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Lina Lahiri studied architecture at the Oxford School of Architecture, Oxford Brookes University, and the Bartlett, University College London, where she obtained her diploma in 2005. As a freelance she designed a private timber house in Nyköping, Sweden and won among others a 1st prize at the Europan Competition in 2009.
Since 2005 she is an architect at Sauerbruch Hutton where she has been in charge of crucial projects such as Jessop West, a building for the University of Sheffield, and Low2No, a low-carbon mixed-use district with housing in Helsinki. Currently she is the project leader for the Stockholm One Tower, a landmark project for Stockholm's new district of Hammarby Sjöstad.


text:
Home Crypto Strategies
Reddit Confidentially Submits Draft Registration Statement for Proposed IPO
in Crypto Strategies
Reddit has confidentially submitted a draft registration statement with the Securities and Exchange Commission (SEC) for a proposed initial public offering of its common stock, the company announced on Twitter.
The numbers of shares to be offered and the price range for the proposed offering have yet to be determined, the announcement said.
The IPO is expected to occur after the SEC completes its review process, subject to market conditions.
Reddit is in a "quiet period," and for regulatory reasons cannot disclose more information, the announcement said.
The social media platform announced in January it was expanding its work with the Ethereum Foundation to provide development resources to scaling tools. In the announcement, posted to the Ethereum subreddit, Reddit employee u/jarins said the move increases the company's commitment to the technology and echoes its long-held "decentralized ethos."
Wednesday's release does not constitute an offer to sell or the solicitation to buy any securities, Reddit said in a disclaimer.
Hi, we have some news. pic.twitter.com/DjkjBRloHF
— Reddit (@Reddit) December 16, 2021
Read more: Reddit Rolls With Arbitrum to Scale Its Ethereum-Based Community Points System
On Thursday, Metaco, an enterprise cryptocurrency platform for securing, trading, issuing, and managing digital assets, announced the addition of...
The percentage of the global Bitcoin mining industry running on renewable power increased by 1% to 58.5% in the...
IreneDAO NFTs causing a stir on crypto Twitter
The story of the IreneDAO NFT craze started in October when 28-year-old Chinese crypto influencer and model Yuqing Irene...
It appears that retail giant Walmart Inc. is getting ready to enter the Metaverse with patent filings indicating that...
22-year-old Indonesian boy makes $1M by selling NFT selfies on OpenSea
An Indonesian college student has reportedly become a millionaire by selling nonfungible token (NFT) versions of his selfies on...

redpajama_set_name: RedPajamaCommonCrawl


By now, you know of Tony Romo's psychic abilities.
The former Cowboys quarterback, now the main color analyst on CBS who will call Super Bowl LIII alongside Jim Nantz, was asked to predict the final score of the game.
Romo's propensity for accurately predicting plays before they happen was on full display during the Patriots' 37-31 AFC title game win over the Kansas City Chiefs. But that was hardly the first time Romo has done that.
In his very first game as a broadcaster (Week 1 of the 2017-2018 season), he accurately predicted a handful of plays before they happened and it became a national story. He is far and away the best at explaining the X's and O's in a way that anyone can understand and has a level of enthusiasm that is unmatched by any broadcaster.
That, at least for me, is going to make watching this Super Bowl a lot more fun.


Can I Sue My Employer If I'm Injured At Work?
Since 1959, The Floyd Law Firm has brought thousands of personal injury and workers' compensation claims to a successful resolution in St. Louis and the surrounding areas. It is just one of the many benefits of hiring our attorneys, but why is their dual experience so important?
An on-the job-injury can leave you unable to work temporarily or permanently. It can leave you with disabilities that affect your entire life. The consequences can have an impact on your family's future. In a nutshell, you have a lot riding on the outcome of your case. It is a reason many people want to know if they can sue their employer. The better question is, "How do I maximize my recovery?"
In some situations, such as those involving a car accident or unsafe machinery, you may have two claims: one under workers' compensation insurance and one under personal injury law. We recently obtained a $5 million verdict for a FedEx employee injured at a warehouse owned by a third party. Alone, his workers' compensation claim would have been worth approximately $70,000.
Knowing the difference between a workers' compensation and personal injury claim can help you understand how we can maximize your overall recovery.
Workers' compensation claims: The general rule is that you can recover compensation for any injury or illness you suffer at work or while completing a task for your employer. You do not have to prove fault and could even have been responsible for your own injuries. For this reason, your employer is protected from personal injury liability and the amount of compensation and types of benefits you can recover are limited by the laws of the state in which you file.
Personal injury claims: A personal injury claim (also known as a third-party liability claim) is a civil lawsuit for damages. You must prove that a third party was at fault, which is one reason your request for damages is not limited to medical or lost wage benefits. It can also include compensation for pain and suffering as well as punitive damages for situations involving gross negligence or intentional acts.
Don't Jeopardize Your Rights. Contact Us Immediately For Your Free Consultation.
The types of damages you can recover and whether you must prove fault are not the only differences between the two claims. Things like filing deadlines and notice requirements also vary significantly from one claim to another and depend on whether your accident occurred in Missouri or Illinois.
We can help you navigate this complex situation, but waiting could harm your case. The sooner you contact us, the better. Call 314-492-6958 or send us your information for a free consultation with a lawyer.


People watch comedies to do one thing and one thing only; laugh. There are some people who want to laugh and revel in how clever they are when they get the high brow political jokes or those who want to laugh until their sides aches. There are even some who just want to 'turn their brains off' and laugh away their daily struggles. Comedy is valid whether you're a fan or not. It's good to see Game Night so sharp in its comedy. A film that is smartly written, with creative characters, inventive filmmaking and tight pacing.
In other words it's a film that is easy to like, it's not the funniest film ever but it is the funniest film of 2018 (so far). Game Night stars Rachel McAdams and Jason Bateman as ultra competitive couple, Annie and Max. They host a weekly board game night with best friends Ryan (Billy Magnussen), Sarah (Sharon Horgan) and the happily married Michelle and Kevin (Kylie Bunbury and Lamorne Morris). On the sidelines having never been invited to such an event is creepy next door neighbour Gary (Jesse Plemons).
After the stage is set with all the character pieces in place, Max's older brother Brooks (Kyle Chandler) breaks the status quo by announcing the ultimate: A game night to end all game nights. Unbeknownst to the group the events which follow aren't necessarily how they appear to be, and consists of a collection of set-pieces and life or death moments which the gang take as part of 'the game'. Once the characters realise what is going on, this game can't simply be put down or leave in a cupboard to get dusty. This is one game night where they realise if they lose, there's a whole lot more at stake than just bragging rights.
Game Night excels in its approach. Directors John Francis Daley and Jonathan Goldstein (Vacation) mould real life peril with out there situations. The characters (created within Mark Perez's screenplay) treat the situation seriously, making you care for the characters and their stories. This is a film shot like a thriller wrapped in a comedic script. Think Adam McKay directing through David Fincher's eye for good lighting and intense camera angles.
An Unforgettable Evening - Making 'Game Night' - a good EPK but as with all EPKs is far too short at under four minutes.
Gag Reel - no explanation required.
Shot digitally by Barry Peterson, a cinematographer of note within the comedy genre, having worked on the Jump Street films and Zoolander. There is a fluorescent look to the film akin to the plastic pieces of a board game and the 'hyper reality' the film is set in. With it being shot digitally the film has a superior sharpness with solid blacks throughout.
Game Night is available on digital download, DVD, Blu-ray and VoD now.
Game Night isn't particularly groundbreaking but with a high calibre cast and a behind the camera team who understand the material, this is one film you might come back to time and again.


text:
Culoptila jamapa är en nattsländeart som beskrevs av Bueno-soria och Santiago de Fragoso 1996. Culoptila jamapa ingår i släktet Culoptila och familjen stenhusnattsländor. Inga underarter finns listade i Catalogue of Life.

Källor

Externa länkar

Stenhusnattsländor
jamapa

redpajama_set_name: RedPajamaWikipedia


text:
\section{Introduction}

In the Simon Problem with parameter~$k$, we are given an oracle for a function~$f : {\mathbb Z}_2^k \rightarrow S$, for some co-domain~$S$ with~$\size{S} \ge 2^k$. The function~$f$ is either injective, or there is an unknown non-zero element~$s \in {\mathbb Z}_2^k$ such that for all~$x,y \in {\mathbb Z}_2^k$, we have~$f(x) = f(y)$ if and only if~$x + y \in \set{0, s}$. In the latter case, we say that the function ``hides'' the element~$s$, and call a pair of distinct inputs~$x,y$ a ``collision'' if~$f(x) = f(y)$. The task is to determine which of the two cases holds.

Simon designed a bounded-error quantum algorithm with query complexity~$\mathrm{O}(k)$ for the eponymous problem, and showed that any bounded-error classical (i.e., randomized) algorithm for the problem requires~$\Omega(2^{k/2})$ queries~\cite{S97-simon-algorithm}. (The query lower bound stated in the paper is $\Omega(2^{k/4})$ for classical algorithms with error at most~$\tfrac{1}{2} - 2^{-k/2}$. However, the proof can be modified in a straightforward manner to show that if the algorithm makes error at most~$\tfrac{1}{4}$, at least~$2^{k/2 - 1}$ queries are required.) The lower bound for classical algorithms is optimal up to a constant factor, as is implied by the ``Birthday Paradox'': if we pick~$t \coloneqq \ceil{2^{k/2 + 1}}$ elements~$X_1, X_2, \dotsc, X_t$ independently and uniformly at random from~${\mathbb Z}_2^k$, if~$f$ hides some non-zero element, with probability at least~$3/4$ we find a collision (i.e., a pair~$X_i, X_j$ with~$i,j \in [t]$ such that~$X_i \neq X_j$ but~$f(X_i) = f(X_j)$).

We present a simple, \emph{deterministic\/} algorithm for the Simon problem that achieves the asymptotically optimal classical query complexity (Theorem~\ref{thm-simon-det}). Since posting the first version of this article~\cite{Nayak21-hsp-classical}, we have learnt that algorithms achieving the same asymptotic query complexity, including the same algorithm, were known before~\cite{CQ18-simon-problem,WQTLC21-generalized-simon,YHLW21-generalized-simon}. Nevertheless, it is instructive to understand the algorithm underlying Theorem~\ref{thm-simon-det}, as it forms the basis of the generalizations that we describe next.

The Simon Problem is a special case of the Hidden Subgroup Problem (see, e.g., Ref.~\cite{CvD10-quantum-algorithms}). In the Hidden Subgroup Problem, we are given the description of a finite group~$G$, and an oracle for a  function~$f : G \rightarrow S$, where~$\size{S} \ge \size{G}$. The function~$f$ is either injective, or there is an unknown non-trivial subgroup~$H$ of~$G$ such that for all~$x,y \in G$, we have~$f(x) = f(y)$ if and only if~$x^{-1} y \in H$. In other words, the function~$f$ is constant on left cosets of a possibly trivial subgroup of~$G$, and takes distinct values for distinct left cosets of the subgroup. We say that the function ``hides'' the subgroup~$H$. The task is to determine whether the subgroup is trivial, i.e., the function~$f$ is injective, or not. When~$G \coloneqq {\mathbb Z}_2^k$ and the hidden subgroup~$H$ is of the form~$\set{0, s}$ for some unknown non-zero element~$s \in {\mathbb Z}_2^k$, we get the Simon Problem.

The randomized algorithm for the Simon Problem generalizes immediately to the Hidden Subgroup Problem when the order~$m$ of the hidden subgroup~$H$ is known, and the resulting algorithm has query complexity~$\mathrm{O}(\sqrt{ n/m }\, )$, where~$n$ is the order of the group~$G$. When the order~$m$ is not known, we can use this basic algorithm to design a new algorithm. The new algorithm has query complexity~$\mathrm{O}(\sqrt{n} \, )$ when the function is injective, and \emph{expected\/} query complexity~$\mathrm{O}(\sqrt{ n/m }\, )$ when the function~$f$ hides some non-trivial subgroup~$H$ with unknown order~$m$. Namely, we run the basic randomized algorithm above, with error at most~$1/4$, assuming that the order of the hidden subgroup is at least~$r$, starting from~$r \coloneqq n/2$. If we do not succeed, we halve~$r$, and repeat (until~$r \le 1$).

We may ask if the Hidden Subgroup Problem also admits a deterministic algorithm that is as efficient in the worst-case as the best randomized algorithm. We answer this in the affirmative when the underlying group is abelian (\Cref{cor-ahsp}). For non-abelian groups, we present a deterministic algorithm that comes within a multiplicative factor of $\mathrm{O}(\sqrt{\log n} \,)$ of the optimal randomized query complexity (\Cref{cor-non-abelian-hsp}). Both the algorithms are based on the construction of a \emph{generating pair\/} of subsets for a group (\Cref{def-gen-pair}), with optimal or nearly optimal size.

The optimal randomized algorithm for the Hidden Subgroup Problem described above has expected query complexity that is sensitive to the input, namely~$ \mathrm{O}(\sqrt{ n/m}\, )$. It is natural to ask if there is a deterministic algorithm that has query complexity with a similar dependence on the order of the hidden subgroup. Algorithms coming close to this bound were known prior to this work for the case of~$G \coloneqq {\mathbb Z}_p^k$ for prime~$p$, when the subgroup~$H$ has order~$p^l$ and~$l$ is known; see Ref.~\cite{WQTLC21-generalized-simon} for the~$p = 2$ case, and Ref.~\cite{YHLW21-generalized-simon} for the general case. The algorithm in Ref.~\cite{YHLW21-generalized-simon} has query complexity~$ \mathrm{O}( \sqrt{ n (\log m)  / m} + \log m ) $.
\suppress{
of the order of
\[
\max\set{\log m, ~ \sqrt{ n (\log m) / m } \, } \enspace.
\]
}

The above question was posed as an open problem in the first version of this article~\cite[Sec.~5]{Nayak21-hsp-classical}. Prompted by this question, Ye and Li~\cite{YL21-hsp-classical} present deterministic algorithms for \emph{abelian\/} groups which solve the problem with query complexity~$ \mathrm{O}(\sqrt{ n/m }\, )$, and find the hidden subgroup with query complexity~$ \mathrm{O}( \sqrt{ n (\log m)  / m} + \log m ) $. These algorithms build on a construction of generating pairs similar to the one in Ref.~\cite{Nayak21-hsp-classical}. Moreover, Ye and Li exhibit instances which show that in general, the deterministic query complexity of the problem may be~$\mathrm{o}(\sqrt{ n/m } \,)$, and that of \emph{finding\/} the entire subgroup may also be~$\mathrm{o}(\sqrt{ n/m } \,)$ or even~$\upomega(\sqrt{ n/m } \,) $. In more detail, for the group~${\mathbb Z}_{p^k}$ for a given prime~$p$ and~$k \ge 2$, the deterministic query complexity of the problem is~$2$ and that of finding the hidden subgroup is~$\mathrm{O}( \log (n/m))$, while~$ n/m $ may be~$ \upomega(1)$. For~$G \coloneqq {\mathbb Z}_p^k$ and a hidden subgroup~$H$ of order~$p^{k-1}$, any deterministic algorithm that \emph{finds\/} the subgroup has query complexity~$\Omega(k \log p)$, while~$\sqrt{ n/m }$ is~$\sqrt{p}$. This gives an arbitrarily large separation as~$k$ grows for a fixed prime~$p$.

We present a different deterministic algorithm for the Hidden Subgroup Problem that also has query complexity~$ \mathrm{O}(\sqrt{ n/m }\, )$ for abelian groups (Algorithm~\ref{alg-fc}, \Cref{thm-fc}). The algorithm is arguably simpler. Furthermore, it also works for non-abelian groups, and has query complexity~$ \mathrm{O}(\sqrt{ (n/m) \log (n/m) }\,) $ for classes of instances which include those over supersolvable groups. We observe that for abelian groups the problem of \emph{finding\/} the hidden subgroup may be reduced to that of finding a single collision, and obtain an~$ \mathrm{O} \big( (\log m) \sqrt{ n/m }\, \big)$-query deterministic algorithm for it (Algorithm~\ref{alg-find-ahs}, \Cref{thm-abelian-find-hs}). We build on these results to design an algorithm that finds the hidden subgroup in certain non-abelian instances with~$ \mathrm{O} \big( (\log m) \sqrt{ (n/m) \log (n/m)  }\, \big) $
\suppress{
\[
\mathrm{O} \! \left( (\log m) \sqrt{ (n/m) \log (n/m)  }\, \right)
\]
}
queries (Algorithm~\ref{alg-fnc}, \Cref{thm-find-hs}). The instances are precisely the ones in which the underlying group is a bicrossed product of the hidden subgroup with another subgroup~\cite{ACIM09-bicrossed-product}. All these algorithms again rest on generating pairs of (near-) optimal size.

\paragraph{Acknowledgements.}
A.N.\ is grateful for the opportunity to teach quantum computation at the undergraduate level, which prompted this work. He is also grateful to Kanstantsin Pashkovich for several helpful discussions, especially for a course-correction early in this work. He thanks William Slofstra for a pointer to relevant literature, and Zekun Ye for bringing Ref.~\cite{YHLW21-generalized-simon} to his attention. This research is supported in part by a Discovery Grant from NSERC Canada.

\section{The Simon Problem}
\label{sec-simon}

We start with a simple, deterministic algorithm for the Simon problem that matches the query complexity of the asymptotically optimal classical algorithm.
\begin{theorem}
\label{thm-simon-det}
There is a deterministic algorithm that makes~$2^{\floor{k/2}} + 2^{\ceil{k/2}}$ queries and solves the Simon Problem with parameter~$k$. Moreover, if the input function hides the non-zero element~$s$, the algorithm finds~$s$. Finally, any deterministic algorithm for the problem requires at least~$q$ queries, where~$q$ is the smallest positive integer such that~$\binom{q}{2} \ge 2^k - 1$.
\end{theorem}
\begin{proof}
Let~$l \coloneqq \floor{k/2}$. Viewing elements of~${\mathbb Z}_2^k$ as~$k$-bit strings, we query the function~$f$ at all~$2^l$ elements of the form~$u \, 0^{k - l}$, where~$u \in {\mathbb Z}_2^l$, and at all~$2^{k-l}$ elements of the form~$0^l \, v$, where~$v \in {\mathbb Z}_2^{k-l}$. If the function is distinct at all these points, we say~$f$ is injective. Otherwise, we say that~$f$ is not injective, and output~$x + y$, where~$x \neq y$ and~$x,y$ are a colliding pair (i.e., are such that~$f(x) = f(y)$).

If the function is injective, the above algorithm outputs the correct answer. Suppose~$f$ hides a non-zero element~$s \in {\mathbb Z}_2^k$. Let~$a$ be the projection of~$s$ onto the first~$l$ coordinates, and~$b$ be the projection of~$s$ onto the last~$k-l$ coordinates. Then~$a \, 0^{k-l} + 0^l \, b = s$. As~$f(a \, 0^{k-l}) = f(0^l \, b)$, the above algorithm detects a collision  and computes~$s$ correctly.

Now consider any deterministic algorithm for the problem that makes~$t$ queries such that
\[
\binom{t}{2} \quad \le \quad 2^k - 2 \enspace.
\]
Then we argue that there is an injective function~$f_0$, and a function~$f_1$ that hides a non-zero element~$s$, such that~$f_0$ and~$f_1$ agree on all the~$t$ elements queried. Suppose the~$t$ queries that the algorithm makes are~$x_1, x_2, \dotsc, x_t$. We take~$f_0$ to be any injective function. Consider the set of elements~$S \coloneqq \set{ x_i + x_j : i, j \in [t], i \neq j}$. We have~$\size{S} \le \binom{t}{2} \le 2^k - 2$, so there is at least one non-zero element in~${\mathbb Z}_2^k \setminus S$. Let~$s$ be one such element. By definition of~$s$, we have~$x_i + s \neq x_j$ for any~$i,j \in [t]$. Thus there is a function~$f_1$ that equals~$f_0$ at all the points~$x_i$, $i \in [t]$, and also hides~$s$.
\end{proof}
Note that for~$k \ge 2$, if~$\binom{q}{2} \ge 2^k - 1$, we have~$q^2 \ge 2^{k+1} - 2$, and therefore
\[
q \quad \ge \quad 2^{(k+1)/2} (1 - 1/2^k)^{1/2} \quad \ge  \quad 2^{(k+1)/2} - 2^{- (k-1)/2} \enspace.
\]
Since~$q$ is integral, $q \ge 2^{(k+1)/2}$. The deterministic upper bound in the theorem comes within a factor of~$3/2$ of this lower bound. Also note that the upper bound in the theorem is within a factor of~$3\sqrt{2}$ of the query lower bound for randomized algorithms with error at most~$1/4$.

\section{The Abelian Hidden Subgroup Problem}
\label{sec-abelian}

We now turn our focus to the Hidden Subgroup Problem when the underlying group is abelian. It is not clear how the use of projections in the algorithm presented in Section~\ref{sec-simon} may be extended to this case, let alone to the non-abelian case. The issue is that the given abelian group may be the direct product of cyclic groups with vastly different orders. We give a different extension of the algorithm, which combines several ways of expressing an abelian group as a sum of two subsets.

More formally, the basic idea behind the algorithm in Theorem~\ref{thm-simon-det} is to find a pair of sets~$S_1, S_2$ of elements of the group~$G$ such that~$\size{S_1}, \size{S_2} \in \mathrm{O}(\sqrt{\size{G}} \, )$ and~$S_1 + S_2 = G$, where
\[
S_1 + S_2 \quad \coloneqq \quad \set{x + y : x \in S_1, ~ y \in S_2} \enspace.
\]
(Here, `$+$' denotes the group operation.) We explain how such a pair of sets may be constructed in a few cases. Together, they yield a construction for a general abelian group.

\begin{definition}
\label{def-gen-pair}
For any group~$G$, we say a pair~$S_1, S_2 \subseteq G$ is a \emph{generating pair\/} for~$G$ if the set~$S_1 S_2$ defined as~$S_1 S_2  \coloneqq \set{ xy : x \in S_1, ~ y \in S_2 } $ equals~$G$.
\end{definition}

We start with the construction of a generating pair for a cyclic group.
\begin{lemma}
\label{lem-cyclic}
Let~$n$ be an integer greater than~$1$, and let~$m \coloneqq \ceil{\sqrt{n} \,}$. There is a generating pair~$S_1, S_2$ for~${\mathbb Z}_n$ such that~$\size{S_1} = m$ and~$\size{S_2} \le \floor{ n / m} + 1$. If~$n$ is a perfect square, the pair further satisfies~$\size{S_1} = \size{S_2} = \sqrt{n}\,$.
\end{lemma}
\begin{proof}
We use the Division Algorithm to find such subsets. If we choose a divisor~$m$ that roughly equals~$\sqrt{n}$, then the number of different remainders and quotients we get when we divide non-negative integers less than~$n$ are both roughly~$\sqrt{n}$.

Let~$m \coloneqq \ceil{\sqrt{n}\, }$, let~$S_1 \coloneqq \set{ 0, 1, 2, \dotsc, m - 1}$, and let~$S_2 \coloneqq \set{ m i : 0 \le i < n/m }$. The subsets~$S_1, S_2$ satisfy the required properties.
\end{proof}

Next, we consider a direct product of two cyclic groups whose orders are both odd powers of the same prime number.
\begin{lemma}
\label{lem-odd-power}
Consider the group~$G \coloneqq {\mathbb Z}_n \times {\mathbb Z}_m$, where~$n \coloneqq p^k$, $m \coloneqq p^l$, $p$ is prime, and~$k,l$ are positive odd integers. There is a generating pair~$S_1, S_2$ for~$G$ such that~$\size{S_1} = \size{S_2} = \sqrt{nm} \,$.
\end{lemma}
\begin{proof}
W.l.o.g., assume that~$k \ge l$. Let~$q \coloneqq p^{(k + l)/2}$. Note that~$0 < q = \sqrt{nm} < n$. Consider
\begin{align*}
S_1 \quad & \coloneqq \quad {\mathbb Z}_q \times \set{ 0 } \enspace, \quad \text{and} \\
S_2 \quad & \coloneqq \quad \set{ iq : 0 \le i < n/q} \times {\mathbb Z}_m \enspace.
\end{align*}
By the Division Algorithm, we have
\[
{\mathbb Z}_q + \set{ iq : 0 \le i < n/q} \quad = \quad {\mathbb Z}_n \enspace.
\]
So~$S_1 + S_2 = G$. Moreover, we have~$\size{S_1} = q = \sqrt{nm} = (n/q) \times m = \size{S_2}$.
\end{proof}

We may combine the generating pairs for two groups to obtain one for their direct product. While we present the proof of the lemma below with the notation for abelian groups, it also holds for non-abelian groups.
\begin{lemma}
\label{lem-product}
Consider a direct product group~$G \coloneqq G_1 \times G_2$. Suppose~$S_1, S_2$ is a generating pair for~$G_1$, and~$T_1, T_2$ is a generating pair for~$G_2$. Then~$(S_1 \times T_1), (S_2 \times T_2)$ is a generating pair for~$G$.
\end{lemma}
\begin{proof}
Any element~$g \in G_1$ may be expressed as~$g_1 + g_2$, where~$g_1 \in S_1$ and~$g_2 \in S_2$. Similarly, any element~$h \in G_2$ may be expressed as~$h_1 + h_2$, where~$h_1 \in T_1$ and~$h_2 \in T_2$. Then~$(g, h) = (g_1, h_1) + (g_2, h_2) \in (S_1 \times T_1) + (S_2 \times T_2)$.
\end{proof}

The construction for a general finite abelian group combines the above building blocks.
\begin{theorem}
\label{thm-abelian}
For any finite abelian group~$G$ with order~$n$, there is a generating pair~$S_1, S_2$ for~$G$ such that~$\size{S_1}$ and~$\size{S_2}$ are both at most~$2\sqrt{n}$.
\end{theorem}
\begin{proof}
By the fundamental theorem of abelian groups, any non-trivial finite abelian group~$G$ may be expressed as a direct product of cyclic groups of prime power order:
\begin{equation}
\label{eq-decomp}
G \quad \cong \quad {\mathbb Z}_{p_1^{k_1}} \times {\mathbb Z}_{p_2^{k_2}} \times \dotsb \times {\mathbb Z}_{p_l^{k_l}} \enspace,
\end{equation}
where the integers~$p_i$ are primes, not necessarily distinct, and the integers~$k_i \ge 1$ for all~$i \in [l]$. We prove the lemma by strong induction on~$l$, the number of cyclic groups in a decomposition of~$G$.

If~$l = 1$, the statement follows from Lemma~\ref{lem-cyclic}.

Suppose the statement holds for all non-trivial finite abelian groups which have a decomposition as above with at most~$m$ cyclic groups, for some~$m \ge 1$. Suppose~$l \coloneqq m+1$, and consider a group~$G$ with order~$n$ and a decomposition with~$l$ cyclic groups as in \cref{eq-decomp}.

Suppose for some~$i \in [l]$, the integer~$k_i$ is even. Let~$r$ be such an index. We let~$G_1 \coloneqq {\mathbb Z}_{p_r^{k_r}}$ and~$G_2$ the direct product of the remaining cyclic groups so that~$G \cong G_1 \times G_2$. By \Cref{lem-cyclic}, there is a generating pair~$S_1, S_2$ for~$G_1$ such that~$\size{S_1} = \size{S_2} = \sqrt{n_1}\,$, where~$n_1 \coloneqq {p_r}^{k_r}$. By the induction hypothesis, there is a generating pair~$T_1, T_2$ for~$G_2$ such that~$\size{T_1}, \size{T_2} \le 2 \sqrt{n_2}\,$, where~$n_2 \coloneqq \size{G_2}$. By \Cref{lem-product}, we get a generating pair~$(S_1 \times T_1), (S_2 \times T_2)$ for~$G$ such that~$ \size{S_1 \times T_1}, \size{S_2 \times T_2} \le 2 \sqrt{n_1 n_2} = 2 \sqrt{n}\,$.

Suppose for all~$i \in [l]$, the integers~$k_i$ are odd. Suppose for some~$i,j \in [l]$, $i \neq j$, we have~$p_i = p_j$. Let~$r,s$ be such a pair of indices. We let~$G_1 \coloneqq {\mathbb Z}_{p_r^{k_r}} \times {\mathbb Z}_{p_s^{k_s}}$ and~$G_2$ the direct product of the remaining cyclic groups so that~$G \cong G_1 \times G_2$. By \Cref{lem-odd-power}, there is a generating pair~$S_1, S_2$ for~$G_1$ such that~$\size{S_1} = \size{S_2} = \sqrt{n_1}\,$, where~$n_1 \coloneqq {p_r}^{k_r} {p_s}^{k_s}$. By the induction hypothesis, there is a generating pair~$T_1, T_2$ for~$G_2$ such that~$\size{T_1}, \size{T_2} \le 2 \sqrt{n_2}\,$, where~$n_2 \coloneqq \size{G_2}$. By \Cref{lem-product}, we get a generating pair~$(S_1 \times T_1), (S_2 \times T_2)$ for~$G$ such that~$ \size{S_1 \times T_1}, \size{S_2 \times T_2} \le 2 \sqrt{n_1 n_2} = 2 \sqrt{n}\,$.

Suppose for all~$i \in [l]$, the integers~$k_i$ are odd, and the primes~$p_i$ are all distinct. Then, by the Chinese Remainder Theorem, we have~$G \cong Z_n$. The statement now follows from \Cref{lem-cyclic}, by noting that for~$n \ge 2$ we have~$\ceil{ \sqrt{n} \,} \le 2 \sqrt{n}$ and~$\floor{ \sqrt{n} \,} + 1 \le 2 \sqrt{n} \,$.
\end{proof}

The algorithm for the abelian Hidden Subgroup Problem follows directly from the existence of a suitable generating pair for the underlying group.
\begin{corollary}
\label{cor-ahsp}
There is a deterministic algorithm with query complexity at most~$4 \sqrt{n} \,$ that solves the Hidden Subgroup Problem over an abelian group~$G$ with order~$n$. Moreover, if the input function hides the non-trivial subgroup~$H$, the algorithm finds all the elements of~$H$.
\end{corollary}
\begin{proof}
By \Cref{thm-abelian}, there is a generating pair~$S_1, S_2$ for the group~$G$ such that~$\size{S_1}, \size{S_2} \le 2 \sqrt{n}\, $. We query the oracle function~$f$ at~$-x$ for all~$x \in S_1$ and at all elements~$y \in S_2$. If the function is injective on the set of points queried, we say~$f$ is injective. Otherwise, we say that~$f$ is not injective, and output~$\set{ x + y : x \in S_1, ~ y \in S_2, ~ f(-x) = f(y) }$.

If the function is injective, the above algorithm outputs the correct answer. Suppose~$f$ hides the non-trivial subgroup~$H$. Let~$h \in H$ be any element of the hidden subgroup. We have~$h = h_1 + h_2 $ for some~$h_1 \in S_1$ and~$h_2 \in S_2$. By definition of~$H$ we have~$f(- h_1) = f( h_2)$. When~$h$ is not the identity, we have~$ - h_1 \neq h_2$ and the algorithm detects a collision. Moreover, the algorithm computes~$H$ correctly.
\end{proof}
Note that we can get upper bounds with better constants for certain abelian groups by appealing to the properties of the generating pair constructed in \Cref{thm-abelian}. As the Simon Problem is a special case, \Cref{thm-simon-det} implies that in the worst case, these upper bounds are tight up to a constant factor.

\section{The General Hidden Subgroup Problem}
\label{sec-general}

Finally, we consider the Hidden Subgroup Problem for a general finite group. Non-abelian groups are more varied in structure than abelian ones, and it appears challenging to extend the ideas used in the abelian case to them. Instead, we resort to a probabilistic argument to show the existence of a generating pair of small size. The construction comes within a poly-logarithmic multiplicative factor of optimal, and leads to our final algorithm.

\begin{theorem}
\label{thm-non-abelian}
For any finite group~$G$ with order~$n$ (with~$n > 1$), there is a generating pair~$S_1, S_2$ for~$G$ such that~$\size{S_1}$ and~$\size{S_2}$ are both at most~$\big\lceil { \sqrt{n \ln n} \,} \big\rceil$.
\end{theorem}
\begin{proof}
We let~$S_1 \coloneqq \set{ g_1, g_2, \dotsc, g_t}$, any subset consisting of~$t$ distinct group elements, where~$ t \coloneqq \big\lceil{ \sqrt{n \ln n} \,}\big\rceil$. Let~${\bm R} \coloneqq \set{ {\bm h}_1, {\bm h}_2, \dotsc, {\bm h}_t}$ be a set of~$t$ distinct group elements chosen uniformly at random from the collection of~$t$-element subsets of~$G$. Recall that the set of products of elements from~$S_1$ and~${\bm R}$ is denoted as~$S_1 {\bm R}$, i.e., $S_1 {\bm R} \coloneqq \set{xy : x \in S_1, ~ y \in {\bm R}}$.

We show that for any fixed element~$g \in G$, the probability that~$g \not\in S_1 {\bm R}$ is less than~$1/n$.
\begin{align*}
\Pr( g \not\in S_1 {\bm R}) \quad & = \quad \Pr( \forall i \in [t], ~ g_i^{-1} g \not\in {\bm R} ) \\
    & = \quad \binom{n - t}{t} \binom{n}{t}^{-1} \\
    & = \quad \frac{(n - t) (n - t - 1) (n - t - 2) \dotsb (n - 2t + 1) }{ n (n - 1) (n - 2) \dotsb (n - t + 1)} \\
    & = \quad \left( 1 - \frac{t}{n} \right) \left( 1 - \frac{t}{n - 1} \right) \left( 1 - \frac{t}{n - 2} \right)        \dotsb \left( 1 - \frac{t}{n - t + 1} \right) \enspace.
\end{align*}
Since~$t > 1$, we get
\[
\Pr( g \not\in S_1 {\bm R})
    \quad  < \quad \left( 1 - \frac{t}{n} \right)^t
    \quad < \quad \exp \! \left( - \frac{t^2}{n} \right)
    \quad \le \quad \frac{1}{n} \enspace.
\]
By the Union Bound, the probability that~$S_1 {\bm R} \not= G$ is strictly less than~$1$. So there is a subset~$S_2$ of size~$t$ such that~$S_1 S_2 = G$.
\end{proof}
A similar result may be derived from a generalization of the Erd{\H{o}}s-R{\'e}nyi Theorem~\cite{ER65-prob-method-group-theory} due to Babai and Erd{\H{o}}s~\cite{BE82-short-products}. However, this gives us a generating pair in which the size of a set may be as large as~$4 \sqrt{n \ln n} \,$.

As before, the algorithm for the general Hidden Subgroup Problem follows directly from the existence of a suitable generating pair for the underlying group.
\begin{corollary}
\label{cor-non-abelian-hsp}
There is a deterministic algorithm with query complexity at most~$2 \big\lceil \sqrt{n \ln n}\, \big\rceil$ that solves the Hidden Subgroup Problem over an arbitrary group~$G$ with order~$n$. Moreover, if the input function hides the non-trivial subgroup~$H$, the algorithm finds all the elements of~$H$.
\end{corollary}
\begin{proof}
By \Cref{thm-non-abelian}, there is a generating pair~$S_1, S_2$ for the group~$G$ such that~$\size{S_1}, \size{S_2} \le \big\lceil \sqrt{n \ln n}\, \big\rceil$. We query the oracle function~$f$ at~$x^{-1}$ for all~$x \in S_1$ and at all elements~$y \in S_2$. If the function is injective on the set of points queried, we say~$f$ is injective. Otherwise, we say that~$f$ is not injective, and output~$\set{ xy : x \in S_1, ~ y \in S_2, ~ f( x^{-1} ) = f(y) }$.

If the function is injective, the above algorithm outputs the correct answer. Suppose~$f$ hides the non-trivial subgroup~$H$. Let~$h \in H$ be any element of the hidden subgroup. We have~$h = h_1 h_2 $ for some~$h_1 \in S_1$ and~$h_2 \in S_2$. By definition of~$H$ we have~$f(h_1^{-1}) = f( h_2)$. When~$h$ is not the identity, we have~$h_1^{-1} \neq h_2$ and the algorithm detects a collision. Moreover, the algorithm computes~$H$ correctly.
\end{proof}

We can obtain explicit, optimal algorithms for certain classes of non-abelian groups. For example, if a group~$G$ of order~$n$ has a subgroup~$H$ of order~$\Theta( \sqrt{n} \,)$, then we can construct a generating pair~$S_1, S_2$ of size~$\Theta( \sqrt{n} \,) $ by taking~$S_1 \coloneqq H$ and~$S_2$ to be a complete set of coset representatives of~$H$. In fact, in this case, the group satisfies a stronger property; see, for example, Ref.~\cite{BW20-cayley-sudoku}. Not all groups have a subgroup of such size. For example, the abelian group~${\mathbb Z}_p$ for a prime~$p$ does not have such a subgroup, and yet admits a suitable generating pair.

\section{Subgroup-dependent query complexity}
\label{sec-subgroup-dependent}

There is a randomized algorithm for the Hidden Subgroup Problem that has expected query complexity~$ \mathrm{O}(\sqrt{ \size{G} / \size{H}}\, )$ when the oracle~$f$ hides~$H$, and has worst-case query complexity~$\mathrm{O}( \sqrt{ \size{G}} \,) $. In this section we use the notion of a generating pair in a more sophisticated manner to match this performance with a deterministic algorithm for all abelian groups, and some classes of non-abelian groups.

The algorithm rests on the following observation.
\begin{lemma}
\label{lem-pigeon-hole}
Suppose~$G_1$ and~$H$ are subgroups of the possibly non-abelian finite group~$G$ such that~$G_1$ has cardinality at least~$\size{G}/ \size{H}$. Then either~$G_1 H = G$ or~$\size{G_1 \cap H} > 1 $.
\end{lemma}
\begin{proof}
The intersection~$G_1 \cap H$ is a subgroup of~$G$ and contains the identity element. There are exactly~$\size{G}/ \size{H}$ distinct left cosets of~$H$. If~$G_1 H \neq G$, by the Pigeon-Hole Principle, there are two distinct elements~$g_1, g_2 \in G_1$ such that~$g_1 H = g_2 H$, i.e., $ g_2^{-1} g_1 \in H$. Since~$g_2^{-1} g_1$ is also an element of~$G_1$, and is not the identity element, we have~$\size{G_1 \cap H} > 1$.
\end{proof}

Suppose that~$G$ is abelian, we know the order~$m$ of the hidden subgroup~$H$, and~$m > 1$. Then we may find a non-trivial element of~$H$ as follows. Consider any subgroup~$G_1$ of~$G$ of order~$n/m$, where~$n \coloneqq \size{G}$. Such a subgroup exists since~$G$ has a subgroup with order~$d$ for any positive divisor~$d$ of~$n$~\cite[Corollary~2.4, page~77]{H74-algebra}. Let~$S_1, S_2$ be a generating pair for~$G_1$, and let~$g$ be any element of~$ G \setminus G_1$. By \Cref{lem-pigeon-hole}, either~(i) there is a non-identity element~$h \in H$ that is also in~$G_1$, or (ii)~$G_1 H = G$. In case~(i), let~$h = a + b$, where~$a \in S_1$ and~$b \in S_2$. If we query the oracle~$f$ at the elements~$ -x$ and~$y$, for all~$x \in S_1$ and~$y \in S_2$, we will find~$f( -a) = f(b)$, and can compute~$h$. In case~(ii), we have~$g = g_1 + h$ for some~$g_1 \in G_1$ and~$h \in H$. Since~$g$ is not in~$G_1$, the element~$h$ is not the identity. Suppose~$g_1 = a + b$, with~$a \in S_1$ and~$b \in S_2$. Then~$g - a = b + h$, and~$f(g - a) = f(b)$. If we query the oracle~$f$ at the elements~$g - x$ and~$y$, for all~$x \in S_1$ and~$y \in S_2$, we will find~$f(g - a) = f(b)$, and can compute~$h$. This is the key idea underlying the algorithm.

For sets~$S_1, S_2 \subseteq G$, define~$ S_1^{-1}$ as the set~$ S_1^{-1} \coloneqq \set{ x^{-1} : x \in S_1 }$. Following our notation for the product of sets of group elements, $ S_1^{-1} S_2 = \set{ x^{-1} y : x \in S_1, ~ y \in S_2}$. Algorithm~\ref{alg-fc} (Find-Collision) implements the above idea with a geometrically decreasing sequence of guesses for the order of~$H$. We show in \Cref{thm-fc} that the algorithm is correct and has the query complexity we seek for a large class of groups.
\begin{algorithm}[ht]
\caption{Find-Collision($G, f$) \label{alg-fc}}

\SetKwInOut{Input}{Input}
\SetKwInOut{Output}{Output~}
\SetKwData{Inj}{injective}
\SetKwData{Coll}{collision}

\Input{group~$G$ of order~$n$, with~$n > 1$; oracle for~$f : G \rightarrow S$ that hides a subgroup}
\Output{\Inj, or \Coll~$a,b \in G$}

\BlankLine
Let~$k$ be the smallest integer~$l$ such that~$n \in (2^l, 2^{l+1}]$ \;
\lIf{ $G$ is abelian }{ $k_0 \leftarrow 0$ }
\lElse{ $k_0 \leftarrow -1$ }
\While{ $k \ge k_0$ }{
  Find, if there is one, a subgroup~$G_1 \le G$ with the largest order in~$\big[ n/2^{k + 1}, ~ n/(\floor{2^k} + 1) \big]$ \;
  \uIf{ such a subgroup~$G_1$ exists}{
    Find a generating pair~$S_1, S_2$ for~$G_1$ which minimizes~$\max \set{ \size{S_1}, ~ \size{S_2} }$ \;
    \lIf{ $ G_1 = G$ }{ $ g \leftarrow e$, the identity element of~$G$ }
    \lElse{ $ g \leftarrow $ any element in~$ G \setminus G_1$ }
    Let~$R \leftarrow S_1^{-1} \cup S_2 \cup \big( S_1^{-1} \set{g} \big) $ \;
    Query~$f$ at all the elements in~$R$ \;
    \lIf{ $f(z) = f(y)$ for some~$z,y \in R$ such that~$z \neq y$}{\Return{\Coll~$z,y$} }
  }
  $k \leftarrow k - 1$ \;
}
\Return{\Inj}
\end{algorithm}

\begin{theorem}
\label{thm-fc}
Algorithm~\ref{alg-fc} (Find-Collision) solves the Hidden Subgroup Problem over any finite group~$G$. If the order of the group is~$n$ and that of the hidden subgroup~$H$ is~$m$, the algorithm has query complexity as stated below.
\begin{enumerate}
\item
\label{part-abelian}
If~$G$ is abelian, the algorithm has query complexity~$ \mathrm{O}( \sqrt{n/m} \,)$.
\item
\label{part-non-abelian}
If~$G$ is not abelian, the algorithm has query complexity~$\mathrm{O}( \sqrt{n \ln n} \,)$. Further, if~$G$ has a subgroup of order~$n_1$ such that~$n/m \le n_1 \le \kappa n/m$ for some~$\kappa  \ge 1$, then the query complexity is~$ \mathrm{O}( \sqrt{(\kappa n/m) \ln (\kappa n/m)} \,)$.
\end{enumerate}
\end{theorem}
\begin{proof}
Since the algorithm outputs a collision only when it finds one, it gives the correct answer when the oracle function~$f$ is injective. Suppose the function~$f$ hides a non-trivial subgroup~$H$, so that~$m \ge 2$.

If the group is abelian, it has a subgroup of order~$n/m$~\cite[Corollary~2.4, page~77]{H74-algebra}, and~$n/m \le n/2$. If it does not find a collision in an earlier iteration, Algorithm~\ref{alg-fc} finds a collision in an iteration with~$k = \ell \ge 0$, where~$\ell$ is such that~$n/m \in \big[ n/2^{\ell + 1}, ~ n/(2^\ell + 1) \big]$. This is due to the reasoning given after \Cref{lem-pigeon-hole}. It thus outputs the correct answer. We have~$\ell = \ceil{ \log_2 m} - 1$, and initially, $k = \ceil{ \log_2 n} - 1 \ge 0$. By \Cref{thm-abelian}, we have~$\size{S_1}, \size{S_2} \le 2 \sqrt{ \size{G_1}}$ in every iteration with queries. So the query complexity of the algorithm is at most
\begin{align*}
\sum_{k = \ell}^{\ceil{ \log_2 n} - 1} 3 \cdot 2 \sqrt{ n/ 2^k} \quad
    & \le \quad 6 \sqrt{ \frac{n}{ 2^\ell}} ~ \sum_{i \ge 0} \frac{1}{ \sqrt{2^i}} \\
    & \le \quad 12 (1 + \sqrt{2} \,) \sqrt{n/m} \enspace,
\end{align*}
as~$\ell \ge \log_2 m - 1$. The bound on the query complexity when~$m = 1$ is the same as that for~$m = 2$, as the algorithm executes all the iterations until~$k = 0$. Part~\ref{part-abelian} of the theorem thus follows.

Suppose~$G$ is non-abelian. If the algorithm does not find a collision in earlier iterations, when~$k = -1$, we have~$G_1 = G$, and correctness follows as in \Cref{cor-non-abelian-hsp}. If~$G$ has a proper subgroup of order~$n_1$ with~$n/m \le n_1 \le \kappa n/m$ for some~$\kappa $, the algorithm finds a collision as in the abelian case in an iteration with~$k \ge \ell \ge 0$, where~$\ell$ is such that~$\kappa n/m \in \big[ n/2^{\ell + 1}, ~ n/(2^\ell + 1) \big]$. It thus outputs the correct answer. Further, we have~$\ell = \ceil{ \log_2 (m/\kappa )} - 1$, and initially, $k = \ceil{ \log_2 n} - 1 \ge 0$. By \Cref{thm-non-abelian}, we have~$\size{S_1}, \size{S_2} \le \sqrt{ \size{G_1} \ln \size{G_1}} + 1$ in every iteration with queries. So the query complexity of the algorithm is at most
\begin{align*}
\lefteqn{ \sum_{k = \ell}^{\ceil{ \log_2 n} - 1} 3 \left( 1 + \left( \frac{n}{ 2^k} \ln \frac{n}{ 2^k} \right)^{1/2} \right) } \\
    & \quad \le \quad 3( \ceil{ \log_2 n} - \ceil{ \log_2 (m/\kappa )} + 1 ) + 3 \left( \frac{n}{ 2^\ell} \ln \frac{n}{ 2^\ell} \right)^{1/2} \sum_{i \ge 0}  \frac{1}{ \sqrt{2^i}} \\
    & \quad \le \quad 6 + 3 \log_2 (\kappa n/m) + 3 (2 + \sqrt{2} \,) \left( \frac{2\kappa n}{m} \ln \frac{2\kappa n}{m} \right)^{1/2} \enspace,
\end{align*}
as~$\ell \ge \log_2 (m/\kappa ) - 1$. The bound on the query complexity when~$m = 1$ is~$2 \ceil{ \sqrt{ n \ln n} \,}$ more than that for~$m = 2$, as the algorithm executes all the iterations until~$k = -1$. Part~\ref{part-non-abelian} of the theorem thus follows.
\end{proof}
Unlike abelian groups, a non-abelian group of order~$n$ may not have a subgroup of order~$n/m$ when it has a proper subgroup of order~$m$. For example~$A_4$, the alternating group of degree~$4$, has order~$12$, has several subgroups of order~$2$, but does not have a subgroup of order~$6$. However, in large classes of instances of the Hidden Subgroup Problem, subgroups of suitable size exist. An immediate example is the class of CLT groups. (A group~$G$ is called a \emph{converse Lagrange Theorem (CLT) group\/} if it contains a subgroup of order~$d$ for every positive divisor~$d$ of~$\size{G}$.) CLT groups include supersolvable groups; see, e.g., Ref.~\cite{McLain57-subgroup-order}. For such instances, Find-Collision achieves query complexity~$\mathrm{O}( \sqrt{ (n/m) \ln(n/m)} \,)$.

\section{Finding the hidden subgroup}
\label{sec-finding-subgroup}

Unlike the algorithms in \Cref{cor-ahsp} and \Cref{cor-non-abelian-hsp}, Algorithm~\ref{alg-fc} may find only one non-trivial element from the hidden subgroup. In this section, we show how to extend Algorithm~\ref{alg-fc} to find the entire hidden subgroup.

When the group~$G$ is abelian, the problem of finding the entire subgroup may be reduced to that of finding one non-trivial element of the subgroup. This allows us to identify the subgroup by repeatedly using Algorithm~\ref{alg-fc} to find a set of generators.

\begin{algorithm}[ht]
\caption{Find-Abelian-Subgroup($G, f$) \label{alg-find-ahs}}

\SetKwInOut{Input}{Input}
\SetKwInOut{Output}{Output~}
\SetKwData{Inj}{injective}
\SetKwData{Coll}{collision}
\SetKwData{Gen}{generators}

\Input{group~$G$ of order~$n$, with~$n > 1$; oracle for~$f : G \rightarrow S$ that hides a subgroup}
\Output{\Inj, or \Gen~$S \subset G$ of the hidden subgroup}

\BlankLine
$S \leftarrow \emptyset$ \;
\Repeat{ outcome $=$ \Inj }{
  $H_1 \leftarrow \langle S \rangle$, the subgroup generated by~$S$ \;
  $G_1 \leftarrow G/H_1 $ \;
  \uIf{ $ \size{G_1} > 1$ }{
    Let~$f_1$ be the function defined by~$f$ and~$H_1$ in the proof of \Cref{thm-abelian-find-hs} \;
    \textit{outcome\/} $\leftarrow$ Find-Collision($G_1, f_1$) \;
    \lIf{ outcome $=$ \Coll~$a,b$ }{ $S \leftarrow S \cup \set{ g_1^{-1} g_2}$, where~$a = g_1 H_1$ and~$b = g_2 H_1$ }
  }
  \uElse{ \textit{outcome\/} $ \leftarrow$ \Inj }
}
\lIf{ $S = \emptyset$ }{ \Return{\Inj} }
\lElse{ \Return{ \Gen~$S$} }
\end{algorithm}

\begin{theorem}
\label{thm-abelian-find-hs}
There is a deterministic algorithm that solves the Hidden Subgroup Problem over any finite abelian group~$G$, and finds the hidden subgroup with~$ \mathrm{O}( (\log m) \sqrt{n/m} \, )$ queries when the order of~$G$ is~$n$ and that of the hidden subgroup is~$m$.
\end{theorem}
\begin{proof}
Suppose the oracle is~$f$, the hidden subgroup is~$H$, and we know a set of generators for a subgroup~$H_1 \le H$.

Define a function~$f_1$ on the quotient group~$G/ H_1$ as~$f_1( g H_1) \coloneqq f(g)$ for any~$g \in G$. The function~$f_1$ is well-defined as left cosets of~$H_1$ in~$G$ are subsets of left cosets of~$H$ in~$G$, and the function~$f$ is constant on left cosets of~$H$. Moreover, the function~$f_1$ hides the subgroup~$H/H_1$ of~$G/H_1$, as the left cosets of~$H$ in~$G$ correspond to left cosets of~$H/H_1$ in~$G/H_1$. Finally, the function~$f_1$ may be evaluated with one query to the oracle for~$f$.

Using this reduction, we may find the hidden subgroup~$H$ using Algorithm~\ref{alg-find-ahs}. The correctness of the algorithm follows by observing that in any iteration, if~$f_1$ is injective, then~$H_1$ equals the hidden subgroup. If~$f_1$ is not injective, by \Cref{thm-fc}, Find-Collision($ G_1, f_1$) returns a collision~$g_1 H_1, g_2 H_1 \in G/H_1$ for~$f_1$. Note that~$g_1, g_2$ is a collision for~$f$. We also have~$ g_1^{-1} g_2 \not\in H_1 $, so along with~$ g_1^{-1} g_2 $, the set~$S$ generates a larger subgroup of~$H$. Thus, the size of the subgroup~$H_1$ increases by a factor of at least~$2$ in every iteration a collision is found, and the algorithm terminates after at most~$\log_2 m$ iterations.
\end{proof}

It is not clear how to extend Algorithm~\ref{alg-find-ahs} to the non-abelian case, since the group $H_1$ may not be normal in general, and the corresponding quotient groups may not be defined. We present a different algorithm, Algorithm~\ref{alg-find-hs}, that works for some non-abelian instances, although it does not necessarily work for all abelian instances. Algorithm~\ref{alg-find-hs} builds on a variant of Algorithm~\ref{alg-fc}, and is also based on \Cref{lem-pigeon-hole}.
\begin{algorithm}[ht]
\caption{Find-New-Collision($G, H_1, f$) \label{alg-fnc}}

\SetKwInOut{Input}{Input}
\SetKwInOut{Output}{Output~}
\SetKwData{Nocoll}{no-new-collision}
\SetKwData{Coll}{collision}

\Input{group~$G$ of order~$n$, with~$n > 1$; subgroup~$H_1 \le G$; oracle for~$f : G \rightarrow S$ that hides a subgroup containing~$H_1$}
\Output{\Nocoll, or \Coll~$a,b \in G$ such that~$a^{-1}b \not\in H_1$}

\BlankLine
Let~$k$ be the smallest integer~$l$ such that~$n \in (2^l, 2^{l+1}]$ \;
\lIf{ $G$ is abelian }{ $m_1 \leftarrow \max \set{ 2, \size{H_1} }$ }
\lElse{ $m_1 \leftarrow \size{H_1}$ }
$k_0 \leftarrow \ceil{ \log_2 m_1 } - 1$ \;
\While{ $k \ge k_0$ }{
  Find, if there is one, a subgroup~$G_1 \le G$ with the largest order in~$\big[ n/2^{k + 1}, ~ n/(\floor{2^k} + 1) \big]$ such that~$G_1 \cap H_1 = \set{e}$, where~$e$ is the identity element of~$G$ \;
  \uIf{ such a subgroup~$G_1$ exists}{
    Find a generating pair~$S_1, S_2$ for~$G_1$ which minimizes~$\max \set{ \size{S_1}, ~ \size{S_2} }$ \;
    \lIf{ $ G_1 H_1 = G$ }{ $ g \leftarrow e$, the identity element of~$G$ }
    \lElse{ $ g \leftarrow $ any element in~$ G \setminus (G_1 H_1)$ }
    Let~$R \leftarrow S_1^{-1} \cup S_2 \cup \big( S_1^{-1} \set{g} \big) $ \;
    Query~$f$ at all the elements in~$R$ \;
    \lIf{ $f(z) = f(y)$ for some~$z,y \in R$ such that~$z^{-1}y \not\in H_1$}{\Return{\Coll~$z,y$} }
  }
  $k \leftarrow k - 1$ \;
}
\Return{\Nocoll}
\end{algorithm}
\begin{theorem}
\label{thm-fnc}
Let~$(G,f)$ be an instance of the Hidden Subgroup Problem, $H$ the subgroup that~$f$ hides, and~$H_1$ a subgroup of~$H$. Let the orders of~$G, H, H_1$ be~$n, m, m_1$, respectively. Suppose~$G$ has a subgroup~$G_0$ of order~$n_0$ such that~$n_0 \ge n/m $, and~$G_0$ intersects~$ H_1$ only in the identity element. Then Find-New-Collision($G, H_1, f$) (Algorithm~\ref{alg-fnc}) returns ``no new collision'' if~$H_1 = H$, and returns a collision~$a,b \in G$ such that~$a^{-1} b \not\in H_1$ otherwise. Further, if~$ n_0 \le \kappa n/m$ for some~$\kappa \ge 1$, the algorithm has query complexity as stated below:
\begin{enumerate}
\item
\label{part-abelian-fnc}
if~$G$ is abelian, the algorithm has query complexity~$ \mathrm{O}( \sqrt{ \kappa n/m} \,)$, and
\item
\label{part-non-abelian-fnc}
if~$G$ is not abelian, the algorithm has query complexity~$\mathrm{O}( \sqrt{( \kappa n/m) \ln ( \kappa n/m) } \,)$.
\end{enumerate}
\end{theorem}
\begin{proof}
When~$H_1 = \set{e}$, the algorithm is identical to Algorithm~\ref{alg-fc}, and its correctness follows by \Cref{thm-fc}. Suppose~$H_1$ is not the trivial subgroup.

Since the algorithm reports a collision~$a,b$ only when~$a^{-1} b \not\in H_1$, it gives the correct answer when~$H_1 = H$. Suppose~$H_1 \neq H$. By hypothesis, $G$ has a subgroup~$G_0$ of order at least~$n/m$ such that~$G_0$ intersects~$ H_1$ only in the identity element. Then~$\size{G_0} \le n/m_1 \le n/2$; otherwise, we would have two distinct elements of~$G_0$ in the same coset of~$H_1$, which implies that $\size{ G_0 \cap H_1} > 1$.

If it does not find a collision in earlier iterations, the algorithm finds a subgroup~$G_1$ with~$\size{G_1} \ge n/m$ and~$G_1 \cap H_1 = \set{e}$ in an iteration with~$k \ge k_0 \ge 0$ (recall that~$k_0 \coloneqq \ceil{ \log_2 m_1 } - 1 \ge 0$, since~$m_1 \ge 2$).  Let~$S_1,S_2$ be the generating pair for~$G_1$ computed by the algorithm.

Suppose~$G_1 H_1 = G$. Consider any element~$h \in H \setminus H_1 $. We have~$h = g_1 h_1 $ for some~$g_1 \in G_1$ and~$h_1 \in H_1$. So~$g_1 = h h_1^{-1} \in G_1 \cap H $, and~$g_1 \not\in H_1 $. We also have~$ g_1 = ab $ for some~$a \in S_1$ and~$b \in S_2$, so~$f(a^{-1}) = f(b)$. The algorithm queries the oracle~$f$ at the elements~$x^{-1}$ and~$y$, for all~$x \in S_1$ and~$y \in S_2$, so it finds and returns a collision.

Now suppose~$G_1 H_1 \neq G$, so that the algorithm also finds an element~$g \in G \setminus (G_1 H_1)$ in the same iteration.

If~$G_1 H = G$, we have~$g = g_1 h$ for some~$g \in G_1$ and~$h \in H$. Since~$g \not\in G_1 H_1$, $g_1 h \not\in G_1 H_1$, and~$h \not\in H_1$. Suppose~$g_1 = ab$, with~$a \in S_1$ and~$b \in S_2$. Then~$a^{-1} g = b h$, and~$f(a^{-1} g) = f(b)$. The algorithm queries the oracle~$f$ at the elements~$x^{-1} g$ and~$y$, for all~$x \in S_1$ and~$y \in S_2$, so it finds~$f(a^{-1} g) = f(b)$, and returns a collision.

If~$G_1 H \neq G$, as in \Cref{lem-pigeon-hole}, there are two distinct elements~$g_1, g_2 \in G_1$ such that~$ g_2^{-1} g_1 \in H$. Since~$ G_1 \cap H_1 = \set{e}$, and~$g_1 \neq g_2$, we have~$ g_2^{-1} g_1 \in H \setminus H_1$. We also have~$g_2^{-1} g_1 = ab$, for some~$a \in S_1$ and~$b \in S_2$. The algorithm queries the oracle~$f$ at the elements~$x^{-1}$ and~$y$, for all~$x \in S_1$ and~$y \in S_2$, so it finds~$f(a^{-1}) = f(b)$, and returns a collision in this case as well.

Assuming~$n/m \le \size{G_0} \le \kappa n/m \le n$ for some~$\kappa \ge 1$ with~$G_0$ as in the statement of the theorem, the algorithm executes iterations with~$k \ge \ell$, where~$\ell$ is such that~$\kappa n/m \in \big[ n/2^{\ell + 1}, ~ n/(\floor{2^\ell} + 1) \big]$. So the query complexity of the algorithm follows by same kind of analysis as in \Cref{thm-fc}.
\end{proof}

Under certain conditions, we may use Algorithm~\ref{alg-fnc} repeatedly to find a set of generators for the hidden subgroup, using Algorithm~\ref{alg-find-hs}.
\begin{algorithm}[ht]
\caption{Find-Subgroup($G, f$) \label{alg-find-hs}}

\SetKwInOut{Input}{Input}
\SetKwInOut{Output}{Output~}
\SetKwData{Inj}{injective}
\SetKwData{Coll}{collision}
\SetKwData{Gen}{generators}

\Input{group~$G$ of order~$n$, with~$n > 1$; oracle for~$f : G \rightarrow S$ that hides a subgroup}
\Output{\Inj, or \Gen~$S \subset G$ of the hidden subgroup}

\BlankLine
$S \leftarrow \emptyset$ \;
\Repeat{ outcome $=$ \Inj }{
  $H_1 \leftarrow \langle S \rangle$ \;
  \textit{outcome\/} $\leftarrow$ Find-New-Collision($G_1, H_1, f$) \;
  \lIf{ outcome $=$ \Coll~$a,b$ }{ $S \leftarrow S \cup \set{ a^{-1} b}$ }
}
\lIf{ $S = \emptyset$ }{ \Return{\Inj} }
\lElse{ \Return{ \Gen~$S$} }
\end{algorithm}
\begin{theorem}
\label{thm-find-hs}
Algorithm~\ref{alg-find-hs} (Find-Subgroup) solves the Hidden Subgroup Problem over a group~$G$ and finds the hidden subgroup~$H$ when~$G$ has a subgroup~$G_0$ of order~$n/m$ such that~$\size{G_0 \cap H} = 1$, where~$n$ and~$m$ are the orders of~$G$ and~$H$, respectively. Moreover, the algorithm makes
\begin{itemize}
\item $ \mathrm{O}( (\log m) \sqrt{n/m} \, )$ queries when~$G$ is abelian, and
\item $ \mathrm{O}( (\log m) \sqrt{(n/m) \log (n/m)} \, )$ queries when~$G$ is non-abelian.
\end{itemize}
\end{theorem}
\begin{proof}
The existence of a subgroup~$G_0$ as in the statement of the theorem implies that all the hypotheses of \Cref{thm-fnc} are satisfied for every subgroup~$H_1$ of~$H$. So, starting with~$H_1 = \set{e}$, where~$e$ is the identity element of~$G$, Find-New-Collision($G, H_1, f$) returns a collision~$a,b$ in each iteration of Algorithm~\ref{alg-find-hs} in which~$H_1 \neq H$. Since~$ a^{-1} b \not\in H_1 $, the set~$S \cup \set{ a^{-1} b}$ generates a larger subgroup of~$H$. Thus, the size of the subgroup~$H_1$ increases by a factor of at least~$2$ in every iteration in which a collision is found, and the algorithm terminates after at most~$\log_2 m$ iterations. The query complexity follows from \Cref{thm-fnc}.
\end{proof}
A subgroup as in the statement of \Cref{thm-find-hs} exists if~$G$ is the semidirect product of~$H$ with another subgroup, i.e., $H$ is normal and there is a subgroup~$K$ such that~$G = H \rtimes K$, or there is a normal subgroup~$K$ such that~$G = H \ltimes K$. We may then take~$G_0 \coloneqq K$. Not all groups have  a semidirect product structure, even if the hidden subgroup is normal. For example, every proper subgroup~$H$ of the abelian group~${\mathbb Z}_{p^k}$ with~$k > 1$ is normal, but~$ {\mathbb Z}_{p^k}$ cannot be expressed as a semidirect product of~$H$ with another subgroup. On the other hand, a group need not have a semidirect product structure for a subgroup with the properties in \Cref{thm-find-hs} to exist. For example, consider~$S_n$, the symmetric group of degree~$n$, and~$S_{n-1}$ as its subgroup consisting of permutations that map~$n$ to itself. Then~$S_n = S_{n - 1} H$, where~$H$ is the subgroup generated by the cycle~$(1 ~ 2 ~ 3 ~ \dotsb ~ n)$, and neither~$S_{n - 1}$ nor~$H$ is normal in~$S_n$ for~$n \ge 4$. Such groups are known as the \emph{bicrossed\/} products (also as \emph{Zappa-Sz{\'e}p} or \emph{knit} products); see, e.g., \cite{ACIM09-bicrossed-product,Brin05-ZS-product}. Thus, Algorithm~\ref{alg-find-hs} finds the hidden subgroup~$H$ with query complexity as in \Cref{thm-find-hs} whenever~$G$ is the bicrossed product of~$H$ with another group. A description of groups arising as bicrossed products is a matter of ongoing research~\cite{ACIM09-bicrossed-product}.

\section{Open problems}
\label{sec-open}

We conclude with a few open problems.
There are a number of variants of the Hidden Subgroup Problem, for example, when the underlying group is specified implicitly. These may also admit deterministic algorithms with optimal classical query complexity. The precise characterization of the deterministic query complexity of the Hidden Subgroup Problem for explicitly specified groups, especially in the non-abelian case, is perhaps the most interesting problem left open by this work. Related questions are whether there is a generating pair of size~$\mathrm{O}( \sqrt{n} \,)$ for any non-abelian group of order~$n$, for what instances of the problem Algorithms~\ref{alg-fc} and~\ref{alg-fnc} give the correct output with query complexity~$\widetilde{\mathrm{O}}( \sqrt{ n/m} \,)$, where~$m$ is the order of the hidden subgroup, or whether there are similar ``generic'' algorithms that achieve this query complexity for larger classes of groups.

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


An updated RSS Data Users Guide (Version 1.1) is available.
The PDS Rings node has posted version 2 of the dataset of Saturn ring radial profiles derived from RSS radio occultation data.
You may browse the dataset CORSS_8001, or download the entire dataset CORSS_8001 tar.gz.
Version 2 includes reprocessed versions of the material contained in the earlier version, plus all subsequent RSS occultations obtained prior to the loss of the Cassini UltraStable Oscillator (USO) in late 2011.
In addition, this version includes 'DLP' files for each observation. These contain higher resolution calibrated, but diffraction-limited, optical depth and phase shift profiles of Saturn's rings, that is, calibrated profiles before reconstruction to remove diffraction effects.
While Version 1 contained only X band data files, version 2 contains K, S, and X band data.
Details of the data processing involved in the production of this data set are provided in MAROUFETAL1986.
NOTE: Raw RSS data may be obtained from the PDS Atmospheres Node.


Joe Nocera was an Op-Ed columnist for The New York Times Opinion pages between April 2011 and November 2015. Before his Opinion column, he wrote the Talking Business column for The New York Times each Saturday and was a staff writer for The New York Times Magazine. He joined the paper in 2005.
For more than three decades, Mr. Nocera has chronicled the world of business at magazines like Fortune, GQ, Esquire and Texas Monthly. He has won three Gerald Loeb awards, including the 2008 award for commentary, and three John Hancock awards for excellence in business journalism. A 2007 Pulitzer finalist, he has written books including "A Piece of the Action: How the Middle Class Joined the Money Class," which won the New York Public Library's 1995 Helen Bernstein Award; "Good Guys and Bad Guys: Behind the Scenes With the Saints and Scoundrels of American Business (and Everything in Between)," and "All the Devils Are Here: The Hidden History of the Financial Crisis," which he co-authored with Bethany McLean.
Mr. Nocera received a B.S. in journalism from Boston University in 1974. He lives in New York City.
Full video and edited transcript; "Surviving Gun Violence"; February 10, 2017.


text:
Home Arts Two memorable season closing films: Repertory Dance Theatre's Homage, Ririe-Woodbury Dance Company's...
Two memorable season closing films: Repertory Dance Theatre's Homage, Ririe-Woodbury Dance Company's Cadence
Les Roka
Two of Salt Lake City's historically significant dance institutions — Repertory Dance Theatre (RDT) and Ririe-Woodbury Dance Company — are concluding this unusual season with new concert films. Both films end on an encouraging note, signaling a hopeful return next fall to their regular performing venue in the Rose Wagner Center for Performing Arts. Both companies also are saying farewell to dance artists with the close of the season. Notably, this was the first filmed performance for both companies, where the dance artists were seen without masks. Each company followed strict testing and monitoring protocols to ensure all persons involved were safe and healthy.
REPERTORY DANCE THEATRE: HOMAGE
When Jaclyn Brown joined the Repertory Dance Theatre (RDT) in 2014, she did not anticipate just how much she would be able to stretch her wings, leaving behind the comfort and safety of her Utah Valley University days as an aspiring dance artist. One of her earliest challenges was learning the technique associated with José Limón, one of the modern masters of the 20th century whose work also is part of the RDT historical repertoire. One of the technique's signature phrases was fall and recovery, representing the physical extremes in movement. "I struggled with it because I had no experience with it," Brown says. "I have a Type A personality so I am used to controlling everything. Working with Nina Watt [who has carried forward the technique and work of Limón as well as his mentor Doris Humphrey] was such an uncomfortable transition. I even would have nightmares about it."
Jaclyn Brown, Dan Higgins. Invention, by Doris Humphrey. Repertory Dance Theatre. Photo Credit: Sharon Kain.
But, Brown says it was a turning point in her maturing as a dance artist in a company where there are so many opposites and counterpoints in the historical and contemporary repertoire. She has grown artistically confident with classical pieces of modern dance, including Limón's Suite of Mazurkas (1958) and Doris Humphrey's Invention (1949). And, her experience with the Limón technique has nurtured her connection to the Gaga movement language, as epitomized in contemporary works developed by Noa Zuk, Ohad Naharin and Danielle Agami and performed by RDT.
Homage, the filmed concert to close RDT's 55th season produced and edited by Wonderstone Films, features four works created between 1949 and 1959, signifying the artistic roots of a company created to preserve what now encompasses 120 years of modern dance history. Brown also is making her final appearance as an RDT artist after a seven-year tenure. She will live in Orem with her husband, Terry, and daughter, Layla, while Brown pursues a master's degree at the University of Wisconsin-Milwaukee. The program is flexible with low residency requirements. Brown also will teach jazz dance classes at UVU.
The works in Homage include Limón's Mazurkas and Humphrey's Invention. Only two companies have rights to include the Humphrey piece in their repertoire; RDT and the Limón Company.
Castor and Pollux, Elizabeth Waters. Repertory Dance Theatre. Photo Credit: Sharon Kain.
Featured in a short interview introducing the performance of Invention, Watt, who toured nearly three decades with the Limón Dance Company and has been described as "the perfect Limón dancer," restaged the work for RDT. As previously explained in The Utah Review, Invention is a character study for three dancers (one male, two female). It arises from the male's point of view and the unique experiences he has had with two women at different stages in his life and how those experiences are synthesized in his memory. Humphrey commissioned music by Norman Lloyd, which is resourceful for how he blends his melodic lines into the rhythms.
Brown is joined by Dan Higgins and Lauren Curley for the performance, which was so effortlessly sculpted and elegant as the trio demonstrated the physical intensity of the movement language crafted by Humphrey and Limón.
Seven company dancers were featured in the solos, duets, trios, and ensemble sections of Limón's Mazurkas. Created in 1958, the suite is the choreographer's interpretation of Chopin's mazurkas from a century before. Chopin's music embodied the Polish national spirit and one could see Limón's intent to remind artists and audiences that even as Poland was an Iron Curtain satellite state of the Soviet Union, the pride, elegance, vivacity, dignity, valor and lyricism of the music and the dance form of the mazurka will never vanish. The RDT dancers did superb justice to demonstrating how deeply Limón could extract from the music and dance form in their creative interpretation.
Donald McKayle's Rainbow Round My Shoulder, a 1959 work that is introduced in the Homage film with an excerpt from a documentary interview with the choreographer (who died in 2018), is a fully realized narrative of a Southern chain gang, told from the male point of view. The work features seven male dancers: Higgins, Daniel Do, Jonathan Kim, and Kareem Lewis from RDT, who are joined by former RDT artist Tyler Orcutt and guest artists Brendan Rupp and Austin Hardy. RDT artist Ursula Perry reprises her 2018 performance as the female dancer, who represents freedom, first as a vision then as a remembered sweetheart, mother, and wife. The songs, curated from folk music that Alan Lomax catalogued scrupulously in his work as an ethnomusicologist, accompany their arduous labor, are rich in polyphony and tell a bitter, sardonic and tragic story. Leslie Watanabe, who performed as a soloist with McKayle's Inner City Repertory Company, reset the work for RDT.
As this was the first filmed concert to feature the dance artists without masks, Rainbow Round My Shoulder is invigorating to watch because we are not just watching the movement but also the performers' faces and emotions. In the interview clip shown, McKayle recalled how quickly and effortlessly everyone responded to the work when it premiered in 1959, just as the Civil Rights Movement became part of every nightly news broadcast. The same intensity of passion carried through in this latest performance, just as news about the Black Lives Matter movement has become so integral to the contemporary news landscape, along with immigrants being detained in tent cities or the disenfranchisement of citizens subjected to the abuses of private prisons.
Ursula Perry, Suite of Mazurkas, José Limón. Repertory Dance Theatre. Photo Credit: Sharon Kain.
The film closes with Castor & Pollux, a work created in the 1950s by Elizabeth Waters, a vaudeville performer-turned-dance artist, teacher and choreographer. Waters, who died in 1993, moved to New Mexico, where she observed and was inspired by the Pueblo tribes' dance rituals. Prior to moving to the Southwest, Waters had become absorbed with the technique of Hanya Holm, popularly known as one of the Big Four founders of modern dance. The technique emphasizes geometric concepts of planes, space and patterns. RDT first performed the work 50 years ago and its sense of timelessness comes through the incredible music that Harry Partch wrote for it. Partch, sometimes described as the Magellan of sound, built his own instruments for the music, including bells from glass bottles and a customized marimba.
It's in two sections: the first, Castor and the second, Pollux, each with three duets and then the company ensemble. Simply, it is a magical fertility rite culminating at the end in the miracle of birth. One could not imagine a more apt selection to the end of a season where the original plans were so disrupted and revamped, thanks to the pandemic. In Castor & Pollux, we see total harmonization and all eight dancers in perfect conjunction with each other. It is a hopeful sign for RDT's 56th season and the promising return of live stage performances.
The 95-minute film is available for streaming, tickets required, through May 23. For more information, see the RDT website.
RIRIE-WOODBURY DANCE COMPANY: CADENCE
One of Bashaun Williams' earliest memories in his ten-year tenure with Ririe-Woodbury Dance Company comes from a tour appearance in France, where the company was presenting work by Alwin Nikolais, one of the 20th century's most forward-looking choreographers, in honor of the centennial of his birth. The Nikolai technique has been a central element in the company since its founding 57 seasons ago. "Being new, I didn't realize how much we practiced to do bows," Williams recalls. "They told me, 'wait until you see the audience.' It was a spectacular performance. The theater was sold out. We did eight curtain calls that night."
As the veteran of the company, in later years, Williams has shepherded the others through the practice of bowing after a performance. This included the company's spectacular cultural ambassador tour in Mongolia and South Korea. When he joined ten years ago, Williams says that he would not have imagined that he would be with Ririe-Woodbury so long. But, as the pandemic unfolded and other life changes occurred, Williams has decided to leave after this season to prepare for the next chapter as a dance artist, teacher and choreographer. And, of course, to spend time with Fallon and their daughter, Eko, who will soon celebrate her first birthday.
For Cadence, Ririe-Woodbury's filmed concert to close an unprecedented yet artistically profound season, Williams also counts the highlights of two world premieres — Charles O. Anderson's Rites and Yin Yue's In The Moment Somehow Secluded — and the company's premiere of Andrea Miller's I can see myself, restaged from the 2010 original titled Pupil Suite. Williams had injured his hip during dress rehearsal but rather than face the prospect of missing his final company performance, he managed to deal with the circumstances based on the lessons of the past decade.
The company also marks the departure of two other artists who are pursuing new paths after two seasons: Dominica Greene and Nicholas Jurica. Greene, who came to Ririe-Woodbury from freelancing as a collaborative artist, says that her tenure has led to fresh discoveries about herself, with the "multiplicities of a 360-degree experience in digesting it through the body, mind and spirit." In fact, all three departing dancers look fondly about the experience of making the lecture demonstrations for kids as inspiring and fun as possible. Greene plans to stay in Utah for the moment but also looks toward Seattle and New York City for collaborative opportunities. She will continue to work with the Dance Church movement, led by professionally trained dancers and open to non-dancers, who are encouraged to trust their own movement instincts and enjoy the 'dance party' at their own pace. The unprecedented pandemic impact has become for Greene a channel to expand beyond the notions of conventional space for creating art and dance. "I plan to slow down and take my time as an artist."
Bashaun Williams. In The Moment Somehow Secluded, Yin Yue. Ririe-Woodbury Dance Company. Courtesy: Tori Duhaime.
Like his colleagues, Jurica experienced his own epiphanies about participating in the company's educational outreach programs, especially about the intertwining and reciprocating effects of being the teacher and the student in cultivating expertise. Among his most memorable performances was his first season opener in 2019, which featured Ann Carlson's Elizabeth, the dance. "She is a brilliant human being," Jurica says, "and she reaffirmed that you can have it all." And, the last concert film earlier this year, Home Run, a uniquely framed documentary performance that included works created by the six dancers, is a highlight for him "It was a kind of magnum opus" reflecting the COVID-19 experience, he explains, not just for its constraints but also for the chance to make creative decisions outside of the boundaries of the conventional live theatrical experience.
Jurica and his wife will move to Beaverton, Oregon, where he will pursue teaching and dancing. "It's time to reconnect since it has been six years when I lived there and the pandemic made it very obvious how important family was for me," he adds.
The closing concert film for Ririe-Woodbury's 57th season is the company's winner of the year and features some of the best cinematography by Wonderstone Films, relative to recent filmed productions. In Cadence, each work receives a flattering cinematic presentation.
Rites, Charles Anderson. Ririe-Woodbury Dance Company. Courtesy: Tori Duhaime.
Rites, a world premiere, also is the first work since the pandemic slammed the doors on live performances, in which the guest artist could work in studio with the Ririe-Woodbury dancers. Anderson, who lives and works in Austin, Texas, was on a creative residency, which coincided with the Utah Museum of Fine Arts' special exhibition Black Refractions: Highlights from the Studio Museum in Harlem.
The filmed performance makes evident just how passionately invested the dancers were in the experience of presenting Rites, an exceptional meta-artistic commentary with precise, concise and incisive political, sociological, cultural and historical threads sewn onto a choreographic canvas enhanced by multimedia elements. The cohesion and cogency of this work are impressive, as every element fell seamlessly into its place.
Prior to the performance, a title card with Anderson's artistic statement about Rites is displayed long enough for the viewer to read without pushing the pause button on whatever video platform they are using. Anderson's statement reads: "Drawing upon contemporary African Diasporic movement practices, Rites is a movement meditation inspired by civil rights movements (Black, Woomen, Hispanic, LGBTQIA). It is a ritual to contextualize the pain of marginalization and disenfranchisement — the underbelly of the American Dream. And by contextualize I mean process, experience, imagine, embody and ultimately accept. Ultimately, it is a work about resilience and resistance in the face of ongoing oppression. It is about leveraging individuality in service to collective solidarity."
The work and performance fulfill this statement to the letter. The dance artists are clad in acolyte white robes, amplifying the significance of the last sentence in Anderson's statement. The backdrop adds textures with images from historical and contemporary scenes. Their facial expressions and gestures punctuate the context of their movement. Anderson's curated sound design includes Frank London's Blessed is The Eternal, an exemplar of New Jewish Renaissance music; Babatunde Olatunji's Watusi, with its long string of hypnotic drum rhythms, and Leon Bridges' River, an outstanding song that could have been written as gospel music a century ago but, composed in 2015, it is about Bridges' own spiritual reckoning during a period of depression and the Baltimore protests after the murder of Freddie Gray. The score includes sound composition by Benjamin Galvan.
I can see myself, Andrea Miller. irie-Woodbury Dance Company. Courtesy: Tori Duhaime.
The other two works, presented in abridged form for the film, were originally slated to be on the spring concert in 2020. Yin Yue's world premiere In the Moment Somehow Secluded evokes a cinematic painting, mesmerizing yet in a different visual impact than with Anderson's Rites. Yin, who directs YY Dance Company, developed the FoCo technique that comprises the expression of five elements (root, wood, water, metal and fire) and three stages (pulse, drop and flow). Indeed, this dazzling work epitomizes this technique, especially as one observes both dancers' posture and positioning throughout the work. The music includes Rutger Zuydervelt's (a/k/a Machinefabriek) Wheeler and Session I by Peter Broderick and Machinefabriek, excellent examples of the oscillating moods of electro-acoustic music which defy conventional ambient categorization.
For the closing work, Ririe-Woodbury chose wisely in performing a company premiere, a restaged version of Miller's 2010 original — a rollicking, witty, tantalizing dance composition to look back at the past year's struggles and to see encouragement ahead, with deserved satisfaction for managing through the challenges of the last 13 months. I can see myself pops with the music of Balkan Beat Box, an Israeli band, and the dancers are clearly enjoying themselves. Ririe-Woodbury always handles mad cap energy and physical comedy well and the six dancers deliver their personalities with genuine pride and flair (Williams, Greene, Jurica, Fausto Rivera, Corinne Lohner and Megan McCarthy). And, don't miss McCarthy's solo, which brings the zany pace to an abrupt halt in the middle, as she struggles to recover her stance after her legs go limp and she contorts herself, only to fall back once again. Oh, to have McCarthy's flexibility.
The 58-minute film is available for streaming, tickets required, through May 22. For more information, see the Ririe-Woodbury website.
arts in the COVID-19 pandemic
Bashaun Williams
Charles O. Anderson
Dominica Greene
jaclyn brown
Nicholas Jurica
nina watt
Repertory Dance Theatre
Ririe-Woodbury Dance Company
yin yue
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I am a native of Toledo, Ohio, having received my Ph.D. in journalism and mass communication from Ohio University's Scripps School of Journalism in 2002. In addition to teaching at Utah State University and the University of Utah, I have worked extensively in public relations for a variety of organizations including a major metropolitan university, college of osteopathic medicine, and community college. When it comes to intellectual curiosity, I venture into as many areas as possible, whether it's about music criticism, the history of journalism, the practice of public relations in a Web 2.0 world and the soon-to-arrive Web 3.0 landscape, or how public debates are formed about many issues especially in the political arena. As a Salt Lake City resident, I currently write and edit a blog called The Selective Echo that provides an entertaining, informative, and provocative look at Salt Lake City and its cosmopolitan best. I also have been the U.S. editorial advisor for an online publication Art Design Publicity based in The Netherlands. And, I use social media tools such as Twitter for blogging, networking with journalists and experts, and staying current on the latest trends in culture and news. I also have been a regular monthly contributor to a Utah business magazine, and I have recently conducted a variety of editing projects involving authors and researchers throughout the country and the world, including Dubai, Saudi Arabia, Canada, Lebanon, Cyprus, the United Kingdom, France, and Japan. I'm also a classically trained musician who spent more than 15 years in a string quartet, being involved in more than 400 performances.
Sundance 2023: Curious questions raised in the futuristic The Pod Generation are never fully settled but the film has plenty of heart to absorb...


text:
Q: Bad pointer type with a typedef I'm having troubles when calling a function taking a pointer to a string as a parameter. I need to get an Element's name.
// method
void getStringFromCsv( char ** str );

Let me introduce the structures I'm working with (not written by me and part of a much bigger project, I can't modify them).
// typedefs
typedef char      T_CHAR64[64];
typedef T_CHAR64  T_SYMBOL;

// generic element
typedef struct Element
{
  T_SYMBOL  name;
} T_Element;

// csv element
typedef struct CsvElement
{
  Element * pElement;
  int   id;
} T_csvElement;

So, basically, I thought I would call the function like this :
T_Element * pData; // Not null, filled earlier
getStringFromCsv( &pData->pElement->name );

But this doesn't work (warning: passing argument 1 of 'STR_toCsv' from incompatible pointer type). I'm using gcc with NetBeans 6.8.
I tried many things...
T_SYMBOL foo = "foo";
T_SYMBOL * pFoo = &foo;

getStringFromCsv( pDef->name, &pFoo ); // error : passing from incompatible pointer type

T_CHAR * pBar = &foo;      // error : init from incompatible pointer type
T_CHAR * pBaz = &(foo[0]); // OK

getStringFromCsv( pDef->name, &pBaz ); // OK

T_SYMBOL * pFooTest = &(foo[0]); // error : init from incompatible pointer type

...but ended up casting name to a char ** :
getStringFromCsv( (char**) &pData->pElement->name );

What is wrong with my code ?
Basically, SYMBOL = CHAR *, right ? Why is SYMBOL* != CHAR** ?
I'm pretty sure I'm missing something simple but right now... Nothing came.
EDIT
Here is getStringFromCsv :
void getStringFromCsv( char ** data )
{
  // pDesc is defined and not null
  csvDescriptorCat( pDesc, *data);
  csvDescriptorCat( pDesc, "\t");
}

void csvDescriptorCat( CsvDescriptor * pDesc, char* str)
{
  int len;
  if( str != NULL)
  {
    len = strlen(str);
    strcpy( &pDesc->line[pDesc->pos], str);
    pDesc->pos += len;
  }
}

A: name is an array of chars, so &name gives you a pointer to char[64], as Vicky already answered. But casting makes things worse, because it tells the compiler to treat the first chars of the array as a pointer to the real array.
See the C-FAQ: http://c-faq.com/aryptr/aryptr2.html
I think you can use a temporary char* here:
char *tmp = pData->pElement->name; // array decays to pointer
getStringFromCsv(&tmp);

If this is expected by the function. Expecting a char**, make sure that it doesn't try to reallocate the memory. For simply filling it, a char* would be enough.

A: If you wish to pass &pData->pElement->name to the function, the function must be declared as:
void getStringFromCsv(T_SYMBOL * str);

Alternatively you can use a temporary char * as Secure offered - but there's not much point in doing this, because any updates to that char *'s value can't be used - the ->name member can't be modified, as it's an array.
You might as well just declare the function as:
void getStringFromCsv( char * str );

...and call it as:
getStringFromCsv( pData->pElement->name );

(In this case, the function can still change the contents of the ->name array.  What you can't do is to change the position of the array itself).

As well as Secure's option, there's another way if your compiler supports C99 compound literals:
getStringFromCsv( &(char *){ pData->pElement->name } );

A: Alas, one of the little secrets of C that people fail to tell you, an array is not the same thing as a pointer.  if x is defined as int x[5] or whatever, &x == x.  Try out this code below:
#include <stdio.h>
int main(int argc, const char *argv[])
{
   char x [5];
   char *y;

   printf("%08x\n", x);
   printf("%08x\n", &x);
   printf("%08x\n", y);
   printf("%08x\n", &y);

   return 0;
}

A: Considering this : http://c-faq.com/decl/strlitinit.html
char  a[4] = "hello";
char* p = "hello";

Aren't the same thing (even if they seem to be).
So my SYMBOL and CHAR* cannot be exchanged, right ?
Is there a workaround, or another solution ?

A: Yes, under the covers T_SYMBOL is handled like a char *. But you've declared it as a char[64], so you're passing in a pointer to a char[64] not a pointer to a pointer to a char. The compiler is keeping track of that for you.
Personally in this situation I would just cast it as you did at the end of your question.

redpajama_set_name: RedPajamaStackExchange


Before the introduction of Christianity in the Western world, villages in Europe were filled with women who performed "nature rituals" in their villages and towns. These rituals kept the people who performed them, and those who participated in them, emotionally and spiritually connected to the earth and sky and fields and rivers. Humanity at that time was seen as living in sacred partnership with the earth.
With the advent of the early Christian Church, a new dispensation was introduced into humanity's thinking: the idea that we were not in divine partnership with the earth, but rather that the earth was given to us to be used for our purposes. While this thought is not inherent to Christian teaching, it reigned for centuries and created a fundamental rift in Man's relationship with nature.
That change in thinking - the spiritual rift between humanity and nature - was the beginning of our environmental crisis.
Today, centuries later, that rift has reached grotesque proportions.
From the destruction of our topsoil to the desecration of our wilderness, from the carbonization of our atmosphere to the removal of mountaintops and the destruction of millions of acres of farmland, humanity's obeisance to corporate profits has replaced our devotion to what is morally good. And this must stop, or we will kill ourselves. No species survives that systematically destroys its own habitat.
As president, one of my first actions will be to appoint a world class environmentalist to head our Environmental Protection Agency. Climate change will be seen as an existential threat to the survival of our species. No longer will the agency set up to be the earth's protector be led by former oil and chemical company executives whose goal is only to protect short term profits of fossil fuel companies. We will restore the Clean Water Act and the Clean Air Act. We will decarbonize quickly. We will transition to green sources of energy. We will permanently ban pesticides that are known to cause damage to a developing child's brain. And yes, we will pass a Green New Deal.
None of this will happen because I "go to Washington and fight for you." It will happen because I go to Washington and co-create with you a new era in our nation's ongoing story, a course-correction in our thinking, and a rededication to our shared moral responsibility to take care of the earth for our children and our children's children.
We need a moral and spiritual awakening in America, which alone will awaken us to our responsibility as citizens to override the soulless forces of a sociopathic economic system.
Our Earth, and possibly our lives, depend on it.
Let's speed up the evolution.


text:
Weekly Topic
Imam Khomeini Highlighted Moral Values
The founder of the Islamic Republic called on the whole humanity to comprehend the moral and Islamic values promoted by the holy prophet of Islam.
The holy prophet said in one of his traditions that he was sent and assigned to complete and promote the best moral and ethical values. Imam Khomeini revived these genuine teachings.
The spiritual and religious leader of the Muslim world, invited the whole Islamic community to follow footsteps of the honorable prophet of Islam Hadrat Mohammad (peace be upon him) and his infallible successors.
The founder of the Islamic Republic used to issue special congratulatory messages on this occasion.
Imam Khomeini used to extend his greetings to the millions strong Muslim Ummah and all the lovers of the holy Ahlul Bayt (the people of the household of prophet) on these sad events.
The honorable prophet of Islam left behind the "Thaqalayn" (the holy Qur'an and his pure progeny) for the guidance of the Muslims and the whole world.
When Hadrat Muhammad (peace be upon him) was 38 years of age, he spent most of his time in meditation and solitude. The cave of the mount Hira was his favorite place. It is there that he used to retire with food and water and spend days and weeks in remembrance of Allah. Nobody was allowed to go there except Khadijah and 'Ali (peace be upon them). He used to spend the whole month of Ramadhan therein.
The period of waiting had come to a close. His forty years of life had varied experiences, and from the world's point of view, he had developed a maturity of mind and judgment, although in reality he was the embodiment of perfection from the very beginning.
Highlighting the significance of 27 Rajab, history recounts a very important event that took place on this date. It is Mab'ath (Holy Prophet Muhammad's, peace be upon him and his progeny, official appointment to the Prophetic Mission).
Mab'ath (Holy Prophet Muhammad's, peace be upon him and his progeny, official appointment to the Prophetic Mission) signals the turning point in the history of the world as the Seal of Messengers with the final revelation was finally brought into this material world.
Prophet Muhammad Ibn Abdellah Al-Mustafa, peace be upon him and his progeny, says: "Kuntu Nabiyan Wa Adam Bain Al-Maa' Wat Teen" (I was Prophet even when (Prophet) Adam was in between Water and Clay). However, the time and day when Prophet Muhammad, peace be upon him and his progeny, was ordered by Allah to declare his Prophethood to the world was 27 Rajab, 40th year of elephant (610 C.E.). At that time he was 40 years old.
On that day he was in the cave of Hira, and the Angel Gabriel came to him conveying to him the following message of Allah: "Read in the name of thy Lord Who created, Created man from a clot (of congealed blood): Read and thy Lord is most Bountiful, He Who taught with the pen, taught man that which he knew not." (The Holy Qur'an, 96:1-5).
These were the first verses to be revealed. The flow of the Divine message which continued for the next 23 years had begun, and the Holy Prophet, peace be upon him and his progeny, had arisen to proclaim the Unity of God and the Unity of Mankind, to demolish the edifice of superstition, ignorance, and disbelief, to set up a noble conception of life, and to lead mankind to the light of faith and celestial bliss.
Muslims in Iran and across the world have held celebrations to commemorate the auspicious occasion of Eid al-Mab'ath -- the anniversary of the appointment of Prophet Mohammad (PBUH) to prophethood.
The joyous occasion is annually celebrated by Muslims marked by various ceremonies and rituals, on the 27th Rajab in the lunar calendar (May 16th), the day of Muhammad's appointment as the last and greatest prophet of the Al-Mighty.
The Iranian people from all walks of life went to mosques on Saturday to participate in celebrations marking Eid al-Mab'ath, which is one of the most important religious holidays in the country.
Mohammad (PBUH) often retreated to the Hira cave outside the holy city of Mecca, Saudi Arabia, to pray to God, to reflect on the true meaning of life and to ponder over life and death.
The angel Gabriel came to Mohammad in the Hira cave bringing him the message of God Almighty, telling him he was appointed as the last prophet to guide mankind out of ignorance.
In the same pattern, Imam Khomeini also followed footsteps of the holy prophet and coordinated the struggles of the Iranian people against the Shah's despotic regime to advance the goals of Islam.
The great leader continued the struggle until establishment of the sacred Islamic system, which guarantees the freedom of the people, civil liberties, independence and implementation of social justice.
Several of Imam Khomeini works emphasized on following the holy prophet as a role model for the whole humanity.
It is noteworthy that Imam Khomeini established a comprehensive system based on divine and human values in contemporary history. Imam Khomeini frequently said that Islam was fully capable of establishing progressive democratic system.
US troops wounded in Iranian missile attack despite earlier denials ... Friday, January 17, 2020 10:07
Revolution under Imam Khomeini leadership gained momentum, Shah fled ... Thursday, January 16, 2020 01:06
Pure Islam from viewpoint of Imam Khomeini Saturday, January 11, 2020 02:45
Oman's Sultan Qaboos bin Said dies, Muscat declares national mourning Saturday, January 11, 2020 02:19
Imam Khomeini restored identity to Christians, sent messages on ... Sunday, December 22, 2019 08:25
Imam Khomeini highly trusted late Ayatollah Rafsanjani Thursday, January 09, 2020 09:29
Iran marks the anniversary of Imam Khomeini's message to Mikhail ... Wednesday, January 01, 2020 11:00
Reza Khan acted against cultural values, religious sentiments Monday, January 06, 2020 12:06
Holy prophet of Islam
infallible successors
27th of Rajab
Mab'ath


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Salami, pepperoni, Italian sausage, mushroom.
Olives, mushrooms, tomatoes, green peppers, onions.
Chicken, tomatoes, feta, gorgonzola, basil, provolone, roasted garlic.
Provolone, roma tomatoes, fresh basil.
Chicken, artichoke hearts, sun-dried tomatoes, pesto sauce.
Feta, gorgonzola, cheddar, jack and mozzarella.
Beef, tomatoes, lettuce, olives, onions.
Gorgonzola, Mushrooms, Black Olives, Tomatoes, Pineapple, Pepperoncini, Italian Sausage, Artichoke Hearts, Pepperoni, Salami, Spinach, Ground Beef, Meatballs, Green Olives, Canadian Bacon, Bacon, American Bacon, Chicken, Broccoli, Zucchini, Green Peppers, Onions, Feta, Garlic, *Prosciutto, *Sun-Dried Tomatoes, *Anchovies, *Kalamata Olives or *Bay Shrimp.
Rigatoni tossed with our house meat sauce and topped with alfredo sauce, then baked with your house cheese blend and romano cheese.
Shrimp and mushrooms sauteed in a garlic and tossed with a white wine sauce, with fresh basil, roma tomatoes and served over angel hair pasta.


but had done nothing to be blamed and accused of.
Then I noticed the fact there was no time.
and was asking whether I would be alive.
or judge me to be guilty of so many incurable sins.
Tell me why, or why not.
maybe I overlooked something fatal for me.
and I was in fetters at the mercy of the mob.
The silent warning became louder every day.
Then I kept pretending not to hear.
Its meaning had been in the eyes of beholder all along.
It had grown dark before I found a sign.
"Among the nonsense tragedies, what on earth you are looking for?
You only have to be honest to yourself and your own fate."
Ⅲ : There is nobody who knows there will be nobody.
Except for me, all the world has gone mad.
So what is forgiveness you are willing to withhold?
What is the well-being you are willing to make?
Now what? So what? Don't you come interrupt me, oh please.
while I am interrupting myself.


This walk starts at Liverpool St Station and loops through the East End following a trail of independent cafes, market stalls and vintage stores. The natural habitat of the hipster. If you're visiting London then you'll be amazed at the density of vintage rangefinders and Lomo cameras you'll see in use along this route. Half the time you'll be photographing photographers.
This walk starts from Oxford Circus and lets you stalk the West End hunting tourists, crowds and street life in it's densest form. You'll see screaming kids in front of Topshop, lonely pensioners having a cup of tea in a street cafe and Middle Eastern billionaires taking in the sights.
What are your favourite locations to shoot the people of London?
This entry was posted in London and tagged Locations, London, Map by Peter Thomson. Bookmark the permalink.
– Along the Strand to Trafalgar Square.
Great walk Kit! That would add in lots of 'city' characters that the other walks miss out. Shooting around the Thames is always a good idea because people are feeling a bit more chilled out.
I have been following your posts since I have stayed in U.K. It's really great to read your articles from time to time, and mindset is walking along your view sights to observe the walk of life in England.
I personally am fond of street photography as you are and your works do inspire me a lot. At the moment, my gears are GH2 with 25mm/1.4F, 45mm-175mm X lens, 14mm/2.5f +GWC1 and my "pocket camera" –Fuji X10.
I had dropped by in London couple times, I will plan to bend sometime to visit London again even though I have just traveled from Germany last month with family. So your "shooting guide" is very helpful. Thanks! keep in touch and please post more.
thank you so much for this post. I am always amazed at the generousity of my fellow street photographers. I haven't been to London for 20 years but will be there soon for a day or so to do street and this is a HUGE HUGE help..thank you very much!!
just the information i need, coming to london in a week.


Shalom ladies! Here is a presentation Abba helped me to make regarding the mitzvah of wearing tzitzit. It includes the Messianic prophecy references, a brief Hebrew word study, the fulfillment of prophecy when Yeshua used His tzitzit as a healing ministry for the multitudes, etc. Abba really helped me encapsulate and summarize their purpose, which I think is helpful for a lot of believers who question the validity of tzitzit or would like a deeper understanding about this commandment. Please feel free to post your comments and questions below. You will, if viewing from a mobile phone, download the Prezi app in order to view! This is relevant and current for our generation. Let's help share the message of love once again! Click on the link below to get started, and if you like it please share with your loved ones!