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Tag Archives: Vivienne Westwood
Runaway Brides: Tartan Wedding Dresses and Scottish Rebellion
June 15, 2021 Posted by:
lucycorkish
Commentary Dress in Focus From the Collections Work in Progress No Comments
Tartan is a fabric of rebellion, and it has long held appeal with those who consider themselves to be outsiders.
In 1745, the Scottish House of Stuart led the Jacobite Army in an unsuccessful attempt to seize the British throne from King George II. Following the uprising, a series of laws were enacted to subdue the fiercely independent Jacobites. The Dress Act of 1746, one of these laws, made tartan dress illegal in the United Kingdom. Anyone who wore tartan or other signifiers of traditional Scottish dress could face fines, imprisonment or exile.
However, The Dress Act of 1746 seemed only to strengthen the power of tartan. People across the United Kingdom began wearing full tartan outfits in defiance of the British government. Artists painted influential figures dressed in tartan but left their paintings unsigned, fearing that they would be punished for these public displays of dissent. Tartan became a signifier of anti-establishment attitudes, a very punk choice in the 18th century.
In 1974, Vivienne Westwood and Malcolm McLaren opened their boutique SEX to cater to the burgeoning Punk rock scene in London. Westwood stocked the shop with every type of plaid imaginable, and soon the Sex Pistols were singing out against the British government in full tartan suits, just like the Jacobites nearly two hundred years before them.
Vivienne Westwood, Wedding Ensemble, Anglomania, Autumn/Winter 1993, National Gallery of Australia, NGA 94.278.1-4.A-B.
This 1993 wedding dress by Vivienne Westwood is a vision of tartan excess, crafted from yards of plaid silk that cascade over a bloom of tulle in coordinating colours. Though the concept of a tartan wedding dress is unorthodox, it is not purely a whim of Westwood's wild imagination and is rooted deeply in the history of Scottish fashion. The Isabella MacTavish Fraser Wedding Gown provides an example – a rare one, albeit – of a tartan wedding gown dating back to 1785. Though many details about the spectacular Isabella MacTavish Fraser Wedding Gown remain shrouded in mystery, the tartan of the dress can be identified as one woven between 1775 and 1784. This means that it was likely created while The Dress Act of 1746 was still in place, making this wedding dress an illegal creation.
Isabella MacTavish Fraser Wedding Gown, 1785, Inverness Museum and Art Gallery.
The headpiece that accompanies Westwood's dress makes reference to a traditional piece of women's Highland dress known as the earasaid. The earasaid is a length of pleated fabric that would be wrapped around the head like a veil and affixed at the waist. Although there is little information available on women's dress in the Highlands before the turn of the 18th century, some historical evidence suggests that the wearing of the earasaid could date back to Pictish times. By the 1800s, written accounts and sketches of working-class women in earasaids were circulated across the United Kingdom, solidifying the trend as a hallmark of Scottish brides, even though the accuracy of these accounts remains contestable. Westwood recalls the earasaid with her veil's gentle pleats and billowing volume but elevates its humble origins by pairing it with a regal gown of matching plaid.
A Victorian interpretation of how the earasaid may have looked. Robert Ronald McIan, plate from The Clans of the Scottish Highlands, 1845.
In a thoroughly contemporary interpretation of bridal traditions, Kate Moss first wore this gown down the runway at Westwood's Autumn/Winter 1993 show with the bodice lowered to reveal one of her breasts. Preserving her modesty was a handful of flowers that once again harken back to tartan's origins. Moss's bouquet is studded with white roses, the symbol of the Jacobite army. Women participants in the Jacobite rebellion would often have their portraits painted with white roses tucked into the bodices of their dresses and their hair to signify their allegiance to the cause.
Vivienne Westwood, Wedding Ensemble, Anglomania, Autumn/Winter 1993. Runway photograph courtesy of Vogue Runway.
The bride who selects this dress to wear on her wedding day, likely one of the most publicly visible events of her life, chooses consciously not to perform the societal role expected of her. This wedding gown eschews the notion of brides dressing in virginal white, and it recalls a raucous national identity far more than a standard wifely one. With its earasaid and Jacobite references, this dress pays homage to the oft-overlooked women participants in the radical movements of Scottish history.
Westwood is one of the foundational contributors to tartan's punk reputation, but she has also worked to ensure the medium's longevity beyond the punk rock movement and, almost certainly conscious of her status as an Englishwoman, to emphasise its unique Scottish heritage. This dress is cut from MacAndreas tartan, a sett of tartan created by Westwood as a romantic tribute to her husband and collaborator Andreas Kronthaler. MacAndreas tartan is now officially listed in the Scottish Register of Tartans. Westwood has also created many of her tartan garments in accordance with the Harris Tweed Act, a 1993 Act of Parliament which seeks to protect and promote the traditional methods of woollen fabric weaving in Scotland.
From the Dress act of 1746 to the Harris Tweed Act of 1993, the lawmaking that surrounds tartan begs the question: why is tartan something institutions feel the need to control? Is it a dangerously influential pattern that incites revolt across centuries? Is it a precious national resource that must be protected at all costs? For the bride who dons a tartan wedding dress, one thing is certain. Tartan is a testament to fierce individuality and national history, suitable to dress herself in for one of the most sacred days of her life.
By Ruby Redstone
Faiers, Jonathan. Tartan. London: Bloomsbury, 2008
Inverness Museum and Art Gallery online. 'The Isabella Project'. Published February 2020.https://www.highlifehighland.com/inverness-museum-and-art-gallery/the-isabella-project/
MacDonald, Peter Eslea. 'Musing on the Arisaid and Other Female Dress.' Scottish Tartans online. Published 2016.https://www.scottishtartans.co.uk/Musings_on_the_arisaid_and_other_female_dress.pdf
Maspero, Ida. 'Tartan Romance,' National Museums Scotland online. Published 26 May 2019. https://blog.nms.ac.uk/2019/05/26/tartan-romance/
Scottish Tartans Authority. 'Tartan and the Dress Act of 1746.' Accessed 28 February 2021. http://www.tartansauthority.com/resources/archives/the-archives/scobie/tartan-and-the-dress-act-of-1746/
The Scottish Register of Tartans. 'Westwood MacAndreas.' Published 1 January 1993. https://www.tartanregister.gov.uk/tartanDetails?ref=5530
V&A online. 'Vivienne Westwood: A taste for the past'. Last accessed 15 February 2021. https://www.vam.ac.uk/articles/vivienne-westwood-a-taste-for-the-past
Watt, Patrick and Rosie Waine. Wild and Majestic: Romantic Visions of Scotland. Edinburgh: National Museum of Scotland, 2019
A Personal Collection of Vivienne Westwood Shoes Exhibition Review
November 9, 2018 Posted by:
A Personal Collection of Vivienne Westwood Shoes juxtaposes Westwood's outlandish designs with York Castle Museum's diverse shoe collection, ranging from the 18th century to the present day. An intriguing aspect is that the Westwood shoes come from a private collector, amassed over 30 years, while Westwood selected shoes from the museum's collection. The exhibition focuses on the craftsmanship involved in making a shoe, as well as allowing the viewer to revel in the style and beauty of footwear.
The exhibition display.
The layout of the main room is particularly eye-catching, with a catwalk display of Westwood's designs interspersed with the museum's collection. This means that the viewer can see the wide range of influences in Westwood's work, as she combines historical knowledge of the shoe with a love of exaggeration. This is exemplified by the Super Elevated Gillie Heel, from the Anglomania collection, 1993, which appears to take a inspiration from ribbon-lace style shoes from the 18th and 19th centuries. Westwood renders her innovation with vibrant colours and the heel extended to lofty (and dangerous) heights. The model, Naomi Campbell, famously wore a version of these platform shoes as she took a tumble on the runway due to the 30.5 cm heels.
Super Elevated Gillie Heel.
Some favourites from the Castle Museum's collection, which illustrate the range of style and dates on show, are a pair of ivory white satin shoes with a Louis heel from around 1730, and a pair of men's Derby boots with a stacked leather heel, from the early 20th century. These shoes are both beautiful and elegant, illustrating the high level of craftsmanship, equally as striking as they are unique. They also demonstrate the evolution of footwear fashion for both genders throughout history.
A pair of ivory white satin shoes with an upturned toe and a Louis heel, circa 1730.
A bespoke pair of men's derby boots with a stacked heel. Made by A.E Marlow for the Saxonia in Northampton. Early 20th century.
The influence of fashion on the development of footwear is further explored through the final room in the exhibition, with historical information about how shoes were made, and a display showing how shoes changed through different decades. By educating the viewer on the history of the shoe, the exhibition shows how Vivienne Westwood bases her designs on tradition and then subverts them through material or features to create something new. I also felt that this exhibition communicates how the function of the shoe was originally pure practicality, and how it has developed through time into an extension of our identity.
For me, shoes are the language of personality. Every shoe represents a different mood, translated through the height, material and colour. Different to clothes, which have to change as we fluctuate shape, shoes are a longer-term investment, and fit no matter what. This exhibition interprets the shoe as more than merely an accessory; instead the viewer is presented with the shoe as the main focus. As an avid lover of shoes, and faced with the continual struggles of storage and display, it is gratifying to see an exhibition where the shoe is given pride of place.
My shoes on display in my York room.
A Personal Collection of Vivienne Westwood Shoes is at the York Castle Museum until 28 April 2019.
Ellen Bhamra
Images 1-4 courtesy of York Castle Museum.
Documenting Fashion Visits NYC, Dec 2016: Masterworks: Unpacking Fashion at the Metropolitan Museum of Art's Anna Wintour Costume Center
Commentary Fashion Now No Comments
The current exhibition on show at the Anna Wintour Costume Center at the MET has been given the title Masterworks: Unpacking Fashion. Rather than exploring a theme, as Judith Clark has so brilliantly done with Vulgar at the Barbican Centre, the MET's assistant curator Jessica Regan presents viewers with a mix of fashion with no unifying theme or trend or feature other that the 50 pieces were acquired over the last decade, and that each may be termed a 'masterwork'.
HOUSE OF WORTH (French, 1858-1956) Jean-Philippe Worth (French, 1856-1926) BALL GOWN, 1898, haute couture | Light blue silk satin brocaded à la disposition with yellow and ivory silk; embroidered with silver sequins, clear rhinestones, and clear and silver seed beads; trimmed with ivory cotton lace, black silk velvet, and light blue silk mousseline | Brooklyn, Museum Costume Collection at The Metropolitan Museum of Art | Gift of the Brooklyn Museum, 2009; Gift of Mrs. Paul Pennoyer, 1965 (2009.306. 1324a, b)
A decade since the MET's last acquisitions show, blog.mode: addressing fashion, in 2007, Masterworks marks a shift in the collecting strategy of the museum when it comes to fashion. The phrase 'Unpacking Fashion' speaks to this. The set of the exhibition is formed from crates, suggestive of the archives in which the garments are stored to best preserve them, from which they are then unpacked for display. But the term also refers to the academic practice of unpacking an idea, a point, a proposition in order to understand its significance. Why is a sculptural, slashed tulle gown by Viktor and Rolf worthy of being exhibited in a museum? What makes it seminal, important, a masterwork?
The dress in question is not part of everyday dress trends seen on women walking down the street; it was not mass produced, indeed it was worn by only a handful of people. It may not warrant a significant space in an encyclopaedia of Western fashion, but the challenging design and painstaking skill of its construction make it worthy of celebration. No one else, quite simply, has made anything like it.
VIKTOR & ROLF (Dutch, founded 1993) Viktor Hosting (Dutch, born 1969) Rolf Snoren (Dutch, born 1969) BALL GOWN, spring/summer 2010 | Blue polyester tulle, and black silk-synthetic moiré embroidered with white plastic sequins | Purchase, Friends of the Costume Institute Gifts, 2011 (2011.8)
The main Lizzie and Jonathan Tisch Gallery is organised chronologically, with each garment accompanied by an in-depth explanation, or rationalisation, of its presence in the exhibition. Designers represented range from Paul Poiret to Yves Saint Laurent and Vivienne Westwood. The design advances of new names – Demna Gvasalia at Balenciaga – and less widely known – Noritaka Tatehana, maker of the extraordinary heel-less shoes favoured by Daphne Guinness – are acknowledged.
Left: MADELEINE VIONNET (French, 1876-1975) EVENING DRESS, 1929, haute couture | Dress: pink silk tulle, embroidered with pink rayon floss; slip: pink silk gauze and crepe de chine | Purchase, Friends of the Costume Institute Gifts, 2009 (2009.248a, b) Right: JOHN GALLIANO (British, born Gibraltar, 1960) EVENING DRESS, spring/summer 1999 | Peach nylon lace | Gift of John Galliano, 2000 (2000.168)
HOUSE OF BALENCIAGA (French, founded 1937) Cristóbal Balenciaga (Spanish, 1895-1972) DRESS, 1967, haute couture | Green silk gazar | Gift of Judith Straeten, 2015 (2015.711)
The Carl and Iris Barrel Apfel Gallery at the end of the exhibition features ensembles donated by designers on the occasion of the retirement of long-serving curator Harold Koda in January this year. These represent specific masterworks long and especially admired by Koda and include a design, re-made for the occasion, from Karl Lagerfeld's first collection for Chanel. This stands alongside an intricately embroidered frock coat by Raf Simons for Dior, across from a screen featuring tributes from the great and good of today's industry.
Karl Lagerfeld (French, born Hamburg, 1938) DRESS, 2015; original design: spring/summer 1983, haute couture | Black silk crepe embroidered with pearls, clear rhinestones, and red, green, gold and orange beads | Gift of CHANEL, in honour of Harold Koda, 2016 (2016.632)
HOUSE OF CHANEL (French, founded 1913) Karl Lagerfeld (French, born Hamburg, 1938) SUIT, spring/summer 2011 Jacket and skirt: navy and cream cotton-acrylic nylon-polyester tweed with ivory nylon net; blouse: ivory rayon twill | Gift of CHANEL, 2013 (2013.157.1a-e, h)
Left: HOUSE OF DIOR (French, founded 1947) Raf Simons (Belgian, born 1968) ENSEMBLE, autumn/winter 2014-15, haute couture | Coat and waistcoat: black silk faille embroidered with polychrome silk and metal thread, silver plastic sequins, and clear and iridescent glass beads; blouse: black brushed-wool twill; trousers: black wool flannel | Gift of Christian Dior Couture in honour of Harold Koda, 2016 (2016.256a-d) Right: HOUSE OF DIOR (French, founded 1947) Christian Dior (French,1905-1957) "MYSTÈRE" COAT, autumn/winter 1947-48 | Black wool melton and dark green silk taffeta | Gift of Irene Stone, in memory of her daughter Mrs. Ethel S. Greene, 1959 (C.I.59.26.2)
The lingering question posited by Masterworks is that age old debate: is fashion art? It is clear what the MET believes. The first work you see as you come down the stairs into the exhibition is an expertly crafted Viktor and Rolf dress which resembles a painting smashed over the head of a mannequin – an attempt, surely, to reinforce the point that each garment should be viewed with the same attitude as that afforded by a Van Gogh upstairs. Andrew Bolton, curator-in-charge, has commented on the Costume Institute's renewed mission 'to present fashion as a living art that interprets history, becomes part of the historical process, and inspires subsequent art.' It is a vow restated by this thoughtful exhibition, with extraordinary skill and innovation displayed and emphasised by curious pairings and dramatic exchanges – no more so than in the vivid red of a John Galliano for Martin Margiela coat in conversation with its 18th century inspiration.
Left: MAISON MARGIELA (French, founded 1988) John Galliano (British, born Gibraltar, 1960) ENSEMBLE, spring/summer 2015, artisanal | Coat: red brushed wool needle-felted with red silk chiffon trimmed with red rayon velvet; bra top: black nylon net embroidered with gold metallic thread and gold plastic sequins; briefs: black polyester rib-knit embroidered with gold metallic thread, gold plastic sequins, and gold glass bugle beads; shorts: black cotton denim | Purchase, Friends of the Costume Insititute Gifts, 2015 (2015.541a-f) Right: French COAT, 1787-92 Red wool broadcloth Purchase | Irene Lewisohn Trust Gift, 1992 (1992.65)
ISSEY MIYAKE (Japanese, born 1938) BUSTIER, autumn/winter 1980-81 | Red moulded polyester resin and cellulose nitrate | Purchase, Friends of the Costume Institute Gifts, 2015 (2015.61)
Masterworks: Unpacking Fashion is at the Metropolitan Museum of Art until 5 February, 2017.


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Applying the physicist rule of thumb suggests that the supernova is brighter. And indeed, it is … by nine orders of magnitude.
That rocked me back on my heels. And it got me thinking: how far away would one have to be for a supernova to be only as bright as an h-bomb pressed against one's eyeball?
So if you're close enough to a supernova that the light takes six months to reach you, it will still be like being nuked at point-blank range.
How far away from a supernova do you need to be to be safe? According to this article, even at a distance of 3000 light years, a supernova could still wreck the ozone layer of an Earth-like world.
Gamma rays from a supernova would induce a chemical reaction in the upper atmosphere converting molecular nitrogen into nitrogen oxides…. In 2009, elevated levels of nitrate ions were found in Antarctic ice, which coincided with the 1006 and 1054 supernovae.
Amazing. The 1054 supernova is near and dear to my heart. Its visible remnant, the Crab Nebula, is also catalogued as Messier 1. I have observed it dozens of times, most notably during my nearly-annual Messier Marathons. I had no idea that it had literally left its mark on Earth.
So, here's something to be thankful for this Thanksgiving: there are no particularly good supernova candidates close enough to Earth to pose a serious threat. All of the contenders are not massive enough yet (if they're white dwarfs) or too far away, or won't blow for millennia, or some combination of the above. So you can tuck in with abandon. We could still be annihilated at any moment by death from space–just ask the folks in Chelyabinsk–but it probably won't come in form of a supernova.
You are currently browsing the 10 Minute Astronomy blog archives for the day Thursday, November 28th, 2013.


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BCU Legend Jack "Cy" McClairen Passes Away at 89
"The world has lost a Hall of Famer."
By HBCU Gameday
Courtesy of Dan Ryan/Senior Writer-Historian BCU Athletics
DAYTONA BEACH, Fla. – Jack "Cy" McClairen, the living embodiment of all things Bethune-Cookman University Athletics, whose contribution to his beloved alma mater included seven decades as a playing, coaching and senior administrative living legend as well as a father figure, passed away on Monday, December 28, 2020. He was 89.
Former Pittsburg Steeler All Pro, Jack "Cy" McClairen has passed. Drafted in 1953, he served in the military for 2 years and played from 1955-1960. He was a graduate of Bethune-Cookman. pic.twitter.com/S1gthCJ9DO
— Kelvin G. Mayner (@g_mayner) December 28, 2020
"The world has lost a Hall of Famer, " said Lynn W. Thompson, Vice President for Intercollegiate Athletics. "I am merely one of the thousands of people whose lives have been significantly impacted directly by this man as a coach, father figure, leader and friend. The Wildcat Nation and beyond pauses in prayer to simply say 'Thank You God for Jack 'Cy' McClairen'."
Known and loved simply as "Coach Cy," McClairen was the cornerstone of the Wildcat athletic program, first as an athlete. The Panama City product earned 12 varsity letters from Bethune-Cookman College during his playing days, picking up letters in football, basketball and track and field.
In 1952, he caught the famed game-winning touchdown pass in the Wildcats' homecoming upset of Florida A&M. Later that year, he and future Basketball Hall of Famer John Chaney would lead the basketball team to a Southern Intercollegiate Athletic Conference championship and a berth in the National Basketball Invitational Tournament, an NAIA regional that served as the defacto Black College National Championship. The Wildcats advanced to the championship game, falling to Tennessee State.
As a student, McClairen's contributions extended beyond the playing field. As one of the few scholars on campus with a driver's license, he would often serve as a chauffeur for College founder Dr. Mary McLeod Bethune.
"His DNA is woven through the history of this great quilt we call Bethune-Cookman," said Thompson.
After graduation in 1953, "Cy" was drafted twice, first by the Pittsburgh Steelers and then by the United States Army for a two-year tour of duty, where his football skills were utilized during his stint at Fort Sill (Oklahoma) for one of the nation's top service teams. McClairen then began a successful six year NFL career with the Pittsburgh Steelers, who in 1955 kept him and cut a quarterback named Johnny Unitas.
Happy BDAY tomorrow to former #Steelers 50's WR Jack "Cy" McClairen
"When I got there no one knew who I was. They didn't know anything about me. The coaches didn't know who I was."
@SteelersUnite #HereWeGo pic.twitter.com/6jNjhF2j9Z
— Steelers Takeaways 🌗 (@PittsburghSport) March 1, 2020
In 1957, he finished third in the league in receptions – finishing ahead of Frank Gifford – and was named to the NFL All-Pro team. His 1958 Pro Bowl roommate was NFL Hall of Famer Jim Brown.
He returned to Bethune-Cookman in 1961, holding down duties as head football/basketball coach and athletic director at the same time. Still, McClairen found a way to record non-losing records in his first combined 27 seasons as head coach of both sports.
His football teams had to contend with a Florida A&M squad in its heyday. However, he inherited a team that lost to Florida A&M 97-0 and got that down to 23-20 in 1968, and McClairen considered that the best game of his first tenure. McClairen would lay the groundwork for a successful program, and his two successors – Charles Wesley Moore and Andy Hinson – would beat the Rattlers in three of the first four meetings after McClairen stepped down in 1972.
From 2016 — 55 years of Bethune-Cookman Men's Basketball Coaches in 1 photo — Gravelle Craig, Horace Broadnax, the Legendary "Cy" McClairen, Tony Sheals and Clifford Reed. pic.twitter.com/HwtoqInrE5
— 🏀BCU MBB (@BCUhoops) August 14, 2020
His second stint as head football coach found "Cy" stepping in to replace Sylvester Collins after the 1993 season. That year he earned his only win over FAMU as a head football coach, beating the Billie Joe-led Rattlers in Tampa in the Florida Classic and winning the 1994 Mid-Eastern Athletic Conference Coach of the Year Honors. After the 1996 season, McClairen stepped down- setting the stage for longtime assistant Alvin Wyatt, Sr., who started the modern day turnaround of Wildcat football that continues to this day.
Jack "Cy" McClairen coached many who would go on to the professional ranks, notably NFL Hall of Famer Larry Little, a member of the 1972 Miami Dolphins. Bethune-Cookman had two players on that perfect Dolphin team – Little and Maulty Moore, and were quarterbacked most of the season by Earl Morrall, McClairen's quarterback at Pittsburgh.
McClairen also coached the late Charles "Boobie" Clark, the 1973 AFC Rookie of The Year, and Wyatt, who would go on to become the winningest football coach in Bethune-Cookman history.
As a basketball coach, McClairen led the Wildcats to SIAC championships and NCAA Division II appearances in 1965, 1968 and 1980, with that 1980 team struggling through a 10-14 regular season only to put together an incredible tournament run. He would lead that program through the transfer to Division I, stepping down in 1993. Of particular note is the fact that he personally sacrificed his career coaching record in basketball by negotiating major guarantee games to generate revenue during the transitional era.
Bethune-Cookman legendary coach Jack "CY" McClairen and current head coach Terry Sims admire the #NewBreed #NSD2016 pic.twitter.com/lbVJoCmk0J
— 🏈 Bethune-Cookman FB (@BCUGridiron) February 3, 2016
He finished his active coaching career with records of 71-60-3 in football and 396-436 in basketball.
Before retiring in 2017, he served as Senior Associate Director of Athletics, providing leadership in several areas of athletics, including assistant golf coach for the men's and women's teams under the late Dr. Gary Freeman. In 1988, he was inducted into the Florida Sports Hall of Fame. He was an inaugural member of the Bethune-Cookman Athletic Hall of Fame in 2000 and was named to the MEAC and SIAC Hall of Fames as well.
He was preceded in death by his wife and college sweetheart, Margaret, and is survived by his children Robin, Michelle and Dwayne, who played basketball for him as well as competing in track and field.
Related Items:Bethune-Cookman, featured, History, Jack "Cy" McClairen
Orangeburg Icon and Naismith Hall of Famer Frank Staley Jr. Passes Away
HBCU product Willie Mack thriving as pro golfer
Bethune-Cookman will not compete in 2020-21 sports

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James Ellison
Larry Ellison
Harlan Ellison
Ellison Onizuka
10177 Ellison
Jennifer Ellison
Chase Ellison
Síndrome de Zollinger-Ellison
Fred P. Ellison
Michael Ellison

Desambiguações de sobrenomes

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Q: Homotopic retract vs deformation retract Let's say that $A \subset X$ is a deformation retract. It follows that $A$ is both a retract and a space homotopically equivalent to $X$. Is the converse true? Probably not, but I couldn't find any example yet.
More specifically the converse would be:
If $A \subset X$ is a retract which is homotopic to $X$ as a topological space then does there exist a homotopy between the retraction and the identity map:
$$H:X \times [0, 1] \to X$$
such that $H(x,0)=x$, $H(x,1)\in A$ and $H(a,1)=a$ for $a\in A$.

A: No. Let $X = \{0,1,2,3,\dots\}$ and $A = \{1,2,3,\dots\}$, both with the discrete topology, and let $i: A \to X$ be the inclusion. Then $i$ has a retraction $r: X \to A, n\mapsto\max\{n,1\}$, and is even a cofibration. $X$ and $A$ are clearly isomorphic. The inclusion, however, is not a homotopy equivalence.

A: If your question is whether every space $Y$ which is homotopically equivalent to a space $X$ must be a retract of $X$ then this is certainly not true. One main requirement for a deformation retract is that $A \subset X$. One can easily produce homeomorphic spaces which aren't subspaces of each other. Consider two disjoint disks of different radii in the plane. Does this answer your question?

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NLCE provided Mechanical and Electrical works to these 60 apartments and the below retail units. A very detailed lighting plan was drawn up for the building including the installation of ground lighting within the steps leading up to the retail units, lighting under the canopy at roof level and much more. External lighting to the surrounding pedestrian footpath areas and car park was also included in NLCE drawings. Due to the building's residential function, detailed life safety systems were also included in the designs.


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Publicerat den 03 maj 2010 19 augusti 2022
The Economist om Labour och BNP
Artikel i The Economist: "When grown men cry: Why the BNP vote is growing".
Labour's vote in Barking [. . .] has declined sharply, from around 65% in 1997 to around 50% in 2005 as the BNP vote has increased. In 2006 the BNP won a score of council seats. And in the European elections last year, for the first time, the party gained two MEPs (one of them Mr Griffin), attracting 6.2% of the popular vote across the country.
Tony Travers, of the London School of Economics, says Labour "is completely split about what to do about the threat of the BNP". In fact all liberal parties find it hard to "put across an offer that goes some way to recognising the aspirations and feelings of those who vote BNP". But if they don't engage with the people who support the BNP out of a sense of powerlessness, the party will advance. If it wins control of the council or a parliamentary seat, it will be time, Mr Travers concludes, "to press the panic button".
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Seductive Subversion opens today and the show takes a look at the impact of women artists on the traditionally male-dominated field of Pop art. The exhibition team wanted to keep things simple in the gallery—a spare look, so the pop art would really pop out at you. At the same time, the team had a plethora of research about each of the 25 artists featured in the show and wanted a way to share that with the public. So, the goals of this endeavor became two-fold. First, how do we share the research and, second, how do we do it in a way that won't overwhelm the visitor experience? Wikipedia + iPads became the answer.
Where's the technology in this gallery? Nicely hidden on the column (at right) with the iPad installed on a shelf so visitors are not drowned in mounted screens upon entry.
Let's start with Wikipedia. To get the research into the hands of the biggest audience possible, updating Wikipedia made the most sense. After all, more people go there for information than any other source, so why not take the information we have and make a contribution where it will count? Over the past several months, Rebecca Shaykin, has been working to update the Wikipedia articles on the 25 artists featured in the exhibition. Rebecca is going to talk a little bit more about this process in the next post.
iPads are installed on shelves in several locations throughout the exhibition and two units are provided in a seating area as well.
Once we updated the Wiki, the question became how to get that back into the gallery in an unobtrusive way. For the first time, we've installed iPads in the gallery and we are using the Wiki API to grab the appropriate data and bring it into the in-gallery interactive. iPads are installed on shelves (much like what you see at the Apple store), so visitors are not surrounded by potentially distracting mounted screens. Hardware is installed in various locations throughout the exhibition, so people can stand near the works and browse the Wiki. In addition, we've got a couple of iPads in a lounge-like seating area for a more comfy browsing experience.
For some technical nitty-gritty, we are using the wKiosk app on the iPads to kiosk-ify the browsing experience. We custom designed and manufactured pieces to sit over the home button and power buttons, so visitors can't accidentally break out of the kiosk environment. We are using the same alarm system (SK-T6X-W from Se-Kure) used in the Apple store to prevent theft, but still allow the devices to be picked up and played with. Beau Sievers worked to create an iPad-like browsing experience using HTML and is going to post about the technical ins and outs and release some code we think may be helpful to others.
Wikipedia on the iPad using Beau's BklynFlow to retain iPad-like navigation.
So, what are we looking to learn from this? First, we'd like to see if visitors want this much information in an exhibition setting. The interactive uses the 25 artist articles as a starting point, but visitors have access to the entire wiki from there. Most educators and interpretation staff will say less is more and tend to favor a more guided learning experience, but that's counter to the web. When providing a web resource in the gallery, do visitors want more control over the information they browse? Second, we'd like to see how visitors react to this type of hardware and how we've installed it. Does this provide a better user experience both for the people who want to use it and those who'd rather not be distracted by tech in the gallery setting? Should we be using this type of low-impact equipment in more places throughout the museum? Third, we are going to be looking at browse statistics and how they differ when people are standing near objects versus when they are sitting. Does a seating area mean visitors spend more time with the devices or do people really want to have the information near the works of art?
We'll report back on our findings after the show closes. In the meantime, come see this fantastic show, play with this new system and tell us what your own experience is when using it. Of course, you can also browse Wikipop online and if you hit that link on your own iPad, you'll see what visitors are seeing in the gallery.
This post is part of a three-part series on Wikipop.


text:
Arabian bites
Warm welcome: Michael Winner with Christine Rio and staff at le Tobsil in Marrakesh (Georgina Hristova)
"The Belgians are buying houses in the medina," said Abdelkrim Temsamani, the chief concierge of La Mamounia, as their Mercedes took me from Marrakesh airport to the hotel. "Belgians, Germans, Swiss, many Europeans," he added. This saddened me greatly. Also, the news that a lot of the old buildings there are being turned into boutique hotels or rooming houses. I discovered Marrakesh 10 years ago. It was magical. Now signs of desecration are looming. More hotels, more building. Extra flights each day pour in tourists who eventually destroy everything.
The medina of Marrakesh is in the ancient, walled town. Some visitors are too scared to go there. They also think they need a guide to protect them from villainy in the souk. Both places are much safer than London. The medina has narrow alleys with high walls either side. Groups of Moroccans gather by small shops as cars squeeze along the tiny streets. Go through an unassuming door and you find yourself in a marvellous mansion with amazing tiles. You can buy a luxurious residence for £30,000. Philippe Cluzel told me that when I went to his French restaurant, Le Pavilion, in one of these old villas. "You mean I could get this for £30,000?" I asked in amazement. "Yes," said Philippe. We sat in the central courtyard listening to the muezzin calling Muslims to prayer. There were orange and fig trees all around. Philippe keeps last year's oranges on them for decoration. "I hope they don't fall on your plate," he said. "You mean on his head," said Georgina.
The food in Marrakesh is usually very good indeed. Some of it is historic. Philippe's tops the "very good" list. I had hot foie gras with mango and ginger, Georgina had john dory with saffron sauce. Then I had confetti de canard maison. "It's crispy duck as in the Perigord," explained Philippe. I've no idea what happens in the Perigord, but I'll find out, I thought. I know I liked it, because when I went and tried to order it again, it was off and changed to chicken tandoori. I also had a memorable chocolate millefueille - which wasn't. It was three crunchy biscuits made of chocolate with cream between them and a chocolate sauce around. Philippe used to be a Parisian antique dealer and looks like it. He's well dressed and neat. He runs a good restaurant and has a boutique hotel, La Maison Arabe, a few doors away.
The best restaurant in Marrakesh is Le Tobsil, owned by another French runaway, Christine Rio. This is one of the greatest places in the world. It's also in an old Arabian mansion, where two local musicians in Berber robes play attractively "stoned" music. The food is totally historic. I had a pastilla of vegetables, a sort of fried-vegetable pie. It was a major taste experience. Then we had a Moroccan salad - many bowls of everything from sheep's brains to mashed aubergine. The setting is deeply romantic, with the roof open to the stars, old balconies around the courtyard, beautiful hanging lights and lovely candelabra. Another time, I had pigeon pastilla and a spicy fish casserole. Christine paces the room marshalling the traditionally dressed Arab staff. She's one of my all-time favourites.
Book for Marrakesh now. You can sunbathe well into November and then again in February. Easter's a good time. All the flowers are out around the pool of La Mamounia. The hotel has a superb Moroccan restaurant. The poolside open-air place offers very tasty food, including my favourite, Moroccan Fanta, which is nothing like the jaded version we get in England.
The Aman group has opened the Amanjena, a few miles out of town in scrub-like surroundings. It's a fake old Moroccan hotel, but luxurious and much visited by Aman groupies and others. A young Swiss couple, Pierre and Marieke Baumgartner, run it very Well. You should also visit Chez Ali, a marvellous circus-cum-cabaret with Berber horsemen firing guns all over the place. You eat in tents. It's touristy, but very good.
The old-time favourite, which hotel concierges still recommend to American tourists, is Yacout. It used to be terrific, but they now seat guests in an odd extension by a pool. They do a classical Moroccan set menu. "This is a very average restaurant," observed Georgina. "One of our guests referred to it as function catering," said Mr Temsamani. He's right.
The best thing about Marrakesh is the Moroccans. They have immense charm, warmth and vitality. They should be imported to teach the English and French staff in London hotels and restaurants how to behave in a welcoming manner. On the other hand, that's a lost cause. So they may as well stay in Morocco.
Next time Michael Winner is en route to Alassio (August 5), he should make time to visit Baia del Sole in Laigueglia. The restaurant is located right by the sea and the chef-patron is a student of Gualtiero Marchesi, one of Italy's leading chefs. Her modern Italian style of cooking is excellent -and great value at about £25 a head.
Ron Zanre, Chislehurst, Kent
As restaurant owners in the deepest Dordogne, we rarely see Michael Winner's column, but a customer inadvertently left us a copy, so we were able to read the recent correspondence concerning children in restaurants. At our own restaurant, as at most in France, children are actively welcomed. The French enjoy eating out en famille. After all, how else will children learn to eat in restaurants? In our experience, French children, who have dined out from infancy, generally behave impeccably, whereas British children do not. Enough said.
Stewart and Sue Edwards, Abjat sur Bandiat, France
I noticed that, in a recent photograph, Michael Winner was wearing a jacket that had only three buttons on the cuff. Surely four buttons are now considered de rigueur?
Eric Downham, Kingston-upon-Thames, Surrey
On a recent visit to Fish! in Guildford, we twice had to send back our chips because they were undercooked. Surely this is the ultimate failure: a fish restaurant that can't cook chips?
Maggie Barnes, London
Lest anyone believe that Mr Winner's statements are in any sense authorative, please allow me to point out that, contrary to his experience (August 19), you can indeed land by helicopter in the Grand Canyon. The art is to book with a helicopter company that has not had its landing rights taken away by the Hualapai Indians for consistent abuse of the privilege. I am surprised - nay, delighted - to have achieved something that has eluded the great man himself.
Ian Murray, Bozeat, Northants
The photograph of Michael Winner being pounced upon by three great danes that accompanied his column on Michael's Nook (August 26) was the best ever. In case you decide to honour this historic shot with a caption competition, here's my entry: "Yum, Winner-lot!"
Thomas Kell, by e-mail


Now, more than ever, having secure passwords is important. Recent developments in the computing field are increasing the potential potency of password cracking programs.
Avoid common words or passwords. A popular, and very successful method for cracking passwords involves trying every word in the dictionary, as well as lists of common passwords. Check this list of common passwords and avoid anything listed.
Avoid names, places and birthdays. For example, Mary1985 would be a bad choice if your name is Mary, and you are born in 1985 because they are easily tied to you.
If possible, include uppercase, lowercase, numbers and special characters. It increases the complexity of your password and can greatly improve the time it would take to 'crack' your password. For example, darlbringer would take 30 hours to crack, but d4rlBr1nger! would take 1 year to crack. These are estimates from here.
Use a passphrase. A passphrase greatly increases the potential time it would take to crack your password. The best way to choose a passphrase is 4 random words of any length. For example, Big b00ks are best! This would take an estimated 4000 years to crack with todays technology. This is an estimate from here. Not all systems support spaces, but if possible, include spaces because it increases the overall character count, thus increasing the length of time to crack the password.
Don't use the same password for more than one site. This is one of the leading causes of hacked accounts. If the same username and password combo are used on multiple sites, when one site is compromised, that username and password combo should be considered compromised on all other sites where it is in use.
Following these tips does not guarantee your passwords are uncrackable, but greatly increases the time investment needed to crack your password.
Please keep in mind that these passwords are examples only and should not be used.


text:
Around the World >
From Our Correspondent >
A little whisky blasphemy
06 July 2016 by Lew Bryson
A group of whisky lovers is breaking all the rules when it comes to drinking Scotch, injecting a high-level dosage of 'fun' back into the category. Lew Bryson reports.
Whisky Blasphemy promotes an ethos of enjoying whisky, not using it to impress
'Blasphemy is what an old mistake says of a newly discovered truth.'
--Robert Ingersoll
When you hear the words 'whisky blasphemy,' what do you picture? Macallan on the rocks? Not blasphemous enough. A Lagavulin Rob Roy? You're getting warmer, but still too cautious.
'How about we do Johnnie Walker Blue jello shots?'
Yeah, that's 'whisky blasphemy'. That's also the name of a fast-growing whisky club in the Philadelphia suburbs, and beyond, as the blasphemy spreads on their Facebook page of more than 2,300 members. I sat down with the founders – Judd Weisgal, Jun Nunez, and Matt Kinson – recently over Aberlour A'Bunadh Old Fashioneds, made with orange and chocolate bitters. Don't make that face; they were delicious.
'We don't have rules,' Nunez says, when I ask what the club is about. Then he grins, and notes two rules. 'We don't allow arguments on the Facebook page, none of that crap about NAS, is it bad, is it okay; just leave that out. And we don't allow any prices, or trading, or selling on the page. Really, the only rule is: don't be a jackass.'
It works; there is no trolling, no flaming and no bullying on the page. They simply don't stand for it. 'No one picks on people, no one judges,' Weisgal says. 'Every hobby has snobs, purists, elitists; they also have a group who doesn't feel bound by rules. We're the first one of those groups that has a voice.' It's loud, and people are hearing it, and they like it.
Before Blasphemy, though, there was Weisgal, who had about 250 bottles in his collection, and Nunez, who had about 1,000. 'I bought big when the prices were still good,' he says, and the older bottles he pours prove it. Weisgal's at his regular liquor store, and the manager tells him there's 'this guy Jun you have to meet'.
Nunez was doing a blending experiment with one-gallon Sherry casks. 'I was doing a solera, ageing and blending cask strength whiskies,' he recalls, as if this was something people do in their homes every day. 'I poured Judd a sample, and he's picking out the whiskies in the blend! I like this guy!' That was about five years ago. Kinson, a friend of Weisgal's, joined in the fun soon after.
Whisky Blasphemy: good drams, good cigars, good company and no 'jackasses'
A couple of years later, the three were at Nunez's house, talking about how to break one of their favourite guys at Holt's Cigars out of his Johnnie Walker Blue rut. He loves Blue, won't drink anything else, won't even think about it. That's when Kinson dropped the jello shot bomb. With grins all around, it wasn't long before the group had a name, 'Whisky Blasphemy', and soon they were meeting at Holt's every Thursday night (they did make the jello shots, with Blue, and it did work).
The cocktails started early. 'We like cocktails,' Judd says, 'and the best cocktails come from the best ingredients.' Maybe the definitive Whisky Blasphemy cocktail story is the one where Nunez had won some money in Las Vegas, and says to his bartender: 'Make me an Old Fashioned with that 25-year-old Macallan.' The bartender calls his manager over, asks him if it's okay. 'Sure,' the manager says, 'and make me one, too.' He wound up putting it on the bar's menu as a 'High Roller'. Score one for Blasphemy; score one for Scotch whisky.
I first met Nunez and Weisgal in early 2015, at a liquor store event. They introduced themselves, talked a lot about Whisky Blasphemy, which I'd never heard of; talked a lot about cocktails made with cask strength Ardbeg and Pappy Van Winkle. They talked a lot, honestly, and sounded crazy, but it was fun. They invited me to join them at a Thursday night meeting.
I didn't get there till almost a year later, and it was an eye-opener. In that year the club had grown explosively, and garnered the attention of Glenmorangie's David Blackmore. I joined him and a very proud trio of 'blasphemers' (and about 30 other members) midday for the simulcast US launch of the Glenmorangie Private Edition Milsean.
There were decorations everywhere, the big screen where Dr. Lumsden would Skype in from Scotland, and even representatives from the Pennsylvania state liquor monopoly were there to take orders for bottles. The other launch locations were in Chicago and San Francisco, and I can tell you they weren't having as much fun as we were. Cocktails were being invented right on the bar, cigars were lit and everyone was friendly. Whisky Blasphemy was on the map.
What put them there was a lack of pretense. Members are from widely varied backgrounds: bankers, mechanics, lawyers, web technicians, salespeople, and bureaucrats. They all mingle at the meetings, they all share good times on Facebook. 'We have nothing in common,' Weisgal says, 'except whisky. People bring $30 bottles, they bring $300 bottles. Don't worry about bringing a great bottle, about impressing us.'
'Yeah,' Kinson says, 'you can just bring cookies!'
'People think we're about cocktails,' Wesigal explains. 'Maybe 5% of the drinks we have at meetings are cocktails. We drink mostly neat. People are learning more about whisky, they're becoming more comfortable with whisky. The best whisky is shared with friends, whatever it is.'
Maybe I nodded too much at that point, because Nunez quickly assured me: 'We're not on a mission! We just want to have fun.'
'Fun' is a word that may have been a bit distant from Scotch whisky lately. The Blasphemy crowd is bringing it back, with cocktails, with cigars and Cognacs, with a generous sharing of great bottles. 'Why not?' Nunez says. 'To me it's a bottle; to some of our members, it's an experience. Am I supposed to save a bottle like that for a special occasion? Every day you're alive is an occasion!'
That doesn't sound blasphemous at all.
Here's the recipe for that A'Bunadh Old Fashioned:
60ml (2oz) Aberlour A'Bunadh
10ml (0.25 oz) Raw Cane Liquid Turbinado Sugar
3 dashes of Fee Brothers Aztec Chocolate Bitters
2 dashes of AZ bitters Orange Sunshine
10ml (0.25 oz) Woodford Reserve Cherry syrup (the liquid in the jar of cherries)
Method: Shake and pour into a Double Old Fashioned glass with a big ice cube. Garnish with a Woodford Reserve Cherry.
Cocktails 08 November 2017
London's best Scotch Old Fashioneds
Move over Bourbon; it's time for Scotch whisky to take on the Old Fashioned.
Features 04 December 2018
The whisky gift guide for every budget
From stocking fillers to blowout experiences, we've got presents for whisky fans on every level.
Cocktails 19 February 2019
Cocktail Basics: Old Fashioned
Despite endless interpretations, the attraction of the Old Fashioned lies in its simplicity.
From the editors 08 January 2018
Why aren't bartenders excited by Scotch?
Scotch brands and classic cocktails are failing to inspire bartenders, says Becky Paskin.
From the editors 09 October 2019
The Glenlivet capsules: why the hysteria?
What's more outrageous: cocktails served in capsules or the fact it's a Scotch? asks Becky Paskin.


text:
Q: how to make checkbox visible on multiple selects "I have a common controller. In that there are 2 fields. Text field and a checkbox." My question is that "when user selects single item,the check box should not be visible and when the user selects multiple items, the check box should be visible." This is in android. Please guide me through this.
the code is as given below:
public class ProfileItemDialogFilterableAdapter extends RecyclerView.Adapter {
List<SelectableItem> mItemList;
private LayoutInflater mLayoutInflater;
public static int mSingleSelection = 1;
public static int mMultipleSelection = 2;
private int mSelectedPosition;
private int mSelectionMode;
private Context mContext;

public ProfileItemDialogFilterableAdapter(Context context, List<SelectableItem> itemList) {
    mItemList = itemList;
    mLayoutInflater = LayoutInflater.from(context);
    mContext = context;

}

@Override
public ProfileItemDialogFilterableAdapter.ViewHolder onCreateViewHolder(ViewGroup parent, int viewType) {
    View itemLayoutView = mLayoutInflater.inflate(R.layout.list_item_profile_item_dialog, parent, false);

    return new ViewHolder(itemLayoutView);
}

@Override
public void onBindViewHolder(ProfileItemDialogFilterableAdapter.ViewHolder holder, int position) {

    SelectableItem current = mItemList.get(position);

    if (current == null)
        return;

    holder.mLabelName.setText(current.getName());

    if (mSelectionMode == mMultipleSelection){

        holder.mCheckBoxItem.setVisibility(View.VISIBLE);
        holder.mCheckBoxItem.setChecked(current.isSelected());
    }
    else {
        holder.mCheckBoxItem.setVisibility(View.GONE);

        if (position == mSelectedPosition) {
            holder.mLabelName.setTextColor(mContext.getResources().getColor(R.color.colorBackground));
            holder.mLabelName.setBackgroundColor(mContext.getResources().getColor(R.color.colorPrimary));

        } else {
            holder.mLabelName.setTextColor(mContext.getResources().getColor(R.color.textColorFullBlack));
            holder.mLabelName.setBackgroundColor(mContext.getResources().getColor(R.color.colorBackground));
        }
    }
}

@Override
public int getItemCount() {
    return (mItemList != null ? mItemList.size() : 0);
}

public void setResult(List<SelectableItem> itemList) {
    mItemList = itemList;
    notifyDataSetChanged();
}

public class ViewHolder extends RecyclerView.ViewHolder {

    CheckBox mCheckBoxItem;
    TextView mLabelName;

    public ViewHolder(View itemView) {
        super(itemView);

        mCheckBoxItem = (CheckBox) itemView.findViewById(R.id.checkbox_item);
        mLabelName = (TextView) itemView.findViewById(R.id.label_name);
        mCheckBoxItem.setOnCheckedChangeListener(onCheckedChangeListener);
        mLabelName.setOnClickListener(onClickListener);
    }

    CompoundButton.OnCheckedChangeListener onCheckedChangeListener = new CompoundButton.OnCheckedChangeListener() {
        @Override
        public void onCheckedChanged(CompoundButton buttonView, boolean isChecked) {

            mItemList.get(getAdapterPosition()).setSelected(isChecked);

        }
    };

    View.OnClickListener onClickListener = new View.OnClickListener() {

        @Override
        public void onClick(View view) {

            mSelectedPosition = getAdapterPosition();
            notifyDataSetChanged();
        }
    };
}

}

A: Add a listener to your selection (single or multiple) and inside the listener change the visibility of the checkbox.


The burglary and criminal mischief involved items worth $8,000 in a safe that was stolen, Assistant Falabella Replica Bags District Attorney Joe Scalzo said. Niagara County Sheriff's deputies said they recovered $5,164 in cash; two crack pipes, a marijuana pipe; a canvas tote bag, a damaged cash register, a check for $1,300; checks from the safe; two debit cards; a door frame; a liquor room key; a pocket knife; food, raffle tickets and a safe..
falabella bag stella mccartney Although there is no cure for a sore throat caused by a cold virus, there are ways to help you feel more comfortable. Drinking warm liquids, gargling with warm salt water, sucking on ice chips, or taking an over the counter medicine may relieve symptoms of pain or fever. If you have a particular rose that you love and want a second or third for your landscape or want to give a great gift, consider https://www.stellamccartneysoutlet.com propagating with a potato. This project is easy enough for the novice gardener and attractive to anyone who wants to save money on their garden budget falabella bag stella mccartney.


text:
Q: Does calling dispatch_sync(dispatch_get_global_queue()) on main thread cause app to "hang"? // Method called when a button is clicked

- (void)handleClickEvent {
      dispatch_sync(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{
         [self backgroundProcessing];
      });
      // Some code to update the UI of the view
      ....
      [self updateUI];
      ....
}

1) handleClickEvent is called on the main thread when a button on the view is pressed.
2) I  used dispatch_sync() because the following code that updates the UI of the view cannot be done until a variable in the backgroundProcessing method is calculated.
3) I used dispatch_get_global_queue in order to get the backgroundProcessing off the main thread. (following the rule: generally put background processing off main thread and generally put code that affect the UI on the main thread).
My question is: Does the backgroundProcessing method "hang" the main thread until it is complete since I am using dispatch_sync?
EDIT:
Based on the answer below i have implemented this solution:
- (void)handleClickEvent {
      dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{
         [self backgroundProcessing];
         dispatch_async(dispatch_get_main_queue(), ^{
            [self updateUI];
         });
      });
}

solution from this link: Completion Callbaks

A: Yes, dispatch_sync will block until the task is complete. Use dispatch_async and when the job is complete have it post a block back to the main queue to update the view.


text:
\section{Introduction}
\label{S: Introduction}

Let $T$ be a rooted tree (whose edges are directed from fathers to children). Assume that the vertices of $T$ are labeled with independent and identically distributed continuous random variables; given a vertex $v\in T$, we denote by $w(v)$ its label.  Let \[P=v_0\rightarrow v_1\rightarrow \cdots \rightarrow v_h\] be a path in $T$ from the root to a leaf. We say that $P$ is a $k$\textit{-accessible} if there is a subsequence  $S:=v_{r(0)},v_{r(1)},v_{r(2)},\ldots,v_{r(t)}$ of the vertices in $P$  such that: \[w({v_{r(0)}})<w({v_{r(1)}})<w({v_{r(2)}})<\cdots<w({v_{r(t)}}),\]
each $k$ consecutive vertices of $P$ contains at least one vertex in $S$, $v_0=v_{r(0)}$ and $v_h=v_{r(t)}$. In Figure~\ref{fig:kAccessiblilityInT} we illustrate a labelled tree and  its $2$-accessible paths.  We denote by $\theta_k(T)$ the probability of having a $k$-accessible path in $T$.
\begin{figure}[htb]
	\centering
	\includegraphics[width=.6\linewidth]{kAccessiblilityInT}
	\caption{An example of a labelled tree without $1$-accessible paths, and for which the only $2$-accessible paths  are labeled as: $53$,$99$,$68$,$4$,$71$; and $53$,$65$,$13$,$78$,$26$,$91$.}
	\label{fig:kAccessiblilityInT}
\end{figure}%

\subsection{Motivation.}
Fitness landscapes are by construction a static concept that assigns fitness values to the points of an underlying configuration space \cite{DefLandscapes2003}. This concept was first proposed in 1932 by Sewall Wright \cite{LandscapeDef1932} (as a mapping from a set of genotypes to fitness); since then, it has attracted particular interest in evolutionary biology and in evolutionary computation, because it offers an approach to conceptualize and visualize how an evolving population may change over time, in various population genetics models \cite{FitnessLandscape2014Book}.

For the case of an haploid asexual population on a given fitness landscape, framing by the 'strong selection, weak mutation' (SSWM) regime, a mutations path is considered selectively accessible if the fitness values encountered along it are monotonically increasing \cite{JKAJ2011,WDDH2006,WWC2005}. Motivated by the concept of selective accessibility, a new kind of percolation was introduced by Nowak and Krug in \cite{NK2013}, in which they studied the  probability of having a monotonically increasing path in an graph whose vertices has been labelled with random numbers \cite{NK2013,RZ2013,ApHc2016,BBS2017,Li2017}.

The model of accessibility percolation was introduced by Nowak and Krug \cite{NK2013} as follows. Imagine a population of some life form endowed with the same genetic type (genotype). If a mutation occurs, a new genotype is created which can die out or replace the old one. Provided that natural selection is sufficiently strong, the latter only happens if the new genotype has larger fitness. As a consequence, on longer timescales the genotype of the population takes a path through the space of genotypes along which the fitness is monotonically increasing \cite{G1984}.

The hypercube is a graph whose vertices are all possible $N$-tuples in $\lbrace 0,1 \rbrace^N$ (for some positive integer $N$), where two vertices are connected by an undirected edge, if the number of coordinates at which they differ is one. In many basic mathematical models of genetic mutations, the genotype sequence space is represented by the hypercube: each genome is represented as a node of the hypercube and each mutation involves the flipping of a single bit from 0 (the "wild" state) to 1 (the "mutant" state) \cite{JKAJ2011,WWC2005, Martinsson2014}. An $f$-ary (complete $f$-ary) tree with height $h$, is a rooted tree (whose edges are directed from fathers to children) whose vertices at distance at most $h-1$ to the root, have $f$ children. Being a simpler case and motivated by the hypercube case, Nowak and Krug in \cite{NK2013} study the problem of determining the growth of $f$ as a function of $h$ for having accessibility percolation in $f$-ary trees.

According to the ruggedness or smoothness of the landscape, there are some topological features (as peaks, valleys, ridges  etc.) that block the selectively accessible mutations paths towards the highest peak in the landscape. The evolutionary process that may allow escaping those topological features is known as valley crossing, which is not allowed in the selectively accessible mutation paths. However, in natural populations of sufficient size, a number of double mutants is present at all times, and the crossing valleys can be relatively facile \cite{JKAJ2011,WC2005,WDFF2009}; the SSWM assumption may therefore seem very restrictive.

A simple way of allowing peaks and valleys on the accessibility percolation model, consists in `` allowing holes'' in the monotonicity of the path that the genotype population takes.
In this paper, as a variation of the concept of accessibility percolation introduced by Nowak and Krug in \cite{NK2013},  we introduce the concept of $k$-accessibility percolation in which crossing small valleys is allowed. Imagine a population of some organisms  endowed with the same genotype. If a mutation occurs, a new genotype is created. The genotype die, when in the path $P$ of consecutive mutations, there is not a subsequence $S$ with increasing fitness and small holes.

\subsection{Previous results}

Given a function $f:\mathbb{Z}_+\rightarrow \mathbb{Z}_+$, we denote by $T_h(f)$ the $f(h)$-ary tree with height $h$ and we denote by $\mathcal{T}_f$ the sequence of trees $\lbrace T_h(f)\rbrace_{h\in \mathbb{Z}_+}$. We denote by $\theta_k(\mathcal{T}_f):=\lim_{h\rightarrow \infty} \theta_k (T_{h}(f))$ provided the limit exists, when $\theta_k(\mathcal{T}_f)>0$  we say that there is $k$-percolation in $\mathcal{T}_f$.

In the first work on accessibility percolation, Nowak and Krug studied the problem of determining the growth of $f$, with respect to $h$, such that there is $1$-accessibility percolation  on $\mathcal{T}_f$ \cite{NK2013}. Nowak and Krug prove that: if $f$ is a super-linear function, then there is $1$-accessibility percolation in $\mathcal{T}_f$; and if $f$ is a sub-linear function, then there is not $1$-accessibility percolation in $\mathcal{T}_f$. They also studied the linear case, $f(h)=\alpha h$, establishing the existence of a threshold between $e^{-1}$ and $1$ on the scaling constant factor; there is $1$-accessibility percolation in $\mathcal{T}_f$ when $\alpha>1$, and there is not $1$-accessibility percolation in $\mathcal{T}_f$ when $\alpha<e^{-1}$. See Theorem~\ref{thm:NK}

\begin{thm}[Nowak-Krug~\cite{NK2013}] \label{thm:NK}
$\theta_1 (\mathcal{T}_f)>0$, if $f(h)\geq  \left\lfloor \alpha h \right\rfloor $ for $h$ large enough and $\alpha>1$. $\theta_1 (\mathcal{T}_f)=0$, if $f(h)\leq  \left\lfloor \alpha h \right\rfloor$  for $h$  large enough and $\alpha \leq e^{-1}$.
\end{thm}

Continuing with the work of Nowak and Krug, Roberts and Zhao \cite{RZ2013} determined that $\alpha=e^{-1}$ is a threshold on the scaling constant factor for $1$-accessibility percolation on $\mathcal{T}_f$, see Theorem~\ref{thm:NK1p} and Theorem~\ref{thm:NK1perc}.

\begin{thm}[Roberts-Zhao \cite{RZ2013}]\label{thm:NK1p}
	\[\theta_1 (\mathcal{T}_f)=
	\begin{cases}
	1,       & \quad \text{if } f(h)\geq  \left\lfloor \alpha h \right\rfloor \text{ for $h$ large enough and } \alpha>e^{-1},\\
	0,       & \quad \text{if } f(h)\leq  \left\lfloor \alpha h \right\rfloor \text{ for $h$ large enough and } \alpha \leq e^{-1}.\\
	\end{cases}  \]
\end{thm}

\begin{thm} [Roberts-Zhao \cite{RZ2013}]\label{thm:NK1perc}
If $f(h)=\left( \frac{1+\beta_h}{e} \right)h$ where $\beta_h \rightarrow 0$ as $h \rightarrow \infty$, then
\begin{equation*}
 \theta_1 (\mathcal{T}_f)=
	\begin{cases}
	1,      & \quad \text{if } h\beta_h/\log h \rightarrow \infty. \\
	0,      & \quad \text{if } \log h -2h\beta_h \rightarrow \infty,\\
	\end{cases}
\end{equation*}
\end{thm}

Accessibility percolation has been also studied,  recently, for the case when the underlying graph is the hypercube; see \cite{Martinsson2014,martinsson2015,ApHc2016,BBS2017,Li2017,BBS2017}.

\subsection{Our results.}

We denote by $\Omega(g)$ the functions $t$ such that $\lim_{h\rightarrow\infty}(t(h)/g(h))\geq k>0$.
We denote by $\omega(g)$ the functions $t$ such that $\lim_{h\rightarrow\infty}(t(h)/g(h)) = \infty$ (for a more formal definition of $\Omega$ and $\omega$ the reader may change $\lim_{h\rightarrow\infty}$ by $\liminf_{h\rightarrow\infty}$).
Therefore $\omega\left(\sqrt[k]{\log(h)}\right)$ denotes a function $t$ such that, for any constant $C>0$, $t(h)\geq C\sqrt[k]{\log(h)}$ for $h$ large enough; similarly $\Omega(h^c)$ denotes a function that, for some constants $c,C>0$, it is above $Ch^c$ for $h$ large enough.

In this paper we  determine the growth of $f$, as a function of $h$, for which there is $k$-accessibility percolation on $\mathcal{T}_f$. Our main result is the following.

\begin{thm}\label{thm:main}
Let $\mathcal{T}_f$ be the sequence of $n$-ary trees $\lbrace T_h(f)\rbrace_{h\in \mathbb{Z}_+}$. Then
\begin{equation} \label{eq:mainthm}
 \theta_k(\mathcal{T}_f)=
	\begin{cases}
	1,       & \quad \text{if } f(h)\geq  \sqrt[k]{h/(ek)}+\omega\left(\sqrt[k]{\log(h)}\right), \\
	0,       & \quad \text{if } f(h)\leq  \sqrt[k]{h/(ek)}- \Omega(h^c) \text{ and } c>0.\\
	\end{cases}
\end{equation}
\end{thm}

The reader may note that, from Theorem~\ref{thm:main} it follows that

\begin{cor}
\[\theta_k(\mathcal{T}_f)=
	\begin{cases}
	1,       & \quad \text{if } f(h)\geq \sqrt[k]{h/(ek)}c \text{ for $h$ large enough and } c>1, \\
	0,       & \quad \text{if } f(h)\leq \sqrt[k]{h/(ek)}c \text{ for $h$ is large enough and } c<1.
	\end{cases}  \]
\end{cor}

\section{Preliminaries}

Before proceeding with the proof of Theorem~\ref{thm:main}, we introduce the concept of $k$-transitive closure any we state an equivalent version of Theorem~\ref{thm:NK1perc}.

We define the $k$-transitive closure of $G$, $G^k$, as the graph  obtained from it, by adding new edges from each vertex $u$ to each vertex $v$, with the property that $G$ does not already contain the directed edge from $u$ to $v$ but does contain a directed
path from $u$ to $v$  with length at most $k$.

Although the concept of $k$-accessibility percolation was only introduced for trees, it can be easily extended  to directed graphs where: there is a single vertex distinguished as the source, some vertices distinguished as sinks, and all of the maximal paths are directed paths from the source to some sink; we name the graphs that satisfy the previous statements as monotone graphs. For the case of monotone graphs, we say that a path in it is $1$-accessible if: it starts in the source, ends at some sink and its vertices have increasing labels. Similar to the case of trees, we denote by $\theta_1(G)$ the probability of having a $1$-accessible path in $G$.

In Figure~\ref{fig:1AccessiblilityInT3} we illustrate the $2$-transitive closure of the graph depicted in Figure~\ref{fig:kAccessiblilityInT} and its $1$-accessible paths. Note that the $1$-accessible paths in the graph depicted in Figure~\ref{fig:1AccessiblilityInT3} correspond to the $2$-accessible paths in the graph depicted in Figure~\ref{fig:kAccessiblilityInT}.

\begin{obs}\label{def1}
	 Let $T$ be a rooted tree with height $h$ and $T^k$ be its $k$-transitive closure. Then $\theta_1(T^k)=\theta_k(T)$.
\end{obs}
\begin{proof}
 Let $T$ be a rooted tree with height $h$ and $T^k$ be its $k$-transitive closure.
 If $P$ is a $k$-accessible path in $T$, there is a subsequence $S$ of the vertices in $P$ with increasing labels, that contains at least one vertex in each $k$ consecutive vertices in $P$ and that contains the root and a leaf of $T$; therefore $S$ is a $1$-accessible path in $T^k$.
 On the other direction,
 if there is a $1$-accessible path in $T^k$, the vertices in such path define a subsequence $S$ of the vertices in some path $P$ in $T$ that make $P$ a $k$-accessible path.
\end{proof}

\begin{figure}[htb]
	\centering
	\includegraphics[width=.6\linewidth]{1AccessiblilityInT2}
	\caption{An illustration of the $2$-transitive closure of the graph depicted in Figure~\ref{fig:kAccessiblilityInT}, the $1$-accessible paths in such graph are labeled as: $53$,$68$,$71$; and $53$,$65$,$78$,$91$.}
	\label{fig:1AccessiblilityInT3}
\end{figure}%

For proving Theorem~\ref{thm:main} we use the following proposition, that it is an equivalent version of Theorem~\ref{thm:NK1perc}.

\begin{prop}\label{prop:nk}
\begin{equation*}
 \theta_1 (\mathcal{T}_f) =
	\begin{cases}
	1,      & \quad \text{if } f(h)= \frac{h}{e}+ \omega\left(\log(h) \right),  \\
	0,      & \quad \text{if } f(h)= \frac{h}{e}+ \frac{\log(h)}{2e} - \omega(1).\\
	\end{cases}
\end{equation*}
\end{prop}

\begin{proof}
 Let $f(h)$, $g(h)$ and $\beta_h$ be such that
 \[f(h)= \frac{h}{e}+ g(h)\log(h)= \frac{1+\beta_h}{e}h.\]
 As $\beta_h=g(h)\frac{h}{\log(h)}$, it follows that, $h\beta_h/\log h$ goes to infinity when  $h$ goes to infinity, if and only if $g(h)=\omega(1)$. Therefore, by Theorem~\ref{thm:NK1perc}, if $f(h)= \frac{h}{e}+\omega(\log(h))$ then  $f(h)= \frac{h}{e}+\omega(1)\log(h)$ and $\theta_1 (\mathcal{T}_f)=1$.

 Let $f(h)$, $g(h)$ and $\beta_h$ be such that
 \[f(h)= \frac{h}{e}+\frac{\log(h)}{2e}- g(h)= \frac{1+\beta_h}{e}h.\]
 As $\beta_h=\frac{\log(h)- 2e\cdot g(h)}{2h}$, it follows that, $\log h -2h\beta_h$ goes to infinity when  $h$ goes to infinity, if and only if $g(h)=\omega(1)$. Therefore, by Theorem~\ref{thm:NK1perc}, if $f(h)= \frac{h}{e}+\frac{\log(h)}{2e}-\omega(1)$ then $\theta_1 (\mathcal{T}_f)=0$.

\end{proof}

\section{Proof of Theorem~\ref{thm:main} }

Now we proceed to prove Theorem~\ref{thm:main}. The proof is divided two steps according with the two cases in Equation~\ref{eq:mainthm}.

\subsection*{Step 1}
Let $\mathcal{T}_f$ be the sequence of $n$-ary trees $\lbrace T_h(f)\rbrace_{h\in \mathbb{Z}_+}$. In this section we assume $$f(h)\geq  \sqrt[k]{h/(ek)}+\omega \left( \sqrt[k]{\log(h)} \right)$$ and we prove that $\theta_k(\mathcal{T}_f)=1$.

Let $\mathcal{T}^k_f$ be the sequence of graphs $\lbrace T_h^k(f)\rbrace_{h\in \mathbb{Z}_+}$ where $T_h^k(f)$ is the $k$-transitive closure of $T_h(f)$.
Let $g(h)= h/e + \omega \left(\log(h) \right)$, $T^k=T_h^k(f)$ and $T'=T_{\left\lfloor h/k\right\rfloor }(g)$. By Remark~\ref{def1}, provided the limit exists,
\[ \theta_k(\mathcal{T}_f)=\lim_{h\rightarrow \infty}\theta_k(T_h(f))=
\lim_{h\rightarrow\infty}\theta_1(T_h^k(f)).\]
By Proposition~\ref{prop:nk}
\[\lim_{h\rightarrow\infty}\theta_1(T_h(g))=\theta_1( \mathcal{T}_g)=1.\]
To prove that $\theta_k(\mathcal{T}_f)=1$, it is enough to show that
$\theta_1(T^k)\geq \theta_1(T')$.

Let $G$ be the subgraph of $T^k$, obtained from removing the vertices in $T$, whose distance from the root is not a multiple of $k$. Note that $G$ is a tree contained in $T^k$ with height $\left\lfloor h/k\right\rfloor$.
Also note that, for $h$ large enough,
\[ \text{deg}_G(v)\geq \left( \sqrt[k]{h/(ek)}+\omega \left(\sqrt[k]{\log(h)} \right) \right) ^k \geq {h/(ek)}+\omega \left({\log(h)} \right) \geq  g(\left\lfloor h/k\right\rfloor);\]
thus the non leaves vertices of $G$ have degree at least $g(\left\lfloor h/k\right\rfloor)$. Therefore, for $h$ large enough $T'\subset G \subset T^k.$  Therefore, the probability of having at least one  $1$-accessible path in $T'$, is a lower bound of the probability of having at least an $1$-accessible path in $T^k$.

\subsection*{Step 2}
Let $\mathcal{T}_f$ be the sequence of $n$-ary trees $\lbrace T_h(f)\rbrace_{h\in \mathbb{Z}_+}$. In this section we assume $f(h)\leq  \sqrt[k]{h/(ek)}- \Omega(h^c)$ for some $0<c<1$, and we prove that $\theta_k(\mathcal{T}_f)=0$.

Let $\mathcal{T}^k_f$ be the sequence of graphs $\lbrace T_h^k(f)\rbrace_{h\in \mathbb{Z}_+}$ where $T_h^k(f)$ is the $k$-transitive closure of $T_h(f)$. Let $g(h)=\frac{h}{e}$, $T=T_h(f)$, $T^k=T_h^k(f)$ and $T'=T_{ h/k}(g)$. By Remark~\ref{def1}, provided the limit exists,
\[ \theta_k(\mathcal{T}_f)=\lim_{h\rightarrow \infty}\theta_k(T_h(f))=
\lim_{h\rightarrow\infty}\theta_1(T_h^k(f)).\]
By Proposition~\ref{prop:nk}
\[\lim_{h\rightarrow\infty}\theta_1(T_h(g))=\theta_1( \mathcal{T}_g)=0.\]
To prove that $\theta_k(\mathcal{T}_f)=0$, it is enough to show that
$\theta_1(T^k)\leq \theta_1(T')$. For this we define a tree $H^k$ such that
\[ \theta_1(T^k)\leq \theta_1(H^k) \leq  \theta_1(T').\]

$H^k$ will denote a tree, whose vertices are in correspondence with the paths in $T^k$ that start in the root.  As an outline of the construction of $H^k$ note that, any  path in $T^k$ is characterized by the following two issues. The first issue is the vertex $v$ in which the path ends. Suppose that $v$ is at distance $l$ to the root. The second issue is the information about the levels in which it has no vertices; it can be represented by a subset $s$ of $\lbrace 1,2,\ldots,l-1 \rbrace$ that does not contain $k$ consecutive numbers. Therefore the vertices in $H^k$ will be determined by a vertex $v$ in $T^k$ and a $s$ subset of $\lbrace 1,2,\ldots,l-1 \rbrace$ that does not contain $k$ consecutive numbers. About the edges in $H^k$, two vertices will be adjacent in $H^k$ if: one of the corresponding paths in $T^k$ contains the other corresponding path, and those corresponding only differ in one edge.
As an example of this outline, consider the paths $v_0,v_1,v_3$ and $v_0,v_1,v_3,v_5$ in the graph $T^2$ illustrated in Figure~\ref{fig:ex-H:Tg2}; in $H^2$, illustrated in Figure~\ref{fig:ex-H:H}, those paths are represented by $v_3^{ \{2\}}$ and $v_5^{ \{2,4\}}$, respectively, and they are adjacent.

We define $H^k$ formally as follows.

Let $\mathcal{A}_l$ be the set whose elements are the subsets $s$ of $ \lbrace 1,2,\ldots ,l-1 \rbrace$ that do not contain $k$ consecutive numbers. Let $H^k=(V,E)$ be the graph whose vertices and edges are defined as follows. See Figure~\ref{fig:ex-H}.
\begin{figure} [htb]
	\begin{subfigure}{.2\textwidth}
		\centering
		\includegraphics[width=.26\linewidth]{Tg}
		\caption{}
		\label{fig:ex-H:Tg}
	\end{subfigure}%
	%
	\begin{subfigure}{.2\textwidth}
		\centering
		\includegraphics[width=.5\linewidth]{Tg2}
		\caption{}
		\label{fig:ex-H:Tg2}
	\end{subfigure}%
	%
	\begin{subfigure}{.6\textwidth}
		\centering
		\includegraphics[width=0.9\linewidth]{H-2}
		\caption{}
		\label{fig:ex-H:H}
	\end{subfigure}%
	\caption{An example of the subgraphs of $T$ (\ref{fig:ex-H:Tg}), $T^2$ (\ref{fig:ex-H:Tg2}) and $H^2$ (\ref{fig:ex-H:H}), respectively, corresponding to six consecutive vertices in the path $v_0, v_1, v_2, v_3, v_4, v_5, v_5, v_5$ depicted Figure~\ref{fig:kAccessiblilityInT}. }
	\label{fig:ex-H}
\end{figure}
The set of vertices in $H^k$ is defined as
\[V:=\lbrace v^s:v\in T \textnormal{ and } s\in \mathcal{A}_l \textnormal{, where $l$ is the distance of $v$ to the root} \rbrace. \]
The edges in $H^k$ are defined recursively as follows.
Let $v$ be a vertex of $T$ at distance $l$ to the root and let $s\in \mathcal{A}_l$.
We say that $w^{s'}$ is a son of $v^s$ if:
\begin{itemize}
	\item $w$ is a son of $v$ in $T$ and $s=s'$
	\item $w$ is a descendant of $v$ at distance $1<j\leq k$ in $T$ and $s'=s\cup\lbrace l+1,l+2,\ldots l+j-1 \rbrace$
\end{itemize}

Given a vertex $v$ in $T^k$, we claim that: the paths in $T^k$ from the root to $v$, the elements in $\mathcal{A}_l$, and the paths in $H^k$ from the root to $v^s$ (for some $s\in\mathcal{A}_l$),  are in one-to-one correspondence. For each path $P$ in $T^k$ from the root to $v$, there exist one and only one $s$ in $\mathcal{A}_l$ whose elements are the indices $i$ such that $v_i$ is not in $P$. For each $s$ in $\mathcal{A}_l$, there is one and only one path in $P$ that starts in the root and ends in $v$; such path contains the vertices $v_i$ such that $i$ is not in $s$. For each path in $H^k$ from the root to $v^s$, obviously there is one and only one $s\in \mathcal{A}_l$. Now suppose that $s$ is in $\mathcal{A}_l$, there is one and only one path $P$ in $H^k$ from the root to $v^s$; such path contains the vertices $v_i^{s(i)}$ where $i$ is not in $s$ and $s(i)=s\cap \{1,2,\dots, i-1\}$. As an example consider: In the graph $G^2$ for the graph $G$ in Figure~\ref{fig:kAccessiblilityInT}, the path $v_0,v_1,v_3,v_5$; in $\mathcal{A}_5$, the set $\{2,4\}$; and in $H^2$, the path $v_0^{\phi},v_1^{\phi},v_3^{\{2\}},v_5^{\{2,4\}}$ (See Figure~\ref{fig:ex-H}).

Assume that the vertices in $H^2$ are labelled with independent and identically distributed random variables with the same distribution as the labels in $T$.

 \begin{figure} [htb]
  \begin{subfigure}{.3\textwidth}
  \centering
  \includegraphics[width=.34\linewidth]{H0}
  \caption{}
  \label{fig:ex-H:H0}
 \end{subfigure}%
 %
 \begin{subfigure}{.33\textwidth}
  \centering
  \includegraphics[width=.36\linewidth]{H1}
  \caption{}
  \label{fig:ex-H:H1}
  \end{subfigure}%
  %
  \begin{subfigure}{.36\textwidth}
  \centering
  \includegraphics[width=0.50\linewidth]{H2}
  \caption{}
  \label{fig:ex-H:H2}
 \end{subfigure}%

 \begin{subfigure}{.4\textwidth}
  \centering
  \includegraphics[width=.73\linewidth]{H3}
  \caption{}
  \label{fig:ex-H:H3}
 \end{subfigure}%
 %
 \begin{subfigure}{.6\textwidth}
  \centering
  \includegraphics[width=.85\linewidth]{H4}
  \caption{}
  \label{fig:ex-H:H4}
 \end{subfigure}%
 \caption{An example of a sequence of graphs in which the graph depicted in Figure~\ref{fig:ex-H:H} is obtained from the graph in Figure~\ref{fig:ex-H:Tg2}. It illustrates how to obtain the sequence of graphs used in Lemma~\ref{lem:Tg2-H-2}.}
 \label{fig:ex-H-etapas}
 \end{figure}
 %

\begin{lem}\label{lem:Tg2-H-2}
	$\theta_1(H^k)\geq \theta_1(T^k).$
\end{lem}
\begin{proof}
 In what follows we define a sequence of graphs $H_0, H_1, \ldots, H_t$, such that $H_0=T^k$, $H_t=H^k$ and $\theta_1(H_i)\leq \theta_1(H_{i+1})$, for $i=0,1,\ldots, t-1$.
 The graph $H_0$ is obtained from $T^k$ changing each vertex $v$ for a vertex $v^\phi$, see Figure~\ref{fig:ex-H:H0}. Given some $H_i$, we say that a vertex $v\in H_i$ is divisible if: (see vertex $v_j^s$ in Figure~\ref{fig:divisiones1})
 \begin{itemize}
  \item The subgraph of $H_i$, induced by $v$ and its descendants vertices, is a directed tree. We denote such tree as $T(v)$.
  \item There are at least two edges that starts in an ancestor of $v$ and ends at $v$. From such edges we denote by $e(v)$ the edge that starts in the ancestor $x=u^s$ of $v$ such that $u$ is the nearest to the root in $T$.
 \end{itemize}
 %
  	%
	\begin{figure} [htb]
		\begin{subfigure}{.5\textwidth}
			\centering
			\includegraphics[width=.9\linewidth]{divisiones1}
			\caption{}
			\label{fig:divisiones1}
		\end{subfigure}%
		%
		\begin{subfigure}{.5\textwidth}
			\centering
			\includegraphics[width=.9\linewidth]{divisiones2}
			\caption{}
			\label{fig:divisiones2}
		\end{subfigure}%
		\caption{ Illustration of the proof of $\theta_1(H_i)\leq \theta_1(H_{i+1})$. }
		\label{fig:divisiones}
	\end{figure}
	%
Whenever $H_i$ has at least one divisible vertex, we define $H_{i+1}$ as follows. Let $v^s$ be a divisible vertex in $H_i$. Let $v_0,v_1, \ldots, v_j$ be a path in $T$ from the root to $v$, then $v_0^{s(0)},v_1^{s(1)}, \ldots, v_j^{s(j)}$ is a path in $H_i$ from the root to $v^s$. Let $x=v_{j'}^{s(j')}$ be the vertex where $e(v)$ starts. Let $Q$ be the indices of the vertices that $e(v)$ jumps, $i.e.$ $Q:=\lbrace \{ j'+1, j'+2, \ldots j-1\} \rbrace$. Let $T^Q$ be a copy of $T(v)$ where each vertex $u^{s'}$ is changed by $u^{s'\cup Q }$. $H_{i+1}$ is defined as the graph obtained from $H_i$ by: removing $e(v)$, adding $T^Q$, and adding an edge $e'(v)$ that starts in $x$ and ends at the root of $T^Q$. See Figure~\ref{fig:ex-H-etapas}.

 The reader may notice that, if $H_i$ does not have divisible vertices then $H_i=H^k$. It remains to prove that $\theta_1(H_i)\leq \theta_1(H_{i+1})$.

 Given a vertex $y$, in a labeled directed graph $G$, we denote by $[y\leadsto G]$ the event of having a path in $G$ that starts in $y$,  ends at a sink of $G$ and has increasing labels; we denote by $[y\not\leadsto G]$ its complement. For the case when $y$ is the root it is denoted by $0$.

 Consider in the graph in Figure~\ref{fig:divisiones1}, the event of having an $1$-accessible path  that contains the edge $e(v_j^s)$.  Note that such event occurs, if and only if, there is an increasing path from $0$ to $x$, $w(x)<w(v^s_j)$ and $[v^s_j \leadsto T(v_j^s)]$. Similarly, consider the event in Figure~\ref{fig:divisiones2}, the event of having an $1$-accessible path  that contains the edge $e'(v_j^s)$.  Note that such event occurs, if and only if, there is an increasing path from $0$ to $x$, $w(x)<w(v_j^{s\cup Q})$ and $[v_j^{s\cup Q} \leadsto T^Q]$.
 With the notation introduced in the construction of $H_{i+1}$, define $\Gamma_1$ has the event of having an $1$-accessible path in $H_i$ that contains the edge $e(v_j^s)$, and define $\Gamma_2$ has the event of having an $1$-accessible path in $H_{i+1}$ that contains the edge $e'(v_j^s)$.

 This lemma follows from the following facts:
 \begin{itemize}
  \item Note that $\Gamma_1$ is the event of having an increasing path from $0$ to $x$, $w(x)<w(v_j^s)$ and $[v_j^s \leadsto T(v_j^s)]$; and $\Gamma_2$ is the event of having an increasing path from $0$ to $x$, $w(x)<w(v^{s\cup Q}_j)$ and $[v^{s\cup Q}_j \leadsto T^Q]$. Therefore, as
 \begin{align*}
  \mathbf{P}\left( w(x)<w(v^{s}_j) \right)&=\mathbf{P}\left( w(x)<w(v^{s\cup Q}_j) \right) \text{ and }\\
  \mathbf{P}\left( [v_j^s \leadsto T(v_j^s)] \right)&=\mathbf{P}\left( [v^{s\cup Q}_j \leadsto T^Q] \right) \text{ then }\\
  \mathbf{P}\left( \Gamma_1 \right)&=\mathbf{P}\left( \Gamma_2 \right).
 \end{align*}
  \item   As the only edge that joins $T^Q$ with the vertices in $H_i$ is $e'(v_j^s)$, and $H_i$ and $H_{i+1}$ only differ in $e(v_j^s)$, $e'(v_j^s)$ and $T^Q$ then
  $H_{i} -e(v_j^s)=H_{i+1} -\lbrace e'(v_j^s),T^Q\rbrace$.
  Therefore, as there are no paths in $H_{i+1} -e'(v_j^s)$ from the root to a leaf in $T^Q$ then
  \begin{align*}
   \mathbf{P} \left(  [0\leadsto H_{i+1} -e'(v_j^s)] \right)
   &= \mathbf{P} \left(  \left[0\leadsto H_{i+1} -\lbrace e'(v_j^s),T^Q\rbrace \right] \right)\\
   &=\mathbf{P} \left( [0\leadsto H_{i} -e(v_j^s)]  \right)
  \end{align*}

 \item Let $[0\leadsto x]$ the event of having an increasing path from $0$ to $x$. Then
  \begin{align*}
   \mathbf{P}\left([0\leadsto H_i-e(v)] \big| \Gamma_1\right)
   &\geq \mathbf{P}\left([0\leadsto H_i-e(v)] \big| [0\leadsto x],w(x)<w(v^s_j) \right)\\
   & = \mathbf{P}\left([0\leadsto H_{i+1} -\lbrace e'(v_j^s),T^Q\rbrace] \big| [0\leadsto x],w(x)<w(v^{s\cup Q}_j) \right)\\
   & =\mathbf{P}\left([0\leadsto H_{i+1} -\lbrace e'(v_j^s),T^Q\rbrace] \big| \Gamma_2 \right)\\
   & =\mathbf{P}\left([0\leadsto H_{i+1} - e'(v_j^s)] \big| \Gamma_2 \right)
  \end{align*}

 \item  $\theta_1(H_{i+1})-\theta_1(H_{i})\geq 0$.
 \begin{align*}
 &\theta_1(H_{i+1})-\theta_1(H_{i})= \mathbf{P} \left( [0\leadsto H_{i+1}] \right) - \mathbf{P} \left( [0\leadsto H_{i}] \right) \\
 &=\mathbf{P} \left( [0\leadsto H_{i+1}]\cap [0\leadsto H_{i+1} -e'(v)] \right)  -\mathbf{P} \left( [0\leadsto H_{i}]\cap [0\leadsto H_{i} -e(v)]  \right)\\
 &+\mathbf{P} \left( [0\leadsto H_{i+1}]\cap [0\not\leadsto H_{i+1} -e'(v)] \right)
 - \mathbf{P} \left( [0\leadsto H_{i}]\cap [0\not\leadsto H_{i} -e(v)]  \right)\\
 &=\mathbf{P} \left(  [0\leadsto H_{i+1} -e'(v)] \right)
 -\mathbf{P} \left( [0\leadsto H_{i} -e(v)]  \right)\\
 &+\mathbf{P} \left( \Gamma_2\cap [0\not\leadsto H_{i+1} -e'(v)] \right)
 - \mathbf{P} \left( \Gamma_1\cap [0\not\leadsto H_{i} -e(v)]  \right)\\
 &=\mathbf{P} \left( \Gamma_2 \right) -\mathbf{P} \left( \Gamma_2\cap [0\leadsto H_{i+1} -e'(v)] \right) \\
 &- \mathbf{P} \left( \Gamma_1 \right) +\mathbf{P} \left( \Gamma_1\cap [0\leadsto H_{i} -e(v)]  \right)\\
 &=\mathbf{P} \left( \Gamma_1 \right)\left[
 \mathbf{P} \left( [0\leadsto H_{i} -e(v)] \big| \Gamma_1 \right)
 -\mathbf{P} \left( [0\leadsto H_{i+1}-e'(v) ] \big| \Gamma_2 \right)
 \right] \geq 0
 \end{align*}
 \end{itemize}

\end{proof}

\begin{lem}
	 $\theta_1(T')\geq \theta_1(H^k)$
\end{lem}

\begin{proof}
By definition we require to prove that: the probability of having a $1$-accessible path in $T'$, is an upper bound for the probability of having a $1$-accessible path in $H^k$.

	Let $H'\subset H^k$ be the tree induced by the vertices of $H^k$ at distance at most $h/k$ to the root. Notice that $\theta_1(H') \geq \theta_1(H^k)$. Also notice that, if $v$ is a vertex of $H'$ at distance at most $h/k-1$ to the root then (for $h$ large) \[ \text{deg}_{H'}(v)=\sum_{j=1}^k \left\lfloor f(h) \right\rfloor ^j  \leq \sum_{j=1}^k f(h) ^j = \frac{f(h)^{k+1}-1}{f(h)-1}\]
	We claim that $\text{deg}_{H'}(v)\leq g(\lfloor h/k \rfloor)$ (recall, $g$ was defined as $g(h)=\frac{h}{e}$), from which  $H'\subset T'$ and, as both trees have the same height, then $\theta_1(T') \geq \theta_1(H')$.

	Now, to finish this proof, we prove that the claim holds. Let $A=\sqrt[k]{h/(ek)}$. As $f(h)\leq  \sqrt[k]{h/(ek)}- \Omega(h^c)$ then for $c<x<1$ and $h$ large enough, $ f(h)\leq A-A^{x}$. It is enough to prove that
	\[ \frac{\left(A-A^{x} \right)^{k+1} -1}{\left(A-A^{x} \right)-1}\leq A^k\]
	but it follows from that
	\[\left(A-A^{x} \right)^{k+1} -1= A^{k+1}-(k+1)A^{k+x}+o(A^{k+x})\] and
	\[ A^k\left(A-A^{x} -1 \right)= A^{k+1}-A^{k+x} +A^k.\]
\end{proof}

\noindent\textbf{Acknowledgments:}
The authors are thankful to Ricardo Restrepo for helpful discussions.

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


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The Acer Aspire One model D225E is a 10.1-inch netbook that features an ultra-thin and lightweight design with a powerful battery making it ideal for everyday computing on the go. Also known as Acer Aspire One D255E-13639.
Why won't it load Windows 7 starter?
How to reset to factory setting.
Why has my screen gone dark?
Weighing in at only 2.76 pounds, the netbook features an extremely sleek and lightweight design that is built with plenty of power underneath. The netbook comes with an Intel Atom N455 central processing unit that is capable of delivering speeds of up to 1.66 GHz. with 1GB of random access memory. The graphics unit is a GMA 3150 graphics processor, allowing for better displays and sharper images. The display of the monitor comes in at an impressive 1024 x 600 resolution and includes an LED backlight and aspect ratio of the screen is 16:9.
The hard drive moves at 5400 rotations per minute, allowing for fast loading times. It includes an integrated webcam holstering 1.3 Megapixels that captures screenshots of resolutions 1280 x 1024 or 640x480. There is no optical drive but it makes up for this with its very slim and compact design and fast wireless connectivity supporting LAN, given emphasis on cloud computing. The D225e features a 3-cell lithium ion battery that is capable of receiving 8 hours of web browsing during its full charge.
Other interfaces include a 2-in-1 SD Card or MultiMediaCard reader, as well as 3 USB 2.0 ports, a VGA port, Headphone Output, Microphone Input, and Local Area Network (Ethernet) port. Purchasing a new version of the computer comes with a preloaded version of Microsoft Office, including the entire Office 2010 Suite (Word, Excel, Powerpoint, etc.). The model is preloaded with the Windows 7 Stater OS and can be upgraded to Windows 7 Ultimate Season Pack 2 (Windows 10 upgrade is not recommended).
It is an excellent, fast, and very portable netbook that is suitable for almost any on the go computing job.


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Home biography Jessica Rogan Biography
Jessica Rogan Biography
Facts of Jessica Rogan Biography
Jessica Rogan
Scottsdale, Arizona, United States of America
Long Auburn
Relationship Statistics of Jessica Rogan
What is Jessica Rogan marital status ? ( married,single, in relation or divorce):
How many children does Jessica Rogan have ? (name):
Is Jessica Rogan having any relationship affair ?:
Is Jessica Rogan Lesbian ? No
Jessica Rogan is a married woman. Jessica Rogan and her husband Joe Rogan have been married for a long time since 2009. Her hubby is a TV character and performing artist who facilitated the acclaimed Fear Factor Dare arrangement. The couple got married following a time of their engagement. In May 2008, the couple progressed toward becoming guardians surprisingly. From that point forward, Jessica gives birth their second daughter, soon after a time of their marriage.Moreover, Jessica Rogan has added a little girl from her past relationship. Moving to her Husband's past relationship, he dated Jerri Manthey, before getting connected with to his better half Jessica. The couple was impractically required in a relationship in 1997. In addition, he likewise engaged with Lakana Zee in 2007. The couple is presently living together in Bell Canyon, California with their three kids.Moreover, Jessica Rogan has added a little girl from her past relationship. Moving to her Husband's past relationship, he dated Jerri Manthey, before getting connected with to his better half Jessica. The couple was impractically required in a relationship in 1997. In addition, he likewise engaged with Lakana Zee in 2007. The couple is presently living together in Bell Canyon, California with their three kids.
Who is Jessica Rogan?
Jessica Rogan is an American citizen. She is an excocktail waitress who is eminent as a life partner of a standup comedian, commentor, television host, actor and former Martial Arts Artist Joe Rogan. She is a model and stunt driver also.
Jessica Rogan's Childhood and Education History
Talking about her initial life, Jessica Rogan was born on June 28, 1978, in Scottsdale, Arizona, United States of America. Her original name was Jessica Lynne Schimmel. She is the girl of Vicki and Robert Schimmel who is a Comedian. Jessica went through her youth with her five kin Max Schimmel, Sam, Derek, Aaliyah, and Jacob as well. Jessica conveys an American nationality and has a place with the Caucasian ethnical background. As to her education, Jessica completed her College degree from 'California State University Long Beach'. She majored in Arts and Technical Theatre.
Jessica Rogan's Profession, Net Worth and Salary
Jessica Rogan began her career as a volunteer of shoreline cleanup group at 'Spare Our Beach'. She has furthermore done impermanent position at 'International Creative management'. In addition, Jessica was set apart with Chinese Modeling organization EM Model Management'. Henceforth, she worked in as a model for Chinese office for three months. Later in November 2010, Jessica modeled for Wholesale Company arranged in Korea 'Kingdom B. Whole Sale'. In February 2011, Jessica started working in as a model for Thailand based modeling organization 'The Only Models'. She worked in The Only Models for one year and three months. In May 2013, Jessica delighted in a relief from displaying and worked in as aproduct specialist at 'Volvo Motorsports'. Jessica also acted as office chief at 'Bromley Foods, Inc' for quite a while. At present, Jessica Rogan is working in as item epert at 'Volvo Car Group'. She is working at Volvo Car Group for quite a while and eight months. Jessica Rogan has done numerous shows in her profession, for example, 'Arman Noor FIVE Form Show', 'Amusement Unite Fashion Show', 'LA Fashion Week Project Ethos 'Hatchery', 'LA Fashion Week NCL Couture' and 'LA Fashion Week Lauren Elaine'. All through her demonstrating career, she has never done nude photo shoots. Jessica has gotten a decent salary from her profession work, yet the correct figure is obscure. However, her total net worth is still unrecognizable.
Jessica Rogan's Controversies and rumors
It has been over eight years of Joe's marriage with Jessica, and we don't see any division bits of gossip, extramarital issues, not by any means a debate in their relationship. Henceforth, she is abundantly worried about her career that is the purpose behind this she is far from gossipy tidbits and controversies.
Jessica Rogan's Body Measurements
Talking about her body estimations, Jessica has a flawless tallness remaining of 5 feet 8 inches and weighs around 49 kg. She has a body size of 32-24-34 inches. She has a thin body with green shaded captivating eyes and long reddish hair. References:(earnthenecklace.com,frostsnow.com,whosdatedwho.com) kids. Updated On Thu Oct 04 2018 Published On Thu Oct 04 2018 By Married Bio


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» RZA Comments On Tik Tok "Bing Bong" Trend & Discusses Wu-Tang Clan's Legacy In Today's Generation
RZA Comments On Tik Tok "Bing Bong" Trend & Discusses Wu-Tang Clan's Legacy In Today's Generation
jbanks Posted on December 9, 2021 at 9:38 am
photo credit: Slaven Vlasic l Getty Images
A true icon and pioneering legend in the hip-hop space, who has remained in a lane of his own, Wu-Tang Clan's RZA checked in with the Power 106 Mornings Letty and Teddy show to discuss his instrumental artistry and impact on crafting the staple of the hip-hop genre.
With a hand in portraying the story of Wu-Tang Clan in the captivating Hulu series, RZA speaks with Letty and Javen about the challenges and rewards of producing a monumental project on his group's come-up in the game. Speaking on the art and meaning behind spirituality, as well as the debut album 'Enter The Wu-Tang (36 Chambers),' the multi-faceted icon shares how his connected thinking changed his outlook on life and how it also inspired his own creativity.
RZA goes on to speak on the viral Tik Tok "Bing Bong" trend, which is somewhat sampled by his saying, shares his musical heroes, which includes Stevie Wonder, Issac Hayes, talks his legacy, and much more.
Check out the full interview below!
The Weeknd & Amazon Music Wants Fans To Relive 'DAWN FM' Livestream Experience
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Flo Rida & Miri Ben-Ari Honor MLK Jr. With New Collab + Name Florida's Hip-Hop Mount Rushmore
jbanks January 18, 2022
Tupac Shakur. Wake Me When I'm Free.
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The 26-year-old singer bested fellow musicians One Direction and Adele to claim the top spot on Forbes' 100 List. British boyband One Direction came in at second place with US $110 million while English singer-songwriter Adele earned the ninth spot with US $80.5 million.
Taylor Swift at last year's Billboard Music Awards.
Swift landed the number one spot by smashing the Rolling Stones' North American touring record. The 1989 World Tour grossed US $200 million in the continent alone before climbing to a quarter-billion dollars in total.
Apart from her concert earnings, Swift is also a product endorser for Diet Coke, Keds sneakers, and Apple.
One of Swift's ad campaigns for Keds sneakers.
Swift's 2016 income is the highest total of her career so far. She has more than doubled her earnings from last year's US $80 million sum.
The other top 10 spots are held by American author James Patterson (US $95 million), TV host and personality Dr. Phil McGraw and Real Madrid soccer star Cristiano Ronaldo were tied for fourth (US $88 million), comedian Kevin Hart (US $87.5 million), radio host Howard Stern (US $85 million), Barcelona footballer Lionel Messi (US $81.5 million), and radio host Rush Limbaugh (US $79 million).
Narrowly missing the top ten spots are NBA star LeBron James (US $77 million), pop icon Madonna (US $76.5 million), and TV host Ellen DeGeneres (US $75 million).
Taylor Swift with the highest paid celebrity couple, Beyonce and Jay Z, and singer Justin Timberlake.


Many people rely on walking and transit as their primary way to get around, and without a wide, clear path through snow and ice, it is especially difficult for people with disabilities, seniors, and children to walk safely. According to the Municipal Code of Chicago 10-8-180, property owners and occupants are responsible for keeping sidewalks clear of snow and ice.
Report locations that DO NOT clear their sidewalks by making a "Snow – Uncleared Sidewalk" request with the City of Chicago 311 Service Request line.
Make sure the problem occurs on the sidewalk.
NOT use this category to report snow on streets, parking lots, or alleys.
Provide a specific address where the problem occurs.
Request a reference number from the operator, this will help you track the status and resolution of your request.
Yes, the new sidewalk snow regulations affect all home, business, and property owners in Chicago. Business owners that rent space adjacent to sidewalks are responsible for shoveling snow under the ordinance. Some landlords for residential and commercial property hold tenants responsible for snow clearance as a part of their lease agreements, others don't. Renters who aren't certain of their shoveling responsibilities should check their rental agreements or ask their landlords for clarification.
DO NOT use this category to report snow on streets, parking lots, or alleys.
Request a reference number from the operator. This will help you track the status and resolution of your request.