$r\cdot(i + pt_1)p^{j-1} + (p^e - 1 - r)\cdot(i' + pt_2)p^{j'-1}\equiv{0}\bmod{p^e}.$$
    Suppose for the first case that $i\neq{i'}$, but that $j = j'$. Suppose without loss of generality that $i' = i + s_1$ for some $1\leq{s}\leq{p-1}$. Then the left-hand side of the critical equivalence is
    $$p^{j-1}((p^e-1)i + rs_1 + rpt_2 + (p^e - 1 - r)pt_1)\equiv{p^{j-1}}(-j - pt_1 + r(s_1 + p(t_2 - t_1)))\bmod{p^e}.$$
    Since $\gcd(s_1,p) = 1$, $\gcd(s_1 + p(t_2 - t_1),p^{e-j+1}$ and thus $s_1 + p(t_2 - t_1)$ is invertible modulo $p^{e-j+1}$. Thus if we let
    $$r\equiv\frac{i + pt_2}{s_1 + p(t_2 - t_1)}\bmod{p^{e-j+1}},$$
    $r$ solves the critical equivalence, implying that the desired vertices are not adjacent if $i\neq{i'}$ and $j = j'$.
    \newline\\
    For the second case, assume that $i = i'$ but that $j \neq{j'}$. Assume without loss of generality that $j' = j + s_2$ for some $1\leq{s_2\leq{e-1}}$. The left-hand side of the critical equivalence becomes
    $$r\cdot{p^{j+s_2-1}}(i + pt_2) + (p^e-1-r)(i + pt_1)\equiv{p^{j-1}}(r(p^{s_2}(i+pt_2)-i-pt_1) - (i+pt_1))\bmod{p^e}.$$
    Since $\gcd(i,p) = 1$, $\gcd(p^{s_2}(i + pt_2) - i - pt_1,p^{e-j + 1}) = 1$, and thus $p^{s_2}(i + pt_2) - i - pt_1$ is invertible modulo $p^{e-j + 1}$. If we then let
    $$r\equiv\frac{i + pt_1}{p^{s_2}(i + pt_2) - i - pt_1},$$
    then the critical equivalence is solved, implying that the two vertices are not adjacent.
    \newline\\
    For the final case, assume that both $i\neq{i'}$ and that $j\neq{j'}$. Assume without loss of generality that $i' = i + s_1$ and that $j' = j + s_2$, for $1\leq{s_1}\leq{p-1}$ and $1\leq{s_2}\leq{e-1}$. The left-hand side of the critical equivalence is then
    $$r\cdot{p^{j + s_2 - 1}}(i + s_1 + pt_2) + (p^e - 1 - r)p^{j-1}(i + pt_1)\equiv p^{j-1}(r(p^{s_2}(i + s_1 + pt_2) - i - pt_1) - (i + pt_1)).$$
    Since $\gcd(i,p) = 1$, $\gcd(p^{s_2}(i + s_1 + pt_2) - i - pt_1,p^{e-j+1}) = 1$, and thus $p^{s_2}(i + s_1 + pt_2) - i - pt_1$ is invertible modulo $p^{e - j + 1}$. If we then let
    $$r\equiv\frac{i + pt_1}{p^{s_2}(i + s_1 + pt_2) - i - pt_2}\bmod{p^{e-j+1}},$$
    then the critical equivalence is solved, implying that the vertices are nonadjacent.
    \newline\\
    These facts prove that $\Gamma = \Delta_{p^e,p^e-1}$ is a collection of $e(p-1)$ many simplices, with $e-1$ many $(p^j-1)$-simplices for $0\leq{j}\leq{e-1}$.

    \item Let $\mathcal{L}(\mathcal{A})$ be the corresponding intersection lattice of subspaces. Since $\Delta_{p^e,p^e-1}$ is a collection of $e(p-1)$ many $(p^j-1)$-simplices for each $0\leq j\leq e-1$, then by Proposition \ref{lattice-rank2}, $\mathcal{L}(\mathcal{A})$ is graded of rank $2$.

    \item Furthermore, $\Delta_{p^e,p^e-1}$ has $p^e-1$ vertices, and for any facet $|F|=p^j$ for some $0\leq j\leq e-1$. Notice that for each dimension $p^j-1$ with $0\leq j\leq e-1,$ there are $p-1$ facets, and therefore, $p-1$ atoms in $\mathcal{L}(\mathcal{A}).$ So, by Proposition \ref{char-poly-union}, $$\chi_\mathcal{A}(x)= x^{p^e-1}-(p-1)\sum^{e-1}_{j=0}x^{p^{j}}.$$

\end{enumerate}

\subsection*{Proof of Theorem \ref{total-armslegs}}
In what follows, we say that two vertices $i$ and $j$ are \textbf{adjacent} in a simplicial complex $\Gamma$ if $\{i,j\}\in\Gamma$.

\begin{lemma}
\label{arms-legs-simplices-lemma}
Consider $\Gamma_s = \Delta_{2p,2p-s}$ for prime $p\geq{3}$ and odd $s\leq 2p$. Then the set of odd vertices is a face of $\Gamma_s$.
\end{lemma}
\begin{proof}
Take any collection of $2p-s$ many (not necessarily distinct) odd vertices $\{2t_1 + 1,2t_2 + 1,\cdots,2t_{2p-s} + 1\}$.
Then
$$\sum_{j=1}^{2p-s}(2t_j + 1) = 2\sum_{j=1}^{2p-s}(t_j) + 2p-s = 2\left(\sum_{j=1}^{2p-s}t_j - \frac{s-1}{2}\right) - 1,$$
which is an odd element of $\mathbb{Z}/2p\mathbb{Z}$, and thus nonzero. Hence, any subset of odd vertices must be a face of $\Gamma_s$; in particular, the set of all odd vertices is a face of $\Gamma_s$.
\end{proof}

\begin{lemma}
\label{no-even-vertices-arms-legs}
Consider $\Gamma_s = \Delta_{2p,2p-s}$ for $p$ prime and $s\leq p$. Then no two even vertices are adjacent in $\Gamma_s$; moreover, if $s$ is even, then no two odd vertices are adjacent either.
\end{lemma}
\begin{proof}
Two even vertices $v$ and $w$ are adjacent in $\Gamma_s$ if and only if the critical equivalence
 $$r(v) + (2p-s-r)(w)\equiv{0}\bmod{2p}$$
has no solution $0\leq{r}\leq{2p-s}$. Then if we substitute in two distinct even vertices $v=2t_1$ and $w=2t_2$, the critical equivalence becomes
\begin{align*}
r(2t_1) - (2p-s-r)(2t_2) &\equiv{0}\bmod{2p} \\
\equiv{r(2(t_1-t_2)) - s(2t_2)} &\equiv{0}\bmod{2p} \\
r(2(t_1-t_2)) &\equiv s(2t_2)\bmod{2p} \\
r(t_1-t_2)&\equiv st_2\bmod{p}
\end{align*}
Note that $t_1-t_2\not\equiv{0}\bmod{p}$, otherwise $2t_1$ and $2t_2$ are not distinct modulo $2p$. Thus $t_1 - t_2$ is invertible modulo $p$, and so
$$r\equiv{\frac{st_2}{t_1 - t_2}}\bmod{p}.$$
There is necessarily such an $r$ in the range $0\leq r<p\leq 2p-s$, and so no two even vertices are adjacent.
\newline\\
Now consider the case where $s$ is even, and let $\frac{s}{2} = q$. For any odd vertices $v = 2t_1 + 1$ and $w = 2t_2 + 1$,
\begin{align*}
r(2t_1+1) - (2p-s-r)(2t_2+1) &\equiv{0}\bmod{2p} \\
r(2(t_1-t_2)) - s(2t_2+1) &\equiv{0}\bmod{2p} \\
r(2(t_1-t_2)) &\equiv s(2t_2+1)\bmod{2p} \\
r(t_1-t_2)&\equiv q(2t_2+1)\bmod{p}.
\end{align*}
As before, $t_1-t_2$ is invertible modulo $p$, and hence
$$r\equiv\frac{q(2t_2 + 1)}{t_1-t_2}\bmod{p}.$$
As before, there is such an $r$ in the range $0\leq{r}<p\leq{2p-s}$, and thus no two odd vertices are adjacent in this case.
\end{proof}

\begin{lemma}
\label{arms-lemma}
Fix a prime $p$, a positive integer $s<p$, an odd element $i\in\mathbb{Z}/2p\mathbb{Z}$ and a nonzero even element $j\in\mathbb{Z}/2p\mathbb{Z}$. Then a solution exists to the equivalence
$$r(i) + (2p-s-r)j\equiv{0}\bmod{2p}$$
if and only if $i\not\equiv{j}\bmod{p}$. In this case, all solutions satisfy
$$r\equiv\frac{sj}{i-j}\bmod{2p}.$$
\end{lemma}
\begin{proof}
Begin by observing that
\begin{align*}
r(i) + (2p-r-s)(j) &\equiv{0} \bmod{2p} \\
r(i-j) - sj &\equiv{0}\bmod{2p} \\
r(i-j) &\equiv{sj}\bmod{2p}
\end{align*}
First, suppose that $i\not\equiv{j}\bmod{p}$. Note that since $i-j\not\equiv{0}\bmod{p}$ and is also odd, we have that $\gcd(i-j,2p) = 1$, and thus $i-j$ has a multiplicative inverse modulo $2p$. Then a solution to the critical equivalence exists, and any solution satisfies
$r\equiv\frac{sj}{i-j}\bmod{2p}$.
\newline

To prove that if $i\equiv{j}\bmod{p}$, then there exists no solution to the critical equivalence, suppose that $i\equiv{j}\bmod{p}$. Then then in the critical equivalence, since $i-j$ is odd, this implies that $r$ must be even. Suppose that $\frac{r}{2} = q$. Then the critical equivalence becomes
$$q(i-j)\equiv{sj}\bmod{p}.$$
Since $i-j\equiv{0}\bmod{p}$, and since $j\not\equiv{0}\bmod{p}$ and $s\not\equiv{0}\bmod{p}$ (as $s<p$), it follows that there exists no solution to the critical equivalence when $i\equiv{j}\bmod{p}$. These facts show that a solution to the critical equivalence exists if and only if $i\not\equiv{j}\bmod{p}$.
\end{proof}

\armslegs*

\begin{enumerate}[(a)]
    \item
    \begin{itemize}
        \item The set of odd vertices is a face of $\Gamma_1$ by \textbf{Lemma \ref{arms-legs-simplices-lemma}}.

    A vertex $i$ is adjacent to an vertex $j$ if and only if the critical equivalence
    $$r(i) + (2p-1-r)(j)\equiv{0}\bmod{2p}$$
    has no solution $1\leq{r}\leq{2p-1}.$ By \textbf{Lemma \ref{arms-lemma}}, $i$ odd and $j$ even, a solution to this equivalence exists if and only if $i\equiv{j}\bmod{p}$. Thus, an odd vertex $i$ is adjacent to an even vertex $j$ if and only if $i\equiv{j}\bmod{p}$.
    Note that by \textbf{Lemma \ref{no-even-vertices-arms-legs}}, we also have that no even vertices are adjacent in $\Gamma_1$.

    \item Two vertices $i$ and $j$ are adjacent in $\Gamma_2$ if and only if the critical equivalence
 $$r(i) + (2p-2-r)(j)\equiv{0}\bmod{2p}$$
 has no solution $1\leq{r}\leq{2p-2}.$ Observe that if $r\equiv{-1}\bmod{2p}$, then
 $$-1(i-j)\equiv{2j}\bmod{2p}\implies j-i\equiv{2j}\bmod{2p}$$
 and so $i\equiv{-j}\bmod{2p}$. In particular, $r\not\equiv{-1}\bmod{2p}$ is never a solution if $i$ is odd and $j$ is even. Therefore, by \textbf{Lemma \ref{arms-lemma}}, an odd vertex $i$ is adjacent to an even vertex $j$ if and only if $i\equiv{j}\bmod{p}$.
 \newline
Finally, by \textbf{Lemma \ref{no-even-vertices-arms-legs}}, no two even vertices are adjacent in $\Gamma_2$ and no two odd vertices are adjacent in $\Gamma_2$ either.
\item By \textbf{Lemma \ref{arms-legs-simplices-lemma}}, the set of odd vertices forms a face in $\Gamma_3$, and by \textbf{Lemma \ref{no-even-vertices-arms-legs}}, no even vertices in $\Gamma_3$ are adjacent.
\newline\\
Note that any vertex $i$ is adjacent to a vertex $j$ if and only if the critical equivalence
$$r(i) + (2p-3-r)(j)\equiv{0}\bmod{2p}$$
has no solution $1\leq{r}\leq{2p-3}$. This equivalence holds only when
$$r(i - j)\equiv 3j\bmod{2p}.$$
For $i$ odd and $j$ even, by \textbf{Lemma \ref{arms-lemma}} these vertices are adjacent when $i\equiv{j}\bmod{p}$ since no solution to the equation exists in this case. Now, suppose that $i\not\equiv{j}\bmod{p}.$ Then by \textbf{Lemma \ref{arms-lemma}}, it follows that a solution to the critical equivalence exists. Note that if a solution satisfies, for any two vertices $i$ and $j$, $1\leq{r}\leq{2p-3}\bmod{2p}$, then by \textbf{Lemma \ref{arms-lemma}} it follows that $i$ and $j$ are disconnected. Now, suppose our solution satisfies
$$r\equiv{-1}\bmod{2p}.$$
Then this implies that
\begin{align*}
-1(i-j)&\equiv{3j}\bmod{2p}\\
j-i&\equiv{3j}\bmod{2p}\\
i&\equiv{-2j}\bmod{2p}
\end{align*}
which is a contradiction as $i$ is assumed to be odd. For the case that a solution exists that satisfies $r\equiv{-2}\bmod{p}$, this implies that
\begin{align*}
    -2(i-j)&\equiv{3j}\bmod{2p}\\
    2j - 2i&\equiv{3j}\bmod{2p}\\
    -2i&\equiv{j}\bmod{2p}
\end{align*}
Thus, a solution satisfies $r\equiv{-2}\bmod{2p}$ if and only if $j\equiv{-2i}\bmod{2p}$. Thus, since an odd vertex $v$ is disconnected from an even vertex $w$ only when a solution exists satisfying $1\leq{r}\leq{2p-3}\bmod{2p}$, it follows that the vertices $i$ and $j$ are connected when $j\equiv{-2i}\bmod{2p}.$ Thus, an odd vertex $i$ is connected to an even vertex $j$ if and only if either $i\equiv{j}\bmod{p}$ or $j\equiv{-2i}\bmod{2p}.$
\end{itemize}

Now consider the lattice $\mathcal{L}(\mathcal{A}_1).$ Recall that $\Delta_{2p,2p-s}$ $s\in\{1,3\}$ contains exactly one facet, $F_0,$ such that $F_0$ is a $(p-1)$-simplex with odd vertices, $v\in\{1,3,\ldots, 2p-1\}.$ Now, for all facets $F\neq F_0,$ recall that $F=\{\l_1,\l_2\}$ where $\l_1$ odd and $\l_2$ even. The corresponding subspace for each facet is $S_F=\{\bm{x}\in\mathbb{K}^{2p-1}:x_i=0,\forall i\not\in F\}.$ Furthermore, notice that for any $F\neq F_0$, $F_0\cap F=\{\l_1\},$ $\l_1$ odd and $\l_1\neq p.$ So, the corresponding subspace is $S_{\l_1}=\{\bm{x}\in\mathbb{K}^{2p-1}:x_i=0,\forall i\in V\setminus\{\l_1\}\},$ with $\l_1$ odd and $\l_1\neq p.$

For $s=1,$ there are $p$ many facets, thus there are $p$ many $S_F.$ Since all $S_F\neq S_{F_0}$ are disjoint, we have that there are $p-1$ $S_{\l_1}.$ Now, since each $\l_1$ is distinct, the intersection of any $S_{\l_1}$ is $\{0\}=\hat{1},$ so we are done.

For $s=2,$ we also have that for each even $\l_2\in V,$ there exists facets $F_1,F_2\neq F_0$ such that $F_1\cap F_2=\{\l_2\}$ where $\l_2\equiv -2\l_1\mod{2p}.$ So, along with the previous subspaces $S_{\l_1},$ we have that there are $p-1$ subspaces $S_{\l_2}=\{\bm{x}:x_i=0,\forall i\in V\setminus\{\l_2\}\}$ with $\l_2$ even. Since for each even $\l_2\in V$ and each odd $\l_1\in V\setminus\{p\},$ there exists correspodning subspace $S_\l,$ for all $\l\in V\setminus\{p\},$ then the intersection of any two subspaces $S_\l$, is $\{0\}=\hat{1}.$

Therefore, any maximal chain of $\mathcal{L}(\mathcal{A}_s)$ for either $s=1$ or $s=3$
has the form
\begin{align*}
    \mathbb{K}^{2p-1}\supset S_F
    \supset S_\l
    \supset\{0\},
\end{align*}
for some facet $F$ and some $\l\in F$. Moreover, every $\mathcal{L}(\mathcal{A}_s)$ has the same length, thus $\mathcal{L}(\mathcal{A}_s)$ is graded of rank $3$ when $s\in\{1,3\}$.

Note that $\mu(\hat{0},\hat{0})=+1.$ Since each $S_F$ is an atom in $\mathcal{L}(\mathcal{A}_s)$, $s\in\{1,3\},$ then $\mu(\hat{0},S_{F_0})=-1$ and $\mu(\hat{0},S_{F})=-1.$ Moreover, each $S_\l$ is the intersection of any two $S_F.$ So, $\mu(\hat{0},S_\l)=+1.$ Finally, the intersection of any $S_\l$ is $\{0\}=\hat{1}.$ Thus, $\mu(\hat{0},\hat{1})=-1.$

Since for $s=1$ there are $p$ many facets, then there are $p$ many $S_F$, with $p-1$ of dimension $2$ and one of dimension $p$, which corresponds to the $(p-1)$-simplex. Furthermore, each $S_\l$ has dimension $1$ and there are $(p-1)$ odd vertices in $V,$ then there are $p-1$ $S_\l.$ Thus, $$\chi_{\mathcal{A}_1}=x^{2p-1}-x^{p}-(p-1)x^2+(p-1)x.$$

For $s=3$ in addition to the facets and intersections contained in the $\mathcal{L}(\mathcal{A}_1)$ there are $p-1$ many $S_F$ with dimension $2$, which gives us $p-1$ more $S_\l$ of dimension $1.$ Therefore, $$\chi_{\mathcal{A}_3}=x^{2p-1}-x^p-2(p-1)x^2+2(p-1)x$$

Finally, let $\mathcal{L}(\mathcal{A}_2)$ be the corresponding intersection lattice of subspaces. Since $\Gamma$ on a vertex set $V$ is a collection of $p$ many edges, then by Proposition \ref{lattice-rank2}, $\mathcal{L}(\mathcal{A}_2)$ is graded of rank $2$.
Now, notice that $|V|=2p-2$ and for each facet $F\in\Gamma,$ $|F|=2.$ So, by Corollary \ref{char-poly-union-eqdim},  $$\chi_\mathcal{A}(x)=x^{2p-2}-p\cdot x^{2}+p-1.$$

\end{enumerate}

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


Here we are using JUnit and embedded Jetty for testing different rest methods (getName-> GET & changeName-> POST). We are attaching CXFRestfulTutorial.war programmatically and this war contains all the rest methods(getName & changeName) we are going to test.
2 Responses to "CXF Webservice Using Embedded Jetty"
org.eclipse.jetty.http.HttpBuffers cannot be resolved. It is indirectly referenced from required .class files, any idea?
This issue is related to HttpBuffers class, which internally used by your program.
2) You can download jetty-all.jar, which contains all the required class in a single package as mentioned in my example.


text:
Burris' Justin Xia has broken several records but he's chasing more, including Lary Schulhof's from 1960
Mobile apps offer last-minute deals
Stacy A. Anderson
NEW YORK – While new technology and mobile applications have made old-fashion road trips easier, the thrill of spontaneity and surprise still remains. If travelers are flexible, apps can offer ease with last-minute deals, especially good for day trips and overnight stays.
Need a hotel room immediately? There's an app for that. Or for a quick turnaround— when you just need a space to rest for a few hours—there's an app for that too.
Here are three handy apps for mobile device users that I often use on weekend trips from Washington, D.C., to the Big Apple, which can be used for other destination cities.
HotelTonight offers discounted rooms at more than 10,000 hotels in nearly 400 destination cities including Miami, San Francisco, Los Angeles, Chicago and Boston to international hotspots like London, Paris and Barcelona. Reservations are for the night of the day you select the discounted rate and can be booked up to five nights.
The app recently launched Express Check-in, which allows guests to immediately pick up their keys at the front desk without a lengthy process. This zip feature is only available for select hotels using the app on iPhones, and will roll out to Android users later this month.
Over a recent holiday weekend in New York— when locals often vacate the city and tourists take over— HotelTonight still offered substantial deals, especially for the early worm checking the app when new postings go live at 9 a.m. Rooms can be booked until 2 a.m.
For example, the Element by Westin in Times Square, a solid-rated Starwood Hotels and Resorts property, offered one king bed for $129 on HotelTonight, compared to $169 on its own website.
For a step up, the Soho House New York, a lux-rated hotel in the trendy Meatpacking District, posted a small room with a queen size bed for $298 on the app, a nearly 15 percent savings compared to $350 on its own website.
If you've checked out of the hotel room already and don't depart until later in the day, or take a day trip and have some leisure down time, there may be a private space for relief in your city— or coming soon.
The Breather app launched in New York City earlier this year lets you check-in to a commercial space complete with a table and couch to "meet, work or rest" for 30 minutes to several hours at a time. Some rooms are available as early as 6 a.m. to 10 p.m. for $25 an hour, and may accommodate up to 10 people. Reservations can be made via the app on iPhone and Android devices, or online.
Co-founder and CEO Julien Smith said Breather is a welcome alternative to loud, bustling coffee chains where travelers often fight for space and outlets.
TKTS
New York is best known for big lights and live shows on Broadway, but that doesn't mean you have to pay hundreds for top-notch seats.
The Theatre Development Fund has TKTS Discount Booths with locations in Times Square, South Street Seaport and downtown Brooklyn that sell Broadway and off-Broadway show tickets for 20 to 50 percent off.
Though discounted tickets must be purchased in person, a TKTS app lets customers see availability in real-time for evening shows that day and matinees.

redpajama_set_name: RedPajamaCommonCrawl


text:
Posts Tagged 'clouds'
On December 7th Eve and I flew from Austin to Seattle, then changed planes for Taipei, and finally changed once more to get to Cebu City, whose metropolitan area has the second largest population in the Philippines. While much of our 19-day trip went for family matters on the island of Cebu, including a wedding, I'd brought along a reduced version of my usual photo kit and hoped to get in some nature photography.
One Philippine province Eve (and therefore I) had never visited was Palawan, and so on the morning of December 12th we flew to the island of Busuanga in the very northern part of Palawan. That afternoon we joined a tour of the main town, Coron. The last place the tour took us was the base of Mount Tapyas, whose heights we reached by climbing 724* steps (and by enduring sore leg muscles when we had to climb more steps the next day). I see on the internet that Mount Tapyas is known for its sunsets, and it didn't let us down.
In the first photograph the sun was still so bright that I underexposed by 3, 4, and even 5 f/stops to keep from blowing out the highlights in the solar disk. By the time of the second picture, which came 13 minutes later, I got away with an underexposure of only 1.33 f/stops, though you'll notice some flaring on the hills beneath the sun. Just chalk it up to my usual flair as a photographer.
— — — — — — — — — — — — — — — — — — — — — — — — — — — — — —
* When our tour guide told us that there are 724 steps my immediate reaction was to think that 7 and 24 happen to be the lengths of the legs of a right triangle with a hypotenuse of 25 (you can do the arithmetic to verify that 7 squared plus 24 squared equals 25 squared).
Tagged with clouds, landscape, Philippines, sun, sunset, water
Enchanted Rock, part 5
A couple of years ago I was shown a photograph taken by Brian P. Barnes of a geological structure at Enchanted Rock I'd never seen or even heard of. Eventually I learned that it's called Window Rock, and that's where I most wanted to go during our November 1st visit. No one in the park's office could point out on a trail map exactly where the structure is, but one of the staff marked off a stretch of the Loop Trail and told me that Window Rock is located a short distance off that section of the trail. After trying several side paths and not finding Window Rock, I finally came to one that took me to what I'd been looking for.
That path led to the rock but not initially to the best photographs. The picture above shows how the side of the formation that greeted us was shadowed, given that the sun was in front of us. I got around that difficulty by literally working my way around to the other side for better lighting.
As with the jug-like boulders in the previous post, I spent time portraying
Window Rock from various angles and in different degrees of abstraction.
The view below strongly reminded me of the moai on Easter Island.
And so ends the series of posts devoted to Enchanted Rock.
Tagged with abstract, clouds, Enchanted Rock, geology, rocks, Texas, Texas Hill Country
Artists in general and photographers in particular sometimes like to depict the same person or thing in various ways. The Cubists got excited about showing multiple views of a subject simultaneously, as in Marcel Duchamp's famous "Nude Descending a Staircase." At Enchanted Rock on November 1st I took a more-conventional approach, making separate photographs showing different aspects of an intriguing boulder formation that looked like huge jugs or flasks with short hoodoos in lieu of stoppers. The first photograph gives you an overview of the formation.
The second view isolates part of the formation that was central in the first image.
I wouldn't be me if I didn't include some closer and more-abstract takes on these formations,
the first of which gives you a better look at the orange and yellow lichens on the boulder above.
The two abstractions below continue playing up the rough texture of the weather-sculpted rocks.
You might think you're looking at the ruins of some ancient civilization in a desert.
The wispy clouds that stayed with us the whole time made for excellent backdrops.
Tagged with abstract, clouds, Enchanted Rock, geology, landscape, rocks, Texas, Texas Hill Country


Lion Janine Helman (#16) left too early and Lions Cade Wilson (#27) and Brad Carson (#44) were both absent and again one of our Tail Twisters won the pot, Lion Bill Dickens (#38).
Only 15 members were present. Our Attendance and Membership Committees need to call those missing.
We had 7 members which attended the District 7-0 Convention last Saturday. Thanks to those members who helped make the convention enjoyable.
We put out the flags Monday for Columbus Day. Thanks to all members who worked a flag route. Our next flag day will be Election Day, Tuesday, November 4th.
Michelle Gibson, owner of The Mortgage Source, was our speaker this week. She gave a most informative talk about the mortgage lending business and current trends affected by the turn of events on Wall Street. Mrs. Gibson has been in the mortgage lending business since 1981. She also talked about the reverse mortgage loans and who is eligible for a reverse mortgage loan.
Next Tuesday is our regular monthly business meeting but, we will also have as a speaker, Miss Craighead County Rice Jillian Russell.
Our District Governor John Blackley will make his official visit to our club on October 28. Thanks to Lion Ron Erwin for the excellent programs during the month of October. Lion Charles Hartwig will be our program chairman during the month of November.


Jazz singer Mary Jenson has gathered some intriguing songs for this new record and she sings with real class.
The opening stab at Tom Waits, "Temptation" has a fresh style arrangement that really brings out the best in her voice. The title song is one of Jenson's own song and it's a dreamy adventure that features some strong vocals. "Flying, Falling", also an original song, is a sweetly arranged ballad with dreamy vocals and a nice piano part by Frank Martin.
"Moon at the Window" by Joni Mitchell gets a relaxed and confident reading that sees Jenson at her best. It's a truly strong album.


text:
Q: Why current date is showing next date from the current date in php? I want to get current date in php but it is displaying one day advance.
I am surprised why this is happening as I am using very simple code:
date('Y-m-d')

A: Try to put this function date_default_timezone_set("Asia/Bangkok")

A: Please set date time zone as IST. Then try. May your server config time zone is USA. So please add below lines in your code and check it
date_default_timezone_set('Asia/Kolkata');
echo date('d-m-Y');

For more about PHP date time please visit here

redpajama_set_name: RedPajamaStackExchange


Venture Heads is our series about Silicon Valley VCs and Kool-Aid drinking startups. In this episode, a paranoid startup founder tries to pitch his idea without revealing what it is.
Venture Heads is our new series about Silicon Valley VCs and Kool-Aid drinking startups. In this episode, Dunbar pitches a solar bikini and other crazy ideas to his venture partners.
Venture Heads is our new series about Silicon Valley VCs and Kool-Aid drinking startups. In this episode, a young startup figures out how to manipulate the stock market using Twitter.
Venture Heads is our new series about Silicon Valley VCs and Kool-Aid drinking startups. In this episode, a startup pitches their scheme to get filthy rich off of Bitcoins.
Venture Heads is our new series about Silicon Valley VCs and Kool-Aid drinking startups. In this episode, a startup pitches their concept for starting the first colony on Mars.
Venture Heads is our new series about Silicon Valley VCs and Kool-Aid drinking startups. In this episode, Chad ventures into the deep web — the dark side of the Internet no one talks about and few dare enter.
Venture Heads is our new series about Silicon Valley VCs and Kool-Aid drinking startups. In this episode, a startup can't decide if they want to save the world, make money or spy on celebs.


Rootsgear - For The Culture - Black Zip Up Hoodie - Rootsgear Clothing Co.
From our Fall/Winter 18 Collection. This design is an ode to the rich tradition of the Phulkari textiles from Punjab. Traditionally the Phulkari is hand embroidered in vibrant colors and patterns. This design is our futuristic digital take on the Phulkari.
Printed on cotton/poly fleece zip up hoodie. Unisex fit.