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18 of the world's sickest waterslides
Galleries Family Travel
Alex Scola
GROWING UP IN THE NEVADA DESERT, nothing cut through the 110-degree stupor like a family outing to Wet and Wild, the local water park. Despite my debilitating fear of heights, precious few things kept me from trudging up the trillions of stairs, looking down, and immediately regretting my decision before being forcibly pushed into the tube by the burly guy regulating human pileups. It was all worth it for the rush of shooting out the bottom of the slide, flying a few feet, and performing a spectacular belly flop into the pool below.
But as I've grown older, the slides have seemed smaller, the ride less an epic adventure and more a soggy disappointment — so after my last sad, wet journey to some tiny water park in the middle of nowhere, I vowed to find a slide that lived up to the excitement, the fear, and the joy of my childhood.
Here are 18 of the world's most insane, epic, and terrifying waterslides.
Scorpion's Tail - Noah's Ark Waterpark, Wisconsin
The Scorpion's Tail at Noah's Ark Waterpark is a frightening waterslide, with its huge loop and clear plastic tubing. Adding to the intimidation factor is the entrance to the slide—rather than sliding into the tube on your ass like a normal waterslide, you step into a vertical capsule and the floor drops out beneath you with no warning whatsoever. You immediately drop TEN STORIES before entering the loop at roughly 50 feet per second, and you're done with the 400-foot-long slide in six seconds flat. Holy shit.
Master Blaster - Atlantis The Palm Resort, Dubai
These slides are actually kind of beautiful, which is pretty unusual for waterslides. Each of the four slides starts from a different level of a Mayan pyramid, and shoots riders off all over the park. The big feature of Master Blaster coasters (which seems to be more of a brand than a ride name) is their powerful streams of water that actually propel riders uphill. And as if all that wasn't cool enough, the whole park is located on the Palm Jumierah—the man-made island formation in the shape of a date palm.
Epic Plunge - Norwegian Cruise Line
This one is double cool, because not only is it an epic waterslide, but it's also on a boat. Yep, one of the best waterslides in the world is on a ship belonging to the Norwegian Cruise Line. The slide extends over almost half of the ship's length before dropping riders down into a huge bowl and eventually ejecting them into a top-deck swimming pool.
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King Cobra - Six Flags, New Jersey
If you lined up the pictures of every slide on this list and told me to choose "the one," the King Cobra would be my pick every time. Though it's only about 60 feet tall, the Cobra is a one-of-a-kind raft racer, where riders hurtle through the snake's coils and shoot down a huge drop before flipping up into its mouth. The raft then slides back and forth in the gigantic half-pipe until finally coming to a stop at the base of the animal's hood. The winner of the race is well-rewarded, too, because the faster you're going, the further you'll be able to fly into the mouth of the beast.
Bulletbowl - The Watercube, Beijing
The Beijing National Aquatics Center, constructed for the 2008 Summer Olympics, is known to the locals simply as "The Watercube," and if you've ever seen it, it's pretty obvious why. But since the Olympics, the Watercube has undergone massive renovations, and rather than a series of Olympic swimming and diving pools, it now contains a huge water park. The Bulletbowl is one of the crown jewels of the magic-themed water park, and contains one of the smallest bowls (for the fastest speeds) of any water slide in the world.
Boeing 747 - Evergreen Wings and Waves Waterpark, Oregon
Air-and-space museums were my absolute favorite when I was a kid. No number of classical Renaissance paintings could ever compare to seeing the Spirit of Saint Louis and the Lunar Lander, or walking through the inside of an actual space shuttle. But the Evergreen Aviation and Space Museum in Oregon has done all others one better—it's got its own waterpark. The Boeing 747 water slide is crafted from a real retired airplane, and riders get a great view of the surrounding area as they follow the curves of the slide across the museum's roof.
Leap of Faith - Atlantis Paradise Island, Bahamas
This might be my favorite waterslide on the list. Leap of Faith in the Bahamas begins at the top of a life-sized Mayan pyramid and propels riders over a 60ft freefall. The drop ends in a translucent, underwater tunnel, which travels through a one-of-a-kind shark-filled lagoon, in what has to be the single coolest aquarium experience of all time. (The company assures prospective riders that the sharks are not interested in chewing on the tunnel tube.)
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Insano - Beach Park, Brazil
Insano is the tallest, fastest waterslide ever built. It stands over 130 feet high, but the ride only lasts about four or five seconds because your body is rocketing back towards the earth at a brain-melting 65mph. Firsthand accounts say that the ride is so steep, and you're moving so fast, that your body doesn't actually make contact with the slide until you get to the bottom—you just fall through the air.
Aquaduck - Disney Cruise Ships
Thought it's technically classified as a water coaster and not a waterslide, the Aquaduck is one of the most awesome rides on this list. Installed on several of the Disney Cruise ships, the 'duck covers four decks and propels raft-riders up and over hills through crystal-clear tubing (which, at one point in the ride, extends 12 feet over the side of the ship, 150 feet above the open ocean).
Head First Racer - Chimelong Waterpark, China
Eight identical side-by-side slides start 65 feet up, complete a full circle, and straighten out for a big splash-down following an arrow-straight drop. You just pick your favorite slide at the top and see if you can beat seven of your friends (or seven strangers, if none of your friends are brave enough to ride with you). As implied by the name, you ride this slide penguin-style.
Toboggan - Città del Mare, Sicily
Let's face it—most waterslides aren't all that pretty. More often than not, they look like a circus clown's balloon failure, all lumpy plastic and loop-the-loops. But not the Toboggan slide at the Città del Mare resort in Italy. Located on the coast of Sicily, this slide is the same bright blue as the ocean surrounding it. It descends three tiers of sheer, rocky cliffs before dropping riders into the sparkling waters of the Mediterranean.
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The Mammoth - Splashin' Safari, Indiana
The Mammoth at Splashin' Safari is aptly named—it is HUGE! The coaster itself covers over three acres and is the longest water ride in the world, but the raft ride is where it gets its massive reputation. First, you ride your little round boat up a seven-story ramp on a gigantic conveyor belt. Then you travel just under half a mile, over hills, through dark tunnels, and under the open sky.
Jumeirah Sceirah - Wild Wadi Waterpark, Dubai
Jumeirah Sceirah, in Wild Wadi Waterpark in Dubai, was remodeled just this year and is one of the tallest and fastest vertical-drop waterslides outside of the US. From the 105ft tower, riders drop almost straight down and travel the full 350 feet of the slide in a matter of seconds, at top speeds of over 50mph. The slide tower itself is famed for its fantastic views of some of Dubai's best-known structures, including the Burj Al Arab.
The Cyclone - World Waterpark, Canada
World Waterpark in Canada's West Edmonton Mall is open year round, which must be nice when outdoor temperatures dip to 40 below. The Cyclone was the first slide with a 360-degree loop built in Canada, which has made it quite a popular attraction. It drops riders into a 55ft freefall before shooting them up into the nearly-vertical loop and then dropping them into the pool.
Cliffhanger / F5 Twin Twisters - Schlitterbahn, Texas
This pair of waterslides is as different as night and day. The Cliffhanger is an open-air, nearly vertical chute that plunges riders 80 feet toward the earth at speeds of up to 35mph before depositing them in the pool. The F5 Twin Twisters (currently being remodeled as the "Screaming Serpents"), are a pair of enclosed tubes that wind around one another like DNA, hurling riders through fog and lights on their way to the serpents' mouths.
Twister and Speedy - Sonnentherme Lutzmannsburg Thermal Spa, Austria
These slides are evenly matched, though with different strengths: Speedy has, of course, speed, while Twister has distance. Speedy starts 75 feet in the air and traverses over 460 feet of tubing, while Twister is a little more gentle but extends over 660 feet. Plus, it's pretty fantastical how they wind around that tower like big cartoon snakes.
Tantrum Alley - Wild Wadi Waterpark, Dubai
Another ride from Wild Wadi in Dubai, Tantrum Alley was the first waterslide in the world to contain three funnels. Four-person rafts are subjected to two huge drops, and then tumbled in all directions in what the park's website calls the "tornadoes." As intense as that sounds, it's not as scary as it looks, as those big striped bowls aren't part of Tantrum Alley but instead belong to its sister slide, the Burj Surj.
Summit Tower - Calypso Park, Canada
Another slide found in Canada, Summit Tower in Calypso Park is the tallest free-standing waterslide in North America. The tower provides the base for no fewer than 10 waterslides, all of which look totally amazing. Four slides known as the "Accelerators" start all the way at the top of the tower (90 feet up) and travel straight down over bumps and drops. Four more, called the "Family Twisters," start at a friendlier 30ft height and wind their way back to the ground for a gentler ride. And finally, two matching and intertwined "AquaLoops" start 55 feet up and shoot the most daring watersliders through 360-degree loops.
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Simon Furman has kindly updated his new blog with some information on the second (The Gathering being the 1st) Beast Wars comic series, "Beast Wars The Ascending". He mainly focuses on the first issue, with which he has just finish the script for (and the reason for this blog post). The first issue is out in September.
Being released at the same time are the Beast Wars Profile Books which feature ALL Beast Wars characters (yes, including the Japanese ones).
Simon's full blog post can be read here.


Jane Clayton is a corporate finance lawyer based in Dubai.
Jane focuses on cross border mergers and acquisitions in the Middle East and North Africa (MENA) region, investment funds, private equity transactions, joint ventures, inward investments and DFSA regulatory advice.
Jane recently advised Saudi Arabian Airlines Corporation on the privatization of its private jet business carried out through its wholly owned subsidiary, Saudia Private Aviation Limited (SPA). The transaction involved the sale of 30 per cent of the share capital of SPA to PrivatAir Saudi Arabia Limited, an international aviation group based in Saudi Arabia and Switzerland.
Transactions on which Jane has also advised include advising Islamic Development Bank on the establishment of a US$2 billion infrastructure fund focusing on certain member countries of the Islamic Development Bank, and advising Islamic Corporation for the Development of the Private Sector on the establishment of a Shari'a compliant renewable energy fund (which also focused on certain member countries of the Islamic Development Bank).
According to Chambers Global Jane has "excellent depth of experience in the region, having practised in Dubai for 12 years. Her investment funds knowledge forms part of her wider corporate practice and she is well regarded for cross-border advice".
Jane is described in Chambers Global 2014 as having "a broad swathe of experience of M&A and investment funds mandates, as well as DIFC and regulatory law matters".
Jane was listed in the 2014 edition of Expert Guide to the World's Leading Women in Business.
Jane was awarded Private Investment Funds – Lawyer of the Year – UAE in 2016 by Lawyer Monthly.
Jane has over 18 years of experience working in Dubai on MENA related transactions.


text:
Letter from SVP
The Annual Fund for GH
Clinical Growth Plan
CARES Act and Giving
Thanks To Donor Support
Partners in Care
The Benefit for GH
HomeGivingCARES Act and Giving
How the New CARES Act May Affect Your Gift Planning
Legislation known as the CARES Act, designed to rescue the economy from the effects of the coronavirus pandemic, was passed by Congress and signed into law by the president on March 27, 2020. Here is a summary of provisions applicable to charitable giving included in the Act, officially named the Coronavirus Aid, Relief, and Economic Security Act.
New Charitable Deduction for Non-Itemizers
Taxpayers who take the standard deduction rather than itemizing their deductions will nevertheless be able to claim a charitable deduction of up to $300 for cash donations made in 2020. Those who itemize may continue to claim an income tax charitable deduction subject to applicable limits.
Higher Deduction Limits
Individuals in 2020 will be entitled to a 100% income tax charitable deduction for cash gifts to the extent of their adjusted gross income. The charitable deduction limit for corporations has been raised from 10% to 25% of taxable corporate income.
Required Minimum Distributions Waived in 2020
For the year 2020, there will be no mandatory distributions from retirement accounts (whatever the age of the owner), thus allowing those accounts to recover. The minimum age for making a tax-free transfer from an IRA to a charity remains at 70½, and the annual limit remains at $100,000. However, since cash gifts are deductible in 2020 to the extent of adjusted gross income, a person could withdraw and then contribute a larger amount—with the deduction offsetting the taxable withdrawal.
Waiver of Penalties When Retirement Funds Are Used for Coronavirus Purposes
If you are under the age of 59½ and withdraw money in 2020 from your retirement plan to cover expenses incurred by you or a family member related to treatment of the coronavirus, the 10% tax penalty will not apply, taxation of the distribution can be spread over three years, and the amount withdrawn can later be added without regard to contribution limits.
Contact Us and Your Advisors
The CARES Act is several hundred pages long and includes numerous provisions that could benefit you financially. Consult your advisors about these provisions, and contact us if we can help you with any gift-planning opportunities.
Greenwich Hospital Office of Development
Photos from Past Events
Growth Plan Giving Opportunities
Honor & Causes
COVID-19 Support Funds
COVID-19 Employee Disaster Relief
Cos Cob, CT 06807
© 2021 Greenwich Hospital Office of Development, All Rights Reserved.

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Tesla market value crosses $500 billion in meteoric rally
Tesla is Wall Street's seventh most valuable company, just behind Berkshire Hathaway , and its shares have rallied over 30 per cent since Nov. 16, when it was announced Tesla would join the S&P 500 benchmark
Tesla Inc blew past $500 billion in market value on Tuesday as investors snapped up its shares in the run-up to its debut in the S&P 500, extending a meteoric rally that has seen it surge over 500 per cent this year.
The California electric carmaker's stock rose nearly 5 per cent, putting its market capitalization at $519 billion.
Tesla is Wall Street's seventh most valuable company, just behind Berkshire Hathaway , and its shares have rallied over 30 per cent since Nov. 16, when it was announced Tesla would join the S&P 500 benchmark.
Index funds that replicate the S&P 500 will have to buy more than $50 billion worth of Tesla's stock ahead of its inclusion to the index on Dec. 21. Additionally, Goldman Sachs estimated last week that actively managed mutual funds could buy another $8 billion of Tesla shares after it is added.
Tesla has become by far the world's most valuable automaker, despite production that is a fraction of Toyota Motor Corp, Volkswagen or General Motors Co.
Shares of other electric vehicle (EV) makers have also rallied in recent months as President-elect Joe Biden made boosting EVs a top priority during his campaign.
Chinese electric carmaker Nio Inc fell 4.9 per cent on Tuesday, trimming its gain in November to 72 per cent.
"One of the core underpinnings of the Biden platform will be around pushing clean energy and zero-emissions vehicles with hopes of accelerating the deployment of electric vehicles and public charging outlets by 2030," Wedbush analyst Daniel Ives wrote in a research note.
(Reporting by Subrat Patnaik in Bengaluru and Noel Randewich in San Francisco; Editing by Tom Brown)
elect joe biden
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The term jus post bellum is used increasingly to refer to the legal frameworks applied in post-conflict peacebuilding projects. This thesis considers the recent application of three jus post bellum frameworks in states emerging from conflict to determine which framework has the greatest potential for success in terms of securing lasting peace and security in the post-conflict state. The three frameworks considered are: the law of occupation applied in Iraq, the United Nations-led interim administrations applied in Kosovo and East Timor, and the light footprint approach applied in Afghanistan. The thesis concludes that the light footprint approach, with its focus on local ownership over the peacebuilding process, should be considered for future post-conflict states, but with enhanced attention to security and coordination. A light footprint "plus" approach that includes increased international support and mentorship is advocated as the clearest route to lasting peace and security. This thesis concludes that the law of occupation is not an effective tool for post-conflict peacebuilding because it restricts the types of changes that can be made within the post-conflict state and it only arises in rare instances of international armed conflict. In Kosovo and East Timor, the UN-led interim administrations took control of all aspects of governance and made significant changes. While UN-led interim administrations can bring about significant post-conflict change, the lack of popular consultation and perceived lack of accountability makes them less desirable as post-conflict peacebuilding frameworks. In Afghanistan, peace builders were wary of the risks of imposing change on the Afghan people and adopted a light footprint approach that allowed Afghan authorities to lead post-conflict rebuilding efforts. Unfortunately, the international community did not provide sufficient support to the Afghans, the result of which was a poor security environment, an uncoordinated approach, and a failure to incorporate existing judicial frameworks into the new institutions of government. Although the light footprint approach is considered a failure in Afghanistan, a light footprint "plus" approach cannot be discounted for future peacebuilding initiatives.


Onze winkel in Apeldoorn is gesloten op Koningsdag, Bevrijdingsdag, Hemelvaartsdag en vrijdag 11 mei.
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Red Hat Solutions Provide Reliability and Performance Gains for Munich...
Red Hat Solutions Provide Reliability and Performance Gains for Munich Airport
Airport migrates servers and desktops to Red Hat Enterprise Linux, Red Hat Cluster Suite and Red Hat Network
노스캐롤라이나주 롤리 - 2008년 8월 6일 —
Red Hat (NYSE: RHT), the world's leading provider of open source solutions, today announced that Munich Airport, recognized for three consecutive years as Europe's "Airport of the Year" by air transport research company Skytrax, has migrated a large portion of its UNIX servers and desktops to Red Hat Enterprise Linux. The Airport has since realized heightened performance, cost savings and reliable support.
After experiencing consistent performance issues, Munich Airport conducted detailed cost-analysis research on alternative technology solutions that could boost performance for its IT systems. Results revealed that Red Hat Enterprise Linux, already in small use in the Airport's IT systems, could provide both the savings and performance benefits desired. In 2005, Munich Airport migrated a portion of its servers to Red Hat Enterprise Linux and in March 2006, it also migrated most of its desktop workstations to Red Hat Enterprise Linux Desktop. Since September 2006, Munich Airport has equipped more than 30 servers and 40 desktops to run using Red Hat Enterprise Linux. Red Hat-based systems are also used by Airport's external mail gateways providing anti-virus and anti-spam functionality.
Munich Airport's Red Hat Enterprise Linux Desktop solution plays an important role in airport operations, providing the control tower with technology to ensure smooth, automated aircraft dispatch and passenger-handling sequences. The system coordinates aircraft parking spaces, taking raw data from the radar system as input, and using this data to calculate aircraft arrival times and coordinate logistical activities on the ground.
For server and desktop administration, Munich Airport also uses Red Hat Network in conjunction with Red Hat Satellite Server. "Red Hat was also able to score with Red Hat Network against other Linux vendors, who had nothing comparable to offer at the time when the decision to select Red Hat was made," said Hubert Bosl, Unix system administrator at Munich Airport and a Red Hat Certified Engineer (RHCE). "In the mid-run, we will also use Red Hat Network to manage the systems which still run on proprietary Unix. Moreover, this management tool was extremely important for our ISO 20000 certification."
In 2007, Munich Airport served 34 million passengers and was ranked the 28th busiest airport worldwide. It aims to become the most efficient and attractive European airport hub by 2010 and expects that the implementation of Red Hat solutions will help expedite this goal. Currently, Munich Airport is also running another large-scale project together with Red Hat solutions. The main components of the Universal Display System in the Airport's Terminal 1 has more than 1,400 display devices providing passengers with up-to-date information on arrivals and departures, and these servers are now running on Red Hat Enterprise Linux.
"Munich Airport is a worldwide market leader in many respects. We are therefore especially delighted that it deploys Red Hat Enterprise Linux, particularly on its traffic-critical systems," said Werner Knoblich, vice president, EMEA at Red Hat. "We are proud that the reliability, manageability and security of our operating system have contributed to the airport's top rank for service and customer satisfaction."
For more news about Red Hat, visit www.redhat.com. For more news, more often, visit www.press.redhat.com.
Red Hat, the world's leading open source solutions provider, is headquartered in Raleigh, NC with over 50 satellite offices spanning the globe. CIOs have ranked Red Hat first for value in Enterprise Software for four consecutive years in the CIO Insight Magazine Vendor Value study. Red Hat provides high-quality, affordable technology with its operating system platform, Red Hat Enterprise Linux, together with applications, management and Services Oriented Architecture (SOA) solutions, including the JBoss Enterprise Middleware Suite. Red Hat also offers support, training and consulting services to its customers worldwide. Learn more: http://www.redhat.com.
Certain statements contained in this press release may constitute "forward-looking statements" within the meaning of the Private Securities Litigation Reform Act of 1995. Forward-looking statements provide current expectations of future events based on certain assumptions and include any statement that does not directly relate to any historical or current fact. Actual results may differ materially from those indicated by such forward-looking statements as a result of various important factors, including: risks related to the integration of acquisitions; the ability of the Company to effectively compete; the inability to adequately protect Company intellectual property and the potential for infringement or breach of license claims of or relating to third party intellectual property; risks related to data and information security vulnerabilities; ineffective management of, and control over, the Company's growth and international operations; adverse results in litigation; and changes in and a dependence on key personnel, as well as other factors contained in our most recent Quarterly Report on Form 10-Q (copies of which may be accessed through the Securities and Exchange Commission's website at http://www.sec.gov), including those found therein under the captions "Risk Factors" and "Management's Discussion and Analysis of Financial Condition and Results of Operations". In addition to these factors, actual future performance, outcomes, and results may differ materially because of more general factors including (without limitation) general industry and market conditions and growth rates, economic conditions, and governmental and public policy changes. The forward-looking statements included in this press release represent the Company's views as of the date of this press release and these views could change. However, while the Company may elect to update these forward-looking statements at some point in the future, the Company specifically disclaims any obligation to do so. These forward-looking statements should not be relied upon as representing the Company's views as of any date subsequent to the date of the press release.
LINUX is a trademark of Linus Torvalds. RED HAT® and JBOSSreg&; are registered trademarks of Red Hat, Inc. and its subsidiaries in the US and other countries.
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Utah state law requires that any athlete suspected of a concussion must be removed from the event and cannot return to play until cleared by a qualified healthcare provider.
Establishing a baseline of your congnitive function before starting a sport season makes it much faster and easier for a healthcare provider to accurately diagnose your level of injury. We use a computerized neurocognitive assessment tool called ImPACT testing along with other tests to collect a complete picture of your mental state. Once this data is collected, it can be used as a comparison following a brain injury to determine the extent of the damage.
Testing takes about 30 minutes for each athlete. We offer group discounts for whole teams.
Find concussion-related articles on the Intermountain Blog Network.


Copenhagen Wolves was a Danish eSports organization founded in August 2012. The organization was best known for its team in League of Legends, where some of their former players have now become the star player on stage. The star players such as Søren "Bjergsen" Bjerg, Dennis "Swede" Johnsen, Maurice "Amazing" Stückenschneider and Mike "Wickd" Petersen have been well sought after.
Copenhagen Wolves was active in the divisions League of Legends, Starcraft 2, Counter-Strike: Global Offensive and FIFA. The organization was active for four seasons before it was dissolved in June 2016.


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

Let  $G=(V,E)$ be a graph. Given a set $X\subset V$ of vertices, we denote by
$$
\partial X=\{ y\in V\setminus X: d(X,y)=1\},\; B(X)=\{ y\in V: d(X,y)\le 1\} =X\cup \partial X,
$$
the {\it boundary} and the {\it ball} of $X$ respectively, where $d(X,y)$ denotes $\min\{d(x,y):x\in X\}$.

We write $\partial_G$ and $B_G$ when the reference to $G$ has to be made explicit. The {\it vertex-isoperimetric function} (we will call it simply {\it isoperimetric function}) of $G$ is defined as
$$
\mu_G (k)=\min \{|\partial X|: X\subset V,\; |X|=k\},
$$
that is, $\mu_G (k)$ is the size of the smallest boundary among sets of vertices with cardinality  $k$.

The isoperimetric function is known only for a few classes of graphs. One of the seminal results is the exact determination of the isoperimetric function for the $n$--cube obtained by Harper \cite{harper66} in 1966 (and by Hart with the edge--isoperimetric function at \cite{Hart76} in 1976.) Analogous results were obtained for cartesian products of chains by Bollob\'as and Leader \cite{BL90} and Bezrukov \cite{bezrukov90}, cartesian products of even cycles by Karachanjan \cite{Karachanjan82} and Riordan \cite{Riordan98} (see also Bezrukov and Leck at \cite{bezrukovleck09}) and some other cartesian products by Bezrukov and Serra \cite{bezrukovserra02}.

The Johnson graph $J(n,m)$ has  the $m$--subsets of $[n]=\{1,2,\ldots,n\}$ as  vertices and two $m$--subsets are adjacent in the graph whenever their symmetric difference has cardinality $2$.  It follows from the definition that, for $m=1$, the Johnson graph $J(n,1)$ is the complete graph $K_n$. For $m=2$ the Johnson graph $J(n,2)$ is the line graph of the complete graph on $n$ vertices, also known as the triangular graph $T(n)$. Thus, for instance, $J(5,2)$ is  the complement of the Petersen graph, displayed in Figure \ref{fig:j52}. Also, $J(n,2)$ is the complement of the Kneser graph $K(n,2)$, the graph which has the $2$--subsets of $[n]$ as vertices and two pairs are adjacent whenever they are disjoint.

\begin{figure}[h]
	\begin{center}
		\begin{tikzpicture}[scale=0.3]

		\foreach \i in {0,...,9}
		{
			\path (36*\i:5) coordinate (P\i);
			\path (36*\i:6) coordinate (Q\i);
			\draw[fill] (P\i) circle (3pt);
		}

		\node  at (Q0) {$01$};
		\node  at (Q1) {$02$};
		\node  at (Q2) {$12$};
		\node  at (Q3) {$13$};
		\node  at (Q4) {$23$};
		\node  at (Q5) {$24$};
		\node  at (Q6) {$34$};
		\node  at (Q7) {$30$};
		\node  at (Q8) {$40$};
		\node  at (Q9) {$41$};

		\foreach \i / \j  in {0/1,1/2,2/3,3/4,4/5,5/6,6/7,7/8,8/9,9/0}
		{
			\draw (P\i) --(P\j);
		}

		\foreach \i / \j  in {0/2,2/4,4/6,6/8,8/0,0/3,3/6,6/9,9/2,2/5,5/8,8/1,1/4,4/7,7/0}
		{
			\draw (P\i) --(P\j);
		}

		\foreach \i / \j  in {1/5,1/7,3/7,3/9, 5/9}
		{
			\draw (P\i) --(P\j);
		}

		\end{tikzpicture}
		\caption{The Johnson graph $J(5,2)$.}

	\end{center}
\end{figure}\label{fig:j52}

Johnson  graphs arise from the association schemes named after Johnson who introduced them, see e.g. \cite{Cam99}.The Johnson graphs are one of the important classes of distance--transitive graphs; see e.g.  Brouwer, Cohen, Neumaier \cite[Chapter 9]{browercohen1989} or  Godsil  \cite[Chapter 11]{godsil1993}.

Given a family $S$ of $m$--sets of an $n$--set, its \textit{lower shadow} $\Delta (S)$ is the family of $(m-1)$--sets which are contained in some $m$--set in $S$. The \textit{upper shadow} $\nabla (S)$ of $S$ is the family of $(m+1)$-sets which contain some $m$--set in $S$. The \textit{ball} of $S$ in the Johnson graph $J(n,m)$ can be written as
\begin{equation}
\label{deltas}
B(S)=\nabla(\Delta (S))=\Delta (\nabla (S)).
\end{equation}

These equalities establish a connection between the isoperimetric problem in the Johnson graph with the Shadow Minimization Problem (SMP) in the Boolean lattice, which consists in finding, for a given $k$, the smallest cardinality of $\Delta (S)$ among all families $S$ of $m$--sets with cardinality $k$. The latter problem is solved by the well--known Kruskal--Katona theorem \cite{Kruskal63,Katona68}, which establishes that the initial segments in the colex order provide a family of extremal sets for the SMP.

Recall that the {\it colex order} in the set of $m$--subsets of $[n]$ is defined as $X\le Y$ if and only if $\max ((X\setminus Y)\cup(Y\setminus X))\in Y$ (we follow here the terminology from Bollob\'as \cite[Section 5]{bollobas}; we also use ${[n]\choose m}$ to denote the family of $m$--subsets of an $n$--set, and $[k,l]=\{k,k+1,\ldots ,l\}$ for integers $k<l$.) The computation of the boundary of initial segments in the colex order (the family of the first $m$-- subsets in this order) provides the following upper bound for the isoperimetric function of Johnson graphs:

\begin{proposition}\label{prop:upper} Let $\mu_{n,m}: [N]\to {\mathbb N}$ denote the isoperimetric function of the Johnson graph $J(n,m)$, where $N= {n\choose m}$. Let
	$$
	k={k_0\choose m}+{k_1\choose m-1}+\cdots +{k_r\choose m-r},\;  k_0> \cdots > k_r\ge m-r>0,
	$$
	be the $m$--binomial representation of $k$. Then
	\begin{equation}\label{eq:ub}
	\mu_{n,m}(k)\le f(k,n,m),
	\end{equation}
	where
	\begin{equation}\label{eq:ub1}
	f(k,n,m)=
	\binom{k_0}{m-1}(n-k_0)+\sum_{i=1}^{r}\left(\binom{k_i}{m-i-1}(n-k_0-1)-\binom{k_i}{m-i}\right).
	\end{equation}
\end{proposition}

\begin{proof} The initial segment $I$ of length $k$ in the colex order is the disjoint union
	$$
	I=I_0\cup\cdots \cup I_r,
	$$
	where $I_0$ consists of all $m$--sets in ${[k_0]\choose m}$ and, for $j>0$, $I_j$ consists of all sets containing $\{k_{j-1}+1,\ldots, k_0+1\}$ and $m-j$ elements in $[k_j]$. The right hand side of \eqref{eq:ub1} is the cardinality of $\partial I$ as can be shown by induction on $r$. If $r=0$ then $\partial I$ consists of the ${k_0\choose m-1}(n-k_0)$ sets obtained by replacing one element in $[k_0]$ by one element in $[k_0+1,n]$ from a set in $I$. Suppose that $r>0$ and write $I=I'\cup I_r$ as the disjoint union of $I'=I_0\cup \cdots \cup I_{r-1}$ and $I_r$. We have $\partial I=(\partial I'\setminus I_r)\cup (\partial I_r\setminus B(I'))$, the union being disjoint. Since $I_r\subset \partial I'$ we have,
	$$
	|\partial I'\setminus I_r|=|\partial I'|-|I_r|=|\partial I'|-{k_r\choose m-r},
	$$
	while
	$$
	|\partial I_r\setminus B(I')|={k_r\choose m-r-1}(n-k_0-1),
	$$
	since the only sets in $\partial I_r\setminus B(I')$ are those obtained from a set in $I_r$ by replacing one element in $[k_r]$ by one element in $[k_0+2,n]$.
\end{proof}

The family of initial segments in the colex order  does not provide in general a solution to the isoperimetric problem in $J(n,m)$. A simple example is as follows.

\begin{example}
	\label{example1}
	Take $n=3(m+1)/2$. The ball $B(\{{\mathbf x}\})$ of radius one in $J(3(m+1)/2,m)$ has cardinality
	$$
	|B_1|=1+m(n-m)=\frac{(m+2)(m+1)}{2}={m+2\choose m},
	$$
	and its boundary has cardinality
	$$
	|\partial B_1|={m\choose 2}{n-m\choose 2}=\frac{m(m-1)(m+3)(m+1)}{16}.
	$$
	On the other hand, according to \eqref{eq:ub1} and the $m$--binomial decomposition of $|B_1|$, the initial segment $I$ of length $|B_1|$ has cardinality
	\begin{align*}
	|\partial I|&={m+2\choose m-1}\frac{m-1}{2}=\frac{(m+2)(m+1)m(m-1)}{12}\\&=|\partial B_1|+\frac{(m+1)m(m-1)^2}{48},
	\end{align*}
	which shows that the unit ball can have, as a function of $m$, an  arbitrarily smaller boundary than the initial segment  in the colex order
\end{example}

In his monograph on discrete isoperimetric problems  Leader \cite{leader91}  mentions   the isoperimetric problem for Johnson graphs as one of the intriguing open problems in the area. Later on, in his extensive monograph on isoperimetric problems, Harper \cite{Harper04} atributes the problem to Shapozenko, and recalls that it is still open. Recently, Christofides, Ellis and Keevash \cite{keevash} have obtained a lower bound for the isoperimetric function of Johnson graphs  which is asymptotically tight for sets with cardinality $\frac{1}{2}{n\choose m}$. The Johnson graphs $J(n,2)$ provide a counterexample to a conjecture of Brouwer on the $2$--restricted connectivity of strongly regular graphs, see Cioab\^a, Kim and Koolen  \cite{cioaba2012} and  Cioab\^a,  Koolen  and Li\cite{cioaba2014}, where the connectivity of the more general class of strongly regular graphs and distance--regular graphs is studied. It is also worth mentioning that the edge version of the isoperimetric problem, where the minimization is for the number of edges leaving a set of given cardinality, has also been studied, see e.g.  Ahlswede and Katona \cite{ahslwedekatona78} or Bey \cite{bey2006}. We will only deal with the vertex isoperimetric problem in this paper and refer to it simply as the isoperimetric problem.

Our main purpose in this paper is to show that the initial segments in the colex order still provide a solution to the isoperimetric problem in $J(n,m)$  for many small values of $k$, thus providing the exact value of the isoperimetric function in these cases.

We call a set $S$ of vertices of $J(n,m)$ {\it optimal} if $|\partial (S)|=\mu_{n,m}(|S|)$.   Our first result shows that initial segments in the colex order are optimal sets  in $J(n,2)$.

\begin{theorem}\label{thm:m=2} For each $n\ge 3$ and each $1\le k\le {n\choose 2}$ we have
	$$
	\mu_{n,2}(k)=f(k,n,2).
	$$
	In particular, the initial segments in the colex order are optimal sets of $J(n,2)$ for each $n\ge 3$.
\end{theorem}

The following theorem allows one to show that  the inequality \eqref{eq:ub} is also tight in $J(n,m)$ for very small sets.

\begin{theorem}\label{thm:m+1} For $k< m-1$ and $n\ge 2(m-1)$ the initial segment of length $k$ of the colex order in $J(n,m)$ is an optimal set.
\end{theorem}

Our last result, Theorem \ref{thm:asymptotic}, extends Theorem \ref{thm:m+1} in an asymptotic way, by showing that the inequality \eqref{eq:ub} is tight for  a large number of small cardinalities and gives a lower bound for all small cardinalities.

\begin{theorem}\label{thm:asymptotic} Let $k, m$ be positive integers and let
	$$
	k={k_0\choose m}+{k_1\choose m-1}+\cdots +{k_r\choose m-r},\;  k_0> \cdots > k_r\ge m-r>0,
	$$
	be the $m$--binomial representation of $k$.

	There is $n(k,m)$ such that, for all $n\ge n(k,m)$, the following holds.
	\begin{enumerate}
		\item[{\rm (i)}] If $r<m-1$ then
		$$
		\mu_{n,m}(k)=f(k,n,m),
		$$
		and  the initial segment in the colex order with length $k$ is the only (up to automorphisms) optimal set with cardinality $k$ of the Johnson graph $J(n,m)$.
		\item[{\rm (ii)}]
		If $r=m-1$ then
		$$
		\mu_{n,m}(k)\le f(k-k_r,n,m)+k_r.
		$$
	\end{enumerate}
\end{theorem}

The proof of Theorem \ref{thm:asymptotic} provides the  estimation
$$
n(k,m)\le m+k+1-\mu_{m+k+1,m}(k)+f(k,m+k+1,m)
$$
for the value of $n(k,m)$ above for which the statement of Theorem \ref{thm:asymptotic} holds. This upper bound for $n(k,m)$ is not tight but we  make no attempt to optimize its value in this paper.

Example \ref{example1} shows that the initial segment in the colex order of length ${m+2\choose m}$ can fail to be an optimal set in $J(n,m)$ if  $n=3(m+1)/2$. In the last section  we describe another infinite family of examples for which the initial segment in the colex order  fails again to be  an optimal set in $J(n,m)$ for every fixed $m$  and all $n$ large enough.

\begin{proposition}\label{prop:nocolex} Let $m$ be a positive integer. For each  integer $k$ of the form
	$$
	k={t\choose m}+3{t\choose m-1}
	$$
	with $t$ sufficiently large with respect to $m$
	there is a set $S$ with cardinality $k$  such that
	$$
	|B(S)|<f(k,n,m)
	$$
	for all $n\geq t+3$.
\end{proposition}

When $n=t+3$ the set $S$ in Proposition \ref{prop:nocolex} can be easily described as the ball in $J(n,m)$ of ${[t]\choose m}$, the family of $m$--subsets of the first $t$ symbols. Such sets are clear candidates to be optimal sets. The examples described in Proposition \ref{prop:nocolex} are closely related to the non--unicity of solutions to the Shadow Minimization Problem in the Boolean lattice (see Theorem \ref{thm:fg} below.)

Standard compression techniques are used to prove the above results. These tools fall short to solve the isoperimetric problem of Johnson graphs in full mainly because, as pointed out in \cite{keevash}, for instance, optimal sets in Johnson graphs do not have the nested property (the ball of an optimal set is not optimal.) However these techniques are still useful to show that the colex order provides a sequence of extremal sets for small cardinalities.

The paper is organized as follows. Section \ref{sec:comp} recalls the shifting techniques and compression of sets. The proofs of theorems \ref{thm:m=2}, \ref{thm:m+1} and \ref{thm:asymptotic} are given in sections \ref{sec:m2}, \ref{sec:m+2} and \ref{sec:nlarge} respectively. In the proof of Theorem \ref{thm:asymptotic} we use a result by F\"uredi and Griggs \cite{furedigriggs} which characterizes the cardinalities for which the Shadow Minimization Problem for the Boolean lattice has unique solution. The statement below is a rewriting of a combination of Proposition 2.3 and Theorem 2.6 in \cite{furedigriggs}.

\begin{theorem}[\cite{furedigriggs}]\label{thm:fg} Let
	$$
	k={k_0\choose m}+{k_1\choose m-1}+\cdots +{k_r\choose m-r},\;  k_0> \cdots > k_r\ge m-r>0,
	$$
	be the $m$--binomial representation of $k$.

	The initial segment in the colex order is the unique (up to automorphisms) solution to the Shadow Minimization Problem in the Boolean lattice  if and only if $r<m-1$.
\end{theorem}

Finally in Section \ref{sec:nonoptimal} we prove Proposition \ref{prop:nocolex}. The result  describes an infinite family of examples which show that the initial segments in colex order may fail to be optimal sets. The nature of this example shows that the isoperimetric problem in Johnson graphs still has many intriguing open questions to be solved.

\section{Shifting techniques}\label{sec:comp}

Shifting techniques are one of the key tools in the study of set systems. They were initially introduced in the original proof of the Erd\H os--Ko--Rado theorem \cite{ekr} and have been particularly used in the solution by Frankl and F\"uredi \cite{franklfuredi81}  of the isoperimetric problem for hypercubes.

In what follows we identify subsets of $[n]$ with their characteristic vectors   ${\mathbf x}=(x_1,\ldots ,x_n)\in \{0,1\}^n$ where  $x_i=1$ if $i$ is in the corresponding set and $x_i=0$ otherwise. We denote the {\it support} of ${\mathbf x}$ by
$$
\s{{\mathbf x}}=\{ i: x_i=1\}.
$$
and the $\ell_1$-{\it norm} of ${\mathbf x}$ by
$$
|{\mathbf x}|=\sum_i x_i.
$$
The support $\s{S}$ of   $S\subset \{0,1\}^n$ is the union of the supports of its vectors. We often identify  a set $S\subset \{0,1\}^n$ with the subset in $\{0,1\}^{n'}$, $n'>n$, obtained  by adding zeros to the right in the coordinates of its vectors. Thus, the initial segment  of length $k$ is considered to be a subset of $\{0,1\}^n$ for each sufficiently large $n$.

The sum
$
{\mathbf x}+ {\mathbf y}=(x_1+y_1 \pmod 2,\ldots ,x_n+y_n\pmod 2),
$
of characteristic vectors is meant to be performed in the field ${\mathbb F}_2^n$ and it corresponds to the symmetric difference of the corresponding sets.  We also denote by ${\mathbf e}_1,\ldots ,{\mathbf e}_n$ the unit vectors with $1$ in the $i$--th coordinate and zero everywhere else.

With the above notation, the set of vertices of the Johnson graph $J(n,m)$ are all vectors of $\{0,1\}^n$ with norm $m$, and the neighbors of ${\mathbf x}$ in $J(n,m)$ are the vectors
$$
{\mathbf x}+{\mathbf e}_i+ {\mathbf e}_j,
$$
for each pair $i, j$ such that $x_i+x_j=1$.

We next recall the definition of the shifting transformation.

\begin{definition} Let $i,j\in [n]$. For a set $S\subset \{0,1\}^n$ define
	$$
	S_{ij}=\{ {\mathbf x}\in S: x_i=1 \;\mbox{ and } \; x_j=0\},
	$$
	and
	$$
	T_{ij}({\mathbf x},S)=\left\{\begin{array}{ll} {\mathbf x}+{\mathbf e}_i+{\mathbf e}_j, &\mbox{if}\;  {\mathbf x}\in S_{ij} \; \mbox{and }\; {\mathbf x}+{\mathbf e}_i+{\mathbf e}_j\not\in S\\
	{\mathbf x}&\mbox{otherwise}\end{array}\right.
	$$
	The $ij$--shift of $S$ is defined as
	$$
	T_{ij}(S)=\{ T_{ij}({\mathbf x},S):{\mathbf x}\in S\}.
	$$
\end{definition}

It follows from the definition that the shifting  $T_{ij}$ of a set preserves its cardinality and the norm of its elements. Moreover,  it sends every vertex to a vertex at distance at most $1$.  The main property of the shifting transformation is that it does not increase the cardinality of the ball of a set. This property follows from the analogous ones for upper and lower shadows. We include a direct proof here for completeness.

\begin{lemma}\label{lem:contrac}
	Let $i,j\in [n]$  and write $T=T_{ij}$. For each set $S$ of vertices in the Johnson graph $J(n,m)$ we have
	\begin{equation}\label{eq:contrac2}
	B(T(S))\subseteq T(B(S)).
	\end{equation}
	In particular,
	\begin{equation}\label{eq:contrac3}
	|B(T(S))|\leq |T(B(S))|=|B(S)|.
	\end{equation}
\end{lemma}

\begin{proof} We will show that,
	\begin{equation}\label{eq:contract4}
	\mbox{ for each } \; {\mathbf y}\in T(S), \; \mbox{  we have }  B({\mathbf y})\subseteq T(B(S)),
	\end{equation}
	which is equivalent to  (\ref{eq:contrac2}). We observe that then (\ref{eq:contrac3}) follows since $|B(X)|=|X|+|\partial X|$ for every subset $X$ and
	$$
	|\partial (T(S))|=|B(T(S))|-|T(S)|\le |T(B(S))|-|T(S)|=|B(S)|-|S|=|\partial S|.
	$$

	Let ${\mathbf x}$ be the element in $S$ such that ${\mathbf y}=T({\mathbf x},S)$. We consider two cases.

	{\it Case 1.} $(x_i,x_j)\neq (1,0)$. In this case we certainly have ${\mathbf y}={\mathbf x}$. Moreover, if ${\mathbf z}\in \partial{\mathbf x}$ such that $(z_i,z_j)=(1,0)$ then it is readily checked that ${\mathbf z}+{\mathbf e}_i+{\mathbf e}_j\in B({\mathbf x})$. Therefore the transformation $T(\cdot,B(S))$ leaves  $B({\mathbf x})$ invariant. Hence, $B({\mathbf x})\subseteq T(B(S)).$

	{\it Case 2.} $(x_i,x_j)= (1,0)$.  Then ${\mathbf z}={\mathbf x}+{\mathbf e}_i+{\mathbf e}_j$  is the only neighbour of ${\mathbf x}$ with $(z_i,z_j)=(0,1)$.

	{\it Case 2.1} ${\mathbf y}={\mathbf x}$.  Then, by the definition of $T(\cdot,S)$, we have  ${\mathbf z}\in S$ and $T({\mathbf z},S)={\mathbf z}$. Observe that every neighbour ${\mathbf z}'$ of ${\mathbf x}$ is  left invariant by $T(\cdot,B(S))$.
	This is clearly the case if   $(z'_i,z'_j)\neq (1,0)$ and, if $(z_i,z_j)= (1,0)$, because  we then have ${\mathbf z}''={\mathbf z}'+ {\mathbf e}_i+{\mathbf e}_j\in B({\mathbf z})\subset B(S)$. Hence
	$$B({\mathbf y})=B({\mathbf x})=\cup_{{\mathbf z}\in B({\mathbf x})} T( {\mathbf z}, B(S))\subseteq T(B(S)).$$

	{\it Case 2.2} ${\mathbf y}\neq {\mathbf x}$. Then ${\mathbf y}\not\in S$ but ${\mathbf y}\in B({\mathbf x})\subseteq B(S)$. Each neighbour ${\mathbf z}$ of ${\mathbf y}$ distinct from ${\mathbf x}$ is of the form ${\mathbf z}={\mathbf z}'+ {\mathbf e}_i+ {\mathbf e}_j$ for some neighbour ${\mathbf z}'$ of ${\mathbf x}$ and therefore it belongs to $T(B(S))$. For ${\mathbf x}$ itself we have $T({\mathbf x},B(S))={\mathbf x}$ because ${\mathbf y}={\mathbf x}+{\mathbf e}_i+{\mathbf e}_j\in B(S)$. Thus we again have $B({\mathbf y})\subset T(B(S))$. This completes the proof of \eqref{eq:contract4}
\end{proof}

The weight of a vector ${\mathbf x} \in \{0,1\}^n$ is
$$
w({\mathbf x})=\sum_{i=1}^n ix_i,
$$
and the weight of  a set $S$ is
$$
w(S)=\sum_{{\mathbf x}\in S} w({\mathbf x}).
$$
We note that, if $i>j$ then $w(T_{ij}(S))\le w(S)$. Moreover, equality holds if and only if $T_{ij}(S)=S$. Thus,  successive application of transformations $T_{ij}$ using pairs  $i,j$  with $i>j$ eventually produces a set which is stable by any of such transformations. This fact leads to the following definition.

\begin{definition}\label{def:compresed}  We say that a set $S$ is \textit{compressed} if $T_{ij}(S)=S$ for each pair $i,j\in [n]$ with $i>j$.
\end{definition}

Every  set can be compressed by keeping its cardinality and without increasing its boundary. Therefore, in what follows  we can restrict our attention to compressed sets in our study of optimal sets.

\section{The case $m=2$}
\label{sec:m2}

Theorem \ref{thm:m=2} follows from the following proposition which characterizes compressed optimal sets in $J(n,2)$.

\begin{proposition} A set $S$ of vertices of the graph $J(n,2)$ with cardinality $\binom{t-1}{2}<|S|\leq \binom{t}{2}$  is optimal if and only if, up to isomorphism,  $S\subseteq\binom{[t]}{2}$.
\end{proposition}

\begin{proof} Write $V(J(n,2))$ as the disjoint union
	$$
	V(J(n,2))=S\cup \partial S\cup \tilde{S},
	$$
	where $\tilde{S}$ is the set of vertices at distance two from  $S$. Let $t=\left|\s{S}\right|$ be the amount of elements of $[n]$ in the support $\s{S}$ of $S$. The only vectors in $\tilde{S}$ are the ones which have both nonzero coordinates in $[n]\setminus \s{S}$. Therefore
	$$
	|\tilde{S}|={n-t\choose 2},
	$$
	and
	$$
	|\partial S|={n\choose 2}-|S|-{n-t\choose 2}.
	$$
	Hence, for a given cardinality $|S|$, $|\partial S|$ is a increasing function of $t$ alone. The optimal value is therefore obtained when $t$ is smallest possible, which is the smallest $t$ such that $|S|\le {t \choose 2}$. This is achieved for any subset $S\subset {[t]\choose 2}$ if $|S|>{t-1\choose 2}$
\end{proof}

As a consequence of the above proposition, we can see that the solution to the isoperimetric problem in $J(n,2)$  is unique (up to isomorphism) for sets of cardinality ${t\choose 2}$ (and also for sets of cardinality $\binom{t}{2}-1$).

\section{Small sets}
\label{sec:m+2}

In this section we prove Theorem \ref{thm:m+1}.
Consider the partition
$$
V(J(n,m))=\{\mathbf{x}\in V(J(n,m)) |  x_n=0\} \cup \{\mathbf{x}\in V(J(n,m)) |  x_n=1\}=V_0\cup V_1.
$$
The subgraph of $J(n,m)$ induced by $V_0$ is isomorphic to $J(n-1,m)$ and the subgraph induced by $V_1$is isomorphic to $J(n-1,m-1)$. There is an edge in $J(n,m)$ joining ${\mathbf x}\in V_0$ with ${\mathbf y}\in V_1$ if and only if $\s{{\mathbf y}}\setminus\{n\}\subset \s{{\mathbf x}}$.

\begin{lemma}\label{lem:section} Let  $S$ be a   set of vertices  in $J(n,m)$. Let $S_0=S\cap \{ x_n=0\}$ and $S_1=S\cap \{ x_n=1\}$. If $S$ is compressed then
	$$
	B(S)=B(S_0)\; \mbox{ and } |B(S)|= |B'(S'_0)|+ |\Delta (S'_0)|,
	$$
	where $S'_0=\{ {\mathbf x}\in \{0,1\}^{n-1}: ({\mathbf x}, 0)\in S_0\}$ and $B'$ denotes the ball in $J(n-1,m)$.
\end{lemma}

\begin{proof}  Let ${\mathbf x} \in S_1$ and $i\not\in  \s{{\mathbf x}}$. Since $S$ is compressed, we have
	$$
	{\mathbf y}={\mathbf x}+ {\mathbf e}_n + {\mathbf e}_i\in S_0,
	$$
	which implies ${\mathbf x}={\mathbf y}+ {\mathbf e}_n+ {\mathbf e}_i\in B(S_0)$. Hence, $S_1\subset B(S_0)$. Moreover, if $j\in \s{{\mathbf x}}$ then
	$$
	{\mathbf x}+ {\mathbf e}_i + {\mathbf e}_j=({\mathbf x}+ {\mathbf e}_n + {\mathbf e}_i) + {\mathbf e}_n + {\mathbf e}_j\in B(S_0),
	$$
	so that $B({\mathbf x})\subset B(S_0)$.
	Hence $B(S)=B(S_0)\cup B(S_1)= B(S_0)$. This proves the first part of the sentence.

	For the second part, we just note that $B(S_0)$ is the disjoint union $(B(S_0)\cap V_0)\cup (B(S_0)\cap V_1)$. Since the subgraph induced by $V_0$ is isomorphic to $J(n-1,m)$, we have  $|B(S_0)\cap V_0|=|B'(S'_0)|$. On the other hand,  there is an edge in $J(n,m)$ joining $x\in V_1$ with $y\in V_0$ if and only if $\s{x}\setminus\{n\}\subset \s{y}$. It follows that $B(S_0)\cap V_1=\Delta (S'_0)$
\end{proof}

The next proposition considers the case of Johnson graphs $J(n,m)$ when $n=2m-2$.

\begin{proposition}\label{prop:2m-2}
	For each $k=1,\ldots ,{2m-2 \choose m}$ the initial segment of length $k$ in   the colex order  is an optimal set of the graph  $J(2m-2,m)$.
\end{proposition}

\begin{proof}
	Let us  recall \eqref{deltas} and express the ball of a set $S$ as
	$$B(S)=\nabla(\Delta (S)).$$
	By the Kruskal--Katona theorem, $\Delta S$ is minimized by the initial segment $I_k$, $k=|S|$,   in the colex order of ${[2m-2]\choose m}$. Moreover, $\Delta I_k$ is also an initial segment of the colex order in ${[2m-2\choose m-1}$.

	The automorphism of the Boolean lattice given by taking complements sends  the middle level ${[2m-2\choose m-1}$ to itself, and the initial segments in the colex order are exchanged by the initial segments in the lexicograhic order. It follows that the initial segments of the colex order are also a solution to the minimization of the {\it upper} shadow (as well as the initial segments in the lexicographic order). Hence, $I_k$ minimimizes $|\nabla(\Delta (S))|$.
\end{proof}

We use the Proposition \ref{prop:2m-2} as the base case of the induction for  the proof of Theorem \ref{thm:m+1}.

\begin{proof}[of Theorem \ref{thm:m+1}.] The proof is by induction on $n$. For $n=2m-2$ the result follows from Proposition \ref{prop:2m-2}. Let $S$ be a compressed set of cardinality $k\le m-1$ in $J(n,m)$, $n\ge 2m-1$ and consider its decomposition $S=S_0\cup S_1$. Since $S$ is compressed, every element in  $S_1$ gives rise to at least $m-1$ elements in $S$. Since  $k<m-1$ we have $S_1=\emptyset$.

	By Lemma \ref{lem:section}, the cardinality of the ball of $S$ is $$|B(S)|= |B'(S'_0)|+ |\Delta (S'_0)|. $$ By the induction hypothesis, the initial segment in the colex order minimizes the ball $|B'(S'_0)|$ in $J(n-1,m)$ as well as, by teh Kruskal--Katona theorem, the lower shadow  $|\Delta (S'_0)|$.
\end{proof}

\section{Optimal sets for large $n$}\label{sec:nlarge}

In this Section we give the proof of Theorem \ref{thm:asymptotic}. In what follows we call a positive integer $k$ {\it critical} if its $m$--binomial representation has length $m$ (namely, it has $r=m-1$).

\begin{proof}[of Theorem \ref{thm:asymptotic}.]  Let $S$ be an optimal set with cardinality $k$ in $J(n,m)$. We may assume that $S$ is compressed. Let $n_0$ be such that the support of every element in $S$ is contained in $[n_0]$. Since $S$ is compressed, if the support of ${\mathbf x}\in S$ contains $n_0$ then  we have $T_{n_0i}({\mathbf x})\in S$ for each $i\in [n_0-1]\setminus \s{{\mathbf x}}$. It follows that $n_0\le m+k+1$.  For each $n\ge n_0$ every element in $\Delta (S)$ gives rise to $n-n_0$ distinct vectors in $\partial S$ which have a coordinate in $[n_0+1,n]$ and therefore are disjoint from the ball $B_0(S)$ in $J(n_0,m)$. Moreover every two such vectors which only differ in their coordinate from $[n_0+1,n]$ come from a unique element in $\Delta(S)$. Therefore, we have
	$$
	|B(S)|=|B_0(S)|+(n-n_0)|\Delta S|,
	$$
	where $B_0$ denotes the ball of $S$ in $J(n_0,m)$ and $B$ denotes the ball of $S$ in $J(n,m)$. Similarly, for $I$ be the initial segment of length $k$ in the colex order. We have
	$$
	|B(I)|=|B_0(I)|+(n-n_0)|\Delta I|
	$$
	Hence,
	\begin{equation}\label{eq:compare}
	|B(S)|=|B(I)|+(|B_0(S)|-|B_0(I)|)+(n-n_0)(|\Delta (S)|-|\Delta (I)|).
	\end{equation}
	If $|\Delta (S)|>|\Delta (I)|$ then we have $|B(S)|>|B(I)|$ for each sufficiently large $n$. By  Theorem \ref{thm:fg}, if the $m$--binomial representation of $k$ has less than $m$ terms then the initial segment in the colex order is the unique solution to the Shadow Minimization Problem. It follows that, if $S\neq I$ then $|B(S)|>|B(I)|$ for all
	$n>n_0-|B_0(S)|+|B_0(I)|$. This proves the first part of Theorem \ref{thm:asymptotic} and gives the estimate
	$$
	n(k,m)\le m+k+1-\mu_{m+k+1,m}(k)+f(k,m+k+1,m).
	$$

	On the other hand, we have $\mu_{n,m}(k)\le \mu_{n,m}(k+1)-1$, since otherwise an optimal set $X$ with cardinality $k+1$ satisfies $|\partial (X\setminus \{x\})|<\mu_{n,m} (k)$ for every $x\in X$, a contradiction with the definition of $\mu_{n,m}$. Suppose that $k$ is a critical integer and let $k,k-1,\ldots ,k-\ell+1$ be the longest decreasing sequence of critical integers. By the above remark we have  $\mu_{n,m} (k)\le \mu_{n,m}(k-\ell)+\ell$ and $\mu_{n,m} (k-\ell)=f(k-\ell,n,m)$, the cardinality of the boundary of an initial segment in colex order with length $k-\ell$. The value of $\ell$ is clearly
$k_r$. This proves the second part of the statement.
\end{proof}

\section{Initial segments which are not optimal}\label{sec:nonoptimal}

We conclude the paper by proving Proposition \ref{prop:nocolex}, which shows that there are values of $k$  for which  the initial segment of length $k$ in the colex order fails to be an optimal set of $J(n,m)$ for all sufficiently large $n$.

We prove first that, for each $m$ and each integer $t$ sufficently large with respect to $m$,
$$
g(t,m)={t\choose m}+3{t\choose m-1}
$$
is a critical cardinality, namely, the $m$-binomial expansion of $g(t,m)$  has $m$ terms. This means that the solution of the Minimal Shadow Problem  is not unique for $k=g(t,m)$. This fact is  used in the proof of Proposition \ref{prop:nocolex}.

\begin{lemma}\label{lem:gtm} There is an infinite strictly increasing integer sequence $\{\lambda_i\}_{i\geq 0}$, $\lambda_i+1<\lambda_{i+1}$ such that, for each $t$ and each $m\ge 1$,
	\begin{equation}\label{eq:gtm}
	g(t,m)=\sum_{i=0}^{m-1} {t-\lambda_i\choose m-i}+1.
	\end{equation}
\end{lemma}

\begin{proof} By induction on $m$. For $m=1$ we have
	\begin{equation}\label{eq.first_ind_step}
	g(t,1)=t+3={t+2\choose 1}+1,
	\end{equation}
	and for $m=2$,
	\begin{equation}\label{eq.second_ind_step}
	g(t,2)={t+2\choose 2}+{t-2\choose 1}+1,
	\end{equation}
	giving $\lambda_0=-2$ and $\lambda_1=2$.
	By using ${n\choose m}=\sum_{j=0}^{n-1}{j\choose m-1}$ and induction, for   $m\ge 3$ we have
	\begin{align*}
	g(t,m)=&{t\choose m}+3{t\choose m-1}\\
	=&\sum_{j=0}^{t-1}{j\choose m-1}+3\sum_{j=0}^{t-1}{j\choose m-2}\\
	=&\sum_{j=0}^{t-1}g(j,m-1)\\
	=&\sum_{j=0}^{t-1} \left(\sum_{i=0}^{m-2}{j-\lambda_i\choose m-1-i}+1\right)\\
	=&\sum_{i=0}^{m-2}\sum_{j=0}^{t-1}{j-\lambda_i\choose m-1-i}+t.
	\end{align*}
	By using $\lambda_0=-2$ and $\lambda_i>0$ for $i\in[1,m-2]$, we can write
	\begin{align*}
	g(t,m)=&\sum_{j=0}^{t-1}{j-\lambda_0\choose m-1}+\sum_{i=1}^{m-2}\left(\sum_{j=\lambda_i}^{t-1}{j-\lambda_i\choose m-1-i}+\sum_{j=0}^{\lambda_i-1}{j-\lambda_i\choose m-1-i}\right)+t\\
	&={t-\lambda_0\choose m}+\sum_{i=1}^{m-2}{t-\lambda_i\choose m-i}+t+\sum_{i=1}^{m-2}\sum_{j=0}^{\lambda_i-1}{j-\lambda_i\choose m-1-i},
	\end{align*}
	which shows that \eqref{eq:gtm} holds with
	\begin{align}
	\lambda_{m-1}=&-\sum_{i=1}^{m-2}\sum_{j=0}^{\lambda_i-1}{j-\lambda_i\choose m-1-i}+1\nonumber \\
	=&-\sum_{i=1}^{m-2}\sum_{\ell=1}^{\lambda_i}{-\ell\choose m-i-1}+1 \nonumber \\
	=&-\sum_{i=1}^{m-2}\sum_{\ell=1}^{\lambda_i}(-1)^{m-i-1}{m-i-2+\ell\choose m-i-1}+1\nonumber \\
	=&\sum_{i=1}^{m-2}(-1)^{m-i}{m-i-1+\lambda_i\choose m-i}+1\nonumber \\
	=&\sum_{j=2}^{m-1}(-1)^{j}{\lambda_{m-j}+j-1\choose j}+1 \label{eq.l_def}
	\end{align}
	We observe that the sequence is uniquely determined once $\lambda_1$ is fixed.
	The first values of the sequence are
	$$
	-2,2, 4,7,14,51,928, 409625, \ldots
	$$
	It remains to show that the sequence is increasing. We will in fact show that $\lambda_m\ge \max\{ \lambda_{m-1}+2,\lambda_{m-1}^2/4\}$ for all $m\ge 2$.
	The above inequality holds for $m\le 7$ as shown by the first values of the sequence. By using \eqref{eq.l_def} (with $\lambda_m$ instead of $\lambda_{m-1}$), we have,
	$$
	\lambda_{m}\ge \sum_{j=2, j \text{ even}}^{2\lfloor m/2\rfloor -1}\left({\lambda_{m-j+1}+j-1\choose j}-{\lambda_{m-j}+j\choose j+1}\right).
	$$
	For $j=2$ we have
	\begin{align*}
	{\lambda_{m-1}+1\choose 2}-{\lambda_{m-2}+2\choose 3}&=\frac{\lambda_{m-1}^2}{4}+\frac{\lambda_{m-1}(\lambda_{m-1}+2)}{4}-{\lambda_{m-2}+2\choose 3}\\&\geq \frac{\lambda_{m-1}^2}{4}+\frac{\lambda_{m-2}^4}{64}+\frac{2\lambda_{m-2}^2}{16}-{\lambda_{m-2}+2\choose 3}\\
	&>\frac{\lambda_{m-1}^2}{4}
	\end{align*}
	where the last inequality holds  (for $m\ge 4$) because the largest root of the polynomial $x^4/64+x^2/8-{x+2\choose 3}$ is smaller than $\lambda_4=14$.

	On the other hand, for $j\ge 4$,
	\begin{align*}
	{\lambda_{m-j+1}+j-1\choose j}&-{\lambda_{m-j}+j\choose j+1}=\frac{1}{j!}\left(\prod_{t=0}^{j-1} (\lambda_{m-j+1}+t)-\frac{\prod_{t=0}^j(\lambda_{m-j}+t)}{j+1}\right)\\
	&\stackrel{ \lambda_{m-j+1}> \lambda_{m-j}+1}{>}\frac{1}{j!}\prod_{t=2}^j(\lambda_{m-j}+t)\left(\lambda_{m-j+1}-\frac{\lambda_{m-j}(\lambda_{m-j}+1)}{j+1}\right)\\
	&\stackrel{ \lambda_{m-j+1}\ge \lambda_{m-j}^2/4}{\ge}\frac{1}{j!}\prod_{t=2}^j(\lambda_{m-j}+t)\left(\frac{\lambda_{m-j}^2}{4}-\frac{\lambda_{m-j}(\lambda_{m-j}+1)}{j+1}\right).
	\end{align*}
	The right-hand side is nonnegative if $m-j\ge 2$, as then  $\lambda_{m-j}\ge 4$. If $m-j=1$ then it follows by induction on $j\ge 1$ that
	$$
	{\lambda_2+j-1\choose j}-{\lambda_1+j\choose j+1}={3+j\choose j}-{2+j\choose j+1}>0.
	$$
	This completes the proof.
\end{proof}

For $t$ larger than $\lambda_{m-1}$ the equality \eqref{eq:gtm} in Lemma \ref{lem:gtm} provides the $m$--binomial expansion of $g(t,m)$. Hence this binomial  expansion has length $m$ and, by Theorem \ref{thm:fg}, $g(t,m)$ is a critical cardinality.
The proof of  Proposition \ref{prop:nocolex} uses this fact by choosing two distinct optimal sets for the SMP problem which have different boundaries in the Johnson graph.

\begin{proof} {\it of Proposition \ref{prop:nocolex}}
	Let  $S=B_0({[t]\choose m})$, the ball of ${[t]\choose m}$ in the Johnson graph $J(n_0,m)$, $n_0=t+3$. The cardinality of $S$ is
	$$
	k={t \choose m}+3{t\choose m-1}=g(t,m).
	$$
	Let $I(k)$ denote the initial segment of length $k$ in the colex order.

	By Lemma \ref{lem:gtm} the $m$-binomial expansion of $k$ has $m$ terms and can be written as
	\begin{equation}
	{t \choose m}+3{t\choose m-1}={t-\lambda_0 \choose m} + {t-\lambda_1 \choose m-1} + \ldots + {t-\lambda_{m-1} \choose 1}.
	\label{expansion}
	\end{equation}
	We note that the shadows of $S$ and of $I(k)$ have the same cardinality:
	\begin{align*}
	|\Delta S|&={t \choose m-1}+3{t\choose m-2}\\&={t-\lambda_0 \choose m-1} + {t-\lambda_1 \choose m-2} + \ldots +{t-\lambda_{m-2}\choose 1}+ {t-\lambda_{m-1} \choose 0}\\&=|\Delta (I(k))|.
	\end{align*}

	It can be readily checked that the boundary of $S$ in $J(n_0,m)$ has cardinality
	$$
	|\partial S|=\left|\partial^2 {[t] \choose m}\right|=3{t \choose m-2},
	$$
	On the other hand, the boundary in $J(n_0,m)$ of the initial interval $I(k)$ as given by the function $f(k,n_0,m)$ in \eqref{eq:ub1} is
	\begin{align*}
	|\partial I(k)|=&{t-\lambda_0 \choose m-1}+\sum_{i=1}^{m-1} \left({t-\lambda_i\choose m-i-1}(n-t+\lambda_0-1)-{t-\lambda_i \choose m-i}\right)\\
	=&{t+2 \choose m-1}-\sum_{i=1}^{m-1} {a_i \choose m-i}={t+2 \choose m-1}+{t+2\choose m} -{t\choose m}-3{t\choose m-1}\\
	=&{t+1\choose m-2}+2{t\choose m-2},
	\end{align*}
	which is strictly larger than $3{t\choose m-2}$. Thus $|B_0(I(k))|>|B_0(S)|$. Moreover, by \eqref{eq:compare}, for all $n\ge n_0$, we have
	$$
	|B(I(k))|=|B(S)|+(|B_0(I(k))|-|B_0(S)|)>|B(I(k)|.
	$$
	Therefore the intial segment in colex order $I$ fails to be an optimal set for all $n\ge t+3$.
\end{proof}

\section*{Acknowledgements}
 The authors are grateful to the comments and remarks of the referees, pointing out some inaccuracies in the original manuscript and helping to improve the readability of the paper.

\bibliographystyle{abbrv}

redpajama_set_name: RedPajamaArXiv


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text:
O2 to furnish Dell Streak with Android 2.1 in lieu of Froyo
The Dell Streak will be getting an update to its Android OS but it won't be version 2.2, or Froyo, as previously announced on the company's official blog.
Written by Ben Woods, Contributor on Aug. 6, 2010
The update will be provided by O2, which currently has an exclusive on the Streak, and will bring the device up to version 2.1 of the Android operating system, according to the operator.
"O2's software testing team is currently working on bringing Android 2.1


Courtesy The Plush Family (CINCINNATI) — Kyle Plush, the Ohio teen who died after becoming trapped in his minivan last week, will be remembered as a "beloved" friend "whose heart was just as big as his smile," his high school said in a statement.
Students at The Seven Hills School in Cincinnati created a collection of hand-painted stones on Sunday, honoring the 16-year-old with colorful drawings, special messages and words of inspiration, according to a statement posted on the school's Facebook page.
"We are all grieving the loss of this beloved member of our school family — a young person of keen intelligence, good humor, great courage — and we feel this loss profoundly," the school said in a statement obtained by ABC affiliate WKRC on Sunday.
Plush suffocated last Tuesday while trapped under the third-row seat of his Honda Odyssey minivan, according to police. Investigators believe the seat may have flipped and pinned him as he was trying to reach for items in the back of the vehicle.
The Cincinnati teen called 911 at least twice on the afternoon of April 10, saying he was trapped and in need of immediate assistance.
"I am trapped inside a gold Honda Odyssey van in the parking lot of Seven Hills. … Send officers immediately. I'm almost dead," Plush said in the second phone call he made.
"I probably don't have much time left, so tell my mom I love her if I die," he said in another call.
An operator repeatedly asked Plush where he was, but the teen could not hear the dispatcher and wasn't able to answer the operator's questions, police said. According to the audio recordings of the 911 calls it appears as though Plush was using Siri, Apple's voice command system, to make the calls.
Police were dispatched to Seven Hills High School, where the teen said he was located, but authorities could not find him. A family member found the teenager unresponsive in the minivan hours later.
The teen's parents have demanded more information about how he died.
Plush's parents also thanked members of the media for asking questions about their son's death.
The Cincinnati Police Department said it plans to conduct a full investigation of the incident.


text:
NSW Production Sports Racing Cars Association Australia
Mawer/Lazaris win Hour Enduro
JACK PHELAN IS LORD OF THE RIVER DANCE
Substantial Growth For NSW Production Sports Car Championship In 2014
Campbell and Sparks clean sweep Morgan Park
Campbell and Sparks outshine opposition in Queensland
There can only be one. NSW ProdSports heads into the highlands for round five of the 2014 Championship.
AMSF NSW Drive Day for 2014
NEWS NEWS NEWS...!!!!
A Fitting tribute
The Allan Simonsen Pole Position Award
IMAGE: Jack Phelan wouldn't allow the wet conditions to stop him taking three victorys, clean sweeping the final round of the NSW Production Sports Car Championship.
Credit: Gary Rogers www.garyrogers.com.au
In absolutely miserable conditions at Gouldburn's Wakeifeld Park circuit, Jack Phelan has taken home a round victory as the NSW CAMS Production Sports Car championship state series came to a close last weekend. With a rounds racing condensed into a one day format at Wakefield, the Production Sports Car category was unlucky with its draw, taking to the circuit on Sunday morning right as rain began to fall.
However, Phelan refused to allow the conditions to get the better of him, a masterful wet weather performance securing him pole position for the opening race. Qualifying second was Glenn Townsend, half a second off Phelan's best time, in his Lotus Elise. Rounding out the top three was Mark Stinson, who was looking to have a quieter weekend, standing on the verge of taking home the championship crown.
IMAGE: Phelan leads through turn two as Townsend tries to pressure the Porsche driver into a mistake.
Credit: Gary Rogers, www.garyrogers.com.au
In race one a determined Townsend, who missed out on a certain class win in the dying stages during the last round at Sydney Motorsport Park, was intent on making life as difficult as possible for Phelan. "He didn't get me by much in qualifying, I'm going to make it as hard as possible on him in this race," said Towsend before the race. Off the line the Lotus driver got the jump and managed to lead out of turn two, however, soon after, Phelan re-asserted his intentions, taking the lead and driving away to what would ultimately be the first of three comfortable victories on the day.
VIDEO: Ride onboard with Glenn Townsend as he goes for a wild trip onto the infield in race two at Wakefield Park.
Credit: Blend Line TV
As the day wore on, track conditions continued to deteriorate, and with each return to the track for NSW Production Sports Car competitors, racing became more and more difficult. Both Lotus of Townsend and Magaitis struggled in the rain, with each taking trips onto the infield during the second and third races. Townsend suffering the biggest moment of the weekend on his way through turn six, losing traction before completing three rotations on the infield. If Production Sports had decided to undertake a diving competition, Townsend's perfect execution of the rotations while splashing through puddles, would have earnt him full marks. Unfortunately, the race to the chequered flag was the goal, and Townsend's effort handed Phelan a comfortable victory.
IMAGE CREDIT: Gary Rogers, www.garyrogers.com.au
Despite race three seeing the worst conditions of the weekend, Phelan yet again kept his composure to drive a controlled effort to victory. Not done with showing he was the class of the field all weekend, Phelan punched out the fastest lap of the race on the last lap. Quicker than his qualifying effort by 0.8s it served to show that he is a force to be reckoned with in Production Sports Cars. "It was great to come away with the victory today. Really hard out there, but found a rhythm and brought it home. Huge thanks to the crew and all my sponsors with the Bob Jane T-Marts Blacktown car, looking forward to the rounds ahead now," enthused Phelan after the race.
IMAGE: Mark Stinson did enough in the first two races to secure the 2014 Production Sports Car State Championshiip.
While he didn't take to the course for the final race of the weekend, Mark Stinson had done enough to take home the 2014 NSW Production Sports Car State Championship. The Nissan Silvia driver proving that consistency and strong class results are enough to stay at the top of the table against more fancied machinery.
The NSW Production Sports Car Championship heads to Queensland's Morgan Park circuit for the next round of the Endurance Championship. To see all the action from Wakefield Park tune in to Blend Line TV in the coming weeks. Television times will be advised closer to the air date.


text:
Photo by: Doug Howell/5 On Your Side
Jacob Long
Born just across the river in Granite City, Jacob knew at an early age he wanted to tell the stories of St. Louis.
Author: KSDK Staff
Published: 6:18 AM CST February 16, 2016
Updated: 10:18 AM CST December 11, 2018
Jacob Long is an Emmy and Associated Press Award-winning reporter at 5 On Your Side in St. Louis, Missouri, and specializes in investigations.
This is Jacob's second stint at 5 On Your Side and he is thrilled to be back in his hometown. He originally joined the team in 2015 as a general assignment multi-skilled journalist. In 2017 he moved to Boston where he freelanced as a general assignment reporter for a year at WFXT Boston 25 News. He returned to 5 On Your Side in 2018 as an investigative reporter.
Jacob was born just across the river in Granite City and is a proud graduate of the public school system (Go Warriors!). He knew at an early age he wanted to tell the stories of the St. Louis region.
"I knew as a child I wanted to become a reporter because I liked talking and was generally curious about everything," said Jacob. "As I grew older, I realized as journalists, we get to be the first draft of history. My favorite part is serving and informing my community by reporting on the issues and events that impact them every day."
Throughout his time in news he's had many memorable moments in front of and behind the camera. One story that sticks out is from his time reporting in Florida. He told the story of the birth and successful separation of conjoined twin boys.
"It was such a high-risk and extremely rare situation for this small family, and they trusted me to tell their story from start to finish," said Jacob. "Being at the hospital the day they were delivered and successfully separated was one of the most emotional and meaningful days of my life."
Jacob is a graduate of Illinois State University where he earned a B.S. in broadcast journalism. Before moving home, he was a reporter and anchor at WMBD-TV in Peoria, Illinois and our TEGNA sister station First Coast News in Jacksonville, Florida.
"After every story I tell, no matter how long or short, I find something special about it that sticks with me and makes me a better reporter," he said. "I'm in love with St. Louis and a major champion for this city and region. It's such a privilege to be here to tell my community's story."
Jacob is newly married, and the proud father of a miniature golden retriever named Tucker. In his spare time, he enjoys working out, traveling, and eating and drinking his way around the city (hence the working out part).
You can follow Jacob on Facebook, Instagram, and Twitter.


text:
The Golden Ring - Great Scenes from Der Ring des Nibelungen Hybrid SACD
Release Date 28 October 2022
The Golden Ring - Great Scenes from Der Ring des Nigelungen on Hybrid SACD
Decca's landmark recording of Wagner's 15-hour epic, Der Ring des Nibelungen conducted by Sir Georg Solti is the most celebrated classical recording in history.
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The perfect marriage of art and technology, this massive undertaking would not have been greeted with such unprecedented enthusiasm and rave reviews were it not for the skill of Decca's dream team: Producer John Culshaw and Chief Engineer Gordon Parry. They created in sound the ideal 'Theatre in the Mind.' As Culshaw said: "It is precisely because the work is so huge and demanding that one should not pass lightly over any detail."
The recording set new standards for opera in stereo and is still regarded as the benchmark over six decades later
For the first time on Hybrid SACD, Decca proudly present a selection of the greatest scenes from Wagner's Der Ring des Nibelungen conducted by Sir Georg Solti and includes a 24pp booklet, including introductory texts on each opera and many original session photographs.
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Georg Solti, Chicago Symphony Orchestra
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Die Walküre 5LP
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I saw this pop up in my neck of the internet woods recently for like the half dozenth time and was hoping to illuminate some people without googling.
It's Alan Bean. I read one of his books when I was 8, because I loved space and I couldn't believe his last name was real.
His name is pretty easily remembered, so you can stash that in the back of your head forever like I did.


Author's Notes: Another 10-character/15-word fic meme offering. It was supposed to be the "romance" prompt, but that meme has clearly never met Greg House, so this is Gen ("Partners") instead.
"Who did you say you were you again?" House asked the tall young man at the door. He didn't actually care, but he'd found people got annoyed at having to repeat themselves. He figured he was good to ask that question again in about ten minutes.
"Sam Winchester," the guy answered.
"From the phone. You're the ghostbuster."
"No," Sam sighed impatiently, then reconsidered. "Kind of," he amended. "You told the police your neighbor was killed by a ghost."
"Or boredom. That'll definitely do it."
Sam gritted his teeth, thinking Dean would've punched this guy by now. "Did you or did you not—"
House lurched across the room, waving at Sam to follow him. "Playing the piano usually brings him out of the woodwork," he said, starting something with a freeform, jazzy sound. He shifted into a new tempo: "But he hates ragtime like you wouldn't believe," he yelled over the music.
There was a loud banging from the floor.
"That's just your downstairs neighbor, complaining about the noise!" Sam protested.
"Well how about this?" House raised his head toward the ceiling and began howling an accompaniment.
A painting suddenly flew across the room, where House ducked it expertly and kept on playing. "Well?"
Sam grinned in spite of himself, knowing he would probably regret it later. "Yeah," he nodded. "I think I can work with that..."
LOL but where was Dean?
Dean was off in a different part of the same meme, meeting up with Cameron (from Terminator:SCC), and then Iron Man (which I swear I will post up separately as soon as I can think of a title for the darn thing).


text:
Biografia
Lasciata la scuola a 16 anni, Wynne (Winifred) Gibson trovò lavoro come ballerina di fila a Broadway e poi divenne la partner del ballerino Ray Raymond nella sua compagnia itinerante, che si esibì anche in Europa. Nel 1929 riuscì a entrare nell'ambiente del cinema con il film Nothing But the Truth, il cui successo le garantì un contratto dalla Paramount e una carriera più che ventennale, per quanto svoltasi interpretando film di non elevata qualità. Lasciò il cinema nel 1956 e divenne una dirigente sia della Actors' Equity, il sindacato degli attori cinematografici degli Stati Uniti, che dell'American Federation of Television and Radio Artists.

Nel 1927 aveva sposato l'attore John Gallaudet, dal quale divorziò nel 1930. Nel 1938 Wynne Gibson andò a vivere con l'attrice Beverly Roberts, in un sodalizio che durò tutta la sua vita, conclusa nel 1987 per un attacco di trombosi cerebrale. È sepolta nel Forest Lawn Memorial Park di Glendale, accanto all'amica Roberts.

Filmografia parziale
 Nothing But the Truth (1929)
 Amore bendato (Children of Pleasure), regia di Harry Beaumont (1930)
 The Fall Guy (1930)
 Luna di giugno  (1931)
 Lo stroncatore di gang  (1931)
 Man of the World, regia di Richard Wallace e Edward Goodman (1931)
 Le vie della città  (1931)
 I gioielli rubati (1931)
 Two Kinds of Women, regia di William C. de Mille (1932)
 Night After Night, regia di Archie Mayo (1932)
 Il delitto di Clara Deane (The Strange Case of Clara Deane), regia di Louis J. Gasnier, Max Marcin  (1932)
 Se avessi un milione  (1932)
 Cuore d'amanti (Lady and Gent), regia di Stephen Roberts  (1932)
 La guardia del corpo (Her Bodyguard), regia di William Beaudine (1933)
 Troppa armonia  (1933)
 The Crime of the Century  (1933)
 The Crosby Case (1934)
 Admirals All (1935)
 The Crouching Beast (1935)
 Come Closer, Folks (1936)
 La miniera misteriosa  (1937)
 Racketeers in Exile (1937)
 Michael O'Halloran (1937)
 The Falcon Strikes Back (1943)
 Mystery Broadcast (1943)

Altri progetti

Collegamenti esterni

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text:
1911 Encyclopædia Britannica/Sofala
< 1911 Encyclopædia Britannica
←Sofa
1911 Encyclopædia Britannica, Volume 25
Soffioni→
See also Sofala on Wikipedia; and our 1911 Encyclopædia Britannica disclaimer.
19630751911 Encyclopædia Britannica, Volume 25 — Sofala
SOFALA, a Portuguese seaport on the east coast of Africa, at the mouth of a river of the same name, in 20° 12′ S. Pop. (1900), about 1000. The town possesses scarcely a trace of its former importance, and what trade it had was nearly all taken away by the establishment of Beira (q.v.) a little to the north in 1890. Sofala Harbour, once capable of holding a hundred large vessels, is silting up and is obstructed by a bar. Ruins exist of the strong fort built by the Portuguese in the 16th century. Previous to its conquest by the Portuguese in 1505 Sofala was the chief town of a wealthy Mahommedan state, Arabs having established themselves there in the 12th century or earlier. At one time it formed part of the sultanate of Kilwa (q.v.). Sofala was visited by the Portuguese Jew, Pero de Covilhão, in 1489, who was attracted thither by the reports of gold-mines of which Sofala was the port. The conquest of the town followed, the first governors of the Portuguese East African possessions being entitled Captains-General of Sofala. (See Portuguese East Africa.) Thomé Lopes, who accompanied Vasco da Gama to India in 1502 and left a narrative of the voyage (first printed in Ramusio, Viaggi e Navegationi) , identifies Sofala with Solomon's Ophir and states that it was the home of the Queen of Sheba. This identification of Sofala with Ophir, to which Milton alludes (Par. Lost, xi. 399–401) is untenable.
The small island of Chiloane, with a good harbour, 40 m. S. of Sofala, has been colonized from Sofala (the township being named Chingune) as has also the island Santa Carolina, in the Bazaruto archipelago.
See Bull. Geogr. Soc. Mozambique (1882) for an account of the Sofala mines; and, generally, Idrísí, Climate, i. § 8, O. Dapper, Description de l'Afrique (Amsterdam, 1686); T. Baines, The Gold Regions of South Africa (1877); G. McC. Theal's Records of South Eastern Africa (1898–1903); Sir R. Burton's notes to his edition of Camoens.
Retrieved from "https://en.wikisource.org/w/index.php?title=1911_Encyclopædia_Britannica/Sofala&oldid=5990950"
EB1911:Cities:Africa


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Father : How are you?
Daughter: Fine. Thank you papa.
Father : Don't say thank you baby, say quite well.
Father : Where is your hair?
Daughter: (Pointing) It's here papa!
Father : What's the colour of your hair?
Father : How many eyes do have you?
Father : How many hands do have you?
Father : How many noses do have you?
Daughter: Noses! no papa nose. I do have only one nose.
Father : How many teeth do have you?
Daughter: I can't count them.
Father : Can you see them?
Daughter: No papa, I can feel them.
Father : Alright show your tongue.
Father : How many ears do have you?
Father : How many fingers do have you?
Father : How baby? There ten fingers.
Daughter: No papa, there are two thumbs.
Father : But there are fingers in your feet also.
Daughter: No Papa they are all toes.
Father : Your are a clever girl.
Daughter: Of course, I am.
Father : Won't you go to school today?
Daughter: No papa I am not feeling well.
Father : Have you given an application?
Daughter: Papa please write on for me.
Father : Why didn't inform your principal on the telephone?
Daughter: She will not believe me Papa.
Father : Where is your mummy?
Daughter: She is in the kitchen.
Daughter: How do I know Papa?
Father : Go and see what she is doing?
Daughter: O.K. Papa wait I am coming.
Father : She comes back.
Daughter: She is preparing for breakfast table.
Father : Then let's go for breakfast.


The Cabinet Committee on Economic Policy supported in its meeting of 28 April 2009 the Ministry of Transport and Communications' proposal of incorporating the Finnish Civil Aviation Administration, Finavia, as of 1 January 2010. The business activities currently managed by the Civil Aviation Administration will be transferred to a new company limited by shares that will be wholly owned by the state. The new company will be responsible for the basic infrastructure of Finland's airports and the provision of air navigation services.
Provisions concerning the basic duties of the company will be laid down in aviation laws. A special act will be enacted on the national network of airports, taking into account the new requirements of the EU Directive on airport charges. A special act will also be enacted on pricing and special services in military aviation. Necessary legislative amendments are proposed to enter into force simultaneously with the incorporation process.
The next stage in the incorporation process is for the Ministry of Transport and Communications to draw up a government proposal for an act about the incorporation. The legislative proposal concerns the Government budget for 2010 that will be submitted in September 2009. The incorporating process will not have any direct impacts on central government finances. Measures related to the establishment of the new company will be completed by the end of 2009. Responsibility for corporate governance concerning the new company will rest with the Ministry of Transport and Communications.
Personnel employed by Finavia on 31 December 2009 will continue as employees of the new company under the old terms. Public service employment relationships of Finavia employees will be changed into contractual employment relationships in accordance with the Finavia incorporation act. The retirement plans of the personnel will remain at the same level as before and supplementary pension benefits will be funded through the new company's balance sheet. The process of incorporation will not affect the ongoing restructuring measures at Finavia.
The balance sheet of the new company will be based on Finavia's assets and liabilities of 31 December 2009. The new company will continue Finavia's investment programme as planned, including the construction of a railway station for the ring rail line in the area of the Helsinki-Vantaa airport (estimated cost 30 million euros). In order to improve transport connections to the airport the Finnish Rail Administration will also prepare for the construction of a railway station in Viinikkala.
The process of incorporating the unincorporated state enterprises is based on the European Commission decision of 2007 concerning the Finnish Road Enterprise, which criticised the compatibility of unincorporated state enterprises with the common market. In 2008 the Finnish Ministry of Finance appointed a working group to examine the organisation of unincorporated state enterprises and to find out alternatives for the present model. The working group will complete its work at the end of 2009.


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Q: delete and deallocate vs delete without deallocate I'm reading the documentation of vector. Here I found that:

~vector(); calls allocator_traits::destroy on each of the contained elements, and deallocates all the storage capacity allocated by the vector using its allocator.

I don't understand how can you delete an element without deallocating it.
Going further with destroy (allocator_type& alloc, T* p); :

Destroys the element object pointed by p without deallocating its storage. In the non-specialized definition of allocator_traits, this member function calls alloc.destroy(p) if such a call is well formed. Otherwise, it invokes p->~T().

*

*I don't understand what they mean by destroy the element without deallocating its storage (again).

*How would this involve p->~T() if my vector<T> obj {T(), T()} consists of automatically objects, not pointers?

*How could it call destroy on a T** if my vector is vector<T*>...?

I'm trying to do a parallel between calling the destructor on objects like:

*

*MyClass obj() respectively

*MyClass obj = new MyClass()
vs

*

*vector<T> obj {T(), T()}

*vector<T*> obj {new T(), new T()}
and I can't see how they are similar, because they should be as one contains the other.

A: A vector<T> holds a chunk of memory, and that memory has some number of T objects in them. The vector manages the storage separately from the objects.
When you call v.pop_back(), it destroys the last element in the vector, but does not release the storage occupied by that element. If you then call v.push_back(), it will place the new (last) element in the location previously occupied by the element that you erased.

A:
I don't understand how can you delete an element without deallocating it.

As the next quotation mentions, this is ultimately done through calling its destructor manually. The important part is to do so correctly, as delete will already call it for you.
Minus some details and layers of abstraction, the way vector works is through placement new, which can construct an object in existing storage, separating the memory allocation from the construction. This allows vector to manage its storage separately when resizing as well as to avoid the need for a default constructor (or any other) when there is extra memory with no object there:
char memory[2 * sizeof(T)]; // No Ts yet. Keep in mind alignment is important too.
new(memory) T{/* constructor arguments */}; // One T at the start of the memory, rest unused.
new(memory + sizeof(T)) T{}; // Two Ts next to each other in the memory.
static_cast<T*>(memory)->~T(); // First part of memory now unused again.
// Destroy any other objects appropriately.
// Free store memory would be explicitly deallocated here.

How would this [invoke] p->~T() if my vector obj {T(), T()} consists of automatically objects, not pointers?

The vector provides this function a pointer to that value. The vector knows where in memory the element is and knows there's a T object there, so it can obtain a proper pointer. destroy takes the pointer and falls back to calling the destructor through that pointer. This is safe because the object was constructed in an appropriate manor instead of through new T(…).

How could it call destroy on a T** if my vector is vector...?

It works the same way when the element type is a pointer. The vector obtains a pointer to that element (so a pointer to a pointer in this case) and passes it to the destroy function. The destroy function calls the destructor through that pointer. Now you might be wondering why that works when pointers don't have destructors, but there's a special rule saying that a destructor call on a fundamental type is valid through an aliased type (the name T in this case) and does nothing. This keeps generic code working without different code for different Ts.

A: I'd like to give another view here and focus on the vector of pointers
vector<T*> obj {new T(), new T()}

and the difference between destroying and deallocating the pointers in it. (That's how I interpret your question, but maybe you know all of this already :D)
First of: this vector contains only pointers. A pointer is usually 4 or 8 bytes long. Let's assume it's 8 and the vector contains 2 pointers, so it has to allocate 16 bytes, regardless of sizeof(T). The two objects allocated with new are somewhere else on the heap.
Let's assume this vector get's destroyed, for example because it goes out of scope in some function. It will destroy every element inside the vector and deallocate the memory that was allocated by the vector (e.g. 16 bytes in this example).

*

*Destroying a pointer, doesn't do anything, since a pointer is not much more than a numeric value that specifies a location in memory. Though a T might have a destructor, a T* doesn't have one.

*Deallocating means only to release the memory that is used to store the pointers. The memory those pointers point to is completely untouched.

It's simmiliar to the other example you gave:
MyClass *obj = new MyClass();

If obj goes out of scope the destructor of MyClass will not be called, obj itself (4 or 8 bytes as above) will be deallocated, but the memory obj pointed to won't be touched. This results in a memory leak and the same happens for the vector above.
Only the pointers are gone, the memory they pointed to remains.

It's almost the same for a vector with values
vector<T> obj {T(), T()}

Same scenario as above, but now the objects are stored inside the memory allocated by the vector. Thus the vector must now allocate 2 * sizeof(T) bytes to contain the two objects. When the vector gets destroyed the same as above happens:

*

*Each element get's destroyed. If T has a destructor it is called.

*The memory allocated by the vector (the 2 * sizeof(T) bytes) is deallocated

It's simmiliar to
MyClass obj;

When this goes out of scope the destructor ~MyClass will be called and the memory of obj (sizeof(MyClass) bytes) will be deallocated. No memory leak happens, just like in the vector above.

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