text:
Author: John Nagy '00M.A.
Published: Winter 2017-18
Late one glowing summer morning when I was an uncomplicated, cheerful 10 or 11 and Tom was a crusty, jaded 15, and we'd been out adventuring with our mother on a summer road trip, we stopped in Ohio to visit ancient Aunt Marie, who walked us boys over to the house next door where we would spend the day with our 60-something cousin Ernie.
I know. Lame. A wasted summer day. Marie and Mom had something family to do and Ernie could've been playing 18 holes — I could see it from the golf shirt when he opened his door.
But it wasn't a brush-off at all, and Mom and Aunt Marie knew it.
See, Ernie was a retired ballplayer. He fed us sliced steak on paper plates and we munched on corn chips as we watched re-runs of M*A*S*H and some movie that made kid-me feel like a teenager. But that ballplayer part: my brother didn't care about baseball, so at one point Ernie invited just me outside to have a catch.
You can look him up in The Baseball Encyclopedia. Ernie Kish got his shot in 1945. He was 27 and handsome, played 43 games in sun field for Connie Mack's Philadelphia Athletics and hit a respectable .245. And there we were, throwing a baseball in his backyard in Solon, Ohio.
Ernie stuck around Philadelphia after his playing days and would visit the clubhouse when he went to games at Shibe Park. In 1949, he got the team to sign a ball for my father, who'd have been 11 at the time. Years later, Dad gave that ball to me. Mostly, it's rested in a spherical plastic display case on a shelf in my closet, waiting for the day when my youngest children are old enough to look without whacking it through a neighbor's window.
I spent one day of my life with dad's cousin Ernie, one day on a road trip when I was maybe 10 or 11. I didn't say it was a great story. But in August I took Ernie's baseball to Antiques Roadshow, so I'm not done telling it yet.
The item
Here's a better story.
Sometime in the frigid late winter of 1968, an economics major at Notre Dame named Daniel Farrell bought a white Pontiac Tempest from his statistics professor, Dr. Gregory Curme. It was six years old — not yet a classic, though years later it sure would be — but it was cool and well-kept, sporty, a convertible. Dan would use the car to speed through the Indiana countryside on warm-weather weekends, stopping in at antique shops to pass the time and furnish his South Bend apartment.
If you had to pick a story to get acquainted with Dan Farrell '69, I think this is the one. Because just as important as the car itself was the sticker Farrell told Curme to leave on the window as part of the arrangement — the sticker that secured Farrell a place in the faculty parking lot. Already at age 20, Dan Farrell had an eye for the deal.
To understand this talent, you could delve into the details of the movies Farrell financed as a young representative of the First National Bank of Chicago's media division, mastering the art of what he calls "special financial effects" over the next 10 years while moving from Chicago to London to Los Angeles. But what we're getting to here is the deal he cut in 1980, when he was burning out on the film industry and its punishing requirements for success. Looking for new opportunities, he bought the rights to reproduce in the United States a popular new BBC show.
Antiques Roadshow's army of appraisers describes consulting producer Dan Farrell '69 (above) as generous, amazing, "steady as a rock," kind, honest, quiet, smart, "the Dos Equis guy." Adds sculpture authority Eric Silver: "A lot of people do it because they have this wonderful feeling about Dan. And we've all benefited from it." Photos throughout by Leah Nash
Antiques Roadshow, the original BBC version, got its start when the wife of a producer named Robin Drake dragged him along to an Heirloom Discovery Day at Sotheby's auction house and the chain pulled on the Tiffany lamp inside Drake's brain. What if you could get ordinary Britons to bring out their old but everyday things and tell their stories on camera to experts who'd know something about their value? You'd see loads of kitsch, true, but you'd find plenty of rare, fascinating and gorgeous items as well, and now and then you might discover something of genuine historical importance when, for instance, it turns out that the portrait in grandmum's will is a Gainsborough. You could capture reactions in the moment and film it in the manicured gardens of English estates. And of course no one would mind learning history that way.
Farrell caught the show sitting in his London hotel room one Sunday evening and thought, Here's something worth taking home with me. It was the Eighties: cable television was opening new markets, All in the Family and Sanford and Son had proven the crossover potential for British shows repackaged for American viewers, and Farrell knew people who knew people. The BBC was receptive, and before he knew it Dan Farrell was flying home with the rights to remake the antiques-appraisal hit in America and an agreement that the BBC couldn't compete with him there. All for cheap: about $1,250.
There you go, Nation. It would take Farrell another 15 years and several false starts to get it going, but he eventually struck another deal to produce an all-American Antiques Roadshow with WGBH in Boston. And for the past 22 years, some 8 to 10 million of us have tuned in to our local PBS affiliates on Monday nights and watched strangers talk about cool beautiful stuff for an hour, meeting each other on camera, sharing what they know and cooperating to tell brand-new stories that we all get to hear and see together for the first time. It's vignette history. It's our weekly dose of something we didn't already know; it's mysteries solved and new ones afoot. It's a household word, it's our first — and arguably our only genuine, honest-to-god unscripted and unstaged — reality TV show. It's a unique square in the patchwork quilt of American popular culture. And every once in a while, the familiar punchline discovery at the end, the audience-payoff moment with the dollar signs attached to it, is a piece of information that changes someone's life.
Little thrills. All the feels. Pure entertainment with a dash of education and no commercial breaks. And if you want one person to thank for it, Dan Farrell is your man.
"Good morning! Welcome to Antiques Roadshow. If you have a firearm, please check in with police at the security table before entering the hall."
It's 8 a.m. on a Saturday in August and I'm walking down the wide, red-carpeted corridors of the Oregon Convention Center in Portland. I'm one of the few guests who isn't wheeling a dolly, stroller, rolling suitcase or child's wagon laden with eye-catching curios or a cardboard-concealed treasure chest of one size or another — my autographed baseball is nestled safe in a brown cotton sack inside my backpack. Down the way from the greeter's tables and through an open door I spy Roadshow's neighbor for the day, the 2017 IFPA World Footbag Championships and small circles of alert, happy people kicking tiny beanbags or juggling pins.
The better spectacle by far is inside Exhibit Hall C, where an orderly but lengthening queue of Roadshow guests is coiling through a maze of stanchions that would make even the most seasoned Disney World ridegoer gasp. Maggie, my volunteer escort, whisks me past the maze, past the 18-wheeler studio truck, the curtained green room and the four "triage tables" where guests first present the two items they're allotted to bring in for appraisal, and we blow right by it all into Exhibit Hall D, where a cherry red midcentury love seat is rolling off the first of three brightly lit studio sets and several hundred people are making that familiar, conversational din of Antiques Roadshow background noise that I recognize from the TV broadcasts.
Portland is PBS's No. 1 market for everything, and Roadshow is no exception. This is the show's third visit to the city in 22 seasons, not including the 2012 episodes that were taped two hours down Interstate 5 in Eugene. Some 23,000 fans, I'm told, applied for the pairs of tickets that would entitle them to spend hours waiting in line for their turns with professional appraisers, and the show is 100 percent committed to making sure that each guest leaves with at least a smidgen of new knowledge about their prized possessions and a "verbal approximation" — not a formal appraisal — of each piece's value.
By the end of the day, 13 hours hence, all 5,078 official guests will have snaked their way through the main triage line into the shorter lines that lead to the 26 category tables, from Ancient Art to Musical Instruments to Watches, learned something worthwhile and walked out the door. That's more than 10,000 spot-appraisals performed by 70 or 75 appraisers, roughly 10 informational transactions per auctioneer or antiques dealer per hour. About 120 of them get taped for television.
"It's a lovefest," says Suzanne Perrault, a glass and tile expert from Rago Arts & Auction Center in New Jersey who's been a Roadshow appraiser since the first tour in 1996. "It's about a relationship with people. If you don't like people, this is hell for a day. If you do, it's pure unadulterated love from people who are happy to be there, happy to reach the front of the line, happy to tell you stories."
Standing in the geographical center of this whirling universe of material culture, logistical efficiency and delectable hope is calm Dan Farrell, Roadshow's consulting producer. Now silver-haired and 70, two-way radio on his belt, he's still smiling as big as he was 20 years ago when the show was catching fire. "They did finally radio and tell me you were coming," he says, "but I was out solving a little problem with triage."
The pearly gates of the triage tables were Farrell's domain for years, and this is an important point because the single most valuable asset he brought to WGBH beyond the rights he signed over in 1995 was his firsthand knowledge of the antiques world. Not so much the artifacts themselves, for which he had boundless curiosity and a generalist's grasp, but the people who handle, research and evaluate them to earn their daily bread. Farrell, who in his second career prior to Roadshow built an art and antiques book-distribution business and read a lot of his own stock, knew these appraisers as his customers, and he knew what they knew and what they didn't know. This asset was much more than a Rolodex (speaking of antiques). It was a living, growing, personal network of expertise and goodwill, and on Roadshow it meant that when guests stepped forward with something he'd never seen before, as they do, thousands of times a day, he knew where to send them — and who to send them to.
Past triage, the longest lines typically build up at the Collectibles, Paintings and Arms & Militaria tables. In some cities, Portland being one, the lines for Asian Arts and Tribal Arts might be out the door. Photography is usually light, Farrell says, and Pottery & Porcelain has declined over time. But as the show's producers repeatedly point out, they never know what will turn up on any given Saturday. "Good stuff has feet," executive producer Marsha Bemko likes to say — which helps explain a thousand counterintuitive discoveries like the 1776 French instrument made by Marie Antoinette's favorite harp-maker that turned up in Tucson. All of which means that the preseason homework Farrell does with Bemko and supervising producer Sam Farrell (no relation, though Dan refers to the expat Brit as his "illegitimate cousin") to assign particular appraisers to each of the show's six summer tour stops is vitally important to success, even if much of the who-goes-where comes down to carefully considered guesses.
Dan Farrell left triage some years ago for the out-in-the-open backstage area where the production team huddles to hash out decisions and issues throughout the day. Opposite where he greets me beneath the large square lighting truss are the three on-camera sets, arranged in a horseshoe, where the formal appraisals are filmed. Surrounding all of this is the wide circle of blue background drapes, appraisal tables and milling guests.
Backstage is where some of Farrell's other loosely defined roles emerge. A big one is flow control — relaying reports on ticket numbers to producers and security, and keeping the master clipboard on which appraiser is taping what item, where and when. Then, as the guy in the black AR staff polo shirt whom the talent knows best, Farrell serves both as ombudsman for appraisers' concerns and as disciplinarian when they quibble with producers over items they're pitching — or, less commonly, with unhappy guests. "I'll take their side if I think it's right," he says of the appraisers. "And I'll tell them to grow up if I think they're wrong."
He has one other "gameday" job: second-string security coordinator, a serious business that supervises in-house personnel, plainclothes and uniformed police officers and a K-9 unit, and covers everything from gun checks to crowd control to escorting guests from the hall when an appraisal exceeds a top-secret dollar amount, shielding them from undesirable attention. Farrell's qualification? Nine years as a mayor-appointed police commissioner in his Massachusetts hometown. Sean Quinn, the former NYPD detective and head of security whose fluency in Irish and familiarity with Ireland's history occasionally draws him into unofficial consultations on items from the Emerald Isle, calls Farrell "Commissioner Gordon" to his "Chief O'Hara."
I remember one dealer, early on, he said, 'If Antiques Roadshow comes anywhere near me I'm going to go to the hall and roll a hand grenade into it.'
This morning, though, it's business as usual, which means that when a guest arrives with a whale's inner-ear bone mounted on another piece of whale bone it's Farrell's genius for the appraisers network that's indirectly on display. Ken Farmer, an independent consultant from Charlottesville, Virginia, receives the strangely sculptural piece at the Folk Art table, and when his own intuitions run short, he calls over to Ancient Art for some spot insights from Anthony Slayter-Ralph, a Santa Barbara-based international antiquities and modern fine arts dealer who happens to be an avocational naturalist. Slayter-Ralph gives Farmer enough background to pitch the piece to a producer, who bites on the idea and sets Farmer and the guest up for an informal, over-the-shoulder appraisal. Bang: 60 seconds of potential prime-time television born of an unlikely collaboration between two professionals who'd have no reason to know each other's expertise if not for their long-term involvement with Antiques Roadshow. Farrell's recruits strike again.
Back in 1996 as Roadshow was feeling its way through its first tour of 13 U.S. cities, Farrell cold-called David Rago, a ponytailed, young-turk ceramics authority from New Jersey who came recommended by a mutual friend. They'd spent 10 minutes on the phone, Farrell making his appeal, when the would-be recruit asked for clarification. "Let me get this straight," Rago recalls saying. "We gotta pay our own way. You don't pay us. You don't guarantee we're gonna tape. And, if we tape, you don't guarantee you're going to use it. Is that right?"
Rago had heard correctly. This was public television after all. There was no budget to underwrite expenses, no promise yet of compensation in the form of a star turn on national TV as an authority in one's field. It was all risk with a fingers-crossed hope of payoffs for everyone in the end — if they all gave it their best shot. Which explains why to this day, Farrell is unwilling to take credit as the one guy who brought Antiques Roadshow to America. "I don't think anybody's more aware than I am of how much collaboration it takes to do this," he says.
Rago initially turned Farrell down. But he was curious enough that he and his wife and business partner, Suzanne Perrault, decided to visit a colleague and friend in Kansas City and drop in on the tour as it passed through town. As he recalls, "Within three minutes, we said 'OK! This thing's what they say it is. This thing's gonna work.' We just knew it."
You must understand that the antiques world at that time was — Rago's words — a bunch of "unprofessional misfits." And not everyone felt the same way as Rago and Perrault suddenly did about the show's potential to take the industry in a better direction. "I remember one dealer, early on, he said, 'If Antiques Roadshow comes anywhere near me I'm going to go to the hall and roll a hand grenade into it.' They hated it because they felt we were giving up all the secrets.
"But for me? Number one, it raises your game. It's collegial. I'm sitting at tables with people I've been going hammer and tong with competitively, and now I'm colleagues with them and we're sharing information and loving it. So it's definitely brought the world to a smaller, better place. It's forced us to be more professional.
"It also allows me to go out and spend money on really nice clothes," Rago adds. "I'm into natural fibers. You can quote me."
And how did Dan Farrell the Dealmaker manage to bring auctioneers, shop owners and independent appraisers out of their silos and get national and regional competitors to play nice in the proverbial sandbox? "I'll tell you what overcame it," he says. "Friday night at the bar overcame it. What group of people could have more in common to talk about?"
The appraisers were the easy sell. Far more difficult had been the television executives who'd turned Farrell down for 15 years before the agreement with WGBH. He'd been thinking the likes of A&E or Discovery Channel, but they weren't thinking much about him. "As soon as the word 'antiques' came out of my mouth, I could virtually hear their eyes glaze over."
Early on, he'd enlisted the help of Don Taffner, the Anglophile independent TV producer who'd translated the ITV sitcom Man About the House into Three's Company and delivered comedian Benny Hill to delighted American audiences. Taffner had a New York office full of Mackintosh furniture and a posh townhouse off London's Russell Square. No one was better suited to take Farrell's dream under his wing, but then nothing came of it. In the 1990s, thinking he needed to show rather than tell, Farrell pulled host Monty Hall of Let's Make a Deal fame out of retirement to create a pilot that's never been seen except by Farrell and a few prospective partners.
All the while, he says, he avoided public television "because I entertained the idea of making money off this thing." But by 1995, WGBH executives had seen the BBC show and felt the same tickle Farrell had years before.
The public hardly needed convincing. Publicity, not lack of interest, was the major issue during the 1996 launch tour. In College Park, Maryland, maybe two dozen people showed up all day. "We were thinking of trawling on the street," jokes Peter Cook, one of the show's early producers, just to get the 15 appraisals they needed to fill a show. But broadcast audiences that winter loved it. By Season 2, PBS affiliates in host cities needed to do little to lure guests. Farrell remembers waking up in Pittsburgh and opening the curtain to see crowds lined up around the convention center. In Los Angeles, Season 3, Cook says, Roadshow tied up the freeway. By the time Season 5 aired in February 2001, CBS's 60 Minutes had caught on to the phenomenon and Farrell could fairly comment to Dan Rather that Roadshow, and not the network's own Survivor, was America's original foray into reality TV.
WGBH wanted to call it This Old Stuff. Farrell's choice was The Great American Treasure Hunt. "What I never could seem to sell them on was what I truly believed about the show, which is that it isn't about antiques at all. It's about treasure hunting." People want to know about their showpieces, whether they'd found their prized possession in a Dumpster, bought it for $10 at a flea market or inherited it from a tall-talking uncle. To that last point, Farrell says that Stuart Whitehurst, a Roadshow regular, has threatened to write a book he'd call Your Grandma Lied, exposing the all-too-human credulity that accompanies so many artifacts he sees in his Florida business and the typical tour stop.
The fascination driving the show, Farrell argues, is whether someone will get lucky. "We see a lot of two-figure pieces," Rago says, though for him the excitement is not about dollar signs — not even after he finishes a mainstage appraisal in Portland of a piece of Picasso pottery that he values at more than $100,000. The pot is "large, impressive, aesthetically important," yes, but Rago can't connect with it.
By contrast, folk-art authority Allan Katz couldn't believe his eyes 13 years ago when a schoolteacher approached him in St. Paul, Minnesota, with a two-foot-high, cobalt blue stoneware jug, stamped as made for Philadelphia's Centennial Exposition in 1876, that he'd inherited from his grandmother. Quick research showed the 80-pound piece was the work of the fair's gold-medal winner. Encouraged by Katz's appraisal, the owner sold it through an auction house for $110,000. He bought a used Volkswagen and got a vanity plate that read "JUGBUG." Better yet, Katz says, the teacher finally had the money he needed to adopt a child. It was Katz's first show, and this remains his favorite story.
Behind these moments is the alchemy of the "pick," the producers' vetting of item, guest, story and appraiser enthusiasm that determines whether a piece gets taped for consideration back in Boston. So, midafternoon in Portland, we meet Paul, looking anxious as he sits on a foldout chair behind the Metalwork & Sculpture table with a crumpled paper bag in his lap. He's casting glances at appraiser Jason Preston and fatigued producer Sam Farrell, who has tasked a runner to retrieve some "anti-inflammatories."
Some things have a value that transcends mere dollars. But how would you appraise a television program that's become part of our national vocabulary?
Farrell is holding a smooth wooden block the size and shape of a brick, on which is mounted the hammered bronze image of a horse lying on its side. Paul uses it as a bookend and knows little about it — a plus — and Preston is telling Farrell that the late artist, Susan Hertel, was once an assistant to Millard Sheets (he of Touchdown Jesus fame, a fact that here in Portland interests no one but me). Hertel's paintings go for $10,000 to $15,000, Preston says, but the mini-bronze appears to be a unique work, done early in her career. He thinks it could pull $2,000 at auction.
"It's not knocking my socks off," Farrell replies. He notes some upside: the artist is a woman, new to Roadshow; the piece is done in a Western-modern style, somewhat abstract but rippling with energy. They're desperate for formal appraisal items in the middle, he concedes, but he isn't sold. Maybe a shorter over-the-shoulder appraisal amidst the bustle of the floor?
The runner returns with four painkillers and a cup of water. "Oh, thank you," Farrell says cheerfully. "I don't need four, but are you trying to kill me? Anybody want some ibuprofen?"
He turns to Preston. "I think I'll go with the horse."
Paul's sculpture does better than my baseball, which gets the polite but thorough attention of appraiser Jasmani Francis over at the Sports Memorabilia table. Several long-suffering guests roll their eyes at my media credential as I cut the line.
They saved the ball's sweet spot for Connie Mack, Francis notes, turning the ball in his hands as he talks. That's a point of pride for me, the signature of baseball's longest serving and winningest manager still large and bold in the year before he retired at age 87. "I'm looking for the Fox," he continues, meaning Hall of Fame second baseman Nellie Fox, who is on the ball. Amazingly, within seconds, Francis can tell the whole team is there (plus Cousin Ernie).
The 1949 Athletics finished fifth in the American League, and Francis has memorized the roster? He's seen baseballs like this one before. He confers with another appraiser. It's a souvenir, not an official game ball. Still, the potency of the signatures is strong. No game wear, which would be better, but a good baseball. "$300 to $500 at auction; $750 at most in insurance."
That's the number I care about, because the ball ain't going anywhere.
Francis grins — a genuine, friendly smile. "Thank you for bringing it in."
The final appraisal
Roadshow is reinventing itself. Next stop is Newport, Rhode Island's Rosecliff mansion, and the American show's first-ever indoor/outdoor taping. "The show needed freshening," Dan Farrell explains one morning over breakfast.
I tell him about my ball and ask what he'd pick if he could bring in two items of his own. He doesn't hesitate.
He'd take a watercolor by the late Richard Yarde, an African American painter "not well known but maybe regionally," who taught for many years at the University of Massachusetts. It's a New Orleans parade scene, a marching band in deep blues, reds and blacks, "very vibrant for a watercolor."
And he'd grab his autographed 1966 national champion Fighting Irish football. "I personally went around and got everybody on the team, including Ara, to sign." The last signature he collected was Alan Page's. Farrell knocked on Page's dorm room door and tossed him the football — a story he recounted to the all-American defensive end, Purple People Eater of Minnesota Vikings fame and Minnesota Supreme Court justice years later when Page '67 was a guest on Roadshow.
Farrell gave the ball to his dad, then got it back when his father died. "It sits in a closet wrapped in tissue, and all of that. I've talked to some people about it. Yeah, it's worth some money, but. . . ." His voice trails off.
Sentimental value.
Dan Farrell never took an art class at Notre Dame. He'd started in biology and switched to econ when a course in physical chemistry blew him into the liberal arts. It was a life pattern established early, deciding they were having more fun over there than he was having here and wanting a piece of that — while making a few shrewd deals along the way. It led him to banking, to the movies, to a lark on the rights to a TV show concept that looked like it would never get made. Later in life, it led to his own art consultancy and a partnership with Allan Katz to buy the 1700-piece collection that General Electric wanted to offload before moving its corporate headquarters in 2016. "One of the guys who helped me move, he said, 'What are you going to do with all this art?' I said, 'We're going to sell it and make a lot of money.'"
But some things have a value that transcends mere dollars. His work on Antiques Roadshow comprises 40 percent of his business, Farrell estimates, and if WGBH decides to cut the show, the rights to make it anew would revert to him. But how would you appraise a television program that's endured 22 years of changing tastes and maintained its audience of millions, that's brought professionalism and camaraderie to an industry that never valued those things before, that's become an American institution, part of our national vocabulary?
"I do still have these 'pinch me' moments," he says. "We've entered the popular culture. And that's kind of fun to think about."
John Nagy is managing editor of this magazine.


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text:
\section{Introduction}
Considerable attention in recent years was devoted to the study of error
correction with codes on graphs. In this paper we are interested in
estimating the number of errors correctable with codes on
graphs constructed as generalizations of LDPC codes. LDPC
codes are constructed on a bipartite graph $G(V,E), V=V_1\cup V_2$
by associating
code's coordinates with the vertices in one part of $G$,
replicating the values of each vertex on the edges incident to it,
and imposing a parity-check constraint at each vertex of the
other part of $G$.
The generalization that we have in mind is concerned with replacing
the repetition and single-parity-check codes as local codes at the
graph's vertices with other error-correcting codes.

Error correction with codes on graphs has been studied along two lines,
namely, by computing the average number of errors correctable with some
decoding algorithm by codes from a certain random ensemble of graph codes,
or by examining explicit code families whose construction involves graphs
with a large spectral gap.
The first direction originates in the works of Gallager
\cite{gal63} and Zyablov and Pinsker \cite{zya75} who showed
that random LDPC codes of
growing length can correct a nonvanishing fraction of errors.
Recently the decoding algorithm of \cite{zya75} was studied by
Burshtein \cite{bur08}
who derived an improved
estimate of the number of correctable errors compared to \cite{zya75}
and by Zyablov et al. \cite{zya09} who provided estimates of the number
of errors
under the assumption of local single error-correcting (Hamming) codes.
The second line of work, initiated in Tanner's paper \cite{tan81} and in
Sipser and Spielman's \cite{sip96},
pursues estimates of error correction with codes on regular graphs with a small second
eigenvalue and ensuing expansion properties.
Presently it is known that such codes under iterative decoding
can correct the number of errors equal to a half of the designed distance
of graph codes  \cite{bar05}. This estimate fits in a series of analogous results
for various ``concatenated'' coding schemes and has prompted a view
of graph codes as parallel concatenations of the local codes \cite{bar05}.
However, this result relies on certain restrictive assumptions
discussed below.

An extension of Tanner's construction
from graphs to hypergraphs was proposed
by Bilu and Hoory \cite{bil04} who showed that such codes (for high code rates)
can have minimum distance greater than the best known bipartite-graph constructions.
Interestingly, the codes considered in \cite{bil04} are a direct
extension of a construction in \cite{gal63} in the same way as
Tanner's graph codes extend LDPC codes.

As is well known, graphs with high expansion and random graphs share many
properties that can be used to prove estimates of error correction.
This similarity in the coding theory context
was emphasized in our recent work \cite{bar08b} where we showed
that ensembles of codes on random graphs
and explicit expander-like constructions share many common features
such as properties of the minimum distance and weight distribution.

Regarding the proportion of errors corrected by graph codes under
iterative decoding, we note one difference between
(generalized) LDPC codes on random graphs and explicit constructions based on
the graph spectrum. The explicit constructions based on regular graphs
depend on the difference between the largest and the second largest eigenvalue
of the graph (the ``spectral gap''). For this reason,
one is forced to rely on local codes with rather large minimum distance $d_0$,
for instance, $d_0$ greater than the square root of the degree $n$ of the graph.
Even though in the construction of \cite{sip96} and later works $n$ is kept
constant, this effectively rules out of consideration
local codes with small minimum distance such as the Hamming codes and the like.
The square root restriction is implied by the spectral gap of regular
bipartite graphs, and is the best possible owing to the Alon-Boppana
bound for graph spectra \cite{nil91}.
The purpose of the present work is to lift this limitation on the distance
$d_0$ by switching from
graphs with a large spectral gap to random graphs.

In this paper we obtain new estimates of the number of correctable errors
for random ensembles of bipartite-graph and hypergraph codes under iterative
decoding.
The first part of the paper is devoted to codes on regular
bipartite graphs. To construct long graph codes, we assume that the
degree of the graph is fixed and the number of vertices in both parts
approaches infinity.
Assuming that local constraint codes are used to correct 2 or more errors,
we show that almost all codes in the ensemble of graph codes
are capable of correcting all error patterns of weight that forms a constant
fraction of the code length.
This is a much less restrictive assumption on the local codes than
the one taken in earlier works on decoding of graph codes
\cite{zem01,bar05}. The proof of this result employs some ideas
of \cite{bar08b} introduced there for the analysis of the weight
distribution of graph codes.

We then observe that if the degree of the graph is allowed to increase
then graph codes with local codes of constant distance
do not correct a linearly growing number of errors under the proposed iterative decoding.
This motivates us to study graph codes with long local codes correcting a
growing number of errors that forms a fixed proportion of the degree. The results obtained
in this case parallel earlier theorems for product codes and graph codes
based on the spectral gap.

In the second part of the paper we establish similar results for codes
on hypergraphs, showing that a constant proportion of errors is corrected
by an iterative decoding algorithm that combines some ideas of
\cite{bar08b} with the results proved for codes on bipartite graphs in the
first part of the paper.
Constructing the code ensemble based on regular hypergraphs of a fixed
degree, we show that they contain codes capable of correcting a constant
proportion of errors. The proof involves no assumptions on the
distance of the local codes; in particular, we show that networks of
Hamming codes correct a fixed proportion of errors under iterative decoding.
This fact was previously proved by Tanner \cite{tan81} under the
assumption that the underlying graph is a tree. This assumption
is not needed in our results. As in the case of the graph ensemble, we also
perform the analysis of the decoding algorithm for the case of growing degree, finding
the proportion of errors correctable with hypergraph codes based on
long local codes.

This paper is dedicated to the memory of Ralf Koetter.
The first-named author discussed the problem of estimating the performance
of codes on graphs with Ralf in the beginning of 2004.
Ralf's idea at that point was to investigate the error correcting
capability of codes defined on some distance-regular graph, with local
constraints imposed at the vertices of the graph.
Presently it is understood that the setting most amenable to analysis
is that of codes defined on a regular bipartite graph.
Ralf himself made an initial attempt to analyze such codes in
a joint paper with Xiangyu Tang \cite{tan07}. The emphasis in \cite{tan07}
is on the estimation of the largest channel error rate tolerated by graph codes
under such decoding.
In the present paper, similarly
to \cite{sip96,zem01} and later works, the local codes are decoded up to their
correction radius guaranteed by the minimum distance.

\section{Code ensembles}
An $[N,K]$ binary linear code is a linear subspace of $\{0,1\}^N$ of dimension
$K.$ To construct an $[N,RN]$ binary linear graph code $C,$
consider an $n$-regular bipartite graph $G(V=V_1\cup V_2, E),$
where the set of vertices $V$ consists of two disjoint parts $V_1,V_2$
of size $m$ each, all the edges are of the form $(u,v), u\in V_1,v\in V_2,$
and the degree of every vertex $v$ in $V$ is $n.$ Let
$A[n,R_0n,d_0]$ be a linear binary code of length $n$ called the local code below.
We identify the coordinates of $C$ with the set $E$ and for a vertex $v \in V$
denote by
$\bfx(v)\in \{0,1\}^n$ the projection of a vector
$\bfx\in \{0,1\}^N, N=nm,$ on the edges incident to $v.$
A graph code $C(G)$ is defined as follows:
    \begin{equation}\label{eq:def}
      C=\{\bfx\in \{0,1\}^N:\; \forall_{v\in V}\bfx(v)\in A\}.
    \end{equation}
The ensemble of codes $\sG(A,m)$ is constructed by associating a code
$C(G)$ with a graph $G$ sampled from the set of graphs defined by a random
permutation on $N$ elements which establishes how the edges originating
in $V_1$ are connected to the vertices in $V_2.$

Generalizing this construction,
consider an $l$-partite $n$-regular uniform hypergraph $H=(V,E)$
i.e., a finite set $V=V_1\cup\dots \cup V_l,$ where $|V_1|=\dots=|V_l|=m,$
and a collection $E$
of $l$-subsets (hyperedges) of $V$ such that every $e\in E$
intersects each $V_i, 1\le i\le l$ by exactly one element and
each vertex $v\in V$ appears in exactly $n$ different subsets of $E$.
Aiming at constructing an $[N,RN]$ binary linear code $C$ by imposing local
constraints at the vertices, we again identify the coordinates
of $C$ with the (hyper)edges of $H$.
By definition, the code $C$ is formed of the vectors $\bfx$ that
satisfy condition (\ref{eq:def}) for every vertex in $V.$
The ensemble of codes $\sH(A,l,m)$ in this case is constructed by sampling a random hypergraph from the set of hypergraphs defined by $l-1$ independent
random permutations on $N$ elements. For $i=1,2,\dots, l-1,$ the $i$th permutation accounts for the placement
of edges between parts $V_1$ and $V_{i+1}$ of $H$. Of course, $\sH(A,l,m)$ becomes
$\sG(A,m)$ for $l=2$.

The following is known about the parameters of codes in the
graph and hypergraph ensembles considered here.
It is easy to see that the rate $R$ of the codes $C\in \sH(A,l,m)$ satisfies
$R \ge lR_0 -(l-1), l=2,3,\dots .$
Denote by $d(\sH)=d(\sH(A,l,m))$ the average value of the minimum distance
of codes in the hypergraph ensemble and let
  \begin{equation}\label{eq:distance}
   \delta=\delta(\sH)\triangleq \liminf_{N\to\infty}\frac{d(\sH)}N.
  \end{equation}
A way to bound the value of
$\delta$ below using the distribution of distances in the local code $A$
was suggested in \cite{bou99,len99}. More explicit results in this direction
were obtained in \cite{bar06,bar08b}. In particular, \cite{bar08b} shows
that $\delta(\sH)>0$ if the local distance $d_0$ satisfies $d_0>l/(l-1).$
For the bipartite graph ensemble $\sG(A,m)$ (i.e., for $l=2$) this implies that $d_0\ge 3,$
i.e., with high probability codes in ensemble are asymptotically good
(have nonvanishing rate and relative distance) when the local
codes correct one or more errors. For hypergraphs with $l=3$ or more
parts any local codes (without repeated vectors) account for an
asymptotically good ensemble.
An explicit lower bound for $\delta(\sH)$ that depends only on $l$ and
$d_0$ is given by \cite{bar08b}, see Theorem \ref{thm:distance} below.
For the case when $n$ is large and $d_0=\delta_0 n,$ a lower
estimate of $\delta(\sH)$ is given by the solution for $x$
of the following equation \cite[Cor.~6]{bar08b}:
  \begin{equation}\label{eq:delta}
  \frac{h(x)}{x} =\frac{l}{l-1}\frac{h(\delta_0)}{\delta_0}.
  \end{equation}

Finally, if the local codes are chosen randomly as opposed to a fixed
code $A$ used at every vertex of $H,$ then the codes in the (hyper)graph
ensemble match the best known linear codes, i.e., reach the asymptotic
Gilbert-Varshamov bound on the minimum distance \cite{bar08b}.

{\em Remarks.} 1. An equivalent description of the bipartite code
ensemble is obtained
by considering an edge-vertex incidence graph of the graph $G(V,E)$, i.e., a
bipartite graph $D=(D_1\cup D_2, \bar E)$ where
$D_1=E, D_2=V_1\cup V_2,$ each vertex in $D_1$ is connected to one vertex
in $V_1$ and to one vertex in $V_2,$ and there are no other edges
in $\bar E.$ Thus, for all $v\in D_1, \deg(v)=2$ and for all $v\in D_2,$
$\deg(v)=n.$ The local code constrains are imposed on the vertices in
$D_2$.
By increasing the number of parts in $D_2$ from two to $l$, we then
obtain the hypergraph codes defined above. This gives an alternate description of
the hypergraph code presented in Fig.~\ref{fig1}.

The ensemble of hypergraph codes
with local constraints given by single parity-check codes was introduced
by Gallager \cite[p.12]{gal63}. The proportion of errors correctable with these codes using the so-called ``flipping'' algorithm
was estimated in \cite{zya75}. Several generalizations
of this ensemble were studied in \cite{bil04,bar08b}.

\begin{figure*}[tH]
\centering
\includegraphics[scale=0.8]{hg.eps}
\vspace*{-9pt}
\caption{Alternate construction of the hypergraph code: The set
$D_1=\{e_1,\dots,e_N\},$ where $\deg(e_i)=l$ for all $i$,
represents the coordinates of the code (hyperedges of $H$);
the sets $V_1,\dots, V_l,$ where $|V_j|=m$ for all $j$, represent the vertices of
the hypergraph $H$.
Each vertex $v_{i,j}, 1\le i\le l, 1\le j\le m$ carries a codeword of the
local code $A$ of length $n$. }\label{fig1}
\end{figure*}

2. The derivations of this paper are not specific to binary codes:
any local linear codes such as Reed-Solomon codes can be used in the
construction with no conceptual changes to the analysis and the conclusions.

\section{Decoding algorithms for graph (Generalized LDPC) codes}
Even though the ensemble $\sG(A,m)$ forms a particular case of the ensemble
$\sH(A,l,m),$ in our analysis we employ different decoding algorithms for the
cases $l=2$ and $l\ge 3.$ The reason for this is that edge-oriented
procedures commonly used for bipartite-graph codes do not generalize
well to hypergraphs.

\emph{A. Decoding for the ensemble $\sG(A,m)$.}
In our estimates of the number of correctable
errors for the ensemble $\sG$ we rely
upon the algorithm of \cite{zem01} which iterates between decoding
all the vertices in parts $V_1$ and $V_2$ in parallel using some
decoding algorithm of the code $A$. Let $C\in \sG(A,m)$ be a code.
For the ease of analysis we
assume that the local codes are decoded to correct up to $t$ errors,
where $t\ge 0$ is an integer that satisfies $2t+1\le d_0$ and $d_0$ is
the distance of the code $A.$
Formally, define a mapping $\psi_{A,t}:\{0,1\}^n\to \{0,1\}^n$ such that
$\psi_{A,t}(\bfz)=\bfx\in A$ if $\bfx$ is the unique codeword that
satisfies $d(\bfz,\bfx)\le t$ and $\psi_{A,t}(\bfz)=\bfz$ otherwise.
Let $\bfy^{(i)}$ be the estimate of the transmitted vector before the $i$th
iteration, $i\ge 1,$ where $\bfy=\bfy^{(1)}$ is the received vector.
The next steps are repeated for a certain number
of iterations.

{\em Algorithm I} $(\bfy^{(1)})$\\[1mm]
 \hspace*{3mm} $\bullet$ $i$ odd: for all $v\in V_1$ put $\bfy^{(i+1)}(v)=\psi_{A,t}(\bfy^{(i)}(v));$\\
  \hspace*{3mm} $\bullet$ $i$ even: for all $v\in V_2$ put $\bfy^{(i+1)}(v)=\psi_{A,t}(\bfy^{(i)}(v)).$

\remove{Then in odd iterations the algorithm performs decoding of all the local
codes in part $V_1,$ subsequently replacing the current estimate
of the transmitted vector by the result of this decoding. If a
particular vertex fails to find a codeword within distance $t$
from its current estimate, then no changes are made from the current estimate.
In even iterations this procedure is repeated for the part $V_2.$
Below we call this decoding Algorithm I.}

\vspace*{2mm}\emph{B. Decoding for the ensemble $\sH(A,l,m)$.}
For the hypergraph ensemble $\sH(A,l,m)$ we use the decoding
algorithm proposed in \cite{bar08b}. It proves to be the best choice in terms
of the number of correctable errors among several possible
algorithms for these codes such as the one in \cite{bil04}
and procedures analogous to Algorithm I above.

Let $C\in \sH(A,l,m)$ be a code and let $H(V,E), V=V_1\cup\dots\cup V_l$
be the graph associated with it.
For every $i=1,2\dots,l$ we will define an $i$-th subprocedure that
decodes the local code $A$ on every vertex in the part $V_i.$ Suppose
that a vector
$\bfu\in\{0,1\}^N$ is associated with the edges $e\in E.$
Let $v_{i,1},\dots,v_{i,m}$ be the vertices in the part $V_i$ of $H$
and let $\bfu_{i,1}=\bfu(v_{i,1}),\dots,\bfu_{i,m}=\bfu(v_{i,m})$ be the $m$ subvectors
obtained from $\bfu$ upon permuting its coordinates according to the
order of edges in $V_i$ and projecting it on the vertices
$v_{i,1},\dots,v_{i,m}.$ In other words, the vector
$(\bfu_{i,1},\dots,\bfu_{i,m})$ is obtained from $\bfu$ using the
permutation that establishes edge connections between parts $V_1$ and $V_i.$
The $i$th subprocedure replaces the vector
$(\bfu_{i,1},\dots,\bfu_{i,m})$ with the vector
$(\psi_{A,t}(\bfu_{i,1}),\dots,\psi_{A,t}(\bfu_{i,m})).$

The algorithm proceeds in iterations.
Let $\bfy\in\{0,1\}^N$ be the received vector.
Denote by
$Y^{(j)}$ the set of estimates of the transmitted
codeword (i.e., the set of $N$-vectors) stored at the vertices of $H$
before the $j$th iteration $j=1,2,\dots.$ After each iteration, this set
is formed as the union of the vectors obtained upon decoding of the
vertices in the $i$th part, $i=1,\dots, l.$
Decoding begins with setting $Y^{(1)}=\{\bfy\}.$
After the first iteration we obtain $l$ potentially different vectors
(one for each subprocedure)
which form the current estimates of the transmitted vector.
These vectors form the sets $Y_i^{(2)}, i=1,\dots, l.$
In the next iteration each subprocedure will have to be applied to each
of the $l$ outcomes of the preceding iteration.
Proceeding in this way, we observe that $|Y_i^{(j)}|\le l^{j-1}.$

This algorithm, called Algorithm II below, will only be applied for
a constant number $s$ of iterations until we can guarantee that at
least one subprocedure has reduced the number of errors to a specified
proportion, say from $\gamma_0 N$ to some $\gamma_1 N, \gamma_1<\gamma_0.$
We then let another algorithm take over and decode all the $l^s$ candidates.
Any low-complexity
decoder of graph codes that removes an arbitrarily small positive fraction
of errors $\gamma_1$ will do at this stage. This is because taking the proportion of
errors from $\gamma_0$ to $\gamma_1>0$ can be accomplished in a
constant number $s$ of steps, so the number of candidates that this decoder has
to handle is at most $l^{s}$ and does not depend on $N.$

For the case of local codes correcting $t\ge 2$ errors
we let this algorithm to be the decoding algorithm of bipartite-graph
codes (Algorithm I), making sure that $\gamma_1$ is below the proportion
of errors that are necessarily corrected by this algorithm for the
ensemble $\sG(A,m).$ This is possible because, leaving any two parts of
the original hypergraph $H$ to form a bipartite graph $G$, we obtain a
random code from the ensemble $\sG(A,m)$ which with high probability
(over the ensemble) will
remove all the residual errors from at least one candidate estimate.
For $t=1$ this approach fails for the reasons discussed in the next section,
so we resort to a procedure in \cite{zya09} that corrects a small linear fraction
of errors for single-error-correcting Hamming codes.

Upon performing the described procedure we obtain a list of at most $l^s$
candidate codewords of the code $C.$ The final decoding result is
found by choosing the codeword from this list closest to $\bfy$ by
the Hamming distance.

Though the last step of the decoding algorithm described is different
from \cite{bar08b}, the main idea is similar to that paper,
so we refer to it for a more detailed description and a
discussion of the algorithm.

\section{Number of correctable errors for the ensemble $\sG(A,m)$}

Let $C\in\sG(A,m)$ be a code and let $G(V,E)$ be the graph
associated with it. For a given subset of vertices
$S\subset V_i,i=1,2$ and a vertex $v$ denote by $\deg_S(v)$ the
number of edges between $v$ and $S.$
Let $T_r(S)=\{v\in V: \deg_S(v)\ge r+1\},$
where $r\in\{0,\dots,n-1\}$ is an integer.

Below $h(\bfz)$ denotes  the entropy of the probability vector
$\bfz\in \reals^{n+1}$.
In the particular case of $n=1$ we write $h(z)$ instead of $h(z,1-z).$

Let $t\ge 0$ be any integer such that $2t+1\le d_0.$
The calculation in this section is based on the following simple observation.
\begin{proposition}\label{prop:decoding}
Suppose that for all
$S\subset V_i, i=1,2, |S|\le \sigma m, \sigma\in(0,1),$ there exists $\epsilon>0$
such that $|T_t(S)|\le|S|-\epsilon m.$ Then any $\sigma t m=\sigma t(N/n)$
errors will be corrected by Algorithm I in $O(\log m)$ iterations.
\end{proposition}
\begin{IEEEproof}
Suppose that no more than $\sigma t m$ errors occurred in the channel.
Let $S_i$ be the set of vertices that are decoded incorrectly in iteration $i$
of Algorithm I.
The assumption of the proposition implies that $|S_{i+1}|\le |S_i|(1-\epsilon/\sigma),$
so $O(\log m)$ iterations suffice to remove all the errors.
\end{IEEEproof}
Define
\begin{align}
     F_{n,t}(\sigma)&=h(\sigma)-\sigma n\log x\nonumber\\
         &+\sigma \log\sum_{i=t+1}^n\binom ni x^i
      +(1-\sigma)\log\sum_{i=0}^t\binom ni x^i, \label{eq:m}
   \end{align}
where $x>0$ is found from the equation
   \begin{equation}\label{eq:x}
     \sum_{i=0}^t\sum_{j=t+1}^n\binom ni\binom nj(\sigma(n-j)-i(1-\sigma))
   x^{i+j-t-1}=0.
   \end{equation}
Let $\cZ_n=\{\bfz\in[0,1]^{n+1}:\, \sum_{i=0}^nz_i=1\}$ be
the $(n+1)$-dimensional probability simplex.

The main result of this section is given by the next theorem.
\begin{theorem}\label{thm:bg} Let $A[n,R_0n, d_0]$ be the local code,
let $m\to\infty$, and let $2\le t<d_0/2.$
All codes in the ensemble $\sG(A,m)$
except for an
exponentially small (in $N$) proportion of them correct any combination of errors
of weight  $\sigma tm$ in $O(\log m)$ iterations
of Algorithm I, where $0<\sigma<\sigma_0$ and $\sigma_0$ is the
smallest positive root of the equation
    $$
       F_{n,t}(\sigma)=(n-1)h(\sigma).
    $$
\end{theorem}
{\em Remark.} The case of local codes with $t=1$ is excluded from this
theorem because $G$ with high probability contains a large number of
4-cycles, which means that correcting single error at every vertex
does not ensure overall convergence of the decoding. Indeed, if
two vertices are affected by two errors each, and the corresponding 4 edges
form a cycle, then the decoder will loop indefinitely without approaching
the correct decision. The theorem
is still valid in this case, but gives $\sigma_0=0.$

\begin{IEEEproof} We need to verify the assumption of Proposition
\ref{prop:decoding}.
Let $S\subset V_1,|S|=\sigma m$ and let $m_i=|\{v\in V_2: \deg_{S}(v)=i\}|,
i=1,\dots, n.$  Clearly,
   $$
        \sum_{i=1}^n m_i\le m,\; \sum_{i=t+1}^n m_i=|T_t(S)|,\;
\sum_{i=1}^n im_i=|S|n.
   $$
Let us compute the probability (over the choice of $G$) that
$|T_t(S)|\ge(\sigma-\epsilon)m.$ Let ${\boldsymbol\mu}=(m_1,\dots,m_n)$
be a vector with nonnegative integer components, let
   \begin{multline*}
    M_\epsilon(t,\sigma)=\{{\boldsymbol\mu}: \;\sum_{i=1}^n m_i\le m,\,
                    \sum_{i=1}^n im_i=\sigma N,\;\\[-1mm]
             \sum_{i=t+1}^n m_i\ge
     (\sigma-\epsilon) m\},
   \end{multline*}
and let $\binom m{\boldsymbol\mu}$ denote the number of choices of subsets
of size $m_1,\dots, m_n$ out of a set of size $m$. We have
  \begin{equation}\label{eq:pT}
   P(|T_t(S)|\ge|S|-\epsilon m)=\frac1{\binom N{\sigma N}}
        \sum_{{\boldsymbol\mu}\in M_\epsilon(t,\sigma)}\binom m{\boldsymbol\mu}\prod_{i=1}^n\binom ni^{m_i}.
  \end{equation}
Let $\sL_1(s)$ denote the event that $V_1$ contains a subset
$S, |S|=s$ for which $|T_t(S)|\ge |S|-\epsilon m.$ We have
  $$
  P(\sL_1(\sigma m))\le \binom m{\sigma m}P(|T_t(S)|\ge|S|-\epsilon m)
  $$
and
  $$
  P\Big( \bigcup\limits_{i=1}^{\sigma m}\sL_1(i)\Big)\le
    m P(\sL_1(\sigma m)).
  $$
Denote by $\sL_2(\sigma)$ an analogous event with respect to $V_2.$
Then
  \begin{equation}\label{eq:P}
  P\Big(\bigcup\limits_{i=1}^{\sigma m}(\sL_1(i)\cup\sL_2(i))\Big)
    \le  \frac{2m\binom m{\sigma m}}{\binom N{\sigma N}}
     \!\sum_{{\boldsymbol\mu}\in M_\epsilon(t,\sigma)}\!\binom m{\boldsymbol\mu}\prod_{i=1}^n\binom ni^{m_i}.
  \end{equation}
Letting $L$ to be the logarithm of the left-hand side divided by $m$ and
omitting $o_m(1)$ terms,
we obtain the estimate $L\le  n^{-1} \bar F_{n,t}(\sigma),$ where
   $
  \bar F_{n,t}(\sigma)= -(n-1) h(\sigma
+\max_{\bfz\in \sM'_\epsilon(t,\sigma)}
   \Big(h(\bfz)+ \sum_{i=1}^n z_i\log\binom ni\Big),
   $
where
   $$
 \sM'_\epsilon(t,\sigma)=\Big\{\bfz\in\cZ_n:
\sum_{i=1}^n iz_i=\sigma n, \,\sum_{i=t+1}^n z_i\ge\sigma-\epsilon\Big\}
   $$
and $z_i=m_i/m, z_0=(m-\sum m_i)/m.$

The rest of the proof is concerned with the evaluation of the above maximum.
Define
  \begin{equation}\label{eq:g}
     g(\bfz)=h(\bfz)+ \sum_{i=1}^n z_i\log\binom ni
  \end{equation}
  $$
 \bar\sigma=\sup\{\sigma>0: \bar F_{n,t}(y)<0 \text{ for all } 0\le y<\sigma\}.
  $$
As long as $\sigma<\bar\sigma,$ the probability of not being able to correct
$\sigma tm$ errors with a random code from the considered ensemble approaches zero.
Thus, we need to
find the maximum $\max_{\bfz\in \sM'_\epsilon(t,\sigma)}g(\bfz)$
for all $\sigma\in[0,\bar\sigma).$
The proof will be accomplished in the next three steps.
Since $\epsilon$ will be assumed arbitrarily small, we will omit it from our
considerations and write $\sM'$ instead of $\sM'_\epsilon$.

1. We find the point $\bfz^\ast$ that gives the maximum of $g(\bfz)$ without the
constraint $\sum_{i=t+1}^n z_i\ge \sigma.$

2. Next we show that for $0\le\sigma<\bar\sigma,$ the point $\bfz^\ast\not \in
\sM',$ and therefore the maximum over $\sM'$ is attained on the boundary, i.e.,
we can replace $\sM'$ with
   $$
   \sM(t,\sigma)=\Big\{\bfz\in\cZ_n:
\sum_{i=1}^n iz_i=\sigma n, \,\sum_{i=t+1}^n z_i=\sigma\Big\} .
   $$

3. Finally we compute the value of the maximum.

\vspace*{.05in}Step 1. Without the constraint $\sum_{i=t+1}^n z_i\ge \sigma$
the maximum is easily computed.
Indeed, the proportion of edges incident to the vertices in $S$
out of the $N$ edges of $G$ is $\sigma,$
so the fraction of vertices with $S$-degree $i$ should be close to
$z_i^\ast(\sigma)=\binom ni \sigma^i (1-\sigma)^{n-i}.$ Thus, the coordinates of
the maximizing point
$\bfz^\ast=\bfz^\ast(\sigma)$ are $z_i^\ast, i=1,\dots,n; z_0=1-\sum_iz_i^\ast,$ and
 \begin{equation*
  g(\bfz^\ast)=n h(\sigma).
 \end{equation*}
Slightly more formally, note that $\bfz^\ast$ is the unique stationary
point of the function $g(\bfz)$, and that this function is strictly
concave in $\bfz.$  Therefore, $\bfz^\ast$ is a unique maximum of
$g(\bfz)$ on $\cZ_n$, and the function  $g(\bfz)$ grows in the
direction $\bfz^\ast-\bfz$ for any $\bfz\in \cZ_n.$

\vspace*{.05in}Step 2. Suppose that $0\le\sigma\le\bar\sigma.$
Observe that $p(\sigma)\triangleq\sum_{i=t+1}^n z_i^\ast=P(X\ge t+1),$
where $X$ is a $(\sigma,1-\sigma)$ binomial random variable.
This probability is monotone increasing on $\sigma$ for $\sigma\in[0,1],$
and $p(0)=p'(0)=0.$
Thus for $\sigma\in [0,\alpha)$ where $\alpha$ is the smallest positive root of
$\sum_{i=t+1}^n z_i^\ast(\sigma)=\sigma,$  we have
   $$
    \sum_{i=t+1}^n z_i^\ast=\sum_{i=t+1}^n \binom ni\sigma^i(1-\sigma)^{n-i}
       <\sigma,
   $$
and so the point $\bfz^\ast(\sigma)\not\in \sM'(t,\sigma).$
Our claim will follow if we show that $\bar \sigma<\alpha.$ This is indeed
the case because for $0\le\sigma<\bar \sigma,$
  $$
     \max_{\bfz\in \sM'(t,\sigma)} g(\bfz^\ast(\sigma))<(n-1)h(\sigma).
  $$
On the other hand, $g(\bfz^\ast(\alpha))=nh(\alpha).$
This establishes that the maximum of $g(\bfz)$ on $\bfz\in \sM'$ is attained
on the hyperplane $\sum_{i=t+1}^n z_i=\sigma.$

\vspace*{.05in}Step 3.
To compute the maximum of $g(\bfz)$ on $\bfz,$ let us form the
Lagrangian
 \  \begin{multline*}
    U(\bfz,\tau_1,\tau_2)=h(\bfz)+\sum_{i=1}^n z_i\log\binom ni\\+
    \tau_1\Big(\sum_{i=1}^n iz_i-\sigma
    n\Big)+\tau_2\Big(\sum_{i=t+1}^n z_i-\sigma\Big).
  \end{multline*}
Setting $\nabla U=0$ and $\tau_1=\log x, \tau_2=\log y,$ we find that
  $$ z_i=\begin{cases} {\displaystyle \binom ni
  x^iD}&\quad\text{if }0\le i\le t\\[.2in] {\displaystyle \binom
  ni yx^iD}&\quad\text{if }t<i\le n,
     \end{cases}
     $$ where we have denoted
     $$ D=\Big[\sum_{i=0}^t\binom ni x^i+y\sum_{i=t+1}^n\binom ni x^i\Big]^{-1}.
     $$
Adding these equations together, we find conditions for $x$
and $y$:
   $$ \sigma=Dy\sum_{i=t+1}^n\binom ni x^i
   $$
   $$ \sigma n=D\Big(\sum_{i=0}^ti\binom ni x^i+y\sum_{i=t+1}^ni\binom
   ni x^i\Big).
   $$
Once $y$ is eliminated from the last two equations, we obtain
the condition (\ref{eq:x}) for $x.$ Finally, substituting the found
values of $z_i,i=1,\dots,n$ into $g(\bfz),$ we find that
the maximum evaluates to the expression
$F_{n,t}(\sigma)$ given in (\ref{eq:m}) (and therefore, $\bar\sigma=\sigma_0).$
Since we seek to obtain a
value $L<0,$ the boundary condition for the proportion of correctable
errors is obtained by setting $L=0.$ This concludes the proof.  \end{IEEEproof}

\vspace*{.1in}
\begin{example} Using Theorem \ref{thm:bg} together with (\ref{eq:m})
we can compute the proportion of errors corrected by codes in the
ensemble $\sG(A,m), m\to \infty$ for several choices of the local code $A$.
For instance, taking $A$ to be the binary Golay code of length $n=23$
we find $\sigma_0\approx0.0048586$ and therefore, the proportion of
correctable errors is $\frac{\sigma_0 t}n\approx0.00063.$ Similarly,
for the 2-error-correcting $[n=31,k=21]$ BCH code we find
$\sigma_0\approx 0.000035$ and $\frac{\sigma_0 t}n\approx 0.0000023.$
\end{example}

\vspace*{.1in}
To underscore similarities with the results obtained for
product codes and their later variations including graph codes
(e.g., \cite{zem01}) we compute the proportion of errors correctable with
codes from the ensemble $\sG(A,m)$ in the case of large $n$.
\begin{proposition} Let $t=\tau n.$ Then the ensemble $\sG(A,m)$ contains
codes that correct  $\sigma\tau N$ errors for any $\sigma\le\sigma_0,$ where
$\sigma_0$ is given by
   \begin{gather*}
    \sigma_0=\sup\Big\{\sigma>0: \forall_{0<x<\sigma}\;(1-x)
         h\Big(\frac{x(1-\tau)}
     {1-x}\Big)\\+x h(\tau)+\varepsilon_n<h(x)\Big\}
   \end{gather*}
where $\varepsilon_n=(1+\log n)/n.$
\end{proposition}
\begin{IEEEproof} Referring to the notation of the previous proof,
let us evaluate the asymptotic behavior
of the exponent $L$ of the probability in (\ref{eq:P}).
  Since $h(\bfz)\le\log n,$ we have
  \begin{multline*}
    n^{-1}\bar F_{n,t}(\sigma)\le -h(\sigma)+n^{-1}
\max_{\bfz\in \sM(\tau n,\sigma)} \sum_{i=0}^nz_i\log\binom ni
   \\+n^{-1}(1+\log n).
  \end{multline*}
Next,
  $$
    \frac 1n\sum_{i=0}^nz_i\log\binom ni\le \sum_i z_ih\Big(\frac in\Big)
  $$
\begin{align*}
   &=(1-\sigma)\sum_{i=0}^t\frac{z_i}{1-\sigma}h\Big(\frac in\Big)
  +\sigma \sum_{i=t+1}^n \frac {z_i}{\sigma}h\Big(\frac in\Big)\\
   &\le (1-\sigma)h\Big(\frac{\sum_{i=1}^t iz_i}{(1-\sigma)n}\Big)
     +\sigma h\Big(\frac{\sum_{i=t+1}^n iz_i}{\sigma n}\Big).
 \end{align*}
Let $y=n^{-1}\sum_{i=t+1}^n iz_i,$ then for any $\bfz\in\sM(\tau n,\sigma)$ we have
  $$
\frac 1n\sum_{i=0}^nz_i\log\binom ni\le \max_{\tau\sigma\le y\le\sigma}\Big\{
      (1-\sigma)h\Big(\frac{\sigma-y}{1-\sigma}\Big)
     +\sigma h\Big(\frac{y}{\sigma}\Big)   \Big\}.
  $$
The function on the right-hand side of this inequality is concave. Its
global maximum equals $h(\sigma)$ and is attained for $y=\sigma^2.$
Thus, assuming that $\sigma<\tau,$
we conclude that the constrained maximum occurs for $y=\tau \sigma,$ which gives
the following bound on $n^{-1}\bar F_{n,t}(\sigma):$
   $$
   n^{-1}\bar F_{n,t}(\sigma)\le -h(\sigma)+(1-\sigma)h\Big(\frac{\sigma(1-\tau)}{1-\sigma}\Big)
  +\sigma h(\tau)+\varepsilon_n.
   $$
As long as the right-hand side of the this inequality is negative,
the previous proof implies that the code corrects all errors of multiplicity
up to $\sigma\tau N.$
\end{IEEEproof}

From the expression of this proposition we observe that (as $n\to\infty$)
the value of $\sigma_0$ approaches $\tau,$
so the ensemble $\sG$ contains codes that correct up to a $\tau^2$
proportion of errors, where $\tau n=d_0/2$ is the error-correcting capability
of the code $A.$ This result parallels the product bound on the error-correcting
radius of direct product codes.
As in the case of product and expander codes (e.g., \cite{bar05}), the
proportion of correctable errors can be improved from $\tau^2=(d_0/(2n))^2$
by using a more powerful decoding algorithm.

\section{Number of correctable errors for the ensemble $\sH(A,l,m)$}
\label{sect:hypergraph-decoding}
In this section we first state a sufficient condition for the existence
of at least one subprocedure within each step of Algorithm II that reduces the
number of errors, and then perform the analysis of random hypergraphs
to show that with high probability this condition is satisfied. Overall
this will show that the number of errors in at least one of the candidates
in the list generated after a few iterations is reduced to a desired
level.

Denote by $E(v)$ the set of edges incident to a vertex $v\in V.$
Let $C\in\sH(A,l,m)$ be a code and let $H(V,E)$ be its associated graph.
Let $\cE\subset E$ be the set of errors at the start of some iteration
of the algorithm. The next set of arguments will refer to this iteration.
Let $G_i=\{v\in V_i: |E(v)\cap \cE|\le t\}$ be the set of vertices such
that each of them
is incident to no more than $t$ edges from $\cE$ (such errors will be
corrected upon one decoding). Let
$B_i=\{v\in V_i: |E(v)\cap \cE|\ge d_0-t\}$ be the set of vertices that
can introduce errors after one decoding iteration. Note that each of such
vertices introduces at most $t$ errors.

The main condition for successful decoding is given in the next lemma.
\begin{lemma}\label{lemma:red} Assume that for every
$\cE\subset E, |\cE|\le \gamma N$ there exists
$i=i(\cE), 1\le i\le l$ such that $|\cE(G_i)|\ge t |B_i|+\epsilon N,$ where
$\cE(G_i)$ is the set of edges of $\cE$ incident to the vertices of $G_i$ and
$\epsilon>0.$  Then for any $0<\beta<\gamma,$
Algorithm II will reduce any $\gamma N$ errors in the received vector to
at most $\beta N$ errors in $c(\beta,\gamma,\epsilon),$ iterations where
$c$ is a constant independent of $N$.
\end{lemma}
\begin{IEEEproof} We need to prove that at least one of the subprocedures
will find a vector with no more than $\beta N$ errors after a
constant number of iterations. In any given iteration by the assumption of
 the lemma
there exists a component $V_i$ for which the $i$th subprocedure will
decrease the count of errors by $|\cE(G_i)|-tB_i\ge
\epsilon N.$ Thus, in each iteration
there exists a subprocedure that reduces the number of errors by
a positive fraction. \end{IEEEproof}

Next we show that the assumption of Lemma \ref{lemma:red} holds
with high probability over the ensemble. Consider the function
        \begin{multline*}\label{eq:barF}
  \tilde F_{n,t}(\gamma)= \max_{\bfz\in \sM(t,\gamma)}
   \Big(h(\bfz)+ \sum_{i=0}^n z_i\log\binom ni\Big),
   \end{multline*}
where in this section the region $\sM(t,\gamma)$ will be as follows:
      \begin{equation}\label{eq:mts}
    \sM(t,\gamma)=\Big\{\bfz\in \cZ_n: \sum_{i=1}^n iz_i=\gamma n,
         \sum_{i=1}^t iz_i=\sum_{i=d_0-t}^n tz_i\Big\}.
     \end{equation}
\begin{lemma}\label{prop:hg} Let $m\to\infty$ and let
    \begin{equation}\label{eq:gast}
    \gamma_0=\sup\{x>0: \forall_{0<\gamma\le x}\,
   (l/n) \tilde F_{n,t}(\gamma)<(l-1)h(\gamma)\}.
    \end{equation}
A hypergraph from the ensemble of $l$-partite uniform
$n$-regular hypergraphs with probability $1-2^{-\Omega(N)}$ has the
property that for all $\cE\subset E, |\cE|<\gamma_0 N,$ and some $\epsilon>0$,
the inequality $|\cE(G_i)|\ge t |B_i|+\epsilon N$ holds for at least one
$i\in\{1,\dots, l\}.$
\end{lemma}
\begin{IEEEproof} Let $\cE\subset E, |\cE|=\gamma N.$
Let $m_i=|\{v\in V_1: |E(v)\cap \cE|=i\}|, i=1,\dots,n.$ Clearly
$|\cE(G_1)|=\sum_{i=0}^t i m_i$ and $|B_1|=\sum_{i=d_0-t}^n m_i.$
We have
   \begin{multline*}
    p\triangleq P(|\cE(G_i)|\le t |B_i|+\epsilon N)\\=\frac 1{\binom N{\gamma N}}
      \sum_{{\boldsymbol\mu}\in M_\epsilon(t,\gamma)}
         \binom m{{\boldsymbol\mu}}\prod_{i=0}^n\binom ni^{m_i},
   \end{multline*}
where $\boldsymbol\mu=\{m_1,\dots,m_n\},$
  \begin{multline*}
   M_\epsilon(t,\gamma)=\{{\boldsymbol\mu}\in (\integers_+\cup0)^{n}:
      \sum_{i=1}^n m_i\le m, \\[-2mm]\sum_{i=1}^n im_i=\gamma N,
       \sum_{i=1}^t im_i\le\sum_{i=d_0-t}^n t m_i+\epsilon N\}.
  \end{multline*}
Denote by $\sL(\cE)$ the event that for a given subset $\cE\subset E,
|\cE|=\gamma N$ no part $V_i$ of $H$ satisfies the assumption of
Lemma \ref{lemma:red}. Then $P(\sL(\cE))=p^l$ and
  $$
    P\{\exists \cE:\; (|\cE|\le \gamma N)\wedge (\sL(\cE))\}
   \le N \binom N{\gamma N} p^l.
  $$
Letting $L$ to be the logarithm of the left-hand side of this inequality
divided by $N$ and omitting $o_N(1)$ terms, we obtain
  \begin{equation}\label{eq:L}
    L\le -(l-1)h(\gamma)+\frac ln\max_{\bfz\in \sM'(t,\gamma)} g(\bfz),
  \end{equation}
where $g(\bfz)$ is defined in (\ref{eq:g}),
  $$
   \sM'(t,\gamma)=\{\bfz\in \cZ_n: \sum_{i=1}^n iz_i=\gamma n,\,
     \sum_{i=1}^t iz_i\le \!\!\sum_{i=d_0-t}^n tz_i\}
  $$
and $z_i=m_i/m$ (as in the previous section, we have omitted $\epsilon$
which can be made arbitrarily small).

The proof will be complete if we show that the optimization region
$\sM'$ can be replaced by $\sM.$ For that we follow the logic of the second part
of the proof of Theorem \ref{thm:bg}. As before, the maximum of $g(\bfz)$ without
the constraint $\sum_{i=1}^t iz_i\le \sum_{i=d_0-t}^n tz_i$ is attained
at the point $\bfz^\ast(\gamma)=(z_0^\ast,z_1^\ast,\dots,z_n^\ast)\in \cZ_n,$  where
   $$
     z_i^\ast=z_i^\ast(\gamma)= \binom ni\gamma^i(1-\gamma)^{n-i}, \quad
   i=1,\dots, n.
   $$
We need to show that as long as $0\le\gamma <\gamma_0,$ the point
$\bfz^\ast\not\in\sM'(t,\gamma).$ By concavity of the objective function
and the optimization region, this will imply that the maximum is on the boundary.
As before, it is possible to show that in the neighborhood of $\gamma=0,$
  $$
   \sum_{i=1}^t iz_i^\ast> \sum_{i=d_0-t}^n tz_i^\ast.
  $$
and thus for $\gamma<\beta,$ where $\beta$ is the smallest positive root
 of $\sum_{i=1}^t iz_i^\ast=\sum_{i=d_0-t}^n tz_i^\ast,$ the point
$\bfz^\ast(\gamma)\not\in \sM'(t,\gamma).$ Let
  $$
  \bar\gamma=\sup\{\gamma:\;\forall\, {0<x<\gamma}, \;\text{rhs of (\ref{eq:L})}
     <0\}.
  $$
We note that for all $\gamma\le\bar\gamma,$
  $$
    \max_{\bfz\in \sM'(t,\sigma)} g(\bfz)<(l-1)nh(\gamma).
  $$
On the other hand, $g(\bfz^\ast(\beta))=n h(\beta).$ This implies that
$\bar\gamma<\beta,$ and so for all $\gamma<\bar\gamma,$ the point
$\bfz^\ast(\gamma)\not\in \sM'(t,\gamma).$ Thus the region
$\sM'$ in the maximization can be replaced with $\sM$ (and $\bar\gamma=\gamma_0$).
\end{IEEEproof}

This lemma establishes that
the number of errors in at least one of the candidates
in the list generated after a few iterations is reduced to a desired
level. After that the residual errors can be removed by another procedure
as described above. In this situation we say that the errors are correctable
by Algorithm II, without explicitly mentioning the second stage.

In the next theorem, which is the main result of this section, $\delta$
refers to the lower estimate of the average relative distance of the
hypergraph code ensemble $\sH$ from Theorem  \ref{thm:distance} below.
\begin{theorem} Let $t\ge 2$ be the number of errors correctable by the local code
$A.$
Algorithm II corrects any combination of up to $N(\min(\gamma_0,\delta/2))$
errors for any code $C\in \sH(A,l,m)$ except for a proportion of codes
that declines exponentially with the code length $N=nm, m\to\infty.$
\end{theorem}
\begin{IEEEproof} With high probability over the ensemble of hypergraphs
considered, for a given hypergraph $H(V,E)$
a constant number $s$  of iterations of the algorithm will
decrease the weight of error from $\gamma_0 N$ to any given positive
proportion $\beta$ for at least one of the $l^s$ candidates in the list
$Y^{(s+1)}_1.$ Take $\beta=\sigma_0,$ where $\sigma_0$ is the quantity
given by Theorem \ref{thm:bg}. Next consider the bipartite
graph $G(V_G=V_1\cup V_2, E_G)$ where $V_1,V_2$ are the parts of $H$
and where $(v_1,v_2)\in E_G$ if $v_1,v_2\in e$ for some edge $e\in E.$
By the previous section, with high probability these $\sigma_0 N$
errors can be corrected with $O(\log m)$ iterations of Algorithm I.
Finally, the correct codevector will be selected from the list of candidates
because the proportion of errors is assumed not to exceed $N\delta/2.$
\end{IEEEproof}

The complexity of this decoding is $O(N\log N)$ where the implicit constant
depends on the code $A.$

In the following theorem we extend the results of this section to the case
of $A$ being a perfect single-error correcting Hamming code of length
$n=2^r-1$ for some $r=3,4,\dots.$
In this case the maximum on $\bfz$ in the above proof can be computed in a
closed form. As remarked above, in this case
in the last part of the error correction procedure
we use the decoding algorithm of \cite{zya09} to remove residual errors
from the candidate vectors.
\begin{theorem} Suppose that the local codes $A$ are taken to be
one-error-correcting Hamming codes and let
$\delta=\delta(\sH)$ be the relative average distance (\ref{eq:distance})
of the ensemble $\sH(A,l,m).$
Then almost all codes in the ensemble $\sH(A,l,m)$ can be decoded
to correct $N\min(\gamma_0,\delta/2)$ errors, where $\gamma_0$ is given by (\ref{eq:gast})
and
  \begin{equation}\label{eq:tf}
   \tilde F_{n,1}(\gamma)=
-\gamma n\log x+\log\biggl(1+2\sqrt{n\sum_{i=2}^n\binom ni
    x^{i+1}}\biggr)
  \end{equation}
where $x$ is the only positive root of the equation
  $$
   \frac { \sum_{i=2}^n(i+1)\binom ni x^{i+1}}
  {2n\sum_{i=2}^n\binom ni x^{i+1}+\sqrt{n\sum_{i=2}^n\binom ni x^{i+1}}}
   =\gamma .
  $$
\end{theorem}
\begin{IEEEproof} It is obtained by maximizing the function $g(\bfz)$ over the region
  $$
    \sM(1,\gamma)=\{\bfz\in \cZ_n: \sum_{i=1}^n iz_i=\gamma n, z_1=\sum_{i=2}^n
   z_i\}.
  $$
The Lagrangian takes the form
  $$
h(\bfz)+\sum_{i=2}^n z_i\Big(\log n+\log\binom ni\Big)+\lambda
\Big(\sum_{i=2}^n(i+1)z_i-\gamma n\Big),
  $$
where $\bfz=(z_1,z_2,\dots,z_n,1-\sum_i z_i)$ and
  $
  z_1=\sum_{i=2}^n z_i
  $
and $\lambda$ is an arbitrary multiplier.
Setting the partial derivatives to zero, we find the value $\lambda$ to satisfy
$2^x=\lambda,$ where $x$ is given above.
The calculations are tedious but straightforward and
will be omitted.
\end{IEEEproof}

The last theorem enables us to find the proportion of correctable errors
for the case when $A$ is the Hamming code of length $n=2^r-1, t=1. $
Since the examples below rely on the value of the ensemble-average distance,
we quote the corresponding result from \cite{bar08b}.
\begin{theorem}\label{thm:distance} \cite[Thm.5]{bar08b}
Let $\delta(\sH)$ be the asymptotic average relative distance of codes
in the $l$-hypergraph ensemble constructed from the local code $A$ of
length $n$ and distance $d_0.$ Then
   \begin{multline*}
   \delta(\sH)\ge \sup_{\omega>0}\Big\{\omega: \frac ln
      \log\frac{1+\sum_{i=d_0}^n\binom ni x_0^i}{x_0^{\omega n}}
  <(l-1)h(\omega)
\Big\}
  \end{multline*}
where
$x_0=x_0(\omega)$ is the positive solution of the equation
   $$
    \omega n+\sum_{i=d_0}^n\binom ni(\omega n-i)x^i=0.
   $$
\end{theorem}

For instance, for the case $n=31,l=5$ this theorem gives the value of
the relative distance $\delta(\sH)\ge 0.01618$ (the rate of codes
$R\ge 6/31$).
Performing the calculation in (\ref{eq:tf}), we find that the average
code from the ensemble $\sH(A, 5,m)$ the proportion of
errors correctable by codes in the ensemble using Algorithm II
to be at least $\gamma_0=1.2\times 10^{-5}.$

We include some more examples. In the following table $n=2^9-1.$

\vspace*{1mm}{\em Example 2:}

\vspace*{2mm}{\small\begin{tabular}{|c|c|c|c|c|}
\hline $l$  & 17 & 23 & 28 & 34 \\
\hline
Rate  & $0.7006$ & $0.5949$ &
$0.5069$ & $0.4012$  \\
\hline
$\gamma_0$ &
$0.000235$ & $0.000401$  & $0.000521$ & $0.000644$  \\
\hline $\delta(\sH)$  &0.00415  &0.00504  &0.00558 &0.00608  \\
\hline
\end{tabular}\\[1mm]
\hspace*{3.5mm}\begin{tabular}{|c|c|c|c|}
\hline
$l$  & 40 & 45 & 51\\\hline
Rate  & $0.2955$ & $0.2074$ & $0.1018$\\\hline
$\gamma_0$ & $0.000747$  & $0.000821$  &  $0.000898$ \\
\hline $\delta(\sH)$  &0.00648  &0.00676  &0.00704 \\
\hline
 \end{tabular}}

\vspace*{2mm}
It is also of interest to compute the values of $\gamma_0$ for code
rate $R(C)\approx 0.5.$

\vspace*{2mm}{\small
\begin{tabular}{|c|c|c|c|c|}
\hline
$n$ & $127$ &  $255$ & $511$ & $1023$ \\
  \hline
  $l$  & $9$ & $16$ & $28$ & $51$ \\
  \hline
  Rate  & $0.5039$ & $0.4980$ & $0.5068$ & $0.5015$ \\\hline
  $\gamma_0$ & $0.0002012$ & $0.0004873$  & $0.0005207$ & $0.0004227$ \\
\hline $\delta(\sH)$  &0.01157  &0.008658  &0.005581 & 0.003394  \\\hline

\end{tabular}}

\vspace*{2mm} These estimates are at least an order of magnitude
better than the corresponding results
in \cite{bur08,zya09}
obtained for LDPC codes and their generalizations based on
the ``flipping'' algorithm of \cite{zya75}.

\vspace*{.1in}
\emph{The case of large $n$.} As in the previous section, it is interesting to examine the case of long local
codes $A$ because it reveals some parallels with the analysis of the decoding
algorithm in the case of nonrandom hypergraphs \cite{bar08b}.
We begin with the observation that the proportion $\gamma_0$ of
correctable errors for the ensemble $\sH(A,t,m)$
computed above is a function of the number of errors $t$ that each local
code corrects in each iteration.
\begin{lemma}\label{lemma:hl}
Let $t=\tau n$, $d_0=\delta_0 n.$ The ensemble $\sH(A,t,m)$ contains codes that correct $\gamma N$ errors
for any $\gamma < \gamma_0(\tau) \triangleq \min(\tau,x_0(\tau))$
where
\begin{align*}
x_0(\tau) =& \sup\{x>0:\Big(1-\frac x{\delta_0}\Big)
    h\Big(\frac{x\tau}{\delta_0-x}\Big)
        +\frac{x}{\delta_0}h(\delta_0-\tau)\\
   &+\varepsilon_n <(1-1/l)h(x)\}
\end{align*}
and $\varepsilon_n=\log n /n.$
\end{lemma}
\begin{IEEEproof}
Referring to the proof of Lemma \ref{prop:hg}, we aim at
establishing conditions for the exponent $L$ of the event $\sL(\cE)$
to be negative as $m$ approaches infinity.
We assume that $\gamma\le \tau$ (otherwise
our estimates do not imply that
the convergence condition of Lemma \ref{lemma:red} holds with high
probability over the graph ensemble).

From (\ref{eq:L}), (\ref{eq:g}) we have
    $$
    L\le -(l-1)h(\gamma)+l\max_{\bfz\in\sM(t,\gamma)}
           \sum_{i=0}^nz_ih\Big(\frac in\Big)+\frac{l\log n}{n},
    $$
where $\sM(t,\gamma)$ is defined in (\ref{eq:mts}).
Next, write
  \begin{align}
   \sum_{i=0}^t z_ih\Big(\frac in\Big)\le \lambda
 h\Big(\frac {\sum_{i=1}^tiz_i}{\lambda n}\Big) =\lambda h
\Big(\frac {\mu_1}{\lambda}\Big), \label{eq:conv}
   \end{align}
where we have denoted $\sum_{i=0}^t z_i=\lambda,$ $\sum_{i=1}^t iz_i=\mu_1 n.$
In addition let us put $\sum_{i=d_0-t}^n iz_i=\mu_2 n,$ then
the values of the sums $\sum_i z_i$ and $\sum_i iz_i$ over each of the
three intervals $I_1=[0,t],\,I_2=[t+1, d_0-t-1],\, I_3=[d_0-t,n]$ can be found
from the following table:

\vspace{.1in}\begin{tabular}{cccc}
&$I_1$&$I_2$&$I_3$\\[1mm]
$\sum z_i$& $\lambda$& $1-\lambda-{\mu_1}/{\tau}$& $\mu_1/\tau$\\[1mm]
$\sum \frac i n z_i$&$\mu_1$&$\gamma-\mu_1-\mu_2$&$\mu_2$.
\end{tabular}

\vspace{.1in}
\noindent The variables introduced above depend on the point $\bfz$
and satisfy the
following natural constraints: for any $\bfz\in\sM(t,\gamma)$,
   \begin{align}
     \mu_1&\le\tau\lambda\nonumber\\
\tau\Big(1-\lambda-\frac{\mu_1}{\tau}\Big)\le\gamma-&\mu_1-\mu_2\le
    (\delta_0-\tau)\Big(1-\lambda-\frac{\mu_1}{\tau}\Big)\nonumber\\
  (\delta_0-\tau)\frac{\mu_1}{\tau}\le&\mu_2 \le \frac{\mu_1}{\tau}.\label{eq:con}
  \end{align}
Proceeding as in (\ref{eq:conv}), we can estimate the sum on $z_i$ in
$L$ as follows:
   \begin{equation}
   \label{eq:u1}

  \sum_{i=0}^nz_ih\Big(\frac in\Big)\le
 f(\lambda,\mu_1,\mu_2)
   \end{equation}
where
   \begin{align*}
   f(\lambda,\mu_1,\mu_2)=&
 \lambda h\Big(\frac{\mu_1}{\lambda}\Big)+
\Big(1-\lambda-\frac{\mu_1}{\tau}\Big)h\Big(\frac{\gamma-\mu_1-\mu_2}
  {1-\lambda-(\mu_1/\tau)}\Big)\nonumber \\&
  +\frac{\mu_1}\tau h\Big(\frac{\mu_2\tau}{\mu_1}\Big).
    \end{align*}
Our plan is to prove that some of the inequalities in (\ref{eq:con})
can be replaced by equalities, thereby
expressing the variables $\lambda,\mu_1,\mu_2$ as functions of $\gamma,\tau.$
We will rely on the fact that the function $f$ is concave in its domain,
proved in the end of this section.

Note that for all $\bfz\in \cZ_n$ the sum
  $$
   \sum_{i=0}^nz_ih\Big(\frac in\Big)\le h(\gamma)
  $$
and that it equals $h(\gamma)$ at the point
$\tilde\bfz$ such that $z_i=1$ for $i=\lceil\gamma n\rceil$ and $z_i=0$ elsewhere.
Also note that
since $\gamma<\tau,$ the point $\tilde\bfz$ is outside the region
$\sM(t,\gamma)$ and thus, by concavity,
  $$
    a:=\max_{\bfz\in\sM(t,\gamma)}
           \sum_{i=0}^nz_ih\Big(\frac in\Big)< h(\gamma).
  $$
Let $\bfz_1$ be the point at which this maximum is attained, and let
$\bfx_1=(\lambda,\mu_1,\mu_2)$ be the corresponding point for the
arguments of $f.$ By construction, the point $\bfx_1$ satisfies the
inequalities of (\ref{eq:con}).
At the same time, consider the function $f(\cdot)$ on the line
$\lambda=\mu_1=\mu_2.$ As the variables approach 0 along this line, the
value $f(\lambda,\mu_1,\mu_2)$ approaches $h(\gamma).$

To summarize, we have found two points, $\bfx_1$ and
$\bfx_2=(0,0,0)$ that are located on different sides of the hyperplane
   $$
    \tau\Big(1-\lambda-\frac{\mu_1}{\tau}\Big)=\gamma-\mu_1-\mu_2
   $$
such that $f(\bfx_1)\ge a, f(\bfx_2)>a.$ Invoking concavity of the function
$f,$ we now conclude that there is a feasible point $\bfx'$ on this hyperplane
such that $f(\bfx')\ge a.$

Therefore, put $\mu_2=\gamma-\tau(1-\lambda)$ and write
  \begin{align*}
    f_1(\lambda,\mu_1)=&\lambda h\Big(\frac{\mu_1}{\lambda}\Big)+
  \Big(1-\lambda-\frac{\mu_1}{\tau}\Big)h(\tau)\\
   &+\frac{\mu_1}\tau h\Big(\frac{\tau(\gamma-\tau(1-\lambda))}{\mu_1}\Big)
  \end{align*}
where the variables are constrained as follows: for any $\bfz\in\sM(t,\gamma),$
  \begin{align}
     \mu_1&\le\tau\lambda\nonumber\\
    \tau(1-\lambda)&-\mu_1\ge 0\label{eq:e1}\\
  (\delta_0-\tau)\frac{\mu_1}{\tau}\le\gamma-&\tau(1-\lambda)
                 \le \frac{\mu_1}{\tau}.         \label{eq:e2}
  \end{align}
Since $f_1$ is a restriction of $f$ to a hyperplane, it is still
concave.
Now notice that $f_1(1,\tau)=h(\gamma)$ and that the point $(1,\tau)$
does not satisfy inequality (\ref{eq:e1}) and the left of the inequalities
(\ref{eq:e2}). Repeating the above argument, we claim that
the function $f$ in (\ref{eq:u1}) can be further restricted to the intersection of the
planes $\tau(1-\lambda)=\mu_1$ and $(\delta_0-\tau)(\mu_1/\tau)=
\gamma-\tau(1-\lambda).$ Altogether this gives:
  $$
  \lambda=1-\gamma/\delta_0, \;\;\mu_1=\gamma\tau/\delta_0.
  $$
Let us substitute these values into the expression for $f_1$ and
rewrite (\ref{eq:u1}) as follows: for any $0\le\gamma<\tau,$
  \begin{equation}\label{eq:2}
    \max_{\bfz\in\sM(t,\gamma)}\sum_{i=0}^nz_ih\Big(\frac in\Big)\le
     \Big(1-\frac{\gamma}{\delta_0}\Big)h\Big(\frac{\gamma\tau}{\delta_0-\gamma}
    \Big)+\frac\gamma{\delta_0}h(\delta_0-\tau).
  \end{equation}
Thus if the condition in the statement is fulfilled then $L<0$.
This concludes the proof.
\end{IEEEproof}

\vspace*{.1in}
{\em Remark.} The main part of the proof is estimating the solution
of the following linear program
  \begin{align*}
   \max_{\bfz} &\sum_{i=1}^n z_i h\Big(\frac in\Big)\\
   \bfz=(z_0,z_1,&\dots,z_n)\in \sM(t,\gamma)
  \end{align*}
where the variables define a probability distribution
on $\{0,1,\dots,n\}.$
It is clear from concavity that the maximum is attained at the point
where among all the indices $i\in I_1$ at most one value $z_i$ is nonzero,
and the same applies to $I_2$ and $I_3.$ We have shown that the value of the
program is bounded above by the right-hand side of (\ref{eq:2}). The following
point gives this value and is therefore a maximizing point:
  $$
   z_{i_1}=1-\frac\gamma{\delta_0},\; z_{i_2}=\frac\gamma{\delta_0}, \;
  z_i=0 \text{ otherwise},
  $$
where $i_1=n\gamma\tau/(\delta_0-\gamma), i_2=n(\delta_0-\tau).$
Since
  $$
   \frac{\gamma\tau}{\delta_0-\gamma}\le \tau,
  $$
this shows that the worst-case allocation of errors to vertices in a given
part of the graph assigns no edges to vertices that are neither good
nor bad. This also confirms the intuition suggested by Lemma \ref{lemma:red}
that bad vertices (vertices assumed to add errors) should each be assigned
the smallest possible number of error edges $d_0-t.$

The next proposition is now immediate.
\begin{proposition}\label{prop:th}
The ensemble $\sH(A,l,m)$ with long local codes
contains codes that can be decoded using
Algorithm II to correct all error patterns whose weight is less than
$\gamma_0 N,$ where
  \begin{equation}\label{eq:gt}
   \gamma_0=\max_{0<\tau\le\delta_0/2} \gamma_0(\tau).
  \end{equation}
\end{proposition}

Estimating the number of correctable errors for the ensemble $\sH(A,l,m)$
from Proposition \ref{prop:th}
analytically is difficult because it involves
optimization on $\tau$ (generally, the local codes
should be used to correct a smaller than $\delta_0/2$ proportion of errors).
We note that in the particular case of $\tau=\delta_0/2$ the proof of
Lemma \ref{lemma:hl} can be considerably simplified, although the resulting
value of $\gamma$ is not always optimal.

{\em Example 3.} Let  $l=3.$
Using local codes with $\delta_0=0.05$ we can construct hypergraph codes
of rate $R\ge 0.19$. From \cite[Cor.~6]{bar08b}, the ensemble-average
relative distance is at least $\delta\approx0.0112$ and the proportion of errors
correctable by Algorithm II is found from (\ref{eq:gt}) to be
$\gamma_0\approx 0.0035.$

{\em Example 4.} Let $\delta_0=0.01$ and $l=10.$
In this case, we find from \cite[Cor.~6]{bar08b} the value of the
relative distance
$\delta\approx 0.00599.$
The code rate satisfies $R\ge 0.14.$
Performing the computations in (\ref{eq:gt}) and Lemma \ref{lemma:hl}
we find the estimate of the proportion of correctable errors to be
$\gamma_0\approx 0.002198$.

\emph{Proof that $f(\lambda,\mu_1,\mu_2)$ is concave.} First we prove
that the function
  $$
    \phi(x,y)=(1-x)h\Big(\frac{\gamma-y}{1-x}\Big)
  $$
is concave (not necessarily in the strict sense) for $0<x,y<1,0<\gamma-y<1-x.$
For that, let us compute its Hessian matrix:
  $$
   H=\frac1{\ln 2}\begin{pmatrix}\frac{\gamma-y}{(1-x)(\gamma-y+x-1)}&
- \frac1{\gamma-y+x-1}\\[2mm]
      - \frac1{\gamma-y+x-1}&\frac{1-x}{(\gamma-y)(\gamma-y+x-1)}
   \end{pmatrix}
  $$
The eigenvalues of $H$ are
  $$0,\;\;\frac{(\gamma-y)^2+(1-x)^2}{(1-x)(\gamma-y)(\gamma-y-(1-x))}<0,
  $$
so
$H\preceq 0,$ and so $\phi$ is concave.
Next observe that the function
  $$
    \Big(1-\lambda-\frac{\mu_1}{\tau}\Big)h\Big(\frac{\gamma-\mu_1-\mu_2}
  {1-\lambda-(\mu_1/\tau)}\Big)
  $$
can be obtained from $\phi$ by a linear change of variables
  $$x=\lambda+\mu_1/\tau, \;\;y=\mu_1+\mu_2$$
and therefore is also concave.
Finally, the functions $\lambda h(\mu_1/\lambda)$ and
$(\mu_1/\tau)h(\mu_2\tau/\mu_1)$ are also concave, and thus so
is the function $f(\lambda,\mu_1,\mu_2).$

\section{Conclusion}
We have estimated the proportion of errors correctable by codes from
ensembles defined by random $l$-partite graphs, $l\ge 2.$
In contrast to the case of expander codes \cite{sip96},
\cite{zem01}, \cite{bar05},  \cite{bil04}, \cite{bar08b}
our calculations cover the case of local codes of arbitrary given length
and distance, including small values of the distance.
The behavior of code ensembles considered here was examined from a
different perspective in \cite{bar08b} where we computed estimates
of the expected distance and weight distribution of these codes.
The paper \cite{bar08b} and the present work together
provide answers to the set of basic questions regarding random networks of
short linear binary codes and extend our perspective of concatenated
code constructions to the case of sparse regular graphs.

\vspace*{2mm}{\em Acknowledgment.} The authors are grateful to J{\o}rn Justesen
and to an anonymous reviewer for useful comments on this work.

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\href}[2]{#2}

redpajama_set_name: RedPajamaArXiv


text:
>> Other Broadband Technologies
Pages in this thread: 1 | 2 | 3 | 4 | 5 | (show all) Print Thread
jossmer
(newbie) Wed 25-Feb-09 12:15:37
World's largest single EU funded municipal project
From the digital region South Yorkshire web site
http://www.digitalregion.co.uk/news.html
World's largest single EU funded municipal project on track to bring high bandwidth broadband access to all in South Yorkshire (Doncaster, Rotherham,Sheffield,Barnsley)
Digital Region will enable citizens, business and public sector organisations within South Yorkshire to benefit from services delivered over a world-class broadband infrastructure that will provide a level of access unprecedented in the UK over such a large geographical area.
The project is set to pick up pace as this landmark initiative seeks full ratification from all partners, including Europe, UK government, Yorkshire Forward and the four local authorities in South Yorkshire during October and November 2008.
(has now been ratified )
Set to provide South Yorkshire with an unrivalled level of high bandwidth broadband access for all households and businesses in the region, Digital Region will represent the EU�s biggest municipal project in the world with financial backing of Yorkshire Forward and the four local authorities in South Yorkshire. The contract with Thales Transport and Security Limited, who will deliver and manage the project infrastructure, will be ready for signature in October 2008.
( Has now been signed )
The first customers are likely to access the new broadband network in each of the four authorities in Autumn 2009, with the full infrastructure being completed by the end of 2010. Digital Region will put South Yorkshire at the forefront of access to digital technology with a service that will outperform other the majority of UK and rest of Europe.
Read more. http://www.digitalregion.co.uk/index.htm...
This is well under way and will be fibre to cabinet.
So if you live in South Yorkshire its all good stuff
Zen Internet Active pro 8000
[IMG]http://www.speedtest.net/result/41013470...[/IMG]
(staff) Wed 25-Feb-09 13:02:37
Re: World's largest single EU funded municipal project
[re: jossmer] [link to this post]
Still many questions, i.e. is it really going to provider fibre to the cabinet, as in a BT street cabinet. If so what is the impact with regards to BT's own plans, and which broadband providers will the consumer have access to.
If it ends up as a single provider, then people may not find it appealing, i.e. home/hobby customers after a static IP address that costs say £15 a month extra.
The project has lots of potential but the lack of info is of concern before I can say things like Yorkshire is the place to be. Places like Sheffield already have Metro Ethernet options in some new build flats
http://www.thinkbroadband.com - formerly known as ADSLguide.org.uk
Ignitionnet
(committed) Wed 25-Feb-09 13:19:46
Re: World's largest single EU funded municipal pro
Looks interesting, hope that the infrastructure is delivered better than the text on the website though. It looks like it was written in a bit of a hurry...
Is an odd phrase, you shouldn't really be discussing 'world's largest' and 'EU funded' as the EU is hardly likely to fund anything outside the EU and there are far larger non-EU funded municipal projects. Maybe drop the 'world's largest' stuff and replace with 'Single largest EU funded municipal project' easier to read.
with a service that will outperform other the majority of UK and rest of Europe.
You are mistaken thinking FTTC outperforms the rest of Europe, there are projects running FTTP at 1Gbit and there is 100Mbit+ FTTP live too.
Other than that very, very good luck with this. Competition is all good and neither BT nor Virgin Media have enough of it which is why they are so content to sit on their laurel cushioned bottoms.
Edited by Ignitionnet (Wed 25-Feb-09 13:20:39)
The very same questions you raise I have already emailed to them Andrew. Still waiting on a reply...Thales the main consortium have just built a 20 million pound hi tech centre in Doncaster as the central link.
It remains to be seen what the public consumer will actually get out of this. I have in the past sat on a few committees so will use those links to press for more detail as to what this means for the residential customer.
If and when I get more info I will post.
12eason
(committed) Thu 26-Feb-09 20:44:32
Wow, what a waste of EU funds. BT should have been forced to do this.
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s0nic
(learned) Fri 27-Feb-09 12:50:29
[re: 12eason] [link to this post]
Living in rural South Yorkshire I'm obviously interested in this, but I wish they would update their site more often and let us know what the latest is.
This is probably my only chance of a fast connection.
bookey
(committed) Tue 03-Mar-09 09:25:30
Re: Digital Region using Sub-LLU
[re: s0nic] [link to this post]
While economically challenging, sub-loop unbundling in its current form may
still have a role to play
6.17 Despite their limitations, we believe the current sub-loop unbundling products may
play a potentially important role, particularly in areas where BT chooses not to
upgrade its network and no other next generation access networks exist. In these
areas, sub-loop unbundling remains the only option for other communications
providers wishing to offer super-fast broadband services without having to build a
completely new network from scratch. While today, the costs are high and the
revenues uncertain, this may not always be the case. The cost side of the equation
may be improved for communications providers who have access to public funding,
such as the Digital Region project planning to use sub-loop unbundling in South
Yorkshire. Moreover, future cost reductions and a developing market for super-fast
broadband services with new revenue streams may further improve the economic
case for entry using these products.
This is taken from the most recent ofcom statement on BT rolling out FTTx
Edited by bookey (Tue 03-Mar-09 09:46:59)
(experienced) Thu 23-Apr-09 22:25:00
Rereading this there's something that concerns me:
http://www.digitalregion.co.uk/summary.html
The DRL Network is offering 25mb+ of bandwidth which is both guaranteed and can be delivered with a range definable Quality of Service attributes. In simple terms, this means the end user gets �what they pay for�; if the end user requires 10mb they receive a guaranteed 10mb; if they want 5mb or 25mb this is guaranteed. Some services require more than just high-speed connections: this is true of live entertainment such as sports, and critical services such as aspects of tele-medicine. The DRL Network can configure specific connections with Quality of Service attributes that allow these services to be delivered in a seamless way to the end consumer - so no stop-starting or picture-jitter for example when watching a live football match.
Do these guys have any idea how expensive it is to guarantee bandwidth?
How come this is costing 100mill+ to do 546,000 homes? That's a very heavy cost for FTTN, it's well within boundaries of FTTO P2P costing considering the demographics!
(staff) Fri 24-Apr-09 07:50:50
[re: Ignitionnet] [link to this post]
Guaranteed over the digital region fibre network, or it will be priced such that a consumer wanting 25Mbps is paying >£50 a month.
Or these are just QoS options, and consumers will get the usual 25Meg, with 50:1 or more contention.
NOTE: It is a wholesale network
www.thinkbroadband.com - formerly known as ADSLguide.org.uk
(experienced) Fri 24-Apr-09 10:53:08
In reply to a post by MrSaffron:
Yes I realised the stuff about a wholesale network, that makes things much less painful on them and brings costs down much more, depending on the wholesale bitstream pricing.
Well then, go South Yorkshire! While I would have preferred to see FTTO/P as I'm weird like that it's definitely a firm step in the right direction and hopefully will provide a model for other municipal stimulated networks in the UK.

redpajama_set_name: RedPajamaCommonCrawl


I am Williams class of '06, and an occasional reader of Ephblog. As someone who just graduated with a PhD in Finance this June, I just wanted to say thanks for standing up to academic finance as a liberal art on an ephblog discussion today.
Over time, I had many similar discussions, and I have found that persuading people to change their prior opinions on this topic is, well, challenging. I hope that Williams will beef up its finance faculty in the future (e.g., prof Bill Gentry at the econ department at Williams was a big influence in my decision to go to finance phd program- he used to be a faculty at Columbia BS prior to coming to Williams).
I need it. If Williams has a major in computer science, then it ought to have a major in finance. Perhaps Professor Gentry would be willing to head up a committee to look into the topic? Note that Adam Falk played a role in creating a similar program at Hopkins, although I believe that his predecessor as dean deserves most of the credit.
By the way, to all the liberal arts traditionalists who complained the last time around: Do you have an objection to a computer science major and department?
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One thing these maps point out is the urban/rural divide. It is difficult to believe that land mass and population are so out of whack until you examine these maps. The land mass Trump drew from involves just about the entire interior of America. But the popular vote gave the Democrats 3 million more votes than Trump got. If we voted by land mass the Republicans would win every time. The fact that we vote by individuals and that the Democrats got so many more votes shows how much of our population is concentrated along our coasts and waterways.
This also might explain the more "cosmopolitan" outlook of people in our largest cities. They have more exposure to travelers and traders from other nations. They are centers of cultural diversity. It would be almost impossible for these coastal folks to be exactly like those who inhabit America's interior – its more land-locked places. Conservatives have nothing good to say about coastal Americans. They overvalue intellect. They treat "aberrant" behavior as normal. They welcome immigrants and they are bleeding hearts for immigrants and the poor. Conservatives would be quite happy if there were no liberals and they are working overtime to make it so.
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