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Mike Pence's America
Posted on January 19, 2017 by Stacey S.
Since the election of Donald Trump in November, countless people have reveled in the hope that perhaps some obscure constitutional gambit or criminal indictment would take place preventing him from taking office on January 20.
Mike Pence's legislative record stands in opposition to his self-proclaimed reverence for life.
The sentiment is understandable to those of us who abhor this man and all that he stands for, but such a scenario would present an awful alternative in the form of Vice President-elect Mike Pence, who would take Trump's place in the Oval Office as our new president.
While Trump has spoken about his frightening and detestable political views, he has no legislative record to back them up. Former congressman and current Indiana governor, Mike Pence, however, has a lengthy one.
And it is positively horrifying.
Mike Pence on Reproductive Rights
This is a man who signed nearly every anti-choice bill that arrived on his desk while he served as governor. But making abortion less accessible wasn't enough: Mike Pence's Indiana was the first state ever to charge, convict, and jail a woman (Purvi Patel, who was suspected of self-aborting) for feticide.
Mike Pence wants to criminalize embryonic stem cell research and is a believer in "personhood." He co-authored the "Life at Conception Act," which seeks to "implement equal protection … for the right to life of each born and preborn human person … at all stages of life, including the moment of fertilization."
He believes a newly fertilized egg, which is nearly impossible to view with the naked eye, is a human person with rights. Personhood bills, were they to become law, would make all abortion (even when a fetus is non-viable and cannot survive outside the womb) illegal.
Contraceptive birth control pills would also be illegal.
"Personhood" has the potential to make even IVF procedures illegal.
In 2016, Mike Pence signed a bill into law that bars women from aborting a fetus that would be born with a life-altering disability.
In 2011, the self-proclaimed "pro-life" vice president-elect spearheaded efforts to defund Planned Parenthood in Indiana and succeeded. In 2013, the lack of funding led to the closure of several Planned Parenthood clinics, one of which served as the lone HIV testing center in Scott County — and, it should be noted, did not provide abortions.
In 2015, following the closures of those Planned Parenthood clinics, a severe and unprecedented HIV outbreak in Indiana led Gov. Pence to declare a public health emergency in the state. Did Pence stop to reflect that perhaps this significant public health emergency might have been much less grave had his defunding of Planned Parenthood not closed the sole HIV testing center in Scott County?
No. And, in fact, quite the opposite: He diligently supported the recent efforts of GOP leaders to repeal the Affordable Care Act (which would strip health insurance from 30 million Americans) and defund Planned Parenthood.
Mike Pence on Sexual Health
In 2000, Mike Pence — a staunch opponent of LGBTQ rights — wanted to add a stipulation to the Ryan White Care Act (which provides extensive funding for HIV/AIDS treatment) that stated:
Congress should support the reauthorization of the Ryan White Care Act only after completion of an audit to ensure that federal dollars were no longer being given to organizations that celebrate and encourage the types of behaviors that facilitate the spreading of the HIV virus. Resources should be directed toward those institutions which provide assistance to those seeking to change their sexual behavior.
Mike Pence didn't want funds going to "organizations that celebrate and encourage" the types of behaviors that facilitate the spread of HIV.
Such as …?
Organizations that celebrate homosexuals and queer individuals as human beings with hopes, dreams, and, yes, CIVIL RIGHTS?
And which institutions provide assistance in "changing" the sexual behavior of individuals? Conversion therapy institutions do.
So is it reasonable to conclude Mike Pence supports the shaming and literal shocking of gay people? In November 2016, when asked about this matter, his representative stated it was "patently false" that Mr. Pence supported or advocated this practice. He stated he simply wanted federal funds "to be directed to groups that promoted safe sexual practices."
What an interesting and blatantly false lie. When it truly mattered, Mike Pence helped to defund one of the premier organizations in this country dedicated to promoting safe sexual practices — Planned Parenthood.
Additionally, Mike Pence does not support comprehensive sex ed, which actually promotes "safe sexual practices" such as STD testing and condom use. In fact, despite the vast amount of medical research to the contrary, Gov. Pence believes condoms are "very, very poor protection against sexually transmitted diseases."
The CDC, more of an authority on the matter, disagrees.
Mike Pence on Civil and LGBTQ Rights
Mike Pence voted against the Employment Non-Discrimination Act in 2007, which would have banned workplace discrimination on the basis of sexual orientation. In defending his vote, he predicted same-sex marriage would induce "societal collapse."
Mike Pence also voted against the Matthew Shepard and James Byrd Jr. Hate Crimes Prevention Act.
Mike Pence opposed the repeal of "Don't Ask, Don't Tell" and made the statement that "Homosexuality is incompatible with military service because the presence of homosexuals in the ranks weakens unit cohesion."
Mike Pence's Family Values
During his time in Congress, Pence opposed parental leave for federal employees — both paid and unpaid.
Mike Pence's Junk Science
Pence, who fancies himself a fan of "small government" (except apparently when it concerns our uteri), is an anti-science troglodyte who once wrote an op-ed stating "smoking doesn't kill."
He argued against the Tobacco Master Settlement Agreement (which funded anti-smoking educational campaigns and permitted states to settle their Medicaid lawsuits against the tobacco industry for the high medical costs of beneficiaries related to illnesses caused by smoking) on the grounds of "big government = bad."
Mike Pence on the Environment
During his tenure as governor, Mike Pence promised to defy all rules relating to energy efficiency and the initiative to reduce greenhouse gas emissions in Indiana just last year! He vowed to be in violation of the law if it were held up in court.
Mike Pence on Gun Control
Mike Pence is such a fan of firearms that he signed a bill allowing guns to be kept in vehicles on school property.
He also signed a bill limiting lawsuits against gun retailers who sell firearms illegally.
Mike Pence's War Record
While Donald Trump was excoriating Hillary Clinton for voting to authorize military action against Iraq, he willfully neglected to mention that his running buddy, Mike Pence, did the exact same thing.
Welcome to Mike Pence's America
In Mike Pence's America, women can be imprisoned over their right to choose. Vice President-elect Pence simply doesn't believe women should be able to dictate whether or not they give birth. He doesn't believe that if you are pregnant with a what would be a severely disabled child, you should be able to make a merciful decision that would potentially save you, your family, and the child from suffering. And he also doesn't support the affordable health care you would need to raise such a child.
Or any child, really.
And, when those children grow up, he would prefer they not learn how to use condoms to protect themselves from potentially deadly diseases. He condones conversion therapy for LGBTQ youth, despite the fact that it can lead to depression, drug use, and suicide.
In Mike Pence's America, keeping money from lifesaving HIV-related research, unless it properly shames and shuns gay people, is acceptable. Protecting and preserving the lives of LGBTQ individuals suffering from HIV and AIDS clearly means very little to this self-proclaimed "pro-life" man.
In Mike Pence's America, hate crimes don't deserve special consideration or prosecution. Protecting the lives of those vulnerable to hate crimes is not a priority for Gov. Pence.
And if you happen to be a federal employee, Mike Pence doesn't support you receiving paid or unpaid time off to give birth or tend to your parenting duties.
In Mike Pence's America, it's not worthwhile to try to protect people's lives in the fight against tobacco, one of the world's top killers.
In Mike Pence's America, a governor violating the law in order for businesses to keep pumping the air full of greenhouse gas emissions is the norm. Protecting the lives of Americans from pollution apparently isn't high on his "pro-life" to-do list.
In Mike Pence's eyes, your child has no right to go to a school that doesn't allow guns near the premises or in the parking lot. In Mike Pence's America, the rights of gun owners take priority over the safety of children. Obviously, it's just not worthwhile for Mike to protect the lives of your already born children from gun violence. He doesn't even think retailers who sell guns illegally should be subject to legal consequences when deaths occur.
Yet he and anti-abortion groups blather on and on about how "pro-life" he is. Do you detect any legitimate passion or concern for people's lives in the policies Mike Pence has supported throughout his lengthy political career?
Donald Trump is a menace of unknown magnitude. Mike Pence, however? He is a menace we fully know and his America is one we should be incredibly afraid to live in.
This entry was posted in National News and tagged abortion, Affordable Care Act, birth control pills, climate change, comprehensive sex education, condoms, conversion therapy, defunding, Don't Ask Don't Tell, Donald Trump, Employment Non-Discrimination Act, environmental justice, fetal anomalies, feticide, gay rights, global warming, gun control, hate crimes, HIV, in vitro fertilization, Indiana, LGBT rights, Life at Conception Act, maternal leave, Matthew Shepard and James Byrd Jr. Hate Crimes Prevention Act, Mike Pence, military, parental leave, personhood, pro-life, Purvi Patel, reproductive rights, Ryan White Care Act, safer sex, same-sex marriage, Scott County, self-abort, self-induced abortion, sexual orientation, smoking, tobacco, Vice President, Vice President-elect by Stacey S.. Bookmark the permalink.
About Stacey S.
Stacey S. works in the wonderful world of fintech and has been an active Planned Parenthood volunteer since 2009. A lifelong feminist and advocate for choice, she has served Planned Parenthood as a clinic escort, public policy advocate, community outreach volunteer, and blogger.
View all posts by Stacey S. →


Some of the toughest decisions you'll ever make are people decisions — made even tougher by the costs associated with a bad hire, which can cost your company $25,000 to $50,000!
As with anything in business, onboarding calls for a detailed, comprehensive plan. Otherwise, you just leave new hires —short- or long-term — to their own devices, which isn't all that efficient or effective.
Compliance. Compliance is just as it sounds. You want to give new temporary employees the opportunity to read and review company-specific policies and complete all the necessary government documents for their employment. An online employee onboarding portal provides easy access and allows HR to monitor and secure the whole process.
Clarification. With clarification, the goal is to provide as much information as possible about the role, its responsibilities, and how it relates to the company. Think of it as giving the new person access to your knowledgebase — and for the entirety of his or her lifecycle at your organization. You may even want to handle the first few days as a hands-on FAQ session to get the candidate up and running as quickly and smoothly as time allows.
Culture. Culture deals mostly with the norms and expectations of any employee at your company. If you've got a company mission or value statement, share it with the new hire. Explain what these things mean to the company and get the person excited about contributing to your goals — even if it's only for a short period of time.
Connection. Being new (and short-term, at that), it's important to make temporary employees feel like their part of the organization. Encourage seasoned employees to develop relationships with new hires. Show them where to access information or who to contact with questions. You want to give them all the necessary resources to thrive while under your employ.
Onboarding often inundates new hires with information from day one. And while it's important to explain the role, company, culture, and other details, you really want to create a two-way channel for communication. This not only shows your interest in the employees' input and feedback but can help identify any issues or gaps in training when they're left guessing what to do. As soon as you begin onboarding, clarify which colleagues to go to with questions. That way, people always feels heard and know when they're doing things right.
Most companies have a fairly standardized onboarding process, which is critical to setting new hires up for success. But we should never lose sight of the person during the process, especially with how much more diverse the workforce has become. Consider and address the needs of every new hire, even when that new hire is a temporary employee. Something as straightforward as pairing diverse new hires with peer mentors can help. They can turn to them for tips and advice on ways to be successful within the organization.
Explain. Describe exactly how to do the task, and then share why it needs to be done in that particular way.
Demonstrate. Roll up your sleeves and show new employees how to do the task as you explain the details.
Observe. Ask the trainees to perform the task as you watch, making sure to note any mistakes and progress.
Advise. Give the new hires feedback on their performance. But don't just offer tips on how to improve. Highlight their accomplishments as well.
Flooding new hires with information for hours on end just increases the chances that they'll tune out after an hour or two. Instead of a marathon session, break up onboarding into blocks. Thirty minutes for one, 60 minutes for another, and so on. In between each session, you can then make introductions, tour the office, or get a jump-start on some of the easier tasks.
Onboarding doesn't stop when all the information has been shared. It's always important to follow up with new hires as they get more familiar with their work, your company, and its culture. Include regular check-ins as part of onboarding and pay close attention to their questions. You may discover a pattern in what needs to be discussed, which can inform your talking points going forward.
If you'd like to learn more about onboarding, or need advice on how to best create an orientation and training process, please let us know. Our goal is to be just as much of a resource for information as we are for finding the right talent for the job — be it temporary, part-time, or full-time.


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Howard Law Group
Redesign of logo
Artful photography to convey strong, confident and tough.
OBJECTIVE: Build brand awareness and WordPress website for one of the nation's top litigation boutiques.
BACKGROUND: The Howard Law Group is a leading business and civil litigation boutique, handling some of the biggest, ugliest, toughest, most-complicated lawsuits. Individuals, Fortune 500 companies, and insurance companies call them to:
Sue the bastards, or
Defend them after they've been sued, in mostly nasty, big-dollar, bet-the-company type cases.
For example, an insurance company may call HLG to defend their factory-owner insured after an explosion at the factory killed some workers, or a major truck crash and pile-up on the expressway, or fatal train derailment, or major theft of trade secrets.
Or they'll represent the tragically injured plaintiff after a terrible accident.
Basically, individuals and businesses call the Howard Law Group when it's all on the line.
When it's their personal or business reputation. When they'll go bankrupt if they lose. Or when a loss would set a negative precedent for the entire industry. When their good name is on the line because it's a family business.
Companies should call Howard when fighting against a company that never settles, or when you need to show rapacious plaintiffs' lawyers that your company never settles.
They're a dynamic, effective trial law firm. Bill Howard almost never loses, he's won over 300 large, complex trials and lost only four in his entire career. He's roughly 300-4. Jean Treece Howard is a master story teller and excellent trial lawyer as well.
Unfortunately, the firm's FindLaw template website was bland and generic, without focus or emotion. It made them look like every other small firm, which they certainly were not.
There was an opportunity here to use powerful photography to help tell the firm's remarkable story, using a powerful Google-friendly, mobile-ready WordPress website. We also needed a new logo and stationery to support the brand message.
Small firms typically have name recognition and awareness issues, so we wanted to use the HOW in Howard as a hook.
Every law firm claims to have a winning record using puffy promises of "successful representations." In marketing, you can say that you win, or you can prove it, with facts and evidence.
When you have the facts, use them.
WHO ARE WE TALKING TO?
Mid- to upper-level executives and in-house lawyers and middle-market and larger companies nationwide.
Other lawyers in private firms for referrals and co-counsel cases.
WHAT ARE WE SAYING TO THEM?
When it's really on the line, you need to hire the Howard Law Group.
When you need to defend hard, you need Bill Howard. When you can't afford to lose or really don't want to, call HLG.
When you're up against an opponent whom you know will try this case, or you must try the case, call us to help.
MARKETING/WEBSITE
WordPress Website Features
Integrated Marketing Campaigns
Responsive Design for Mobile/Tablets
Fishman Marketing's Legal Platform
Integrated video backgrounds
WHERE ARE WE TALKING TO THEM?
On their website, primarily. Also, print ads and direct mailers.
HOW ARE WE TALKING TO THEM?
Strong, confident, tough.
They are serious lawyers doing serious work for serious companies.
Bill Howard is tough and physically intimidating. We wanted to leverage his appearance in the marketing.
The HLG lawyers are very nice, fun, and friendly people, with a good sense of humor. But that's not why clients come here. They get hired because they win.
WHAT ATTRIBUTES COULD WE FEATURE IN THE ADS?
Bill is ~300-4 in big, complex, fiercely contested cases against outstanding trial lawyers.
Bill hasn't lost a big case in over a decade.
We wanted to prove our story using real cases. We built a marketing campaign around real case studies, and an astonishing but provable win-loss record.

redpajama_set_name: RedPajamaCommonCrawl


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La gastronomía sino-estadounidense es un tipo de cocina fusión que surge de los inmigrantes chinos que viajaron a los Estados Unidos y Norteamérica durante el siglo XIX. Esta cocina fue evolucionando en gustos e ingredientes hasta el punto de que los platos servidos en ciertos restaurantes chinos de los Estados Unidos alcanzaron una identidad propia. Algunas de estas preparaciones se hicieron muy populares, y por extensión, se extendieron a otros muchos países occidentales, adaptando su estilo a los gustos locales, y difiriendo así levemente de la gastronomía de China.

Historia
En el siglo XIX los restauradores chinos desarrollaron la gastronomía chino-estadounidense al modificar sus recetas para adaptarlas al paladar occidental. Destinados primeros a los trabajadores del ferrocarril, estos restaurantes se ubicaban en ciudades donde la comida china era completamente desconocida, de forma que sus propietarios tuvieron que adaptarse a los ingredientes disponibles localmente y a los gustos de sus clientes. Los platos del menú solían recibir números, y a menudo se ofrecía como acompañamiento pan y mantequilla.

En este proceso de adaptación, los cocineros inventaron platos tales como el pollo General Tso diseñado en Nueva York. Como resultado, desarrollaron un estilo de comida china que no está presente en China. Los restaurantes y las lavanderías proporcionaron un nicho étnico para los pequeños negocios en una época en la que los chinos eran excluidos de la mayoría de los trabajos por discriminación racial o falta de fluidez en el idioma.

Diferencias con la cocina china original
La gastronomía chino-estadounidense suele tratar la verdura como guarnición, como puede verse en el empleo de zanahoria y tomate, mientras la cocina china enfatiza su uso como ingrediente. La cocina china original emplea frecuentemente verdura de hoja asiática, como el bok choy y el kai-lan, y hace mayor hincapié en la carne y el marisco fresco. Como resultado, la cocina chino-estadounidense suele ser menos amarga que la cocina china.

La gastronomía chino-estadounidense tiende a cocinar muy rápidamente con gran cantidad de aceite y sal. Muchos platos pueden prepararse rápida y fácilmente, necesitando ingredientes baratos. Las frituras y salteados tienden a ser las técnicas culinarias más frecuentes, pudiéndose hacer fácilmente en un wok. La comida también tiene fama de tener niveles altos de glutamato monosódico para acentuar el sabor. Los síntomas del llamado síndrome de restaurante chino se han atribuido a la sensibilidad al glutamato, pero estudios científicos cuidadosamente controlados no han hallado estos efectos negativos del glutamato. La publicidad negativa y las exigencias de los clientes han hecho que muchos restaurantes ofrezcan platos libres de glutamato monosódico.

La mayoría de los restaurantes chino-estadounidenses ofrecen menús escritos en el idioma local o con fotografías para los clientes que no leen chino. Cuando existen menús específicos para el público chino, suelen incluir delicias típicas como hígado, patas de pollo y otros platos exóticos que no suelen ser del gusto occidental.

La cocina chino-estadounidenses suele emplear ingredientes que no son originarios de China y rara vez se usan en este país. Un ejemplo es el frecuente uso del brécol occidental (xi lan, 西蘭) en lugar del brécol chino (gai lan], 芥蘭).

Platos chino-estadounidenses
Algunos platos que aparecen frecuentemente los menús chino-estadounidenses son:

 Pollo General Tso: trozos de pollo fritos con brécol condimentados con jengibre, ajo, aceite de sésamo, cebolleta y guindilla.
 Pollo al sésamo: pollo deshuesado, rebozado y frito que se aliña con una salsa de soja china traslúcida pero roja oscura, dulce y ligeramente agripicante hecha de maicena, vinagre, caldo de pollo y azúcar, servido a menudo con brécol al vapor.
 Ensalada de pollo china: suele contener verdura de hoja cruda y fideos crujientes (o piel de wonton frito), aliñada con sésamo. Algunos restaurantes sirven esta ensalada con mandarinas.
 Chop suey ('sobras' en chino): suele ser una mezcla de verduras y carne en salsa marrón, aunque también puede servirse en salsa blanca.
 Chow mein ('fideos fritos'): fideos fritos con trocitos de carne y verdura. Puede hacerse con pollo, cerdo, gamba o ternera, acompañándose a menudo este último con tomates rojos.
 Chao fen (o chow fun): parecido al chow mein, pero hecho con fideos gruesos y anchos.
 Rangoon de cangrejo: pieles de wonton fritas rellenas de sucedáneo de cangrejo (surimi) y queso crema.
 Galletas de la suerte: inventadas en San Francisco como versión occidental del omikuji senbei japonés, se han vuelto dulces y encontrado hueco en muchos restaurantes chino-estadounidenses. Se han hecho tan populares que incluso alguno restaurantes chinos auténticos las sirven al final de la comida como postre.
 Arroz frito: en sartén, normalmente con trozos de carne, verdura y a menudo huevo.
 Ternera con jengibre (生姜牛肉): cortada en trozos y mezclada con jengibre y diversa verdura china.
 Ternera frita con jengibre (乾炒牛肉絲): cortada en tiras, rebozada, frita y salteada en el wok con salsa dulce. Es una variante de un popular plato del norte de China.

Platos regionales chino-estadounidenses:
 Sándwich de chow mein: con gravy (sur de Massachusetts, Rhode Island).
 Sándwich de chop suey: sobre un pan de hamburguesa, tradicionalmente envuelto en un cono de servilleta y comido con tenedor (costa norte de Massachusetts).
 Sándwich St. Paul: huevos fu yung en un pan de sándwich blanco normal (San Luis, Misuri).
 Pollo Springfield: un tipo de pollo con anacardo que combina pollo empanado frito, anacardos y salsa de ostra (Springfield, Misuri).

Versiones americanizadas de platos chinos

 Huevos fu yung: tortilla china con verdura y carne, servida habitualmente con gravy marrón.
 Rollo de huevo: mientras los rollitos de primavera chinos tienen una corteza fina y crujiente y van rellenos de champiñón, bambú y otras verduras, la versión chino-estadounidense (específicamente la presente en zonas metropolitanas del noreste, como Boston y Nueva York) tiene una corteza gruesa y va rellena de repollo y trocitos de carne o marisco (como cerdo y gamba), pero sin huevo.
 Arroz frito: los platos de arroz frito son populares en los restaurantes chino-estadounidenses gracias a su preparación fácil y rápida y su adecuación a los gustos occidentales. Suele prepararse con arroz frío del día anterior, lo que permite aprovechar el sobrante. Suele usar más salsa de soja que la versión original.
 Pollo Kung Pao: el plato original de Sichuan es muy picante, y la versión occidental suele serlo menos.
 Lo mein ('fideos salteados'): habitualmente hechos con huevo y harina, lo que los hace más masticables que los hechos solo con agua. Los fideos con forma de espaguetis gruesos se fríen en sartén con carne y verdura. A veces este plato se denomina chow mein ('fideos fritos' en cantonés).
 Mei fun: vermicelli de arroz.
 Cerdo mu shu: la versión china emplea ingredientes más típicamente chinos (incluyendo hongos oreja de madera y capullos de lirio de día) y panqueques finos de harina, mientras la versión chino-estadounidense usa verduras más familiares para los occidentales y panqueques más gruesos. Este plato es bastante popular en los restaurantes chino-estadounidenses, pero no mucho en China.
 Sopa de wonton: en la mayoría de restaurantes chino-estadounidenses solo se sirven dumplings wonton en caldo, mientras la versión china puede venir con fideos. La sopa de wonton auténtica cantones es una comida completa consistente en fideos de huevo finos y varios wontons de cerdo y gamba en un caldo de sopa de cerdo o pollo o caldo de fideos. Los wontons chino-estadounidenses, especialmente los de restaurantes de comida para llevar, se hacen a menudo con masa más gruesa.
 Pollo con anacardo: véase más arriba.
 Carne con verdura: ejemplos frecuentes de variantes de este plato son cerdo, pollo, ternera o gamba cocinado con champiñón, arveja china u otra verdura. Este plato se sirve a veces con salsa de soja o de ajo. Son principalmente variantes del salteado al estilo cantonés.

 Cerdo Pekín: este plato existe en la cocina china, pero emplea gai-lan (brécol chino) en lugar de brécol occidental.
 Tomates: debido a que los tomates son plantas americanas, resultan nuevas para la cocina china tradicional. Las salsas de tomate pueden encontrarse en algunos platos chino-estadounidenses como la popular ternera con tomate.

Variantes regionales en la cocina chino-estadounidenses

San Francisco
Desde la década de 1990, muchos restaurantes chino-estadounidenses influenciados por la cocina californiana han abierto en San Francisco y el Área de la Bahía. Los platos típicos de la cocina chino-estadounidense permanecen en el menú, pero se hace más hincapié en las verduras frescas y la carta es apta para vegetarianos.

Esta nueva cocina tiene ingredientes exóticos como el mango y el champiñón común. El arroz integral se ofrece a menudo como alternativa al arroz blanco.

Algunos restaurantes sustituyen las tortillas de harina de trigo a la plancha por tortitas de arroz en los platos mu shu. Esto ocurre incluso en algunos restaurantes que de otro modo no serían identificados como chinos californianos, tanto los más americanizados como y los más auténticos.

Además, existen muchos restaurantes que sirven cocina china de estilo más auténtico, debido al elevado número y proporción de personas de origen chino en San Francisco y el Área de la Bahía. Restaurantes especializados en cocina cantonesa, de Sichuan, hunanesa, del norte de China, Shanghái, Taiwán y Hong Kong están ampliamente disponibles, al igual que restaurantes más especializados, como en mariscos, cocina de Hong Kong al estilo occidental (cha chaan teng, 茶餐廳), dim sum y hot pot. Muchos barrios chinos también cuentan con panaderías chinas, tiendas de té burbuja, siu mei, cocina budista china y reposterías especializadas. El chop suey no está muy extendido en San Francisco, y el chow mein de esta ciudad es diferente al de otras zonas.

Los restaurantes auténticos con menús en chino puede ofrecer 黃毛雞 (pinyin: huángmáo ji, literalmente 'pollo con plumas amarillas'), esencialmente un pollo criado en libertad. El pollo amarillo es apreciado por su sabor, pero necesita ser cocinado adecuadamente para quedar tierno, debido a que su menor contenido en grasa y mayor en músculo. Este plato no suele aparecer en el menú en inglés.

El dau miu () es una hortaliza china que se ha hecho popular desde principios de los años 1990, y ahora no solo aparece en los menús en inglés, por lo general como «brotes de guisante», sino que se sirve a menudo en los restaurantes exclusivos no asiáticos. Originalmente solo estaba disponible durante algunos meses del año, pero ahora se cultiva en invernaderos y está disponible todo el año.

Hawái
La comida chino-hawaiana se desarrolló de forma algo diferente a la del continente estadounidense. Debido a la diversidad étnica en Hawái y a la historia de la influencia china en Hawái, la cocina china local supone un componente de la cocina hawaiana, que es una fusión de diferentes tradiciones culinarias. Algunos platos chinos se sirven típicamente como parte del pā mea ʻai hawaiano. Los nombres de los alimentos también son diferentes, como es el caso de manapua, del hawaiano que significa 'cerdo masticado', para el dim sum bao, aunque la carne no es necesariamente cerdo.

Véase también
 Gastronomía de China
 Gastronomía de los Estados Unidos
 Cocina chifa
 Plato pupu
 Cerdo agridulce

Notas

Enlaces externos

 Chinese Restaurant Project, proyecto para documentar los restaurantes chino-estadounidenses (inglés)
 Chinese food made easy, de la BBC2 (inglés)

Gastronomía de China
Gastronomía de Hawái

redpajama_set_name: RedPajamaWikipedia


This collection is so girly. It includes two pinks: Cosmic Kiss, the obligate Summer pink and Pink Flip-Flops, a light sandal-ready pink. I liked Flip-Flops better with glitter, so I chose to do a reverse gradient, and it made it even more girly. It also includes a stunning green, I was shocked to see how much I liked Sweet Addiction. Again, a very girl/womanly color. I don't wear too much greens, but this one is so worth checking out; my nails looked very sophisticated and effortlessly chic. Lemondrops was ok, I thought it was going to be a two-coater, and for a yellow that's big, but it was three easy coats instead. Blue Capri is super pretty, an almost blurple blue. This one, although not advertised this way, is a color changing nail polish. When you add topcoat, it mutates into a royal blue. I have showed it here without topcoat, but you can see the cool effect of Wacky Laki's swatches here. Overall, the formula was great on all of them. My favorite was Sweet Addiction, it won me over because it is a one-coater. I love how Barielle saves us time with these sets, all of their sets are no-brainers. There is no need to spend an hour thinking which nail lacquers to take on your vacation trip, just toss these in your luggage and go, there is sure to be one color for every occasion.
I love the glitter gradient!
This is such a gorgeous collection!!
Isn't it? I love it too!


People don't change cars nearly as often as they upgrade handsets, and even if you're not happy with your car's stereo head unit, it may not be practical simply to change it for a new one.
Increasingly, audio and other in-car functionality is built in and essentially non-upgradeable. At least not in the sense of ripping out the head unit and banging in a new one.
On the other hand, a lot of add-on kit from aftermarket manufacturers can be downright messy. Added cables, patchy reliability, the usual drill.
In theory, the Pure Highway 300Di offers a solution. You get DAB radio, iPod support and USB connectivity in a single, neatly integrated package. No wires, no mess. The fully fitted price from Halfords is currently £180.
It's certainly a slick-looking package at a reasonable price. And brief acquaintence will convince you it's worth considering. But living with this kind of gadget long term often throws up a few niggles. So what's six months with the Pure Highway 300Di really like?
The answer is mixed. In terms of features, it's mostly good news. The mix of DAB, USB, iPod and line-in pretty much covers all your audio bases. You also get to choose where to locate the line-in and USB connections, so you've no-one but yourself to blame if you're not happy with that.
One major shortfall in this area is telephony support. There is none. What the Pure Highway 300Di really needs is a Bluetooth interface. We suspect an upcoming model will get just that.
The other issue in this area involves the way the Pure Highway 300Di hooks up to your car's audio equipment. In some cases, you can loose analogue radio functionality in the process. Hold that thought.
When it comes to USB, line-in and iPod / iPhone functionality, the answer is simple. It's a yes. We have had occasional glitchiness involving random refusal to recognise an iPhone, but reconnecting has always sorted things out.
Once connected, the Pure Highway 300Di's screen and rotary control make browsing music libraries reasonably ergonomic. OK, it's a little slow to respond to requests for new tracks. But it's tolerable. The ability to pause live radio for up to 15 minutes is very nice, too.
Much more of an issue is the radio reception. Our installation involved the windscreen- rather than roof-mounted aerial. The roof-mounted option provides the best reception, but isn't possible with convertible cars.
As is happens, our test car's head unit also lacked a line-in. In this scenario, the Pure Highway 300Di is plumbed into the the radio aerial input and means you lose analogue radio functionality.
In theory that's fine, because now you have many more channels via DAB. In practice, DAB coverage remains patchy, so you can be left with nothing.
This was exacerbated in our test car by the use of the screen aerial. Put another way, the screen aerial combined with a lack of line-in is the worst possible setup for the Pure Highway 300Di, and one we'd advise against.
Other niggles include sluggish switching of radio stations and infuriating management of the station list. By the latter we mean the way you're never quite sure what's going to happen when you twist the rotary dial to select a new station.
You might move to the next station in the list alphabetically. Or you might find yourself half way down the list. It's probably related to the way stations drop in and out of range. But it needs sorting.
Finally, during a recent trip in Europe, the Pure Highway 300Di failed to pick up any local stations, despite claims that it's Euro-compatible out of the box.
Two thirds of the Pure Highway 300Di experience is a joy. It's a nicely presented unit and mostly a pleasure to use. We reckon the fully fitted price is right and in terms of functionality only wants for Bluetooth.
But there are issues to be aware of. If your car isn't compatible with the roof aerial or you're not willing to have it fitted, we'd say don't bother. Similarly, if your car's audio system doesn't have a line-in, we'd advise against the Pure Highway 300Di because you'll lose standard analogue radio, and that's not acceptable.
However, if you have a tin-top car with line-in, we reckon you'll really enjoy the Pure Highway 300Di. Highly recommended.


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Woibex aspires to highlight international strategies on how to empower women to lead the corporate and business sector and government and explore the dimensions of leadership and knowledge that supports and enables them to hold international positions. Also, the conference will review the most important experiences of leaders in international positions and business. Read More
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Market systems development theory is well documented, but there is relatively little information about how it works in practice, particularly in thin or fragile markets, or those transitioning from relief to development.
Given its relatively recent conflict with Indonesia, unreliable distribution systems, weak market actors and limited supporting services, Timor-Leste provides an interesting place to explore what market development in a fragile environment looks like in practice.
Access to energy represents a great opportunity to test market development in high-risk, low-capacity environments because advances in technology mean a wide range of products are available for households at all levels of income, and energy is a service that even the poorest households pay for — even if that service is low quality, harmful, and expensive.
This learning study looks at the challenges faced by Mercy Corps' Energy For All (E4A) program in Timor-Leste, which facilitated existing actors to build a sustainable market for solar energy products and clean cookstoves.
The learning study provides practical examples of how the E4A team has struggled, learned and adapted. Many of the key lessons will be useful for market development programs in similar settings.


Twenty five people may participate.
Your addy must be on the Geocoin Forum Mailing List.
I will send out two coins to someone (chosen at random) on the Pick and Pass Mission List.
The receiver will keep one of the coins, add one coin to the mission and send it on to the person of their choosing from the list.
That receiver, will in turn, choose a coin, add a coin and send it on.
So each participant receives two coins to choose from.
In the end, the last person, sends the remaining two coins back to me.
I choose 25 people because at 2 weeks a person.... 25 people will take approximately one year.
Nice simple straight forward mission.
Nice and easy. Count me in.
Me likey. I'm in for sure!
I'd love to participate in your mission.
I'm in, this sounds neat.
Sounds like fun, I'd like to participate!
We would love to play!!
Sounds like fun Count me in.
Great idea!! I would like to participate.
Sounds interesting -- add me to the list.
I think it is a good idea for everyone to check their adresses in the googlespreadsheet. I looked through it now and noticed some errors, so fairyhoney, please check it out, yours is not right.
Butterfly_lady also if you signed up here.
There may be more aslo that I do not know.
I'm not sure if you have enough participants yet, but this sounds like a mission that I could be involved with. I checked myself in the google spreadsheet, and my info is correct.
If there is a slot, I am interested!
I think I am too late also but am interested in anyone changes there mind.
This sounds nice and simple, hope I can get in.
Okay, that's 27-28 people. I'll cut it off there and will generate a list of names tomorrow morning. Thanks for the interest!
Cool idea, please put me on an alternates list if any openings appear.
Put me on the alternates list too, this sounds fun.
Sure. I will try it, just walk me thru it.
This sounds like a fantastic idea and would love to participate.
Just found this topic today.
Probably to late, but woud like to join.
Hey, Chicka! Hope you're weathering the cooler temperatures ok!
I had a thought about your mission. Given the interest, have you considered turning this mission into a cointest/race?
I think you're up to (if not already past) 30 participants. These 30 could be divided into two groups of 15, and the "race" portion could be a friendly competition among us coiners. This way, everyone still gets to participate, and it doesn't necessarily take an entire year? The prize for the winning group would be bragging rights, as I'm not sure anyone has 15 coins ready to offer as prizes!
Just a thought...looking forward to seeing this mission in action!
Hmmm... let me see if I understand. Or rather, let me change the mission to this.
Two groups of 15, with myself serving as the beginning person and ending person of the cointest.
I will mail out two packages of three coins to someone on Team 1 and someone on Team 2.
I will randomly pick names to be on the two teams. I reserve the right to "arrange" the teams so that equal number of overseas players (if there are any) will be on each team.
The package will contain a third coin that each participant will have to log as discovered in order to get credit for their team.
Okay, we do have some alternates, so if you believe that for any reason, you will not be able to fulfill the mission requirements, please speak now.
I will divide the names into two teams. We will have the packages going from East coast, to West coast, to Canada, to Australia, to the UK... it is going to be fun to watch! Thank you for your interest.
Don't think so, Team #1 IS the winning team!
I beg your pardon, but Laval was correct the first time...TEAM 2 WILL BE THE VICTORIOUS GROUP!!!
Since you have this East/West/North/South thing set up, does that mean that instead of picking a person off the list, we are supposed to send it to the next person on the list? Doesn't really matter to me, just want the clarification.
I couldn´t have said it better myself. Go team #2!!!!!!
My team name will be "WINNER" That is personal.


As all you Pine Nuts remember, the ruggedly handsome pilot Steve Trevor heroically sacrificed himself at the end of the first Wonder Woman movie, shattering the hearts of people around the globe.
Well. As you can see, it looks like Mr. Trevor makes a miraculous recovery. But anyone who knows Pine signed a multi-picture deal could have predicted that. Also, because you ALWAYS need a love interest. Is my sarcasm coming through?
We can also deduce from the tweet that the second Wonder Woman installment will be set in 1984. Or at least will have flashes to 1984. The sequel is slated to be released Nov. 1, 2019 and has some big shoes to fill. The first movie grossed $817 million at the box office and is universally acclaimed.
More: get your Wonder Woman t-shirts here, and check out these amazing Wonder Woman tattoos.


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

The origin of this paper was our dissatisfaction with the standard definition of the uniform circuit complexity.
In the circuit computation model, a computational problem is associated with a family $C_1, C_2, \dots$ of circuits where each circuit $C_n$ handles $n$-bit inputs in the Boolean case or $n$-qubit inputs in the quantum case.
The most common uniformity condition requires that a single logspace\footnotemark\ Turing machine computes every $C_n$ given $n$ in unary notation.
\footnotetext{Another popular uniformity condition uses polynomial-time rather than logspace, but logspace uniformity is more common.}

This complexity-theoretic uniformity is important for theoretical purposes but is not used for practical programming.
Instead, circuits are usually described parametrically, and this suffices to guarantee uniformity.

A natural problem arises to formalize the notion of parametric uniformity so that one can deal with it theoretically.
In order to address this problem, we looked around for challenging examples.
In this connection, we came across a recent paper \cite{MW} by  Abel Molina and John Watrous describing circuits that simulate quantum Turing machines.
We tried to achieve such simulations by parametric circuits.
This paper reports our results in this direction as well as analogous results for reversible classical Turing machines.

We should emphasize though that the idea of parametric uniformity is just a motivation as far as this paper is concerned.
We intend to formalize parametric uniformity for Boolean and quantum circuits elsewhere \cite{G260}.
For the time being we use the notion of parametrized circuits informally.

It is well known that Turing machines are efficiently simulated by (uniform families of) Boolean circuits. In extended abstract \cite{Yao}, Andrew Yao sketched a proof that quantum Turing machines are efficiently simulated by (rather sophisticated) quantum circuits.
In paper \cite{MW}, Molina and Watrous give an elegant proof of Yao's result, but their circuits and their proofs are still rather involved.

A question arises whether simpler quantum circuits can simulate quantum Turing machines.
A related question is whether reversible Turing machines can be efficiently simulated by reversible Boolean circuits.
\begin{comment}
Since every deterministic one-tape Turing machine is polynomial-time simulated by a deterministic one-tape reversible Turing machine \cite[\S5]{Morita}, this is the same question whether Turing machines and reversible Boolean circuits are polynomial-time equivalent.
\end{comment}

In this paper, we construct reversible Boolean circuits which efficiently simulate reversible Turing machines.
Both the circuits and the simulation proof are rather simple.
Then we give a fairly straightforward generalization of the circuits and the simulation proof to the quantum case.

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

The reader may want to skip this section and consult it only if and when needed.

Reversible Boolean circuits and quantum circuits are mathematical notions requiring precise definitions, and so are Boolean circuit algorithms and quantum circuit algorithms.
Not finding such definitions in popular textbooks, we produced definitions in \cite{G244} and \cite{G250}. To make this paper self-contained, we reproduce those definitions here with minor changes.
An added advantage is useful terminology that arises in the course of the definitions.

\begin{definition}
\label{def:syncirc}
A \emph{syntactic circuit} consists of the following components.
\begin{enumerate}
\item Disjoint finite sets of \emph{input nodes}, \emph{output nodes}, and \emph{gates}.
\item For each gate $G$, two disjoint nonempty finite sets, the set $\Entries(G)$ of the \emph{entries} of $G$ and the set $\Exits(G)$ of the \emph{exits} of $G$.\\
    The sets associated with any gate are disjoint from those associated with any other gate and from the sets in (1).
    The input nodes and gate exits will be called \emph{producers}. The gate entries and output nodes will be called \emph{consumers}.
\item A set of \emph{wires}.  Each wire begins at a producer and ends at a consumer.
    Each producer is the beginning of at least one wire.
    Each consumer is the end of exactly one wire.
\item A gate $G$ is a \emph{direct prerequisite} of a gate $G'$ if some wire goes from an exit of $G$ to an entry of $G'$.
    The direct prerequisite relation is acyclic.
    Its transitive closure is called the \emph{prerequisite} relation (without ``direct''). \qef
\end{enumerate}
\end{definition}

\begin{definition}
Consider a syntactic circuit.
\begin{itemize}
\item A \emph{gate bout} of the circuit is a nonempty set $B$ of gates which form an antichain (so that no gate $G\in B$ is a prerequisite of another gate $G'\in B$).
\item A \emph{schedule} over the circuit is a sequence
  $(B_1; B_2; B_3; \dots; B_T)$
  of disjoint gate bouts such that every gate $G$ belongs to some bout $B_t$ and all prerequisites of $G$ belong to earlier bouts $B_s$ where $s<t$.
\item A schedule $(B_1; B_2; B_3; \dots; B_T)$ is \emph{eager} if every $B_t$ comprises all gates whose prerequisites are all in $\bigcup_{s<t} B_s$ (so that in particular $B_1$ comprises all gates with no prerequisites). \qef
\end{itemize}
\end{definition}

\begin{definition}
\label{def:bcirc}
  A \emph{Boolean circuit} is a syntactic circuit together with an assignment, to each gate $G$, of a Boolean function
  \[ \beta_G: \set{0,1}^{\Entries(G)} \to \set{0,1}^{\Exits(G)}. \]
  An \emph{algorithm} over the Boolean circuit is given by a schedule over the underlying syntactic circuit.
  An algorithm is \emph{eager} if the schedule is eager. \qef
\end{definition}

Intuitively, the algorithm given by a schedule $(B_1; B_2; B_3; \dots; B_T)$ first fires all $B_1$ gates in parallel, then it fires all $B_2$ gates in parallel, and so on. Thus $T$ is the number of steps of the algorithm.
All algorithms over the same Boolean circuit \C\ compute the same Boolean function
  \[ \beta_{\C}: \set{0,1}^{\Inputs(\C)} \to \set{0,1}^{\Outputs(\C)} \]
but may differ in the number of steps.
Below, for simplicity, we restrict attention to eager algorithms.

\begin{proviso}
Below, by default, a Boolean circuit is identified with its eager algorithm.
\end{proviso}

\begin{definition}
\label{def:rcirc}
A Boolean circuit is \emph{reversible} if
\begin{enumerate}
\item each gate function $\beta_G$ is a bijection (so that in particular $G$ has equally many entries and exits), and
\item each provider has exactly one outgoing wire (so that the wires determine a bijection from producers to consumers). \qef
\end{enumerate}
\end{definition}

\section{Turing machines}
\label{sec:tm}

We presume a basic familiarity with Turing machines, in short TMs, and specify a particular TM species.
In this paper, by default, a Turing machine is deterministic and has one two-way infinite tape whose cells are indexed by integers.
We also assume that the tape head moves at every step.
A Turing machine is given by a finite set $Q$ of (control) states, a tape alphabet \Gm\ including a special blank symbol, and a partial transition function
\[
\delta : Q \x \Gamma \to Q \x \Gamma \x \set{-1,+1}. \qefhere
\]
If the machine is in state $p$, its head is scanning symbol $a$ in tape cell $i$, and $\delta(p,a) = (q, b, D)$, then in one step the machine will change state to $q$, write $b$ in the tape cell $i$, and move the tape head to cell $i+D$. (The change is vacuous if $q=p$ and the writing is vacuous if $b=a$.)
If $\delta(p,a)$ is undefined then the machine halts.
Initially, the tape is almost everywhere blank, i.e., the number of nonblank cells is finite.

A \emph{configuration} of a Turing machine $(Q,\Gm,\d)$ is a triple
\[ (p,i,X) \in Q\x\Z\x\Gm^\Z\]
where \Z\ is the set of integers and the function $X:\Z\to\Gm$ is almost everywhere blank.
The configuration $(p,i,X)$ means that the state is $p$, the tape head is in tape cell $i$, and $X$ is what is written on the tape.
(Of course, the blanks far to the left or the right
can be ignored.
But notationally it simpler to use all of \Z\ as the domain of $X$.)
Notice that the notion of configuration depends only on $Q$ and $\Gm$, not on \d.

Obviously, the transition function \d\ gives rise to a partial successor function on the configurations.
In detail, if $\d(p,X(i)) = (q,b,D)$, then the \emph{successor} of configuration $(p,i,X)$ is the configuration
\[ \Big(q,i+D,X(b\At i)\Big) \]
where the symbol \At\ indicates that symbol $b$ is written at cell $i$, so that $X(b\At i)(i)=b$ and $X(b\At i)(j) = X(j)$ for all $j\ne i$.
Accordingly $(p,i,X)$ is a \emph{predecessor} of $\Big(q,i+D,X(b\At i)\Big)$.
If $\d(p,X(i))$ is undefined, then $(p,i,X)$ has no successor.

\subsection{Reversible Turing machines}
\label{sub:rtm}

\begin{definition}\label{def:crev}
A Turing machine is \emph{reversible} if every configuration has at most one predecessor. \qef
\end{definition}

Arbitrary Turing machines are efficiently simulated by reversible ones; see \cite[\S5]{Morita}.
The criterion of reversibility of TM $(Q,\Gm,\d)$ in terms of \d\ is found in \cite[\S5.1.1.2]{Morita}. We present it in a form convenient for our purposes, and we use the following notation:
\begin{align*}
Q_{\mo} = \set{q: \d(p,a)=(q,b,-1)\ \text{ for some }\ p,a,b},\\
Q_{\po} = \set{q: \d(p,a)=(q,b,+1)\ \text{ for some }\ p,a,b}.
\end{align*}

\begin{proposition}\label{prp:crev}
A Turing machine $(Q,\Gm,\d)$ is reversible if and only if \d\ has the following two properties.
\begin{description}
\item[\tt Separability] $Q_{\mo}$ and $Q_{\po}$ are disjoint.
\item[\tt Injectivity] For every pair $(q,b)\in Q\x\Gamma$, there is at most one pair $(p,a)\in Q\x\Gamma$ such that $\d(p,a) = (q,b,D)$ for some $D$.
\end{description}
\end{proposition}

\begin{proof}
Let $X:\Z\to\Gamma$ be almost everywhere blank.

If the separability requirement fails, then $M$ is irreversible. Indeed, if
$\delta(p_1,a_1) = (q,b_1,\po)$ and $\delta(p_2,a_2) = (q,b_2,\mo)$, then configurations
\[ \Big(p_1,\mo,X(a_1\At\mo)(b_2\At\po)\Big) \ne
   \Big(p_2,\po,X(b_1\At\mo)(a_2\At\po)\Big), \]
have the same successor $(q,0,X(b_1\At\mo)(b_2\At\po))$.

If the separability requirement holds but the injectivity requirement fails, then $M$ is irreversible.
Indeed, if $\delta$ maps distinct pairs $(p_1,a_1), (p_2,a_2)$ to triples $(q,b,D_1), (q,b,D_2)$ respectively, then $D_1 = D_2$ by the separability requirement, and therefore configurations $(p_1,0,X(a_1\At0)) \ne (p_2,0,X(a_2\At0))$ have the same successor $(q,D_1,X(b\At0))$.

If both requirements hold, then $M$ is reversible. Indeed suppose that configurations $(p_1,i_1,X_1)$ and $(p_2,i_2,X_2)$ have the same successor $(q,j,X_3)$ where $q\in Q_D$.
Let $i=j-D$, so that $i_1 = i_2 = i$.
Then $X_1, X_2$ coincide with $X_3$ except possibly at $i$.
Then $\delta(p_1,X_1(i)) = \delta(p_2,X_2(i)) = (q,X_3(i),D)$. By injectivity, $p_1 = p_2$ and $X_1(i) = X_2(i)$, and so $(p_1,i_1,X_1) = (p_2,i_2,X_2)$.
\end{proof}

Taking into account the separability property, states in \Qmo\ and \Qpo\ will be called \emph{negative} and \emph{positive} respectively when the Turing machine in question is reversible.
If $Q_{\mo}\cup Q_{\po} \neq Q$, extend $Q_{\mo}, Q_{\po}$ arbitrarily so that they remain disjoint and $Q_{-1} \cup Q_{+1} = Q$.

\begin{lemma}\label{lem:total}
The transition function \d\ of a Turing machine $M = (Q,\Gm,\d)$ can be extended to a total function $\d^*$ in such a way that
\begin{enumerate}
\item if, starting with an initial configuration $(p,i,X)$, $M$ never halts, then the modified machine $M^* = (Q,\Gm,\d^*)$ works exactly as $M$ on that initial configuration, and
\item if $M$ is reversible then $M^*$ is reversible.
\end{enumerate}
\end{lemma}

\begin{proof}
Claim~(1) is obvious no matter how \d\ is extended to a total function $\d^*$.
To prove claim~(2), assume that $M$ is reversible. By the separability property and the convention just before the lemma, $Q_{\mo}$ and $Q_{\po}$ partition $Q$.

If $\d(p,a) = (q,b,D)$, let $\alpha(p,a) = (q,b)$.
By the injectivity property, $\alpha$ is an injective map from $Q\x\Gamma$ to $Q\x\Gamma$. Extend $\alpha$ to a bijection $\alpha^*$ from $Q\x\Gamma$ to $Q\x\Gamma$ in an arbitrary way.
If $\alpha^*(p,a) = (q,b)$ and $q\in Q_D$, define $\d^*(p,a) = (q,b,D)$.
Clearly $(Q,\Gm,\d^*)$ is reversible.
\end{proof}

\subsection{Looped Turing machine}

This paper is devoted to easy circuit simulation of reversible Turing machines and of quantum Turing machines.
Following the example of Molina and Watrous \cite{MW}, we separate concerns and work primarily with machines $(Q,\Gm,\d)$ where \d\ is total, so that the machine never halts.
We address the halting-related issues in \S\ref{sub:hang}.

Fix an arbitrary Turing machine $M_\infty = (Q,\Gm,\d)$ with total transition function \d.
$M$ never hangs.
Further, fix a positive integer $T$, and restrict attention to $T$-step computations of $M_\infty$ where the initial configuration is blank outside the segment $[-T,T]$ of the tape. During such a computation the head of $M_\infty$ is confined to the segment $[-T,T]$.

Truncate the tape of $M_\infty$ to the segment $[-T-1, T+1]$, bend that segment into a loop, and identify the end cells $-T-1$ and $T+1$, resulting in $N = 2T+2$ cells altogether.
Let $M = (Q,\Gm,\d)_T$, be the resulting looped-tape Turing machine.

View the cell indices as residues modulo $N$, and by default interpret index modifications $i\pm1$ modulo $N$.
If the state of $M$ is $p$, its head is scanning symbol $a$ in cell $i$ (modulo  $N$ of course), and $\delta(p,a) = (q, b, D)$,
then in one step $M$ changes its state to $q$, writes $b$ in cell $i$, and moves the head to cell $i+D$.
As far as our $T$-step computations are concerned, there is no difference between $M_\infty$ and $M$.

A \emph{configuration} of $M$ is a triple
\[ (p,i,X) \in Q \x (\Z/N) \x \Gm^{\Z/N}; \]
the role of \Z\ is played by the set $\Z/N$ of residues modulo $N$.
As in the case of $M_\infty$, \d\ induces a successor function on the configurations.
As in the case of $M_\infty$, $M$ is \emph{reversible} if every configuration has at most one predecessor.
Proposition~\ref{prp:crev} remains valid for $M$, with the same proof.
It follows that $M$ is reversible if and only if $M_\infty$ is reversible.

\section{Boolean circuit simulation}
\label{sec:bc}

We will simulate $T$-step computations of the looped-tape Turing machine $M = (Q,\Gm,\d)_T$ with a Boolean circuit of depth $2T$ where rows of gates are staggered like the bricks in the brick wall of Figure~\ref{fig:bwall}.

\begin{figure}[h]
\includegraphics[scale=.4,trim=40 480 40 54,clip]{bwall.png}
\caption{A brick wall (American bond). Source: Wikimedia \cite{WikiBwall} }\label{fig:bwall}
\end{figure}

For our purposes, it is convenient to view the brick wall as being composed of half bricks, intended to represent tape cells of $M$ at various stages of computation.
Each half brick is uniquely determined by two coordinates.
\begin{description}
\item[\texttt{Horizontal coordinate}]
The brick wall comprises $N = 2T+2$ vertical columns of half bricks.
Number the vertical columns left to right using the residues $0,1,\dots,N-1$ modulo $N$.
This gives us the horizontal coordinate $h$.
\item[\texttt{Vertical coordinate}]
The wall has $2T$ rows of bricks.
Number them top down: $0,1, \dots, 2T-1$.
This gives us the vertical coordinate $v$.
\end{description}

A brick comprising the half bricks $(h,v)$ and $(h+1,v)$ represents a circuit gate denoted $G(h,v)$.
Notice that half brick $(h,v)$ is the left half of its brick if and only if $h+v$ is even.
Thus, if $h+v$ is even, then half brick $(h,v)$ belongs to $G(h,v)$; and, if $h+v$ is odd, then half brick $(h,v)$ belongs to gate $G(h-1,v)$.

The use of residues modulo $N$ for numbering the columns hints at a property of our brick wall which is not reflected in Figure~\ref{fig:bwall}. Our wall is bent into a cylinder, reflecting the loop structure of $M$.
The two half bricks at the ends of row 1 (or any other odd-numbered row) are the two halves of the same brick.

Think of the brick wall as a greedy circuit algorithm where each brick is a gate.
The algorithm is greedy in the sense that, at each step, it executes all the gates that can be executed.
All the gates of row 0 fire at the first step of the algorithm, all the gates of row 1 fire at the second step, and so on.

For every $t = 0, 1, \dots, T-1$, the subcircuit composed of rows $2t$ and $2t+1$ will transform any configuration
\[ (p,i,X) \in Q\x \Z/N \x \Gm^\Z \]
of the looped Turing machine $M$ into the successor configuration.
However, for technical reasons, the circuit works with richer data.
We introduce annotated versions $\qn, \qs$ of every state $q\in Q$.
It is presumed that the sets
\[ Q,\quad \Qn=\set{\qn: q\in Q},\quad
     \Qs=\set{\qs: q\in Q},\quad \set{0} \]
are disjoint. Let $Q^* = Q\sqcup \Qn\sqcup \Qs$.
Every row of the brick wall circuit will compute a unary operation on the set
\[ Q^*\x \Z/N \x \Gm^{\Z/N} \]
of \emph{extended configurations} of $M$.

\subsection{The brick function}
\label{sub:bf}\mbox{}

\noindent
All bricks (that is all brick gates) compute the same function $f$, the \emph{brick function}.
A \emph{cell datum} is an element of the set $(Q^*\sqcup\set{0}) \x \Gamma$.
Thus, a cell datum has the form $(q^*,a)$ or $(0,a)$ where $q^*\in Q^*$ and $a\in\Gm$.
In the first case, it is \emph{scanned}, and, in the second, it is \emph{unscanned}.
A \emph{brick datum} is a pair of cell data where at most one cell datum is scanned.
Thus, a brick datum has the form
$(q^*,a,0,b)$, $(0,a,q^*,b)$, or $(0,a,0,b)$, where $q^*\in Q^*$ and $a,b\in\Gm$.
In the first case, it is a \emph{one-head} datum, and, in the second, it is \emph{no-head} datum.

A brick (or brick gate) computes the brick function $f$ transforming any input brick datum to an output brick datum. Such inputs and outputs can, if desired, easily be coded as tuples of bits to make the circuits Boolean.

In the rest of this subsection, we define the brick function $f$. To begin, unsurprisingly
\begin{equation}\label{f0}
f(0,a,0,b) = (0,a,0,b).
\end{equation}
In other words, a no-head input is simply copied into the output.
(For brevity, we'll write $fx$ for $f(x)$.)
Suppose that the input brick datum $x$ includes the scanned cell. Three cases arise depending on whether the state symbol is in $Q$, \Qn, or \Qs.

First suppose that the state symbol is in $Q$, so that the input datum $x$ has the form $(p,a,0,c)$ or $(0,c,p,a)$. Let $\d(p,a) = (q,b,D)$. Then
\begin{equation}\label{f1}
\begin{split}
q\in Q_{+1} \implies f(p,a,0,c) = (0,b,\qn,c) \ \land\
                      f(0,c,p,a) = (0,c,\qs,b),\\
q\in Q_{-1} \implies f(p,a,0,c) = (\qs,b,0,c) \ \land\
                      f(0,c,p,a) = (\qn,c,0,b).
\end{split}
\end{equation}
Thus $\uparrow$ indicates successful execution of the \d-transition, and $\downarrow$ indicates incomplete execution when the brick updated the state and tape symbol but couldn't move the head.

Second, after a successful execution, all that remains is to remove $\uparrow$. We must, however, also define $f$ on 2-data with $\uparrow$ that cannot arise in the Turing machine simulation. In such ``garbage'' cases, we simply interchange the two cell data in the input brick datum.
\begin{equation}\label{f2}
\begin{split}
q\in Q_{+1} \implies f(\qn,c,0,b) = (q,c,0,b) \ \land\
                            f(0,b,\qn,c) = (\qn,c,0,b),\\
q\in Q_{-1} \implies f(0,b,\qn,c) = (0,b,q,c) \ \land\
                            f(\qn,c,0,b) = (0,b,\qn,c).
\end{split}
\end{equation}

Finally, after an incomplete execution, what remains is to move the head in the appropriate direction and remove $\downarrow$. Again, we define $f$ on ``garbage'' cases by interchanging the two cell data in the input brick datum.
\begin{equation}\label{f3}
\begin{split}
q\in Q_{+1} \implies f(\qs,b,0,c) = (0,b,q,c) \ \land\
                            f(0,c,\qs,b) = (\qs,b,0,c),\\
q\in Q_{-1} \implies f(0,c,\qs,b) = (q,c,0,b) \ \land\
                            f(\qs,b,0,c) = (0,c,\qs,b).
\end{split}
\end{equation}
This completes the definition of the brick function $f$.

\subsection{Reversibility}
\label{sub:rns}\mbox{}

\begin{theorem}
If Turing machine $M$ is reversible, then the brick wall circuit is reversible.
\end{theorem}

\begin{proof}
It suffices to prove that the brick gate is reversible, i.e., that the function $f$ defined above is a bijection.
Obviously, $f$ is bijective on no-head inputs.
So we may restrict attention to one-head inputs.
Since every injection from a finite set to a finite set of the same cardinality is a bijection, it suffices to prove that
$ fx = fy \implies x = y $
on one-head inputs.  To this end, assume $fx=fy$.

First we suppose that $fx, fy$ are defined by different equation systems \eqref{f1}--\eqref{f3}. By symmetry, we may assume that $fy$ is defined by a later equation system and that $fx$ is defined by one of the upper equations in \eqref{f1}, in \eqref{f2}, or in \eqref{f3}.
By inspection, there are only three cases:
\begin{enumerate}
\item $fx$ has the form $(0,b,\qn,c)$ and is defined the first upper equation in \eqref{f1}, while $fy$ is defined by the second lower equation in \eqref{f2}.
\item $fx$ has the form $(0,b,\qs,c)$ and is defined the second upper equation in \eqref{f1}, while $fy$ is defined by the second lower equation in \eqref{f3}.
\item $fx$ has the form $(q,c,0,b)$ and is defined by the first upper equation in \eqref{f2}, while $fy$ is defined by the first lower equation in \eqref{f3}.
\end{enumerate}
None of the three cases is possible because of the separability property of \d.

Thus $fx,fy$ are defined by the same equation system.
If $fx, fy$ are defined by system \eqref{f1}, then they are defined by the same equation in \eqref{f1}, and then we have $x=y$ by the injectivity property of \d. If $fx, fy$ are both defined by \eqref{f2} or are both defined by \eqref{f3}, then obviously $x=y$.
\end{proof}

\subsection{Simulation}

\begin{theorem}\label{thm:csim}
The brick wall circuit algorithm simulates any $T$-step computation of $M$.
\end{theorem}

\begin{proof}
It suffices to prove that, for every $t = 0, 1, \dots, T-1$, the two-row subcircuit, composed of rows $2t$ and $2t+1$,  transforms an arbitrary configuration $(p,i,X)$ of $M$ into the successor configuration. Since all these two-row circuits are isomorphic, we may restrict attention to the two-row circuit composed of rows 0 and 1.

Let $a=X(i)$.
Two cases arise depending on the parity of $i$. If $i$ is even, then the scanned cell $(p,a)$ is handled by brick $G(0,i)$ which works with input $(p,a,0,c)$ where $c=X(i+1)$; otherwise $(p,a)$ is handled by brick $G(0,i-1)$ which works with input $(0,d,p,a)$ where $d=X(i-1)$. By symmetry, we may assume that $i$ is even.

According to \eqref{f0}, no-head inputs are just copied into outputs. So we pay attention only to one-head inputs. Let $\d(p,a) = (q,b,D)$.

If $D=+1$, then brick $G(0,i)$ transforms $(p,a,0,c)$ to $(0,b,\qn,c)$, and thus row 0 transforms configuration $(p,i,X)$ to an extended configuration $(\qn,i+1,X(b\At i))$,
so that the scanned cell $(\qn,c)$ is handled by brick $G(1,i+1)$ which works with input $(\qn,c,0,X(i+2))$.
That brick transforms this input into $(q,c,0,X(i+2))$, and
therefore the resulting configuration is $(q,i+1,X(b\At i))$ as required.

If $D=-1$, then gate $G(0,i)$ transforms $(p,a,0,c)$ into $(\qs,b,0,c)$, and thus row 0 transforms configuration $(p,i,X)$ to an extended configuration $(\qs,i,X(b\At i))$, so that the scanned cell $(\qs,b)$ is handled by brick $G(1,i-1)$ which works with input $(0,d,\qs,b)$.
That brick transforms this input into $(q,d,0,b)$, and
therefore the resulting configuration is $(q,i-1,X(b\At i))$ as required.
\end{proof}

\subsection{Premature halting problem}\mbox{}
\label{sub:hang}

In the preceding part of this section, we have assumed that the looped Turing machine does not have halting computations. Every computation will proceed for (at least) the $T$ steps that we want to simulate with a Boolean circuit.
In the present subsection, we comment on the situation where the TM's computations could halt.

Of course, if $T$ steps are completed before the TM halts, then our
previous results apply to those $T$ steps, and subsequent halting
is irrelevant.  The real issue concerns premature halting, before
step $T$. We cannot simply cut off our circuit at the moment of TM's halting, as that moment may depend on the input.
Is circuit simulation of such a TM nevertheless possible?

First, let us agree on a reasonable definition of Boolean circuit simulation in this situation.
A simulation of a TM by a circuit \C\ includes an encoding of initial TM states as inputs to \C\ and a decoding of outputs of \C\ as information about TM configurations.
If the TM arrives at a halting configuration $C = (p,i,X)$, then the circuit's output should code $C$ or, more generally, a required amount of information about $C$, e.g.\ a version of $C$ where $X$ is replaced by its restriction to a vicinity of the cell $i$.
Whatever it is, the required amount of information as well as a coding scheme should be agreed on ahead of time.

In the absence of any reversibility considerations, this situation is easy to address.
Define an extension $M' = (Q',\Gm',\d')_T$ of our Turing machine $M = (Q,\Gm,\d)_T$ so that any halting $M$ configuration $C$ gets a successor $M'$ configuration $C'$ which preserves the required information about $C$.
For example, $C'$ could be $C$ itself.
(This requires allowing the TM's head to stay at the same cell, but our work above can easily be extended to cover this option.)
Instead of halting, $M'$ is idling, pretending to compute without doing much.
Then let the circuit simulate $M'$.

But making the extended machine $M'$ reversible is not obvious.
Let $\Sigma$ be the set of halting $M$ configurations and all their $M'$ descendants, i.e., successors, successors of successors, etc.
Then $\Sigma$ is closed under the successor operation.
Assume that $M'$ is reversible, so that the successor operation is injective.
Since the total number of $M'$ configurations is finite, the successor operation is a bijection on $\Sigma$.
Accordingly, the successor of a nonhalting configuration cannot be halting.
Thus, every nonhalting initial configuration yields a nonhalting computation.

What can one do? Well, keep $\Sigma$ from being closed under the successor operation. Starting from a halting $M$ configuration $C$, the extended machine $M'$ will eventually halt, but not before marching around for sufficiently many steps.

Let\ $M' = (Q',\Gm',\d')_T$ where $Q'$ and $\Gm'$ are disjoint unions $Q\,\sqcup\, \set{p': p\in Q}$ and $\Gm\, \sqcup\, \set{a':a\in\Gm}$ respectively, and let $\d'$ be a function from $Q'\x\Gm$ to $Q' \x \Gm' \x \set{-1,+1}$ such that if $p\in Q$ and $a\in\Gm$ then
\begin{align*}
\d'(p,a) &=
\begin{cases}
  \d(p,a)    &\text{if }\ \d(p,a)\ \text{ is defined},\\
  (p',a',+1) &\text{if }\ \d(p,a)\
              \text{ is undefined and }p\in Q_{+1},\\
  (p',a',-1) &\text{if }\ \d(p,a)\
              \text{ is undefined and }p\in Q_{-1},\\
\end{cases} \\
&\text{and}\\
\d'(p',a) &=
\begin{cases}
  (p',a',+1) &\text{if }p\in Q_{+1},\\
  (p',a',-1) &\text{if }p\in Q_{-1},\\
\end{cases}
\end{align*}
Recall that, just before Lemma~\ref{lem:total}, we arranged that $Q_{-1}$ and $Q_{+1}$ partition $Q$.

$M'$ is reversible because \d\ has the separability and injectivity properties. It remains to check that, on no input, $M'$ halts before step $T$. Fix some input $I$. If $M$ doesn't halt before step $T$ on input $I$, we are done. So suppose that, after $t<T$ steps, $M$ arrives at a halting configuration $(p,i,X)$ on input $I$.

By symmetry, we may assume without loss of generality that $p\in Q_{+1}$.
Abbreviate $X(j)$ to $a_j$; we have that $\d(p,a_i)$ is undefined.
The $M'$ successor of $(p,i,X)$ is $(p',i+1,X(a_i'\At i))$, whose $M'$ successor is $(p', i+2, X(a_i'\At i, a_{i+1}'\At(i+1))$, and so on. Eventually, after $N$ steps (counting from the halting of $M$), $M'$ has converted every $a_j$ into $a_j'$ and halts.

It follows that, on input $I$, $M'$ will perform at least $T$ steps without halting. Its configuration after $T$ steps is $(p',i + (T-t),X')$ where
\[X'(j) =
\begin{cases}
  a_j' &\If\ i\le j < i+(T-t),\\
  a_j  &\text{otherwise}.
\end{cases} \]
The halting configuration $C=(p,i,X)$ of $M$ is easily recoverable from the configuration  $C'=(p',i + (T-t),X')$; just move the head to the cell of the first primed symbol and then remove all the primes.

Extend the transition function $\d'$ of $M'$ to a total function
\[ \d^*: Q'\times \Gm' \to  Q'\times \Gm'\times \set{-1,+1}\]
as in the proof of Lemma~\ref{lem:total}. The resulting TM $M^* = (Q',\Gm',\d^*)$ is reversible and so, by Theorem~\ref{thm:csim}, an appropriate brick wall circuit \C\ simulates $T$-step computations of $M^*$. But $M^*$ works exactly as $M'$ on the initial configurations of $M'$, and so \C\ simulates $T$-step computations of $M'$.

\section{Quantum Turing machine}
\label{sec:qc}

Quantum Turing machines, in short QTMs, were introduced by David
Deutsch \cite{Deutsch}, thoroughly studied by Ethan Bernstein and Umesh Vazirani \cite{BV}, and recently presented elegantly by Abel Molina and John Watrous in article \cite{MW} which is our main reference on QTMs.

Like a classical Turing machine, a quantum Turing machine is also given by a triple $(Q,\Gm,\d)$ where $Q$ and \Gm\ are disjoint finite sets, but now \d\ is a function
\[ \d: Q\x\Gm \to \Co^{Q\x\Gm\x\set{-1,+1}}. \]
Quantum Turing machines are also subject to a unitarity requirement to be introduced below.
We abbreviate $\d(p,a)(q,b,D)$ to $\dpa(q,b,D)$.

A \emph{classical configuration} of $M$ is a triple
\[ (p,i,X) \in Q \x \Z \x \Gm^\Z. \]
This notion coincides with the notion of configuration in \S\ref{sec:tm} for the same $Q$ and $\Gm$; the change in \d\ makes no difference.
Initially a QTM is in a classical configuration, just like a classical TM with the same states and alphabet (but of course later the QTM can enter a quantum superposition of classical configurations).

Our goal is a simple quantum-circuit simulation of $T$-step computations of a given QTM
\[ M = (Q,\Gm,\d).\]
Here $T$ is an arbitrary but fixed positive integer.
It is presumed that, initially, the head is in cell 0, and the tape is blank outside the segment $[-T,T]$.
As in the classical case, we may assume without loss of generality that $M$ has a looped tape whose cells are numbered by residues modulo $N = 2T+2$. Accordingly, a \emph{classical configuration} of the looped-tape QTM $M$ is a triple
\[ (p,i,X) \in Q \x (\Z/N) \x \Gm^{\Z/N}. \]

To simplify the exposition, we separate concerns and abstract from two issues, as in \cite{MW}.
\begin{enumerate}
\item The transition function \d\ of our looped-tape QTM is total.
\item We place no computability restrictions on the complex numbers appearing in \d. ``Any computationally offensive properties possessed by \d\ will be inherited by the quantum circuits that result from the simulation,'' \cite[\S3]{MW}.
\end{enumerate}

\subsection{The Hilbert space of the QTM}
\mbox{}

For any finite set $S$, let $\lBrack S\rBrack$ be the Hilbert space with an orthonormal basis \set{\ket{s}: s\in S}, the \emph{canonical basis} for $\lBrack S\rBrack$.
If $B_1, B_2$ are canonical bases for Hilbert spaces $\H_1, \H_2$ respectively then the \emph{canonical basis} for $\H_1 \ox \H_2$ is the orthonormal basis composed of vectors $\ket{b_1b_2} = \ket{b_1}\ox \ket{b_2}$ where \ket{b_1}, \ket{b_2} range over $B_1, B_2$ respectively.

Let CELL be the Hilbert space $\big\lBrack Q\sqcup\set{0}\big\rBrack \ox \lBrack\Gm\rBrack$, and consider
the tensor product $\CELL^{\ox\, \Z/N}$.
Each classical configuration $(p,i,X)$ of the looped-tape QTM $M$ is represented by a $\CELL^{\ox\, \Z/N}$ vector
\begin{equation}\label{canonvec}
\begin{aligned}
\ket{p,i,X} =\ &\ket{r_0a_0} \ox \ket{r_1a_1} \ox \cdots\ox \ket{r_{N-1}a_{N-1}}\ \text{ where }\\
&r_i=p,\ \text{ all other $r_j=0$,
and all } a_j = X(j).
\end{aligned}
\end{equation}
Intuitively, these vectors are the ones where exactly one cell contains a head.
The Hilbert space \H\ of $M$ is the subspace of
$\CELL^{\ox\, \Z/N}$ generated by vectors \ket{p,i,X}
which form the \emph{canonical basis} for $\H$.
Thus, as in \cite{MW}, the canonical orthonormal basis for the Hilbert space of the QTM of interest consists of vectors \ket{p,i,X} representing classical configurations of the QTM.

Finally, we can complete the definition of quantum Turing machine  $M = (Q,\Gm,\d)$. It is required that the transformation
\[\U_M\ket{p,i,X} = \sum_{q,b,D}\d_{pX(i)}(q,b,D)\,\ket{q,i+D,X(b\At i)}\]
be unitary. This $U_M$ represents any one step in the computation of $M$.

\subsection{Transition function}\mbox{}

View every $\d_{pa}$ as a vector
\[ \dpa = \sum_{q,b,D} \d_{pa}(q,b,D) \ket{q,b,D} \]
in the Hilbert space $\lBrack Q\rBrack\ox \lBrack\Gm\rBrack\ox \lBrack\set{-1,+1}\rBrack$. Let
\[ \ketb{\Lpab} = \sum_q \dpa(q,b,-1)\ket{q,b,{-1}}, \quad
\ketb{\Rpab} = \sum_q \d_{pa}(q,b,+1)\ket{q,b,{+1}}, \]
so that $\d_{pa} = \sum_b\ketb{\Rpab} + \sum_b\ketb{\Rpab}$.

\begin{proposition}
\d\ has the following three properties.
\begin{align*}
&\text{Unit length}:
  &&|\d_{pa}| = 1
  &&\text{for all } p,a.\\
&\text{Orthogonality}:
  &&\d_{p_1a_1}\ort \d_{p_2a_2}
  &&\text{for all } (p_1,a_1)\ne(p_2,a_2).\\
&\text{Separability}:
  &&\ketb{L_{p_1a_1b_1}} \ort \ketb{R_{p_2a_2b_2}}
  &&\text{for all } p_1,a_1,b_1,p_2,a_2,b_2.
\end{align*}
\end{proposition}
\noindent
The three claims%
\footnote{Orthogonality and separability here are the analogs of injectivity and separability in Proposition~\ref{prp:crev}.}
form a part of Theorem~5.3 in \cite{BV}, but it seems useful to spell out a detailed proof.

\begin{proof}
Let $X$ be an arbitrary element of $\Gm^{\Z/N}$; for example, $X$ may be everywhere blank.

\texttt{Unit length.} If $C$ is a classical configuration of the form $(p,0,X(a\At0))$, then $\U_M\ket C = \sum_{q,b,D}\d_{pa}(q,b,D)\ket{q,D,X(b\At0)}$.
Since $\U_M$ is unitary, we have
\[
 1 = |\U_M\ket C|^2 = \sum_{q,b,D} \big(\d_{pa}(q,b,D)\big)^*\d_{pa}(q,b,D)  = |\d_{pa}|^2.
\]

\texttt{Orthogonality}.
If $(p_1,a_1)\ne(p_2,a_2)$, then canonical basis vectors \ket{p_1,0,X(a_1\At0)} and \ket{p_2,0,X(a_2\At0)} of $\H$ are orthogonal, and so are their $\U_M$ images:
\[\sum_{q,b,D}\d_{p_1a_1}(q,b,D)\ket{q,D,X(b\At0)} \ort
\sum_{q,b,D}\d_{p_2a_2}(q,b,D)\ket{q,D,X(b\At0)}.
\]
Hence
\[
 0 = \sum_{q,b,D} \big(\d_{p_1a_1}(q,b,D)\big)^* \d_{p_2a_2}(q,b,D) =
 \braket{\d_{p_1a_1}}{\d_{p_2a_2}}.
\]

\texttt{Separability.}
For any $p_1,a_1,b_1,p_2,a_2,b_2$, the canonical basis configurations \ket{p_1,1,X(a_1\At1,b_2\At{-1})} and \ket{p_2,-1,X(b_1\At1,a_2\At{-1})} are orthogonal, and therefore so are their $\U_M$ images:
\begin{align*}
&\sum_{q,b,D}\d(p_1,a_1)(q,b,D)\ket{q,1+D,X(b\At1,b_2\At{-1})} \ort\\
&\sum_{q,b,D}\d(p_2,a_2)(q,b,D)\ket{q,-1+D,X(b_1\At1,b\At{-}1)}.
\end{align*}
A summand in the upper sum are orthogonal to every summand in the lower sum unless $b=b_1$ and $D=-1$. Similarly, a summand in the lower sum is orthogonal to every summand in the upper sum unless $b=b_2$ and $D+1$.
Setting $Y = X(b_1\At1,b_2\At{-}1)$, we have
\[ \sum_q\d(p_1,a_1)(q,b_1,-1)\ket{q,0,Y} \ort
   \sum_q\d(p_2,a_2)(q,b_2,+1)\ket{q,0,Y}.
\]
The separability property follows.
\end{proof}

\section{Quantum circuit simulation}
\label{sec:qc}

We convert the reversible Boolean brick wall circuit, used to simulate $T$-step computations of a reversible Turing machine, to a quantum circuit simulating $T$-step computations of the quantum Turing machine $M$ of the previous section.
As in \S\ref{sec:bc}, we use annotated versions of states of $M$. Again, sets
\[ Q,\quad \Qn=\set{\qn: q\in Q},\quad
     \Qs=\set{\qs: q\in Q},\quad \set{0} \]
are disjoint and $Q^* = Q\sqcup \Qn\sqcup \Qs$.
The obvious bijections $q\mapsto \qn$ and $q\mapsto \qs$ give rise to obvious isomorphisms from $\lBrack Q \rBrack$ to $\lBrack \Qn \rBrack$ and to $\lBrack \Qs \rBrack$ respectively.

The quantum brick circuit works with an extension $\H^*$ of the Hilbert space $\H$ of our QTM $M$.
Let $\CELL^*$ be the Hilbert space $\big\lBrack Q^*\sqcup\set{0}\big\rBrack \ox \lBrack \Gm\rBrack$.
The extended space $\H^*$ is the subspace of the tensor product
$ \left(\CELL^*\right)^{\ox\, \Z/N} $
generated by vectors
\begin{equation}\label{hmstar}
\begin{aligned}
\ket{q^*,i,X} =\ &\ket{r_0a_0} \ox \ket{r_1a_1} \ox \cdots\ox \ket{r_{N-1}a_{N-1}}\quad\text{where}\\
& q^*\in Q^*,\ i\in\Z/N,\ X:\Z/N\to\Gm,\\
& r_i = q^*,\ \text{ all other $r_j=0$, and all }\ a_j=X(j).
\end{aligned}
\end{equation}
Every row of the quantum brick wall circuit will compute a linear operator on $\H^*$.

To simplify the exposition, our quantum circuit works  with qudits in Hilbert spaces $\lBrack Q^*\rBrack$ and $\lBrack\Gm\rBrack$. Obviously such quantum circuits can be easily simulated by qubit-based circuits.

\subsection{The brick operator}\mbox{}

Let BRICK be the subspace of tensor product $\CELL^* \ox \CELL^*$ generated by vectors
\[
\ket{q^*,a,0,b},\ \ket{0,a,q^*,b},\ \ket{0,a,0,b}\quad
\text{where }\ q^*\in Q^*\ \text{and } a,b\in\Gm.
\]
Vectors \ket{q^*,a,0,b} and \ket{0,a,q^*,b} are \emph{one-head vectors}, and vectors \ket{0,a,0,b} are \emph{no-head vectors}.

All gates of our quantum circuit are copies of a single quantum gate, called the \emph{(quantum) brick gate}, which computes a linear operator $U$ on BRICK.
By linearity, it suffices to define $U$ on a set of vectors spanning BRICK. We do that in equations \eqref{u0}--\eqref{u3}.

We start with no-head inputs.
$U$ just copies any such input into its output.
\begin{equation}\label{u0}
 U\ket{0,a,0,b} = \ket{0,a,0,b}
\end{equation}

Let $p$ range over $Q$.
The operator $U$ updates the state and the tape symbol according to \d, but it may or may not be able to move the head.
Processing any input \ket{0,c,p,a}, a gate $G(h,k)$ succeeds with the left but not the right moves; processing \ket{p,a,0,c}, it succeeds with the right but not the left moves.
In either case, successes are marked with $\uparrow$, and failures are marked with $\downarrow$:
\begin{equation*}
\begin{aligned}
U\ket{0,c,p,a}
&= \sum_{qb} \dpa(q,b,\mo)\,\ketb{\qn,c,0,b}
 + \sum_{qb} \dpa(q,b,\po)\,\ketb{0,c,\qs,b}, \\
U\ket{p,a,0,c}
&= \sum_{qb} \dpa(q,b,\po)\,\ketb{0,b,\qn,c}
 + \sum_{qb} \dpa(q,b,\mo)\,\ketb{\qs,b,0,c}.
\end{aligned}
\end{equation*}

Recall that $\ketb{L_{pab}} = \sum_q \dpa(q,b,\mo)\ket{q}$ and
$\ketb{R_{pab}} = \sum_q \dpa(q,b,\po)\ket{q}$.
Let \ketb{\Ln_{pab}}, \ketb{\Rn_{pab}} and \ketb{\Ls_{pab}}, \ketb{\Rs_{pab}} be the images of \ketb{L_{pab}}, \ketb{R_{pab}} under the obvious isomorphisms from $\lBrack Q \rBrack$ to $\lBrack \Qn \rBrack$ and to $\lBrack \Qs \rBrack$ respectively.
Then the two equations can be rewritten thus:
\begin{equation}\label{u1}
\begin{aligned}
U\ket{0,c,p,a}
&= \sum_b \ketb{\Ln_{pab},c,0,b}
 + \sum_b \ketb{0,c,\Rs_{pab},b},\\
U\ket{p,a,0,c}
&= \sum_b \ketb{0,b,\Rn_{pab},c}
 + \sum_b \ketb{\Ls_{pab},b,0,c}.
\end{aligned}
\end{equation}

Let $L,R$ be the subspaces of $\lBrack Q\rBrack$ generated by  vectors \ketb{L_{pab}} and by vectors \ketb{R_{pab}}.
Let subspaces \Ln, \Rn of $\lBrack \Qn\rBrack$ and subspaces \Ls, \Rs\ of $\lBrack \Qs\rBrack$ similarly be the images of $L, R$ under the obvious isomorphisms.
By the separability property of \d, $L$ and $R$ are orthogonal.
It follows that $\Ln \ort \Rn$ and $\Ls \ort \Rs$.
Let \ket{v} range over the unit vectors of $\lBrack Q\rBrack$.

The gates $G(h-1,k+1)$ and $G(h+1,k+1)$ complete $G(h,k)$'s successful execution by removing the success marks. To achieve that, we define

\begin{equation}\label{u2}
\begin{aligned}
U\ket{0,d,\vn,c} &= \ket{0,d,v,c}   &&\text{if }\ket v\in L,\\
U\ket{\vn,c,0,d} &= \ket{v,c,0,d}   &&\text{if }\ket v\in R.
\end{aligned}
\end{equation}

We must also define $U\ket{0,d,\vn,c}$ when $\vn\ort L$ and define $U\ket{\vn,c,0,d}$ when $\vn\ort R$. As in \S\ref{sub:bf}, we swap cell data.

\begin{equation}\label{u2'}
\begin{aligned}
U\ket{0,d,\vn,c} &= \ket{\vn,c,0,d} &&\text{if }\ket v \ort L,\\
U\ket{\vn,c,0,d} &= \ket{0,c,\vn,d} &&\text{if }\ket v \ort R.
\end{aligned} \tag{\ref{u2}$'$}
\end{equation}

The gates $G(h-1,k+1)$ and $G(h+1,k+1)$ perform the moves that gate $G(h,k)$ couldn't accomplish or swap cell data.
\begin{equation}\label{u3}
\begin{aligned}
U\ket{0,d,\vs,c} &= \ket{v,c,0,d}   &&\text{if }\ket v\in L,\\
U\ket{\vs,c,0,d} &= \ket{0,d,v,c}   &&\text{if }\ket v\in R.
\end{aligned}
\end{equation}

\begin{equation}\label{u3'}
\begin{aligned}
U\ket{0,d,\vs,c} &= \ket{\vs,c,0,d} &&\text{if }\ket v \ort L,\\
U\ket{\vs,c,0,d} &= \ket{0,c,\vs,d} &&\text{if }\ket v \ort R.
\end{aligned} \tag{\ref{u3}$'$}
\end{equation}

\subsection{Unitarity}

\begin{proposition}
There is a unique linear operator $U$ on BRICK, satisfying equations \eqref{u0}--(\ref{u3}$'$), and this $U$ is unitary.
\end{proposition}

\begin{proof}
Let $B$ be the set of vectors which appear as arguments of $U$ in equations \eqref{u0}--(\ref{u3}$'$).
The uniqueness of $U$ is obvious because $B$ spans BRICK.
Inspection of the equations shows the existence of $U$ because the equations don't clash when extended linearly.
(In more detail, choose orthonormal bases for $L$, for $R$, and for the orthogonal complement of $L+R$ in $\lBrack Q\rBrack$. Restrict \ket{v} in \eqref{u1}--(\ref{u3}$'$) to range over these bases.
Then $U$ is defined on an orthonormal basis for BRICK, and the existence of a linear extension is clear.
And of course \eqref{u1}--(\ref{u3}$'$) continue to hold as written.)
It remains to prove the unitarity of $U$.

Since $B$ spans BRICK, it suffices to prove that $U$ preserves the inner product on $B$, i.e., that $\braket{Ux}{Uy} = \braket xy$ holds for all $x,y\in B$.
This is obvious if $x$ or $y$ is a no-head vector; in that case $\braket{Ux}{Ux} = \braket xx =1$ and $\braket{Ux}{Uy} = \braket xy = 0$ for every $y\ne x$ in $B$.
In the rest of the proof, we assume that both, $x$ and $y$, are one-head vectors. By symmetry, we may assume that $Uy$ is defined by the same equation as $Ux$ or by a later equation.

Case~1: $Ux$ is defined by \eqref{u1}.
By the symmetry, we may assume that $Ux$ is defined by the upper equation.

If $Uy$ is defined by one of the following equations, then trivially \braket xy = 0. In all these cases, trivially \braket{Ux}{Uy} = 0 except for the following two cases.
One is the equation in (\ref{u2}$'$) where $\ket{v} \ort L$; we have $\ketb{\vn} \ort \Ln$ and therefore \braket{Ux}{Uy} = 0.
The other is the equation in (\ref{u3}$'$) where $\ket{v} \ort R$; we have $\ketb{\vs} \ort \Rs$, and therefore \braket{Ux}{Uy} = 0.

So suppose that $Uy$ is also defined by the upper equation in \eqref{u1}.
If $x=y$ then, by the unit-length property of \d, we have \braket{Ux} {Ux} = 1 = \braket xx.
Otherwise, by the orthogonality property of \d, \braket{Ux}{Uy} = 0 = \braket xy.

Case~2: $Ux$ is defined by \eqref{u2}.
By the symmetry, we may assume that $Ux$ is defined by the upper equation, where $\ket{v}\in L$.

If $Uy$ is defined by one of the following equations, then trivially \braket xy = 0.
In all these cases, trivially \braket{Ux}{Uy} = 0 except for the case of the lower equation in \eqref{u3}, where $\ket{v}\in R$. Use the fact that $L\ort R$.

So suppose that $Uy$ is also defined by the upper equation in \eqref{u2}.
We have \braket{Ux}{Uy} = \braket xy by the obvious isomorphism from $\lBrack Q \rBrack$ to $\lBrack \Qn \rBrack$.

Case~2$'$: $Ux$ is defined in (\ref{u2}$'$).
By the symmetry, we may assume that $Ux$ is defined by the upper equation.
If $Uy$ is defined by one of the following equations, then trivially \braket xy = 0 = \braket {Ux}{Uy}.
If $Uy$ is also defined by the upper equation in (\ref{u2}$'$), then obviously \braket xy = 0 = \braket {Ux}{Uy}.

Case~3: $Ux$ is defined in \eqref{u3}.
By the symmetry, we may assume that $Ux$ is defined by the upper equation.
If $Uy$ is defined by one of the following equations, then trivially \braket{Ux}{Uy} = \braket xy.
So suppose that $Uy$ is also defined by the upper equation in \eqref{u3}.
We have \braket{Ux}{Uy} = \braket xy by the obvious isomorphism from $\lBrack Q \rBrack$ to $\lBrack \Qs \rBrack$.

Case~3$'$ is similar to case~2$'$.
\end{proof}

A quantum circuit may give rise to a number of quantum algorithms. Semantically, all these algorithms are equivalent \cite{G250}. In particular, if one of them computes a unitary transformation then they all compute the same unitary transformation. But the number of computation steps may differ from one algorithm to another. The greedy algorithm, at each step, executes all the gates that can be executed. Below by circuit computations we mean the computations of the greedy algorithm of the circuit in question.

\subsection{Simulation}

\begin{theorem}
The brick wall circuit simulates any $T$-step computation of our quantum Turing machine $M$.
\end{theorem}

\begin{proof}
It suffices to prove that, for every $t = 0, 1, \dots, T-1$, the two-row subcircuit of our quantum brick circuit, composed of rows $2t$ and $2t+1$,  computes the unitary transformation $\U_M$ on Hilbert space $\H$.
Since all these two-row circuits are isomorphic, it suffices to prove that the two-row circuit $C$ composed of rows 0 and 1 computes $\U_M$.
By linearity, it suffices to prove that $C$ transforms any canonical basis vector \ket{p,i,X} of $\H_M$ into
\[
\U_M\ket{p,i,X} = \sum_{q,b,D}\dpa(q,b,D)\ket{q,i+D,X(b\At i)}
\]
where $a = X(i)$.

Let $c=X(i+1)$ and $d=X(i-1)$.
Two cases arise depending on the parity of $i$.
If $i$ is even, then the crucial cell $(p,a)$ is handled by
gate $G(0,i)$ which works with input $(p,a,0,c)$, and, if $i$ is odd, then $(p,a)$ is handled by gate $G(0,i-1)$ which works with input $(0,d,p,a)$. By symmetry, we may assume that $i$ is even.

Recall that $\ket{\Lpab} = \sum_q\dpa(q,b,\mo)\ket{q}$ and $\ket{\Rpab} = \sum_q\dpa(q,b,\po)\ket{q}$.
Let $\lambda = \set{b: \braket{\Lpab}{\Lpab}>0}$ and $\rho = \set{b: \braket{\Rpab}{\Rpab}>0}$.
Then $\U_M\ket{p,i,X}$ is the sum
\[  \sum_{b\in\lambda}\ketb{\Lpab,i-1,X(b\At i)} +
    \sum_{b\in\rho}\ketb{\Rpab,i+1,X(b\At i)}\]
of vectors \ketb{\Lpab,i-1,X(b\At i)} and \ketb{\Rpab,i+1,X(b\At i)}.

According to \eqref{u0}, given any canonical basis vector, only one gate of the upper row does real work. The other gates just copy their inputs to their outputs.
By the lower equation in \eqref{u1}, the first row of gates transforms \ket{p,i,X} into
\[ \sum_{b\in\lambda} \ket{\Ls_{pab},i,X(b\At i)} +
   \sum_{b\in\rho}     \ket{\Rn_{pab},i+1,X(b\At i)}.
\]
which, according to \eqref{u0},\eqref{u2} and \eqref{u3}, is transformed by the second row of gates into
\[ \sum_{b\in\lambda} \ket{\Lpab,i-1,X(b\At i)} +
   \sum_{b\in\rho}     \ket{\Rpab,i+1,X(b\At i)}
\]
as required.
\end{proof}

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redpajama_set_name: RedPajamaArXiv