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A supermanobrabilidade é a habilidade de uma aeronave em manter-se sob o controle do piloto e realizar manobras, em situações e modos que excedem aquelas usadas em voo convencional ou realizadas por objetos puramente aerodinâmicos.

Tal capacidade foi inicialmente desenvolvida em 1983, e as primeiras aeronaves a apresentá-la foram os caças russos Mikoyan MiG-29 e Sukhoi Su-27. Desde então, tornou-se padrão em suas aeronaves de quarta e quinta gerações de tal matriz. Existe certa especulação, mas o mecanismo detrás da supermanobrabilidade das aeronaves russas permanece desconhecido. Contudo, análises de post-stall tornaram-se mais comuns nos últimos anos, devido ao avanço da capacidade de manobrabilidade com empuxo vetorial nos motores.

A ênfase no combate aproximado e baixa velocidade permitidos pela supermanobrabilidade, contraria a concepção ocidental de "energia-manobrabilidade", desenvolvida pelo coronel americano John Boyd, que favorece a retenção de energia cinética para aumento de ganho, em possibilidades de manobras em embates duradouros.

Manobrabilidade aerodinâmica x supermanobrabilidade

A manobrabilidade tradicional de aeronaves está atrelada ao fluxo de ar que passa pelas superfícies de controle da aeronaves, caso dos ailerons, profundor, flaps, air brakes e leme (navegação). Algumas dessas superfícies de controle podem ser combinadas com os chamados "ruddervators" de lemes V-tail, mas as propriedades básicas continuam inalteradas. Quando a superfície de controle é movida para apresentar um ângulo com o fluxo de ar em sentido contrário, isso altera o fluxo de ar sobre a superfície, mudando a pressão distribuída, aplicando forças sobre o profundor, leme e aileron. O ângulo de controle de superfície deflete e resulta em força direcional sobre a aeronave sendo controlada tanto pelo piloto quanto pelos sistemas de controle dispostos sobre o avião para manter a altitude desejada, assim como o "pitch", "roll" e "heading". Para controle de manobrabilidade convencional ser mantida, a aeronave precisa manter velocidade suficiente e baixo ângulo de ataque para ter fluxo de ar sobre as asas e controle sobre as superfícies de controle. O fluxo de ar diminui assim como a efetividade as superfícies de controle e manobrabilidade. Em outro ponto, se o ângulo de  ataque exceder seu valor crítico, o avião entrará em stall. Os pilotos são treinados para evitar stalls ao realizarem acrobacias aéreas e, especialmente, em combate, já que uma aeronave em stall pode permitir que o oponente ganhe grande vantagem na tentativa de recuperação da perda de velocidade de controle do avião.

A velocidade da aeronave tem ao realizar sua máxima aerodinâmica de manobrabilidade ainda é conhecida como corner airspeed; que em altíssimas velocidades as superfícies de controle não conseguem operar em sua máxima eficiência devido ao estresse elevado e induzida instabilidade devido ao alto fluxo de ar sobre as superfícies de controle. Em baixas velocidades ocorre o redirecionamento do ar sobre elas, o que diminui a força necessária aplicada sobre a aeronave, não levando a taxa máxima. Dessa forma, é ideal manutenção de velocidade na "corner airspeed" para melhor manobrabilidade.

Em uma aeronave com capacidade de supermanobrabilidade, o piloto pode manter a alta taxa de manobrabilidade mesmo abaixo da "corner airspeed", ao menos com controle de limitação de altitude sem perda devido a velocidades abaixo de "stall". Tais aeronaves são capazes de realizar manobras que são impossíveis a sistemas de aerodinâmica convencionais. Recentemente, devido ao aumento de aviões com propulsão a jato, drones tiveram melhorias no potencial de voo com angulos de ataque superiores a 90º, assim como domínios sobre características "post-stall".

Evidência
Não existe um manual apontando quais exatas características são necessárias para uma aeronave ser classificada como supermanobrável. De qualquer modo, a supermanobrabilidade possui definição própria: a habilidade da aeronave realizar manobras com alto ângulo de ataque que são impossíveis de serem realizadas com aeronaves convencionais. Tais manobras incluem o Pugachev's Cobra e o Herbst maneuver (também conhecido como o "J-turn").

Algumas aeronaves são capazes de realizar o Pugachev's Cobra sem ajuda de mecanismos que ajudam em manobras post-stall, caso do empuxo vetorial; Caças avançados da quarta geração como o Su-27, MiG-29, juntamente com suas variantes, são capazes de realizar tais manobras utilizando motores convencionais sem empuxo vetorial. A habilidade de tais aviões para realiza-las está relacionada a inerente instabilidade de tais em comportamento post-stall.

Aeronaves supermanobravéis atualmente em uso
Mig-29
Sukhoi Su-27
Mikoyan MiG-35
Lockheed Martin F-22 Raptor
Sukhoi Su-35
Sukhoi PAK FA

Acrobacias aéreas


Alice, your friend, Mika, is outside. She said you two are going out to watch a movie. What movie is that?
We are going to watch Spider-Girl. It is showing today and we have been waiting for it since last year.
I heard about that movie too. Many people are excited to watch Spider-Girl.
It is exciting to watch, especially that they got the perfect actress to play as Spider-Girl. You know you can come with us to watch it.
I will watch Spider-Girl with my friends tomorrow. Right now, I will just watch my favorite television show, The Lost Island.
Okay then. Enjoy watching. See you later, brother!
1. What is the latest movie you watched so far? Did you like it?
2. What is your most favorite English and Japanese movie?
3. Who is your favorite English movie actor and actress?
4. Name one movie that you would like to happen in real life.
5. What are your favorite TV shows? Why do you like it?
7. Is there a TV show that everyone seems to like but you don't like? Why?
8. Suggest a good TV show (Japanese and English) that I should watch.
9. How many hours a day do you watch television?
10. Would you ever join a talent competition show on television? Why or why not?


The Algeria Erasmus+ National Office team conducted a follow-up meeting at the University of Mostaganem, on the 2nd of October, 2018. This monitoring concerned the Capacity Building in the Field of Higher Education (CBHE) project "e-LIVES" (e-Learning InnoVative Engineering Solutions).
Prof. Benachenhou is in charge of the project at the University of Mostaganem. On October 3, 2018, he organized a visit to the Electromagnetism and Guided Optics laboratory, which includes material acquired through the Tempus and Erasmus+ Projects in which the University of Mostaganem previously participated as a partner.


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Luke Bond
Luke Bond, whose single 'On Fire' (feat. Roxanne Emery) was nominated for 'Best Trance Track' at the 30th annual International Dance Music Awards in 2014, has been on quite the roll since he returned from a sabbatical year spent honing his craft. Considering the tidal wave of stand-out singles, gigs on international stages at Electric Zoo, Godskitchen, Hollywood Palladium, Groove Cruise and Marquee Las Vegas, remixes for the likes of Cosmic Gate and Gareth Emery, and the huge mainstage debut of his track 'U' by Armin van Buuren at the mainstage of Ultra Music Festival 2018, the U.K. wonderworker counts as one of Trance's most exciting acts of the moment. 'Clockwork' shows fans from all over the world exactly why he's regarded as such.
All PostsProgressive | Big Room | ElectroReleasesTrance
Armin van Buuren x Luke Bond drop monumental collab 'Revolution' (Feat. KARRA)
Want to turn the entire Trance scene on its head?Better let one of its mainfigures team up with one of...
Luke Bond – Clockwork signed exclusively with Armada Music
After releasing a slew of acclaimed Trance tunes on a variety of Armada-related labels, U.K. DJ and producer Luke Bond...

redpajama_set_name: RedPajamaCommonCrawl


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Q: Why is there a derivative with respect to psi of potential there in klein gordon equation? In the Wikipedia article on the Klein Gordon equation there is a section titled Klein–Gordon equation in a potential that gives the equation for a field $\psi$ in  potential $V$ as:
$$ \Box\psi + \frac{dV}{d\psi} = 0 $$
Why is there a derivative of the potential in this equation? What does it mean?

A: Because this is the way, the equation of motion is derived from the action principle. You have some action functional and look for the stationary points of this functional, such that for any infinitesimal in some sense function $\delta \phi$, satisfying certain conditions, like differentiability up to given order, and vanishing on the boundaries of the integration domain.
More precisely, action for a Klein-Gordon theory with some potential is:
$$
S = \int d^{D} x \left(\frac{1}{2}\partial_\mu \phi \partial^\mu \phi - V(\phi) \right)
$$
We look for the stationary points of the functional, such that for $\phi + \delta \phi $, variation vanishes in the first order:
$$
\delta S = \int d^{D} x \left(\partial_\mu \delta \phi \partial^\mu \phi - \frac{\partial V(\phi)}{\partial \phi} \delta \phi \right)
$$
Here note, that the leading order term in $V(\phi + \delta \phi) - V(\phi)$ is proportional to derivative of $V(\phi)$ with respect to field $\phi$. Then after integration by parts, one has:
$$
\delta S =  \int d^{D} x \left( \partial_\mu \partial^\mu \phi + \frac{\partial V(\phi)}{\partial \phi} \right) \delta \phi
$$
And in order for this variation $\delta S$ to be zero for all admissible variation of $\phi$, the following equation has to hold:
$$
\partial_\mu \partial^\mu \phi + \frac{\partial V(\phi)}{\partial \phi} = 0
$$
Which is the Klein-Gordon equation with potential

redpajama_set_name: RedPajamaStackExchange


So she put together a team of colleagues and knitting instructors, and a weekly pilot project launched recently at the Ontario Science Centre. This group will also include young adults with other neurodevelopmental disorders, physical disabilities or an acquired brain injury. Another KneuroKnits group is planned to begin meeting at Holland Bloorview in January. The team will be studying participants' experiences to examine the program's benefits, and may even publish a manual so other sites can organize KneuroKnits, too.
Insights like this are a part of what makes Dr. Penner so valued at Holland Bloorview, where she is also the Holland Bloorview Kids' Rehabilitation Hospital Foundation Chair in Developmental Pediatrics. One of her key research interests is service delivery. In 2016, she made media waves as the lead author of a study in JAMA Pediatrics showing that eliminating wait lists for intensive autism treatment, then averaging 2.7 years, could save the Ontario government some $53,000 per person over their life to age 65. Moreover, society at large would save $267,000.
Holland Bloorview "has been a great partner," explains Dr. Penner. In 2017 she published a study after conducting in-depth interviews with Toronto pediatricians, finding there were many possible reasons children were waiting for an autism diagnosis. "We found that for some cases of suspected autism, it wasn't so much the doctors' ability to make a diagnosis as their confidence in being able to manage all the different resources and referrals after making the diagnosis," she says. As a result, Holland Bloorview used provincial funding to pilot a program allowing community pediatricians to send families of children they diagnose with ASD directly to its own social workers. "It really does take someone whose full-time job it is to help families navigate the system," Dr. Penner notes.


In the early nineteen twenties, small boatyards dotted the shores of the Kieler Bucht (Bay of Kiel). The cousin of George Dibbern's wife Elisabeth, Günther Niemeyer (a younger brother of George's childhood best friend Hans), was drawn into boat building and passed his master boat builder exam in September 1922. He had already established a reputation as a boat designer and boat builder, winning praise for the ease of handling and the seaworthiness of his yachts. His motto was Simplicitas sigillum veritatis, "Simplicity is the sail of truth." He was interested in the aesthetics as well as the technology of boat building and produced a thesis setting out aesthetic norms for yacht building.
It was in the Tetsche Müller boatyard in Mönkeberg, a little town adjacent to the Niemeyer estate in Heikendorf on the Kieler Bucht, that George first saw the hull of what would become Te Rapunga. In October 1923, Günter Niemeyer completed the design for the 10 metre (31 ft. 11 in.) Baltic double-ender with the lines of a Norwegian pilot cutter, with a beam of 3.32 metres (~ 11 ft.), a draught of 1.5 metres (~5 ft.). Tonnage was listed as 7 tons.
Constructed out of 2.8 centimetre (1 1/8 in.) oak planking, she was originally rigged as a sloop, but has spent most of her life with a ketch rig which Dibbern found suited her better. While he often referred to her as his "cradle" and his "fatty" he did so with affection and admiration for her sailing performance. "Niemeyer had an eye and hand for designing; there was no doubt about that," he wrote in Quest, describing how he was moved to invest in Günter's boat building business.
Unfortunately, there were not enough sailors who could afford the luxury of a yacht in those post war years and the enterprise failed. All that George got out of his partnership was the hull that Günter uncannily had insisted all along was destined to be his.
Te Rapunga is currently undergoing restoration in Tasmania, Australia.


In 1881, F. Edgeworth defines the first principle of economics such that every agent is motivated only by self-interest. The idea of self-interest motivation is conceptualized with a hypothetical economic agent named as "homo-economicus" who behaves rationally and in accord with logical principles, thus seeks to maximize his/her own benefits. Even though the use of homo-economicus, or the "money-making animal" and "money-hungry monomaniac" in the words of John Kells Ingram, as a rational representative agent in economic modeling becomes the norm in economic studies, the presence of irrational and quasi-rational people in the society and the actuality of irrational behaviors (in economic sense) in rational people pose some question about the use of this constructed man to understand the people's economic behaviors.
Pioneered by Adam Smith, one of the founders of capitalism, the idea of homo-economicus is used as a base to construct theories on consumption and investment. In his famous book The Wealth of Nations, Adam Smith emphasizes the importance of self interest in human behaviors and describes the self-interest as the key driver of social prosperity. An economic agent has been constructed as a representative one which typifies the economic behavior of a typical person in the society and it is assumed that this agent systematically behaves rationally when deciding on any issue, especially the economic issues. If we define rationality, from the economics standpoint, as an act that aims for the maximum possible degree of acquirements with given limited resources, this constructed agent will be rational by behaving in favor of his/her own self-interest, hence he/she will always be optimizing.
Would you accept this gamble?
They realized that, even though the expected value of gain in the above gamble (50% x $150= $75) is higher than the expected value of loss (50% x $100= $50), there are fewer takers of the gamble. Instead of showing their other experiment results, we can easily look at some of the marketing tricks to have an idea about the irrational choices we make in our daily life.
In a marketing strategy called Decoy marketing, marketers observe a boost in their sales of an expensive offer when they offer a similar item that is inferior in quality but about the same price of the expensive item. In his book Predictably Irrational Don Areily conducts an experiment using magazine subscription offers where subjects were offered two alternative subscription choices to The Economist magazine, one print only subscription (the inferior option which serves to make the combined offer look like a better value) and one internet and print subscription and they observe huge jump in sales due to the decoying effect.
In another experiment in Predictably Irrational, the author shows that people would walk 15 minutes to save $7 on an $18 item but wouldn't take the same 15-minute trip to save the same amount on a $455 item. In fact, it is the same amount of savings, and a rational person is expected to behave consistently at each case.
The current financial turmoil is a typical example for how people can make irrational decisions. Currently, the world is experiencing a catastrophic financial and economic meltdown started at the subprime mortgage markets in United States and spread out to other financial markets. Some people claim that the irrational behavior of sophisticated investors was the root cause of the recent meltdown. To get a better understanding of the issue, we need to look at the balance of greed and fear in the investors' behavior. While greed leads to overconfidence in the economy and excessive risk taking, the fear in the market leads to inactivity in the market and excessive aversion to risk. When investors make a decision about whether buying a stock, bond or any sophisticated security, they are expected to evaluate the expected reward versus the risk of possible expected loss. As predicted in Kahneman and Tversky study, the wisdom in the Wall Street is that "Fear always trumps greed." Neuroscientists describe the situation when greed dominates the fear as a "moral meltdown," and this is the situation where human brain loses control of reward and loss comparison and completely ignores risks. Take the trading in collateralized debt obligations (CDOs) which was perceived as one of major problems that leads to the recent financial crises. Since the security was rated excellent by the rating agencies and all losses were insured by AIG insurance company, the investors observed too much reward opportunities while no weight was given to the risk of a collapse in the market. That is the point when greed dominates fear in the market and fear fails to balance greed in the normal investor's brain. In the aftermath of the financial collapse, what we see right now is just the opposite of what was the case in the buildup of the financial bubble. Now, both investors and bankers are in a mode of excessive fear where negative news significantly decrease the stock prices and instead of providing credit to the markets, investors sit on a huge pile of cash.
If the excessive greed and overconfidence in the economy results in catastrophic events, should we ask investors to be less greedy? According to the Adam Smith, all humans have sympathy for the rich and famous. We can even find this in Holy Scriptures as well. There is always a sympathy for, and envy of, the rich. Hence, asking people to act against their innate nature may not be a wise decision but we should also keep in mind that this inherent human tendency can lead to moral corruption in the society. What we need is the balance of fear and greed in our behaviors and if one dominates the other, there will be misery for the whole.
Devrim Bulut holds a PhD degree in economics. Having taught at several universities in the United States for the last 5 years, he conducts research on market irrationality.


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I want to share my excitement with this amazing work. I wish If there is such a beauty in my church .
Could I order a prints from your work please?
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I recently purchased Sicario on iTunes in 4k because what the heck, even though I already owned it in Digital HD on Vudu I really wanted to see it in 4k with HDR. And, it was on sale for only $6.99!
But when I went back to Vudu to see if it was upgraded to UHD (the acronym Vudu and Google Play use for 4k resolution), it wasn't. Although, I thought it might have been because Apple and Vudu are now part of that whole Movies Anywhere thing.
Oh, but wait. Lionsgate isn't partners with Movies Anywhere. No wonder why there was no upgrade to the movie. Hmm, what about UltraViolet? Aren't iTunes and Vudu connected that way? They are (see clarification in comments below), but Lionsgate isn't an UltraViolet partner either.
Not to pick on Lionsgate. Paramount and MGM aren't Movies Anywhere partners either. And, Universal titles are no longer supported on UltraViolet. So, in a nutshell, I purchased Sicario twice just to get the movie in 4k.
The lack of consistency across digital movie services doesn't just end with studios not supporting MA or UV though. Sometimes the service just doesn't have the title in 4k. For example, neither Google Play or iTunes sell Avengers: Infinity War in UHD, while Vudu and FandangoNow do.
That's true for many Disney and Disney/Marvel titles. A Wrinkle in Time and Black Panther are two other examples. Vudu and FandangoNow sell them in UHD but Apple and Google miss out.
There's also inconsistency across franchise films. As an example, Vudu sells Sicario and the Sicario: Day of the Soldado in UHD, but the sequel doesn't appear to be encoded with HDR10 like the original. Is that just a mislabeled SKU on the VUDU website? Or, was it really digitally mastered without HDR? It's hard to tell because HDR really isn't something you can be sure of unless you're able to do a direct comparison.
Could there ever be one standardized digital packet of video/audio versions that is distributed across all platforms? Maybe someday. But for now we still have to think about output before deciding on where to buy a digital movie. Much of the inconsistency stems from MA and UV not having universal support. But as far as 4k, one might wonder if Disney is holding back how much 4k is released in anticipation of their own digital service.
I think the deal with Disney is refusing to give in to Apple's plan to sell 4K movies no higher than $19.99 – the same price as new releases for 1080p. Because Disney stubbornly keeps its digital prices higher than the rest of the studios and because Apple wants to leverage "free" 4K upgrades to iTunes users who have existing films in their library and buy a 4K AppleTV (which is a great incentive), Disney and Apple appear to be at a stalemate.
I think part of it is Disney hoarding their content. They won't release most movies to 4k Blu-ray, they don't pre-sell BDs on Amazon. I think they are waiting for their streaming service to launch maybe.
You made a mistake in your article. iTunes is not Ultraviolet. Lionsgate is Ultraviolet, provided it's not acting as distributor for an Independent studio. Sicario is UV on Vudu & FandangoNOW.
Carl thanks for the comment. What I mean to say is iTunes and Vudu are connected through Movies Anywhere, but yeah, iTunes and UV are not the same. Anyway, the point of the article is to express confusion about where to buy 4k.


text:
Jason Williams signs one-year professional deal with Blues
Striker signs contract until end of 2016/17 season.
Southend United are delighted to announce that Jason Williams has signed a new one-year professional contract with the club, which will run until the end of the 2016/17 season.
Williams, a product of the Blues academy has made six substitute appearances for the First Team; two coming earlier this season against Millwall and Chesterfield.
The frontman spent the second half of last season on loan at National League South side Chelmsford City, scoring an impressive seven goals in 13 appearances.


text:
Culture Review: Beard
Review: Beard
McNelis' new musical is gripping and raises the important issue of homophobia in sport.
Tatiana Gilfillan
McNelis's Beard is a musical with a message – and a message that is powerfully delivered despite the light-hearted tone and quickfire quips that make each scene fizz. It starts in a Premier League team locker room, and it soon becomes clear that the banter and macho culture of football has its dark side.
Beard follows the story of Chris Prince (Emilio Campa), the captain of the team, as he attempts to accept his sexuality and make it known to the world, despite the rampant homophobia present in the football culture he is at the centre of. Intertwining with his story are the struggles of Sam (Hero Douglas), who thinks she's found her ideal man and the solution to all her problems in Chris.
The story creates the opportunity for plenty of comic moments, with Chris trying to perform the role of the 'macho' footballer when around his best friend Westy, a scene-stealing Luke Buckley-Harris. Buckley-Harris has a gutsy voice that matches the overbearing yet likeable character he plays, and the fast pace humour of Chris and Westy's scenes makes for great entertainment. Campa has perfect comic timing, switching from genuine enjoyment at being with his mate to not-quite-concealed nerves when he is expected to give a certain answer or take part in 'locker-room banter'. Despite the convincing subtlety of his performance his nervous persona could at times have been more varied. The friendship between these two is one of the most problematic yet intriguing elements of the play, particularly in the fraught scene where Chris reveals his sexuality to Westy. The homophobic response to this revelation jars, yet underneath his baffled fury there is also a well conveyed sense of betrayal at the secrets that have been kept from him as he accuses Chris 'I didn't know the real you at all'.
The musical portrays with great clarity the struggles and difficulties of having a career in such an intensely prejudiced world, leading to interesting questions about the ethics of Chris's behaviour, particularly as involves his relationship towards Sam. In a praiseworthy performance Douglas tugs on the heartstrings as a softly sympathetic woman deeply in love with a man she can't have. In one particularly heartrending scene she perfectly conveys the hurt of Chris's betrayal, passionately exclaiming 'I was his fucking beard'. Praise should also be given to Math Roberts and Elise Busset for their performances as Andy Price and Sam's friend Gabby. Roberts has a star turn in the hilarious cabaret scene, while Busset's occasional lack of projection in the songs is more than made up for by her slick dancing and compelling stage presence.
Throughout the music is catchy and lively, with swinging syncopation and smart, sharp lyrics. There are also some intensely emotional songs, delivered with real feeling by the cast. One song that was particularly effective was the layering of all the different voices of the cast at the end, as the musical works up to an emotional climax. However, at times the tenor of the music felt a little similar, with occasional lack of variety in the melodic line.
Overall though Beard is a gripping musical that raises important issues in a script where humour and emotional substance is perfectly blended.
lgbt theatre
Review: Spring Awakening
Preview: Spring Awakening
"All My Loving"- a love letter to the Beatles' uncompromising "A Hard Day's Night"


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text:
Artículo principal: ISO 3166-2

ISO 3166-2:IT es la entrada para Italia, parte de los códigos de normalización ISO 3166 publicados por la Organización Internacional de Normalización (ISO), que define los códigos para los nombres de las principales subdivisiones (p.ej., provincias, regiones, etc) de todos los países codificados en ISO 3166-1.

En la actualidad, para Italia los códigos ISO 3166-2 se definen para dos niveles de subdivisiones:

 15 regiones y 5 regiones autónomas
 80 provincias, 2 provincias autónomas, 4 entidades regionales descentralizadas, 6 consorcios municipales libres y 14 ciudades metropolitanas

Cada código consta de dos partes, separadas por un guion. La primera parte es IT, el código para Italia en ISO 3166-1 alfa-2. La segunda parte tiene, según el caso:

 dos cifras: regiones

 dos letras: provincias autónomas, entidades regionales descentralizadas, consorcios municipales libres y ciudades metropolitanas

Para las regiones, la primera cifra indica la región geográfica donde se sitúa la subdivisión:

 2, 3, 4: Italia del Norte
 5: Italia central (excepto Lacio)
 6, 7: Italia Sur (incluye Lacio)
 8: Italia insular

107 provincias. Cada código comienza con «IT-», seguido de dos letras, que corresponden generalmente a la sigla automovilística italiana con algunas excepciones: la ISO hasta ahora ha mantenido sin cambio los códigos correspondientes a la antigua provincia de Forlì («FO») —ahora provincia de Forlì-Cesena, matrícula «FC»—, y Pésaro («PS») —ahora provincia de Pesaro y Urbino, matrícula «PU», así como Medio Campidano «VS» cuyo nombre se basa en sus dos capitales, Villacidro y Sanluri —.

Los códigos provinciales de dos letras se usaban en las placas de matrículas entre 1905 y 1994 (excepto para Roma, cuyo código siempre ha sido ROMA en vez de RM).

En 1994, se introdujo una nueva demarcación única para toda la nación, eliminando totalmente los códigos provinciales. Esto se demostró impopular, y en 1999 se permitió añadir el antiguo código de dos letras (o "ROMA") en una franja azul a la derecha de la placa, del estilo de la EU, aunque unos años después se eliminó de nuevo, aunque se conserva la franja azul a la derecha.

Códigos actuales
Los nombres de las subdivisiones se ordenan según el estándar ISO 3166-2 publicado por la Agencia de Mantenimiento ISO 3166 (ISO 3166/MA).

Los códigos ISO 639-1 se usan para representar los nombres de las subdivisiones en los siguientes idiomas oficiales:

 (it): Italiano
 (de): Alemán

Estos códigos son definidos en la norma oficial sin el prefijo ISO 3166-1 correspondiente al país, y por lo tanto no garantizan la singularidad en un contexto global por sí solos. Pueden ser completados añadiendo «IT-».

Los códigos y los nombres de las subdivisiones figuran están listados como en la norma oficial publicada por la Agencia de Mantenimiento de la norma ISO 3166 (ISO 3166/MA).

Las columnas se pueden ordenar en orden alfabético haciendo clic en sus respectivos botones.

Regiones

Regiones autónomas

Ciudades metropolitanas

Provincias

Provincias autónomas (Trentino-Alto Adigio)

Entidades regionales descentralizadas (Friul-Venecia Julia)

Consorcios municipales libres (Sicilia)

Actualizaciones
En el momento de la primera publicación de la 3166-2 en 1998 los códigos identificaban las 20 regiones italianas y las 100 provincias existentes. Se hicieron dos actualizaciones, anunciadas por la ISO 3166/MA en las siguientes publicaciones:

Códigos suprimidos el 9 de abril de 2019

 ^ Volver a:a b c d Estas provincias se han transformado en uniones de territorios inter-comunales, representados  colectivamente en ISO 3166-2:IT como la región de Friul-Venecia Julia.

Código suprimido el 22 de noviembre de 2019

Notas

Véase también
 ISO 3166-2, tabla de referencia para los códigos regionales de los países
 ISO 3166-1, tabla de referencia para los códigos de los países
 NUTS:IT

Enlaces externos
En inglés:
 Cambios en la ISO 3166-2
 Plataforma en línea de la ISO: IT
 Provincias italianas, Statoids.com

IT
Organización territorial de Italia

redpajama_set_name: RedPajamaWikipedia


text:
Famine Early Warning Systems Network
& Regions
& Topics
Agroclimatology
Markets & Trade
Integrated Phase Classification
FEWS NET Data Center
Food Security Classification
Administrative Boundaries
Livelihood Zones
Remote Sensing Imagery
Price & Cross-Border Trade
Agroclimatic Monitoring
From NOAA
From USGS
Key Message Update
Security incidents have occurred almost daily in the last three months
February - May 2020
IPC v3.0 Acute Food Insecurity Phase
1: Minimal
2: Stressed
3+: Crisis or higher
Would likely be at least one phase worse without
current or programmed humanitarian assistance
FEWS NET classification is IPC-compatible. IPC-compatible analysis follows key IPC protocols but does not necessarily reflect the consensus of national food security partners.
FEWS NET Remote Monitoring countries use a colored outline to represent the highest IPC classification in areas of concern.
3: Crisis
4: Emergency
5: Famine
Would likely be at least one phase worse without current or programmed humanitarian assistance
Presence countries:
Since the beginning of the year, more than 500 security incidents causing 1,643 deaths have been recorded in 34 of the country's 45 provinces (ACLED). The Sahel and North Central regions account for 63 percent of incidents and 72 percent of fatalities. Over the past three months, there has been an increase in attacks by non-state armed groups: 21 in August with 109 deaths; 25 in September with 106 deaths; and 29 in October with 176 deaths (WANEP). These incidents continue to increase the number of IDPs, estimated at more than 486,000 in early October.
Crops from the recent harvest are the main source of food in accessible areas. In contrast, in areas most affected by insecurity, internally displaced populations (IDPs) who have been unable to cultivate or have abandoned their fields are dependent on host communities and humanitarian assistance. This assistance, that will be available until December, remains limited to more accessible urban centers (31 percent of IDPs are still living in areas that are difficult to access). Assistance should reach 20 to 30 percent of the populations in Soum province and neighboring municipalities and meet at least 65 percent of their needs. Populations in these areas are therefore experiencing Stressed (IPC Phase 2!).
Usually during this period, sales of agricultural products and livestock are the main sources of household income. However, insecurity is leading to a decrease in livestock market attendance in the north with a 10 to 25 percent decrease in small ruminant supplies in Youba, Djibo and Gorom-Gorom markets. Nevertheless, prices are stable overall or have decreased by 5 to 10 percent compared to the average. Among agricultural producers, the price of peanuts is up 8 percent, but the price of cowpeas and corn are down 11 percent and 18 percent respectively compared to the average. Overall revenues from the sale of agricultural products and livestock are likely below average.
Region Contact Information
Email: fewsinquiry.burkina@fews.net
A concerning nutritional situation in areas most affected by insecurity
Remote Monitoring Report
Increased areas facing Crisis (IPC Phase 3) acute food insecurity
Food Assistance Outlook Brief
Global Weather Hazards
Story Map Gallery
About FEWS NET
The Famine Early Warning Systems Network is a leading provider of early warning and analysis on food insecurity. Created by USAID in 1985 to help decision-makers plan for humanitarian crises, FEWS NET provides evidence-based analysis on some 28 countries. Implementing team members include NASA, NOAA, USDA, and USGS, along with Chemonics International Inc. and Kimetrica. Read more about our work.


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Hand & Wrist Conditions
Injuries and Fractures
The wrist is comprised of two bones in the forearm, the radius and ulna, and eight tiny carpal bones in the palm. The bones meet to form multiple large and small joints. A wrist fracture refers to a break in one or more of these bones.
Fractures of the Hand and Fingers
The hand is one of the most flexible and useful parts of our body. Because of overuse in various activities, the hands are more prone to injuries, such as sprains and strains, fractures and dislocations, lacerations and amputations while operating machinery, bracing against a fall and sports-related injuries.
Distal Radius Fractures
The forearm consists of two bones, the radius and ulna. The radius is the larger of the two forearm bones, and the region towards the wrist is called the distal end. Fractures in this end are most common.
Injuries caused due to stretching or tearing of the ligaments in the wrist are called wrist sprains. These injuries are usually caused by a fall during daily activities or sports activities. Sprains can range from mild to severe based on the extent of injury to the ligament.
Flexor Tendon Injuries
Tendons are the bands of fibrous connective tissue that connect muscles to bone. Tendons aid in movement of the fingers, hand and all other body parts.
There are two types of tendons present in the hand- extensor tendons and flexor tendons. Extensor tendons present on top of the hand help with straightening the fingers. Whereas, flexor tendons that lie on the palm side of the hand help in bending the fingers.
Finger fracture
The hand is made up of 27 bones that form the wrist, palm, and fingers. Fingers can easily injure from daily activities, and fractures are common injuries that can occur. Injury to the metacarpal bone that connects to the little finger are the most common fractures, accounting for about one-third of all hand fractures.
Mallet Finger
Mallet finger is a condition where the end of the finger is bent and does not straighten. It occurs when the extensor tendon on the back of the finger is damaged. The finger joint is a hinge-joint that allows bending and straightening of the fingers. Each finger is composed of 3 phalanges bones, joined by 2 interphalangeal joints (IP joints).
Finger and Thumb Sprain
Injuries that involve tearing or stretching of the ligaments of your fingers are termed as sprains. Sprains in the fingers are most often caused from a fall when you extend your arms to reduce the impact of the fall, or from overuse or repetitive activity of the thumb such as with texting.
Thumb Fracture
A break or a crack in the bones of the thumb is known as a thumb fracture. Thumb fractures can occur from a direct blow, a fall, and muscle contractions or twisting during sports such as football, hockey, skiing and wrestling. Fractures may occur anywhere on the thumb, but a fracture at the base of the thumb, near the wrist, is considered the most serious.
Scaphoid Fracture
The scaphoid bone is a small, boat-shaped bone in the wrist, which, along with 7 other bones, forms the wrist joint. It is present on the thumb side of the wrist causing it to be at a high risk for fractures. A scaphoid fracture is usually seen in young men aged 20 to 30 years. They can occur at two places: near the thumb or near the forearm.
Arthritis of the Hand and Wrist
Arthritis is an inflammatory condition of joints. There are several types of arthritis and the most common type is osteoarthritis or wear-and-tear arthritis. Arthritis affects various joints in the body and the arthritis in hand affects the joint at the base of the thumb.
Arthritis of the Thumb
Arthritis is an inflammatory condition of joints. There are several types of arthritis; the most common type is osteoarthritis or wear-and-tear arthritis that affects the joint at the base of the thumb. Thumb arthritis is more common in women than men, and usually occurs after the age of 40 years.
Ganglion Cyst
Ganglion cysts are swellings that most commonly develop along the tendons or joints of wrists or hands. They can be found either at the top of the wrist, palm side of the wrist, end joint of a finger, or at the base of a finger.
Carpal Tunnel Syndrome is a common, painful, progressive condition that is caused by compression of the median nerve at the wrist area.
De Quervain's Tendinosis
The muscles and bones of the hand are connected by thick flexible tissue called tendons. Tendons are covered by a thin soft sheath of tissue known as synovium. Extensor pollicis brevis and abductor pollicis longus are two tendons located on the thumb side of the wrist.
The ability to bend the fingers is governed by supportive tendons that connect muscles to the bones of the fingers. The tendons run along the length of the bone and are kept in place at intervals by tunnels of ligaments called pulleys.
Hand pain is characterized by distress in the joints and tissues of the hand or fingers. Hand pain can be depicted as pulsating, aching, increased warmth, prickling, irritation and inflexibility.
The wrist is a commonly seen injured joint in the body. Problems include sprains and strains as well as fractures which can occur with lifting and carrying heavy objects, while operating machinery, bracing against a fall, or from sports-related injuries.
Gamekeeper's Thumb
Gamekeeper's thumb, also known as skier's thumb, is a tear of the ulnar collateral ligament, a band of tissue that supports the joint at the base of the thumb. Damage to the ulnar collateral ligament may lead to chronic instability of the thumb, creating problems in its normal functioning.
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text:
24 March, 2021 24 March, 2021 Darryl Chamberlain
Silvertown Tunnel pollution monitoring undecided despite work having started, Khan admits
Piling work is taking place underneath the Thames Cable Car, from where this photo was taken yesterday
Levels of air quality monitoring for the Silvertown Tunnel have not been decided – even though work is already well under way on the £2billion new toll road under the Thames.
London mayor Sadiq Khan took four months to answer an assembly member's question on how Transport for London will watch levels of pollution after the tunnel opens. At present, just one form of pollution, nitrogen dioxide, is due to be monitored.
Khan also took three months to answer a question about pollution monitoring around schools.
The Liberal Democrat candidate for mayor, Luisa Porritt, said it showed that the tunnel was Khan's "dirty little secret".
Piling work is already taking place on the north side of the Thames to make room for the tunnelling machine that will dig the new road from the Royal Docks to the Greenwich Peninsula, where it will join the Blackwall Tunnel approach road. A large area of trees has also been cleared in preparation for work on the tunnel's northern entrance.
Both Khan and TfL have insisted that the tunnel, which is due to open in spring 2025, will not lead to an overall rise in pollution levels. TfL will only concede that a housing development close to the north exit – the Hoola towers – risks seeing a specific increase, and that mitigation measures are planned as part of the permission for the scheme.
The piling work is to create a launch chamber for the tunnelling machine which will head under the Thames to Greenwich. The Crystal – the new City Hall – is to the left of the tower
But while they plan to monitor for nitrogen dioxide – associated with burning petrol and diesel – after opening, there are no plans to monitor for PM10 and PM2.5, microscopic particles which can come from tyres and brake linings and pose a threat even from electric vehicles.
Research carried out in 2016 showed that all of Greater London failed to meet World Health Organisation guidelines on PM2.5 levels, with the areas around the Blackwall Tunnel's approaches being among the worst hit in the capital outside central London.
While Khan and TfL have insisted the tunnel will "eliminate" persistent queueing at neighbouring Blackwall Tunnel, and that the twin crossings would be able to cope with more traffic. Opponents say that the crossings – which would be tolled – and neighbouring roads will simply become more congested.
Construction crews have taken over a large area of riverside. The tunnel would emerge at Tidal Basin Roundabout, to the top of the picture. The twin Hoola towers are the other side of the viaduct
In October, the Liberal Democrat London Assembly member Caroline Pidgeon, an opponent of the tunnel, challenged Khan on the issue, and was told that "my air quality team is considering options for further PM 2.5 monitoring in support of the work already under way".
She asked for an update at November's mayor's question time, but one did not arrive until last Thursday – four months late, and just before City Hall wound down for elections. Khan said his team was still deciding whether not to monitor PM2.5 pollution.
"TfL does not anticipate that Silvertown [Tunnel] will have a significant impact on particulate matter, however, given the significance of this pollutant my air quality team is considering options for further PM2.5 monitoring in support of the work already underway on NO2," he said in a written response.
Khan also took three months to answer a question on air quality monitoring around schools from Conservative assembly member Andrew Boff, another opponent of the tunnel. Of 38 air quality monitoring sites across east and southeast London, 18 are within 400 metres of schools. "TfL does not expect there to be any significant air quality impacts at any schools as a result of the scheme," Khan said last Friday in response to the question submitted in December.
TfL's main forum for engaging with boroughs about the project is a body called the Silvertown Tunnel Implementation Group, which includes council officers from most east and southeast London boroughs – including Greenwich, Lewisham and Bexley – as well as Highways England. It is next due to meet at the end of May, when it will hear more about the monitoring of the scheme.
However, elected councillors play no role in the group – the job is delegated to council officers, and the meetings are not held in public.
A Google Streetview image from September 2020 showing Tidal Basin Roundabout before the crews moved in
Trees have been cut down to make way for the area where the tunnel will emerge
Pidgeon told 853: "It is quite staggering that the mayor has taken four months to answer such a basic question about the monitoring of air pollution from the Silvertown Tunnel. It says everything about the wrong priorities in place and also the lack of transparency of the Silvertown Tunnel Implementation Group that we are in the incredible situation that work has already started on the Silvertown Tunnel, but the actual level of air monitoring is still being considered.
"It is self evident that any new road building project, which will inevitably generate further road traffic, should have extensive air monitoring in place on all approach roads covering both Nitrogen Dioxide and harmful particulate matter. To only monitor one type of air pollution would be indefensible.
"If the advocates of the Silvertown Tunnel hope that their record on air pollution will be overlooked they are seriously mistaken."
The answers came just before City Hall wound down for elections on 6 May. Both Khan and his Conservative opponent Shaun Bailey support the tunnel, while Lib Dem Luisa Porritt and Green candidate Sian Berry have pledged to cancel the scheme.
This stump is all that remains of an area of vegetation and trees next to the busy Lower Lea Crossing dual carriageway
Porritt told 853: "The Silvertown Tunnel project is the Mayor's dirty little secret, so it's no wonder he's also trying to hide the true extent of the new road's impact on air pollution.
"No matter how Khan tries to spin it, the fact is a new road in the city will cause more air pollution and make meeting our climate change targets much harder. The only responsible thing to do is to abandon the project and invest in greener forms of transport instead."
Last week the officials in charge of the tunnel project were told by a member of a TfL committee that they needed to "rebuild trust" with local people. Lewisham, Hackney and Southwark councils all submitted formal objections to the scheme at planning hearings in 2016, while Newham council later changed its mind to oppose it.
Greenwich & Woolwich MP Matt Pennycook – the shadow climate change minister – also opposes the scheme, as does West Ham MP Lyn Brown, whose constituency's entire riverfront is now taken up by the tunnel worksite.
Greenwich's Labour councillors asked for a rethink in 2019 but leader Danny Thorpe rejected a proposal to join a cross-borough campaign against the scheme.
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Крючково () — присілок в Дєдовицькому районі Псковської області Російської Федерації.

Населення становить 3 особи. Входить до складу муніципального утворення Шелонська волость.

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text:
\@startsection{section}{1}{\z@}{24pt plus 2 pt minus 2 pt{\@startsection{section}{1}{\z@}{24pt plus 2 pt minus 2 pt}
{12pt plus 2pt minus 2pt}{\large\bf}}
\def\subsection{\@startsection {subsection}{2}{\z@}{12pt
plus 2pt minus 2pt}{12pt plus 2pt minus 2pt}{\@setsize\subsize{12pt}\xipt\@xipt\bf}}

\makeatother
\raggedbottom

\begin{document}
\title{\bf Secure Communication Using Qubits\footnote{Submitted to Crypto 2005
on February 14, 2005}}
\author{Saied Hosseini-Khayat\footnote{University of Wollongong (Dubai Campus), UAE,
and Ferdowsi University, Iran, \it{saiedk@uow.edu.au}}
, Iman Marvian\footnote{Sharif University of Technology, Iran}
}
\date{}

\maketitle

\begin{abstract}\it
A two-layer quantum protocol for secure transmission of data using
qubits is presented. The protocol is an improvement over the BB84 QKD protocol.
BB84, in conjunction with the one-time pad algorithm, has been shown to
be unconditionally secure. However it suffers from two drawbacks:
(1) Its security relies on the assumption that Alice's qubit
source is perfect in the sense that it does not inadvertently
emit multiple copies of the same qubit. A multi-qubit
emission attack can be launched if this assumption is violated.
(2) BB84 cannot transfer predetermined keys; the keys it can
distribute are generated in the process.
Our protocol does not have these drawbacks.

As in BB84, our protocol requires an
authenticated public channel so as to detect an intruder's interaction
with the quantum channel, but unlike in symmetric-key cryptography,
the confidentiality of transmitted data does not rely on a shared secret key.\\

{\bf Key words}: quantum cryptography, key distribution, data security.
\end{abstract}

\@startsection{section}{1}{\z@}{24pt plus 2 pt minus 2 pt{Introduction}\label{sec:intro}

Since the appearance of the BB84 protocol \cite{bb84},
a growing class of quantum
cryptographic protocols has emerged.
BB84 and its variants \cite{gisin2002,lomo2000} aim at
providing perfect security in transferring classical data
between two parties.
However, these protocols are key establishment protocols rather
than data transfer protocols.
They provide ultimate data security by generating a random
sequence of bit which is shared between Alice and Bob.
The sequence is then used as a one-time pad or as a
symmetric key. Because this sequence is generated in the process of protocol
execution, it cannot be known in advance. This can be a drawback in
applications where a predefined sequence of bits is to be distributed
securely. Our protocol can be used to avoid this drawback.
Also, though expensive, in some applications it may be
justifiable to use the quantum channel to send confidential data
as opposed to using the classical channel.
Our protocol can be used in those situations as well.

It has been shown that BB84 is unconditionally secure
\cite{biham2000,gottesman02}. However this fact depends on the assumption
that the qubit source does not emit multiple replicas of the same qubit.
In BB84, Alice generates a sequence of qubits which have been chosen at random
from one of two predefined bases.
She then sends a random sequence of bits using those qubits. Bob
makes measurement on the qubits upon receipt and then waits for the
basis information from Alice. The basis information comes through a
public authenticated channel. If the qubit source emits multiple replicas
for each intended qubit, an eavesdropper (Eve) on the quantum channel can
capture and preserve the qubits until the basis is announced by Alice.
At that time, Eve can perform measurement in the correct basis and obtain
the bits send by Alice. This is called a \emph{multi-qubit emission attack}.
Our protocol is not vulnerable to this kind of attack because information
crucial to correct measurement of the qubits is not transmitted or broadcast.

In our protocol, as in BB84 and other key exchange protocols,
the availability of an authenticated
(though not private) channel is an indispensable element.
This channel is the very essential means by which Bob can identify
Alice and differentiates her from Eve. In key exchange protocols
without an authenticated channel, Eve can launch a successful
man-in-the-middle attack. In our protocol, qubits are exchanged between
Alice and Bob, and they must make sure that the qubits are not altered or
inserted
by Eve. An authenticated channel is required for them to ensure this.

Our protocol has two major advantages over BB84 (explained above),
but it comes at a cost: The qubits must make a round trip
instead of a one-way trip. At a time when sending qubits over
long distances is a technical challenge, this requirement may sound too
troublesome.
However, there is a trade-off: A round trip must be weighted
against the challenge of making a single-qubit source.

This paper is organized as follows: In the next section, we
present the key ideas and properties used in our protocol.
Then in Section \ref{sec:prot}, the proposed protocols are presented.
In Section \ref{sec:anal}, we discuss the correctness and
security of our protocol.
At last in Section \ref{sec:con}, we summarize the results of
our paper. The appendix contains some of the technical points that
are necessary but are not crucial to understanding the concepts.

\@startsection{section}{1}{\z@}{24pt plus 2 pt minus 2 pt{Preliminaries}\label{sec:basics}

In this paper, we have borrowed some key concepts from BB84.
One key concept which is not from BB84 is the following:
Suppose Bob wants to send one bit to Alice.
Alice sends a random bit to Bob and she remembers this bit.
Bob performs an exclusive-or between Alice's bit and his bit
and sends the result to back Alice.
Alice can recover Bob's bit.
This is perfectly secure if
Eve cannot inspect Alice's bit on the way to Bob. If Eve inspects the bit
being sent from Bob to Alice, she will not know Bob's bit.
Of course, we should also prevent the possibility of a man-in-the-middle
attack by Eve.

It is hard to realize the above idea in classical cryptography.
However, in quantum cryptography this can be realized at least in
one way that is presented in this paper.

The whole idea of quantum cryptography revolves around the concept of
a \emph{qubit}. A qubit $\ket{\psi}$ is a quantum state vector in a two
dimensional Hilbert
space $\mathbb{H}_2$. A qubit can be measured with respect to any given basis
in $\mathbb{H}_2$. In quantum computation, the basis $\{\ket{0},\ket{1}\}$ is
called the \emph{computational basis}. Measuring a qubit
$\ket{\psi}=a\ket{0}+b\ket{1}$ in the computational basis will change it
to $\ket{0}$ or $\ket{1}$ with probabilities $|a|^2$ and $|b|^2$, respectively.
The observed outcome is logic
value '0' with probability $|a|^2$ or logic value '1' probability $|b|^2$.
Hereafter, whenever a qubit is "measured", we mean it is measured in the
computational basis.

In protocols discussed in this paper, Alice and Bob rotate qubits around a known
fixed axis in the Bloch sphere \cite[page 15]{nielsen2000}.
Without loss of generality and for the sake of convenience,
we choose this axis to be the $y$-axis.
To perform a rotation around the $y$-axis by an angle $\theta$, a given qubit
$\ket{\psi}$ must be operated upon by the following operator:
\[
R (\theta) = \left(
\begin{array}{cc}
  \cos{\frac{\theta}{2}}   &   -\sin{\frac{\theta}{2}} \\
  \sin{\frac{\theta}{2}}   &    \cos{\frac{\theta}{2}} \\
\end{array}
\right).
\]
It is easy to verify that for all angles
$\alpha$, $\beta$ and $\theta$, the following statements hold:
\[
R(\alpha) R(\beta) = R(\beta) R(\alpha) = R(\alpha+\beta),
\qquad R^{\dagger}(\theta) = R(-\theta).
\]
We will work with the following family of qubits:
\[
\ket{\psi(\theta)} = R(\theta) \ket{0} = \cos\frac{\theta}{2}\ket{0} +
\sin\frac{\theta}{2}\ket{1}.
\]
Since $\ket{0}$ is the state vector coinciding with the unit vector along the
$z$-axis in the Bloch sphere, then every $\ket{\psi(\theta)}$ is a
rotated version of the qubit $\ket{0}$ rotated by an angle $\theta$ around the
$y$-axis in the \emph{xz}-plane.
Also be reminded that any two qubits whose Bloch sphere representations are
colinear and point in opposite directions, e.g.\ $\ket{\psi(\theta)}$ and
$\ket{\psi(\theta+\pi)}$, are orthogonal to each other and hence they can be
detected and distinguished with perfect certainty.

Suppose $\ket{\psi(\theta)}$ is known by Alice to encode one of two logic values
0 or 1. If she makes measurement in the computational basis on this qubit, she
will obtain 0 with probability $\cos^2\frac{\theta}{2}$ and will obtain
1 with probability $\sin^2\frac{\theta}{2}$.
However if Alice knows the value of $\theta$, and if she operates
$R^{\dagger}(\theta)$ on this qubit before making measurement in the
computational basis, then she will obtain 0 with probability 1 (perfect
certainty).

Now suppose that Alice receives $\ket{\phi}$ and she knows this qubit
is either
$\ket{\psi(\theta)}$, or $\ket{\psi(\theta+\pi)}$.
Without knowing $\theta$, if she makes measurement, the outcome will
be 0 or 1 at random, the probability of each depending on the actual state
she received.
On the other hand, if Alice knows the value of $\theta$,
she can perform the unitary operation
$R^{\dagger}(\theta)\ket{\phi}$ and then followed by measurement in the
computational basis. With probability one, she will obtain 0 if
$\ket{\phi}=\ket{\psi(\theta)}$ or obtain 1 if
$\ket{\phi}=\ket{\psi(\theta+\pi)}.$
Therefore with known angle $\theta$, the states $\ket{\psi(\theta)}$ and
$\ket{\psi(\theta+\pi)}$ represent 0 and 1, respectively, and
can be distinguished with perfect certainty.

In what follows, we use the random variable $X$ to denote the binary information
transmitted, the random variable $Y$ to denote the binary information
received by an intended party, and the random variable $Z$ to denote the
information bit obtained by an intruder.
Adopting Shannon's definition, we deem a protocol as
unconditionally secure if
$H(X|Y)=0$ and $H(X|Z)=H(X)$. The first condition means that $Y$ reveals
everything about $X$.
The latter means that $Z$ reveals nothing about $X$; this is true if and only if
$X$ and $Z$ are independent.

We now prove some propositions that represent our key ideas and will be used
later. Consider a set of distinct angles
$\{\theta_0, \theta_1, \ldots, \theta_{n-1}\}$, all in $[0,2\pi].$
Suppose Alice prepares a qubit $\ket{\psi} = R(\theta_k+\pi X)\ket{0}$,
where $\theta_k$ was selected with probability $p_k$, and
$X \in \{0,1\}$ is a binary random variable (which is Alice's data).

\begin{proposition}\label{prop1}
If Alice sends the qubit to Bob. If Bob knows $\theta_k$, he
can recover $X$ without error.
\end{proposition}
{\flushleft\it Proof.}\\
Bob can perform a unitary operation and obtain $\ket{\phi}$ as follows:
$\ket{\phi}=R^{\dagger}(\theta)\ket{\psi}$
He then makes measurement to obtain $Y$. We have:
\[
\ket{\phi} = \cos(\pi X/2) \ket{0} + \sin(\pi X/2) \ket{1}
\]
\[
\mbox{Prob}(Y=0 \mid X=0) = 1, \qquad
\mbox{Prob}(Y=1 \mid X=0) = 0
\]
\[
\mbox{Prob}(Y=0 \mid X=1) = 0, \qquad
\mbox{Prob}(Y=1 \mid X=1) = 1.
\]
This obviously implies: $H(X|Y)=0$. Hence the desired conclusion follows.
{$\Diamond$}

\begin{proposition}\label{prop2}
Alice sends the qubit to Bob. Eve, who does not knows
$\theta_k$, intercepts and rotates it by an arbitrary angle of her choice
$-\alpha$, makes measurement on the qubit and obtains a binary value $Z$.
There exist a probability distribution $p_k$ and a set
$\{\theta_0, \theta_1, \ldots, \theta_{n-1}\}$ such that
$H(X|Z)=H(X)$.
\end{proposition}
{\flushleft\it Proof.}\\
We have:
\[
\ket{\psi} =
\cos[(\theta_k+\pi X)/2 ]\, \ket{0} +
\sin[(\theta_k+\pi X)/2 ]\, \ket{1}
\]
Eve's rotation produces:
\[
R(-\alpha) \ket{\psi} = \cos[(\theta_k+\pi X-\alpha)/2 ]\, \ket{0} +
\sin[(\theta_k+\pi X-\alpha)/2 ]\, \ket{1}.
\]
The probabilities for Eve's measurement outcome is:
\[
\mbox{Prob}(Z=0 \mid k,X=0) = \cos^2\frac{\theta_k-\alpha}{2}, \qquad
\mbox{Prob}(Z=0 \mid k,X=1) = \sin^2\frac{\theta_k-\alpha}{2},
\]
\[
\mbox{Prob}(Z=1 \mid k,X=0) = \sin^2\frac{\theta_k-\alpha}{2}, \qquad
\mbox{Prob}(Z=1 \mid k,X=1) = \cos^2\frac{\theta_k-\alpha}{2}.
\]
Therefore:
\[
\mbox{Prob}(Z=0 \mid X=0) = \mbox{Prob}(Z=1 \mid X=1) =
\sum^{n-1}_{k=0} p_k \cos^2\frac{\theta_k-\alpha}{2},
\]
\[
\mbox{Prob}(Z=0 \mid X=1) = \mbox{Prob}(Z=1 \mid X=0) =
\sum^{n-1}_{k=0} p_k \sin^2\frac{\theta_k-\alpha}{2}.
\]
The variables $X$ and $Z$ are independent if and only if:
\[ \mbox{Prob}(Z=0 \mid X=0) = \mbox{Prob}(Z=0 \mid X=1), \]
\[ \mbox{Prob}(Z=1 \mid X=0) = \mbox{Prob}(Z=1 \mid X=1). \]
Therefore $H(X|Z)=H(X)$ if and only if:
\begin{equation}\label{eq:33}
\sum^{n-1}_{k=0} p_k \cos^2\frac{\theta_k-\alpha}{2} =
\sum^{n-1}_{k=0} p_k \sin^2\frac{\theta_k-\alpha}{2}.
\end{equation}
We must find $p_k$'s and $\theta_k$'s such that
Equation (\ref{eq:33}) holds regardless of the choice of $\alpha$.
In Appendix A, Claim 1, it is shown that this is possible by setting
\[
n \ge 2, \quad p_k = \frac{1}{n}, \quad \theta_k = \frac{2k\pi}{n},
\quad \mbox{for } k =0,1,2,\ldots,n-1.
\]
Thus the conclusion follows. {$\Diamond$}

\begin{proposition}\label{prop3}
Suppose Alice sends the qubit to Bob.
Eve who does not knows $\theta_k$ intercepts the qubit and
performs a rotation by an angle of her choice $\alpha$. She then
makes measurement on
the qubit and then transmits the resulting qubit, denoted $\ket{\phi_1}$,
to Bob. If Bob knows $\theta_k$, and if he performs the following
unitary operation $\ket{\phi_2} = R^{\dagger}(\theta)\ket{\phi_1}$ followed
by a measurement of $\ket{\phi_2}$, then the channel between Alice and Bob
has an error probability equal to:
\[
\mbox{Prob}(\mbox{error}) =
\sum_{k=0}^{n-1}  p_k \left[
\sin^2\frac{\theta_k}{2} \, \cos^2\frac{\theta_k-\alpha}{2} +
\cos^2\frac{\theta_k}{2} \, \sin^2\frac{\theta_k-\alpha}{2} \right].
\]
\end{proposition}
{\flushleft\it Proof.}\\
We are looking for $P(Y|X)$ and do this by conditioning on $Z$ and $k$.
\[
P(Y|X) = \sum_k \sum_Z P(Y|X,Z,k)\, P(Z|X,k)\, P(k|X).
\]
Since $k$ is chosen with probability $p_k$ independent of $X$, then
\[
P(k|X) = p_k, \quad k=0,1,\ldots,n-1.
\]
After Eve's measurement, there are two cases to consider:
\begin{enumerate}
\item $Z=0$. In this case: $\ket{\phi_1}=\ket{0}$. After Bob's rotation. we have
\[
\ket{\phi_2}= R^{\dagger}(\theta) \ket{\phi_1} =
\cos\frac{\theta_k}{2}\,\ket{0} - \sin\frac{\theta_k}{2}\,\ket{1}.
\]
After Bob's measurement we have:
\[
\mbox{Prob}(Y=0 \mid k,X,Z=0) = \cos^2\frac{\theta_k}{2}, \qquad
\mbox{Prob}(Y=1 \mid k,X,Z=0) = \sin^2\frac{\theta_k}{2}.
\]
\item $Z=1$. In this case: $\ket{\phi_1}=\ket{1}$. After Bob's rotation. we have
\[
\ket{\phi_2}= R^{\dagger}(\theta) \ket{\phi_1} =
\sin\frac{\theta_k}{2}\,\ket{0} + \cos\frac{\theta_k}{2}\,\ket{1}.
\]
After Bob's measurement we have:
\[
\mbox{Prob}(Y=0 \mid k,X,Z=1) = \sin^2\frac{\theta_k}{2}, \qquad
\mbox{Prob}(Y=1 \mid k,X,Z=1) = \cos^2\frac{\theta_k}{2}.
\]
\end{enumerate}
The conditional probabilities for $Z$ given $X,k$ were obtained in
Proposition (\ref{prop2}). Using those, we can compute the following:
\[
\mbox{Prob}(Y=0 \mid X=0) = \mbox{Prob}(Y=1 \mid X=1) =
\sum_{k=0}^{n-1} p_k \left[
\sin^2\frac{\theta_k}{2} \, \sin^2\frac{\theta_k-\alpha}{2} +
\cos^2\frac{\theta_k}{2} \, \cos^2\frac{\theta_k-\alpha}{2} \right],
\]
\[
\mbox{Prob}(Y=0 \mid X=1) = \mbox{Prob}(Y=1 \mid X=0) =
\sum_{k=0}^{n-1}  p_k \left[
\sin^2\frac{\theta_k}{2} \, \cos^2\frac{\theta_k-\alpha}{2} +
\cos^2\frac{\theta_k}{2} \, \sin^2\frac{\theta_k-\alpha}{2} \right].
\]
Note that:
\[
\mbox{Prob}(\mbox{error}) = P(X=0)P(Y=1|X=0)+P(X=1)P(Y=0|X=1)
\]
Therefore regardless of $P(X)$ we find:
\begin{equation}\label{eq:71}
\mbox{Prob}(\mbox{error}) =
\sum_{k=0}^{n-1}  p_k \left[
\sin^2\frac{\theta_k}{2} \, \cos^2\frac{\theta_k-\alpha}{2} +
\cos^2\frac{\theta_k}{2} \, \sin^2\frac{\theta_k-\alpha}{2} \right].
\end{equation}
The proof is complete.
{$\Diamond$}\\

\begin{proposition}\label{prop4}
In Proposition \ref{prop3}, suppose we set
\[
n > 2, \quad p_k = \frac{1}{n}, \quad \theta_k = \frac{2k\pi}{n},
\]
then
\[
\mbox{Prob}(\mbox{error}) \ge \frac{1}{4},
\]
with equality when $\alpha=0$.
\end{proposition}
{\flushleft\it Proof.} \; See Appendix A, Claim 2. {$\Diamond$}\\

This proposition means that by certain choice of $p_k$'s and $\theta_k$'s
(as prescribed above) we can be sure that Eve's interaction will be detected
with probability at least 1/4.
Also, from the point of view of Eve, who wants to minimize her
probability of being detected, this means that her best strategy is
to do measurement only without any rotations.
This result will be used in the analysis
of our qubit authentication protocol (Protocol 1).

\@startsection{section}{1}{\z@}{24pt plus 2 pt minus 2 pt{Proposed Protocol}\label{sec:prot}

First we present a protocol for sending authenticated qubits. Then we present a
protocol for sending secure data using authenticated qubits.

\subsection{Sending Authentic Qubits}\label{sec:saq}

We define an \emph{authentic qubit} as a qubit which has been sent by Alice
and received by Bob such that Eve has not altered it in any way along the way.
Suppose Alice wants to send authentic qubits to Bob. For convenience, we
restrict our discussion to the family of qubits residing in the \emph{xz}-plane.

\begin{protocol}\rm
Suppose Alice has a sequence of $N$ qubits
$ \ket{\psi_1}, \ket{\psi_2}, \ldots, \ket{\psi_N}, $
and wishes to send them to Bob in an authenticated way. We
assume that a classical bidirectional authenticated channel exists between
Alice and Bob. Also assumed is a set of publicly-known
distinct angles $ \theta_k = \frac{2k\pi}{n},
k=0,1,\ldots,n-1.$
\begin{enumerate}
\item Alice selects $M$ integers uniformly at random in $\{0,1,\ldots,n-1\}$,
denoted $k_1, k_2, \ldots, k_M$.
\item Alice creates qubits $ \ket{\phi_1}, \ket{\phi_1}, \ldots, \ket{\phi_M}, $
such that for each $i \in \{1,2,\ldots,M\}$
\[
\ket{\phi_i}= R(2k_i\pi/n)\,\ket{0}.
\]
We call these the \emph{check qubits}.
\item Alice generates $M$ distinct random integers $j_1, j_2, \ldots, j_M$ in
$\{1,2,\ldots,N+M\}$. She then creates a frame of length $N+M$ and inserts
each $\ket{\phi_i}$ at location $j_i$ in the frame.
Then the sequence of qubits
$ \ket{\psi_1}, \ket{\psi_2}, \ldots, \ket{\psi_N}, $
are inserted at the empty locations in the frame preserving their order.
\item Alice sends the frame to Bob. Bob receives the frame.
\item Alice sends the following data to Bob:
\[ (j_1,k_1),(j_2,k_2),\ldots,(j_M,k_M). \]
These data are sent via the classical authenticated channel.
Eve can read these data.
\item Knowing $j_i$'s, Bob extracts the sequence of
$\ket{\phi_i}$'s from the frame, and for each $i=1,2,\ldots,M$, he performs
$R^\dagger(2k_i\pi/n) \ket{\phi_i}$. Then he
measures the resulting state. The outcome must be a logical '0' for
all $i \in \{1,2,\ldots,M\}$.
If this condition does not hold for any $\ket{\phi_i}$, the frame
is said to have an \emph{authentication error}.
\item If there is an authentication error, Bob notifies Alice and both
drop the frame. Otherwise, the frame is considered unaltered and
the sequence of $\ket{\psi_i}$'s is deemed to be authentic.
\end{enumerate}
\end{protocol}

At a given level of certainty (determined by parameters $N$ and $M$),
this protocol uses a classical authenticated channel to create an authenticated
quantum channel. This protocol is analyzed in Section \ref{sec:anal}.
We use this protocol as a tool to prevent the man-in-the-middle attack
in our confidential data transfer protocol (Protocol 2).

\subsection{Sending Confidential Qubits}\label{sec:ssq}

In this section, we present a protocol that enables sending confidential
data over the quantum channel. In this protocol, qubits make a round trip
from Alice to Bob to Alice and undergo a unitary operation by Bob along the way.
The protocol is described below.

\begin{protocol}\rm
Suppose Bob has a sequence of $N$ data bits $x_1,x_2,\ldots,x_N$ where
$x_i \in \{0,1\}$.
The existence of a classical bidirectional authenticated channel exists
between Alice and Bob is assumed.
\begin{enumerate}
\item Alice generates a sequence of random integers $k_1,k_2,\ldots,k_N$
where $k_i \in \{0,1,\ldots,n-1\}$. She keeps these integers confidential
to herself.
\item Alice creates $n$ qubits
$ \ket{\psi_{11}}, \ket{\psi_{12}}, \ldots, \ket{\psi_{1N}}, $
such that $\ket{\psi_{1i}} = R(\frac{2 \pi k_i}{n})\ket{0}.$
\item Alice sends the $N$ qubits to Bob using Protocol 1.
\item Bob receives the $N$ authenticated qubits from Alice.
On each $\ket{\psi_{1i}}$, without making measurement, he performs the unitary
operation $R(\pi x_i)$ to produce the qubit sequence
\[ \ket{\psi_{21}}, \ket{\psi_{22}}, \ldots, \ket{\psi_{2N}} \]
where $\ket{\psi_{2i}} = R(\pi x_i) \ket{\psi_{1i}}. $
This sequence is sent to Alice in an authenticated way using Protocol 1.
\item Alice receives the $N$  authenticated qubits.
On each qubit $\ket{\psi_{2i}}$, she perform unitary operation
$R^\dagger(\frac{2\pi k_i}{n}).$ The resulting sequence denoted
\[ \ket{\psi_{31}}, \ket{\psi_{32}}, \ldots, \ket{\psi_{3N}} \]
is then measured to produce bit sequence $y_1,y_2,\ldots,y_N$.
This sequence is deemed to be the sequence sent by Bob.
\end{enumerate}
\end{protocol}

As shown in Propositions 1 through 4, there seems to be no advantage
in selecting $n>3$. Therefore we propose setting $n=3$ in both protocols.

\@startsection{section}{1}{\z@}{24pt plus 2 pt minus 2 pt{Analysis}\label{sec:anal}

In this section, we analyze the two protocols to show that they indeed
serve their intended purposes. In our analysis, we take advantage of
the propositions set forward in the preliminary section.
Since our primary purpose in this paper is introducing a protocol for
confidential data transfer, We analyze Protocol 1 first while assuming Protocol
2 serves its purpose perfectly (i.e., sending authentic qubits).

\subsection{Analysis of Protocol 2}\label{sec:anal2}

We analyze this protocol in two steps: First we show it is correct,
second, we show it is secure.

\subsubsection*{Correctness}
We show that Protocol 2 works correctly by invoking Proposition 1.  Notice
that Alice prepares each qubit $\ket{\psi_{1i}}$
at a certain angle $\frac{2 \pi k_i}{n}$ which she only knows.
Bob encodes his data bit in the qubit by performing
$R(\pi x_i) \ket{\psi_{1i}}$. Doing so means that Alice's qubit either
remains the same or is rotated by an angle $\pi$. In Proposition 1, Alice
encodes her data and sends to Bob. If Bob knows the original qubit's angle
can recover Alice's data. In the case of Protocol 2 (which is slightly
different), it is Bob who encodes his data, but since he does not know the
qubit's angle, he sends it to Alice
who (by the reason of Proposition 1) can recover Bob's data without error.
Therefore, Protocol 2 is correct.

\subsubsection*{Security}
We show that Protocol 2 is secure.
This means that an intruder (Eve)
cannot recover Bob's data. We do this in the following:

\begin{itemize}
\item[a)] Note that if Eve can find out the angle
$\frac{2 \pi k_i}{n}$ for each qubit $\ket{\psi_{1i}}$, she, like Alice, can
recover Bob's data.
An invocation of Proposition 2 will prove that this is not possible.

First notice that Protocol 2 uses the specific set of $p_k$'s and $\theta_k$'s
that satisfies Proposition 2 for security. (Thus Equation (\ref{eq:33})
is satisfied.) Next, notice that the first trip of a qubit from Alice to Bob
corresponds to the same trip in Proposition 2 with the exception that Alice's
data is set to a constant logic '0'.
Since the condition for Equation (\ref{eq:33}) is met by our specific choice
of $p_k$'s and $\theta_k$'s, therefore:
\[
\mbox{Prob}(Z=0 \mid X=0) = \mbox{Prob}(Z=1 \mid X=0) = \frac{1}{2}.
\]
This means that when Eve (intercepts and ) measures each $\ket{\psi_{1i}}$ (
possibly after a rotation of her choice), then she observes logic '0' or '1'
with equal probability, regardless of Alice's choice of angle
$\frac{2 \pi k_i}{n}$. Therefore, she can't get any information about the
qubit's angle
in its trip from Alice to Bob.

\item[b)] Although the conclusion in item b) is good news, note that if there
are only two angles for Alice to choose from (i.e., $n=2$) then that conclusion
will break down because Eve know Alice's data is '0', and since Alice can only
prepare one of the two states $R(0) \ket{0}$ or $R(\pi)\ket{0}$, then
Eve can find out the angle with certainty because the two states are orthogonal.
Therefore we have to impose the condition $n>2$ for the sake of security.

\item[c)] To show that performing Protocol 1 is essential in the first leg
of the trip, we assume that Alice sends the qubits to Bob without performing
Protocol 1 (thus modifying Step 3 in Protocol 2).
While Eve can get no information about the specific choice of
qubit angles by Alice, Eve is still able to launch the following
man-in-the-middle attack:\\
Eve intercepts and performs measurement on Alice's qubits.
This allows her to know the
state of each qubit (they will be either $\ket{0}$ or $\ket{1}$.)
She then sends these to Bob. (Or she might as well drop Alice's qubits and send
her own.) Bob, who is not aware, encodes his data in Eve's qubits and send them
to Alice. Eve intercepts, makes measurement (thus perfectly recovering Bob's
data), and then sends arbitrary qubits to Alice. Alice's measurements (after
appropriate rotations that she knows) will produce gibberish. But Eve has
managed to steal Bob's data.
\emph{Thus} it is absolutely essential that Bob knows that the qubits he
receives are authentic; that is, they come from Alice without Eve having the
opportunity to intervene.
This is the job of Protocol 1 (assuming it works flawlessly.)

\item[d)] On the return trip for qubits (from Bob to Alice), Eve can intercept
and make measurement on the qubits. These qubits carry data, however a direct
invocation of Proposition 2 shows that no data can be gained by Eve if she
intercepts (Apply Proposition 2).
However, if Bob does not use Protocol 1 (in Step 4)
in transmitting the qubits back to Alice,
a malicious Eve can intercept, drop the qubits and insert her own.
In this case, Alice will receive gibberish instead of meaningful data.
This is a disruption of communication as opposed to compromised security.

\item[e)] In this and the next items, we show security against multi-qubit
emission attack. Suppose a qubit
source used by Alice is imperfect and for each intended qubit it creates
multiple replicas in exactly the same state as the original.
This can create a severe security problem in BB84: Eve can capture and preserve
a replica qubit and wait until the angle information is announced on the
public channel. She then can perform measurement in the correct basis and
recover the data.
This cannot happen in Protocol 2 because no such information concerning the
data carrying qubits is ever sent.  Only in Protocol 1, angle information about
the check qubits are sent on the public channel. This cannot give away
information about Bob's data. We will deal with the check qubits when we
analyze Protocol 1.

\item[f)] Another attack using multiple qubits is the following:
In Step 3, Protocol 2, Alice sends qubits to Bob. Because of the multiple-qubit
imperfection, Eve can capture and preserve a qubit replica for each original
qubit Alice sends. During the execution of protocol 1, the check qubits are
identified,
therefore Eve can throw them away. She keeps the non-check qubits which
are intended for carrying data. In the return trip, Bob sends the data carrying
qubits encoded with his data.  Not knowing the qubit's angles,
Eve cannot extract any information from the
replicas she received from Alice nor from the originals or replicas
she receives from Bob. (Now assume Protocol 1 is not perform in the
return trip at Step 4.)
Eve can drop Bob's qubits and insert her own. These are
the replicas she kept now encoded with her fake data.
This data can be received and understood by Alice. This attack affects
the integrity of data rather than its confidentiality.

However, this attack cannot succeed because of using Protocol 1 at Step 4.
At Step 4, Bob sends his qubits ($\ket{\psi_i}$'s) along with the check
qubits ($\ket{\phi_i}$'s) he inserts
in the frame. Eve who receives all these qubits has no information about
the position of these qubits. Therefore she cannot
replace the non-check qubits with her own. She can at best make a random
guess about the position of $N$ non-check qubits and attempt to replace them.
Eve must be at least 1 in ${{N+M}\choose N}$ lucky to succeed.
This means that by a choice of sufficiently large $N$ and $M$, her
luck can be made completely insignificant.
Remember that these two parameters can be set to exceed any given numbers.
If necessary, an appropriate padding algorithm can extend the length $N$
of Bob's data to a required minimum.
Therefore, we have justified the use of Protocol 1 in Step 4, and thus
removed the possibility of using multiple-qubit attack.

\end{itemize}

Up to now, we have discussed the various ways Protocol 2 can be attacked,
and have shown that none can succeed. As it was seen, the security of
Protocol 1 was a key assumption. In the next section, we turn to analyzing
this assumption.

\subsection{Analysis of Protocol 1}\label{sec:anal1}

We show that Protocol 1 allows Alice to send Bob authenticated qubits.
Authenticity of these qubits are essential for guaranteeing the security
of Protocol 2. It is easy to see that Protocol 1 works correctly if Eve
is not present.

Suppose Alice sends to Bob a frame of $N+M$ qubits using Protocol 1. The idea
is that Eve cannot interact with the qubits with vanishing probability of
being detected by Bob.
According to Propositions 3 and 4, because of the special choice of
$p_k$'s and $\theta_k$'s, if Eve performs a measurement (possibly after an
arbitrary rotation) on any one of the check qubits, the probability
that her action produces an error detectable by Bob is at least $P_e = 1/4$.

Now suppose Eve decides to inspect $L$ qubits at random in the frame.
The probability that out of $L$ qubits, $k$ are check qubits is as follows:
\[
\mbox{Prob}(\mbox{Eve picks $k$ check qubits}) =
\frac{ {N \choose {L-k}} {M \choose k }  }{ {{M+N} \choose L }}.
\]
The probability that Eve goes undetected given she inspects $k$ check qubits is
as follows:
\[
\mbox{Prob}(\mbox{Eve undetected} \mid k) = (1-P_e)^k.
\]
Therefore the probability that Eve's interaction is detected when she inspects
$L$ qubits in the frame at random is:
\begin{equation}\label{eq:13}
\mbox{Prob}(\mbox{Eve undetected when she inspects $L$ qubits}) = \sum_{k=0}^L
(1-P_e)^k \,
\frac{ {N \choose {L-k}} {M \choose k }  }{ {{M+N} \choose L }}.
\end{equation}
It is easy to compute, either by using
Equation (\ref{eq:13}) or directly, the following extreme cases:
\begin{itemize}
\item[I.] Eve decides to inspect all qubits in a frame.
The probability that her action is not detected is:
\[
\mbox{Prob}(\mbox{Eve undetected}) = (1-P_e)^M.
\]
If she succeeds, she has been able to measure all $N$ qubits that were
supposed to be protected. This event can be made arbitrarily improbable
by increasing the number of check qubits $M$.
\item[II.] Eve decides to inspect only one qubit in a frame.
The probability that her action is not detected:
\[
\mbox{Prob}(\mbox{Eve undetected}) = 1 - \frac{M P_e}{N+M}.
\]
If she succeeds, at best she has been able to measure only one of the $N$
qubits.
The probability of this event can only be made as small as $(1-P_e)$ by
increasing $M$.
\end{itemize}
In both cases, increasing $M$ implies decreasing the probability that
Eve is not detected.

In another attack on this protocol, Eve does not measure the qubits but
attempts to replace them with her own.
Suppose she wishes to send $L$ qubits of her own.
She must choose $L<N$ qubits in the frame to be replaced.
Lack of any knowledge about the position of check qubits makes her
guess randomly. She will be lucky if she selects none of the check qubits.
The probability of this event is
$
{ {N \choose L} } / { {{M+N} \choose L }}.
$
This probability can be made a small as desired by choosing a sufficiently large
number $M$ of check qubits.
In general, by increasing parameter $M$, we can make the probability that Eve's
measurement or replacement of $L$ qubits goes undetected as small as desired.

\@startsection{section}{1}{\z@}{24pt plus 2 pt minus 2 pt{Conclusion}\label{sec:con}

We introduced and analyzed two novel quantum protocols that together
allow a secure transfer of classical data bits.
The first protocol enables Alice and Bob to exchange authenticated qubits.
The second protocol, which makes use of the first one, enables Alice and Bob
to exchange data bits securely.
The combination has the following merits:
\begin{itemize}
    \item[a)] It can send data over the quantum channel securely.
    This is especially useful in key distribution applications where keys are
    generated in advance rather than on the fly.
    BB84 and its variants cannot transfer predetermined keys.
    \item[b)] In BB84, some information leaks out, therefore a process called
    privacy amplification is required by that protocol.
    In our protocol, no information about the data leaks out.
    \item[c)] Our protocol, unlike BB84, is not vulnerable to multi-qubit
    emission attack.
\end{itemize}
One disadvantage of our protocol is its round trip requirement. This is a price
to pay to protect against multi-qubit emission attack.
While we have discussed most probable attack scenarios, there is
more to be done to prove the unconditional security of this protocol.

\@startsection{section}{1}{\z@}{24pt plus 2 pt minus 2 pt*{Appendix A}\label{appx:1}
\begin{claim}\rm
We show that for all $\alpha$, if
\[
n \ge 2, \quad p_k = \frac{1}{n}, \quad \theta_k = \frac{2k\pi}{n},
\quad \mbox{for } k =0,1,2,\ldots,n-1.
\]
then the following holds:
\begin{equation}\label{eq:43}
\sum^{n-1}_{k=0} p_k \cos^2\frac{\theta_k-\alpha}{2} =
\sum^{n-1}_{k=0} p_k \sin^2\frac{\theta_k-\alpha}{2}.
\end{equation}
\end{claim}
{\flushleft\it Proof.} \; Equation (\ref{eq:43}) implies:
\[
\sum^{n-1}_{k=0} p_k \cos(\theta_k-\alpha) = 0.
\]
Note that the left hand side of the above equation can be written as:
\[
\sum^{n-1}_{k=0} p_k \, \BF{r}_{\theta_k} \cdot \BF{r}_{\alpha} = 0,
\]
where $\BF{r}_{\theta_k}$ and $\BF{r}_{\alpha}$ are two units vectors
located at the origin in the $xz$-plane making an angle $\theta_k$ and $\alpha$
with the $z$-axis, respectively.
The above equation must hold for any $\BF{r}_{\alpha}$, therefore we must have:
\[
\sum^{n-1}_{k=0} p_k \, \BF{r}_{\theta_k} = 0,
\]
If we set $ p_k = \frac{1}{n} $ then we must have:
\[
\sum^{n-1}_{k=0} \BF{r}_{\theta_k} = 0,
\]
This can be made true by setting:
\[
\theta_k = \frac{2k\pi}{n},
\quad \mbox{for } k =0,1,2,\ldots,n-1.
\]
The claim is proved. {$\Diamond$}

\begin{claim}\rm
Proposition \ref{prop4} is true.
\end{claim}
{\flushleft\it Proof.} \;
The error probability given in Equation(\ref{eq:71}) can be re-written
using trigonometric identities as:
\[
\mbox{Prob}(\mbox{error}) =
\frac{1}{2} - \frac{1}{2} \sum_{k=0}^{n-1}  p_k \left[
    \cos^2\theta_k \, \cos\alpha +
\cos\theta_k \, \sin\theta_k \sin\alpha \right].
\]
By setting $p_k = \frac{1}{n}$, and $ \theta_k = \frac{2k\pi}{n},$
we arrive at:
\[
\mbox{Prob}(\mbox{error}) =
\frac{1}{2} - \frac{1}{2n} \sum_{k=0}^{n-1}
    \cos^2\frac{2k\pi}{n} \, \cos\alpha
    - \frac{1}{2n} \sum_{k=0}^{n-1}
\cos\frac{2k\pi}{n} \, \sin\frac{2k\pi}{n} \sin\alpha .
\]
Now note that for any $n>2$:
\[
\sum_{k=0}^{n-1} \cos\frac{2k\pi}{n} \, \sin\frac{2k\pi}{n} = 0.
\]
This can be easily verified by noticing that for $i>0$ the
terms corresponding to $k=i$ and $k=n-i$ cancel each other out.
For $k=0$, the term is already zero, and when $n$ is even, the
term corresponding to $k=n/2$ is also zero.

Other the other hand, notice that for any $n>1$:
\[
\sum_{k=0}^{n-1} \cos^2\frac{2k\pi}{n} = \frac{n}{2}.
\]
This is true because:
\[
\sum_{k=0}^{n-1} \cos^2\frac{2k\pi}{n} =
\frac{1}{2} \sum_{k=0}^{n-1} (1-\cos\frac{4k\pi}{n}) =
\frac{n}{2} - \frac{1}{2} \sum_{k=0}^{n-1} \cos\frac{4k\pi}{n}.
\]
(It can be easily shown that $\sum_{k=0}^{n-1} \cos\frac{4k\pi}{n}=0$.)
Therefore, finally:
\[
\mbox{Prob}(\mbox{error}) =
\frac{1}{2} - \frac{1}{4} \cos\alpha \ge \frac{1}{4},
\]
with equality when $\alpha=0$. {$\Diamond$}

redpajama_set_name: RedPajamaArXiv


Home » Blogs » Stefan Schneider's blog » Oracle posts "Oracle Solaris 11 Preflight Checker for Applications"
Oracle just updated their "Oracle Solaris 11 Preflight Checker for Applications". The tool is taking a multi pronged approach to figure out if a Solaris 10 application will work without any hick ups in Solaris 11.
A Binary/ELF analyzer extracts all bindings and symbols in an application which is available as a binary. It then compares them with the symbols which are available in Solaris 11.
A source analyzer for C/C++ and shell scripts is checking for changes as well. Oracle is replacing the old default shell (Bourne /bin/sh) and the korn shell (/bin/ksh) by a more modern ksh93. The shell scanner will find as well files being used in Solaris 10 which are not available on Solaris 11.
usage of files which got moved or removed.
The information listed above can be found in the User Guide which ships with the tool.
It uses a database with all the symbol, file and API information of Solaris 11.
It uses extensively Dtrace to analyze any given running process. This means that it is able to cover all dependent processes which are being used by an application and it is able to cover all resources which will be loaded or accessed dynamically.