An interface definition language (IDL).
A data abstraction (ADT) language.
A statically-typed OO programming language.
A procedural language (i.e., C).
Efficiency of implementation given common compilation techniques.
C compatibility (at both the source and executable levels).
Interactions among these modes and goals lead to some major complexities. Normally, the best way to cope with these complexities is to adhere to a small number of idiomatic design and programming strategies. These notes lay out some of the more common and useful ones. One of the chief goals is to stay away from dark corners of the language that are easier to avoid that to cope with.


text:
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Youth Space - Health in Brazil
The federal government privatization package: New attack on health, basic health units in the sight of the private sector
By Moises Maciel
BRAZIL | A decree of the president of the republic, in Brazil, allows 'studies' of partnerships for the private sector to operate at the front door of the National Health System (SUS), includes the item in the federal government's private program for investment partnership program. While the term 'studies' can bring tranquility to the reader, the government's intention does not seek the same.
The analysis movement made in this field, let us call it that, from the private sector always aims at profit, where it is feasible to put money, close deal!
The logic behind privatization is not so difficult to understand, in a nutshell if the person has money to pay, uses the service, if not, does not use it. If we talk about any object of consumption, the discussion of private property is plausible, the same does not happen when talking about taking rights, selling an inalienable concept.
According to the government, the goal is to find solutions for the number of basic health units, that are unfinished or that are not in operation. However, after negative repercussions, the president granted a note to CNN Brazil and a new decree is pending that edits the term used. All of these movements spell out the objective and clear interest in privatizing, selling, profiting and segregating.
The Bolsonaro government is permeated by inconsistencies and fake news, by successive attacks on the population for their own benefit at the expense of the people. Revoking the decree, editing it, suspending or creating others does not eliminate the concrete facts, the constant attacks on health, to remedy or protect the population on any scale. While seeking to sell primary care, more and more covid deaths are counted. Amanda died. Bruna died. Mariana. João. Paulo. More and more victims of inhumane management
In conversation with the National Health Council, the agency's president reports a worrying situation and confirms "harsh attacks on our SUS".
An image "DEFEND THE SUS" started to circulate on social networks, related to this government measure and a call for activism to defend rights.
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Some of the menu draws directly from Shah's childhood in Curaçao: "Polenta fries are a snack I just grew up eating on the side of any dish, just like French fries," he said. The rest is a mix of traditional fare and "twists on popular American dishes that we made Caribbean," like the jerk-marinated burger with grilled pineapple, and the fish sandwich with cilantro and chipotle mayo.
For their first proper drink menu, Jify and crew put together five original cocktails based on his experiences at Cameo and his "strong ties with the music scene in Brooklyn," he said. Especially dank are the Losing My Edge (jalapeño-infused vodka, grapefruit juice, lime, maraschino, and smoked salt) and Forget (rum, absinthe, Curaçao, Cocchi Americano, and lime).
Much like those tots, Cameo has gotten some spice and a great deal more flavor, but the chill atmosphere remains. "The biggest thing to me is to make sure we have an unpretentious and cool environment," Jify explained. And sure enough, the restaurant's lively island vibe certainly isn't as over-the-top as that other Caribbean newcomer, Miss Lily's at 7A Cafe. Plus it'll continue to be devoted to the local music community, with an Indie Rock Saturday Brunch and a Disco Sunday Brunch, complete with live DJs.
Though Jify's certainly isn't the cheapest Caribbean fare around (at just under $20 for most entrees, $10-ish for apps, and drinks galore, it would be easy for two to push $100) there are tons of options for cheap eats. During happy hour (Monday through Friday, 5 p.m. to 8 p.m) there will be $2 sliders and jerk chicken wings for just a buck, in addition to cheap dranks.
See the Jify's menu below.


A professional travel writer and editor for the past four decades, Don George has explored the farthest corners of the world. In his new book, The Way of Wanderlust, George's adventurous tales over the last 40 years—climbing Mount Kilimanjaro, ascending Yosemite's Half Dome, a moving homestay experience in Cambodia—come together in one inspiring volume. The former travel editor for the San Francisco Examiner and SF Chronicle, George founded the Wanderlust section of Salon.com and was recently the global travel editor for Lonely Planet Publications. George's work has appeared in numerous publications around the globe, and he is a frequent guest on NPR and CNN, among many other TV and radio outlets. Don't miss the chance to hear this award-winning travel expert reflect upon his global expeditions and reignite your wanderlust.


text:
Smart Wallets: UPI drives uptake of digital payments
The move to demonetise high-value currency notes in November 2016 led to the growth of digital payments in the country. The sudden cash crunch saw consumers and markets taking to digital payments in a big way. This also provided the impetus for the government to develop the Unified Payments Interface (UPI) system, which has been providing the biggest push to digital payments.
The uptake of digital payments continued to grow through 2018. Further, transactions processed and facilitated through mobile phones/cellular devices accounted for the majority of digital payments in the country. According to data from eMarketer, India is the world's fastest-growing proximity mobile payment (PMP) market. PMP refers to scanning, tapping or checking in with a cellular device at the point of sale. As of November 2018, approximately 7.6 per cent of the population or 73.9 million people in India were using mobile payments, a significant increase of 39.7 per cent over the previous year.
tele.net takes a look at key developments in the digital payments space…
Growth trends
According to data from the National Payments Corporation of India (NPCI), for the first time, the volume of total transactions over the government's UPI platform crossed the half a billion mark at 524.94 million transactions during November 2018, up fivefold from 104.841 million in November 2017. The total value of the transactions stood at Rs 822.32 billion, up from a meagre 96.4 billion during the same period. A total of 128 banks were live on UPI as of November 2018.
Gauging the potential of UPI following its success, NPCI launched UPI 2.0 in August 2018 with additional features. These include an overdraft facility, which enables customers to link their overdraft accounts with UPI. It assists in simplifying the process of availing of credit for customers with limited credit history. The new UPI version also has provision for a one-time mandate that allows customers to authorise a transaction and pay later. Another key aspect is the invoice in the inbox feature, which allows merchants to send invoices along with a request to collect so that customers can view the invoice before paying. Lastly, the signed intent and QR feature allows customers to verify the authenticity of merchants while scanning QR codes.
Meanwhile, mobile wallet transactions registered a record high in October 2018, in terms of not only value but also volume. According to data released by the Reserve Bank of India (RBI), approximately 368.45 million transactions, grossing a total of Rs 187.86 billion, were conducted in October 2018, registering an impressive increase over the 324.16 million transactions (amounting to Rs 151.02 billion) recorded in September 2018.
Paytm announced that it had registered over 179 million UPI transactions in the month of October 2018, recording a massive increase of 600 per cent in transactions over the past six months. The company also accounted for approximately 37 per cent of the overall UPI transactions. Paytm has been promoting UPI for offline payments and more than 5 million offline merchants out of its total merchant base of 9.5 million now accept payments through UPI. The company had also earmarked a corpus of $27.07 million for promoting UPI transactions.
Flipkart-owned digital payments platform PhonePe announced that it has crossed a billion transactions since its formation in September 2016. The company claims to have recorded an annual total payments volume (TPV) of $38 billion in September 2018, almost double the $20 billion TPV figure announced in June 2018. The company has been able to enrol more than 100 million users in less than a year since it began operations.
Google India, which launched its mobile payments platform Google Pay in September 2017, announced that it had 25 million monthly active users in September 2018. Approximately 1.2 million businesses have registered on the platform and the company has processed 860 million transactions since the commencement of operations.
The boost in e-commerce has partly provided the ground for driving the uptake of various mobile wallets in the country. E-commerce has been growing steadily in the country and several digital payment platforms have been tying up with e-commerce vendors to facilitate payments. Companies such as Flipkart and Amazon have launched their respective payment platforms, PhonePe and AmazonPay. E-commerce companies in partnership with mobile wallets have been offering lucrative discounts and special offers to acquire first-time customers.
The digital wallets and payments space is bound to gain further traction as WhatsApp seeks RBI's approval for the formal launch of its payment services in the country. Currently, NPCI and RBI are holding discussions on whether WhatsApp is fully compliant with India's data localisation rules. With a captive user base of over 200 million, the entry of WhatsApp will make the space more competitive.
With its notification of the interoperability guidelines, the RBI has given the green signal to users of prepaid payment instruments (PPIs) such as mobile wallets for transferring funds between mobile wallets. Interoperability permits payment transactions between different mobile wallets and enables users to transfer those funds to their personal bank accounts.
According to the RBI, interoperability is being introduced in a phased manner. In the first phase, interoperability of wallets will be enabled through UPI. In the second phase, transactions will be allowed between bank accounts and wallets through UPI. This facility will allow users to transfer money from mobile wallets to their bank accounts without incurring any additional charges. Finally, in the third phase, in instances where PPIs are in the form of debit cards, the cards will be affiliated to authorised card networks such as Visa, Rupay and MasterCard. Further, PPI issuers in segments like food and gift vouchers, and transport can be a part of this interoperability.
The ability to issue cards places mobile wallets neck and neck with payments banks, the underlying difference between them being that payments banks have a deposit limit of Rs 100,000 and pay interest on deposits whereas the monthly wallet balance limit for mobile wallets is capped at Rs 10,000 and no interest is paid on the balance amount. However, these guidelines are not mandatory for companies and the final decision rests with them.
The new guidelines, once implemented, will be beneficial for both merchants and users, as they will obviate the need for them to sign up for multiple wallets, and help them to easily receive and make payments via any single wallet.
Aadhaar-based authentication
Initially, the use of e-KYC using Aadhaar reduced the time and cost involved in enrolling new customers. This enabled companies to bring a large number of users on board rather quickly. However, in September 2018, the Supreme Court termed the use of Aadhaar-based authentication by private firms as unconstitutional and held that it can be used only by the government for social welfare schemes. In light of this, mobile wallet companies had to move to an offline verification mechanism, which increased their operational cost for enrolling new customers significantly.
Going forward, the growth in the number of smartphone users will prove to be the single biggest driver of digital payments. According to Cisco's 13th Annual Visual Networking Index, the number of smartphone users in India is expected to double from 404.1 million in 2017 to 829 million by 2022, and account for around 60 per cent of the population, up from 27 per cent in 2017. The report also highlights that India is the fastest growing smartphone market in the world, expanding at a compound annual growth rate of 23 per cent. This presents a huge opportunity for players in the digital payments space to enter the Indian market (WhatsApp being a key example). Further, once interoperability between mobile payment wallets and banks becomes seamless, the digital payments space could see unstoppable growth. According to Credit Suisse, India's digital payments industry, which was worth around $200 billion in February 2018, is expected to grow to around $1 trillion by 2023.
By Aditya Kumar


The prophecies tag has no usage guidance.
What Biblical prophecies appear to have been fulfilled already?
Can bad prophecies be erased?
I've heard that G-d's (or neviim's) good decrees/prophecies can't be erased, but bad ones can if G-d sees that we repent. Is this a well-believed concept? What are the sources of this?
What lines of neviim are generally understood as referring to the final redemption (as opposed to the return in the 2nd temple period)?
I was wondering if the Messiah who is prophesied by many has already to come or are we to continue looking?
Is there a method for evaluating authenticity of revelation to other nations?
Prophecies of the Maccabees uprising are foretold in which ancient book?
Were Shadrach, Meshach, and Abednego told that there would be no miracle?
Is there a way to prove that a prophet is true?
In Sefer Yona 3:4 why was it so easy to convince the city of Ninveh to do teshuva (5 words!) if they were so evil?
When Hashem appeared to Moshe at the burning bush in parshot shamot perek gimil why did he call Moshe's name twice? He could have just said "moshe" ONCE!


text:
Дворец () — деревня, центр Дворецкого сельсовета Рогачёвского района Гомельской области Республики Беларусь.

География

Расположение
В 29 км на запад от районного центра и железнодорожной станции Рогачёв (на линии Могилёв — Жлобин), в 150 км от Гомеля.

Гидрография
На реке Добосна (приток реки Днепр).

Транспортная сеть
На автодороге Бобруйск — Гомель. Планировка состоит из 3 прямолинейных меридиональных улиц, соединённых прямолинейной широтной улицей. Застройка преимущественно двусторонняя, деревянная и кирпичная, усадебного типа. В 1986 году построены 108 домов коттеджного типа, в которых разместились переселенцы из мест, загрязнённых радиацией в результате катастрофы на Чернобыльской АЭС.

История
По письменным источникам известна с XVIII века как селение в Речицком повете Минского воеводства Великого княжества Литовского. После 2-го раздела Речи Посполитой (1793 год) находилась в составе Российской империи. В 1816 году — в Рогачёвском уезде Могилёвской губернии. Помещик владел в 1826 году 104 десятинами земли. Чуть позже была центром одноимённого поместья, его хозяин дворянин Быковский владел в 1858 году 1319 десятинами земли. Во 2-й половине XIX века в центре деревни построена деревянная Троицкая церковь. С 1880 года действовали хлебозапасный магазин, почтовая станция, в Тихиничской волости Рогачёвского уезда Могилёвской губернии. Согласно переписи 1897 года действовали 3 фольварка, школа грамоты, винный магазин, мельница, телеграф. В 1907 году работало народное училище (52 ученика).

С 20 августа 1924 года — центр Дворецкого сельсовета Рогачёвского района Бобруйского (до 26 июля 1930 года) округа, с 20 февраля 1938 года — Гомельской области. В 1930 году организован колхоз «Коммунар», работала ветряная мельница. Во время Великой Отечественной войны оккупанты создали в деревне свой опорный пункт, разгромленный партизанами. В боях за деревню и окрестности 26 июня 1944 года погибли 56 советских солдат и 8 партизан, в их числе Герой Советского Союза Ф. И. Ульянин (похоронены в братской могиле в центре деревни). 21 жителей не вернулись с фронта.

В 1969 году к деревне присоединён посёлок Хуреевка, в 1980 году — деревня Мыза и посёлок Новый Дворэц. Центр совхоза «Дворец». Расположены комбинат бытового обслуживания, средняя школа, Дом культуры, библиотека, больница, аптека, детский сад, ветеринарный пункт, отделение связи, столовая, магазин.

В состав Дворецкого сельсовета входили до 1943 года посёлок Сухое Болото (сожжённый карателями 26 мая 1943 года), до 1969 года — посёлки Марс, Хуреевка, до 1975 года — посёлок Лучинский, до 1980 года — посёлки Заболоцкий, Новый Дворэц, деревни Мыза, Новые Филипковичи, до 1983 года — посёлок Затишье (в настоящее время не существуют).

Население

Численность
 2004 год — 229 хозяйств, 649 жителей.

Динамика
 1816 год — 19 дворов, 114 жителей.
 1897 год — 42 двора (согласно переписи).
 1959 год — 218 жителей (согласно переписи).
 1997 год — 239 дворов, 673 жителя.
 2004 год — 229 хозяйств, 649 жителей.

См. также
 Городские посёлки Белоруссии
 Города Белоруссии

Примечания

Литература

 Гарады і вёскі Беларусі: Энцыклапедыя. Т. 2, кн. 2. Гомельская вобласць / С. В. Марцэлеў; Рэдкалегія: Г. П. Пашкоў (галоўны рэдактар) і інш. — : БелЭн, 2005. — 520 с.: іл. — 4000 экз. — ISBN 985-11-0330-6, ISBN 985-11-0302-0.

Ссылки

Населённые пункты Рогачёвского района

redpajama_set_name: RedPajamaWikipedia


Interested in becoming a member of MOMS?
Membership is open to mothers of multiple birth children or virtual multiple children* or expectant mothers of multiples or virtual multiples residing in Colorado.
*Virtual multiples are defined as unrelated children born within 9 months of each other and who are in the same family through birth or adoption.


As long-time fans of mine know, this frolicsome terrier spent many happy years living in San Pedro. How delightful to return to this quirky port city for my first bookstore signing! Williams Book Store is the oldest bookstore in the Los Angeles area: 97 years in business and going strong. I enjoyed a great afternoon catching up with old friends and making new ones.


Dental implants are mainstream these days. Most people know about this option for replacing a missing tooth. It's easy to assume they've been around a long time, but that's not the case. Various interesting types of implants have been around since ancient times, but they wouldn't exactly qualify as the same technology that goes into the implants we place at Gentle Dental.
Since we like our patients to know as much about their dental health as possible, here's some extra information on these superstars of the dental world.
Archeology has shown us that humans were never big fans of missing teeth, whether the fixes were for function or aesthetics. Archeological digs have unearthed implanted seashells and ivory in the jawbone of ancient Mayans and Egyptians. Who wouldn't want a coaching shell sitting in there between their natural teeth? We could just drive over to the Shore to get all we need!
Those ancient attempts at tooth replacement aren't really in the same category as today's high-tech implants. The modern implant can be traced back to 1952. Swedish orthopedic surgeon, Per-Ingvar Branemark, was studying bone healing and regeneration. He inserted a titanium screw into a rabbit tibia to mend a break but found that when he tried to remove the screw later the bone had fully grown around it and it couldn't be removed. A decade of research followed and the modern dental implant debuted in 1965.
The team at Gentle Dental believes dental implants are the best solution to replace missing teeth, whether it is a tooth that is already gone or a tooth that is so badly damaged or decayed that it requires extraction.
Most of us know someone who has opted to not replace a missing tooth. And it's not just that uncle from Hoboken. Many people are missing at least one tooth. So? What's the big deal? The big deal is the alignment of your teeth because when you don't replace a tooth the teeth on each side tend to slide over into the gap, messing up your alignment and your bite. And that bites!
If you're missing a tooth, let's fill the gap with the star of the dental world, the dental implant. Call us at 732-549-5660 and let's talk about your options.


text:
Newest NO MORE family violence bus wrap unveiled
The newest 'NO MORE' domestic and family violence bus wrap has been unveiled.
Minister for Men's Policy Bess Price said the bus wrap is a great way to raise awareness that domestic and family violence will not be tolerated.
"The bus wrap signage delivers high impact brand awareness not only through sheer size but also thanks to the high visibility of the bus," she said.
NO MORE campaign founder Charlie King said this year we have increased our visual presence in an effort to breakdown the shame that comes with talking about domestic violence issues.
"With a new campaign identity our key focus is to better engage with community. We've opened a Facebook account, and have developed a new website encouraging people to sign up as campaign supporters.
"We want more people, especially sporting clubs to engage in meaningful conversation to break the domestic violence cycle that continues to haunt us.
"Working with over 30 clubs across various sporting codes we are particularly proud of Parramatta Eels for launching the first domestic violence action plan at a national level.
"Action plans encourage clubs to think about what they can do to take a stand against violence.
"I want to thank the Northern Territory Government for their continued support of the NO MORE campaign," Mr King said.


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text:
The week in corruption, 10 May 2019
Transparency Int'l 10 May 2019
Inequality and corruption
This week, we're looking at Panama and South Africa, where corruption and inequality have been at the top of the agenda in recent elections.
While Panama's economic growth is outpacing most of its neighbours, it has one of the highest inequality rates in the Americas. At the same time, the country has been engulfed in various corruption scandals over the past years, including the Odebrecht bribery scheme, wiretapping by former president Martinelli and, of course, the 'Panama Papers'. This historic investigation revealed Panama's role in hiding wealth for the global rich and powerful — while vast parts of its own population, especially in rural areas, continue to live in poverty.
Panama's newly elected president has pledged to address inequality and corruption, and our chapter will be keeping a close eye on whether he is delivering.
In South Africa, 25 years after the end of Apartheid, inequality levels are among the highest in the world: The richest 10 per cent own over two thirds of the wealth, and more than half of the country lives in poverty.
Corruption is rampant across all levels of government, and this only makes matters worse: it affects income distribution, the use of aid flows as well as decisions about public expenditure. Inequality, in turn, can perpetuate corruption, as the wealthy few protect their privilege through dodgy deals and illegal practices.
This week, the party of ex-president Zuma clung on to power in a general election. They should finally deliver concrete anti-corruption measures. But high-ranking members face corruption allegations, and our chapter's pre-election analysis showed that anti-corruption commitments in most parties' manifestos were vague at best.
Governments in Panama, South Africa, and everywhere else for that matter, should prove that they work for the whole population. Tackling corruption is a good place to start.
What do you think? Let us know @anticorruption!


in the agrochemical industry and requirements from federal management bodies.
РН (water, salt), K2O, P2O5, Gum, Mn, Zn, Cu, Co, Mo, B, S, CuT, PbT, ZnT, CdT, etc.
bodies and based on regulatory documents applicable in the agrochemical industry is used.


text:
Tales from the back of the eyelids
Edwin Hubble discovered that galaxies are moving away from each other (*1)
Scientists have assumed this means the universe is expanding
It is more likely, though, that the size of the universe stays the same
Instead, everything inside the universe is shrinking
Our solar system, our planet and its inhabitants will one day shrink to nothing
We don't need to worry as this will happen billions of years from now
The professor was in full flow. He had his back to the class and was vigorously covering the whiteboard with green squiggles (that is to say, mathematical equations) from his marker pen. As he wrote he talked to the board, and here and there between his mumbles we could make out isolated words: integral, tangent, viscosity, parabola. Occasionally he would pause at an equals sign and give a little titter. Once or twice he let out a loud chortle. What the joke was amongst those algebraic proofs none of the students ever knew. We would give each other baffled sidelong glances - but not for long, in case we got left behind in the copying.
To keep up we had to write at a frantic rate. In the rush it was easy to lose our place and often by mistake we would recopy much of what we had already copied. The most precarious moment would come just after the professor reached the bottom right corner of the board. Then, with three or four merciless sweeps of the eraser, he would wipe out all the squiggles to make room for more. In the blink of an eye crucial formulae would be forever lost to us.
Unsurprisingly our lecture notes were completely incomprehensible. Of course we could have tried to decifer them later by cross-referencing with each other. But was there any point? We didn't even know what the course was about (apart from being about maths) as the professor never at any point announced either the subject or the individual topics. Any attempt to make sense of the notes would be a waste of time which would only lead to greater confusion. And there was so little time available to waste, what with drinking and the other enjoyments a student should have. In my view the best strategy was to cram from textbooks a week before the finals.
Even though I might seem sanguine about the prospect of the upcoming exams, this kind of dream is always filled with a heavy sense of foreboding. That's because I know what's coming next and I can't avoid it: soon I will find myself sitting in a big hall with the exam paper in front of me and I'll be trying to read what is written on it. I have three hours to answer five questions. But I am struggling to understand what is being asked of me. The problem is I keep on losing my place because the questions are constantly moving around on the paper. And when I do manage to pin one down, the letters are blurred and difficult to make out. Eventually I recognise some words here and there: integral, tangent, viscosity, parabola. At last I have some hooks on which I can hang my answer. I reel off calculations and draw elaborate diagrams. After a while, though, I realise I am in a hopeless muddle because I have forgotten what it is I'm trying to prove. I tell myself I'll go back and sort out the mess later. In the meantime I'll attempt another question. I look up at the clock and realise I have only half-an-hour left in which to answer four questions. In panic I jump from one question to another, covering the pages in squiggles, including in them everything I know which has anything remotely to do with the subject. Algorithms, functions and graphs swirl around in my head as I write. Everything is getting more and more confused and I am getting more and more frenzied - until the bell rings for the end of the exam, when I wake up covered in sweat.
That, anyway, is what usually happens. However on one particular night a couple of years ago the dream turned out differently. I am in the classroom again but for once I not trying to keep up. It has finally dawned on me that copying out squiggles is pointless. Instead I sit and gaze at the back of the professor as he mumbles into the board. For the first time I notice how scruffy he looks. He is wearing a baggy jumper and ill-fitting trousers and his unkempt white hair is sticking out in all directions. I am surprised to see he isn't wearing any shoes or socks. "He must be cold like that," I think. "If I saw him in the street I'd think he was a vagrant."
Suddenly, as if he could read my wandering thoughts, the professor spun round and wagged his forefinger at me. "Philip," he said. "You must pay attention! Why am I making all this effort to explain this to you, when you can't be bothered to concentrate?"
Astonished, I looked at the man reprimanding me and even more astonished I realised he was Albert Einstein.*2 Seeing that I was taken aback, he softened his tone and continued in a more friendly manner.
"You see," he said. "You must listen because I need you to tell the world about my new theory." In the long years since his death, he told me, he had had plenty of time to reflect in that other place and had come to some conclusions he had never reached while he was alive. It was all to do with the expansion in the first nanoseconds after the Big Bang and was fundamental to understanding all else in our universe from black holes and curved space-time to strings and extra dimensions. In those first moments of expansion lay the unified field theory and therefore, in essence, the link between all theories of the universe.
Einstein was bursting with excitement at his new ideas - and who would not have been? The answer had eluded him all his life and he was desperate to pass it on to me, in order that I, in turn, could pass it on to mankind. He turned to the board again and started to scribble down more equations. This time I made a serious effort to concentrate. Einstein's mathematical argument was a series of complex concepts and yet, despite this, his general drift was beginning to dawn on me. It was an amazing and pleasing feeling: I, who had always longed to understand the mathematics of theoretical physics and yet always found it mystifying was, at last, being let into its secrets. I was getting a glimpse and was beginning to see. I made copious notes, mostly for posterity, so that there would be a record, but also for myself, so that later I could refine my own understanding of the great man's thoughts.
Unfortunately, when the lecture was over and I woke up, all my notes were lost, as you might expect after a dream. I felt agitated, almost feverish: I had to remember what Einstein had told me, to pass his insight on to everyone. I went to the bathroom and sat down on the toilet. With my head in my hands, I recounted the main points to myself and tried to refresh the connections between them. But it was not easy; the dream was already beginning to fade.
"Damn it!" I cursed. "Of all the people he could have chosen, why did he choose me?"
Why indeed? It would not be accurate to call me a mathematical genius. Actually, it would be very inaccurate to call me that. I struggled with the subject throughout my education. That doesn't mean, though, that mathematics is beyond me. In fact Einstein might have picked me because the opposite is true: I am beyond mathematics. My mind is not stuck in a groove like an academic's and I am open to unconventional possibilties. Ask any mathematician what maths is and invariably you will be told it is truth. But ask them to prove that maths is truth, then they are stumped. Not that this failure will shake their faith in their assumptions – which is why, when the rest of us believe there must be a flaw in the theory, they will continue to insist that cats can be alive and dead at the same time.*3
"Ah," you might think to yourself. "If he is sceptical about mathematics, why does he think it's important to remember what Einstein told him? Einstein's theories also seemed barmy when they were first published and even today the notion of bending time is still hard to grasp. What's so special about Einstein?"
It's a valid point: if I didn't believe that Einstein has something significant to say, why would I bother to communicate his ideas? Let's suppose that instead of Einstein, my neighbour Mr Krumpflecker had disturbed me in the middle of the night, all excited, to explain how he had cracked the mysteries of time, space and matter. What would I have done? Of course I'd have called the psychiatric hospital and ask them to come and deal with him. But because it was Einstein I sat up and took note. When it come to theories of the universe, Einstein inspires in me more respect than Mr Krumpflecker does.
But why? If I have never fully understood either Einstein's or Mr Krumpflecker's theories, how come I value Einstein's more? The answer is not reassuring: I believe in Einstein because other people who I believe in, believe in him. In other words: blind faith. The same blind faith which causes societies to choose crooks as leaders. Hardly an adequate recipe for finding the truth. But what else can I do? You can't expect me to become an expert in theoretical physics. In any case, the most important thing is that Einstein's theories of General and Special Relativity have been proved to be true. (At least, that's what the people I believe in believe.)
So I sat on the toilet, trying to piece together what Einstein had told me. I had been given the daunting responsibility of revealing his new theory. Here's what I can tell you: the events of the early universe, in the first nanoseconds after the Big Bang, are determined by a quadratic equation (ie. of the type y = ax2 + bx + c). E is proportional to the component c2-17m. In the quadratic form of the expression this component is found as (c2-17m)2. The number 17 is important because it is one of the early prime numbers.
There was more – quite a lot more – but unfortunately I've forgotten the rest. I'm sorry. All the same, it's a start: contemporary mathematicians and theoretical physicists have got something to ponder on. Perhaps that's Einstein's intention: life would be boring if there were nothing left to prove. That said, he realises scientists are stuck in the grooves of Quantum Theory, String Theory, Chaos Theory and Driving Theory and need a little prod to get them out.
Let's hope it works.
*1. Edwin Hubble (1889-1953) was an American astronomer. He made this discovery in 1929.
*2. Albert Einstein (1879-1955) is the most famous and influential theoretical physicist of the twentieth century. He won the Nobel prize in 1921.
*3. The quantum physicist Erwin Schrodinger (1887-1961) showed in a thought experiment how a cat can be both dead and alive at the same time. Einstein was dismissive of Quantum Theory and said "God does not play dice with the universe." Niels Bohr (1885-1962), who first applied Quantum Theory to atomic structure, is said to have replied "Don't tell God what to do!"
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Home > Archives sector > Case studies and research reports > Case studies > Sustainable services > Derbyshire Record Office
Derbyshire Record Office
Working with an Artist in Residence
As part of a major project to extend Derbyshire Record Office and merge with the county Local Studies Library, the Record Office commissioned an Artist in Residence, Paula Moss.
The Artist in Residence worked with staff and a focus group to inform the look and feel of the new service, including the layout, colours and furniture. She also used the collections to design signage and other special touches, such as window blinds, wallpaper, window vinyls and a timeline.
In the process of creating a bright, welcoming, and inspirational space for customers, she helped to bring the archives and local studies teams together, and enabled them to work more creatively, developing their skills and confidence. As a result, staff developed a new visual aesthetic and creative approach to their work.
The project was delivered in partnership with the County Council's Arts Team, who provided part of the funding and who commissioned Derbyshire's Poet Laureate to write a poem for the Record Office.
The project was funded by Derbyshire County Council's Arts Team with an Arts Council grant of £10,000 and funded the Artist in Residence and bespoke items for the building.
this was the first building project that any of the staff had been involved in which meant staff were learning as they went along
the new public areas of the Record Office were in a Victorian building with a challenging layout that couldn't be changed
there had been no interior design concepts or user consultation on the building interior before building work started
the Artist in Residence project started a month after building work commenced and the timescales dictated by the building programme meant decisions made quickly
finalising the furniture and fittings with the supplier took longer than expected
departmental and corporate rules on signage and publicity materials were very rigid, making it difficult to bring some of the ideas to fruition
staff had very little previous experience of community engagement
the Artist in Residence was well established within the local arts networks, which meant that the Record Office had access to her extensive knowledge and contacts
working together for the first time, staff in both local studies and archives services were open to change and prepared to embrace the artist's ideas
the artist and the staff co-designed the interior using space-planning techniques and colour theory paying great attention to detail
flexibility within the overall building project budget allowed allocation of increased funding for furniture and fittings, to align the building with the overall vision of the Record Office
working with an Artist in Residence enabled the staff at Derbyshire Record Office to address some of their service challenges. For instance, the artist brought a wealth of experience in community engagement and worked to develop and deliver new types of outreach event. By imparting her design skills, she gave staff the confidence to challenge some of the corporate rules on publicity materials
staff were closely involved in all decisions about the building and users were involved in the decisions which would closely affect them for example testing the comfort of the searchroom chairs
positive feedback from readers:
'a wonderful environment for study and research. The displays around the building really add an extra ingredient visually. I love the colour scheme too!'
'it feels more inclusive and welcoming. Well done to all the staff for keeping our history vibrant and relevant for the next generations of local historians.'
'definitely the nicest county archive I've been to… wonderful!'
the archivists' perception of the archives changed – archives are also now recognised for their visual appeal rather than just a source of information. This has led to a change in cataloguing to allow visual or 'quirky' aspects to be recorded so that such documents can be more easily identified
Learning from the project
treating the Artist in Residence as one of the team, with full staff privileges, including access to the repositories was key to the success of the project. This trusted relationship was described by an artist who participated in a tour of the Record Office as 'a marriage made in heaven'
tncluding every member of the staff in all the stages of the project, from the layout of the public areas to the tiles in the toilets, meant everyone felt a sense of ownership and pride in the building. This was an excellent way to build one team from two
while working with an Artist in Residence staff developed the confidence to challenge some of the corporate guidelines within which they normally work, for example, by developing a new approach to notices and posters
due to the timescales, there was limited opportunity to involve the local community with the project on this occasion although this will be redressed during a future project
Developing the work in the future
staff skills and confidence have been embedded as part of the project, and a new culture of creativity within the Record Office means future work will build on the momentum generated by the project
with established links with Arts teams within the County Council and the local artistic community, the Record Office intend to continue to run events based on art and to encourage more artists to use the Record Office as inspiration for their creative work
although the Residency has now ended, the Record Office maintains a strong relationship with Paula Moss and are working with her on other projects
For further information about this case study contact Derbyshire Record Office.
Also in Sustainable services
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Cumbria Archive Service
East Sussex Record Office
Working in a Trust
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Home/Health/FDA authorises two coronavirus vaccines for use in Ghana
FDA authorises two coronavirus vaccines for use in Ghana
The FDA has used emergency use authorization pathway to authorize the two vaccines
he Food and Drugs Authority has indicated that they have authorised two vaccines to be administered in the country through the emergency use of authorization pathway.
Speaking at the COVID-19 vaccination rollout plan forum held by the Ministry Of Information on Friday, February 19, the Chief Executive Officer of the Food and Drugs Authority, Mrs Delese Mimi Darko explained that it was prudent to take that decision based on the authorization pathways.
"This is just to assure you that the FDA has used what we call the emergency use authorization pathway to authorize the two vaccines we have at the moment; that is the Covishield which manufactured from India and the Spotinic-v from Russia. Now, the emergency use authorization is a pathway that we use, that pathway is just to ensure that a needed medical product especially in a pandemic is made available in a timely manner so the only different thing is that we do it quickly so it could be used," she said.
Ghana has recently seen a spike in its Coronavirus cases but authorities say measures are in place to curb the spread of the deadly virus which has claimed over 500 lives so far.
Source: Ghanaweb.com
fourteen − 6 =


Yanni Burton is an Aussie singer, songwriter and bassist. Yanni released his debut EP - SameOldLove in April of 2014, produced by Kieran Kelly, Multi Platinum record producer for Angus and Julia Stone's Album - Down The Way.
Since graduating from The Juilliard School, Yanni has performed around the globe as a bassist and singer songwriter. In his home ground you can find Yanni at New York's music venues and performing alongside pop icons such as John Legend, Michelle Williams, Natasha Beddingfield and Rufus Wainwright. He has written with world renowned song writers Nate Company, Ethan Mentor, Tiger Darrow, Len Pearce and many more.


When it comes to the customer care, strategic approach, and track record of the Jarrod Peterson Real Estate Group, there truly is a difference. As a local real estate team, we talk a lot about Better Service. Better Results. While this is our tagline, it is more than just words. We prove this over and over with the consistent stream of repeat and referral clients that come to us every single day.
How is this different from other real estate teams? It is our commitment to customer care. When your client base is filled with referrals from family and friends, customer care is a priority. We are also extremely strategic at marketing new listings, even before they hit the market. Our team provides exclusive premarketing opportunities, specialized advertising and marketing activities, along with guidance throughout the selling process. When you work with the Jarrod Peterson Real Estate Group, you are also supported by the entire Edina Realty community. This support gives our clients exposure and access to a huge network of realtors, home buyers, and financing experts.
Want to experience the Jarrod Peterson Real Estate Group difference? Give us call. We look forward to hanging your sold sign.


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Bill Nighy: 'You can guess how much I knew about Pokemon before I did Detective Pikachu'
By Rachel Weber 2019-02-06T21:44:43Z
Actor and national treasure on his introduction to the gaming phenomenon
Bill Nighy has a role in the upcoming Detective Pikachu movie. He just can't tell you what is. He can tell our sister magazine Total Film all about discovering Pokemon for the first time though, and what he did with his Pokedex.
"You can guess how much I knew about Pokemon before I did the movie," he tells Total Film in the latest issue. "I know a bit more now. I met the head of Pokemon, the seventh richest man in the world, I think, and the man who drew all the Pokemon. And I bought the Pokedex, which is a fabulous book, actually. It was great fun."
Nighy kindly sent his copy of the sacred Pokemon tome to "a small boy in Canada" and now almost regrets it. "It's such a beautiful book. It's got a plastic ball in the cover, with a Pikachu inside. I got very interested in the thing. In fact, I have two decorative panels made of plaster, to look like stone, with classical figures next to Pokemon figures. They look like ancient Grecian works of art. They were on the set and I admired them, so they gave them to me. I'm the only person in the world who has these things. I might have to Ebay them for charity or something."
Unlike fellow actor Ryan Reynolds, who plays Pikachu, Nighy has to use his own face for his role. "I'm a live-action character, but I'm not at liberty to tell you who." In fact, he's one of several stars in the film with an intriguing blank space next to their names in the cast list: Assassination Nation's Suki Waterhouse and singer Rita Ora are also playing classified characters.
Detective Pikachu will be released May 10, and you can read more about it in this month's Total Film magazine.
(Image credit: Disney/Total Film)
Subscribe to Total Film magazine now and you'll never miss an issue (and you'll save money on the cover price). Get the digital version for your tablet or have an exclusive subs cover through your door every month - or save even more with a print/digital bundle! Grab your first issue FREE direct from Apple Newsstand. Or read our Android version on Google Play.
News ents


text:
'Wear a mask': Trump predicts coronavirus will 'get worse' as he returns to briefings
Courtney Subramanian | USA TODAY
WASHINGTON – President Donald Trump returned to the White House podium on Tuesday and offereda stark shift in tone about the progression of the coronavirus, urging Americans to wear a mask as he predicted the spread of the disease will worsen before it improves.
"It will probably, unfortunately, get worse before it gets better," Trump acknowledged.
After weeks of focusing public events on other issues, Trump urged Americans to wear a mask and encouraged young Americans to avoid crowded bars and to "be smart." The president, who was not wearing a mask himself, previously had said that wearing a mask was a personal choice – though public health officials have increasingly found that mask-wearing is effective in limiting the spread.
"We're asking everybody that when you are not able to socially distance, wear a mask, get a mask," Trump said in perhaps his strongest urging to date on that issue. "Whether you like the mask or not, they have an impact, they have an effect. We need everything we can get."
Despite the warnings, Trump offered a rosy assessment of his administration's response to the virus, saying scientists have "learned so much about this disease" and predicting vaccines would be coming "a lot sooner than anyone thought possible."
Trump returned to the podium Tuesday in an effort to regain control of the narrative surrounding his administration's response to the coronavirus as cases continue to surge in several states. Trump ultimately decided the briefings were a way to tout his administration's effort, despite a debate within the White House about whether the briefings were helping.
Trump also referred briefly to the next phase of coronavirus stimulus being negotiated on Capitol Hill. Lawmakers are considering an extension of unemployment insurance as well as a federal program geared to keep small businesses afloat.
"We're working very hard on it we're making a lot of progress," Trump said. "I also know that both sides want to get it done."
Dr. Anthony Fauci, a frequent presence at earlier briefings who has since faced criticism from senior White House aides, said he was not invited to attend. Also not joining the president on Tuesday: Dr. Deborah Birx, the coordinator of the White House coronavirus response, or other members of the task force.
Asked about Fauci and Birx, Trump said "Dr. Birx is just outside."
He did not comment on Fauci's whereabouts.
Trump said he carries a mask, though he has rarely put it on in public.
"Anything that potentially can help – and that certainly can potentially help – is a good thing," Trump said. "I'm getting used to the mask."
The White House suspended Trump's daily coronavirus briefings in April after the president made a series of controversial statements that some Republicans felt undermined his support. Others argue the briefings offered Trump and public health officials an opportunity to explain the administration's response to voters.
Trump's return to the briefing room was nevertheless a tacit recognition that the virus remains a threat – to public health, to the nation's economy and to the president's own chances for reelection. Though he mostly staged events on other issues in recent weeks, Trump is facing questions about his handling of the crisis as his poll numbers have slip against Democrat Joe Biden in several battleground states.
Trump indicated Monday that he hopes to use the briefing to tout developments in the search for a vaccine and therapeutics to treat the disease. Drug makers told lawmakers Tuesday that a vaccine may be available as early as the beginning of 2021.
But the president is almost certain to field questions about reopening strategies in states like Florida, Texas and Arizona, which have experienced spikes in cases.
"I was doing them, and we had a lot of people watching – record numbers watching," Trump said this week in explaining the decision to bring back the briefings. "It's a great way to get information out to the public as to where we are with the vaccines, with the therapeutics, and, generally speaking, where we are."
Trump suspending the near-daily briefings in April shortly after he suggested the virus could be treated with antiseptics, a statement that drew alarm and warnings from public health officials but that the White House said was misconstrued. Some Republicans thought the briefings, and Trump's penchant for going off-script, were damaging.
White House adviser Kellyanne Conway told Fox News last week that ending the briefings is one of the reasons Trump's approval numbers on handling the coronavirus have gone "maybe a little softer" in recent months. But she also acknowledged there was not universal support with the White House for bringing them back.
She said her position was "my own view, which is different than some people here."
The briefings were also a chance for Americans to hear from public health experts advising the White House, including Dr. Birx and Fauci. While Trump has insisted he has a good relationship with Fauci, he also recently described him as an "alarmist" and top White House aides took to openly criticizing him.
Speculation swirled about whether the White House would invite Fauci to the briefing. Speaking in a series of interviews beforehand, Fauci said he was not invited Tuesday and assumed he would not take part.
"I was not invited up to this point I'm assuming I'm not going to be there," he told CNN.
Trump has sought to downplay the coronavirus pandemic even as more than half of all states, including California and Michigan, have paused reopening plans or are taking steps to halt the spread of COVID-19 amid a surge in cases across the country.


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By Freedom Bunker
To Live and Dine in L.A.
Friday, December 4, 2020 5:09
Los Angeles County Supervisor Sheila Kuehl believes "outdoor dining is probably more dangerous in terms of contagion than any other kind of business," so on November 30 she voted to uphold a ban on outdoor dining in the county. Hours after casting her vote, Supervisor Kuehl dined outside at the upscale Il Forno Trattoria restaurant near her Santa Monica residence. When this lapse came to light, she didn't want to talk about it.
"Supervisor Kuehl has eaten at Il Forno virtually every night for many years, and on the last night that it was permitted by county health orders, she dined there alfresco taking appropriate precautions as recommended by the Department of Public Health," said a statement from Kuehl's office. Missing from news reports was Kuehl's background on the health care front.
As a California state senator Kuehl authored SB-840, a proposal for "single payer" health care. The measure would "establish the California Health Insurance System to be administered by the newly created California Health Insurance Agency under the control of a Health Insurance Commissioner appointed by the Governor." The measure passed in 2006, but Gov. Arnold Schwarzenegger vetoed the measure, with its "vast new bureaucracy" that would "cost the state billions and lead to significant new taxes on individuals and businesses."
Before entering politics, Sheila Kuehl was an actress who appeared in Seven Brides for Seven Brothers in 1954 and gained fame as Zelda Gilroy in the television show "The Many Loves of Dobie Gillis," which ran from 1959-1963. Kuehl's Il Forno Trattoria performance is consistent with both careers.
Actors play roles that differ from their real lives, similar to a hypocrite, a person who "feigns some desirable or publicly approved attitude, especially one whose private life, opinions, or statements belie his or her public statements." Kuehl the politician supports government monopoly health care, which takes away the health plan people want and allows only the care government wants them to have. True to form, Kuehl's recent vote takes away a longstanding practice the ruling-class politician reserves for herself.
The decision to ban outdoor dining in Los Angeles County was based on a CDC study that targeted 11 facilities in 10 states and found patients with coronavirus were twice as likely to have dined out at a restaurant. On the other hand, the study failed to distinguish between infection rates among outdoor versus indoor dining, and the study was not specific to Los Angeles County, home to more than 10 million people.
Supervisors Janice Hahn and Kathryn Barger opposed the decision to shut down outdoor dining. As Barger told Fox News, "to say that they're depending on the CDC when we've been doing this for seven months and, in fact, have been doing inspections that we don't have data, and yet we are targeting one industry and saying with the number of cases rising we are going to set [restaurants] down to me is irresponsible."
The post To Live and Dine in L.A. appeared first on The Beacon.
Source: http://freedombunker.com/2020/12/03/to-live-and-dine-in-l-a/
God's Word - Mark Chapter 7


text:
\section{Introduction}
\label{sec:introduction}

The classification of
$\{1,-1\}$-matrices of order $n$ having largest possible determinants
is one of the most important problems in design theory.
Such matrices are known as \textbf{D-optimal designs}.
D-optimal designs of order $n \equiv 0 \pmod 4$ and $n \equiv 2 \pmod 4$ are, respectively, called Hadamard matrices and EW matrices.
For such matrices, the \textit{Smith normal form} is a useful invariant that is used to distinguish between those of the same order.
The Smith normal form of a skew-symmetric Hadamard matrix is well-known~\cite{MW,HK}.
In this paper we determine the Smith normal form of skew-symmetric EW matrices.

Two $n\times n$ integer matrices $M$ and $N$ are called \textbf{equivalent} if there exist $n\times n$ unimodular integer matrices $P$ and $Q$ such that $N=PMQ$.

Let $M$ be an $n\times n$ integer matrix of rank $r$.
Then $M$ is equivalent to a diagonal integer matrix
\[
		\operatorname{diag}[m_1,m_2,\ldots,m_r,m_{r+1},\ldots,m_n],
	\]
satisfying $m_i \;|\; m_{i+1}$ for all $i \in \{1,\dots, n-1\}$
 and $m_{r+1}=0$.
This diagonal matrix is called the \textbf{Smith normal form} of $M$ and the values $m_1,\ldots,m_n$ are called the \textbf{invariant factors} of $A$.

Throughout, $I_n$, $J_n$, and $O_n$ will (respectively) denote the $n \times n$ identity matrix, all-ones matrix, and all-zeros matrix.
We omit the subscript when the order is apparent and the matrices denoted by $J$ and $O$ may not necessarily be square matrices.
We use $\mathbf{0}_t$ and $\mathbf{1}_t$ to denote the length-$t$ all-zeros and all-ones (column) vectors respectively and we omit the subscript when the length is apparent.

Let $X$ be a $\{1,-1\}$-matrix of order $n$.
Abusing language, we call $X$ \textbf{skew-symmetric} if $X+X^\top = 2I$, i.e., the matrix $X-I$ is skew-symmetric in the usual sense:  $X-I + (X-I)^\top = O$.

A famous inequality due to Hadamard~\cite{Hadamard1893} is the following:
 \begin{equation*}
 	\label{ineq:Hadamard}
 	|\det(X)| \leqslant n^{n/2}.
 \end{equation*}
 If equality holds then $X$ is called a \textbf{Hadamard matrix}.
Hadamard's inequality can be improved if we restrict to matrices whose orders are not divisible by $4$.
Indeed, Ehlich~\cite{Ehlich1964} and Wojtas~\cite{Wojtas1964} independently showed that for a $\{1,-1\}$-matrix $X$ of order $n \equiv 2\pmod{4}$, Hadamard's inequality can be strengthened to
\begin{align}\label{ineq:EW}
|\det(X)| \leqslant 2(n-1)(n-2)^{(n-2)/2}.
\end{align}
Moreover, there exists a $\{1,-1\}$-matrix achieving equality in \eqref{ineq:EW} if and only if there exists a $\{1,-1\}$-matrix $B$ such that
\begin{align}\label{eq:ew}
BB^\top=B^\top B=
\begin{pmatrix}
(n-2)I+2J & O_{n/2} \\
O_{n/2} & (n-2)I+2J
\end{pmatrix}.
\end{align}
A $\{1,-1\}$-matrix  $B$ of order $n$ is called an \textbf{EW matrix} if it satisfies \eqref{eq:ew}.
Hence, in particular, an EW matrix of order $n$ has determinant $\pm 2(n-1)(n-2)^{(n-2)/2}$.
Working up to equivalence, we may assume that an EW matrix has a positive determinant.
Accordingly, throughout this paper, we assume that an EW matrix $S$ of order $4t+2$ has determinant $\det(S) = (4t)^{2t}(8t+2)$.
We refer to the \emph{Handbook of Combinatorial Designs}~\cite{CRCHAndbook2007} for background on Hadamard and EW matrices.

The Smith normal form of skew-symmetric Hadamard matrices of order $4t$ is uniquely determined \cite{MW,HK}.
\begin{theorem}
A skew-symmetric Hadamard matrix of order $4t$ has the Smith normal form
 	\[
		\operatorname{diag}[1,\underbrace{2,\dots,2}_{2t-1},\underbrace{2t,\dots,2t}_{2t-1},4t].
	\]
\end{theorem}

Armario~\cite{A} partially determined invariant factors of a skew-symmetric EW matrix of order $4t+2$.
\begin{theorem}{\rm\cite[Theorem~1.2]{A}}\label{thm:a}
	Let $s_1,\ldots,s_{4t+2}$ be the invariant factors of a skew-symmetric EW matrix of order $4t+2$.
       Then $s_1=1$ and $s_2=\cdots=s_{2t+1}=2$.
\end{theorem}

Further, Armario~\cite{A} conjectured the value of the remaining invariant factors of a skew-symmetric EW matrix.
In this note we prove his conjecture.
The following is our main theorem.
\begin{theorem}\label{thm:main}
	A skew-symmetric EW matrix of order $4t+2$ has Smith normal form
	\[
		\operatorname{diag}[1,\underbrace{2,\dots,2}_{2t+1},\underbrace{2t,\dots,2t}_{2t-1},2t(4t+1)].
	\]
\end{theorem}

Now we provide an outline of the proof.
Let $S$ be a skew-symmetric EW matrix of order $n=4t+2$.
In particular, we have that
\begin{align}\label{eq:2}
SS^\top=\begin{pmatrix}(n-2)I+2J & O \\ O & (n-2)I+2J \end{pmatrix}
\end{align}

\begin{lemma}\label{lem:m1}
Let $S$ be a skew-symmetric EW matrix of order $n=4t+2$ and let $s_1,\ldots,s_{4t+2}$ be the invariant factors of $S$.
\begin{enumerate}
\item $s_{2t+2}=2$.
\item $s_{4t+2}=2t(4t+1)$.
\end{enumerate}
\end{lemma}

We will give the proof of Lemma~\ref{lem:m1}
in subsequent sections.
Using Lemma~\ref{lem:m1} together with Theorem~\ref{thm:a}, we can determine the invariant factors of a skew-symmetric EW matrix.
Indeed, we prove Theorem~\ref{thm:main} as follows.
\begin{proof}[Proof of Theorem~\ref{thm:main}]
Since the other invariant factors have been determined in Theorem~\ref{thm:a} and Lemma~\ref{lem:m1}, it remains to show $s_{2t+3}=\dots=s_{4t+1}=2t$.
Using \eqref{eq:ew}, we see that $\det(S)=s_1\cdots s_{4t+2}=2^{2t+1}(2t)^{2t}(4t+1)$.
Hence, by Theorem~\ref{thm:a} and Lemma~\ref{lem:m1}, we have
\begin{align}\label{eq:m}
s_{2t+3}\cdots s_{4t+1}=(2t)^{2t-1}.
\end{align}
\paragraph{Claim 1:} $s_{4t+1}$ divides $2t$.
Since $s_2=2$ divides $s_{i}$ for each $i\geqslant 2$, we may set $s_i=2s^\prime_i$ for $2\leqslant i\leqslant 4t+2$.
We have $s^\prime_{4t+1}$ divides $s^\prime_{4t+2}=t(4t+1)$ and $\text{gcd}(t,4t+1)=1$.
Furthermore, since $4s^\prime_{4t+1} s^\prime_{4t+2} = 4s^\prime_{4t+1} t(4t+1)$ divides $2(4t+1)(4t)^{2t}$ we must have that $s^\prime_{4t+1}$ divides $t$.
Therefore $s_{4t+1}$ is a divisor of $2t$.

Since $s_i$ divides $s_{i+1}$ for each $i$ and $s_{4t+1}\leqslant 2t$, by Claim~1,
we have that $s_{i}\leqslant 2t$ for $i=2t+3,\dots,4t$, from which it follows that
\begin{align}\label{eq:m1}
s_{2t+3}\cdots s_{4t+1}\leqslant(2t)^{2t-1}.
\end{align}
By \eqref{eq:m}, equality must hold in \eqref{eq:m1}.
Therefore $s_{2t+3}=\dots=s_{4t+1}=2t$.
\end{proof}

Our methods for proving Lemma~\ref{lem:m1} involve a tournament matrix that one associates with an EW matrix, called an \emph{EW tournament matrix} (see Section~\ref{sec:pre} for its definition).
Even though it is not needed for the proof of Lemma~\ref{lem:m1}, we also determine the Smith normal form of an EW tournament matrix (see Theorem~\ref{thm:snfa}).

The organisation of this paper is as follows.
In Section~\ref{sec:pre} we prepare the basic tools for the proof of our main result.
We will prove Lemma~\ref{lem:m1} (i) and (ii) in Section~\ref{sec:2t2} and Section~\ref{sec:4t2}, respectively.
In Section~\ref{sec:ex}, we provide examples of EW matrices with Smith normal forms different from Theorem~\ref{thm:main}.
Note that the EW matrix in Example~\ref{ex:2} is a counterexample to \cite[Corollary 2]{KMS}.
We
attempt to deal with the counterexample to \cite[Corollary 2]{KMS}.
In Section~\ref{appendix:a} we provide a revised result to \cite[Corollary 2]{KMS}, and we pursue the method demonstrated in Section~\ref{appendix:a} to obtain a result for the case that $4t+1$ is a square of a prime in Section~\ref{appendix:b}.
Finally, in Section~\ref{sec:snftournament} we determine the Smith normal form of an EW tournament matrix.

\section{Preliminaries}\label{sec:pre}
In this section we collect some facts about skew-symmetric EW matrices and their invariant factors, which we will use in later sections.

Let $X$ be an EW matrix of order $n$.
It is well-known that $2n-2$ must be a sum of squares~\cite{CRCHAndbook2007}.
Furthermore, Armario and Frau \cite{AF} showed that $2n-3$ must be a square if $X$ is skew-symmetric.
Indeed, they showed the following.
\begin{lemma}{\rm\cite[Proposition~2.4]{AF}}\label{lem:o}
Suppose there exists a skew-symmetric EW matrix of order $4t+2$.
Then $\sqrt{8t+1}$ is an odd integer.
\end{lemma}

A \textbf{tournament matrix} is a $\{0,1\}$-matrix $A$ such that $A+A^\top=J-I$.
A tournament matrix $A$ of order $4t+1$ is called an \textbf{EW tournament matrix} if for some $a\in\mathbb{N}\cup\{0\}$,
\begin{align}
A A^\top&=t(I_{4t+1}+J_{4t+1})+\begin{pmatrix}
-J_t & -J  & -J  &-J\\
-J & J_t  & O  & O \\
-J & O & O_{a}  &-J\\
-J & O & -J  & O_{2t+1-a}
\end{pmatrix},\label{eq:ew1}\\
A^\top A&=t(I_{4t+1}+J_{4t+1})+\begin{pmatrix}
J_t & -J  & O & O \\
-J & -J_t  & -J  & -J \\
O & -J  & J_{a}  & O \\
O & -J  & O & -J_{2t+1-a}
\end{pmatrix}.\label{eq:ew2}
\end{align}
Since $A+A^\top=J-I$,  we have that
\begin{align}
(A+I) (A^\top+I)&=t(I_{4t+1}+J_{4t+1})+\begin{pmatrix}
O_{t} & O & O & O \\
O & 2J_t  & J  & J \\
O & J  & J_{a}  & O \\
O & J  & O & J_{2t+1-a}
\end{pmatrix}, \label{eq:AAt}\\
(A^\top+I)(A+I) &=t(I_{4t+1}+J_{4t+1})+\begin{pmatrix}
2J_{t} & O & J  & J \\
O & O_{t} & O & O \\
J & O & 2J_{a}  & J \\
J  & O & J & O_{2t+1-a}
\end{pmatrix}.\label{eq:AtA}
\end{align}

A skew-symmetric EW matrix $S$ corresponds to an EW tournament matrix $A$ via the equation \cite{A2015}:
\begin{align}\label{eq:s}
S=\begin{pmatrix}1 & \bf{1}^\top \\ -\bf{1} & I+A-A^\top \end{pmatrix}.
\end{align}
The invariant factors of $S$ and those of $A+I$ are related as follows.
\begin{lemma}{\rm\cite[Lemma~2,2, Corollary~2.3]{A}}\label{eq:bm}
Let $S$ be a skew-symmetric EW matrix of order $4t+2$ and let $A$ be the EW tournament matrix of order $4t+1$ obtained by \eqref{eq:s}.
Let $s_1,\ldots,s_{4t+2}$ be the invariant factors of $S$ and $b_1,\ldots,b_{4t+1}$ be the invariant factors of $A+I$.
Then
\begin{enumerate}
\item  $s_{i+1} = 2b_i$ for $i=1,\ldots,4t+1$.
\item $\det(A+I)=t^{2t}(4t+1)$.
\end{enumerate}
\end{lemma}
\begin{proof}
First observe that $S$ is equivalent to $[1]\oplus2(A+I)$.  Indeed, in \eqref{eq:s}, add the first row to all other rows and subtract the first column from all other columns.
Using this equivalence, we see that $s_{i+1} = 2b_i$ for $i=1,\ldots,4t+1$ and $\det(A+I)=(1/2^{4t+1})\det(S)=t^{2t}(4t+1)$.
\end{proof}

Let $M$ be an integer matrix.
The invariant factors of $M$ can be determined from the the greatest common divisor of all minors of $M$.
Indeed, define $d_i(M)$ to be the greatest common divisor of all $i\times i$ minors of $M$ and set $d_0(M)=1$.
Then the next result is standard.

\begin{lemma}[Corollary 1.20 of \cite{CN}]\label{lem:minor}
Let $M$ be an integer matrix of order $n$ and rank $r$ with invariant factors $m_1,\dots,m_n$.
Then $m_i=d_i(M)/d_{i-1}(M)$ for $i=1,\dots,r$.
\end{lemma}

Let $M$ be an invertible $n\times n$ matrix with rows and columns indexed by $\{1,\ldots,n\}$.
For a subset $I \subseteq \{1,\ldots,n\}$, define $I'=\{1,\ldots,n\}\setminus I$.
For nonempty subsets $I,J \subseteq \{1,\ldots,n\}$ with $|I|=|J|$, we denote by $M_{I,J}$ the matrix obtained from $M$ by restricting rows to $I$ and columns to $J$ respectively.

\begin{lemma}[Page 21 of \cite{HJ}]\label{lem:detminor}
Let $M$ be an invertible $n\times n$ matrix with rows and columns indexed by $\{1,\ldots,n\}$.
Suppose $I,J \subseteq \{1,\ldots,n\}$ satisfy $|I|=|J| \ne 0$.
Then
\begin{align*}
\det(M_{I,J})=\pm\det(M) \det((M^{-1})_{J',I'}).
\end{align*}
\end{lemma}

Now we record a result about the determinant of an EW tournament matrix.
It is a corollary of \cite[Theorem 1.1]{GS},

\begin{lemma}\label{lem:snfa1}
	Let $A$ be an EW tournament matrix of order $4t+1$.
	Then $\det(A)=t^{2t}(4t-1)$.
\end{lemma}

Since skew-symmetric EW matrices are in correspondence with EW tournament matrices, the following result implies Lemma~\ref{lem:o}.
We will use Theorem~\ref{thm:quadforma} in Section~\ref{sec:snftournament}.

\begin{theorem}
	\label{thm:quadforma}
	Let $A$ be an EW tournament matrix of order $4t+1$.
	Then the integer $a$ in \eqref{eq:ew1} satisfies the quadratic equation $a^2-(2t+1)a+t(t-1)=0$.
\end{theorem}
\begin{proof}
	Begin with \eqref{eq:ew1}, which gives an expression for $AA^\top$.
Using elementary row operations and properties of the determinant, we find that
\begin{align}
\det( AA^\top) &=\begin{pmatrix}
t I_t+(t-1)J_t & (t-1)J  & (t-1)J  &(t-1)J\\
(t-1)J & t I_t+(t+1)J_t  & t J  & t J \\
(t-1)J & t J & t I_a+t J_{a}  & (t-1)J\\
(t-1)J & t J & (t-1)J  & t I_{2t+1-a}+t J_{2t+1-a}
\end{pmatrix}\displaybreak[0]\nonumber\\
&=\det \begin{pmatrix}
t^2 & (t-1){\bf 1}^\top & (t-1){\bf 1}^\top  & (t-1){\bf 1}^\top  & (t-1){\bf 1}^\top\\
{\bf 0}&t I_{t-1} & O  & O  & O \\
-t{\bf 1}& O & t I_t+2J_t  &  J  &  J \\
-t{\bf 1}& O & J & t I_a+J_{a}  & O\\
-t{\bf 1}& O & J & O  & t I_{2t+1-a}+J_{2t+1-a}
\end{pmatrix}\displaybreak[0]\nonumber\\
&=t^{t-1}\det \begin{pmatrix}
t^2 & (t-1){\bf 1}^\top  & (t-1){\bf 1}^\top  & (t-1){\bf 1}^\top\\
-t{\bf 1}&  t I_t+2J_t  &  J  &  J \\
-t{\bf 1}&  J & t I_a+J_{a}  & O\\
-t{\bf 1}&  J & O  & t I_{2t+1-a}+J_{2t+1-a}
\end{pmatrix}\displaybreak[0]\nonumber\\
&=t^{t}\det \begin{pmatrix}
t & (t-1){\bf 1}^\top  & (t-1){\bf 1}^\top  & (t-1){\bf 1}^\top \\
-{\bf 1}&  t I_t+2J_t  &  J  &  J \\
-{\bf 1}&  J & t I_a+J_{a}  & O\\
-{\bf 1}&  J & O  & t I_{2t+1-a}+J_{2t+1-a}
\end{pmatrix}\displaybreak[0]\nonumber\\
&=t^{t}\det \begin{pmatrix}
t & (3t-1){\bf 1}^\top & (2t-1){\bf 1}^\top & (2t-1){\bf 1}^\top\\
-{\bf 1}&  t I_t  &  O  &  O \\
-{\bf 1}&  -J & t I_a  & -J \\
-{\bf 1}&  -J & -J & t I_{2t+1-a}
\end{pmatrix}\displaybreak[0]\nonumber\\
&=t^{t}\det \begin{pmatrix}
4t-1 & (3t-1){\bf 1}^\top & (2t-1){\bf 1}^\top & (2t-1){\bf 1}^\top \\
{\bf 0}&  t I_t  &  O  &  O \\
(-2){\bf 1}&  -J & t I_a  & -J \\
(-2){\bf 1}&  -J & -J & t I_{2t+1-a}
\end{pmatrix}\displaybreak[0]\nonumber\\
&=t^{2t}\det \begin{pmatrix}
4t-1 &  (2t-1){\bf 1}^\top & (2t-1){\bf 1}^\top \\
(-2){\bf 1}&  t I_a  & -J \\
(-2){\bf 1}&  -J & t I_{2t+1-a}
\end{pmatrix}\displaybreak[0]\nonumber\\
&=t^{2t}(4t-1)\det \left( \begin{pmatrix}
 t I_a  & -J \\
  -J & t I_{2t+1-a}
\end{pmatrix}+\frac{2(2t-1)}{4t-1}J\right)\nonumber\\
&=t^{2t}(4t-1)\det\begin{pmatrix}
 t I_a+\frac{2(2t-1)}{4t-1}J  & -\frac{1}{4t-1}J \\
  -\frac{1}{4t-1}J & t I_{2t+1-a}+\frac{2(2t-1)}{4t-1}J
\end{pmatrix}.\label{eq:ew3}
\end{align}
Next, we use the following general equation obtained by the Schur complement \cite[(0.8.5.1)]{HJ}:
\begin{align}\label{eq:det2x2}
\det\begin{pmatrix}
 \alpha I_a+\beta J_a  & \gamma J \\
  \gamma J & \alpha I_b+\beta J_b
\end{pmatrix}=\alpha^{a+b-2}\left( \alpha^2+(a+b)\alpha\beta+ab\beta^2-ab\gamma^2 \right).
\end{align}
Then, by \eqref{eq:ew3} and \eqref{eq:det2x2}, we have
\begin{align}\label{eq:eq4}
\det( AA^\top) = t^{4t-1}\left((3-4t)a^2+(8t^2-2t-3)a+t(12t^2-t-2) \right).
\end{align}
By Lemma~\ref{lem:snfa1}, we have $\det(A)=t^{2t}(4t-1)$.
Therefore, using the multiplicativity of the determinant we have
\[
\det( AA^\top) = t^{4t}(4t-1)^2.
\]
Therefore we can simplify \eqref{eq:eq4} to obtain
\begin{align*}
a^2-(2t+1)a+t(t-1)=0,
\end{align*}
as required.
\end{proof}

\section{The $(2t+2)$-nd invariant factor of skew-symmetric EW matrices}\label{sec:2t2}
In this section, we provide proof of part (i) of Lemma~\ref{lem:m1}.
For a prime $p$ and an integer matrix $M$, let $\rank_p(M)$ denote the rank of $M$ over the field of integers of modulo $p$.
The following was shown in \cite{A}.
\begin{lemma}{\rm \cite[Lemma~2,4]{A}}
	\label{lem:rankub}
Let $A$ be an EW tournament matrix of order $4t+1$ and let $p$ be a prime that divides $t$.
Then $2t\leqslant \rank_p(A+I)\leqslant 2t+1$.
\end{lemma}

 Next we provide a strengthening of the above lemma, which is a key lemma to determine the invariant factor $s_{2t+2}$.

\begin{lemma}\label{lem:prank}
	Let $A$ be an EW tournament matrix of order $4t+1$ and let $p$ be a prime that divides $t$.
	Then $\rank_p(A+I) = 2t+1$.
\end{lemma}
\begin{proof}
First we show that $\rank_p(A+I) \geqslant 2t+1$.
Let $d = \dim( \mathrm{rowsp}_p(A+I)\cap \mathrm{rowsp}_p(A^\top+I) )$ be the dimension of the intersection of row spaces over the field of integers of modulo $p$.
Then
\begin{align*}
4t&=\rank_p(J+I)=\rank_p((A+I)+(A^\top+I))\\
&\leqslant \rank_p(A+I)+\rank_p(A^\top+I)-d\\
&=2\rank_p(A+I)-d,
\end{align*}
that is, we have that $2t+d/2\leqslant \rank_p(A+I)$.
Hence it suffices to show that $d\geqslant 1$.

Using the assumption that $p$ divides $t$ together with \eqref{eq:AAt} and \eqref{eq:AtA}, it follows that
\begin{align*}
\mathrm{span}\{({\bf 0}^\top_t,{\bf 1}^\top_t,{\bf 1}^\top_{a},{\bf 0}^\top_{2t+1-a}),({\bf 0}^\top_t,{\bf 1}^\top_t,{\bf 0}^\top_{a},{\bf 1}^\top_{2t+1-a})\} &\subseteq \mathrm{rowsp}_p(A+I), \\
\mathrm{span}\{({\bf 1}^\top_t,{\bf 0}^\top_t,{\bf 1}^\top_{a},{\bf 0}^\top_{2t+1-a}),({\bf 1}^\top_t,{\bf 0}^\top_t,{\bf 0}^\top_{a},{\bf 1}^\top_{2t+1-a})\} &\subseteq \mathrm{rowsp}_p(A^\top+I).
\end{align*}
Therefore
\begin{align*}
\mathrm{span}\{({\bf 0}^\top_t,{\bf 0}^\top_t,{\bf 1}^\top_{a},-{\bf 1}^\top_{2t+1-a})\} \subseteq \mathrm{rowsp}_p(A+I)\cap \mathrm{rowsp}_p(A^\top+I),
\end{align*}
and thus $d \geqslant 1$.

The inequality $\rank_p(A+I) \leqslant 2t+1$ follows from Lemma~\ref{lem:rankub}.
However, for the sake of completeness, we provide a proof.
Combine \eqref{eq:AAt} and Sylvester's rank inequality \cite[0.4.5 Rank inequalities (c)]{HJ} to obtain
\begin{align*}
    2\rank_p(A+I)&=\rank_p(A+I)+\rank_p(A^\top+I)\\
    &\leqslant \rank_p((A+I)(A^\top+I))+4t+1\\
    &=4t+3.
\end{align*}
Thus we have $ \rank_p(A+I)\leqslant2t+1$.
\end{proof}

\begin{proposition}\label{prop:b2t1}
Let $A$ be an EW tournament matrix of order $4t+1$ and let $b_1,\ldots,b_{4t+1}$ be the invariant factors of $A+I$.
Then $b_{2t+1}=1$.
\end{proposition}
\begin{proof}
This result follows from the argument\footnote{In Armario's proof, he derived ``$p^{2t+1}$ divides $4t+1$" from the assumption on $b_{2t}=1$. The result obtained from his argument is, in fact, that ``$p^{2t+2}$ divides $4t+1$".} in \cite[Proof of Theorem~1.2]{A} by replacing the role of $b_{2t}$ with $b_{2t+1}$.
For the sake of the reader we provide a proof.

Assume that there exists a prime $p$ dividing $b_{2t+1}$. It follows from Lemma~\ref{lem:minor} and Lemma~\ref{eq:bm} that $b_1\cdots b_{4t+1}=\det(A+I)=t^{2t}(4t+1)$.
Thus $p$ is either
\begin{enumerate}[(1)]
\item a divisor of $t$, or
\item not a divisor of $t$ and a divisor of $4t+1$.
\end{enumerate}
For the case (1), Lemma~\ref{lem:prank} contradicts the fact that
\begin{align*}
\rank_p(A+I)=\max\{i\mid \text{$p$ does not divide $b_i$}\}\leqslant 2t.
\end{align*}
For the case (2), observe that $p$ is odd and hence $p \geqslant 3$.
Furthermore $p$ is a divisor of $b_i$ for $i=2t+1,\ldots,4t+1$.
Thus $p^{2t+1}$ divides $4t+1$.
But this is impossible since $p^{2t+1}>4t+1$ for all $t\geqslant 1$.
Therefore we must have $b_{2t+1}=1$.
\end{proof}
Now we can prove part (i) of Lemma~\ref{lem:m1}.

\begin{proof}[Proof of $s_{2t+2}=2$]
By combining Lemma~\ref{eq:bm} and Proposition~\ref{prop:b2t1} we obtain $s_{2t+2}=2$.
\end{proof}
\begin{remark}
The equality $b_{2t+1}=1$ implies that $b_1=\cdots=b_{2t}=1$. Thus it follows that $s_2=\cdots=s_{2t+1}=2$, which was already shown in \cite[Theorem~1.2]{A}.
\end{remark}

\section{The $(4t+2)$-nd invariant factor of skew-symmetric EW matrices}\label{sec:4t2}
In this section, we will show $s_{4t+2}=2t(4t+1)$ for a skew-symmetric EW matrix of order $4t+2$.

By Lemma~\ref{lem:detminor}, it is enough to calculate the entries of $\det(S)S^{-1}$ explicitly.
By the equation \eqref{eq:2} it is shown that
\begin{align}\label{eq:1}
S^{-1}=S^\top \left(\frac{1}{4t}I-\frac{1}{4t(4t+1)}\begin{pmatrix}J & O \\ O & J \end{pmatrix}\right).
\end{align}
We use the following lemma.
\begin{lemma}\label{lem:1}
Set $S=\left ( \begin{smallmatrix}S_{11} & S_{12} \\ S_{21} & S_{22} \end{smallmatrix} \right )$ where each $S_{ij}$ is a $(2t+1) \times (2t+1)$ matrix.
Then $S_{11}J=S_{22}J=J$ and $S_{12}J=-S_{21}J=\pm\sqrt{8t+1}J$ hold.
\end{lemma}
\begin{proof}
The eigenvalues of $S$ are $1\pm\sqrt{-1}(4t-1),1\pm\sqrt{-8t-1}$, and the eigenvectors corresponding to $1\pm\sqrt{-8t-1}$ are
$\begin{pmatrix}\pm\sqrt{-1}\bm{1} \\ \bm{1} \end{pmatrix}$ or $\begin{pmatrix}\mp\sqrt{-1}\bm{1} \\ \bm{1} \end{pmatrix}$, see the proof of \cite[Lemma~4.5]{NS}.
The result now follows from these eigenvalues and eigenvectors.
\end{proof}
Using Lemma~\ref{lem:1} together with the equation $S+S^\top=2I$, we calculate the right hand side of the equation \eqref{eq:1} as follows:
\begin{align*}
S^{-1}&=(2I-S)\left(\frac{1}{4t}I-\frac{1}{4t(4t+1)}\begin{pmatrix}J & O \\ O & J \end{pmatrix}\right).\\
&=\frac{1}{4t}(2I-S)-\frac{1}{4t(4t+1)}\begin{pmatrix}(2I-S_{11})J & -S_{12}J \\ -S_{21}J & (2I-S_{22})J \end{pmatrix}\\
&=\frac{1}{4t}(2I-S)-\frac{1}{4t(4t+1)}\begin{pmatrix}J & \mp\sqrt{8t+1}J \\ \pm\sqrt{8t+1}J & J \end{pmatrix}\\
&=\frac{1}{4t(4t+1)}\left((4t+1)(2I-S)-\begin{pmatrix}J & \mp\sqrt{8t+1}J \\ \pm\sqrt{8t+1}J & J \end{pmatrix}\right).
\end{align*}
By $\det(S)=(8t+2)(4t)^{2t}$, we obtain
\begin{align*}
\det(S) S^{-1}=2(4t)^{2t-1}\left((4t+1)(2I-S)-\begin{pmatrix}J & \mp\sqrt{8t+1}J \\ \pm\sqrt{8t+1}J & J \end{pmatrix}\right).
\end{align*}

Thus the entries of $\det(S)S^{-1}$ are, up to sign,
\begin{align*}
2(4t)^{2t-1}\times 4t,\quad2(4t)^{2t-1}\times (4t+2),\quad 2(4t)^{2t-1}\times (4t+1\pm \sqrt{8t+1}).
\end{align*}
By Lemma~\ref{lem:o}, the expression $\sqrt{8t+1}$ is an odd integer.
Hence $4t,4t+2$, and $4t+1\pm\sqrt{8t+1}$ are all even.
Moreover, the greatest common divisor among $4t,4t+2,4t+1\pm\sqrt{8t+1}$ is two.
By Lemma~\ref{lem:minor},
we have $d_{4t+1}(S)=4(4t)^{2t-1}$.
Combining Lemma~\ref{lem:minor} and the equalities $d_{4t+2}(S)=\det(S)=(8t+2)(4t)^{2t}$, yields $s_{4t+2}=d_{4t+2}(S)/d_{4t+1}(S)=2t(4t+1)$.

\section{Examples}\label{sec:ex}
In this section we provide a couple of examples of EW matrices that have Smith normal forms different from Theorem~\ref{thm:main}.
\begin{example}\cite[Example 1]{KMS}\label{ex:1}
There exists an EW matrix of order $26$ with the Smith normal form different from Theorem~\ref{thm:main}.
Let $R$ be the circulant $13 \times 13$ $\{1,-1\}$-matrix with
first row
\begin{align*}
    (1, 1, 1, 1, -1, 1, -1, -1, 1, 1, 1, -1, 1).
\end{align*}
Note that $R$ satisfies that $RR^\top=12I+J$.
Then $X=\begin{pmatrix}
R & R\\
-R^\top & R^\top
\end{pmatrix}$
is an EW matrix of order $26$, and its Smith normal form is
\begin{align*}
    \operatorname{diag}[1,\underbrace{2,\dots,2}_{13},\underbrace{12,\dots,12}_{10},60,60].
\end{align*}
Therefore, by Theorem~\ref{thm:main}, the EW matrix $X$ is not equivalent to a skew-symmetric EW matrix of order $26$.
For a result on the Smith normal form of an EW matrix of order $26$ that have a certain block structure, see Section~\ref{appendix:b}.
\end{example}

\begin{example}\label{ex:2}
The following construction is due to Kharaghani \cite{K}.
Let $A$ be the $11\times 11$ circulant $\{0,1,-1\}$-matrix with the first row $$(0, -1, 1, -1, -1, -1, 1, 1, 1, -1, 1).$$
Note that $A$ is skew-symmetric and $AA^\top=11I-J$.
Then $X=\begin{pmatrix}
R_1 & R_2\\
-R_2^\top & R_1^\top
\end{pmatrix}$
is an EW matrix of order $66$, where
\begin{align*}
    R_1&=(A+I_{11})\otimes (J_3-I_3)+(J_{11}-2I_{11})\otimes I_3,\\
    R_2&=(A+I_{11})\otimes (J_3-I_3)+(-A+I_{11})\otimes I_3,
\end{align*}
and its Smith normal form is
\begin{align*}
    \operatorname{diag}[1,\underbrace{2,\dots,2}_{31},\underbrace{8,\dots,8}_{4},\underbrace{32,\dots,32}_{29},2080].
\end{align*}
Therefore by Theorem~\ref{thm:main}, $X$ is not equivalent to a skew-symmetric EW matrix of order $66$.
Note that this EW matrix is a counterexample to \cite[Corollary 2]{KMS}.
We provide a revised result of \cite[Corollary 2]{KMS} in Section~\ref{appendix:a}.
\end{example}

\section{The Smith normal form of an EW matrix of order $4t+2$ where $4t+1$ is square-free} \label{appendix:a}
Here we provide a revised result of \cite[Corollary 2]{KMS}.
\begin{lemma}\label{lem:a1}
Let $X$ be an EW matrix of order $4t+2$.
Let $x_1,\ldots,x_{4t+2}$ be the invariant factors of $X$.
Then $x_1=1$ and $x_2=2$.
\end{lemma}
\begin{proof
The result for $x_1$ is trivial, and that for $x_2$ follows from the fact that some $2\times 2$ minor of $X$ is $\pm2$ and all $2 \times 2$ minors of $X$ are even.
\end{proof}

In the following theorem, we deal with EW matrices that have a block structure.
We use the notion $[x]^n$ to mean $\underbrace{x,\ldots,x}_{n \text{ times}}$.
\begin{theorem}\label{thm:a1}
Let $X$ be an EW matrix $X=\begin{pmatrix} R_1 & R_2 \\ -R_2^\top & R_1^\top \end{pmatrix}
$ of order $4t+2$ such that $R_1J=R_1^\top J=r_1J$, $R_2J=R_2^\top J=r_2J$, and $4t+1$ is square-free.
Let $x_1,\ldots,x_{4t+2}$ be the invariant factors of $X$.
Then
\begin{enumerate}
\item $x_{4t+2}=2t(4t+1)$;
\item $x_{4t+1}=2t$;
\item if $t=2^\ell q$ such that $q$ is a square-free odd integer, we have
\begin{align*}
&(x_2,\ldots,x_{4t})\\
&=([2]^{n_1},\ldots,[2^{k-1}]^{n_{k-1}},[2^k]^{n'_k},[2^k q]^{n''_k},[2^{k+1} q]^{n_{k+1}},\ldots,[2^{\ell+1} q]^{n_{\ell+1}})
\end{align*}
for some positive integer $k$ and non-negative integers
$n_1,\ldots,n_{k-1},n_k'$, $n_k'',n_{k+1},\ldots,n_\ell+1$ such that $1\leqslant k\leqslant \ell+1$, and
\begin{align*}
n_1+\cdots+n_{k-1}+n_k'&=2t+1,\\
n_k''+n_{k+1}+\cdots+n_{\ell+1}&=2t-2,\\
1\cdot n_1+2\cdot n_2+\cdots+(\ell+1)\cdot n_{\ell+1}&=3+2(\ell+2)(t-1),
\end{align*}
where $n_k=n_k'+n_k''$.
\end{enumerate}
\end{theorem}
\begin{proof}[Proof of (i)]
Since $X$ is an EW matrix, $R_1R_1^\top+R_2R_2^\top=4tI+2J$.
Multiplying all-ones vectors from both sides and dividing by $2t+1$ yield
\begin{align*}
r_1^2+r_2^2=8t+2.
\end{align*}
Considering this equation modulo $4$, we have $r_1$ and $r_2$ are both odd.
Since $4t+1$ is square-free, $\text{gcd}(r_1,r_2)=1$.

Next we calculate $X^{-1}$ as follows:
\begin{align*}
X^{-1}&=X^\top \left(\frac{1}{4t}I-\frac{1}{4t(4t+1)}\begin{pmatrix}J & O \\ O & J \end{pmatrix}\right)\\
&=\frac{1}{4t}X^\top-\frac{1}{4t(4t+1)}\begin{pmatrix}R_1^\top J & -R_2J \\ R_2^\top J & R_1 J \end{pmatrix}\\
&=\frac{1}{4t}X^\top-\frac{1}{4t(4t+1)}\begin{pmatrix}r_1 J & -r_2 J \\ r_2J & r_1 J \end{pmatrix}\\
&=\frac{1}{4t(4t+1)}\left((4t+1)X^\top-\begin{pmatrix}r_1 J & -r_2 J \\ r_2 J & r_1 J  \end{pmatrix}\right).
\end{align*}
Since $\det(X)=2(4t+1)(4t)^{2t}$,
\begin{align*}
\det(X)X^{-1}&=2(4t)^{2t-1}\left((4t+1)X^\top-\begin{pmatrix}r_1 J & -r_2 J \\ r_2 J & r_1 J  \end{pmatrix}\right).
\end{align*}
Therefore the entries of $\det(X)X^{-1}$ are $\pm 2(4t)^{2t-1}(4t+1\pm x)$ where $x\in\{r_1,r_2\}$.
Then the greatest common divisor of $4t+1\pm x$ where $x\in\{r_1,r_2\}$ is
\begin{align*}
\text{gcd}(4t+1\pm r_1,4t+1\pm r_2)&=\text{gcd}(4t+1+ r_1,2r_1,4t+1+r_2,2r_2)
=2,
\end{align*}
where in the last equation we used that $r_1$ and $r_2$ are both odd integers and $\text{gcd}(r_1,r_2)=1$.
Therefore $d_{4t+1}(X)=4(4t)^{2t-1}$, and hence we have $x_{4t+2}=\det(X)/d_{4t+1}(X)=2t(4t+1)$.
\end{proof}

Since $x_2=2$ divides $x_{i}$ for each $i\geqslant 2$, we may set $x_i=2x^\prime_i$ for $2\leqslant i\leqslant 4t+2$.

\begin{proof}[Proof of (ii)]

By Lemma~\ref{lem:detminor}, it is enough to calculate $2\times 2$ minors of $X^{-1}$.
It is straightforward to check that each possible $2\times 2$ minor of $4t(4t+1)X^{-1}$ has the form $\pm 2(4t+1)\alpha$, where $\alpha$ is an element of the set $S$ given by
\begin{align*}
	S &= \left \{ \begin{matrix}
		0, &
		   4t, &
		    2 r_1, &
		 2 r_2, &
		     (r_1\pm 1),
		    \\
					 (r_2\pm 1), &
  		    (4t+1\pm r_1), &
  		 (4t+1\pm r_2), & (r_1\pm r_2), &
		   (4t+2\pm r_1\pm r_2)
	\end{matrix} \right \}.
\end{align*}

By the proof of (i), both $r_1$ and $r_2$ are odd integers.
Hence each element of $S$ is divisible by $2$, and therefore each $2\times 2$ minor of $4t(4t+1)X^{-1}$ is divisible by $4(4t+1)$.

Any principal minor induced on the first $(2t+1)$ rows of $X$ is, up to sign, equal to $2(4t+1)(4t+1\pm r_1)$ or $4(4t+1)r_1$.
Also any principal minor induced on a subset of the rows  $\{ 2t+2, \dots,4t+2 \}$ is, up to sign, equal to $2(4t+1)(4t+1\pm r_2)$ or $4(4t+1)r_2$.
Thus the greatest common divisor of the $2\times 2$ minors of $4t(4t+1)S^{-1}$ is $4(4t+1)$.
By Lemma~\ref{lem:detminor}, $d_{4t}(X)=8(4t)^{2t-2}$.
Combining Lemma~\ref{lem:minor} and $d_{4t+1}(X)=4(4t)^{2t-1}$, we obtain $x_{4t+1}=d_{4t+1}(X)/d_{4t}(X)=2t$.
\end{proof}

\begin{proof}[Proof of (iii)]
By (i), (ii), and the equality $\det(X)=x_1\cdots x_{4t+2}$, we have that
\begin{align}\label{eq:a1}
x^\prime_2 \cdots x^\prime_{4t}=t^{2t-2}.
\end{align}
Suppose, for a contradiction, that $x^\prime_{2t+2}$ is divisible by an odd integer greater than $1$.
Let $p$ be any odd prime divisor of $x^\prime_{2t+2}$.
Then $x^\prime_2 \cdots x^\prime_{4t}=t^{2t-2}$ is divisible by $p^{2t}$.
However this is impossible; since $t/2^\ell$ is square-free $t^{2t-2}$ cannot have $p^{2t}$ as a divisor.
Therefore $x^\prime_{2t+2}$ is a power of $2$.

Next let $p'$ be any prime divisor of $q$.
By the above, we see that
$x^\prime_{2t+3} \cdots x^\prime_{4t}=2^r q^{2t-2}$ for some $r$.
Therefore, for $i\in\{2t+3,\ldots,4t\}$, $x_{i}$ equals to a power of $2$ times $q$.
Thus, the invariant factors $x_i$ must have the form given in part (iii).
\end{proof}

\section{The Smith normal form of an EW matrix of order $4t+2$ where $4t+1=p^2$, for $p$ a prime} \label{appendix:b}

\begin{theorem}\label{thm:3}
Let $X$ be an EW matrix $X=\begin{pmatrix} R_1 & R_2 \\ -R_2^\top & R_1^\top \end{pmatrix}
$ of order $4t+2$ such that $R_1J=R_1^\top J=r_1J$, $R_2J=R_2^\top J=r_2J$, and $4t+1=p^2$ where $p$ is a prime.
Let $x_1,\ldots,x_{4t+2}$ be the invariant factors of $X$.
Then
\begin{enumerate}
\item  $\textrm{gcd}(r_1,r_2)$ is $1$ or $p$;
\item $x_{4t+2}$ is $2t(4t+1)$ or $2tp$;
\item $x_{4t+1}$ is a divisor of $2t$ if $x_{4t+2}=2t(4t+1)$, and $x_{4t+1}$ is a divisor of $2tp$ and $p$ divides $x_{4t+1}$ if $x_{4t+2}=2tp$;
\item if $t$ is square-free, then the Smith normal form of $X$ is either
\begin{align*}
		&\operatorname{diag}[1,\underbrace{2,\dots,2}_{2t+1},\underbrace{2t,\dots,2t}_{2t-1},2t(4t+1)], \text{ or }\\
&\operatorname{diag}[1,\underbrace{2,\dots,2}_{2t+1},\underbrace{2t,\dots,2t}_{2t-2},2tp,2tp].
\end{align*}

\end{enumerate}
\end{theorem}
\begin{proof}[Proof of (i)]
By the same argument with Theorem~\ref{thm:a1}, we have
\begin{align*}
r_1^2+r_2^2=8t+2=2p^2.
\end{align*}
Then $\textrm{gcd}(r_1,r_2)$ divides $p$, and it follows from the assumption $p$ being prime that $\textrm{gcd}(r_1,r_2)=1$ or $p$.
\end{proof}

\begin{proof}[Proof of (ii)]
By the same argument as in the proof of Theorem~\ref{thm:a1}, we have
\begin{align*}
\det(X)X^{-1}&=2(4t)^{2t-1}\left((4t+1)X^\top-\begin{pmatrix}r_1J & -r_2 J \\ r_2J & r_1 J  \end{pmatrix}\right).
\end{align*}
Therefore the entries of $\det(X)X^{-1}$ are $\pm 2(4t)^{2t-1}(4t+1\pm x)$ where $x\in\{r_1,r_2\}$.
Then the greatest common divisor of $4t+1\pm x$ where $x\in\{r_1,r_2\}$ is
\begin{align*}
\text{gcd}(4t+1\pm r_1,4t+1\pm r_2)&=2\text{gcd}(4t+1,r_1,r_2)=\begin{cases}
2 & \text{ if } \textrm{gcd}(r_1,r_2)=1,\\
2p & \text{ if } \textrm{gcd}(r_1,r_2)=p.\\
\end{cases}
\end{align*}
Therefore
\begin{align*}
d_{4t+1}(X)=\begin{cases}
4(4t)^{2t-1} & \text{ if } \textrm{gcd}(r_1,r_2)=1,\\
4p(4t)^{2t-1} & \text{ if } \textrm{gcd}(r_1,r_2)=p,
\end{cases}
\end{align*}
and thus
\begin{align*}
x_{4t+2}=\det(X)/d_{4t+1}(X)=\begin{cases}
2t(4t+1) & \text{ if } \textrm{gcd}(r_1,r_2)=1,\\
2tp & \text{ if } \textrm{gcd}(r_1,r_2)=p.\qedhere
\end{cases}
\end{align*}
\end{proof}

\begin{proof}[Proof of (iii)]
For the case $x_{4t+2}=2t(4t+1)$, the proof is same as that of Theorem~\ref{thm:a1}(ii).

For the case $x_{4t+2}=2tp$,  $x_{4t+1}$ divides $x_{4t+2}$. Thus $x_{4t+1}$ divides $2tp$.
By $x_{4t+2}=2tp$ and $\det(X)=x_1\cdots x_{4t+2}$, we have that
\begin{align*}
x^\prime_2 \cdots x^\prime_{4t+1}=pt^{2t-1}.
\end{align*}
Since $\textrm{gcd}(p,t)=1$, $p$ divides $x^\prime_{4t+1}$.
\end{proof}

\begin{proof}[Proof of (iv)]
For the case $x_{4t+2}=2t(4t+1)$, the proof is same as that of Theorem~\ref{thm:a1}(iii).

For the case $x_{4t+2}=2tp$,  the proof is also similar to that of Theorem~\ref{thm:a1}(iii), but we include it here.
By (ii) and $\det(X)=x_1\cdots x_{4t+2}$, we have that $x^\prime_2 \cdots x^\prime_{4t+1}=pt^{2t-1}$.
Furthermore we put $x^{\prime\prime}_{4t+1}=x_{4t+1}/p$, which is integer by (iii).
Then $x^\prime_2 \cdots x^\prime_{4t}x^{\prime\prime}_{4t+1}=t^{2t-1}$.
Suppose, for a contradiction, that $x^\prime_{2t+2}>1$.
Let $q$ be any prime divisor of $x^\prime_{2t+2}$.
Then $x^\prime_2 \cdots x^\prime_{4t} x^{\prime\prime}_{4t+1}=t^{2t-1}$ is divisible by $q^{2t}$.
However this is impossible; since $t$ is square-free $t^{2t-1}$ cannot have $q^{2t}$ as a divisor.
Therefore $x^\prime_{2t+2}=1$.
Furthermore
\begin{align}\label{ineq:31}
x^\prime_{2t+3} \cdots x^\prime_{4t} x^{\prime\prime}_{4t+1}=t^{2t-1}.
\end{align}
By (ii) we have $x^{\prime\prime}_{4t+1}\leqslant t$ and thus $x^\prime_i\leqslant t$ for $i\in\{2t+3,\ldots,4t\}$.
Therefore, by \eqref{ineq:31}, we have $x^{\prime\prime}_{4t+1}= t$ and $x^\prime_{i}= t$ for all $i\in\{2t+3,\ldots,4t\}$.
\end{proof}

Barba~\cite{Bar33} showed that for a $\{1,-1\}$-matrix $X$ of order $n \equiv 1\pmod{2}$, Hadamard's inequality can be strengthened to
\begin{align}\label{ineq:B}
|\det(X)| \leqslant (2n-1)^{1/2} (n-1)^{(n-1)/2}.
\end{align}
Moreover, there exists~\cite{CRCHAndbook2007} a $\{1,-1\}$-matrix achieving equality in \eqref{ineq:B} if and only if $n \equiv 1\pmod{4}$ and there exists a $\{1,-1\}$-matrix $B$ such that
\begin{align}\label{eq:b}
BB^\top=B^\top B=(n-1)I+J.
\end{align}
A matrix $B$ is called a \textbf{Barba matrix} if it satisfies \eqref{eq:b}.
It is known that if  $R$ is a Barba matrix of order $n$, then the matrix $\left (\begin{smallmatrix}
	R & R\\
	-R^\top & R^\top
	\end{smallmatrix} \right )$
is an EW matrix of order $2n$.
We therefore pose the following question about Smith normal forms of EW matrices constructed from Barba matrices.
\begin{problem}
	Does there exist a Barba matrix $R$ such that the Smith normal form of $\left (\begin{smallmatrix}
	R & R\\
	-R^\top & R^\top
	\end{smallmatrix} \right )$ is not equal to
			$\operatorname{diag}[1,\underbrace{2,\dots,2}_{2t},\underbrace{2t,\dots,2t}_{2t-1},2t\sqrt{8t+1}]$.
\end{problem}

\section{The Smith normal form of an EW tournament matrix}
\label{sec:snftournament}

In this section we determine the Smith normal form of an EW tournament matrix.

\begin{theorem}\label{thm:snfa}
The Smith normal form of an EW tournament matrix of order $4t+1$ is
\begin{align*}
		&\operatorname{diag}[\underbrace{1,\dots,1}_{2t+2},\underbrace{t,\dots,t}_{2t-2},t^2(4t-1)].
\end{align*}
\end{theorem}

Recall that for an integer matrix $M$, we denote by $d_i(M)$ the greatest common divisor of all $i\times i$ minors of $M$ and $d_0(M)=1$.
The first lemma we need is elementary.

\begin{lemma}[page 33 of \cite{MN}]\label{lem:snfprod}
Let $M$ and $N$ be invertible $n\times n$ integer matrices, and let $m_i$ be the invariant factors of $M$ and
$p_i$ be the invariant factors of $MN$.
Then $m_i$ divides $p_i$ for each $i$.
\end{lemma}

Now we state and prove two lemmas that we will use to prove Theorem~\ref{thm:snfa}.

\begin{lemma}\label{lem:snfa2}
	Let $A$ be an EW tournament matrix of order $4t+1$.
Suppose that a prime $p$ divides $t$.
Then $\rank_p(A)=2t+2$.
\end{lemma}
\begin{proof}
First we show that $\rank_p(A) \leqslant 2t+2$.
Combine \eqref{eq:ew1} and Sylvester's rank inequality \cite[0.4.5 Rank inequalities (c)]{HJ} to obtain
\begin{align*}
    2\rank_p(A)&=\rank_p(A)+\rank_p(A^\top)\\
    &\leqslant \rank_p(AA^\top)+4t+1\\
    &=4t+4.
\end{align*}
Thus $\rank_p(A)\leqslant 2t+2$.

Next we show that $\rank_p(A) \geqslant 2t+2$.
Let $d = \dim( \mathrm{rowsp}_p(A)\cap \mathrm{rowsp}_p(A^\top) )$ be the dimension of the intersection of row spaces.
Then
\begin{align*}
4t&=\rank_p(J-I)=\rank_p(A+A^\top)\\
&\leqslant \rank_p(A)+\rank_p(A^\top)-d\\
&=2\rank_p(A)-d,
\end{align*}
that is, we have that $2t+d/2\leqslant \rank_p(A)$.
Hence it suffices to show that $d\geqslant 3$.

The equations \eqref{eq:ew1} and \eqref{eq:ew2} together with the assumption that $p$ divides $t$ yields
\begin{align*}
W \subseteq \mathrm{rowsp}_p(A) \text{ and } W \subseteq \mathrm{rowsp}_p(A^\top),
\end{align*}
where
\begin{align*}
W=\mathrm{span}\{({\bf 1}^\top_t,{\bf 1}^\top_t,{\bf 1}^\top_{a},{\bf 1}^\top_{2t+1-a}), ({\bf 1}^\top_t,-{\bf 1}^\top_t,{\bf 0}^\top_{a},{\bf 0}^\top_{2t+1-a}),({\bf 0}^\top_t,{\bf 0}^\top_t,{\bf 1}^\top_{a},-{\bf 1}^\top_{2t+1-a})\}.
\end{align*}
Therefore
\begin{align*}
W \subseteq \mathrm{rowsp}_p(A)\cap \mathrm{rowsp}_p(A^\top),
\end{align*}
and thus $d \geqslant 3$.
\end{proof}

\begin{lemma}\label{lem:a2a}
	Let $A$ be an EW tournament matrix of order $4t+1$ and let $l_1,\dots,l_{4t+1}$ be the invariant factors of $A^2+A$.
Then $l_{4t} = t$ and $l_{4t+1} = t^2(16t^2-1)$.
\end{lemma}
\begin{proof}
	Define $\alpha_\pm := t (8 t+3\pm 2 s)+(1\pm s)/2$,  $\beta_\pm := (2 t+1) (2 t+(1\pm s)/2)$, and $\gamma_\pm := t(4 t+1\pm s)$.
	It is straightforward to deduce that the matrix $t^{2}(16t^2-1)(A^2+A)^{-1}$ is equal to
	\begin{align*}
	-t(16t^2-1)I_{4t+1}+\left(
	\begin{array}{cccc}
	 t (8 t+1)J_t & tJ & \gamma_+ J & \gamma_- J \\
	 (5 t+1)J & t (8 t+1)J_t & \beta_+ J & \beta_- J \\
	 \beta_+ J & \gamma_+ J & \alpha_+ J_{a} & tJ \\
	 \beta_- J & \gamma_- J & tJ &  \alpha_-J_{2t+1-a} \\
	\end{array}
	\right),
	\end{align*}
	where $s=\sqrt{8t+1}$.
Hence, to prove the lemma, it suffices to show that
\begin{align*}
d_1(t^2(16t^2-1)(A^2+A)^{-1})&=1, \\
d_2(t^2(16t^2-1)(A^2+A)^{-1})&=t(16t^2-1).
\end{align*}
It is easy to see that $t$ and $5t+1$ are (coprime) entries of $t^2(16t^2-1)(A^2+A)^{-1}$, belonging to $(1,2)$-block and $(2,1)$-block.
Thus $d_1(t^2(16t^2-1)(A^2+A)^{-1})=1$.

All the possible $2\times 2$ minors of $t^2(16t^2-1)(A^2+A)^{-1}$ have the form $\pm t(16t^2-1) \alpha$ where $\alpha$ is an element of the set $S$ given by
\begin{align*}
S= &\left\{0,t,t+1,2 t,2 t+1,4 t+1,5 t+1,t (8 t+1),8 t^2-3 t-1,t \left(16 t^2-16 t-3\right),\right .\\
&16 t^3-16 t^2+t+1,2 t (4 t\pm s),t (4 t+1\pm s),t (4 t-1\pm s),\\
&2 t \left(8 t^2-8 t-1\pm s\right),16 t^3-16 t^2-7 t-1\pm s(4 t+1),\\
&\frac{2 t+1\pm s}{2},\frac{2 t-1\pm s}{2},\frac{4 t+1\pm s}{2},\frac{6 t+1\pm s}{2},\frac{12 t+1\pm s}{2},\\
&\frac{(2 t-1) (4 t+1\pm s)}{2},\frac{(2 t+1) (4 t-1\pm s)}{2},\frac{(2 t+1)(4 t+1\pm s)}{2},\\
&\left . \frac{16 t^2-4 t-1\pm s}{2},\frac{16 t^2+6 t+1\pm s(4 t+1)}{2},\frac{16 t^2-2 t-1\pm s(4 t+1)}{2}\right\}.
\end{align*}

Note that the greatest common divisor of the numbers in the set $S$ is $1$.
Therefore $d_{4t}(t^2(16t^2-1)(A^2+A)^{-1})=t(16t^2-1)$.
\end{proof}

We are now ready to prove Theorem~\ref{thm:snfa}.

\begin{proof}[Proof of Theorem~\ref{thm:snfa}]
Let $A$ be an EW tournament matrix and let $a_1,\ldots,a_{4t+1}$ be the invariant factors of $A$.

\paragraph{Claim 1:} $a_{2t+2}=1$.
Assume that there exists a prime $p$ dividing $a_{2t+2}$. It follows from Lemma~\ref{lem:minor} and Lemma~\ref{lem:snfa1} that $a_1\cdots a_{4t+1}=\det(A)=t^{2t}(4t-1)$.
Thus $p$ is either
\begin{enumerate}[(1)]
\item a divisor of $t$, or
\item not a divisor of $t$ and a divisor of $4t+1$.
\end{enumerate}
For the case (1), Lemma~\ref{lem:snfa2} contradicts the fact that
\begin{align*}
\rank_p(A)=\max\{i\mid \text{$p$ does not divide $a_i$}\}\leqslant 2t+1.
\end{align*}
For the case (2), observe that $p$ is odd and hence $p \geqslant 3$.
Furthermore $p$ is a divisor of $a_i$ for $i=2t+3,\ldots,4t+1$.
Thus $p^{2t}$ divides $4t-1$.
But this is impossible since $p^{2t}>4t-1$ for all $t\geqslant 1$.
Therefore we have $a_{2t+2}=1$.
This proves Claim 1.

\paragraph{Claim 2:} $a_{4t+1}$ divides $t^2(4t-1)$.
By Lemma~\ref{lem:snfprod}, and Lemma~\ref{lem:a2a}, we have that $a_{4t+1}$ divides $l_{4t+1}=t^{2}(16t^2-1)$.
On the other hand $a_{4t+1}$ divides $\det(A)=t^{2t}(4t-1)$.
Therefore $a_{4t+1}$ divides $t^2(4t-1)$.
This proves Claim~2.

\paragraph{Claim 3:} $a_{4t}$ divides $t$.
By Lemma~\ref{lem:snfprod}, and Lemma~\ref{lem:a2a}, we have that $a_{4t}$ divides $l_{4t}=t$. Therefore $a_{4t}$ divides $t$.
This proves Claim~3.

Now, by Lemma~\ref{lem:snfa1}, we have
\begin{align}\label{eq:cm}
a_{2t+3}\cdots a_{4t+1}=t^{2t}(4t-1).
\end{align}
Since $a_i$ divides $a_{i+1}$ for each $i$ and $a_{4t}\leqslant t$, by Claim~3,
we have that $a_{i}\leqslant t$ for $i=2t+3,\dots,4t-1$, from which with $a_{4t+1}\leqslant t^2(4t-1)$ it follows that
\begin{align}\label{eq:cm1}
a_{2t+3}\cdots a_{4t+1}\leqslant t^{2t}(4t-1).
\end{align}
By \eqref{eq:cm}, equality must hold in \eqref{eq:cm1}.
Therefore $a_{2t+3}=\dots=a_{4t}=t$ and $a_{4t+1}=t^2(4t-1)$.
\end{proof}

\section*{Acknowledgements}
\label{sec:acknowledgements}

We thank the referees for their careful reading of the manuscript and their helpful comments.

redpajama_set_name: RedPajamaArXiv