Iran's clandestine spy network has been threatening and blackmailing scores of journalists, even going so far as to detain and threaten the family members of these reporters, in order to ensure positive coverage in global media outlets, according to a new report that estimates at least 50 international journalists have been threatened in just the past year.
Outlets such as the BBC and Voice of America have been subjected to threats and in some cases have had their computers hijacked by Iran, according to the report, which states that "all international media outlets with Persian-language services are concerned" about the Islamic Republic's often-secret efforts to blackmail reporters in order to gain positive headlines.
The effort is just one part of how Iran blackmails, threatens, and manipulates journalists and dissident voices in order to suppress coverage of its human rights abuses and other illicit activities, such as an ongoing massive military buildup aimed at confronting Western nations such as the United States.
"In the past year, RSF has learned of ten families of journalists who have been summoned to such interviews, usually with intelligence ministry agents," according to the report. "In all, at least 50 journalists based abroad have been threatened in some way in the same period. At least 16 of them have received death threats."
During international nuclear negotiations in Vienna, Lausanne, and elsewhere, the Washington Free Beacon independently witnessed Iranian reporters allied with the country's state-controlled media organizations monitoring reporters and snapping pictures of them as they worked.
This is part of Iran's efforts to monitor and track reporters who cover the country for a range of influential outlets, sources told the Free Beacon.
"It was insane," one writer who is critical of Iran and attended the nuclear negotiations in Vienna told the Free Beacon. "They would openly get in the faces of Iranian reporters from Western outlets and take pictures just to make sure that the journalists knew that the regime knew who they were and they were being watched. It's impossible to imagine this didn't significantly influences the coverage that came out of the nuclear talks, and even more broadly."
Sources who spoke to RSF, often anonymously to avoid repercussions from Iran, such as imprisonment, told the organization that the country's intelligence services have issued threats and in some cases detained the family members of these reporters.
"It is not just BBC Persian employees who are targeted," the report states. "All international media outlets with Persian-language services are concerned, regardless of the country in which the media are based. Journalists with Radio Farda (Radio Free Europe's Persian-language section), with such state-funded broadcasters as Voice of America, Deutsche Welle and Radio France Internationale, and privately-owned broadcasters such as Manoto TV and Radio Zamaneh have also been threatened by Iran's intelligence services or judicial system."
Most of those who spoke to RSF about Iran's activities to monitor and threaten the press "asked not to be identified" out of fear of reprisals from the regime. However, some did speak on record.
Arman Mostofi, the director of Radio Farda, was quoted as telling the organization that at least four of his reports have been targeted with death threats.
"These threats are obviously not signed," Mostofi told RSF. "They sometimes take the form of a comment beneath an article. The journalist may subsequently be contacted in another way but it's exactly the same message that will be transmitted. Sometimes the message includes information that only members of the intelligence services could know."
Family members also have become the subject of threats, according to the report.
Fahimeh Khezr Heidari, a Radio Farda host, recounted efforts by anonymous individuals to force the broadcast's cancellation.
"Ms. Khezr Heidari, Monday will be a horrible day for a member of your family because you did not take our last warning seriously. Thank you, my corrupt sister," read one message sent at the reporter.
Mohammadreza Nikfar, editor-in-chief of Radio Zamaneh, disclosed that the family members of some reporters have been seized by Iranian intelligence agents.
"The family of one of our journalists was summoned by intelligence ministry agents," Nikfar recounted. "After showing articles by him that had been posted on our website, they said: 'Tell him to stop collaborating with Radio Zamaneh.'"
"Another journalist, a former prisoner of conscience, has been threatened several times by telephone. They tell him his family will suffer the consequences if he does not return to Iran," Nikfar said.
The threats and intimidation tactics also extend to the cyber arena, according to others who experienced attempts by individuals to hack their computers.
At least five journalists have been arrested upon landing in Iran and handed harsh sentence's ranging from three to 12 years in prison, according to the RSF report.
The tactics have a tangible effect on these reporters and the stories they file, RSF concluded.
"When your father calls and an intelligence ministry agent takes the phone and says, 'your father is here and we're talking about you,' and you know that your family is being harassed and is in danger of being arrested, how can your write freely?" one reporter for an international outlet who declined to be named was quoted as telling RSF. "After members of my family had been summoned for questioning, I could no longer work as I had before."
RSF, in a statement, formally condemned efforts "by the Iranian judicial system and intelligence services to influence the Persian-language sections of international media outlets by putting pressure on Iranian journalists based abroad and on their families still in Iran."
In addition to the threats and blackmailing, the Iranian regime has used financial tactics to crackdown on Western media outlets.
Iran recently froze the assets of more than 150 BBC reporters, effectively barring them from conducting financial transactions in the country, according to RSF.
"Nowadays, the families of foreign-based journalists are 'politely' summoned to interviews with intelligence officials but the message is still the same: the journalists must 'stop collaborating with enemy media' without delay," according to RSF.


text:
Tag Archives: dystopian future
Fortress (1992)
September 1, 2013 by marklongden
"Fortress" is one of those films I'd have absolutely loved if I'd seen it when it first came out, when I was in my teens. It's a big budget B movie, and Christopher Lambert was one good performance away from joining people like Arnie and JCVD at the top of that straight-to-video tree (at least for a while there).
Lambert and Loryn Locklin, who seems to have not acted on screen since the millennium, play a husband and wife trying to get across the border into Canada. It's immediately obvious we're in a dystopia, and this particular one appears to have something to do with babies – only one allowed per family. Er, wouldn't this eventually kill off the human race? Ah well, turns out our heroes are expecting their second baby after the first one died, which is a big no-no, and they get captured and sent to the super-super-maximum security prison which gives the film its title.
Kurtwood Smith, post Robocop but pre- That 70s Show, plays the villain of the piece, Vernon Wells – the all time great movie bad guy from "Commando" – has a small part, and Lambert's cellmates are people you'll be going "Hey, it's that guy!" at throughout. Luckily, the cellmates have a variety of critical skills and ethnicities, which allows them to formulate their escape plan (spoiler, I guess, although there are few prison movies where everyone just stays put). The security and methods of keeping them in line are all very futuristic (and quite clever, if I say so myself).
There are an awful lot of hefty coincidences in this film. Considering how enormous the prison is, Lambert only really meets the five people in his cell (one of them is the trustee who cleans the warden's rooms, conveniently); and his wife, while being very pleasant looking, certainly isn't the sort of woman who'd make the warden almost ruin his career for, moving her into his quarters so Lambert can easily rescue her later.
FIlms like this just don't get made any more. B-movies with decent sets and casts like this are a thing of the past – digital filmmaking has meant all sets are CGI, and the drop in production costs means good actors are priced out of the market. Its closest relation of recent years is probably the Guy Pearce-starring "Lockout", which cost over $20 million dollars to make (Fortress cost $8 million, and I don't think inflation is running at over 100% for the intervening years). Films need to be either small enough to make money back off home video or very big, and the inbetween stuff like "Fortress" is sadly gone.
But still! We get to watch them whenever we like, and enjoy the products of the good old days. "Fortress" is terrible, make no mistake – Lambert was increasingly revealed by the early 90s as a man who had one acting style, and while it works in "Highlander" it doesn't work here. Kurtwood Smith is too restrained – he needs to chew that scenery!, and of the rest of the cast, Jeffrey Combs probably acquits hmself the best, even if his death scene is one of the stupidest things I've seen in a while.
As corporations take over more and more of our lives, this film seems less and less outlandish, but there's still plenty of good moments and director Stuart Gordon can make a decent film. So, if you're in the mood for a blast from the past, you could do worse, but maybe find something a bit better or a bit cheesier to enjoy.
Fortress on IMDB
Buy Fortress [DVD] [1994]
Posted in Reviews.	Tagged action movie, Christopher Lambert, dystopian future, sci-fi
Tweet Tweet Boom
Youtube Film Club: Night Hunter (1996) iscfc.net/2019/07/09/you… https://t.co/kQUWHSfsxc 6 days ago
Armageddon: The Final Challenge (1995) iscfc.net/2019/07/07/arm… https://t.co/K0lqYaQ2nN 1 week ago
Deadlocked: Escape From Zone 14 (1995) iscfc.net/2019/07/02/dea… https://t.co/CgNtqEGFoH 1 week ago
Twisted Pair (2018) iscfc.net/2019/06/30/twi… https://t.co/oIzDY0oYV3 2 weeks ago
Youtube Film Club: Deadly Ransom (1997) iscfc.net/2019/06/29/you… https://t.co/W533Nr87lx 2 weeks ago
Follow @iscfc
Support the ISCFC
to help cover our site costs
Youtube Film Club: Night Hunter (1996)
Armageddon: The Final Challenge (1995)
Deadlocked: Escape From Zone 14 (1995)
Twisted Pair (2018)

redpajama_set_name: RedPajamaCommonCrawl


You know, almost intrinsically, that owning your own home makes sense – home ownership promotes psychological, emotional and financial well-being. Harvard's Joint Center of Housing Studies recently published a paper highlighting some of the financial reasons that home ownership benefits you. In assessing potential tax advantages, check out this handy Calculator that will give you an idea of how much you are saving each year in taxes through deductions by owning your own home (or how much you could be saving if you choose to buy over renting). While for high income-producing taxpayers the Alternative Minimum Tax may wipe out those tax savings (check with your tax professional to be sure), leverage, appreciation and other financial advantages still tip the scales in favor of home ownership.
It has been reported many times that the American Dream of homeownership is alive and well. The personal reasons to own a home differ for each buyer, but there are many basic similarities.Eric Belsky is the Managing Director of the Joint Center of Housing Studies (JCHS) at Harvard University. He authored a paper on homeownership titled – The Dream Lives On: The Future of Homeownership in America. In his paper, Belsky reveals five financial reasons why people should consider buying a home.


The night before, in social networks began to spread footage from the personal archive of Armen Dzhigarkhanyan and Vitalina Tsymbalyuk-Romanovskaya. On one of the published pictures of the former spouses are sealed at the beach.
In the comments to the posts you can see various reviews: some praise a few for the lack of complexes, and some shamed Dzhigarkhanyan for painful fullness, and Romanov — received thinness.


View Spring 2016 Commencement photos HERE!
(May 7, 2016) – Armstrong State University hosted 2016 Spring Commencement on Saturday, May 7, 2016, conferring degrees on more than 800 students, a record number, in a special ceremony at the Savannah Civic Center.
University President Dr. Linda M. Bleicken warmly welcomed the 2016 graduates and their families to Spring Commencement. Lt. Gen. Edward C. Cardon, the Commanding General for the U.S. Army Cyber Command and Second Army -- who has served with numerous combat units, educated students at the United States Army Engineer School and received nearly a dozen military honors -- delivered the commencement address.
To that end, Cardon encouraged students to remember that their hard-earned degrees are starting points, not end points, in our ever-changing world, directed by the rapid advance of technology and the unrelenting pervasiveness of social media. He challenged them to think about the choices they make in life.
Graduate Johnathan Davis, a military veteran and Texas native who earned a B.S. in Biology, plans on doing just that.
Davis, who plans to attend dental school after graduation, says Armstrong prepared him by offering exceptional veteran services that directly supported his educational goals.
"Armstrong has been great," he noted. "The professors are very personable with students. The university is also very veteran-friendly. Anything you need, they'll give you a helping hand." Sarah Woods, a transfer student who earned a B.A. in Psychology, chose Armstrong specifically for the strength of the Psychology program and its professors.


EAST LIVERPOOL --The Lou Holtz/Upper Ohio Valley Hall of Fame is pleased to announce the winners of this year's Life Improvement Scholarships.
Annually, Ohio Valley College of Technology, Pittsburgh Technical Institute and New Castle School of Trades partner with the Holtz Hall of Fame by offering a matching scholarship for one student attending their school.
This year, OVCT has generously offered to provide up to the $10,000 match as needed all four Life Improvement scholarship students – Grace, Munoz, Rayl and Sabatini – attending their school.
Firm was selected to receive a matching scholarship from Pittsburgh Technical College.
Firm also will receive a one-time $2,000 scholarship from Dr. Marc and Cindy Hoffrichter of East Liverpool. The scholarship will be administered by the Hall of Fame.
DeHaven was selected to receive a matching scholarship from New Castle School of Trades.
Additionally, Platinum Royalties will provide a $500 scholarship to DeHaven. The scholarship will be administered by the Hall of Fame.
To date, more than $645,000 in trade school education funding for Tri-State Area high school seniors has been provided by Coach Holtz and his wife, Beth, through the Holtz Hall of Fame since the scholarship program began in 1999.


Converse often together with your neighbors and friends about your landscaping plans. They could also wish to do work on their yards, and you may be able to go in together to rent things like chippers or tillers. By sharing this equipment, you will all lower your expenses and be capable of get your work carried out. Landscape based on the totally different seasons. Be sure you embody quite a lot of vegetation in your backyard which are at their greatest throughout totally different seasons. This way you'll create a landscape that provides yr-spherical interest. Plant a combination of shrubs, bushes and flowers, and make it possible for there may be at all times one thing colourful to catch the eye, no matter what season.
When you live in part of the nation that is in danger for wildfires, choose landscaping projects accordingly. For example, bushes should be planted not less than 100 toes aside. Additionally, giant timber should not be planted lower than one hundred ft from your home or different constructing structure (garden, shed, workshop, and many others.). When planning a landscaping venture for your own home, make sure to consider the fact that some vegetation are only in bloom for a sure period of time. That is important in your strategy planning stage as a result of you do not need to have awkward trying gaps in your landscaping. Remember to include each perennials and annuals in the same space to provide long lasting coverage.
Trim your bushes and hedges repeatedly. If you happen to trim a hedge,or a bush when it's just starting to turn out to be over grown, will probably be a quick and straightforward course of. For those who wait until the bush is totally over grown, you can count on the venture to take all day. Common maintenance is the best approach to maintain your landscaping trying great. Fertilize your garden typically. If you need your grass to look full and wholesome, fertilizer therapies need to occur usually over time. One software will assist flip your garden a extra vibrant inexperienced colour, and you may be tempted to stop at that time. Nonetheless, lengthy-time period care gives you even higher results, so remain persistent.
You would save a lot of money by simply taking the time to be sociable.
By adding some fastidiously planned landscaping to your property, you have got the flexibility to greatly enhance the appear and feel of your outdoor areas. The bottom line is to arm your self with a great amount of knowledge before you start to work. Take the ideas in this article to heart, and you will have the power to add magnificence and visible attraction to the exterior of your private home. A good way to essentially enhance your landscape is so as to add a pond or waterfall to your design. This can really make your landscaping look way more beautiful and the sound of water flowing adds a sense of peacefulness and serenity to your total landscape. It is a great way to make your panorama more tranquil.
When planning a landscaping challenge for your own home, a fantastic thought is to incorporate any type of water attraction. No matter if it's a pond, creek, or waterfall, it's possible you'll extremely enhance the look, and resale value of your own home. The associated fee is fairly low in comparison to the potential repay. Landscaping is one of the issues that provides a long-lasting first impression of your private home. Some easy issues can make an enormous distinction, in either a superb or dangerous means. This text gives you ideas on the way to make the curb appeal on your private home bounce up just a few share points.
When planning out your landscape design, carefully contemplate the colors of the flowers, bushes and grasses that you just intend to use. You may discover that limiting your palette to only one or two colors has the greatest visible impression. Choosing too many contrasting colours might make your landscaping appear matted and wild. In the event you attempt to panorama whereas budgeting, remember you could complete a project in phases. There's nothing improper with doing a project in steps or phases. This also helps with your budget. Make a list of every challenge you need to complete and rank them in order of importance.
Your property will look much better by doing something so simple as sprucing up your lawn, adding rocks to your landscaping, or putting in flowers or timber. As long as you do a good job, your garden would be the envy of the neighborhood. Make the most of the knowledge from this article to start straight away. For those who worry damaging power traces or other infrastructure the next time you'll want to dig on your landscaping mission, contact a 'dig protected' helpline first. The service, popularized in the northeastern United States, allows you to double check before digging to ensure that you don't damage anything while performing your landscaping work.
Do it yourself landscapers could be smart to scatter annual plants throughout their design rather than solely utilizing perennials. The same could be true for varied bushes and timber.


Opening Reception that Friday, on view 24/7 for 2-months via Union Avenue. Curated by Dan Streible. Organized by Laurie O'Brien.
Walter Forsberg & Catherine A Kane re-imagine Fred Otts Sneeze inside the Black Maria, at the Thomas Edison National Historical Park, West Orange NJ. Shot by Steve Cossman of MONO NO AWARE on ORWO film, June, 2014.


text:
Commission A Post
Support NFG
Glastonbury's Smallest Department Store
Even Glastonbury can seem dull and grey on a cold November morning when the Winter light bleaches all the colour from the Town. On days like these, we need reminders of sunshine. As I pass through the narrow alleyway of the Gauntlet, the rainbow colours of the crockery in Steve and Jan's shop always catch my eye. I'm reminded of a Winter I spent in Andalusia in Southern Spain, where these jugs, bowls and dishes are made.
Popping into the shop, for a pair of tights or a gift, I've always received a friendly welcome and inevitably ended up in a conversation. Steve and Jan display a genuine warmth and interest in their customers and the community. They tell me they started off with one tiny unit in the Gauntlet 'Steve and Jan's Mini Emporium' but have expanded to three shops selling clothes, ceramics, kitchenware, tights, bags and scarves, The new outlets are 'Socks, Crocks and Frocks' and 'Our Other Shop Ltd'. They joke that they are running 'Glastonbury's Smallest Department Store'.
They were keen to offer one essential item – pants. Jan tells me "We're sinking the myth that you can't buy knickers in Glastonbury!". They thought about calling the clothes shop 'Pants on Fire'.
Jan first came to Glastonbury in 1985, as Purchasing Manager for a large printing business. She has also worked in the construction industry, including 7 years at Glastonbury firm Snows Timber. Steve moved to Somerset in 1993. He'd not heard of Glastonbury but found himself drawn to the town. They met in 2001 when Steve was working for a large building materials supplier and fell in love over a conversation about insulation materials. This must have kept the warmth in as they married a year later.
They started off their business doing markets, selling Steve's homemade preserves, bread and cakes, then they added ceramics and kitchenware. Steve loves Spanish ceramics (each village has its own style) and olive wood. They were thinking to do more markets, but Jan's sewing business was taking off and they realised it was cheaper to take on a shop. Facetious as usual, I asked them why they weren't selling crystals, unlike most of the shops in town. They explained they 'aren't really new agers' but are more interested in the 'pretty and the practical'.
The clothes they stock are made locally by Jan, Vivienne and Marie, others are 'pre-loved', fair trade Indian and Italian and French fashion. They cater for everyone, but the layered style of the Italian designs are flattering for the 'larger lady'. They also sell moccasin slippers that are made in Glastonbury, men's clothes and will soon have a range for children.
The Glastonbury Cancer Research Shop 2019 Calendar
Steve appears naked, but for a carefully placed jug, in the Glastonbury Cancer Research shop's 2019 charity calendar. Jan volunteered Steve, and the shop, to appear, then presented it to him as a fait accompli. The photo shoot was at 7.30am on a Sunday morning, it amused Steve that passing locals "didn't bat an eyelid" at the 15 naked men and a Stormtrooper gathered in front of the Cancer Research charity shop.
Exclusive Offer! Subscribers to We Are Normal For Glastonbury get 10% off everything in Steve and Jan's shops in Glastonbury. This is just one of the exclusive deals from Glastonbury businesses available to members, who also get blog posts, news a member's blog and more. Click the box for details oh how to get this offer.
Why Glastonbury?
Steve and Jan love being part of a town with a High Street full of independent retailers, "Most small towns are bland and soulless in comparison". They also feel that Glastonbury's locals and visitors are refreshingly aspirational – "everyone is looking for the solution to their problems, their 'pot of gold'". They are proud to live somewhere where over 70 different belief systems are practised, that has visitors from every corner of the world. Jan speaks German, French, Italian and Russian and is studying Spanish, she enjoys chatting with customers in their own language. Not to be outdone, Steve points out he is fluent in Brummie and doesn't need subtitles for Peaky Blinders.
They are both interested in the spiritual and the unexplained. Steve feels energy and Jan did a course in Parapsychology with Serena Roney-Dougal. Jan is drawn to Dion Fortune, Steve is more drawn to football, cider and Led Zeppelin. With so much to do running the business seven days a week, they don't get a lot of time to develop their interests.
The Glastonbury Community
Steve has been on the Assembly Rooms Committee for 3 years and is now Treasurer, bringing his business acumen to the aid of this important community building. He tells me they are looking to re-launch in 2020, bringing a more diverse programme of entertainment, including theatre productions and classical music to the town.
I spot a tin for the Glastonbury Community Christmas Lunch on the counter, Steve has recently become involved, last year over 240 people who would have otherwise not had a 'Christmas Experience' got together in the Town Hall. Steve is also keen to become actively involved in helping the homeless in the town. We chat about the recent outrage over anti-social behaviour in Glastonbury during Summer and the consequent bad publicity. Steve is forthright on the issue "The BBC presented Glastonbury as an urban wasteland – which is completely and utterly false. I hate the way people are demonized, they may live chaotic lifestyles and be a nuisance, but they are human beings
Jan has volunteered with the Credit Union and the RSPCA but now finds herself too busy with the shops. she's even had to close the sewing business in order to search for quirky stuff to sell. Like other Glastonbury businesses, they are keen to minimise their environmental impact. I notice there is little plastic packaging in the shop, Steve and Jan tell me they recycle everything they can. The three pairs of knickers I buy (which cost me the grand sum of £4.50 and, I can happily report, are very comfortable) are popped into a paper bag.
Glastonbury resident Nikki pops into the shop:
"Oh My God! You've got knickers! Have you anything sensible for children? I nearly left Glastonbury when my 6-year-old refused to wear elf outfits anymore."
It's clear that Steve and Jan's business and their passion for the community are intertwined, "A business is about touching as many people as you can, it doesn't matter if they are spending £1 or £50. We are serving local people what they want, not just chasing the tourist buck. Our local customers are important to us, we need to be trading successfully in January, not just in Summer". There's something reassuringly old-fashioned about Steve and Jan's attitude toward business, customer service and community.
Before I leave Steve tells me:
"I've been normalized to Glastonbury. It happened by degrees. I don't even notice bizarrely dressed people in the High Street anymore, but I notice people noticing".
Enjoy this post? You can now join We Are Normal For Glastonbury, for exclusive content, a personalized membership certificate, discounts from Glastonbury businesses, a members' forum and blog page where you can post your own writing and photos. You'll also be supporting me to carry on writing about and photographing Glastonbury Town and its wonderful creative community. Membership is only £20 a year. Click here to find out more: We Are Normal For Glastonbury.
If you've enjoyed reading this blog please subscribe by email, 'like' the Normal for Glastonbury facebook page and contribute your own stories and comments, and share my blog and facebook posts (this is really important – it's how I reach more readers!). See my 'Hire Me' page if you'd like to pay me to help you with your own projects, you can also check out how to support this blog.
This post was kindly commissioned by Steve and Jan. Text and photographs copyright Vicki Steward.
Free Cannabis in Glastonbury
Glastonbury Frost Fayre 2018
2 thoughts on "Glastonbury's Smallest Department Store"
One Man And His Bear
"Must have kept the warmth in"
👏🏻 kudos for the great pun there.
VickiPurple
Gosh thanks, to be honest that one even made me cringe, and it felt a bit clumsy, but once it occurred to me I had to put it in!
Subscribe to NFG via Email
Follow NFG
Avalon Marshes (4)
Fiction & Humour (9)
Glastonbury Eating Out (2)
Glastonbury Events (43)
Glastonbury Festival (4)
Glastonbury People (41)
Glastonbury Shops & Services (21)
Glastonbury Town (14)
Hemp In Avalon (4)
History & Mythology (12)
Social and Environmental (5)
Visitor Information (19)
A Shed of My Own
Glastonbury Abbey by Andy Brady
Glastonbury Town for Dog Lovers
Glastonbury Fantasy Artist Linda Ravenscroft
Mediums on Honeymoon in Glastonbury's Pagan Bed & Breakfast
Glastonbury Memories from the Seventies
Glastonbury's Rogues Gallery
Glastonbury Festival 2019 - Rubbish and 5G
Accommodation Art Avalon Marshes Beltane Body Art Cannabis Christmas Cider Clothes Commissioned Post Community Cycling Dragons Fancy Dress Fantasy Art Food Free Cannabis Glastonbury Glastonbury Abbey Glastonbury Festival Glastonbury High Street Glastonbury Tor Glastonbury Town Guest Post Hippies Historic Buildings Humour King Arthur Labyrinths Landscapes Magic May Day mythology Pagan Photography Pubs samhain Self Development Shopping Somerset Somerset Levels Starlings St Margarets Chapel Tourism Winter Solstice
© 2020 Normal For Glastonbury • Powered by GeneratePress
Love Glastonbury? Join the Community!
Like the idea of exclusive blog posts, Glastonbury discounts, members forum and more? Fancy supporting Normal For Glastonbury to carry on, spreading the Glastonbury love? Find out more: We Are Normal For Glastonbury.


Do you want to learn more about your blood sugar levels, coping with sweet cravings, healthy snacks and low GI foods? I'll be at Neal's Yard Remedies in Islington this Wednesday 5th June from 3 – 7pm; available to talk to you and answer any questions. Plus I will be serving up yummy samples of healthy snacks. See you there!
This entry was posted in Front Page and tagged Blood sugar balance, Demonstrations, Healthy snacks. Bookmark the permalink.


Published on May 31, 2008 in PHP. 0 Comments Tags: book, enterprise development, Ivo Jansch, PHP, php|arch.
There's a new book available for pre-order on the php|architect website, titled php|architect's Guide to Enterprise PHP Development, by Ivo Jansch of Ibuildings.
This book is great for anyone who's interested in learning how to use PHP in an enterprise environment. It has every step on the way covered, such as planning, development tools, deployment and much more. If you're (planning on) using PHP in your company, this book is going to prove invaluable for everyone involved in the development process, going from developers to development managers or project managers. I believe pretty much everyone involved with software development can benefit from a book like this.


text:
Is a WebGL based 3D engine for my personal experiments.

### Features ###
- Skinning Animation
- Lighting (Direct, Point, Spotlight, Hemispherics)
- Shadow mapping (VSM)
- Bump mapping
- Basic primitives (Cube, Cylinder, Plane, Sphere, Coordinate)
- Camera (Orbit control, First person control)
- Bounding box
- Loaders
 - Ogre loader ([Script](https://github.com/games/OgreToJson))
 - [GLTF](https://github.com/KhronosGroup/glTF)(based spec0.6, Meshes+Textures+Skin+Animation)
 - OBJ Loader(Only support vertices & indices)
 - [BabylonJS](https://github.com/BabylonJS/) loader (Basic)
- Physics engine (Use [Cannon.js](https://github.com/schteppe/cannon.js))
- Fog
- Skybox
- Reflection texture (Spheric, planar, cubic and projection)
- Create mesh from height map
- Water
- Particle System
- Physically based render (not finish yet)
- Selection octrees
- Billboarding
- Scene graph

### TODO ###
- Code refactoring
- Picking
- Post-processes
- Lens flares
- Morph targets
- LOD system
- Sub Mesh
- New physics plugin [ammo.js](https://github.com/kripken/ammo.js)
- Add BabylonJS License

### Thanks ###
- [BabylonJS](https://github.com/BabylonJS/)
- [Cannon.js](https://github.com/schteppe/cannon.js)
- [vector_math](http://pub.dartlang.org/packages/vector_math)
- [GLTF](https://github.com/KhronosGroup/glTF)

### About ###

* [Blog](http://valorzhong.blogspot.com/)

redpajama_set_name: RedPajamaGithub


text:
Anisocarpus scabridus, the leafy raillardiopsis, is a North American species of flowering plants in the family Asteraceae.

Distribution
It is found only in northwestern California, primarily in scree slopes at relatively high elevations in the Coast Ranges of Humboldt, Mendocino, Trinity, Lake, Tehama, and Colusa Counties. There are also a few isolated populations in the southern Cascades of northern Shasta County.

Description
Anisocarpus scabridus is a small plant rarely more than 2 inches (5 cm) high. It has blue-green leaves and flower heads containing both ray florets and disc florets.

The plant is quite different in ecology and appearance from the only other species in the genus, A. madioides. The two were classified in different genera for many years until molecular and anatomical studies in the 1990s demonstrated their close relationship.

References

External links
Calflora Database: Anisocarpus scabridus (Scabrid alpine tarplant)
 Jepson Manual eFlora (TJM2) treatment of Anisocarpus scabridus

Madieae
Endemic flora of California
Flora of the Cascade Range
Flora of the Klamath Mountains
Natural history of the California Coast Ranges
Plants described in 1905
Taxa named by Alice Eastwood

redpajama_set_name: RedPajamaWikipedia


Look through the profiles of Member users that have joined Meet Local Single Men that are tagged with Goth. Dating other members who have like minded interests is an ideal way to find things to do once you are dating. Create a Free Profile to Go out Today!


FTP Password Recovery is a program to recover logins and passwords for all FTP accounts stored by FTP client software. It's very easy to use and works with CuteFTP, SmartFTP, FileZilla, FlashFXP, WS_FTP, CoreFTP, FTP Control, FTP Navigator, FTP Voyager, WebDrive, SecureFX, AutoFTP, 32bit FTP, BulletProof FTP, Far Manager, etc. All passwords are recovered instantly.
* Recover passwords stored by CuteFTP, SmartFTP, FileZilla, FlashFXP, WS_FTP, CoreFTP, FTP Control, FTP Navigator, FTP Voyager, WebDrive, SecureFX and AutoFTP, 32bit FTP, BulletProof FTP, Far Manager.
* Recover forgotten, lost or saved passwords to all FTP accounts.


text:
Michigan Academy is coming up with the annual Christmas classic event to offer a platform to gymnasts
by dsprime	| Posted on December 15, 2021
Westland (Michigan) [USA], December 15: We are in an era when most people are struggling with turbulent times. It is challenging for people in professions that require them to step out of the house. In these times, the Michigan Academy has helped transform lives. With their online platform that focuses on personal growth and development, they have helped bring real and sustainable change to people's lives.
Michigan Academy has now launched the annual Christmas classic event that will give a platform to gymnasts to showcase their skills.
The secret of Michigan academy's success lies in its focus on people's problems. They have emerged as a trusted name because they offered support during stressful times. They offer expert guidance and actionable solutions for physical and mental health issues that have disrupted people's lives. They have also helped people to sail through financial difficulties.
Going ahead on their path of unconditional support, they are now offering a no-strings-attached guarantee to their customers. This offering allows the customers to cancel their subscription anytime if they are not satisfied with the training programs offered.
Michigan Academy's mission is to help people guide their personal development by bringing about a change in their thought processes. It also encourages people to change how they deal with life's challenges. David Michigan, the CEO of Michigan Academy, is a renowned actor, fitness model, entrepreneur, and motivational speaker. He says that he has merged his life lessons in the courses offered at the academy. These are the same techniques that have helped him to excel in life.
David started Michigan academy to offer a judgment-free safe space to people to come and find solutions to their problems. The online portal offers a clean and easy-to-navigate interface with separate physical, emotional, and mental health sections. Apart from providing well-researched articles and videos by doctors, psychologists, and other experts, it emphasizes exercise, diet, and recovery. The range of topics also extends to travel, safety and finance. The vast pool of information is no less than an encyclopedia that has something for everyone wishing to transform their life.
The academy has developed as a positive retreat for people to join and learn in a brief span. Over 1,70,000 people around the globe are already taking benefit from the portal.
"It is just like personal coaching. I always wished to visit a therapist but was scared about being judged. Michigan academy has offered a judgment-free space. It has done wonders for my mental health. I still get anxious, but the actionable insights make it possible for me to feel better soon," says one student. "The fitness section is gold. It made me aware of the amount of misinformation we have been living with," says another.
David feels proud of these testimonials. He says that his goal was only to offer authentic information and bust the myths that people have lived with for so many years, but the way it has touched lives is incredible.
We hope Michigan Academy continues to touch lives and brings about a positive change in the world.
Babubhai Sweets From Surat Sets World Record For Biggest Ghari 15 Kg and Smallest Ghari 15 Gm
Motivation, commitment, and grit- This is how India's First MMA Fighter Alan Fenandes is supporting Indian fighters


For an enjoyable food preparation this Real Thai Black Bean Wok Sauce brings up the peppery taste of the food. Processed from high superiority black bean, this sauce works well with all type of dishes. Arrange and enjoy your food by adding this sauce to your dishes.
Water, Sugar, Black Bean 5%, Salt, Modified Corn Starch, Spices (Garlic Powder, Citric Acid (E330) As Acidity Regulators.


1. Spray your hair with heat protectant.
2. Tie a super high ponytail.
3. Divide it into two parts.
4. Take one strand of hair, curl it .
5. Don't let it fall, catch it and pin it on head to cool.
6. Repeat the same process till all hair are pinned.
7. Let them cool and spry a hairspray.
8. Remove all the pins and there you have beautiful curls.
This is a easiest way to curl hair hair. These 5 minute curls stay all day. If you are looking for the perfect way to curl your hard to curl hair, this quick way would work just perfectly.


text:
Page 13 of 13 First ... 3111213
Thread: Disney+
I'm the problem it's me
Quoting Philip J. Fry (view post)
That show is pretty funny and quite dark. Good stuff.
And it's happened once again
I'll turn to a friend
Someone that understands
And sees through the master plan
But everybody's gone
And I've been here for too long
To face this on my own
Well, I guess this is growing up
Hedonismbot
A Man Called Otto (M. Forster, 2022) ☆
AmbuLAnce (M. Bay, 2022) ☆
The Pale Blue Eye (S. Cooper, 2022) ☆
Drive My Car (R. Hamaguchi, 2021)
In Bruges (M. McDonagh, 2008)
Bullet Train (D. Leitch, 2022) ☆
Chungking Express (Wong K., 1994)
Persona (I. Bergman, 1966) ☆
Le Samouraï (J. Melville, 1967)
Avatar: the Way of Water (J. Cameron, 2022) ☆
First time ☆
Book of Boba Fett starts out surprisingly weak, even amateurish, which is odd considering that the first (and third) episode was directed by Robert Rodriguez. But the second is great stuff, and the whole enterprise picks up steam after the third. It becomes tied very closely to The Mandalorian show, enough that it's fair to accuse it of not being able to stand on its own, but I didn't mind. Was quite happy to see certain characters. As for Boba Fett specifically, I like the people involved and there's some good stuff here (and some weird shit, like the Millennial Vespa Squad), but the finale will have to really pop to tie things together well.
Last edited by Wryan; 02-09-2022 at 03:28 PM.
I saw it in theaters but I would definitely watch that America number again.
Finale was okay. Bryce Dallas Howard's one episode was better than Robert Rodriguez's three, maybe even combined. Overall a missed opportunity.
Kenobi suffers a bit from the by-now-expected obligation to neatly tidy up loose ends and seed various premonitions and easter eggs regarding the future trilogies. Ewan, the only truly great part of the whole enterprise, treads valiantly to keep it afloat and maybe just barely succeeds. Rest of the cast is good, the writing is fair and I liked the new characters, but the little girl who plays Her is better when she has no dialogue to muggingly munch through. Unexpectedly lovely to have [
Show Spoiler
] back. Not really a ton of surprises given, you know, we already know the broad strokes of how things shake out. But it has a few things on its mind beyond fan-fellatio at least.
I thought it was really fun and interesting.
I loved the Obi Wan show. I think it's the best one of the SW bunch from Disney+ so far. BOBF was pretty alright I guess, I could have skipped that one.
First two episodes of Andor are slow-burns, which I normally wouldn't mind much because you can feel Tony Gilory behind them, but they also feel a little too shapeless maybe? More than compensated by the world-building that genuinely feels lived-in and thought-out though, on both sides of the conflict (loved the bell-towering dude hammering the time). And it greatly builds to that incredible third ep, where the world-building makes that town come truly alive against the invading force, with a few great action set-pieces. Just exciting, stirring stuff there.


text:
\section{Introduction}

\subsection{Background and motivation}
Leader election is a classical and  fundamental problem in distributed computing. It originated as the problem of regenerating the ``token'' in a local area \emph{token ring} network~\cite{Lann77-DistSystems} and has since then ``starred'' in major roles in problems across the spectrum, %
providing solutions for reliability by replication (or duplicate elimination), for locking, synchronization, load balancing, maintaining group memberships and establishing communication primitives. As an example, the content delivery network giant Akamai uses decentralized and distributed leader election as a subroutine to tolerate machine failure and build fault tolerance in its systems~\cite{Nygren-Akamai}. In many cases, especially with the advent of large scale networks such as peer-to-peer systems~\cite{Ratnasamy01CAN,Rowstron01Pastry,Zhao04Tapestry}, it is desirable to achieve low cost and scalable leader election, even though the guarantees may be probabilistic.

Informally, the problem of distributed leader election requires a group of processors in a distributed network to elect a unique leader among themselves,
i.e., exactly one processor must output the decision that it is the leader,
say, by changing a special \emph{status} component of its state to the value
\emph{leader}~\cite{Lyn96}.
All the rest of the nodes must change their status component to the value \emph{non-leader}.  These nodes need not be aware of the identity of the leader. This {\em implicit} variant of leader election is rather standard (cf.
\cite{Lyn96}), and is sufficient in many applications, e.g., for token generation in a token ring environment. This  paper focuses  on  implicit leader election (but improves the upper bounds also for the explicit case, by presenting a time and message optimal randomized protocol).

In another variant,  all the non-leaders change their status component to the value \emph{non-leader}, and moreover, every node must also know the identity of the unique leader. This formulation may be necessary in problems where nodes coordinate and communicate through a leader, e.g., implementations of Paxos~\cite{Chandra-PaxosTalk-PODC07,Lamport-Paxos98}.
 %
 %
 %
 In this variant, there is an obvious lower bound of $\Omega(n)$ messages (throughout, $n$ denotes the number of nodes in the network) since every node must be informed of the leader's identity. This {\em explicit} leader election
 %
 %
 %
 can be achieved by simply executing an (implicit) leader election algorithm and then broadcasting the leader's identity using an additional $O(n)$ messages and $O(D)$ time (where $D$ is the diameter of the graph).

The complexity of the leader election problem and algorithms for it,
especially deterministic algorithms (guaranteed to always succeed),
have been well-studied. Various algorithms and lower bounds are known
in different models with synchronous/asynchronous communication and
in networks of varying topologies such as a cycle, a complete graph,
 or some arbitrary topology (e.g., see
\cite{KhanKMPT08,Lyn96,peleg-jpdc,santoro-book,GerardTelDistributedAlgosBook}
and the references therein). The problem was first studied in context of a
ring network by Le~Lann~\cite{Lann77-DistSystems} and discussed for
general graphs in the influential paper of Gallager, Humblet, and
Spira~\cite{GallagerHS1983}. However, leader election in the class of
complete networks has come to occupy a special position of its own
and has been extensively studied~\cite{AG1991:SICOMP,humblet-clique,KorachKuttenMoran-ModularLE-TOPLAS,KorachOptimalTrees87,KorachOptimal89,singh}; see also \cite{sense-of-dir,LouiMW88,singh97} for leader election in complete networks where nodes have a sense of direction.

The study of leader election algorithms is usually concerned with both message and time complexity. For complete graphs, Korach et al.~\cite{KorachPODC1984} and Humblet \cite{humblet-clique} presented $O(n \log n)$ message algorithms.
  Korach, Kutten, and Moran~\cite{KorachKuttenMoran-ModularLE-TOPLAS} developed a general method  decoupling the issue of the graph family from the design of the leader election algorithm, allowing the development of message efficient leader election algorithms for any class of graphs, given an efficient traversal algorithm for that class. When this method was applied to complete graphs, it yielded an improved (but still
  $\Omega(n \log n)$) message complexity. Afek and Gafni~\cite{AG1991:SICOMP} presented asynchronous and synchronous algorithms, as well as a tradeoff between the message and the time complexity of synchronous {\em deterministic} algorithms for complete graphs: the results varied from a $O(1)$-time, $O(n^2)$-messages  algorithm to a $O(\log n)$-time, $O(n\log n)$-messages algorithm.
Singh \cite{singh} showed another trade-off that saved on time, still for
algorithms with a super-linear number of messages. (Sublinear time algorithms were shown in \cite{singh} even for $O(n \log n)$ messages algorithms, and even lower times for algorithms with higher messages complexities).
Afek and Gafni, as well as~\cite{KorachPODC1984,KorachOptimal89} showed a lower bound of $\Omega(n \log n)$ messages  for {\em deterministic} algorithms in the general case.
 One specific case where the message complexity could be reduced (but only as far as linear message complexity) was at the expense of also having a
  linear time complexity, see \cite{AG1991:SICOMP}.
   Multiple studies showed a different case where it was possible to reduce the number of messages to $O(n)$, by using a {\em sense of direction} - essentially, assuming some kind of a virtual ring, superimposed on the complete graph, such that the order of nodes on a ring is known to the nodes \cite{sense-of-dir}.
  %
The above results demonstrate that the number of messages
needed for deterministic leader election is at least linear or even
super-linear (depending on the time complexity).
In particular, existing $O(1)$ time deterministic algorithms require
$\Omega(n^2)$ messages (in a complete network).

At its core, leader election is a symmetry breaking problem. For anonymous networks under some reasonable assumptions, deterministic leader election was shown to be impossible~\cite{AngluinSTOC80} (using symmetry arguments). Randomization comes to the rescue in this case; random rank assignment is often used to assign unique identifiers, as done herein. Randomization also allows us to beat the lower bounds for deterministic algorithms, albeit at the risk of a small chance of error.

A randomized leader election algorithm (for the explicit version) that could
err with probability $O(1 / \log^{\Omega(1)}n )$
was presented in \cite{KrishnaRamanathan:randomized}
with time $O(\log n)$ and linear message complexity\footnote{In contrast,
the probability of error in the current paper is $O(1 / n^{\Omega(1)})$.}.
That paper also surveys some related papers about randomized algorithms in
other models that use more messages for performing leader election
~\cite{Gupta-ProbLE-DISC} or related tasks (e.g., probabilistic quorum systems,
 Malkhi et al~\cite{Malkhi-ProbQSystems}).
    In the context of self-stabilization, a randomized
 algorithm with $O(n \log n)$ messages and $O(\log n)$ time until stabilization was presented in \cite{dmitry}.

\subsection{Our Main Results}
The main focus of this paper is on studying how randomization can help in
improving the complexity of leader election,
especially message complexity in synchronous networks.
We first present an (implicit) randomized leader election algorithm for a
complete network that runs in $O(1)$ time and uses only $O(\sqrt{n}\log^{3/2}n)$
messages to elect a unique leader with high probability\footnote{Throughout,
``with high probability (w.h.p)" means with probability at least
$1 - 1/n^{\Omega(1)}$.}.
This is a significant improvement over the linear number of messages that is
needed for any deterministic algorithm. It is an even larger improvement over
the super-linear number of messages needed for deterministic algorithms
that have low time complexity (and especially compared to the $O(n^2)$
messages for deterministic $2$-round algorithms).
For the explicit variant of the problem, our algorithm implies an algorithm
that uses (w.h.p.) $O(n)$ messages and $O(1)$ time, still a significant
improvement over the $\Omega(n^2)$ messages used by deterministic algorithms.

We then extend this algorithm to solve leader election on any connected
(non-bipartite\footnote{Our algorithm can be easily modified to work
for bipartite graphs as well --- cf. Section 3.}) $n$-node graph $G$
in $O(\tau(G))$ time and $O(\tau(G)\sqrt{n}\log^{3/2} n)$ messages,
where $\tau(G)$ is the mixing time of a random walk on $G$.
The above result implies highly efficient (sublinear running time and messages)
leader election algorithms for networks with small mixing time.
In particular, for  important graph classes such as expanders (used, e.g.,
in modeling peer-to-peer networks \cite{soda12}), which have a logarithmic
mixing time, it implies an algorithm of $O(\log n)$ time and
$O(\sqrt{n}\log^{5/2} n)$ messages, and for hypercubes, which have
a mixing time of $O(\log n \log \log n)$, it implies an algorithm of
$O(\log n \log \log n)$ time and $O(\sqrt{n}\log^{5/2} n \log \log n)$ messages.

 %

For our algorithms, we assume that the communication is synchronous and follows the standard $\mathcal{CONGEST}$ model~\cite{Pel00}, where a node can send in each round at most one message of size $O(\log n)$ bits on a single edge. For our algorithm on general graphs,  we also assume that the nodes have an estimate of the network's size (i.e., $n$)
and the mixing time. We do not however assume that the nodes have unique IDs, hence the algorithms in this paper apply also for anonymous networks.  We assume that all nodes wake up simultaneously at the beginning of the execution. (Additional details on our distributed computation model are given later on.)

Finally we show that, in general, it is not possible to improve over
our algorithm substantially, by presenting a lower bound for randomized
leader election. We show that $\Omega(\sqrt n)$ messages are needed for any
leader election algorithm in a complete network which succeeds with
probability at least $1/e + \varepsilon$ for any constant $\varepsilon > 0$.
This lower bound holds even in the $\mathcal{LOCAL}$ model~\cite{Pel00},
where there is no restriction on the number of bits that can be sent
on each edge in each round.
To the best of our knowledge, this is the first non-trivial lower bound
for randomized leader election in complete networks.

\subsection{Technical Contributions}

The main algorithmic tool used by our randomized algorithm involves
reducing the message complexity via random sampling.
For general graphs, this sampling is implemented by performing random
walks.
Informally speaking, a small number of nodes (about $O(\log n)$), which
are the candidates for leadership, initiate random walks.  We show that if
sufficiently many random walks are initiated (about $\sqrt{n}\log
n$), then there is a
good probability that  random walks originating from different candidates meet
(or collide) at some node which acts as a referee.
The referee notifies a winner among the colliding random walks.  The
algorithms use a birthday paradox type argument to show that a unique
candidate node wins all competitions (i.e., is elected) with high
probability.
An interesting feature of that birthday paradox argument (for general graphs)
is that it is applied to a setting with non-uniform selection probabilities.
See Section 2 for a simple version of the  algorithm that works
on a complete graph.
The algorithm of Section 3 is a generalization of the algorithm of
Section 2 that works for any connected graph; however the algorithm and
analysis are more involved.

The main intuition behind our lower bound proof for randomized leader election
is that, in some precise technical sense, any algorithm that sends fewer
messages than required by our lower bound has a good chance of generating runs
where there are multiple potential leader candidates in the network that
do not influence each other.
In other words, the probability of such ``disjoint'' parts of the network
to elect a leader is the same, which implies that there is a good
probability that more than one leader is
elected. Although this is conceptually easy
to state, it is technically challenging to show formally since our result
applies to all randomized algorithms without further restrictions.

\iffalse
The high level intuition behind our algorithm is as follows.
The nodes use the number of nodes $n$ and the mixing time $\tau(G,n)$ to generate appropriate probabilities and random walk lengths. The algorithm has three main phases. All nodes wake up at the beginning of the election, yet it is not required that every node attempts to be elected. Therefore, with a suitably chosen probability, a large number of nodes drop out of the election, setting their status component to \emph{non-leader}, thereby leaving, w.h.p., a set of only $O(\log n)$ contestants. Each contestant selects a random rank from a sufficiently large space.
In the second phase, each contestant chooses a random sample of $2\sqrt{n \log n}$ \emph{referee} nodes that decide its fate (in the complete network these are neighbors; in general networks, multiple random walks are executed to sample the referees). This selection ensures that w.h.p. every pair of contestants has at least one common referee, and thus, the referee sets act as \emph{quorums}. Finally, each referee declares the  highest rank contestant among those it knows about as the winner and notifies it (directly, in the complete graph, or by retracing a previously executed random walk, in the case of general graphs). If a contestant detects that it has won all of its contests, it declares itself the leader by setting its status to \emph{leader}; w.h.p. there is only one such winner.
\fi

 %

\subsection{Distributed Computing Model} \label{sec:model}
The model we consider is similar to the models of~\cite{AG1991:SICOMP,humblet-clique,KorachKuttenMoran-ModularLE-TOPLAS,KorachOptimalTrees87,KorachOptimal89},
with the main addition of giving processors access to a private unbiased coin.
Also, we do not assume unique identities.
We consider a system of $n$ nodes, represented as an undirected (not
necessarily complete) graph $G=(V,E)$.
Each node %
runs an instance of a distributed algorithm.
The computation advances in synchronous rounds where, in every round, nodes
can send messages, receive messages that were sent in the same round by
neighbors in $G$,
and perform some local computation.
Every node has access to the outcome of unbiased private coin flips.
Messages are the only means of communication; in particular, nodes
cannot access the coin flips of other nodes, and do not share any memory.
Throughout this paper, we assume that all nodes are awake initially and
simultaneously start executing the algorithm.

\subsection{Leader Election}
We now formally define the leader election problem.
Every node $u$ has a special variable $\texttt{status}_u$ that it can set
to a value in $\{\bot, \mbox{\small NON-ELECTED}, \mbox{\small ELECTED} \}$; initially we assume
$\texttt{status}_u = \bot$.
An \emph{algorithm $A$ solves leader election in $T$ rounds}
if, from round $T$ on, exactly one node has its status set to $\mbox{\small ELECTED}$
while all other nodes are in state $\mbox{\small NON-ELECTED}$. This is the requirement
for standard (implicit) leader election.

\iffalse
We also consider a stronger variant of the problem that we call
\emph{explicit leader election} where every node $u$ has a special
variable $\texttt{lid}_u$ that eventually holds the id of the node that
$u$ considers to be the leader.
An algorithm $A$ solves explicit leader election in $T$ rounds if
$\texttt{lid}_u= \texttt{lid}_v$ for all nodes $u$ and $v$ by round $T$.
\\
{\bf DP: Should we also require that this is a legal output, namely,
the ID of some node? OK only for explicit LE...}
\fi

\section{Randomized Leader Election in Complete Networks}
\label{sec: complete}

To provide the intuition for our general result, let us start by illustrating a
simpler version of our leader election algorithm, adapted to complete networks.
More specifically, this section presents an algorithm that, with high
probability, solves leader election in complete networks in $O(1)$ rounds
and sends no more than $O(\sqrt{n}\log^{3/2}n)$ messages.
Let us first briefly describe the main ideas of
Algorithm~\ref{alg:leaderComplete} (see pseudo-code below).
Initially, the algorithm attempts to reduce the number of leader candidates
as far as possible, while still guaranteeing that there is at least one
candidate (with high probability).
Non-candidate nodes enter the $\mbox{\small NON-ELECTED}$ state immediately,
and thereafter only reply to messages initiated by other nodes.
Every node $u$ becomes a candidate with probability $2\log n / n$
and selects a random rank $r_u$ chosen from some large domain.
Each candidate node then randomly selects $2\lceil\sqrt{n\log n}\rceil$
other nodes as {\em referees} and informs all referees of its rank.
The referees compute the maximum (say $r_w$) of all received ranks, and
send a ``winner'' notification to the node $w$.
If a candidate wins all competitions, i.e., receives ``winner'' notifications
from all of its
referees, it enters the $\mbox{\small ELECTED}$ state and becomes the leader.

\begin{algorithm}[h!]
\begin{algorithmic}[1]
\item[]
 \item[\bf Round 1:]
  \STATE Every node $u$ decides to become a candidate with probability
  $2\log n / n$
  and generates a random rank $r_u$ from $\set{1,\dots,n^4}$.
  \\
  If a node $u$ does not become a candidate, then it
  immediately enters the $\mbox{\small NON-ELECTED}$ state;
  otherwise it executes the next step.
  \STATE \textbf{Choosing Referees:}
  Node $u$ samples $2\lceil\sqrt{n\log n}\rceil$ neighbors (the
  \emph{referees}) and sends a message $\msg{u,r_u}$ to each referee.

  \item[]

  \item[\bf Round 2:]
  \STATE  \textbf{``Winner'' Notification:}
  A referee $v$ considers all received messages and sends
  a ``winner'' notification to the node $w$ of maximum rank,
  namely, that satisfies
  $r_w\ge r_u$ for every message $\msg{u,r_u}$.
    \STATE \textbf{Decision:}
     If a node
     receives ``winner'' notifications from all its referees,
     then it enters the $\mbox{\small ELECTED}$ state;
\\
     otherwise it sets its state to $\mbox{\small NON-ELECTED}$.
\end{algorithmic}
\caption{Randomized Leader Election in Complete Networks}
\label{alg:leaderComplete}
\end{algorithm}

\begin{theorem} \label{thm:leaderComplete}
Consider a complete network of $n$ nodes
and assume the $\mathcal{CONGEST}$ model of communication.
With high probability, Algorithm~\ref{alg:leaderComplete} solves leader
election in $O(1)$ rounds, while using $O(\sqrt{n}\log^{3/2} n)$ messages.
\end{theorem}
\begin{proof}%
  Since all nodes enter either the $\mbox{\small ELECTED}$ or $\mbox{\small NON-ELECTED}$
  state after two rounds at the latest, the runtime bound of $O(1)$
  holds trivially.

  We now argue the message complexity bound.
  On expectation, there are $2\log n$ candidate nodes.
  By using a standard Chernoff bound (cf.\ Theorem~4.4 in \cite{upfal}),
  there are at most $7\log n$ candidate nodes with probability at least
  $1-n^{-2}$.
  In step 3 of the algorithm, each referee only sends
  messages to the candidate nodes which contacted it.
  Since there are $O(\log n)$ candidates and each approaches
  $2\lceil\sqrt{n\log n}\rceil$ referees, the total number of messages sent
  is bounded by $O(\sqrt{n}\log^{3/2} n)$ with high probability.

  Finally, we show that Algorithm 1    %
  solves leader election with high probability.
  The probability that no node elects itself as leader is
  $$\left(1 - \frac{2\lceil\log n\rceil}{n}\right)^n \approx
    \exp(-2\log n) = n^{-2}.$$
  Hence the probability that at least
  one node is elected as leader is at least $1-n^{-2}$.
  Let $\ell$ be the node that generates the highest random rank $r_\ell$
  among all candidate nodes; with high probability, $\ell$ is unique.
  Clearly, node $\ell$ enters the $\mbox{\small ELECTED}$ state, since it
  receives ``winner'' notifications from all its referees.

  Now consider some other candidate node $v$. This candidate
  chooses its referees randomly among all nodes.
  Therefore, the probability that an individual referee selected by $v$
  is among the referees chosen by $\ell$, is
  $2\lceil\sqrt{n \log n}\rceil/n$.
  It follows that the probability that $\ell$ and $v$ do not choose any
  common referee node is asymptotically at most
  $$  \left( 1 - 2\sqrt{\frac{\log n}{n}}\right)^{2\sqrt{n\log n}} \le
  \exp\left(-4\log n\right) = n^{-4},  $$
  which means that with high probability, some node $x$ serves as
  common referee to $\ell$ and $v$.
  By assumption, we have $r_v < r_\ell$, which means that node $v$ does
  not receive $2\lceil\sqrt{n\log n}\rceil$ ``winner'' notifications,
  and thus it subsequently enters the $\mbox{\small NON-ELECTED}$ state.
  By taking a union bound over all candidate nodes other than $\ell$,
  it follows that with probability at least $1-1/n$,
  %
  no other node except $\ell$ wins all of its competitions,
  and therefore, node $\ell$ is the only node to become a leader.
\end{proof}

\section{Randomized Leader Election in General Graphs}
\label{sec: general}
In this section, we present our main algorithm, which elects a unique
leader in $O(\tau)$ rounds (w.h.p.), while using
$O(\tau(G,n)\sqrt{n}\log^{3/2} n)$ messages (w.h.p.), where $\tau(G,n)$ is
the mixing time of a random walk on $G$ (formally defined later on,
in Eq. \eqref{eq:mixing}).
Initially, any node $u$ only knows the mixing time (or a constant factor
estimate of) $\tau(G,n)$; in particular $u$ does
not have any a priori knowledge about the actual topology of $G$.

The algorithm presented here requires nodes to perform random walks on the
network by token forwarding in order to choose sufficiently many referee
nodes at random. Thus essentially random walks perform the role of sampling
as done in Algorithm \ref{alg:leaderComplete} and is conceptually similar.
Whereas in the complete graph randomly chosen nodes act as referees,
here any intermediate node (in the random walk) that sees tokens
from two competing candidates can act as a referee and notify the winner.
One slight complication we have to deal with in the general setting is that
in the $\mathcal{CONGEST}$ model it is impossible
to perform too many walks in parallel along an edge.
We solve this issue by sending only the {\em count} of tokens that need
to be sent by a particular candidate, and not the tokens themselves.

While using random walks can be viewed as a generalization of the sampling
performed in Algorithm~\ref{alg:leaderComplete}, showing that two
candidate nodes intersect in at least one referee
leads to an interesting balls-into-bins scenario where balls
(i.e., random walks) have a \emph{non-uniform} probability to be placed
in some bin (i.e., reach a referee node).
This non-uniformity of the random walk distribution stems from the fact
that $G$ might not be a regular graph.
We show that the non-uniform case does not worsen the
probability of two candidates reaching a common referee, and hence an analysis
similar to the one given for complete graphs goes through.

We now introduce some basic notation for random walks.
Suppose that $V=\{u_1,\dots,u_n\}$ and let $d_i$ denote the degree of node
$i$.
The $n\times n$ \emph{transition matrix} $\mathbf{A}$ of $G$ has entries
$a_{i,j}=1/d_i$
if there is an edge $(i,j) \in E$, otherwise
$a_{i,j}=0$.
The entry $a_{i,j}$ gives the probability that a random walk moves from node
$u_i$ to node $u_j$.
The position of a random walk after $k$ steps is represented by a
probability distribution $\pi_k$ determined by $\mathbf{A}$.
If some node $u_i$ starts a random walk, the initial distribution $\pi_0$ of
the walk is an $n$-dimensional vector having all zeros except at index $i$
where it is $1$.
Once the node $u$ has chosen a random neighbor to forward the token, the
distribution of the walk after $1$ step is given by $\pi_1 = \mathbf{A}\pi_0$
and in general we have $\pi_k = \mathbf{A}^k \pi_0$.
If $G$ is non-bipartite and connected, then the distribution of the walk will
eventually converge to the \emph{stationary distribution}
$\pi_*=(b_1,\dots,b_n)$, which has entries $b_i = d_i / (2|E|)$
and satisfies $\pi_* = \mathbf{A}\pi_*$.

We define the \emph{mixing time} $\tau(G,n)$ of a graph G with $n$ nodes
as the minimum $k$ such that, for all starting distributions $\pi_0$,
\begin{equation} \label{eq:mixing}
  |\!| \mathbf{A}\pi_k - \pi_* |\!|_{\infty} \le \frac{1}{2n},
\end{equation}
where $|\!|\cdot|\!|_{\infty}$ denotes the usual maximum norm on a vector.
Clearly, if $G$ is a complete network, then
$\tau(G,n)=1$.
For expander graphs  it is well known that $\tau(G,n) \in O(\log n)$.
Note that mixing time is well-defined only for non-bipartite graphs;
however, by using a lazy random walk strategy (i.e., with probability $1/2$
stay in the current node; otherwise proceed as usual) our algorithm
will work for bipartite graphs as well.

\begin{algorithm}[h!]
\begin{algorithmic}[1]
  \item[]
  \item[\textbf{Variables and Initialization:}]
  \STATE \textbf{VAR} $\texttt{origin} \leftarrow 0$; $\texttt{winner-so-far} \leftarrow \bot$
  \STATE Node $u$ decides to become a candidate with probability
  $2\log n / n$
  and generates a random rank $r_u$ from $\set{1,\dots,n^4}$.
  \item[]
  \item[\textbf{Initiating Random Walks:}]
  \STATE Node $u$ creates $2\lceil\sqrt{n\log n}\rceil$ tokens of type
  $\msg{r_u,k}$.
  \STATE Node $u$ starts $2\lceil\sqrt{n\log n}\rceil$ random walks
  (called \emph{competitions}), each of
  which is represented by the random walk token
  $\msg{r_u,k}$ (of
  $O(\log n)$ bits) where $r_u$ represents $u$'s random rank.
  %
  The counter $k$ is the number (initially 1) of walks
  that are represented by this token (explained in
  Line~\ref{line:counter}).

  \item[]
  \item[\textbf{Disqualifying low-rank candidates:}] (note that any
   intermediate node along the random walk can act as a referee and
   disqualify the token of a low-rank candidate)
  \STATE \label{line:disqualify} A node $v$ discards every received token
  $\msg{r_u,k}$
  if $v$ has received (possibly in the same round) a token $r_w$
    with $r_w>r_u$.
  \IF{a received token $\msg{r_w,k'}$ is not discarded and
  $\texttt{winner-so-far} \neq r_w$}
  \STATE Node $v$ remembers the port of an arbitrarily chosen neighbor that sent
  one of the (possibly merged) tokens containing $r_w$ in variable $\texttt{origin}$
  and sets its variable $\texttt{winner-so-far}$ to $r_w$. \label{line:origin}
  \ENDIF
  \item[]
  \item[\textbf{Token Forwarding:}]
  \STATE Let $\mu=\msg{r_u,k}$ be a token received by $v$ and
  suppose that $\mu$ is not discarded in Line~\ref{line:disqualify}.
  For simplicity, we consider all distinct tokens that arrive in the
  current round containing the same value $r_u$ at $v$ to be merged into a
  single token $\msg{r_u,k}$ before processing where $k$ holds the accumulated
  count.
  Node $v$ randomly samples $k$ times from its neighbors.
  \label{line:counter}
  If a neighbor $x$ was chosen $k_x\le k$ times, $v$ sends a token
  $\msg{r_u,k_x}$ to $x$.

  \item[]
  \item[\textbf{Notifying a winner in round $\tau(G,n)$:}]
  \IF{$\texttt{winner-so-far} \neq \bot$}
  \STATE Suppose that node $v$ has not discarded some token generated by
  a node $w$.
  According to Line~\ref{line:disqualify}, $w$ has generated the largest
  rank among all tokens seen by $v$. %
  %
  %
  \STATE Node $v$ generates a ``winner'' notification
   $\msg{\textsc{WIN},r_w,cnt}$
  for $r_w$ and sends it to the neighbor stored in $\texttt{origin}$ (cf.\
  Line~\ref{line:origin}). The field $cnt$ is set to $1$ by $v$ and contains the
  number of ``winner'' notifications represented by this token.
  \ENDIF
  \STATE If a node $u$ receives (possibly) multiple ``winner'' notifications
    for $r_w$, it simply
  forwards  a token $\msg{\textsc{WIN},r_w,cnt'}$ to the neighbor stored in
  $\texttt{origin}$ where $cnt'$ is the accumulated count of all received tokens.

  \item[]
  \item[\textbf{Decision:}]
  \STATE If a node wins all competitions, i.e., receives $2\lceil\sqrt{n\log
    n}\rceil$
  ``winner'' notifications it enters the $\mbox{\small ELECTED}$ state;
  otherwise it sets its state to $\mbox{\small NON-ELECTED}$.

\end{algorithmic}
\caption{Randomized Leader Election in General Networks}
\label{alg:leader}
\end{algorithm}

\begin{theorem} \label{thm:leader}
Consider a non-bipartite network $G$ of $n$ nodes with mixing time
$\tau(G,n)$, and assume the $\mathcal{CONGEST}$ model of communication.
With high probability, Algorithm~\ref{alg:leader} solves leader election
within $O(\tau(G,n))$ rounds, while using $O(\tau(G,n)\sqrt{n}\log^{3/2}
n)$ messages.
\end{theorem}

\begin{proof}%
  We first argue the message complexity bound.
  As argued in the proof of Thm. \ref{thm:leaderComplete},
  there are at most $7\log n$ candidate nodes with probability at least
  $1-n^{-2}$.
  Every candidate node $u$
  creates $\Theta(\sqrt{n\log n})$ tokens
  and initiates a random walk of length $\tau(G,n)$,
  for each of the $\Theta(\sqrt{n\log n})$
  tokens.
  By the description of the algorithm,
  there are $O(\sqrt{n}\log^{3/2} n)$ random walks of length $O(\tau(G,n))$.
  In addition, at most one notification message is sent at the last step
  of each random walk, and it travels a distance of at most $O(\tau(G,n))$.
  Hence the total number of messages sent throughout the execution is
  bounded by $O(\tau(G,n)\sqrt{n}\log^{3/2} n)$ with high probability.

  The running time bound depends on the time that it takes to complete the
  $2\lceil\sqrt{n\log n}\rceil$ random walks in parallel and the notification
  of the winner.
  By Line~\ref{line:disqualify}, it follows that a node only forwards
  at most one token to any neighbor in a round, thus there is no delay due
  to congestion.
  Moreover, for notifying the winner, nodes forward the ``winner'' notification
  for winner $w$ to the neighbor stored in $\texttt{origin}$.
  According to Line~\ref{line:origin}, a node sets $\texttt{origin}$ to a
  neighbor from which it has received the first token originated from $w$.
  Thus there can be no loops when forwarding the ``winner'' notifications,
  which reach the winner $w$ in at most $\tau(G,n)$ rounds.

  We now argue that Algorithm~\ref{alg:leader} solves leader election
  with high probability.
  Similarly to Algorithm~\ref{alg:leaderComplete}, it follows that there
  will be at least one leader with high probability.

  Now consider some other candidate node $v$.
  Recall that we have that $r_v < r_\ell$ by assumption.
  By the description of the algorithm, node $v$ chooses its referees by
  performing $\rho = 2\lceil\sqrt{n\log n}\rceil$ random walks of length
  $\tau(G,n)$.
  We cannot argue the same way as in the proof of
  Algorithm~\ref{alg:leaderComplete}, since in general, the stationary
  distribution of $G$ might not be the uniform distribution vector
  $(1/n,\dots,1/n)$.
  Let $p_i$ be the $i$-th entry of the stationary distribution.
  Let $X_i$ be the indicator random variable that is $1$ if there is a
  collision (of random walks) at referee node $i$.
  We have
  $$
  \Prob{ X_i = 1 } = (1 - (1-p_i)^\rho )^2.
  $$
  We want to show that the probability of error (i.e., having no collisions)
  is small;
  in other words, we want to upper bound $\Prob{ \bigcap_{i=1}^n (X_i = 0)}$.
  The following Lemma shows that
  $\Prob{\bigcap_{i=1}^n (X_i = 0) }$ is maximized for the uniform
  distribution.
\begin{lemma} \label{lem:poisson}
  Consider $\rho$ balls that are placed into $n$ bins according to some
  probability distribution $\pi$ and let $p_i$ be the $i$-th entry of
  $\pi$.
  Let $X_i$ be the indicator random variable that is $1$ if there is a
  collision (of random walks) at referee node $i$.
  Then $\Prob{ \bigcap_{i=1}^n (X_i = 0) }$ is maximized for the uniform
  distribution.
\end{lemma}
\begin{proof}
  By definition, we have
  $\Prob{ X_i = 1 } = (1 - (1-p_i)^\rho )^2.$
  Note that the events $X_i=1$ and $X_j=1$ are not necessarily
  independent.
  A common technique to treat dependencies in balls-into-bins scenarios is
  the Poisson approximation where we consider the number of balls in each bin
  to be independent Poisson random variables with mean $\rho/n$.
  This means we can apply Corollary 5.11 of \cite{upfal}, which states
  that if some event $E$ occurs with probability $p$ in the Poisson case,
  it occurs with probability at most $2p$ in the exact case, i.e., we only
  lose a constant factor by using the Poisson approximation.
  A precondition for applying Corollary 5.11, is that the probability for
  event $E$ monotonically decreases (or increases) in the number of balls,
  which is clearly the case when counting the number of collisions of
  balls.
  Considering the Poisson case, we get
  \begin{align*}
    \Prob{ \bigcap_{i=1}^n (X_i=0) }
        &= \prod_{i=1}^n \Prob{ X_i = 0 }
        = \prod_{i=1}^n\left( 1 - (1 - (1 - p_i)^\rho)^2\right) \\
        &\leq \prod_{i=1}^n \left(1 - (1 - e^{-p_i \rho})^2 \right)
        \leq \prod_{i=1}^n \left(1 - (p_i \rho)^2 \right)
        \leq \prod_{i=1}^n e^{-p_i^2 \rho^2}
        = \exp\left(-\rho^2 \sum_{i=1}^n p_i^2\right).
  \end{align*}
  To maximize $\Prob{ \bigcap_{i=1}^n (X_i=0) }$, it is thus sufficient to
  minimize $\sum_{i=1}^n p_i^2$ under the constraint $\sum_{i=1}^n p_i =
  1$.
  Using Lagrangian optimization it follows that this is minimized for the
  uniform distribution, which completes the proof of Lemma~\ref{lem:poisson}.
\end{proof}
  By \eqref{eq:mixing}, the probability of such a walk hitting any of the
  referees chosen by $\ell$, is at least $2\sqrt{n \log n} / (2n)$.
  It follows that the probability that $\ell$ and $v$ do not choose a
  common referee node is asymptotically at most
  \[
  \left( 1 - \sqrt{\frac{\log n}{n}}\right)^{2\sqrt{n\log n}} \le
  \exp\left(-2\log n\right).
  \]
  Therefore, the event that node $v$ does not receive sufficiently many
  ``winner'' notifications, happens with probability $\ge 1-n^{-2}$, which
  requires $v$ to enter the $\mbox{\small NON-ELECTED}$ state.
  By taking a union bound over all other candidate nodes, it follows that
  with high probability no other node except $\ell$ will win all of its
  competitions, and therefore, node $\ell$ is the only node to become a leader
  with probability at least $1-1/n$.
\end{proof}

\section{Lower Bound}
\label{sec: lower}

In this section, we prove a lower bound on the number of messages required
by any algorithm that solves leader election with probability at least
$1/e + \varepsilon$, for any constant $\varepsilon >0$.

Our model
assumes that all processors execute the same algorithm and have access
to an unbiased private coin.
So far we have assumed that nodes are \emph{not} equipped with unique ids.
Nevertheless, our lower bound still holds even if the nodes start with
unique ids.

Our lower bound applies to all algorithms that send only $o(\sqrt{n})$
messages with probability at least $1-1/n$.
In other words, the result still holds for algorithms that have small but
nonzero probability for producing runs where the number of messages sent
is much larger (e.g., $\Omega({n})$).
We show the result for the $\mathcal{LOCAL}$ model, which implies the
same for the $\mathcal{CONGEST}$ model.

\begin{theorem} \label{thm:lb}
Consider any algorithm $A$ that sends at most $f(n)$ messages (of
arbitrary size) with high probability on a complete network of $n$ nodes.
If $A$ solves leader election with probability at least $1/e + \varepsilon$, for any
constant $\varepsilon > 0$, then $f(n) \in \Omega(\sqrt{n})$.
This holds even if nodes are equipped with unique identifiers (chosen by
an adversary).
\end{theorem}

Note that Theorem~\ref{thm:lb} is essentially tight with respect to
the number of messages \emph{and} the probability of successfully electing
a leader. To see this,
first observe that our Algorithm~\ref{alg:leaderComplete} can be modified
such that each node becomes a candidate with probability $c/n$, for some
constant $c>0$, and where each candidate only contacts $\Theta(\sqrt{n})$
referee nodes. This yields a message complexity of $O(\sqrt{n})$ and success
with (large) constant probability.
Furthermore, consider the naive randomized algorithm where each node initially
chooses to become leader with probability $1/n$ and then terminates.
This algorithm succeeds with probability
${n \choose 1}(1/n)(1-1/n)^{n-1} \approx 1/e$ without sending any messages
at all, which demonstrates that there has to be a sudden ``jump''
in the required message complexity when breaking the $1/e$ barrier
in success probability.

The rest of this section is dedicated to proving Theorem~\ref{thm:lb}.
We first show the result for the case where nodes are anonymous, i.e.,
are \emph{not} equipped with unique identifiers,
and later on extend the impossibility result to
the non-anonymous case by an easy reduction.

Assume that there exists some algorithm $A$ that solves leader election with
probability at least $1/e  + \varepsilon$ but sends only $f(n) \in o(\sqrt{n})$
messages.
The remainder of the proof involves showing that this yields a contradiction.
Consider a complete network where for every node, the adversary chooses
the connections of its ports as a random permutation on $\{1,\dots,n-1\}$.

For a given run $\alpha$ of an algorithm, define the \emph{communication graph}
$\mathcal{C}^r(\alpha)$ to be a directed graph on the
given set of $n$ nodes where there is an edge from $u$ to $v$ if and only
if $u$ sends a message to $v$ in some round $r'\le r$ of the run $\alpha$.
For any node $u$, denote the \emph{state of $u$ in round $r$ of the run
$\alpha$} by $\sigma_r(u,\alpha)$.
Let $\Sigma$ be the set of all node states possible in algorithm $A$.
(When $\alpha$ is known, we may simply write $\mathcal{C}^r$ and
$\sigma_r(u)$.)
With each node $u \in \mathcal{C}^r$, associate its state $\sigma_r(u)$ in
$\mathcal{C}^r$, the communication graph of round $r$.
We say that node \emph{$u$ influences node $w$ by round $r$} if there is a
directed path from $u$ to $w$ in $\mathcal{C}^r$.
(Our notion of influence is more general than the {causality} based
``happens-before'' relation of \cite{Lam78}, since a directed path from
$u$ to $w$ is necessary but not sufficient for $w$ to be causally
influenced by $u$.)
A node $u$ is an \emph{initiator} if it is not influenced before sending
its first message.
That is, if $u$ sends its first message in round $r$, then $u$ has an outgoing edge in $\mathcal{C}^{r}$ and is an isolated vertex in $\mathcal{C}^{1},\dots,\mathcal{C}^{r-1}$.
For every initiator $u$, we define the \emph{influence cloud} ${\cal
  IC}^r_u$
as the pair ${\cal IC}^r_u = (C^r_u, S^r_u)$, where
$C^r_u = \langle u, w_1,\ldots,w_k\rangle$ is the ordered set of all nodes
that are influenced by $u$, namely, that are reachable along a directed path
in $\mathcal{C}^r$ from $u$. ordered by the time by which they joined\footnote{We say that a node $v$ \emph{joins the cloud of $u$ in $r$} if $v \notin {C}_u^{r-1}$ and $v \in {C}_u^r$.} the cloud (breaking ties arbitrarily),
and $S^r_u = \langle
\sigma_r(u,\alpha), \sigma_r(w_1,\alpha),\ldots, \sigma_r(w_k,\alpha) \rangle$
is their configuration after round $r$, namely, their current tuple of states.
(In what follows, we sometimes abuse notation by referring to the ordered
node set $C^r_u$ as the influence cloud of $u$.)
Note that a \emph{passive} (non-initiator) node $v$ does not send any messages
before receiving the first message from some other node.

Since we are only interested in algorithms that send a finite number of
messages, in every execution $\alpha$ there is some round
$\rho = \rho(\alpha)$ by which no more messages are sent.

In general, it is possible that in a given execution, two influence clouds
$C^r_{u_1}$ and $C^r_{u_2}$ intersect each other over some common node $v$,
if $v$ happens to be influenced by both $u_1$ and $u_2$.
The following lemma shows that the low message complexity of algorithm $A$
yields a good probability for all influence clouds to be disjoint from
each other.

Hereafter, we fix a run $\alpha$ of algorithm $A$.
Let ${N}$ be the event that there is no intersection between
(the node sets of) the influence clouds existing at the end of run $\alpha$, i.e.,
$C^\rho_u \cap C^\rho_{u'} = \emptyset$ for every two initiators $u$ and $u'$.
Let $M$ be the event that algorithm $A$ sends no more than $f(n)$ messages
in the run $\alpha$.

\begin{lemma} \label{lem:noint}
Assume that $\Prob{M} \ge 1-1/n$.
If $f(n) \in o(\sqrt{n})$, then
$\Prob{N \wedge M} \ge 1 - \frac{1}{n}  - \frac{f^2(n)}{n-f(n)} \in 1-o(1)$.
\end{lemma}
\begin{proof}
Consider a round $r$, some cloud $C^r$ and any node $v \in C^r$.
Assuming event $M$, there are at most $f(n)$ nodes that have sent or received a message and may thus be be a part of some other cloud except $C^r$.
Recall that the port numbering of every node was chosen uniformly at random and,
since we conditioned on the occurrence of event $M$, any node knows the
destinations of at most $f(n)$ of its ports in any round.
Therefore, to send a message to a node in another cloud, $v$ must hit upon one
of the (at most $f(n)$) ports leading to other clouds, from among its (at least
$n-f(n)$) yet unexposed ports.
Let $H_v^r$ be the event that a message
sent by node $v$ in round $r$ reaches a node $u$ that is already part of some other (non-singleton) cloud.
(Recall that if $u$ is in a singleton cloud due to not having received or sent
any messages yet, it simply becomes a member of $v$'s cloud.)
We have $\Prob{H_v^r} \le \frac{f(n)}{n-f(n)}$.
During the entire run, $\ell \le f(n)$ messages are sent in total by some nodes $v_1,\dots,v_\ell$ (in possibly distinct clouds) in rounds $r_1,\dots,r_\ell$, yielding events $H_{v_1}^{r_1},\dots,H_{v_\ell}^{r_\ell}$.
Taking a union bound shows that $$\Prob{\bigvee_{i=1}^\ell H_{v_i}^{r_i} \mid M} \le \frac{f^2(n)}{n-f(n)},$$ which is $o(1)$, for
$f(n) \in o(\sqrt{n})$.
Observe that $\Prob{N \mid M} = 1 - \Prob{\bigvee_{i=1}^\ell H_{v_i}^{r_i} \mid M}$.
Since $\Prob{N\wedge M} = \Prob{N\mid M} \cdot \Prob{M}$,
it follows that $\Prob{N \wedge M} \ge \left(1 - \frac{f^2(n)}{n-f(n)}\right)\left(1-\frac{1}{n}\right) \ge 1 - \frac{1}{n} - \frac{f^2(n)}{n-f(n)} \in 1 - o(1)$, as required.
\begin{comment}
Moreover, $v$ sends at most $f(n)$ messages.
Therefore, $\Prob{D_v \mid M} \le \frac{f^2(n)}{n-f(n)}$
Note that ${\bar N}$, the complementary event to $N$, satisfies
${\bar N} = \bigvee_{i=1}^{f(n)}D_{v_i}$, and, considering that at most
$f(n)$ nodes can send a message in total, we have that
$\Prob{{\bar N} \mid M} \le \frac{f^2(n)f(n)}{n-f(n)} = \frac{f^3(n)}{n-f(n)}$.
Clearly, this probability is $o(1)$, since by assumption $f(n) \in o(n^{1/3})$,
thus we get $\Prob{N \wedge M} = 1 - o(1)$.
\end{comment}
\end{proof}

We next consider {\em potential cloud configurations}, namely,
$Z=\langle \sigma_0, \sigma_1, \ldots, \sigma_k \rangle$,
where $\sigma_i \in \Sigma$ for every $i$, and more generally,
{\em potential cloud configuration sequences} ${\bar Z}^r=(Z^1,\ldots,Z^r)$,
where each $Z^i$ is a potential cloud configuration,
which may potentially occur as the configuration tuple of some
influence clouds in round $i$ of some execution of Algorithm $A$
(in particular, the lengths of the cloud configurations $Z^i$
are monotonely non-decreasing).
We study the occurrence probability of potential cloud configuration sequences.

We say that the potential cloud configuration
$Z=\langle \sigma_0, \sigma_1, \ldots, \sigma_k \rangle$ {\em is realized}
by the initiator $u$ in round $r$ of execution $\alpha$
if the influence cloud ${\cal IC}^r_u = (C^r_u, S^r_u)$
has the same node states in $S^r_u$ as those of $Z$, or more formally,
$S^r_u = \langle
\sigma_r(u,\alpha), \sigma_r(w_1,\alpha),\ldots, \sigma_r(w_k,\alpha) \rangle$,
such that $\sigma_r(u,\alpha) = \sigma_0$ and
$\sigma_r(w_i,\alpha) = \sigma_i$ for every $i\in [1,k]$.
In this case, the influence cloud ${\cal IC}^r_u$ is referred to as a
{\em realization} of the potential cloud configuration $Z$.
(Note that a potential cloud configuration may have many different
realizations.)

More generally, we say that the potential cloud configuration sequence
${\bar Z}^r=(Z^1,\ldots,Z^r)$ is realized by the initiator $u$
in execution $\alpha$ if for every round $i = 1,\ldots, r$, the influence cloud
${\cal IC}^i_u$ is a realization of the potential cloud configuration $Z^i$.
In this case, the sequence of influence clouds of $u$ up to round $r$,
${\bar{\cal IC}}^r_u = \langle {\cal IC}^1_u,\ldots,{\cal IC}^r_u\rangle$,
is referred to as a realization of ${\bar Z}^r$.
(Again, a potential cloud configuration sequence may have many different
realizations.)

For a potential cloud configuration $Z$, let $E^r_u(Z)$ be the event that
$Z$ is realized by the initiator $u$ in (round $r$ of) the run
of algorithm $A$.
For a potential cloud configuration sequence ${\bar Z}^r$, let
$E_u({\bar Z}^r)$ denote the event that ${\bar Z}^r$ is realized
by the initiator $u$ in (the first $r$ rounds of) the run of algorithm $A$.

\begin{lemma}
\label{lem:clouds}
Restrict attention to executions of algorithm $A$ that satisfy event $N$,
namely, in which all final influence clouds are disjoint.
Then $\Prob{E_u({\bar Z}^r)} = \Prob{E_v({\bar Z}^r)}$
for every $r \in [1,\rho]$, every potential cloud configuration sequence ${\bar
  Z}^r$, and every two initiators $u$ and $v$.
\end{lemma}

\begin{proof}
The proof is by induction on $r$.
Initially, in round $1$, all possible influence clouds of algorithm $A$ are
singletons, i.e., their node sets contain just the initiator.
Neither $u$ nor $v$ have received any messages from other nodes.
This means that
$\Prob{\sigma_1(u)=s} = \Prob{\sigma_1(v)=s}$ for all $s\in \Sigma$,
thus any potential cloud configuration $Z^1=\langle s\rangle$ has
the same probability of occuring for any initiator, implying the claim.

Assuming that the result holds for round $r-1\ge 1$,
we show that it still holds for round $r$.
Consider a potential cloud configuration sequence
${\bar Z}^r=(Z^1,\ldots,Z^r)$ and two initiators $u$ and $v$.
We need to show that ${\bar Z}^r$ is equally likely to be realized
by $u$ and $v$, conditioned on the event $N$.
By the inductive hypothesis, the prefix ${\bar Z}^{r-1}=(Z^1,\ldots,Z^{r-1})$
satisfies the claim. Hence it suffices to prove the following.
Let $p_u$ be the probability of the event $E^r_u(Z^r)$
conditioned on the event $N \wedge E_u({\bar Z}^{r-1})$.
Define
the probability $p_v$ similarly for $v$.
Then it remains to prove that $p_u=p_v$.

To do that we need to show, for any state $\sigma_j \in Z^r$, that
the probability that $w_{u,j}$, the $j$th node in ${\cal IC}^r_u$,
is in state $\sigma_j$, conditioned on the event $N \wedge E_u({\bar Z}^{r-1})$,
is the same as the probability that $w_{v,j}$, the $j$th node in ${\cal IC}^r_v$,
is in state $\sigma_j$, conditioned on the event $N \wedge E_v({\bar Z}^{r-1})$.

There are two cases to be considered. The first is that the potential
influence cloud $Z^{r-1}$ has $j$ or more states. Then by our assumption
that events $E_u({\bar Z}^{r-1})$ and $E_v({\bar Z}^{r-1})$ hold,
the nodes $w_{u,j}$ and $w_{v,j}$
were already in $u$'s and $v$'s influence clouds, respectively,
at the end of round $r-1$.
The node $w_{u,j}$ changes its state from its previous state, $\sigma'_j$,
to $\sigma_j$ on round $r$ as the result
of receiving some messages $M_1,\ldots,M_\ell$ from neighbors
$x^u_1,\ldots,x^u_\ell$ in $u$'s influence cloud ${\cal IC}^{r-1}_u$,
respectively. In turn, node $x^u_j$ sends message $M_j$ to $w_{u,j}$ on round $r$
as the result of being in a certain state $\sigma_r(x^u_j)$ at the beginning
of round $r$ (or equivalently, on the end of round $r-1$)
and making a certain random choice
(with a certain probability $q_j$ for sending $M_j$ to $w_{u,j}$).
But if one assumes that the event $E_v({\bar Z}^{r-1})$ holds, namely, that
${\bar Z}^{r-1}$ is realized by the initiator $v$,
then the corresponding nodes
$x^v_1,\ldots,x^v_\ell$ in $v$'s influence cloud ${\cal IC}^{r-1}_v$
will be in the same respective states
($\sigma_r(x^v_j) = \sigma_r(x^u_j)$ for every $j$) on the end of round
$r-1$, and therefore will send
the messages $M_1,\ldots,M_\ell$ to the node $w_{v,j}$ with the same
probabilities $q_j$. Also, on the end of round $r-1$, the node $w_{v,j}$ is in
the same state $\sigma'_j$ as $w_{u,j}$ (assuming event $E_v({\bar Z}^{r-1})$).
It follows that the node $w_{v,j}$ changes its state
to $\sigma_j$ on round $r$ with the same probability as the node $w_{u,j}$.

The second case to be considered is when the potential influence cloud
$Z^{r-1}$ has fewer than $j$ states. This means (conditioned on the events
$E_u({\bar Z}^{r-1})$ and $E_v({\bar Z}^{r-1})$ respectively) that the nodes
$w_{u,j}$ and $w_{v,j}$ were not in the respective influence clouds
on the end of round $r-1$.
Rather, they were both passive nodes.
By an argument similar to that made for round 1, any pair of (so far)
passive nodes have equal probability of being in any state. Hence
$\Prob{\sigma_{r-1}(w_{u,j})=s} = \Prob{\sigma_{r-1}(w_{v,j})=s}$
for all $s\in \Sigma$.
As in the former case, the node $w_{u,j}$ changes its state from its previous
state, $\sigma'_j$, to $\sigma_j$ on round $r$ as the result
of receiving some messages $M_1,\ldots,M_\ell$ from neighbors
$x^u_1,\ldots,x^u_\ell$ that are already in $u$'s influence cloud
${\cal IC}^{r-1}_u$, respectively. By a similar analysis,
it follows that the node $w_{v,j}$ changes its state to $\sigma_j$ on round $r$
with the same probability as the node $w_{u,j}$.
\end{proof}

We now conclude that for every potential cloud configuration $Z$,
every execution $\alpha$ and every two initiators $u$ and $v$,
the events $E^\rho_u(Z)$ and $E^\rho_v(Z)$ are equally likely.
More specifically, we say that the potential cloud configuration $Z$
is \emph{equi-probable for initiators $u$ and $v$} if
$\Prob{E^\rho_u(Z) \mid N} = \Prob{E^\rho_v(Z) \mid N}$.
Although a potential cloud configuration $Z$ may be the end-colud of many
different potential cloud configuration sequences, and each such
potential cloud configuration sequence may have many different realizations,
the above lemma implies the following (integrating over all possible choices).

\begin{corollary}
\label{cor:clouds}
Restrict attention to executions of algorithm $A$ that satisfy event $N$,
namely, in which all final influence clouds are disjoint.
Consider two initiators $u$ and $v$ and a potential cloud configuration $Z$.
Then $Z$ is equi-probable for $u$ and $v$.
\end{corollary}

By assumption, algorithm $A$ succeeds with probability at least $1/e + \varepsilon$, for some fixed constant $\varepsilon>0$.
Let $S$ be the event that $A$ elects exactly one leader.
We have
$$1/e + \varepsilon \le \Prob{S} \le \Prob{S \mid M \wedge N} \Prob{M \wedge N} + \Prob{\text{not $(M\wedge N)$}}.$$
By Lemma~\ref{lem:noint}, we know that $\Prob{M \wedge N} \in 1 - o(1)$ and $\Prob{\text{not $(M\wedge N)$}} \in o(1)$, and thus it follows that
\begin{align}
  \Prob{S \mid M \wedge N} \ge \frac{{1}/{e} + \varepsilon - o(1)}{1 - o(1)} > \frac{1}{e},
  \label{eq:succN}
\end{align}
for sufficiently large $n$.
By Cor.~\ref{cor:clouds}, each of the initiators has the same probability $p$
of realizing a potential cloud configuration where some node is a leader.
Assuming that events $M$ and $N$ occur, it is immediate that $0<p<1$.
Let $X$ be the random variable that represents the number of disjoint influence clouds.
Recall that algorithm $A$ succeeds whenever event $S$ occurs.
Its success probability assuming that $X=c$, at most $f(n)$ messages are sent, and all influence clouds are disjoint, is given by
\begin{align}
  %
  %
    \Prob{S \mid M \wedge N \wedge (X=c)} = c p (1-p)^{c-1}. \label{eq:succ}
\end{align}
For any given $c>0$, the value of \eqref{eq:succ} is maximized if
$p=\frac{1}{c}$, which yields that
$\Prob{S \mid M \wedge N \wedge (X=c)} \le 1/e$ for any $c$.
It follows that $\Prob{S \mid M \wedge N} \le 1/e$ as well.
This, however, is a contradiction to \eqref{eq:succN} and completes the proof of
Theorem~\ref{thm:lb} for algorithms without unique identifiers.

We now argue why our result holds for any algorithm $B$ that assumes that
nodes are equipped with unique ids (chosen by the adversary).
Let $S_{B}$ be the event that $B$ succeeds in leader election.
Suppose that $B$ sends only $f(n) \in o(\sqrt{n})$ messages with high probability but $\Prob{S_{B}} \ge 1/e + \varepsilon$, for some constant $\varepsilon > 0$.
Now consider an algorithm $B'$ that works in a model where nodes do not have ids.
Algorithm $B'$ is identical to $B$ with the only difference that before performing any other
computation, every node generates a random number from the range $[1,n^4]$ and uses this
value in place of the unique id required by $B$.
Let $I$ be the event that all node ids are distinct; clearly  $\Prob{I}\ge 1 - 1/n$.
By definition of $B'$, we know that
$\Prob{S_B} = \Prob{S_{B'} \mid I}$ and, from the anonymous case above, we get
$\Prob{S_{B'} \mid I}\Prob{I} \le \Prob{S_{B'}} \le 1/e + o(1),$ since only $o(\sqrt{n})$ messages are sent with high probability by $B'$.
It follows that
$\Prob{S_{B'} \mid I} \le \frac{1/e + o(1)}{1-1/n} < 1/e + o(1),$ and thus also $\Prob{S_B} \le 1/e + o(1)$, which is a contradiction.
This completes the proof of Theorem~\ref{thm:lb}.

\section{Conclusion}
\label{sec: conclusions}
We studied the role played by randomization in distributed leader election.
Some open questions on randomized leader election are raised by our
work:
(1) Can we  improve the message complexity and/or running time for general
graphs?
(2) Is there a separation between the message complexity for algorithms that succeed with high probability versus algorithms that achieve leader election with large constant probability?

\section*{References}

redpajama_set_name: RedPajamaArXiv


taichiform.dk Tai Chi og Qi gong. Mærk den indre ro og livskraften i Tai Chi Chuan.
Tai Chi og Qi gong undervisning med erfaren lærer. Find inspiration: øvelser, video, artikler, downloads og blog om Tai Chi, Qigong og meditation m.m. Kig ind.
tourettenowwhat.tripod.com Tourette Syndrome - Now What? Tourette's syndrome information for the newly-diagnosed.
Information about the optimistic prognosis for those newly-diagnosed with Tourette Syndrome, including many personal stories, definitions, and concepts for understanding Tourette's research.
Tai Chi und Qi Gong in Hamburg - Wu Wei Schule und Akademie - Qualität seit über zehn Jahren mit regelmässigen Gruppen und Seminaren. Pekingform Taichi Yang-Stil Qigong der 18 Bewegungen und acht Brokate in Othmarschen.