text:
\section{$\MenuGap{X}=1$ when $k=1$}\label{app:single}

In this brief section we prove that when $k=1$, for any sequence of $x_i \in \mathbb{R}^+_{\geq 0}$, $\MenuGap{X} = 1$.

\begin{claim}
When $k=1$, for any $X = \{x_i\}_{i=1}^{N}$, $x_i \in \mathbb{R}^+_{\geq 0}$, $\MenuGap{X} = 1$.
\end{claim}

\begin{proof}
Note that when $k=1$, $||x_i||_1 = x_i$. Therefore, $\MenuGap{X, Q} = \sum_i \min_{j < i} (q_i - q_j)$. We make the following observation which allows us to look at structured optimal solutions.

\begin{observation}
Any optimal solution $Q$ to $\MenuGap{X}$ is monotone non-decreasing.
\end{observation}

\begin{proof}
For the sake of contradiction, suppose we are given an optimal solution $Q$ that is not monotone non-decreasing. Let $i$ be the smallest index for which $q_i < q_{i-1}$. Then $\mathrm{gap}_i^{X,Q} = (q_i - q_{i-1}) x_i < 0$. Consider instead a solution $Q'$ where $q'_j = q_j$ for all $j \neq i$ and $q'_i = q_{i-1}$. Now, $\mathrm{gap}_i^{X,Q'} = 0$. Since $Q_{\leq i-1} = Q'_{\leq i-1}$, $\mathrm{gap}_j^{X,Q} = \mathrm{gap}_j^{X,Q'}$ for all $j < i$. Since $q_{i-1} > q_i$, for any $j >i$ it holds that $(q_j - q_{i-1}) < (q_j - q_i)$. Therefore, $q_i$ is not ``setting the gap'' for any point after it. Hence it also holds that $\mathrm{gap}_j^{X,Q} = \mathrm{gap}_j^{X,Q'}$ for all $j > i$. Putting everything together we get that $\MenuGap{X,Q'} - \MenuGap{X,Q} = \mathrm{gap}_{i}^{X,Q'} - \mathrm{gap}_i^{X,Q} > 0$ contradicting the optimality of $Q$.
\end{proof}

With this observation in hand, since the $q_i$ are monotone non-decreasing, without loss of generality it holds that $\mathrm{gap}_{i}^{X,Q} = \min_{j < i} q_i - q_{j} = q_i - q_{i-1}$ ($q_{i-1} \geq q_j$ for all $j < i$). Therefore, we get $\MenuGap{X,Q} = \sum_i q_i-q_{i-1} = q_N - q_0$. Since $q_0 = 0$ and $0 \leq q_N \leq 1$, we get that $\MenuGap{X,Q} \leq 1$.

Finally, note that for any $X$, we can set $q_N = 1$ and $q_i = 0$ for all other $i$, proving that $\MenuGap{X} \geq 1$.

\end{proof}

\section{Omitted Proofs}\label{app:omitted}
\begin{proof}[Proof of Lemma~\ref{lem:align}]
We prove that for all $X,C$, $\ScaGap{X,C} \leq \MenuGap{X}$, which implies the lemma. For a given $X,C$, define:
\begin{itemize}
\item $\vec{q}_i := c_i \cdot \vec{x}_i$, if $\mathrm{sgap}_i^{X,C} > 0$.
\item $\vec{q}_i:=\arg\max_{j < i}\{c_j \cdot \vec{x}_j\}$, if $\mathrm{sgap}_i^{X,C} \leq 0$.
\end{itemize}

Observe first that each $\vec{q}_i \in [0,1]^k$, as each $c_i \vec{x}_i \in [0,1]^k$ (this follows because each component of $\vec{x}_i$ is at most $||\vec{x}_i||_\infty$, and each $c_i$ is at most $1/||\vec{x}_i||_\infty$). Next, observe that if $\mathrm{sgap}_i^{X,C} \leq 0$, then $\mathrm{gap}_i^{X,Q} = 0$. This is by definition in bullet two above. Finally, observe that if $\mathrm{sgap}_i^{X,C} > 0$, then $\mathrm{gap}_i^{X,Q} \geq \mathrm{sgap}_i^{X,C}$. This is because the set of $\{\vec{q}_j\}_{j < i}$ is a subset of $\{c_j \vec{x}_j\}_{j < i}$, and because $\vec{q}_i := c_i \cdot \vec{x}_i$ by bullet one. Therefore, $\mathrm{gap}_i^{X,Q} \geq \max\{0,\mathrm{sgap}_i^{X,C}\}$ for all $i$ and the lemma follows.
\end{proof}

\section{Proof of Corollary~\ref{cor:main}}\label{app:main}
We prove Corollary~\ref{cor:main} by making use of Theorem~\ref{thm:newHN} combined with the sequence $X$ from Section~\ref{sec:description}. The only task is to confirm that $\arev{D} < \infty$ for the resulting $\mathcal{D}$, which essentially requires that we execute and analyze the construction fully. Let us quickly review the~\cite{hart2019selling} construction, given as input a sequence $X$:
\begin{itemize}
\item Let $B$ be a very large constant, to be defined later.
\item Let $\vec{v}_i:=B^{2^i} \cdot \vec{x}_i/||\vec{x}_i||_1$ (for all $i$).
\item Let $\mathcal{D}$ sample $\vec{v}_i$ with probability $1/B^{2^i}$ (for all $i$).
\item Let $\mathcal{D}$ sample $\vec{0}$ with probability $1-\sum_{i\geq 1}1/B^{2^i}$.
\end{itemize}

\cite{hart2019selling} establishes that the above construction yields Theorem~\ref{thm:newHN} (for sufficiently large $B$, as a function of $\varepsilon$). To complete the proof of Corollary~\ref{cor:main}, we just need to relate $\arev{D}$ for this construction to $\ScaGap{X}$.

\begin{proposition}\label{prop:hn} The construction above yields a $\mathcal{D}$ satisfying $\arev{\mathcal{D}} \leq \ScaGap{X}+1/B$.
\end{proposition}
\begin{proof}
Consider any mechanism $M$. We show that $\ScaGap{X} \geq \arev{\mathcal{D},M}-1/B$. To see this, consider the following choice of $C$:
\begin{itemize}
\item If $\vec{v}_i$ is parallel to $\vec{q}^M(\vec{v}_i)$, set $c_i:=||\vec{q}^M(\vec{v}_i)||_2/||\vec{x}_i||_2$.
\item If $\vec{v}_i$ is not parallel to $\vec{q}^M(\vec{v}_i)$, set $c_i:=0$.
\end{itemize}

We now need to lower bound $\mathrm{sgap}_i^{X,C}$, when $i$ satisfies the first bullet. Observe that, because $M$ is truthful, we must have, for all $j < i$:

\begin{align*}
\vec{v}_i \cdot \vec{q}^M(\vec{v}_i) - p^M(\vec{v}_i) &\geq \vec{v}_i \cdot \vec{q}^M(\vec{v}_j) - p^M(\vec{v}_j)\\
\Rightarrow p^M(\vec{v}_i) &\leq p^M(\vec{v}_j) + B^{2^i}\vec{x}_i\cdot (c_i \vec{x}_i - c_j \vec{x}_j)/||\vec{x}||_1\\
\Rightarrow p^M(\vec{v}_i) &\leq 2B^{2^{i-1}}+B^{2^i}\mathrm{sgap}_i^{X,C}/||\vec{x}_i||_1
\end{align*}
Above, the first line follows from incentive compatibility. The second line follows as $\vec{q}^M(\vec{v}_i) = c_i \vec{x}_i$ for all $i$ in the first bullet, and either $\vec{q}^M(\vec{v}_j) = c_j \vec{x}_j$, or $c_j = 0$. The final line follows by taking $j:=\arg\min_{j < i}\{\vec{v}_i\cdot (c_i \vec{v}_i - c_j \vec{v}_j)\}$, and by observing that $\vec{v}_j$ cannot possibly pay more than their value for the grand bundle.

We can then conclude that:

\begin{align*}
\arev{\mathcal{D},M} &\leq \sum_i (2B^{2^{i-1}}+B^{2^i} \mathrm{sgap}_i^{X,C}/||\vec{x}_i||_1)/B^{2^i}\\
&\leq \sum_i 2/B^{2^{i-1}} + \ScaGap{X}\\
&\leq \ScaGap{X}+1/B.
\end{align*}

\end{proof}

Because we can take $B$ as large as we like, we can construct a $\mathcal{D}$ such that $\arev{\mathcal{D}}$ is arbitrarily close to $\ScaGap{X}$, while also maintaining that $\rev{\mathcal{D}}$ is arbitrarily close to $\MenuGap{X}$. Because Theorem~\ref{thm:main2} provides a construction $X$ such that $\MenuGap{X}/\ScaGap{X} = \infty$, the~\cite{hart2019selling} construction, with sufficiently large $B$, yields a $\mathcal{D}$ with $\rev{\mathcal{D}}/\arev{\mathcal{D}} = \infty$, completing the proof of Corollary~\ref{cor:main}.
\section{Minor Extension of Theorem~\ref{thm:main}}
\label{app:ext}

Here, we establish that Theorem~\ref{thm:main} preserves alignment, with the following analogue.

\begin{theorem}\label{thm:ext}
For any distribution $\mathcal{D}$ over $k$ items, there exists a sequence of points $X=(\vec{x}_i)^N_{i=1}$, with each $\vec{x}_i \in \text{supp}(\mathcal{D})$, such that
\[ \ScaGap{X} \geq \frac{\arev{\mathcal{D}}}{9\brev{\mathcal{D}}}. \]

In particular, if $\arev{\mathcal{D}}/\brev{\mathcal{D}} = \infty$, then $\ScaGap{X} = \infty$ as well.
\end{theorem}

The proof of Theorem~\ref{thm:ext} is nearly identical to the proof of Theorem~\ref{thm:main}, and essentially just observes that the entire proof ``preserves alignment''. We include a proof here for completeness, although a significant fraction of the proof is copied verbatim from Section~\ref{sec:converse}. We note that the entire outline is identical, so we omit filler text reminding the reader of the outline.

\subsection{Simplifying the Optimal Mechanism}
\label{sec:simpleext}
\begin{claim}
\label{claim:priceext}
For all $c \in \mathbb{R}_{\geq 0}$, all distributions $\mathcal{D}$, and all mechanisms $M$, there exists a $c$-expensive mechanism $M'$ satisfying $\arev{\mathcal{D}, M'} \geq \arev{\mathcal{D},M}-c$.
\end{claim}

\begin{proof}
Take $M'$ to be exactly the same as $M$, except having removed all entries with price $<c$. For every value in the support of $\mathcal{D}$ with $p^M(\vec{v}) \geq c$ in $M$, we still have $p^{M'}(\vec{v}) \geq c$. This is simply because we in fact have $(\vec{q}^M(\vec{v}),p^M(\vec{v})) = (\vec{q}^{M'}(\vec{v}),p^{M'}(\vec{v}))$: $\vec{v}$'s favorite option in $M$ is still available in $M'$, and all options in $M'$ were also available in $M$. For any value with $p^M(\vec{v}) < c$, we clearly have $p^{M'}(\vec{v}) \geq 0$.

Because all $\vec{v}$ with $p^{M}(\vec{v}) \geq c$ have $\vec{q}^M(\vec{v}) = \vec{q}^{M'}(\vec{v})$, we conclude that for all $\vec{v}$: $p^{M'}(\vec{v})\cdot I(\text{$\vec{v}$ is parallel to $\vec{q}^{M'}(\vec{v})$}) \geq p^M(\vec{v})I(\text{$\vec{v}$ is parallel to $\vec{q}^{M}(\vec{v})$}) - c$, and the claim follows by taking an expectation over $\mathcal{D}$.
\end{proof}

\begin{claim}\label{claim:bucketsext}
For all $c$-expensive mechanisms $M$, and all $\mathcal{D}$, there exists either an oddly-priced or evenly-priced $c$-expensive mechanism $M'$ satisfying $\arev{\mathcal{D},M'} \geq \arev{\mathcal{D},M}/2$.
\end{claim}
\begin{proof}
Simply let $M_1$ denote the set of menu options from $M$ whose price lies in $[c\cdot 2^i,c\cdot 2^{i+1})$ for an odd integer $i$, and $M_2$ denote the remaining menu options (which lie in $[c \cdot 2^i,c \cdot 2^{i+1})$ for an even power of $i$). It is easy to see that $M_1$ is oddly-priced and $M_2$ is evenly-priced. Then for all $\vec{v}$, we must have $p^{M_1}(\vec{v})I(\text{$\vec{v}$ is parallel to $\vec{q}^{M_1}(\vec{v})$}) + p^{M_2}(\vec{v})I(\text{$\vec{v}$ is parallel to $\vec{q}^{M_2}(\vec{v})$}) \geq p^M(\vec{v})I(\text{$\vec{v}$ is parallel to $\vec{q}^{M}(\vec{v})$})$. This is because $\vec{v}$'s favorite menu option from $M$ appears in one of the two menus, and is necessarily $\vec{v}$'s favorite option on that menu (and they pay non-zero from the other menu). Taking an expectation over $\mathcal{D}$ yields that $\arev{\mathcal{D},M_1} +\arev{\mathcal{D},M_2} \geq \arev{\mathcal{D},M}$, completing the proof.
\end{proof}

\subsection{Connecting Structured Mechanisms to MenuGap}
\label{sec:connectext}

\begin{definition}[Aligned Sequences]\footnote{\mattnote{Come up with a better name?}} Let $M$ be a $c$-expensive mechanism which is oddly-priced or evenly-priced, and let $\mathcal{D}$ be any distribution. An $\varepsilon$-\emph{aligned sequence} for $M, \mathcal{D}$ is the following:
\begin{itemize}
    \item Define $c_0:=0$.
    \item Define offset $a$ to be $1$ if $M$ is oddly-priced, and $0$ if $M$ is evenly-priced.
\item Define $A$ to be the subset of the support of $\mathcal{D}$ such that $\vec{q}^M(\vec{v})$ is parallel to $\vec{v}$.
    \item For all $j \in \mathbb{N}_+$, define $B_j:=\{\vec{v} \in A,p^M(\vec{v}) \in [c\cdot 2^{2(j-1)+a},c\cdot 2^{2(j-1)+a+1})$.
    \item For all $j \in \mathbb{N}_+$, let $\vec{x}_j\in B_j$ be such that $||\vec{x}_j||_1 \leq ||\vec{v}||_1\cdot (1+\varepsilon)$ for all $\vec{v} \in B_j$.\footnote{Again note that such $\vec{x}_j$ must exist if $B_j$ is non-empty. If $B_j$ is empty, again skip $\vec{x}_j$ and $\vec{c_j}$ in both lists.}
    \item For all $j \in \mathbb{N}_+$, let $c_j:=||\vec{q}^M(\vec{x}_j)||_1/||\vec{x}_j||_1$.
\end{itemize}
\end{definition}

\begin{proposition}\label{prop:mainali} Let $M$ be a $c$-expensive, and oddly- or evenly-priced. Let $(X,C)$ be an $\varepsilon$-aligned sequence for $M,\mathcal{D}$. Then: $$\ScaGap{X,C} \geq \frac{\arev{\mathcal{D},M}}{4(1+\varepsilon)\brev{\mathcal{D}}}.$$
\end{proposition}

\begin{proof}

The proof will follow from two technical claims. The first, Claim~\ref{claim:ell1}, still holds for aligned sequences. We prove an analog of Claim~\ref{claim:menugap} now.

\begin{claim}\label{claim:menugapali} If $(X,C)$ is $\varepsilon$-aligned with $M,\mathcal{D}$, then $\mathrm{sgap}_i^{X,C} \geq p^M(\vec{x}_i)/2$.
\end{claim}

\begin{proof}
Recall that $\mathrm{sgap}_i^{X,C}:= \min_{j < i}\{\vec{x}_i \cdot (c_i \cdot \vec{x}_i - c_j \cdot \vec{x}_j)\}$, and that $c_i \cdot \vec{x}_i :=\vec{q}^M(\vec{x}_i)$. For any fixed $j < i$, recall that because $M$ was a truthful mechanism, we must have:

\begin{align*}
    &\vec{x}_i \cdot \vec{q}^M(\vec{x}_i) - p^M(\vec{x}_i) \geq \vec{x}_i \cdot \vec{q}^M(\vec{x}_j) - p^M(\vec{x}_j)\\
    \Rightarrow &\vec{x}_i \cdot (c_i \cdot \vec{x}_i - c_j \cdot \vec{x}_j) \geq p^M(\vec{x}_i)- p^M(\vec{x}_j)\\
    \Rightarrow &\vec{x}_i \cdot (c_i \cdot \vec{x}_i - c_j \cdot \vec{x}_j) \geq p^M(\vec{x}_i)/2.
\end{align*}
The first line is simply restating incentive compatibility. The second line is basic algebra, and substituting $c_i \cdot \vec{x}_i :=\vec{q}^M(\vec{x}_i)$. The third line invokes the fact that $p^M(\vec{x}_i) \geq 2^{2(i-1)+a}$, while $p^M(\vec{x}_j) < 2^{2(j-1)+a+1} \leq 2^{2(i-1)+a-1}$.

\end{proof}
With these two claims, we can wrap up the proof of Proposition~\ref{prop:mainali}. We can write the following:

\begin{align*}
    \ScaGap{X,C}&:= \sum_{i=1}^\infty \frac{\mathrm{sgap}_i^{X,C}}{||\vec{x}_i||_1}\\
    &\geq \sum_{i=1}^\infty \frac{p^M(\vec{x}_i)}{2\cdot ||\vec{x}_i||_1}\\
    &\geq \sum_{i=1}^\infty \frac{p^M(\vec{x}_i)\cdot \Pr_{\vec{v} \sim \mathcal{D}}[\vec{v} \in B_i]}{2 \cdot (1+\varepsilon)\brev{\mathcal{D}}}\\
    &\geq \sum_{i=1}^\infty \frac{\mathbb{E}_{\vec{v} \sim \mathcal{D}|\vec{v} \in B_i}[p^M(\vec{v})] \cdot \Pr_{\vec{v} \sim \mathcal{D}}[\vec{v} \in B_i]}{4 \cdot (1+\varepsilon)\brev{\mathcal{D}}}\\
    &\geq \frac{\arev{\mathcal{D},M}}{4\cdot (1+\varepsilon)\brev{\mathcal{D}}}
\end{align*}

Above, the first line is simply the definition of AlignGap. The second line uses Claim~\ref{claim:menugapali}. The third line uses Claim~\ref{claim:ell1}. The fourth line uses the fact that for all $\vec{v} \in B_i$, $p^M(\vec{v}) \leq 2^{2(i-1)+a+1} = 2 \cdot 2^{2(i-1)+a} \leq 2p_i$. The final line is just rewriting the definition of $\arev{\mathcal{D},M}$, and using the fact that every $\vec{v} \in \text{supp}(\mathcal{D})$ with $p(\vec{v}) > 0$ is in some $B_i$.

\end{proof}

Proposition~\ref{prop:mainali}, together with the fact that an oddly- or evenly-priced mechanism is guaranteed to get a good approximation to the optimum, now suffices to prove Theorem~\ref{thm:ext}.

\begin{proof}[Proof of Theorem~\ref{thm:ext}]
The proof will be a simple consequence of the technical lemmas in this section, once we set $c$ and $\varepsilon$ appropriately. In particular, set $c:=\arev{\mathcal{D}}/100$, and $\varepsilon = 1/100$. Note that $\varepsilon$-associated sequences are guaranteed to exist for any $M, \mathcal{D}$, as $\varepsilon > 0$.

Start by letting $M'$ denote the $c$-expensive mechanism promised by Claim~\ref{claim:price}. Let then $M''$ denote the oddly- or evenly-priced mechanism promised by Claim~\ref{claim:buckets}. Finally, let $X,C$ denote the $\varepsilon$-associated sequence for $M'',\mathcal{D}$. We get:
\begin{align*}
   \ScaGap{X,C} &\geq \frac{\arev{\mathcal{D},M''}}{{4(1+\varepsilon)\brev{\mathcal{D}}}}\\
    &\geq \frac{\arev{\mathcal{D},M'}}{8(1+\varepsilon)\brev{\mathcal{D}}}\\
    &\geq \frac{\arev{\mathcal{D}}-c}{8(1+\varepsilon)\brev{\mathcal{D}}}\\
    &\geq \frac{\arev{\mathcal{D}}}{9\brev{\mathcal{D}}}.
\end{align*}
The first line follows immediately from Proposition~\ref{prop:mainali}. The second follows immediately from Claim~\ref{claim:buckets}. The third is immediately from Claim~\ref{claim:price}. The final line follows by substituting in our choice of $c$ and $\varepsilon$.
\end{proof}
\section{Conclusion}
\label{sec:conclusion}

We study the nature of distributions $\mathcal{D}$ with $\rev{\mathcal{D}}/\brev{\mathcal{D}} = \infty$. Prior work established a framework to construct such distributions, and therefore established sufficient conditions~\cite{BriestCKW15, hart2019selling}. Our first main result establishes that the most general of these frameworks is in fact complete (Theorem~\ref{thm:main}). Our second main result establishes that the more restrictive framework, through which all previous constructions arose, is not complete (Theorem~\ref{thm:main2}). Finally, we build upon our main construction to develop a novel distribution $\mathcal{D}$ witnessing $\rev{\mathcal{D}}/\brev{\mathcal{D}}=\infty$, but for which none of the ``aligned'' mechanisms of prior work can possibly witness this (Corollary~\ref{cor:main}).

In terms of future work, it remains open as to whether there is an alternative definition for ``sequences with $\MenuGap{X}=\infty$'' which is easier to parse. Our work establishes that understanding such sequences is necessary and sufficient to understand distributions with $\rev{\mathcal{D}}/\brev{\mathcal{D}} = \infty$, and establishes that ``sequences with $\MenuGap{X} = \infty$'' is not equivalent to ``sequences with $\ScaGap{X} = \infty$.'' But it would be exciting for future work to better understand sequences with $\MenuGap{X} = \infty$.

Interestingly, ongoing work by~\cite{AS22} uses the framework developed in this paper in order to prove \emph{approximation results} for so-called fine-grained buy-many mechanisms, a class of mechanisms which interpolates between buy-one and the recently introduced buy-many mechanisms~\cite{ChawlaTT19, ChawlaTT20}. In general, our techniques make it possible for future work to explore the gap between various simple versus optimal benchmarks without having to reason directly about the underlying mechanisms.
\section{A Converse to Theorem~\ref{thm:newHN}: The~\cite{hart2019selling} Framework is WLOG}\label{sec:converse}
In this section, we prove Theorem~\ref{thm:main}. Our proof has two main parts. First, we will take the optimal auction for $\mathcal{D}$ (or one that is arbitrarily close to optimal) and repeatedly simplify it through a sequence of lemmas, at the cost of small constant-factors of revenue. The second part takes this simple menu and draws a connection to MenuGap.

\subsection{Simplifying the Optimal Mechanism}
\label{sec:simple}

We show that for every $\mathcal{D}$, an approximately-optimal mechanism exists satisfying some useful properties. Our first step simply argues that we may ignore menu options with low prices.

\begin{definition} We say that a mechanism $M$ is \emph{$c$-expensive} if every option has price at least $c$.
\end{definition}

\begin{claim}
\label{claim:price}
For all $c \in \mathbb{R}_{\geq 0}$, all distributions $\mathcal{D}$, and all mechanisms $M$, there exists a $c$-expensive mechanism $M'$ satisfying $\rev{\mathcal{D}, M'} \geq \rev{\mathcal{D},M}-c$.
\end{claim}

\begin{proof}
Take $M'$ to be exactly the same as $M$, except having removed all entries with price $<c$. For every value in the support of $\mathcal{D}$ with $p^M(\vec{v}) \geq c$ in $M$, we still have $p^{M'}(\vec{v}) \geq c$. This is simply observing that $\vec{v}$'s favorite option in $M$ is still available in $M'$, and all options in $M'$ were also available in $M$. For any value with $p^M(\vec{v}) < c$, we clearly have $p^{M'}(\vec{v}) \geq 0$. So for all $\vec{v}$, we have $p^{M'}(\vec{v}) \geq p^M(\vec{v}) - c$, and the claim follows by taking an expectation with respect to $\vec{v}$.
\end{proof}

Our next step will show that we can assume further structure on the prices charged, at the cost of a factor of two.

\begin{definition}
A $c$-expensive mechanism $M$ is \emph{oddly-priced} (respectively, \emph{evenly-priced}) if for all $\vec{v}$, there exists an odd (respectively, even) integer $i$ such that $p^M(\vec{v}) \in [c \cdot 2^i, c \cdot 2^{i+1})$.
\end{definition}

\begin{claim}\label{claim:buckets}
For all $c$-expensive mechanisms $M$, and all $\mathcal{D}$, there exists either an oddly-priced or evenly-priced $c$-expensive mechanism $M'$ satisfying $\rev{\mathcal{D},M'} \geq \rev{\mathcal{D},M}/2$.
\end{claim}
\begin{proof}
Simply let $M_1$ denote the set of menu options from $M$ whose price lies in $[c\cdot 2^i,c\cdot 2^{i+1})$ for an odd integer $i$, and $M_2$ denote the remaining menu options (which lie in $[c \cdot 2^i,c \cdot 2^{i+1})$ for an even power of $i$). It is easy to see that $M_1$ is oddly-priced and $M_2$ is evenly-priced. Then for all $\vec{v}$, we must have $p^{M_1}(\vec{v}) + p^{M_2}(\vec{v}) \geq p^M(\vec{v})$. This is because $\vec{v}$'s favorite menu option from $M$ appears in one of the two menus, and is necessarily $\vec{v}$'s favorite option on that menu (and they pay non-zero from the other menu). Taking an expectation with respect to $\vec{v}$ yields that $\rev{\mathcal{D},M_1} +\rev{\mathcal{D},M_2} \geq \rev{\mathcal{D},M}$, completing the proof.
\end{proof}

This concludes our simplification of the mechanism. In the subsequent section, we draw a connection between MenuGap and the revenue of oddly- or evenly-price $c$-expensive mechanisms.

\subsection{Connecting Structured Mechanisms to MenuGap}
\label{sec:connect}

We begin with the following definition, which describes our proposed $X,Q$ based on a structured mechanism for $\mathcal{D}$.

\begin{definition}[Representative Sequences] Let $M$ be a $c$-expensive mechanism which is oddly-priced or evenly-priced, and let $\mathcal{D}$ be any distribution. An $\varepsilon$-\emph{representative sequence} for $M, \mathcal{D}$ is the following:
\begin{itemize}
    \item Define $\vec{q}_0(\mathcal{D},M):=(0,\ldots, 0)$.
    \item Define offset $a$ to be $1$ if $M$ is oddly-priced, and $0$ if $M$ is evenly-priced.
    \item For all $j \in \mathbb{N}_+$, define $B_j:=\{\vec{v} \in \text{supp}(\mathcal{D}) : p^M(\vec{v}) \in [c\cdot 2^{2(j-1)+a},c\cdot 2^{2(j-1)+a+1})\}$.
    \item For all $j \in \mathbb{N}_+$, let $\vec{x}_j\in B_j$ be such that $||\vec{x}_j||_1 \leq ||\vec{v}||_1\cdot (1+\varepsilon)$ for all $\vec{v} \in B_j$.\footnote{Note that for all $\varepsilon>0$, such an $\vec{x}_j$ exists, even if $B_j$ is not closed (as long as $B_j$ is non-empty). In particular, because $M$ is $c$-expensive, we know that $||\vec{v}||_1 \geq c > 0$ for all $\vec{v}$ who pay $\geq c$. If $B_j$ is empty, instead omit $\vec{x}_j,\vec{q}_j$ from both lists (i.e. decrease all future indices by one).}
    \item For all $j \in \mathbb{N}_+$, let $\vec{q}_j:=\vec{q}^M(\vec{x}_j)$.
\end{itemize}
\end{definition}

\begin{proposition}\label{prop:main} Let $M$ be a $c$-expensive, and oddly- or evenly-priced. Let $(X,Q)$ be an $\varepsilon$-representative sequence for $M,\mathcal{D}$. Then: $$\MenuGap{X,Q}\geq \frac{\rev{\mathcal{D},M}}{4(1+\varepsilon)\brev{\mathcal{D}}}\qquad.$$
\end{proposition}

\begin{proof}
The proof will follow immediately from two technical claims. The first claim relates $||\vec{x}_i||_1$ and $\brev{\mathcal{D}}$.

\begin{claim}
\label{claim:ell1}
$(1+\varepsilon)\cdot \brev{\mathcal{D}} \geq \norm{\vec{x}_i}_1 \cdot  \Pr_{\vec{v} \sim \mathcal{D}}[\vec{v} \in B_i]$.
\end{claim}

\begin{proof}
Recall that $(1+\varepsilon)\cdot ||\vec{v}||_1 \geq ||\vec{x}_i||_1$ for all $\vec{v} \in B_i$. Therefore, if we set a price of $\norm{\vec{x}_i}_1/(1+\varepsilon)$ for the grand bundle, every $\vec{v} \in B_i$ would choose to purchase the grand bundle. This immediately implies the claim, as: \[ \brev{\mathcal{D}} \geq \frac{||\vec{x}_i||_1}{1+\varepsilon}\cdot \Pr_{\vec{v} \sim \mathcal{D}}[||\vec{v}||_1 \geq ||\vec{x}_i||_1/(1+\varepsilon)] \geq \frac{||\vec{x}_i||_1}{1+\varepsilon} \cdot \Pr_{\vec{v} \sim \mathcal{D}}[\vec{v} \in B_i]. \qedhere \]
\end{proof}

The second claim relates $\rev{\mathcal{D},M}$ and $\MenuGap{X,Q}$. Crucially, this claim uses the fact that the mechanism is either oddly-priced or evenly-priced (and therefore $p^M(\vec{x}_i), p^M(\vec{x}_j)$ differ by at least a factor of $2$, for any $i \neq j$).

\begin{claim}\label{claim:menugap} If $X,Q$ is an $\varepsilon$-representative sequence for $M,\mathcal{D}$, then $\mathrm{gap}_i^{X,Q} \geq p^M(\vec{x}_i)/2$.
 \end{claim}
\begin{proof}
Recall that $\mathrm{gap}_i^{X,Q}:= \min_{j < i}\{\vec{x}_i \cdot (\vec{q}_i - \vec{q}_j)\}$, and that $\vec{q}_i:=\vec{q}^M(\vec{x}_i)$. For any fixed $j < i$, recall that because $M$ was a truthful mechanism, we must have:

\begin{align*}
    &\vec{x}_i \cdot \vec{q}^M(\vec{x}_i) - p^M(\vec{x}_i) \geq \vec{x}_i \cdot \vec{q}^M(\vec{x}_j) - p^M(\vec{x}_j)\\
    \Rightarrow &\vec{x}_i \cdot (\vec{q}_i - \vec{q}_j) \geq p^M(\vec{x}_i)- p^M(\vec{x}_j)\\
    \Rightarrow &\vec{x}_i \cdot (\vec{q}_i - \vec{q}_j) \geq p^M(\vec{x}_i)/2.
\end{align*}
The first line is simply restating incentive compatibility. The second line is basic algebra, and substituting $\vec{q}_i:=\vec{q}^M(\vec{x}_i)$. The third line invokes the fact that $p^M(\vec{x}_i) \geq 2^{2(i-1)+a}$, while $p^M(\vec{x}_j) < 2^{2(j-1)+a+1} \leq 2^{2(i-1)+a-1}$.

\end{proof}

With these two claims, we can wrap up the proof of Proposition~\ref{prop:main}. We can write the following:

\begin{align*}
    \MenuGap{X,Q}&:= \sum_{i=1}^\infty \frac{\mathrm{gap}_i^{X,Q}}{||\vec{x}_i||_1}\\
    &\geq \sum_{i=1}^\infty \frac{p^M(\vec{x}_i)}{2\cdot ||\vec{x}_i||_1}\\
    &\geq \sum_{i=1}^\infty \frac{p^M(\vec{x}_i)\cdot \Pr_{\vec{v} \sim \mathcal{D}}[\vec{v} \in B_i]}{2 \cdot (1+\varepsilon)\brev{\mathcal{D}}}\\
    &\geq \sum_{i=1}^\infty \frac{\mathbb{E}_{\vec{v} \sim \mathcal{D}|\vec{v} \in B_i}[p^M(\vec{v})] \cdot \Pr_{\vec{v} \sim \mathcal{D}}[\vec{v} \in B_i]}{4 \cdot (1+\varepsilon)\brev{\mathcal{D}}}\\
    &\geq \frac{\rev{\mathcal{D},M}}{4\cdot (1+\varepsilon)\brev{\mathcal{D}}}
\end{align*}

Above, the first line is simply the definition of MenuGap. The second line uses Claim~\ref{claim:menugap}. The third line uses Claim~\ref{claim:ell1}. The fourth line uses the fact that for all $\vec{v} \in B_i$, $p^M(\vec{v}) \leq 2^{2(i-1)+a+1} = 2 \cdot 2^{2(i-1)+a} \leq 2p^M(\vec{x}_i)$. The final line is just rewriting the definition of $\rev{\mathcal{D},M}$, and using the fact that every $\vec{v} \in \text{supp}(\mathcal{D})$ with $p(\vec{v}) > 0$ is in some $B_i$.
\end{proof}

\subsection{Wrapping Up}
Proposition~\ref{prop:main}, together with the fact that an oddly- or evenly-priced mechanism is guaranteed to get a good approximation to the optimum, now suffices to prove Theorem~\ref{thm:main}.

\begin{proof}[Proof of Theorem~\ref{thm:main}]
The proof will be a simple consequence of the technical lemmas in this section, once we set $c$ and $\varepsilon$ appropriately. In particular, set $c=\rev{\mathcal{D}}/100$, and $\varepsilon = 1/100$. Note that $\varepsilon$-representative sequences are guaranteed to exist for any $M, \mathcal{D}$, as $\varepsilon > 0$.

Start by letting $M'$ denote the $c$-expensive mechanism promised by Claim~\ref{claim:price}. Let then $M''$ denote the oddly- or evenly-priced mechanism promised by Claim~\ref{claim:buckets}. Finally, let $X,Q$ denote the $\varepsilon$-representative sequence for $M'',\mathcal{D}$. We get:
\begin{align*}
   \MenuGap{X,Q} &\geq^{\text{(Proposition~\ref{prop:main}})} \frac{\rev{\mathcal{D},M''}}{{4(1+\varepsilon)\brev{\mathcal{D}}}}\\
    &\geq^{\text{(Claim~\ref{claim:buckets})}} \frac{\rev{\mathcal{D},M'}}{8(1+\varepsilon)\brev{\mathcal{D}}}\\
    &\geq^{\text{(Claim~\ref{claim:price})}} \frac{\rev{\mathcal{D}}-c}{8(1+\varepsilon)\brev{\mathcal{D}}}\\
    &\geq \frac{\rev{\mathcal{D}}}{9\brev{\mathcal{D}}}. \qedhere
\end{align*}
\end{proof}

\section{Introduction}
\label{sec:intro}

Consider a revenue-maximizing seller with $k$ items for sale to a single additive buyer, whose values for the $k$ items are drawn from a known distribution $\mathcal{D}$. When $k=1$, Myerson's seminal work provides a closed-form solution to the revenue-optimal mechanism, and it has a particularly simple form: simply post a price $p:=\arg\max_{p}\{ p \cdot \Pr_{v \leftarrow \mathcal{D}}[v \geq p]\}$, and let the buyer purchase the item if they please~(\citet{Myerson81}). For $k > 1$, however, this \emph{multi-dimensional mechanism design} problem remains an active research agenda forty years later.

While simple, constant-factor approximations are known in quite general settings when $\mathcal{D}$ is a product distribution~(\citet{ChawlaHK07, ChawlaHMS10, ChawlaMS15, KleinbergW12, LiY13, BabaioffILW14, Yao15, RubinsteinW15, CaiDW16, ChawlaM16, CaiZ17}), there may be an \emph{infinite} gap between the revenue-optimal auction and any simple counterpart when values are correlated~(\citet{BriestCKW15,hart2019selling}). More specifically: one simple way to sell $k$ items is to treat the grand bundle of all items as if it were a single item, and sell it using Myerson's optimal auction (which sets price $\arg\max_p\{ p \cdot \Pr_{\vec{v} \leftarrow \mathcal{D}}[\sum_i v_i \geq p] \}$). Letting $\brev{\mathcal{D}}$ denote the revenue achieved by this simple scheme,~\citet{hart2019selling} further show a connection between $\brev{\mathcal{D}}$ and \emph{any} simple mechanism through the lens of menu complexity: any mechanism with menu complexity at most $C$ generates expected revenue at most $C \cdot \brev{\mathcal{D}}$.

These works establishes a strong separation between simple and optimal auctions: \emph{even when $k=2$}, there exist distributions $\mathcal{D}$ such that $\rev{\mathcal{D}} = \infty$ (the optimal revenue) while $\brev{\mathcal{D}}=1$. The fact that $\rev{\mathcal{D}} = \infty$ does not on its own suggest that $\mathcal{D}$ must be ``weird'' (the one-dimensional distribution with CDF $1-1/\sqrt{x}$ has this property). The ``weird'' property is that $\rev{\mathcal{D}}/\brev{\mathcal{D}} = \infty$, which can never occur for a one-dimensional distribution.

\citet{BriestCKW15} and~\citet{hart2019selling} establish sufficient conditions for a distribution $\mathcal{D}$ to satisfy $\rev{\mathcal{D}}/\brev{\mathcal{D}} = \infty$. Simply put, the goal of this paper is to study \emph{necessary} conditions for a distribution to satisfy $\rev{\mathcal{D}}/\brev{\mathcal{D}}= \infty$. We provide two main results. The first establishes that the sufficient condition presented in~\cite{hart2019selling} \emph{is in fact necessary} for $\rev{\mathcal{D}}/\brev{\mathcal{D}} = \infty$ (Theorem~\ref{thm:main}). The second establishes that the sufficient condition used in an earlier version of that work~\cite{HNv1} and~\cite{BriestCKW15} is \emph{not necessary} (Theorem~\ref{thm:main2}). In establishing Theorem~\ref{thm:main2}, we also construct a distribution $\mathcal{D}$ such that $\rev{\mathcal{D}}/\brev{\mathcal{D}} = \infty$ yet the mechanism witnessing this provably falls outside the scope of any previous constructions (Corollary~\ref{cor:main}).

Proper context and formal statements of our results require precise definitions, which we provide in Section~\ref{sec:prelims} immediately below. Section~\ref{sec:results} provides formal statements of our results, along with context alongside related work. Subsequent sections provide proofs.

\section{Preliminaries}\label{sec:prelims}

We consider an auction design setting with a single buyer, single seller, and $k$ heterogeneous items. Note that our positive results hold for arbitrary $k$, while our constructions use only $k=2$ (and $k=1$ is not possible). We use $\mathcal{D}$ to denote a distribution over $\mathbb{R}_{\geq 0}^k$, the (possibly correlated) distribution over the buyer's values for the $k$ items. The buyer is additive, meaning that their value for a set of items $S$ is equal to $\sum_{i \in S} v_i$. We use $\rev{\mathcal{D}}$ to denote the optimal expected revenue achievable by any incentive-compatible mechanism (formally, the supremum of expected revenues, or $\infty$ if the supremum is undefined), and let $\brev{\mathcal{D}}$ denote the revenue achieved by selling the grand bundle as a single item using Myerson's auction.\footnote{We briefly remind the reader that $\brev{\mathcal{D}}$ serves as a proxy for the achievable revenue by any simple mechanism, especially when focusing on the gap between infinite and finite. For example, the revenue achieved by selling the items separately is at most $k\brev{\mathcal{D}}$, the revenue achieved by any deterministic mechanism is at most $2^k\brev{\mathcal{D}}$, and, more generally, the revenue achieved by any mechanism which offers at most $m$ distinct options is at most $m\brev{\mathcal{D}}$~\cite{hart2019selling}.} Finally, a \emph{mechanism} $M$ is a set $\{(\vec{q}_i, p_i)\}_i$, where each $\vec{q}_i \in [0,1]^k$ denotes a vector of probabilities, and $p_i \in \mathbb{R}$ denotes a price. When the buyer's valuation is $\vec{v}$, they pay the auctioneer $p_{i(\vec{v})}$, where $i(\vec{v}):=\arg\max_i\{\vec{v} \cdot \vec{q}_i - p_i\}$.\footnote{How ties are broken is irrelevant to our results --- all results hold for arbitrary tie-breaking. Also, all mechanisms include an all-zero pair $ \vec{q}_0 = (0, \dots, 0)$, $p_0 = 0$ to ensure individual rationality.} $\rev{\mathcal{D},M}$ denotes the expected revenue achieved by a particular mechanism $M$ on distribution $\mathcal{D}$. We will also use the shorthand $\vec{q}^M(\vec{v})$ to denote the allocation vector purchased by a vector $\vec{v}$, and $p^M(\vec{v})$ to denote the price paid (we may drop the superscript of $M$ if the mechanism is clear from context).

\subsection*{Brief Overview of \citet{hart2019selling}}

Below, we formally define two geometric properties of sequences of points, which are the focus of this paper. Many of the ideas below appear in both~\citet{BriestCKW15} and~\citet{hart2019selling}, but we will use the formal definitions from~\citet{hart2019selling} (and an earlier published version~\citet{HNv1}). Below, morally the vectors $\vec{x}_i$ correspond to possible (scaled) valuation vectors, and the vectors $\vec{q}_i$ correspond to possible vectors of allocation probabilities.

\begin{definition}
\label{def:sergiu}
Let $X=(\vec{x}_i)_{i=1}^N$ be an ordered sequence of $N$ points ($N$ may be finite, or equal to $+\infty$), with each $\vec{x}_i \in \mathbb{R}_{\geq 0}^k$. Let $Q=(\vec{q}_i)_{i=0}^N$ be another ordered sequence of $N$ points, with each $\vec{q}_i \in [0,1]^k$, and starting with $\vec{q}_0 = (0,...,0)$. Define the following:
\begin{align*}
\mathrm{gap}_i^{X,Q} := \min_{0 \leq j < i} (\vec{q}_i - \vec{q}_j) \cdot \vec{x}_i && \MenuGap{X,Q} := \sum_{i=1}^N \frac{\mathrm{gap}_i^{X,Q}}{||\vec{x}_i||_1}.
\end{align*}

We will also slightly abuse notation and define $\MenuGap{X}:=\sup_Q \{\MenuGap{X,Q}\}$.\footnote{Observe that $\mathrm{gap}_i^{X,Q}$ and $\MenuGap{X,Q}$ might be negative. Any claims made throughout this paper regarding $\MenuGap{X,Q}$ are vacuously true when $\MenuGap{X,Q} < 0$ (e.g. Theorem~\ref{thm:newHN}). We allow $\mathrm{gap}_i^{X,Q}$ to be negative to match the definition of~\cite{hart2019selling} verbatim (although our work will also show that this peculiarity of their definition is not significant).}
\end{definition}

Intuitively, $\MenuGap{X}$ is some (complicated) measure of how distinct the angles of points in $X$ are. To get intuition for this, one might try to write a short proof that when $k=1$, $\MenuGap{X} = 1$ for all $X$ (or that $\MenuGap{X}=1$ whenever all $\vec{x}_i \in X$ are parallel). We provide such a proof in Appendix~\ref{app:single}.

Still, $\MenuGap{X,Q}$ is just some geometric measure with no obvious intuition for why this quantity should be of interest to auction designers. However, one key result of~\cite{hart2019selling} shows that this quantity has connections to simplicity vs.~optimality gaps. Specifically, they show:

\begin{theorem}[\cite{hart2019selling}, Proposition~7.1]
\label{thm:newHN}
For every pair of sequences $X=(\vec{x}_i)_{i=1}^N, Q=(\vec{q}_i)_{i=0}^N$ starting with $\vec{q}_0 =  (0,...,0)$, and all $\varepsilon > 0$, there exists a distribution $\mathcal{D}$ and mechanism $M$ such that:

\[ \frac{\rev{\mathcal{D},M}}{\brev{\mathcal{D}}} \geq (1-\varepsilon)\cdot  \MenuGap{X,Q}.\]

Moreover, for all $i \in [N]$, the support of $\mathcal{D}$ contains a single point of the form $c_i \vec{x}_i$, for some $c_i \in \mathbb{R}_+$ (and no other points). Additionally, $\vec{q}^M(c_i \vec{x}_i) = \vec{q}_i$.
\end{theorem}

The ``Moreover,...'' portion of Theorem~\ref{thm:newHN} gives some insight to their construction. Further insight can be deduced by observing that the constraint ``$\mathrm{gap}_i^{X,Q} \leq \vec{x}_i \cdot (\vec{q}_i - \vec{q}_j)$'' looks similar (but far from identical) to an incentive compatibility constraint involving a valuation vector $\vec{x}_i$ and two allocation vectors $\vec{q}_i,\vec{q}_j$. We refer the reader to~\cite{hart2019selling} for further details and intuition for this connection. Theorem~\ref{thm:newHN} gives a framework for proving simplicity vs.~optimality gaps, but of course leaves open the question of actually finding a pair of sequences $X, Q$. They approach this through the following observation:

\begin{definition}
\label{def:noam}
Given a sequence of points $X = (\vec{x}_i)^N_{i=0} \in [0,1]^k$ starting with $\vec{x}_0 = (0,...,0)$, define $\SupGap{X}$ (read ``support gap of $X$") as follows:
\begin{align*}
\SupGap{X} := \MenuGap{X,X}.
\end{align*}
\end{definition}

\begin{observation}\label{obs:trivial} For all $X$, $\MenuGap{X} \geq \SupGap{X}$.
\end{observation}

Finally,~\cite{HNv1} propose an explicit construction of a sequence $X$ with infinite SupGap.

\begin{theorem}[\cite{HNv1}]\label{thm:HNgap} There exists an infinite sequence $X$ of points in $[0,1]^2$ such that:
$$\SupGap{X} = \infty.$$
\end{theorem}

Theorems~\ref{thm:newHN} and~\ref{thm:HNgap} together yield a two-dimensional distribution with $\rev{\mathcal{D}}/\brev{\mathcal{D}}=\infty$. It is also worth noting that all prior constructions follow this approach as well. For example,~\cite{BriestCKW15} provides an infinite sequence $X$ of points in $[0,1]^3$ such that $\SupGap{X} = \infty$.

\section{Our Results}\label{sec:results}
Independent of Observation~\ref{obs:trivial} and Theorem~\ref{thm:HNgap}, Theorem~\ref{thm:newHN} \emph{alone} provides a framework for constructing distributions $\mathcal{D}$ so that $\rev{\mathcal{D}}/\brev{\mathcal{D}}$ is large: find sequences $X$ so that $\MenuGap{X}$ is large. Our goal is to understand to what extent this framework is \emph{complete} for constructing such instances. Our first main result establishes that \emph{any distribution with $\rev{\mathcal{D}}/\brev{\mathcal{D}} = \infty$ could have resulted from the framework induced by Theorem~\ref{thm:newHN}}. Specifically:

\begin{theorem}\label{thm:main}
For any distribution $\mathcal{D}$ over $k$ items, there exists a sequence of $N$ points ($N$ can be finite, or equal to $+\infty$) $X=(\vec{x}_i)^N_{i=1}$, with each $\vec{x}_i \in \text{supp}(\mathcal{D})$, such that
\[ \MenuGap{X} \geq \frac{\rev{\mathcal{D}}}{9\brev{\mathcal{D}}}. \]

In particular, if $\rev{\mathcal{D}}/\brev{\mathcal{D}} = \infty$, then $\MenuGap{X} = \infty$ as well.
\end{theorem}

A complete proof of Theorem~\ref{thm:main} appears in Section~\ref{sec:converse}. Observe also that because $\MenuGap{X}$ is monotone (in the sense that adding points to $X$, anywhere, cannot possibly decrease $\MenuGap{X}$), the fact that $X$ is a subset of the support of $\mathcal{D}$ (rather than the entire support) is immaterial.\footnote{Of course, if the support of $\mathcal{D}$ is uncountable, then clearly the entire support of $\mathcal{D}$ cannot be included in $X$.} Put another way, the important aspect in constructing $X$ is how elements in the support of $\mathcal{D}$ are ordered, rather than which points are included.

Observation~\ref{obs:trivial} further provides a framework to construct sequences so that $\MenuGap{X}$ is large: construct sequences $X$ so that $\SupGap{X}$ is large. One may then wonder if $\SupGap{X}$ and $\MenuGap{X}$ are approximately related, for all $X$. For this specific question, the answer is trivially no, due to incompatibility with scaling (multiplying every point in $X$ by $1/2$ will decrease $\SupGap{X}$ by a factor of $2$, but not $\MenuGap{X}$). Therefore, not much insight is gained by studying this precise question.

Instead, we observe that the interesting aspect of constructions resulting through $\SupGap{X}$ is that $\vec{x}_i$ and $\vec{q}_i$ are \emph{aligned} (that is, $\vec{x}_i = c_i \cdot \vec{q}_i$ for some $c_i \in \mathbb{R}_{\geq 0}$). Specifically, even if $\vec{q}_i = \vec{x}_i$, this equality is not maintained through the construction of Theorem~\ref{thm:newHN}. However, if $\vec{q}_i$ and $\vec{x}_i$ are aligned, this alignment property \emph{is} maintained by the construction. We therefore propose the following definition, which captures the maximum value achievable by $\MenuGap{X,Q}$ when $X,Q$ are aligned.

\begin{definition}\label{def:align} Let $X = (\vec{x}_i)_{i=1}^N$ be an ordered sequence of $N$ points in $[0,1]^k$ ($N$ may be finite, or equal to $+\infty$). Let also $C=(c_i)_{i=0}^N$ be an ordered sequence of numbers, with each $c_i \in [0,1/||\vec{x}_i||_\infty]$, starting with $c_0=0$. Define $\mathrm{sgap}^{\cdot,\cdot}$ (read "scalar gap") :$$\mathrm{sgap}^{X,C}_i:= \min_{j < i} \vec{x}_i \cdot (c_i \vec{x}_i - c_j \vec{x}_j),\qquad  \text{ and } \qquad \ScaGap{X,C}:= \sum_{i=1}^N \frac{\max\{0,\mathrm{sgap}^{X,C}_i\}}{||\vec{x}_i||_1}.$$

We will also slightly abuse notation and denote by $\ScaGap{X}:=\sup_C\{\ScaGap{X,C}\}$.
\end{definition}

Recall that we have chosen to let $c_i$ range in $[0,1/||\vec{x}_i||_\infty]$ (rather than be fixed at $1$, or $1/||\vec{x}_i||_\infty$) to give potential constructions flexibility in scaling $\vec{q}_i$. Additionally, by ensuring that the contribution of each $\mathrm{sgap}_i^{X,C}$ is non-negative, we give potential constructions flexibility to ignore points in the sequence. That is, any construction using MenuGap directly can always set $\vec{q}_i:=\arg\max_{j < i}\{\vec{q}_j\cdot \vec{x}_i\}$, which effectively just drops $\vec{x}_i$ from the sequence. Counting $\max\{0,\mathrm{sgap}_i^{X,C}\}$ towards the objective (rather than just $\mathrm{sgap}_i^{X,C}$) gives constructions that arise through AlignGap the same flexibility.

\begin{lemma}\label{lem:align} For all $X$, $\ScaGap{X} \leq \MenuGap{X}$.
\end{lemma}

The proof of Lemma~\ref{lem:align} is in Appendix~\ref{app:omitted}. Lemma~\ref{lem:align} induces a framework to design sequences with large MenuGap: design sequences with large AlignGap. Our second main result establishes that this framework \emph{is not} without loss of generality, even for $k=2$. Specifically:

\begin{theorem}\label{thm:main2}
There exist sequences $X=(\vec{x}_i)_{i=1}^\infty \in [0,1]^2$ such that:

$$\ScaGap{X} \leq 6 \qquad \text{ but } \qquad \MenuGap{X} = \infty.$$
\end{theorem}

A complete proof of Theorem~\ref{thm:main2} appears in Section~\ref{sec:sep}. By the discussion following Definition~\ref{def:align}, the source of this gap is entirely due to the requirement that the sequence $Q$ be aligned with $X$ (and is not due to inability to scale, or inability to ignore difficult points in $X$). We make this crisp with the following definition and corollary, which construct a novel distribution witnessing $\rev{D}/\brev{D} = \infty$ that is provably distinct from all previous approaches.

\begin{definition} For a distribution $\mathcal{D}$ and mechanism $M$, define the \emph{Aligned Revenue} of $M$ on $\mathcal{D}$:
$$\arev{\mathcal{D},M}:=\mathbb{E}_{\vec{v} \leftarrow \mathcal{D}}[p^M(\vec{v}) \cdot I(\text{$\vec{v}$ is parallel to $\vec{q}^M(\vec{v})$})], \qquad \quad \arev{\mathcal{D}}:=\sup_M\{\arev{\mathcal{D},M}\}.$$
\end{definition}

\begin{corollary}\label{cor:main} There exist distributions $\mathcal{D}$ over two items such that $\rev{\mathcal{D}}/\arev{\mathcal{D}} = \infty$.
\end{corollary}

A proof of Corollary~\ref{cor:main} appears in Appendix~\ref{app:main}. It is worth noting that \emph{all} previous constructions establishing $\rev{\mathcal{D}}/\brev{\mathcal{D}} = \infty$ proceeded by producing an $X$ such that $\SupGap{X} = \infty$. Indeed, the~\cite{BriestCKW15} construction provides such an $X$ when $k=3$, the~\cite{HNv1} construction provides an $X$ when $k=2$, and the~\cite{PsomasSW19} construction adapts parameters in that of~\cite{HNv1}. By Theorem~\ref{thm:newHN}, this implies not only that $\rev{\mathcal{D}}/\brev{\mathcal{D}} = \infty$, but also that $\arev{\mathcal{D}}/\brev{\mathcal{D}} = \infty$. Corollary~\ref{cor:main} establishes the existence of a fundamentally different construction,\footnote{On a technical level, our construction certainly borrows several ideas from previous ones, however.} as our $\mathcal{D}$ has a finite ratio between $\arev{\mathcal{D}}/\brev{\mathcal{D}}$, yet still maintains an infinite ratio between $\rev{\mathcal{D}}/\brev{\mathcal{D}}$.

\subsection*{Additional Related Work}
\label{sec:rel}
We've already discussed the most related work to ours, which is that of~\citet{hart2019selling, BriestCKW15}. There is also a large body of work studying \emph{product distributions} specifically, and establishes that simple mechanisms can achieve constant factor approximations in quite general settings~\citet{ChawlaHK07, ChawlaHMS10, ChawlaMS15, KleinbergW12, LiY13, BabaioffILW14, Yao15, RubinsteinW15, CaiDW16, ChawlaM16, CaiZ17}). Recent works have made progress in obtaining arbitrary approximations~(\citet{BabaioffGN17, KothariMSSW19}), which again rely on the assumption that $\mathcal{D}$ is a product distribution.

Three recent lines of work address the~\cite{BriestCKW15,hart2019selling} constructions in a different manner. First,~\citet{ChawlaTT19,ChawlaTT20} consider the related \emph{buy-many} model (where the auctioneer cannot prevent the buyer from interacting multiple times with the auction). \citet{ChawlaTT19} establishes that selling separately achieves an $O(\log k)$-approximation to the optimal buy-many mechanism in \emph{quite} general settings (including the settings considered in this paper). In a different direction,~\citet{PsomasSW19} uses the lens of smoothed analysis~(\citet{SpielmanT04}) to reason about the robustness of the~\cite{hart2019selling} constructions. Finally, \citet{carroll2017robustness} considers a correlation robust framework in which the valuation profile is drawn from a correlated distribution that is not completely known to the seller; the goal is to design a mechanism that maximizes the worst-case (over correlations) seller revenue, when only the items' marginal distributions are known. \cite{carroll2017robustness} shows that selling each item separately is optimal; see \citet{bei2019correlation,gravin2018separation} for further work in this model.

\section{No Converse to Lemma~\ref{lem:align}: Separating MenuGap and AlignGap}
\label{sec:sep}

In this section we prove our second main result: Lemma~\ref{lem:align} does \emph{not} admit a converse, even approximately. We briefly remind the reader that \emph{all} previous constructions witnessing $\rev{\mathcal{D}}/\brev{\mathcal{D}} = \infty$ arose by establishing sequences $X$ with $\ScaGap{X}=\infty$ (in fact, even $\SupGap{X} = \infty$). Theorem~\ref{thm:main2} establishes that constructions exist outside of this more restrictive framework. We now proceed with the proof of Theorem~\ref{thm:main2}, beginning with a description of our sequence $X$.

\subsection{Description of our construction}\label{sec:description}
We describe our infinite sequence $X$, which consists of consecutive \emph{layers} of points, and note that this aspect of the construction is similar to that in~\cite{HNv1}. For $\ell = 2$ to $\infty$, layer $\ell$ will have $n_\ell:=\ell \lceil \ln^2(\ell) \rceil +1$ points. These points/vectors will have $\ell_2$ norm equal to one, and will be evenly spaced (in terms of their angle) between $(1,0)$ and $(0,1)$. If $\ell$ is even, they will go counterclockwise from $(1,0)$ to $(0,1)$. If $\ell$ is odd, they will go clockwise from $(0,1)$ to $(1,0)$. Specifically:
\begin{itemize}
\item Define $n_\ell:=\ell \lceil \ln^2(\ell) \rceil +1$. Define $\theta_\ell:=\frac{\pi}{2(n_\ell-1)}$.
\item Point $\vec{x}_{\ell,j}$ is the $j^{th}$ point in the $\ell^{th}$ layer, and is $(\cos(j\theta_\ell),\sin(j\theta_\ell) )$ when $\ell$ is even, or $(\sin(j\theta_\ell),\cos(j\theta_\ell))$ when $\ell$ is odd.
\item The infinite sequence $X$ is $((\vec{x}_{\ell,j})_{j=0}^{n_\ell-1})_{\ell=2}^{\infty}$.
\end{itemize}

\noindent Figure~\ref{fig:samplelayer} demonstrates two layers of our construction. In the remainder of this section, we may refer to the sequence of points $\vec{x}_{\ell, j}$ by a single indexed sequence $\vec{x}_i$. The latter is the same as the former where points are ordered lexicographically with respect to the original indexing.

\begin{figure}
\centering
  \includegraphics[width=.6\linewidth]{samplelayer2.png}
  \captionof{figure}{An illustration of two layers of our construction. The number of points in each layer increases with $\ell$, but they are always evenly spaced between $(1,0)$ and $(0,1)$. The direction in which the points are placed alternates between clockwise and counterclockwise.}
  \label{fig:samplelayer}
  \end{figure}

\subsection{Upper Bounding $\ScaGap{X}$ via Lagrangian duality}

Now that we have our construction, we first need to upper bound $\ScaGap{X}$ and establish that it's finite. To this end, first observe that for any sequence $X$, $\ScaGap{X}$ is the solution to the following (infinite, if $X$ is infinite) mathematical program, where the variables are $\mathrm{sgap}_i, c_i$ (the sequence $\vec{x}_i$ is fixed, as we're aiming to compute $\ScaGap{X}$):
\begin{align*}
\ScaGap{X} :&\begin{cases}
&\max  \sum_i \max\{0,\mathrm{sgap}_i\}/||\vec{x}_i||_1  \\
\text{subject to: } &\forall i, j < i: \mathrm{sgap}_i \leq \vec{{x}}_i \cdot (c_i \vec{{x}}_i - c_j \vec{{x}}_j) \\
&\forall i: 0 \leq c_i \leq 1/||\vec{x}_i||_\infty
\end{cases}
\end{align*}

We next proceed with a series of relaxations of this program. Some steps are specific to our choice of $X$ from Section~\ref{sec:description}, while others hold for arbitrary $X$. Our first step is specific to this construction, and simply bounds $||\vec{x}||_1$. Consider the following mathematical program:

\begin{align*}
\mathrm{AlignGap'}(X) :&\begin{cases}
&\max  \sum_i \max\{0,\mathrm{sgap}_i\}  \\
\text{subject to: } &\forall i, j < i: \mathrm{sgap}_i \leq \vec{{x}}_i \cdot (c_i \vec{{x}}_i - c_j \vec{{x}}_j) \\
&\forall i: 0 \leq c_i \leq \sqrt{2}.
\end{cases}
\end{align*}

\begin{observation}\label{obs:prime} For the sequence $X$ defined in Section~\ref{sec:description}, $\mathrm{{AlignGap}}'(X) \geq \ScaGap{X}$.
\end{observation}
\begin{proof}
Every $\vec{x}_i$ in the construction has $||\vec{x}_i||_1 \geq ||\vec{x}_i||_2 = 1$. Therefore, the new objective function is only larger. Moreover, every $\vec{x}_i$ in the construction has $||\vec{x}_i||_\infty \geq 1/\sqrt{2}$ (because the $\ell_2$ norm is $1$), so this is relaxing the upper bound on $c_i$.
\end{proof}

We will proceed to upper bound $\text{AlignGap}'(X)$ via a Lagrangian relaxation of the formulation above. Specifically, consider the following Lagrangian relaxation. We put a Lagrangian multiplier of $1$ on every constraint of the form $\mathrm{sgap}_i \leq \vec{x}_i \cdot (c_i \vec{x}_i - c_{i-1}\vec{x}_{i-1})$, for all $i > 1$. We put a Lagrangian multiplier of $0$ on all other constraints involving $\mathrm{sgap}$. We will \emph{not} put a Lagrangian multiplier on constraints binding $c_i$ to $[0,\sqrt{2}]$, and keep those in the program. This yields the following Lagrangian relaxation (for simplicity of notation below, define $c_0 :=0$, and define $\vec{x}_0 = \vec{0}$):

\begin{align*}
\mathrm{\text{LagRel}_1(X)} :&\begin{cases}
&\max  \sum_i \max\{0,\mathrm{sgap}_i\} + \vec{x}_i \cdot (c_i \vec{x}_i - c_{i-1} \vec{x}_{i-1}) - \mathrm{sgap}_i\\
\text{subject to: }
&\forall i: 0 \leq c_i \leq \sqrt{2}
\end{cases}
\end{align*}

\begin{observation}\label{obs:lagrange} For all $X$, $\mathrm{AlignGap}'(X) \leq \LagRelOne{X}$.
\end{observation}
\begin{proof}
This follows immediately from weak Lagrangian duality. For a quick refresher on weak Lagrangian duality, observe that for any feasible solution to the LP defining $\mathrm{AlignGap}'(X)$ we must have $\vec{x}_i \cdot (c_i \vec{x}_i - c_{i-1}\vec{x}_{i-1}) - \mathrm{sgap}_i \geq 0$. Therefore, for any feasible solution to the original LP, that solution is also feasible for $\mathrm{\text{LagRel}}_1(X)$, and the objective is only larger. Therefore, the optimal solution to $\mathrm{\text{LagRel}}_1(X)$ must be at least as large as $\mathrm{AlignGap}'(X)$.
\end{proof}

We now proceed to further simplify $\LagRelOne{X}$. The next step is defined below:

\begin{align*}
\mathrm{\text{LagRel}_2(X)} :&\begin{cases}
&\max  \sum_i  \vec{x}_i \cdot (c_i \vec{x}_i - c_{i-1} \vec{x}_{i-1})\\
\text{subject to: }
&\forall i: 0 \leq c_i \leq \sqrt{2}
\end{cases}
\end{align*}

\begin{observation}\label{obs:lagrange2} For all $X$, $\LagRelOne{X} = \LagRelTwo{X}$.
\end{observation}
\begin{proof}
Observe that for all $i$, $\max\{0,\mathrm{sgap}_i\} - \mathrm{sgap}_i \leq 0$. When $\mathrm{sgap}_i = 0$, the maximum is achieved (and $\mathrm{sgap}_i:=0$ is feasible). Substituting $\max\{0,\mathrm{sgap}_i\} - \mathrm{sgap}_i = 0$ for all $i$ concludes the proof.
\end{proof}

We make one last observation about the relaxation, which simply rewrites the objective function to group all coefficients of $c_i$. For ease of notation below, define $\vec{x}_{N+1}:=\vec{0}$ (if $N$ is finite. If $N = \infty$, there are no notational issues).

\begin{align*}
\mathrm{\text{LagRel}(X)} :&\begin{cases}
&\max  \sum_i c_i \vec{x}_i \cdot (\vec{x}_i -\vec{x}_{i+1})\\
\text{subject to: }
&\forall i: 0 \leq c_i \leq \sqrt{2}
\end{cases}
\end{align*}

\begin{observation}\label{obs:Lagrange3} For all $X$, $\mathrm{AlignGap}'(X) \leq \LagRelOne{X} = \LagRelTwo{X} = \LagRel{X}$.
\end{observation}

Now, we move to analyze $\LagRel{X}$ for our particular sequence $X$.

\begin{claim}\label{claim:X} For the sequence $X$ defined in Section~\ref{sec:description}, $\LagRel{X}=\sqrt{2} \cdot \sum_i 1 - \vec{x}_i \cdot \vec{x}_{i+1}$.
\end{claim}
\begin{proof}
Observe here that $\vec{x}_i \cdot (\vec{x}_i-\vec{x}_{i+1}) \geq 0$ for all $i$, as each $\vec{x}_i$ has $\ell_2$ norm exactly one. This means that the optimal solution for $\LagRel{X}$ sets each $c_i:=\sqrt{2}$. Finally, recalling that $\vec{x}_i \cdot \vec{x}_i = 1$ for all $i$ concludes the claim.
\end{proof}

\begin{proposition}\label{prop:mainduality} For the sequence $X$ defined in Section~\ref{sec:description}, $\LagRel{X} \leq 6$.
\end{proposition}
\begin{proof}
Let us first observe that if $\vec{x}_i$ is the last point in a layer, then in fact $\vec{x}_{i+1} = \vec{x}_i$, and therefore $1-\vec{x}_i \cdot \vec{x}_{i+1} = 0$. Therefore, these terms do not contribute to the sum. We can then rewrite the term to sum over all layers as follows:
\begin{align*}
\LagRel{X}&=\sqrt{2}\cdot \sum_{\ell}\sum_{j=0}^{n_\ell-2} 1-\vec{x}_{\ell,j}\cdot \vec{x}_{\ell,j+1}\\
&=\sqrt{2} \cdot \sum_{\ell}(n_\ell-1) \cdot (1-\cos(\theta_\ell))\\
&\leq \sqrt{2} \cdot \sum_\ell (n_\ell-1) \cdot \theta_\ell^2/2\\
&= \sqrt{2} \cdot \sum_\ell \frac{\pi^2}{8 (n_\ell-1)}\\
&=\sqrt{2} \cdot \sum_\ell \frac{\pi^2}{8 \ell \lceil \ln^2(\ell) \rceil}\\
&\leq 6.
\end{align*}

Above, the first line follows by the reasoning in the first paragraph. The second line follows by observing that the angle between any two points in layer $\ell$ is exactly $\theta_\ell$ (and the two points in question of $\ell_2$ norm equal to one). The third line follows as $\cos(\theta_\ell) \geq 1-\theta_\ell^2/2$ for any $\theta_\ell \in [0,\pi/2]$ (and all $\theta_\ell$ are indeed in $[0,\pi/2]$). The fourth line follows by substituting the definition of $\theta_\ell$ as a function of $n_\ell$. The fifth line follows by definition of $n_\ell$. The final line is just calculation for this particular infinite series.\footnote{This follows as $\sum_{\ell=2}^\infty \frac{1}{\ell \ln^2(\ell)} \leq 3$.}
\end{proof}

Observation~\ref{obs:prime}, Observation~\ref{obs:Lagrange3}, and Proposition~\ref{prop:mainduality} yield the main result of this section:

\begin{proposition}\label{prop:scagap} For the sequence $X$ defined in Section~\ref{sec:description}, $\ScaGap{X} \leq 6$.
\end{proposition}

\subsection{Step 3: Picking a $Q$ to Lower Bound $\MenuGap{X}$.}
Finally, we propose a sequence $Q$ and show that $\MenuGap{X,Q} = \infty$. We describe the sequence again in layers, to match our description of $X$ (that is, the vector $\vec{q}_{\ell,j}$ corresponds to the vector $\vec{x}_{\ell,j}$). In particular, for each even layer $\ell$, the vectors $\vec{q}_{\ell,j}$ will have a fixed $x$-coordinate, and the $y$-coordinate will increase with $j$. For each odd layer, we will introduce no new vectors (i.e. we will just let $\mathrm{gap}_i^{X,Q} = 0$ for all $i$ in an odd layer). Specifically, the construction is as follows:
\begin{itemize}
\item Define $\alpha:= \sum_{\ell=2}^\infty \frac{1}{\ell \ln^2(\ell)}$ (and note that $\alpha < \infty$).
\item Define $z_\ell:= \frac{1}{\alpha} \sum_{j=2}^\ell \frac{1}{j\ln^2(j)}$.
\item Define $\delta_\ell:= z_\ell-z_{\ell-1} = \frac{1}{\alpha \ell \ln^2(\ell)}$.
\item For $j < n_\ell-1$, define $z_{\ell,j}:=1-\delta_\ell \cot((j+1)\theta_\ell)$. For $j = n_\ell-1$, define $z_{\ell,j}:=1$.\footnote{To see that $z_{\ell,j} \geq 0$ observe that $\cot(x) \leq 1/x$ for all $x \in [0, \pi/2]$. Therefore we get $\delta_\ell \cot((j+1) \theta_\ell) \leq \delta_\ell \frac{1}{(j+1) \theta_\ell} \leq \delta_\ell \frac{2 (n_\ell -1) }{\pi (j+1)} \leq \frac{2}{\pi} \delta_\ell n_\ell = \frac{2 \ell \lceil \log^2 \ell \rceil}{\pi \alpha \ell \log^2 \ell}\leq 1$, for all $\ell, j$ ($\alpha > 1.9$). To see that $z_{\ell, j} \leq 1$ simply note that the term we subtract can't be negative.}
\item For all even $\ell$, and all $j$, set $\vec{q}_{\ell,j}:= (z_\ell, z_{\ell,j})$.
\item For all odd $\ell$, and all $j$, set $\vec{q}_{\ell,j}:=\arg\max_{\ell' < \ell,j'}\{\vec{x}_{\ell, j} \cdot\vec{q}_{\ell',j'}\}$.
\end{itemize}

\begin{proposition}
\label{prop:vecgapdiverges}
$\MenuGap{X,Q}=\infty$.
\end{proposition}

\begin{proof}
To ease notation throughout the proof, we'll use the notation $\mathrm{gap}_{\ell,j}^{X,Q}:=\mathrm{gap}_i^{X,Q}$, where $\vec{x}_i:=\vec{x}_{\ell,j}$ ($\vec{x}_i$ is the $j^{th}$ point on layer $\ell$). We will also use the notation $(\ell',j') < (\ell,j)$ if $\ell' < \ell$, or $\ell' = \ell$ and $j' < j$ (that is, if the $j'^{th}$ point in the $\ell'^{th}$ layer comes before the $j^{th}$ point in the $\ell^{th}$ layer). To understand $\mathrm{gap}_{\ell,j}^{X,Q}$, we need to understand which point ``sets the gap'' for $\vec{x}_{\ell,j}$, that is, which $(\ell',j'):=\arg\min_{(\ell',j')< (\ell,j)}\{(\vec{q}_{\ell,j}-\vec{q}_{\ell',j'})\cdot \vec{x}_i\}$.

We first analyze which point sets the gap for $\vec{{x}}_{\ell,j}$ (for even $\ell$; for odd $\ell$ the gap is zero and we don't care which point sets it), and observe that it must either be $\vec{q}_{\ell,j-1}$ or $\vec{q}_{\ell-2,n_{\ell-2}-1}$ (that is, it must be the previous point in the same layer, or the final point in the previous even layer).

\begin{claim}\label{claim:gap}
For all $j$, and all even $\ell$, $\mathrm{gap}_{\ell,j}^{X,Q}= \vec{x}_{\ell,j}\cdot \vec{q}_{\ell,j} - \max\{\vec{x}_{\ell,j} \cdot \vec{q}_{\ell-2,n_{\ell-2}-1}, \vec{x}_{\ell,j} \cdot \vec{q}_{\ell,j-1}\}$.\footnote{For simplicity of notation, define $\vec{q}_{0,j} = \vec{0} = \vec{q}_{\ell,-1}$ for all $\ell, j$.}
\end{claim}
\begin{proof}
First, note that $\mathrm{gap}_{\ell,j}^{X,Q}:=\min_{(\ell',j') < (\ell,j)} \{\vec{x}_{\ell, j} \cdot (\vec{q}_{\ell, j} - \vec{q}_{\ell',j'})\} = \vec{x}_{\ell, j} \cdot \vec{q}_{\ell, j} - \max_{(\ell',j')<(\ell,j)}\{\vec{x}_{\ell, j}\cdot \vec{q}_{\ell',j'}\}$. Observe that the first component of $\vec{q}_{\ell',j'}$
is monotone increasing in $\ell'$ (for fixed $j'$), and the second component
is monotone increasing in $j'$ (for fixed $\ell'$). Moreover, the second component of $\vec{q}_{\ell', n_{\ell'}-1}$ is $1$, and this is the maximum possible. Also, both components of $\vec{x}_{\ell,j}$ are non-negative, and therefore we conclude that $\vec{x}_{\ell,j} \cdot \vec{q}_{\ell-2,n_{\ell-2}-1} \geq \vec{x}_{\ell,j} \cdot \vec{q}_{\ell',j'}$ whenever $(\ell',j') \leq (\ell-2,n_{\ell-2}-1)$ (in fact, this extends even to $(\ell',j') \leq (\ell-1,n_{\ell-1}-1)$ as no new $\vec{q}$ are introduced in layer $\ell-1$). Also, $\vec{x}_{\ell,j} \cdot \vec{q}_{\ell,j-1} \geq \vec{x}_{\ell,j} \cdot \vec{q}_{\ell,j'}$ whenever $j' \leq j-1$.
\end{proof}

Now that we know that the gap is set either by the last point in the previous layer, or the previous point in the current layer, we can nail down $\mathrm{gap}_{\ell,j}^{X,Q}$ exactly.

\begin{lemma}\label{lem:gap} For all even $\ell > 2$, and all $j \in [0,n_\ell-1]$:
\[ \mathrm{gap}^{X, Q}_{\ell,j} \geq \delta_\ell \frac{\sin(\theta_\ell)}{\sin((j+1)\theta_\ell)}. \]
\end{lemma}
\begin{proof}
To prove the lemma, we simply compute the inner product of $\vec{x}_{\ell,j}$ with the three relevant vectors $\vec{q}_{\ell,j}, \vec{q}_{\ell-2,n_{\ell-2}-1}, \vec{q}_{\ell,j-1}$. To this end, recall that:
\begin{align*}
\vec{q}_{\ell,j} &= \left(z_\ell, 1 - \delta_\ell \cot((j+1) \theta_\ell)\right), \\
\vec{q}_{\ell,j-1}&= \left(z_\ell, 1 - \delta_\ell \cot(j\theta_\ell)\right), \\
\vec{q}_{\ell-2,n_{\ell-2}-1}&= \left(z_{\ell-2}, 1 \right).
\end{align*}

Therefore, observe that
\begin{align*}
\vec{x}_{\ell,j} \cdot (\vec{q}_{\ell,j} - \vec{q}_{\ell,j-1}) &= \sin(j\theta_\ell)\cdot\delta_\ell\cdot  \left(\cot(j\theta_\ell) - \cot((j+1)\theta_\ell) \right) \\
&= \sin(j \theta_\ell) \cdot \delta_\ell\cdot \left(\frac{\cos(j\theta_\ell)}{\sin(j\theta_\ell)} - \frac{\cos((j+1)\theta_\ell)}{\sin((j+1)\theta_\ell)} \right) \\
&= \delta_\ell \cdot  \frac{\cos(j\theta_\ell)\sin( (j+1)\theta_\ell) - \sin(j\theta_\ell)\cos( (j+1)\theta_\ell)}{\sin((j+1)\theta_\ell)}\\
&=\delta_\ell \cdot \frac{\sin(\theta_\ell)}{\sin( (j+1)\theta_\ell)}.
\end{align*}

Similarly,
\begin{align*}
\vec{x}_{\ell,j} \cdot (\vec{q}_{\ell,j} - \vec{q}_{\ell-2,n_{\ell-2}-1}) &= (\delta_\ell+\delta_{\ell-1}) \cdot \cos(j \theta_\ell) - \delta_\ell\cot( (j+1)\theta_\ell) \cdot \sin(j\theta_\ell) \\
&\geq \delta_\ell \cdot \cos(j \theta_\ell) - \delta_\ell\cot( (j+1)\theta_\ell) \cdot \sin(j\theta_\ell) \\
&=\frac{\delta_\ell}{\sin((j+1)\theta_\ell)}\left(\sin((j+1)\theta_\ell)\cos(j\theta_\ell) - \sin(j \theta_\ell) \cos((j+1)\theta_\ell) \right) \\
&= \delta_\ell \frac{\sin(\theta_\ell)}{\sin((j+1)\theta_\ell)}.
\end{align*}

This means that no matter which point sets the gap (or if one of the points does not exist), the gap is at least $\delta_\ell \frac{\sin(\theta_\ell)}{\sin((j+1)\theta_\ell)}$.
\end{proof}

Finally, now that we know the gap for each point on an even layer, we just need to sum over each even layer.

\begin{corollary}\label{cor:gap} For any even $\ell > 2$, $\sum_{j=0}^{n_\ell-1} \mathrm{gap}_{\ell,j}^{X,Q} \geq \delta_\ell \cdot \ln(n_\ell)/2$.
\end{corollary}

\begin{proof}
Consider the following sequence of calculations:

\begin{align*}
\sum_{j=0}^{n_\ell-1} \mathrm{gap}_{\ell,j}^{X,Q} &\geq \sum_{j=0}^{n_\ell-1} \delta_\ell \frac{\sin(\theta_\ell)}{\sin((j+1)\theta_\ell)}\\
&\geq \delta_\ell \cdot (\theta_\ell-\theta_\ell^3/6)\cdot \sum_{j=0}^{n_\ell-1} \frac{1}{(j+1)\theta_\ell}\\
&\geq \delta_\ell \cdot (1-\theta_\ell^2/6) \cdot \ln(n_\ell)\\
&\geq \delta_\ell \cdot \ln(n_\ell)/2
\end{align*}

Above, the first line follows from Lemma~\ref{lem:gap}. The second line uses the fact that $\theta_\ell - \theta_\ell^3/6 \leq \sin(\theta_\ell) \leq \theta_\ell$, because $\theta_\ell \in [0,\pi/2]$. The third line follows as the $n$-th harmonic sum is at least $\ln(n)$. The final line follows as $\theta_\ell^2/6 = \pi^2/(24 (n_\ell-1)^2) \leq 1/2$.
\end{proof}

And finally, we can wrap up the proof of the proposition. Here, we just need to recall that $\delta_\ell:=\frac{1}{\alpha n_\ell} = \frac{1}{\alpha \ell \ln^2(\ell)}$. Therefore, we conclude that:
\begin{align*}
\sum_{\ell \text{ even}} \sum_{j=0}^{n_\ell-1} \mathrm{gap}_{\ell,j}^{X,Q} &\geq \sum_{\ell \text{ even}} \delta_\ell \cdot \ln(n_\ell)/2\\
&=\sum_{\ell \text{ even}} \frac{1}{2 \alpha \ell \ln(\ell)}\\
&= \infty
\end{align*}

Above, the first line simply restates Corollary~\ref{cor:gap}. The second line substitutes the definition of $\delta_\ell$ and $n_\ell$. The final line follows as the indefinite integral of $\frac{1}{x\ln(x)}dx$ is $\ln\ln (x)$ (and $\alpha$ is an absolute constant).
\end{proof}

This completes the proof of Theorem~\ref{thm:main2}: Proposition~\ref{prop:scagap} establishes that $\ScaGap{X} \leq 6$, while Proposition~\ref{prop:vecgapdiverges} establishes that $\MenuGap{X} = \infty$.

redpajama_set_name: RedPajamaArXiv


text:
@interface XMGShareLinkeButton : UIButton

@end

redpajama_set_name: RedPajamaGithub


text:
Royal Canin to host webinar exploring industry's trends
With an endless choice of options when it comes to pet food, Royal Canin will be hosting a webinar to find the truth behind fads and trends.
The webinar, aptly named 'Fads and Trends in Today's Pet Food Industry – and how to find the science' will be hosted by Dr Cecilia Villaverde (BVSc, PhD, DACVN, DECVCN) on Wednesday 20th February at 8:00pm. By sifting through marketing messages and client misconceptions, Cecilia will be aiming to come up with the best way to advice clients and patients based on concrete scientific evidence.
Commenting on the upcoming Webinar Dr Cecilia Villaverde said: "As marketing is the source that lures pet owners into buying into fads and trends, I will aim to dissect the common misconceptions that come with these campaigns."
"From raw diets, to homemade treats, fads and trends are rife in the pet food market. This webinar aims to help vets and vet nurses understand the science behind some of the most popular diets, and ultimately help pet owners make well-informed long-term nutrition choices."
Dr Cecilia Villaverde obtained her degree in 2000 and her PhD in animal nutrition in 2005 at the Universitat Autonoma de Barcelona. She worked as a post-doctoral researcher in feline nutrition at the University of California Davis (UCD), where she also completed a residency in small animal clinical nutrition at the Veterinary Medical Teaching Hospital.
Dr Villaverde is board certified in veterinary nutrition by the American College of Veterinary Nutrition (ACVN) and by the European College of Veterinary and Comparative Nutrition (ECVCN) where she has been president since 2016.
To register for the webinar, go to:
PetLondon launches sports-themed dog apparel

redpajama_set_name: RedPajamaCommonCrawl


text:
It's not just São Paulo. In cities across the country – from rainy Rio de Jainero to the capital Brasilia – thousands have turned out to denounce the insurrection: https://t.co/tTxAg0HdTS
twitter@davidrkadler
Oh pleeze. https://t.co/rp1u7jhPt5
Constitution says one must be an American citizen for at least 7 yrs to serve in Congress George Santos (Devolder) lived in Brazil for many years & is wanted there for fraud He also said he was born there Until he can show a US passport it's safe to assume he's a foreign asset
At my bedside when I was sick last winter,before the holidays. Before we knew she had cancer. I miss The Wife so much it's a physical pain in my heart. Grief is such a monstrous thing--the way it creeps up on you and threatens to pull you into the darkness. I love her,oh so much. https://t.co/8EA7b3tNiZ
Keep worrying about Black-on-Black Crime because what you don't see is Black excellence adapting and overcoming; 2022 High school graduation of an all-boys school in New Orleans, with over 9.2 million dollars in college scholarships awarded. The revolution will not be televised! https://t.co/lCwoVU4keP
People will die because of Ron Johnson. Shame on Wisconsin for re-electing him. https://t.co/kB5rqBi5Ow
Trumpers get so excited every time Biden discovers more classified documents But they refuse to see the elephant in the room: Biden is self-reporting while Trump is lying and obstructing. It's really that simple.


Would it be possible for people to subscribe to the calendar and receive new updates or reminders? Also can we create a category?
Generally, this is done by subscribing to the RSS feed for the chosen calendar view.


This study investigates (1) if communication with one?s supervisor is related to empowerment through establishing perceptions of leader integrity, and (2) the extent to which the relationship between leader integrity and empowerment is moderated by intellectual stimulation. Due to the dynamic nature of today?s organizational environment, understanding the nuances among these variables is vital to effective performance at the individual and organizational level. Hierarchical multiple regression tests were performed with a sample of 259 nurses in two regional healthcare facilities in the Midwestern United States. The results support a moderated-mediation relationship such that open communication with one?s supervisor is positively related to empowerment through perceptions of leader integrity, but the relationship between leader integrity and empowerment varies across levels of intellectual stimulation. Specifically, while supervisor integrity mediates the relationship between patient safety communication and empowerment, this mediated relationship is only significant for followers who experience high intellectual stimulation, and is not significant for followers who report low intellectual stimulation. Thus, open communication and leader integrity will only empower followers if the leader is intellectually stimulating. This research clarifies how leaders in health care environments should communicate with their followers to empower them to think and act by their own initiative. Specifically, followers who communicate openly with their supervisor will feel more empowered, but only if they experience high intellectual stimulation which can improve their job performance and patient safety overall.


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text:
Wednesday 8 April 2020
Property and Parking Join Forces to Offer Free Parking for NHS Workers at Over 900 Car Parks Nationwide – 08.04.20
YourParkingSpace.co.uk, the UK's leading on-demand parking platform, has joined forces with property businesses including Hammerson Plc and Canary Wharf Group to provide NHS workers with free parking across the UK.
By registering on YourParkingSpace, NHS workers are provided with a unique code which enables them to use their NHS credentials to reserve spaces in over 900 car parks nationwide.
There are 943 car parks available to book free of charge, including at shopping centres, hotels and supermarkets, with many located near health or medical facilities.
YourParkingSpace.co.uk is in ongoing discussions with its network of space owners and will continue to update its website and Twitter page with additional places to park. It intends to extend the scheme to other key services workers over the coming weeks.
Please visit www.yourparkingspace.co.uk/nhs-parking for more information on how to use the service and the location of car parks.
Harrison Woods, Chief Executive Officer of YourParkingSpace.co.uk, said: "In these unprecedented times, it's important that we do what we can to help.
"With reduced public transport in many places and increased movement of health workers, we're pleased to be working alongside organisations such as Hammerson and Canary Wharf Group to assist key workers in this small way.
"We plan to add new spaces to the platform every day – hopefully it's something that helps to make things easier for NHS staff and other key services workers."
Kathryn Malloch, Head of Customer Experience at Hammerson, said: "We all know that NHS workers are really going the extra mile right now, for all of us, and we are hugely grateful for everything they are doing. Hopefully this initiative will help make their journeys to work a little easier and safer."
Steve Greig, Co-Managing Director, Canary Wharf Management, said: "Through YourParkingSpace, we are delighted to be able to offer free car parking spaces at Canary Wharf to the many brilliant NHS workers who are working hard day and night to support the country."
Chris Box
Red Marlin
chris@redmarlin.co.uk
M 07934 688993
About YourParkingSpace.co.uk:
Founded in 2013, YourParkingSpace.co.uk is an online marketplace that markets and manages over 65,000 parking locations UK-wide, ranging from driveways through to private off-street parking spaces and commercial car parks. The business processes thousands of online booking reservations daily across the UK, generating £24.5million for its clients to-date. Alongside its 60,000 private space providers, the business also works with household names such as Premier Inn, Morrisons and Novotel. YourParkingSpace.co.uk has also been placed 170th in the FT1000 annual list of Europe's fastest-growing companies.
www.yourparkingspace.co.uk
Twitter – @ypsuk
Facebook – YourParkingSpace
Canary Wharf's new district, Wood Wharf, will provide up to 3,600 new homes, 2m sq ft of office space, 350,000 sq ft of retail space and over eight acres of public spaces, squares and parks. There will also be a GP surgery and a two-form primary school for 420 children.


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In the Key of Q
Wuhryn Dumas
"It is possible to be this entity, this identity, this fearless Black, Queer person."
Wuhryn Dumas was born in Atlanta, Georgia and makes beautifully positive music. His style is a blend of funkadelic Pop, R&B, House, Jazz and not to forget an undertone of Soul. His first six-track EP, 'Glitter On Fire' was released in 2020 after 2018's premier single, 'Chemistry'.
Topics covered in the episode include the power of Androgynous identity; the unfortunate absence of joy in Urban Music and Hip Hop; growing up Black and Queer in Atlanta, Georgia (yes the dolls and wearing shirts for wigs!); toxic masculinity with the traditions of the barbershop; the masks that we all wear in life and the magic of Prince and Michael Jackson.
Our Patreon page and forever home.
Say hello at Twitter, Instagram, Facebook or podcast@inthekeyofq.com
Episode transcription available in PDF form.
In the Key of Q composer Paul Leonidou's online home.
Artist's music: Exclusive playlist; Spotify; Apple; SoundCloud;
Artist's social media and online home: Twitter; Instagram; online home
Please subscribe to the podcast, rate and review. And spread the word!
More episodes from In the Key of Q
Blake Mundel - Revisited
Featuring new material made public for the first time, LGBTQ singer Blake Mundel (aka Courier) discusses his five favourite songs plus the one track that he'd chosen as a gateway song into his catalogue. Series 2 of In the Key of Q will begin on 1 March 2022. Links: Blake's http://therealcourier.com/ (forever home) online. Show https://drive.google.com/file/d/1N5N66arp-CGvq7r-CbMLaFhWhTAxwrXT/view?usp=sharing (transcript). Blake (as Courier) and his video in which "https://www.youtube.com/watch?v=okGDwxRfnJY&t=100s (a black man can go out for a jog and make it home)". https://www.casino.org/blog/top-10-gay-nfl-players/ (Inspirational gay NFL players) at Casino.org. Blake's chosen artists: https://www.beyonce.com/ (Beyonce), https://www.paulsimon.com/ (Paul Simon), https://music.sufjan.com/ (Sufjan), https://www.jamestaylor.com/ (James Taylor), https://www.chanceraps.com/ (Chance the Rapper) Blake's chosen song: https://www.youtube.com/watch?v=4hWf8RstqjE (San Francisco). In the Key of Q composer Paul Leonidou's https://www.unstoppablemonsters.com/ (online home). Please subscribe to the podcast, rate and review. And spread the word! You can find us on https://twitter.com/InTheKeyOfQ (Twitter) and https://www.instagram.com/inthekeyofq/ (Instagram) and join the community over on https://www.facebook.com/In-the-Key-of-Q-104781968301237 (Facebook). Support this podcast
Furillostar - new music!
In this bonus mini-episode, Flint-based alumni Furillostar introduces his new single, DWNLD which drops today. And as ever, a London shout-out to Mamma Star! Episode transcription https://drive.google.com/file/d/1X14T9pn9miLpZuoijYWwTI2A8pwpoMyd/view?usp=sharing (here). FURILLOSTAR's linktree can be found https://linktr.ee/Furillostar (here.) His full Key of Q interview can be found https://player.captivate.fm/episode/d9637d94-9907-4e32-911c-6f96236015bc (here). In The Key of Q publishes every Tuesday ('Quesday') and is available at inthekeyofq.com, on Apple, Spotify, Google, Stitcher, Gaana, TuneIn, Overcast, PlayerFM, Deezer, Saavn and most other podcast platforms. Its social media homes are on Twitter, Instagram and Facebook. Series 2 of In the Key of Q drops in March 2022. Support this podcast
Aruan Duval - Revisited
Featuring new material made public for the first time, pop singer Aruan Duval discusses his five favourite songs plus the one track that he'd chosen as a gateway song into his catalogue. Series 2 of In the Key of Q will begin on 1 March 2022. Support the podcast at our https://www.patreon.com/inthekeyofq (Patreon page) for as little as £3 per month. Transcript https://drive.google.com/file/d/1N4RgyDcY8HdQM5bofLLjccA7U1Y65bnz/view?usp=sharing (here). Aruan's band Furiku on https://open.spotify.com/artist/6uJvv7JQmnDJqRBS5JVTq9?si=cL5TqUW1Rm6Nca_0hc7CKA (Spotify) and https://music.apple.com/gb/artist/furiku/362159060 (Apple Music). Aruan on https://open.spotify.com/artist/6uJvv7JQmnDJqRBS5JVTq9?si=tn_G6YRaRfS9D8aRUX1jUQ (Spotify) and https://music.apple.com/gb/artist/furiku/362159060 (Apple Music). Aruan's chosen artists: https://www.prince.com/ (Prince), https://www.davidbowie.com/ (David Bowie), http://www.steviewonder.net/] (Stevie Wonder), https://www.arethafranklin.net/ (Aretha Franklin), https://www.madonna.com/ (Madonna) Aruan's gateway song, https://www.youtube.com/watch?v=_Xcynaciw2w (Go-Go). https://www.youtube.com/watch?v=K_ui701za0w ("A Brief History of Prince") from The Beat Goes On In the Key of Q composer Paul Leonidou's https://www.unstoppablemonsters.com/ (online home). Please subscribe to the podcast, rate and review. And spread the word! You can find us on https://twitter.com/InTheKeyOfQ (Twitter) and https://www.instagram.com/inthekeyofq/ (Instagram) and join the community over on https://www.facebook.com/In-the-Key-of-Q-104781968301237 (Facebook). Support this podcast


The yearly migration of the Humpback whales is one of the highlights along the East coast of Australia. Port Macquarie offers one of the best vantage points to see whales as we are very close to the action. Just a short trip on our boat and our spotter on the hill will direct the skipper to the best sightings. Our Waverider is an open rib boat which has a roof to protect you from sun and rain, at the same time allowing you a real close up to see these beautiful creatures. Sometimes you are able to hear and smell them as they swim on by. Some passengers have got to taste the experience when the splashing gets a bit close.... so be on the lookout!
We are open every day of the season and schedule tours daily from the morning until late. Only bad weather prevents us from going out.


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Lo stemma del Roscommon è rimasto grosso modo sempre invariato. Consiste in uno scudo di colore azzurro nel quale sono situati vari elementi.

Un triangolo verde con la punta verso il basso spicca all'interno dello scudo, e richiama in maniera figurativa il Ros di Comain da cui deriva anche il nome della contea.
Nell'area verde è situata una croce, che rappresenta San Coman, l'Abbazia di Roscommon ed è una stilizzazione della croce di Cong.

L'antica corona riguarda invece Rathcroghan, dove, secondo Mac Firbis, "molti dei re della dinastia Hereman furono seppelliti".

La testa d'ariete rappresenta l'importanza per la contea dell'allevamento ovino, mentre il ramo di quercia è un fatto più recente, ovvero riconducibile alle origini del Dr. Douglas Hyde, primo Presidente della Repubblica d'Irlanda, ricordato nella contea proprio col ramoscello e la scritta "An Craobhin Aoibhinn".

In fondo allo scudo è presente un nastro con il motto in latino, "Constans Hiberniae Cor" (Costante cuore d'Irlanda), che non solo suggerisce la locazione fisica della contea nell'isola, ma armonizza tutti gli elementi rappresentati nell'emblema.

I colori sportivi e culturali della contea sono il giallo e blu.

Bibliografia

Contea di Roscommon

redpajama_set_name: RedPajamaWikipedia


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The bitter is essential in Italian mixing. A strong and bitter taste is produced by a long infusion of sweet and bitter orange peel, gentian, wormwood, carlina and quassio.
Bitter is one of the key components, along with Vermouth, of Italian mixing. Berto wanted to give a classic interpretation of this historic liqueur, adding to the sweet and bitter orange peel also Quassio, Carlina and Assenzio.
MARIA FRANCESCA C. on 07/17/2018 Fantastico prodotto, quello che cercavo!!!!


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Mastercard and Visa Are Making Bold Moves Toward Mass Crypto Adoption
Leading global payment companies Mastercard and Visa have been making moves to accelerate the support of cryptocurrency payment processors by opening up new options for users around the world. Both companies made strong statements in support of the use of cryptocurrencies in July by announcing respective projects and collaborations that are driving the adoption of cryptocurrencies.
The positive attitude toward cryptocurrency exchanges and payment platforms from the world's largest traditional payment processors signals a shift in perception from the traditional financial space. Mastercard has been actively encouraging exchanges and payment service providers to enlist in its recently expanded cryptocurrency card program, becoming partners in just a few weeks as part of its Accelerate program. Meanwhile, Visa outlined its vision of the cryptocurrency space with an overarching theme of positivity toward the market and the role it will play.
The payment service provider noted digital currencies as an exciting avenue to expand existing network-of-networks to support the latest technology powering global commerce. These two global giants are not just offering lip service, either; their payment cards and technology are already powering a number of platforms and service providers within the crypto space. The likes of Coinbase and Binance crypto exchanges use either Visa or Mastercard to power their crypto debit card services.
Visa's and Mastercard's relationship with crypto is growing
Visa's public affirmation of its positive stance toward cryptocurrency payment services reflects its drive to remain a leading player in the global payment network. As highlighted in its "outlook on new digital currency payment flows," the company admits that a growing body of players in the traditional financial sphere has been looking to plug into the crypto space: "It's a concept that is gaining traction beyond fintechs."
The company has already established a working relationship with some leading cryptocurrency-based firms in 2020, including Coinbase and Fold. This is in addition to more than 25 cryptocurrency wallets that are connected to Visa's systems. Visa also has its fintech-focused accelerator program called FastTrack, which allows tech firms including cryptocurrency and blockchain-based companies to access its systems and network.
The company has also been developing its own cryptocurrency projects that include an investment into tech firm Anchorage, which builds security infrastructure for the cryptocurrency ecosystem. Its research team has also been working in the blockchain space for a number of years, culminating in the creation of the white papers for the Zether and FlyClient projects.
Furthermore, Visa has been involved in helping shape regulations and policies toward cryptocurrencies around the world. It has worked with the World Economic Forum to develop recommendations for central banks looking into the use cases of central bank digital currencies. Cointelegraph reached out to Visa for additional insights, but the company declined to provide any further information other than its blog post.
Mastercard has been actively encouraging crypto exchanges and payment service providers to sign up to their Accelerate platform in an effort to expedite the process to become partners by fast-tracking the onboarding of new crypto debit and credit card providers while providing added assistance for market entry and expansion in different countries. Nevertheless, prospective partners need to meet stringent requirements set by Mastercard. This includes high levels of consumer protection and compliance with AML/KYC regulations.
This move to collaborate with the crypto industry comes off the back of the news that Wirex became the first cryptocurrency platform granted a Mastercard principal membership. Part of the functionality allows users the ability to instantly convert cryptocurrencies into conventional fiat currency. An added benefit is a rewards program that gives users 1.5% of purchases made with these cards in Bitcoin.
Bridging the divide
Recently, Binance confirmed that a limited run of its Binance Cards is being shipped to Europe. The move adds real substance to the statements made by Visa and Mastercard, as users are beginning to have access to these card services through some of the biggest players in the cryptocurrency exchange space.
According to Josh Goodbody, the director of European and Latin American growth and institutional business at Binance, traditional banking cards are a "bridge between the crypto and traditional finance," adding: "Crypto debit cards provide a tangible and frictionless way to spend your crypto, and it provides users with the ability to incorporate crypto into their day-to-day lives."
Goodbody declined to go into detail about the direct working relationship with Visa but stated that the acquisition of cryptocurrency payment platform Swipe would allow Binance to tap into an established network of regional service providers where the company hopes to bring new users into the cryptocurrency space. Goodbody believes that mainstream financial firms will have a key role to play in this:
"Visa and other networks' willingness to work with the blockchain industry is a very positive vote of confidence for the further adoption of cryptocurrencies. Not only are traditional technology providers facilitating adoption, they are actively participating in the development of the ecosystem — we see this as an opportunity to further the adoption and accessibility of cryptocurrencies."
Crypto analyst Mati Greenspan also complimented the move by Mastercard and Visa in the Quantum Economics email newsletter from late July, saying: "As far as fundamentals are concerned, this is about as bullish as it gets for Bitcoin and the gang."
Netanel Kabala, the chief analytics officer and co-founder of payment platform Simplex, told Cointelegraph that his company has been working alongside Visa and Mastercard for seven years. The relationship has allowed the company to open up cryptocurrency offerings to new users exploring alternative investment methods: "Crypto adoption is growing globally as people seek out alternative investment avenues." Kabala identified a lag time for new users being introduced into crypto, but the integration of mainstream financial institutions like Mastercard and Visa is a strong signal:
"From an analytics standpoint, we're seeing many new users enter the crypto world. With every new technology or advancement, it often does take some time before the general public incorporates it into the mainstream. While we definitely believed this mainstream adoption would have come earlier, it seems to be starting now."
Simplex CEO Nimrod Lehavi believes that there is a more receptive perception to the potential benefits of cryptocurrency, mainly driven by people who want to wrestle back control of their assets and ability to transact independently: "Anything that lowers friction and helps people gain full control of their assets will spread adoption, and crypto-connected debit and credit cards are a key component in that regard."
Financial industry leaders like Mastercard and Visa vocally supporting and actively working with cryptocurrency and blockchain firms add more credence to the value and utility of these services. Lehavi believes that this will open the door to more users that have not been exposed to digital assets: "The support of major players removes a great deal of the uncertainty people might have regarding cryptocurrencies and will enable them to discover digital assets for what they are."
До этого Elliptic: взломавшие Twitter хакеры пропустили половину биткоинов через два миксера
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Description: Beautiful!! You don't want to miss out on this one!! Come see this AMAZING home in the highly desirable Longleaf Pines neighborhood. This Bill Clark Homes Dawson 2 open floor plan has Maple cabinets, stainless steel appliances, brush nickel hardware, tiled baths, hardwood floors, gas water heater, gas heat and screened in porch. Spacious Bonus Room that can be used for a play room, office or 4th bedroom. Located in the great community just minutes from the back gate of MCAS Cherry Point, beaches and shopping. It is just around the corner from Creekside Elementary and Harris Teeter!


Black Pistol Fire's first Austin show since a coruscating set at Emo's 14 months back gives locals another chance to witness drum animal Eric Owen and high-flying guitarist/vocalist Kevin McKeown bash out cuts from diverse fifth strike Deadbeat Graffiti, which fuzzes out on soul, rock, and garage punk. Phranchyze and Zeale's thrilling rap collab, Blackillac, follows golden-voiced rocker Emily Wolfe.
Street Corner Arts is, in fact, damned proud to present this show – the full title of which is We Are Proud to Present a Presentation About the Herero of Namibia, Formerly Known as Southwest Africa, from the German Sudwestafrika, Between the Years 1884-1915 – and proud they should be, as Jackie Sibblies Drury's "stunning, self-referential theatrical collage" engages with abominable history to tackle the subject of racism and the theatre community's own (sometime self-serving) attempts at engaging with this subject. That (award-winning) Liz Fisher directs a fine cast, but – is this show as powerful good as it sounds? Our reviewer Elizabeth Cobbe sure thinks so.
Through Dec. 15. Thu.-Sat., 8pm. $17-25.
They're creepy and they're kooky, mysterious and spooky … they're Louisianna Purchase and Zane Zena with new Horrorpiece Theatre mistress, Bulimianne Rhapsody. Get ready to scream, queens. Read more about Die Felicia online, and find them in our Best of Austin 2017 winner's list.
Thu., Dec. 13, 10pm-2am. $6.
This is in the original Downtown location, mind you, not in ET's new Linc digs that opened last week … and it's gonna be cozy, all right, with three drafts from Boulevard Brewing's Smokestack Series (Bourbon Barrel Quad, Whiskey Barrel Stout, Tank 7 American Saison Ale) – and their limited release Love Child #9 Sour Ale – and a raffle for Boulevard prize packs of socks and a plaid fleece blanket and more. It's the best of Kansas City right there in Pretzel Heaven.
Here's an informative hour of talks in conjunction with the museums's current "Making Africa: A Continent of Contemporary Design" exhibition.
Thu., Dec. 13, 12:30pm. Free.
Devil in the Garden: Films of Luchino Visconti: Visconti's Palme d'Or winner tells the story of an old world Italian aristocrat in the mid-19th century as he observes his old way of life slowly slipping away.
Nearly a decade after the release of the Lovely Sparrows' debut LP, Bury the Cynics, Shawn Jones returns with new music. Shake the Shadow swoons immediately familiar to local fans, jangly mellow and richly textured folk-bent pop hinging on Jones' literate and intimate narratives – as if he's whispering tales of love and heartbreak in a library. Linen Closet and Dead Recipe set the stage.
Are you ready to snap-crackle-pop with Rebecca Havemeyer and Stanley Roy Williamson's annual holiday hootenanny? Might be the last run, since Reb's leaving us, so make sure not to miss it; plus Silky Shoemaker, Lindsey Taylor, and a helluva lot more are joining in on the stage action.
Thu.-Sat., Dec. 13-15, 7:30pm; Sun., Dec. 16, 6:30pm. $15 online, $20 at door.
Corps Multiple, the first artistic company created by French artists in Austin, unleashes a performance that "opens a discussion on humanity's inherent need for belonging, roots, pairs, and marks." With moving bodies, sounds, text, and lights using the techniques of free improvisation. Bonus: live music.
Dec. 13-15. Thu.-Sat., 8pm. $15-35.
The players of the fictional KPNF radio station re-create Dickens' cherished masterpiece in this Nathan Jerkins adaptation, with all the characters brought to life by a small company of voice actors and live foley sound effects. Directed by Ryan Crowder for Penfold Theatre. And, what's that? You think there's an awful lot of Dickens going on lately? Our Robert Faires tells you why.
Through Dec. 29. Thu.-Sat., 8pm. $15-25.
Cheeki Kahnt and Mary Jane Styles invite you to their sleigh ride with a prancing, dancing crew of queens like Rhonda Jewels, Tanya Lasagna and more. Plus spesh guest DJ Hentaii.
Thu., Dec. 13, 9pm-2am. $10.