How Do We Define Motel?Liittyminen ei maksa mitän." More "Good hotel in an excellent location.
bartender for about 40 customers. Assignments off property Cons The pay is not suffice to the job! " Less TripAdvisor Traveler Rating: Based on 103 reviews Read Recent Reviews 4704 Northeast Park Center Avenue, Lacey, WA miles from Little Creek Casino 14 miles La Quinta Inn Lacey From 99 Enter Dates From 99 Check In 15 00 Check Out 12 00 Rated. My room was on the 3rd floor, and the hotel didn't have an elevator. Casino Player Magazine Best of Gaming: Northwest 1st Place. Show More, cons, high turnover rate in some departments. You work way to hard to only be handed minimum wage. All the employees were nice and helpful. Hotel check-in clerk was rude and inconsiderate to me, but overly gracious to a male guest. Give it a try. I have been working at Little Creek Casino Resort full-time for more than 2 years. " Less TripAdvisor Traveler Rating: Based on 279 reviews Read Recent Reviews 4301 Martin Way East, Olympia, WA miles from Little Creek Casino 14 miles Hampton Inn Suites Olympia From 117 Enter Dates From 117 Check In 15 00 Check Out 12 00 Rated Very. Most Recent HotelGuides Reviews "I had a pleasant stay. All guests must be 21 to attend. Rooms, make yourself at home in one of the 190 air-conditioned rooms featuring refrigerators and flat-screen televisions. Golf Magazine Best New Courses, 2011 Best New Course You Can Play Mehr anzeigen. All the patrons that came threw were super generous I felt like I was at another home when I left. Able to stand or walk for long periods of time Showing 15 of 15 reviews Reviews Little Creek Casino Resort Reviews Little Creek Casino Resort Job Seekers Also Viewed Related Job Search Your response will be removed from the review this cannot be undone. 2011, golfweek Best New Courses, 2011 #8.S. Use this simple form to find hotels, motels, and other lodging near the Little Creek Casino in Shelton, Washington. With minimum wage now at 11 your to say were only a dollar more then minimum now compared to 3 when minimum was at 9?
We had a king bed, best Craps, and max the elevators were casinò dusty. Lenses greater than threeandonehalf inches in length. Casino Player Magazine Best of Gaming.
Little Creek Casino Resort in Washington features casino gaming, luxury hotel rooms, exciting dining destinations, a relaxing spa and robust nightlife.Little Creek Casino Resort sets the standard for gaming excitement in the Pacific Northwest.
More" t affect our stay, get a suite, wA miles from Little Creek Casino 16 miles Days Inn Lacey From 50 Enter Dates From 50 Check In 15 00 Check Out 11 00 Rated Low Not. More" m out of the military I lamore would definitely go back. But it didnapos, theyapos, and the entire place seemed understaffed. The hotel was older and needed some cleaning. And the room was clean and modern. Clean hotel with good customer service.
Pros, outstanding company to work for, they have wonderful health care benefits, very fun work environment, they also provide food in the employee lounges so you don't have to spend money on your meals while working, overtime is offered regularly.The rooms are comfortable, and the common areas are kept clean.
" "Nice hotel in a gorgeous area.
I'd recommend this hotel to family and friends." Less TripAdvisor Traveler Rating: Based on 460 reviews Read Recent Reviews 112 College Street Southeast, Lacey, WA miles from Little Creek Casino 14 miles Super 8 Hotel Lacey From 52 Enter Dates From 52 Check In 1400 Check Out 1200 Rated Very Low Not.Decent breakfast with hot and cold selections.


text:
Our favorite shows 10 top movies 10 top albums Critics loved these 17
'We've lost control!': 'Masked Singer' contestant removes mask before even being eliminated
Erin Jensen
Spoiler alert! This story contains details from the Sept. 30 episode of "The Masked Singer," including the identity of the revealed celebrity contestant.
Group B, the second batch of contestants introduced in the new season of "The Masked Singer," provided a first: an unmasking without an elimination.
The show introduced six new contestants Wednesday, following the costumes introduced in the show's Season 4 premiere last week.
Newcomers Crocodile, Seahorse, Gremlin, Whatchamacallit, Serpent and Baby Alien all made their stage debuts, impressing panelists Ken Jeong, Jenny McCarthy, Nicole Scherzinger and Robin Thicke with their performances. But for one challenger, because of his own decision, it would be his first and final show.
Here's how the evening's history-making episode went.
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Clues: Crocodile revealed that, like his shiny pink stage persona, he is "happiest in water." The star referenced "growing up in Hollywood" and said he was excited to reveal his "hidden talent." Pairing that with his performance of Bon Jovi's "It's My Life" leads us to believe this contestant is celebrated for something other than singing. One of the best parts of his performance? His long pink tail that bopped around as he moved.
Guesses: Jeong delivered the far-fetched guess of "Mad Men" star Jon Hamm. Scherzinger felt her guess was warm with 98 Degrees singer Nick Lachey, while Thicke thought it could be another boy bander, one married to his fellow panelist, McCarthy: Donnie Wahlberg.
Baby Alien
Clues: Baby Alien revealed that in his heyday he was "in a theater every week – even beamed onto the Tony Awards stage, baby." His clue package also included allusions to the 1986 Matthew Broderick comedy "Ferris Bueller's Day Off."
Baby Alien is the first "Masked Singer" costume to have a moving mouth, but his performance of George Michael's "Faith" showed his puppeteer skills could use some work. Baby Alien's mouth wasn't exactly in sync with the song's lyrics, though we can't imagine how hard it must be to focus on singing, your stage presence and moving the mouth of an infant extraterrestrial all at once.
Guesses: McCarthy suspected it could be "Cobra Kai" star Ralph Macchio, while Scherzinger thought it could be "Friends" actor David Schwimmer. Jeong got Freddie Prinze Jr. vibes.
Chris Wallace lost control of the Trump-Biden debate on a historically bad night of TV
Gabrielle Union, NBC reach resolution in dispute over 'AGT' firing, racism allegations
Clues: Seahorse's clue package featured a Western theme, and showed her around a campfire with poodles and teacups. Thicke pointed out a hog with warts, in her package, interpreting it as a reference to Hogwarts, Harry Potter's magical school.
Seahorse said she wanted "to show shiny new sides of myself," leading us to believe she might be a country singer looking to shed her image. Her rendition of Rihanna's "Only Girl (In The World)" revealed her to be the strongest singer so far in the night.
Guesses: McCarthy wondered if the underwater creature could be singer Halsey, and Scherzinger thought of "Pitch Perfect 3" actress Hailee Steinfeld. Thicke threw "I'm a Mess" singer Bebe Rexha into the mix.
Whatchamacallit
Clues: The hair-covered, one-eyed Whatchamacallit said he has been perceived as a shy guy, though he's one of many talents. "I can dance with the stars," he said, "surf candy bars, buzz with the bees and swing from the keys."
Guesses: After hearing Whatchamacallit's version of Skee-Lo's "I Wish," Scherzinger thought of record producer Swizz Beatz. Jeong envisioned basketball player Damian Lillard and Thicke thought the challenger could be musical artist Tyler, the Creator.
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Clues: Serpent let it s-s-slip that he was a mischievous youngster whose dad helped to deter him from a self-destructive path. "I didn't want to cause pain, but heal pain," Serpent said.
His slowed-down rendition of The Proclaimers' "I'm Gonna Be (500 Miles)" was one of the best performances of the night, positioning Serpent as serious competition for Seahorse.
Guesses: McCarthy thought it was possible "Conversations in the Dark" singer John Legend was singing in the darkness of Serpent's mask. Jeong envisioned actor/singer Daveed Diggs and Scherzinger guessed Tony winner Leslie Odom Jr.
Clues: Gremlin's clue package made mention of his "checkered history." He admitted he "can be a bit combative." But the challenger also has a soft spot for animals and enjoys cooking.
His performance of Ben E. King's "Stand By Me" revealed a mature voice – and someone who didn't always play by the rules, the panel would soon find out.
Guesses: Thicke had "Great Balls of Fire" singer Jerry Lee Lewis on his mind, while McCarthy thought of actor Mickey Rourke. Jeong figured former wrestler "Stone Cold" Steve Austin.
The elimination that never was
After their guesses, Gremlin declared, "I want to take this (mask) off right now," causing all sorts of confusion.
"We've lost control! It's the Gremlin show," show host Nick Cannon exclaimed.
And without being eliminated, Rourke, McCarthy's guess, removed his mask and tossed it away.
The panelists erupted in shock.
"I don't know what's happening, ladies and gentlemen," said Cannon. "This is a 'Masked Singer' first ...Mickey said, 'Dammit, it's hot. I'm taking it off.' I love it!"
Gremlin joked that he decided to partake in the show because was simply "in the neighborhood," spurring laughter.
The others in Group B remain, as Gremlin eliminated himself with the reveal.
Episodes of "The Masked Singer" can be watched on Fox's website.

redpajama_set_name: RedPajamaCommonCrawl


text:
Collegio elettorale di Forlì – collegio elettorale della Camera dei deputati del Regno di Sardegna
Collegio elettorale di Forlì – collegio elettorale della Camera dei deputati dal 1861 al 1919
Collegio elettorale di Forlì – collegio elettorale della Camera dei deputati dal 1993 al 2005
Collegio elettorale di Forlì – collegio elettorale del Senato della Repubblica dal 1993 al 2005
Collegio elettorale di Forlì-Faenza – collegio elettorale del Senato della Repubblica dal 1948 al 1993

redpajama_set_name: RedPajamaWikipedia


S3WebLabs provides services in website design, development, & search engine optimization for small businesses, non-profit, and home businesses.
As a small business, S3WebLabs is keenly aware of the limited resources that a small business has available for its online presence.
S3WebLabs can help your business grow by offering you a fresh and dynamic approach to either beginning or improving your website.
If you would like more information on the services we provide please complete the online form found under our contact us link.
Walkersville, Germantown, Myersville, Thurmont, Boonsboro, Hagerstown, Clarksburg.


Does Mase still got it?! Plans to drop a new album!
So Mase, the rapper turned Pastor then rapper again tweeted that he will be releasing a new album. No its not a Gospel album or a greatest hits album, its BRAND NEW! According to his twitter page its going to be titled "Now We Even" Its been 9 years since Mase dropped his last album"Welcome Back". I doubt that his newest offering will drop in 2014, so looks like we'll just have to wait until 2015 to see if Ma$e still has it!


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The Bob Basten Award is given annually to a former St. John's University football player who, in his life as an alumnus, has exhibited outstanding leadership.
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Denny Schleper Receives Basten Award from St. John's University
Denny Schleper
Minneapolis, Minnesota — October 3, 2016 — Denny Schleper, CEO of CliftonLarsonAllen (CLA), is the 2016 recipient of the Bob Basten Excellence in Leadership Award from St. John's University. A 1983 graduate of St. John's and quarterback of the Johnnies football team, Schleper was presented with the award at halftime of the school's homecoming football game in Collegevillle, Minnesota, in September.
The Bob Basten Award is given annually to a former SJU football player who, in his life as an alumnus, has exhibited outstanding leadership traits, is involved in his community and alma mater, has a sense of fair play carried beyond the gridiron, and has an overall commitment to excellence.
Established in 2009, the Basten Award arose from a conversation between three friends from SJU's Class of 1982 who wanted to salute Johnnie football players who represent the program's hallmark traits. One of those friends was Bob Basten, a former Minnesota Viking and business executive who died in 2012 after 10 years of battling ALS/Lou Gehrig's Disease. Schleper is the eighth recipient of the award.
"This is a great honor for me, in part due to the remembrance of the person the award is named for," Schleper said. "The thing I recognize now about my time at St. John's was how they develop young men to become leaders. On the football team we called our own plays, which allowed us to play and win as a team and learn from our mistakes."
Pictured above: Denny Schleper and his wife, Cookie, accept the Bob Basten Award at St. John's University in September.
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Single Family Home: There were 2 Single Family Homes SOLD in Quarry Park in 2012.
View Quarry Park Listings Now or have a look around, fill out the contact Quarry Park REALTOR® form or call (403) 619-1019 with any questions you might have about Quarry Park homes for sale, the Quarry Park Real Estate Market or want to view any of the Quarry Park Calgary listings. No questions is mundane!
Register with todayscalgaryhomes.com By registering with my site, you'll have VIP access to new listings in Quarry Park Calgary properties as soon as they hit the market.
Are you thinking of making a Real Estate move to Quarry Park? Quarry Park is part of the Southeast Calgary Communities. Quarry Park community consist of different types of dwellings from Single Family Home, Condominiums and Townhouses. For Buyers or Sellers you can start your search in Quarry Park Real Estate MLS® search here by viewing all the Quarry Park Homes for Sale with Google interactive maps, Quarry Park Real Estate Statistics, Home Descriptions and seperating Single Family Homes to Condo / Townhouse for sale in Quarry Park.
Contact a Quarry Park REALTOR® or Quarry Park real estate agent today.
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Let local Quarry Park real estate agent guide you through the sale of your current Quarry Park home or purchase your next Quarry Park property. As Quarry Park REALTORS® and real estate agent, Im ready to work with you today!


To fund a fellowship and transition program that enables veterans to explore, discover and go all in on their passions by empowering them to embrace new challenges through leadership, a meaningful education, fitness and employment.
Ultimately, we want to make the University of Nebraska–Lincoln and the great state of Nebraska the most veteran friendly place in America.
Since 2001, the war on terror has become a publicly recognized endeavor and over 50,000 American servicemen and women have been wounded. Each wounded service member and his or her family must now face challenges which are beyond the experiences of most Americans. As these wounded men and women begin the tough transition back to civilian life, it is becoming increasingly apparent that the benefits provided to these individuals and the actual costs of obtaining a college education are distinctly different. The Big Red Challenge wants to provide these individuals the means to successfully complete higher education.


text:
Home Sports Shakib Al Hasan Performance Heads Bangladesh's win against Windies
Shakib Al Hasan Performance Heads Bangladesh's win against Windies
Vastavam web: If his innumerable stunners from the past weren't already enough,Shakib Al Hasan has once again shown his mettle as one of the greatest Bangladeshi cricketers to have played the game. Spearheading his team's cause in its latest encounter against West Indies, the all-rounder shone with both bat and ball to help Bangladesh register a memorable win in the ongoing World Cup.
Chasing a score of 322, Bangladesh had their task cut out against a hostile pace attack. After getting off to a decent start, the men in green lost Soumya Sarkar in the 9th over. The loss brought Shakib to the crease, and the left-hander joined hands with Tamim Iqbal to thrash the Windies bowlers until the 17th over before Iqbal fell prey to a run-out. The experienced Mushfiqur Rahim couldn't trouble the scorers much and was sent back packing after scoring a solitary run.
With game evenly poised at this stage, in-walked Litton Das and there was no looking back for the Asian side. Das provided an able support to Shakib, who looked in red hot form. The southpaw continued with his purple patch in the tournament as he took the bowlers to the cleaners. The 32-year-old carried along his merry way and notched up his second consecutive century of the tournament to bat Windies out of the game. Shakib carried his bat through and remained unbeaten on 124 that came off just 99 deliveries. Interestingly, the left-handed batsman even surpassed Rohit Sharma and Aaron Finch in the process to stand as the tournament's highest run-getter as of now.
Shakib was adjudged as the Man of the Match for his all-round show, which proved to be the cornerstone of Bangladesh's victory. And it was not just the award jury that was in awe of the all-rounder skills.
innumerable stunners
Litton Das
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It was critical of India's performance against West Indies: Kohli
Jhonty Rhodes will be joining Tamil Nadu Premier League as mentor


We're joined by Rachel Ritlop, the owner of The Confused Millennial website and podcast!
What are the biggest challenges with being on that work-from-home grind? And are more people hopping on the freelance train?
The panelists have a classic millennial issue in common: They all worry about the future.
This week's episode is sponsored by Homesick Candles (use promo code MILL for 10% off a $50+ order and free shipping), Lola (use promo code MILL for 40% off a subscription), and Policy Genius!
Pam recently visited Disneyland where it's impossible to find short lines.
In better news for Pam, she's finally settled on a therapist. She talks about how she found the right one.
In other mental health news, Andrew is finally off of Lexapro. However, he wonders if he'd be better off continuing to take small doses.


April 04, 2019 • Volume 18, Issue 6 • A twice monthly newsletter with information about workplace safety and health.
A meeting will be held May 14 in Washington, D.C., to solicit public comments and suggestions from stakeholders on issues facing OSHA in the administration of the whistleblower protection provisions under the Occupational Safety and Health Act. OSHA is seeking input on improving whistleblower customer service and enhancing understanding of whistleblower laws the agency enforces. The meeting will be held at the U.S. Department of Labor's national headquarters. Those interested must register by April 30. For more information, see the Federal Register notice. OSHA previously held stakeholder meetings on whistleblower protection in the transportation and financial sectors.
Governors Laura Kelly of Kansas, John Bel Edwards of Louisiana, Michael L. Parson of Missouri, and Pete Ricketts of Nebraska signed proclamations recognizing March 25-29 as Stand-Up for Grain Safety Week. The purpose of the event is to increase awareness of grain industry hazards, including bin entry, machine guarding, falls, heat, combustible dust, and struck-by. The governors recognized the efforts of participating organizations to ensure worker safety with proper training, supervision, and equipment. More information is on the Stand-Up for Grain Safety website.
Dowa THT America Inc. was cited for exposing workers to atmospheric, thermal, electrical, and mechanical hazards while they performed maintenance inside heat-treating furnaces. The company was cited for 25 violations of confined spaces, fall, machine guarding, respiratory, chemical, and electrical safety and health standards. Dowa THT faces $1,339,596 in penalties and has been placed in the Severe Violator Enforcement Program. Read the news release for more information.
California OSHA issued four citations and $75,750 in penalties to Accurate Comfort Systems Inc. after a worker suffered serious injuries in a fall from a 12-foot-high work area. Inspectors determined that the company allowed workers to stand on the ladder's top cap, and failed to ensure the ladder was secured and extended to the appropriate distance. The company was previously cited for these violations in 2017.
AMTEC Less Lethal Systems Inc. faces $188,290 in penalties after an explosion fatally injured two workers. The company was cited for failing to protect and train workers when it increased the maximum explosive limits of pyrotechnic flash powder. Read the news release for more information.
OSHA cited the operator of a Chili's Grill and Bar restaurant for exposing workers to burns, falls, and other safety and health hazards. A worker was burned after falling from an unguarded platform into a hot water bath. Brinker Florida Inc. was cited for failing to train workers on hazardous chemicals, ensure inspection of portable ladders, and provide suitable facilities for flushing chemicals from workers' eyes. The company faces $62,513 in penalties. Read the news release for more information.
Corley Contractors Inc. faces $106,078 in penalties for exposing workers to excavation hazards while installing water and sewer lines. OSHA initiated an inspection as part of the agency's National Emphasis Program on Trenching and Excavation after observing workers exposed to trenching hazards. Inspectors determined that the company failed to install a protective cave-in system inside the excavation area, and provide a safe means to enter and exit the excavation. For more information, read the news release.
Michigan OSHA issued seven citations and $65,000 in penalties to Powder Cote II for failing to provide fall protection or guardrail systems, guard rotating shafts and machinery, and failing to control the startup of machinery during maintenance. The company was previously cited for these violations in 2014.
Students test their skills working with sheet metal at the CEFGA CareerExpo.
The Construction Education Foundation of Georgia (CEFGA) held its annual CareerExpo March 21-22. The event featured hands-on displays that allowed students to interact with leaders in the construction, energy, and mining industries. This year more than 5,000 students joined teachers, volunteers, and school administrators at the expo. OSHA safety and health professionals and whistleblower investigators provided information and educational material.
OSHA is collaborating with the U.S. Federal Highway Administration, state of Georgia, and local governments and employers to support safety stand-down events during National Work Zone Awareness Week, April 8-12. The event focuses on safety hazards workers face during road improvement and repair projects. The kickoff event will be held at the Frederick Douglass Memorial Bridge in Washington, D.C., on April 9 at 11 a.m. Georgia is encouraging employers to stop work voluntarily from 7-8 a.m., one day during the week, to discuss struck-by hazards and best safety practices. Read the news release for more information, and visit Georgia's Struck-By Alliance Safety Stand-Down page.
Tennessee OSHA is co-sponsoring Safety Fest Tennessee April 29-May 3 in Oak Ridge and Knoxville to provide free workplace safety and health training. Class topics include fall prevention, chain saw safety, recordkeeping, and OSHA hazardous waste operations. Visit the Safety Fest TN website to register for classes.
Carroll Coolers Inc. (CCI), a manufacturer of retail ice equipment and walk-in coolers and freezers, contacted Iowa OSHA's On-Site Consultation Program for help lowering injury and illness rates and improving overall workplace safety. Working with state consultants, CCI implemented new worker training methods and formed a safety committee that includes workers from all levels of the organization. Reducing reportable injuries and improving safety conditions at CCI contributed to lowering workers' compensation premiums. The company is a participant in the Safety and Health Achievement Recognition Program. For more details, read the success story.
OSHA and Lee Lewis Construction recently signed a partnership agreement to help protect workers during construction of the Buddy Holly Hall of Performing Arts and Sciences in Lubbock, Texas. The partnership will focus on preventing fall, electrical, trenching and excavation, struck-by, and caught-between hazards. The partners will also encourage contractors and subcontractors to exchange information on hazards identified during construction activities.
The Brewers Association, Master Brewers Association of the Americas, and the Colorado Brewers Guild signed a two-year agreement to help protect the health and safety of workers in Colorado's brewing industry. Participants will provide information and training resources on preventing exposure to hazards during the brewing, storing, containerizing, handling, and delivery of beer.
The Colorado Contractors Association signed a two-year agreement to help protect the health and safety of construction workers. Participants will provide information and training resources on preventing exposure to work zone, fall, trenching and excavation, electrical, and silica hazards.
The National Association of Minority Contractors-Dallas Fort Worth Chapter signed a two-year agreement to provide information and training resources on preventing exposure to fall and struck-by hazards, and understanding workers' rights and employers' responsibilities.
The Suncoast Utility Contractors Association renewed a five-year agreement to address underground utility construction hazards, such as digging trenches and excavations, and entering confined spaces.
The United Safety Council renewed a five-year agreement to develop safety and health programs related to workplace violence, first aid, CPR, and automated external defibrillators. The participants will continue to communicate these programs to hard-to-reach groups, that include non-English speaking and youth workers.
OSHA requires employers to limit workers' exposure to respirable crystalline silica on the job. A video launched in January 2019 focuses on handheld power saws, showing how an integrated water delivery system controls silica dust created when cutting certain materials.


Can you believe we've heard the annoying yet endearing laughter of SpongeBob SquarePants for ten years now? Today marks the first day of the SpongeBob fan site.
Can you believe we've been hearing the annoying yet endearing laughter of SpongeBob SquarePants for ten years now? Ten! An entire decade of SpongeBob. So what better way to commemorate this big birthday than by finally giving the little tyke a Web site of his own? He will no longer have to share space in the Nickelodeon Web site: Today marks the first day of the standalone SpongeBob fan site.
The site is age-appropriate, although both kids and adults will probably find it entertaining. You can make a personalized profile, chat, and create playlists of your favorite SpongeBob videos. The Web site allows you to search for your favorite SpongeBob moments and comment on your favorite videos. Every week brings a new mini-game, and there are already tons of games to play.
With every new episode that airs on Nickelodeon, the full episode will be available on the site. A selection of full-length episodes are already available.
There's also a backstage area that has a blog from the creators of SpongeBob. The blog has interviews with the crew behind the cartoon, as well as tips on how to draw characters from SpongeBob.
For fans of SpongeBob, this is a haven for you to comment about your favorite and least favorite SpongBob episodes, as well as learn about what goes on behind-the-scenes of your favorite cartoon.


Carson was a guest at Books & Arts Daily in Sidney to discuss Theater and Social Discourse. The show is archived ABC RadioNational.
Read the full Behind the Eye review in CityBeat.
Read an interview with Carson Kreitzer by Rick Pender in CityBeat.
Read the full Behind the Eye review in the Cincinnati Enquirer.
Read an interview with Carson Kreitzer by Jackie Demaline in the Cincinnati Enquirer.
Read an Behind the Eye interview in Dialogue.


Hennessy VS, La Goya Manzallia Sherry, House Falernum, Pink Grapefruit, Lemon. Served Up.
Lavender Infused Beefeater, Bernardin Muscat, Passionfruit, Fresh Citrus. Shaken Sour, Served Up.
Calvados, Pimm's liqueur, peach liqueur, hint of citrus topped with Champagne Foam Served in a flute.
Cranberry, lemon, raspberry, ginger ale Served long.
Pear, apple, cinnamon Served long.
Old & Rare meets traditional cocktail craftsmanship for something truly unique.


text:
require "spec_helper"

describe Story do
  it "should get a short id" do
    s = Story.make!(:title => "hello", :url => "http://example.com/")

    s.short_id.should match(/^\A[a-zA-Z0-9]{1,10}\z/)
  end

  it "requires a url or a description" do
    expect { Story.make!(:title => "hello", :url => "",
      :description => "") }.to raise_error

    expect { Story.make!(:title => "hello", :description => "hi", :url => nil)
      }.to_not raise_error

    expect { Story.make!(:title => "hello", :url => "http://ex.com/",
      :description => nil) }.to_not raise_error
  end

  it "does not allow too-short titles" do
    expect { Story.make!(:title => "") }.to raise_error
    expect { Story.make!(:title => "hi") }.to raise_error
    expect { Story.make!(:title => "hello") }.to_not raise_error
  end

  it "does not allow too-long titles" do
    expect { Story.make!(:title => ("hello" * 100)) }.to raise_error
  end

  it "must have at least one tag" do
    expect { Story.make!(:tags_a => nil) }.to raise_error
    expect { Story.make!(:tags_a => [ "", " " ]) }.to raise_error

    expect { Story.make!(:tags_a => [ "", "tag1" ]) }.to_not raise_error
  end

  it "checks for invalid urls" do
    expect { Story.make!(:title => "test", :url => "http://gooses.com/")
      }.to_not raise_error

    expect { Story.make!(:title => "test", url => "ftp://gooses/")
      }.to raise_error
  end

  it "checks for a previously posted story with same url" do
    Story.count.should == 0

    Story.make!(:title => "flim flam", :url => "http://example.com/")
    Story.count.should == 1

    expect { Story.make!(:title => "flim flam 2",
      :url => "http://example.com/") }.to raise_error

    Story.count.should == 1

    expect { Story.make!(:title => "flim flam 2",
      :url => "http://www.example.com/") }.to raise_error

    Story.count.should == 1
  end

  it "parses domain properly" do
    s = Story.make!(:url => "http://example.com")
    s.domain.should == "example.com"

    s = Story.make!(:url => "http://www3.example.com/goose")
    s.domain.should == "example.com"

    s = Story.make!(:url => "http://flub.example.com")
    s.domain.should == "flub.example.com"
  end

  it "converts a title to a url properly" do
    s = Story.make!(:title => "Hello there, this is a title")
    s.title_as_url.should == "hello_there_this_is_a_title"

    s = Story.make!(:title => "Hello _ underscore")
    s.title_as_url.should == "hello_underscore"
  end

  it "is not editable by another non-admin user" do
    u = User.make!

    s = Story.make!(:user_id => u.id)
    s.is_editable_by_user?(u).should == true

    u = User.make!
    s.is_editable_by_user?(u).should == false
  end

  it "can fetch its title properly" do
    s = Story.make
    s.fetched_content = File.read(Rails.root +
      "spec/fixtures/story_pages/1.html")
    s.fetched_title.should == "B2G demo & quick hack // by Paul Rouget"

    s = Story.make
    s.fetched_content = File.read(Rails.root +
      "spec/fixtures/story_pages/2.html")
    s.fetched_title.should == "Google"
  end

  it "has an underscored url from its title" do
    s = Story.make(:title => "The One-second War (What Time Will You Die?) ")
    s.title_as_url.should == "the_one-second_war_what_time_will_you_die"
  end

  it "sets the url properly" do
    s = Story.make(:title => "blah")
    s.url = "https://factorable.net/"
    s.valid?
    s.url.should == "https://factorable.net/"
  end

  it "calculates tag changes properly" do
    s = Story.make!(:title => "blah", :tags_a => [ "tag1", "tag2" ])

    s.tags_a = [ "tag2" ]
    s.tagging_changes.should == { "tags" => [ "tag1 tag2", "tag2" ] }
  end

  it "logs moderations properly" do
    mod = User.make!(:username => "mod", :is_moderator => true)

    s = Story.make!(:title => "blah", :tags_a => [ "tag1", "tag2" ],
      :description => "desc")

    s.title = "changed title"
    s.description = nil
    s.tags_a = [ "tag1" ]

    s.editor_user_id = mod.id
    s.moderation_reason = "because i hate you"
    s.save!

    mod_log = Moderation.last
    mod_log.moderator_user_id.should == mod.id
    mod_log.story_id.should == s.id
    mod_log.reason.should == "because i hate you"
    mod_log.action.should match(/title from "blah" to "changed title"/)
    mod_log.action.should match(/tags from "tag1 tag2" to "tag1"/)
  end
end

redpajama_set_name: RedPajamaGithub


Replace your branded shampoo and conditioner with homemade shampoo and conditioner. This is a very good hair mask that promotes hair growth. Ask your hair colourist if this can work for you. Including fruits and green leafy vegetables in your diet will keep your body healthy, thereby providing the necessary nourishment to the follicles through the blood vessels.
Learn to reduce stress. Give your treasured mane some love and always get a little snip every six to eight weeks, then the hair will grow strong and healthy.
Constant brushing can cause physical harm to your hair.
To minimize damage, avoid brushing your hair when it's wet, and never use heating irons curling or flattening on wet locks. Over a period of time, your hair will most definitely look stronger and feel healthier.
Instead, if you're looking to work out and want your hair off your neck, gently twist your hair into a bun and clip it with a tiny claw clip, bobby pin, or larger pin. Distribute your hair's natural oil. If you're using a curling or flattening iron, minimize the time it contacts your hair, limiting it to 3-4 seconds per section. And that makes you realize how precious every inch is. Aim for 2.
Styling your hair in loose waves provides an illusion of longer hair. Having long, strong hair doesn't just depend on which products you put on your hair; it also depends on what you put into your body.
LR Liz Rivers Aug 1, 2017.
Unanswered Questions. You want to target where the oils are produced, which is at the hair follicles, located under the scalp. Apply an oil or mask treatment weekly. As chic as platinum hair looks, going from a darker shade to light blonde could stand between you and your longest-possible hair. Proper nourishment through food and required vitamins can boost hair growth and add luster to your hair. We know finding the right shampoo and conditioner is a big task.
Stop doing trendy "cleanses. HAIR 66 photos that prove a fringe can change your entire face 6 Feb 2019. Master at-home entertaining with this recipe for pink gin punch. Oils like castor oil, jojoba oil, and avocado oil boost hair growth. Avoid heat styling tools. Keep in mind there are various kinds of treatments on the market for different hair types, which always require the consumer to read instructions on timing and application.


State Assembly candidate and Palo Alto City Councilmember Marc Berman announced today that he has earned the endorsement of American Federation of State, County, and Municipal Employees, AFSCME Council 57.
An affiliate of the American Federation of State, County, and Municipal Employees, AFSCME Council 57 represents over 25,000 members of 23 locals representing those who work in county health and human service agencies, provide mental health services at the state and local level, and ensure the public safety working as 911 dispatchers, probation officers, and trial court employees. AFSCME Council 57 members serve the people of California through their work in schools and community colleges, transit agencies, public works departments, clinics and hospitals, and water and wastewater facilities and more.
With their endorsement of Berman, AFSCME Council 57 joins numerous state and local leaders in supporting Berman for State Assembly, including incumbent Assemblymember Rich Gordon, Lieutenant Governor Gavin Newsom, Speaker pro Tempore Kevin Mullin, Santa Clara County Sheriff Laurie Smith, and Supervisors Don Horsley and Warren Slocum, as well as former Supervisor Rose Jacobs Gibson. For a full list of Berman's endorsements, please visit www.voteberman.com/endorsements.


The Baker Zorilla Corner Pocket Skateboard Deck is the perfect addition to your skateboard arsenal. It works for everyone, beginner or pro. It measures 8.12" wide x 31.5" long and offers a lot of pop making it ready for every landscape, whether you're in the park, the pipe or on the rail!


text:
A Talk With Michael Thorpe
This talk was setup in the Boston commons, around 10pm after a street photography session with Mike. Random Questions, single strobe, my Canon 5D classic and a bench. Another spontaneous talk session.
OVP: Why do you love photography? MT: ummm I think the reason I love photography is just like knowing everyday you can capture moments that you can keep for the rest of your life. OVP: How do you feel when you capture those moments? MT: It makes me think about the moment and realize the importance of that moment. I started to understand that just recently, when my mom was in the hospital. OVP: Why was your mom in the hospital? MT: She needed surgery done.
OVP: What is the hardest thing you have gone through recently? MT: Losing my good friend who died on the basketball court while we were playing basketball. OVP: You witnessed his death? MT: Yeah, It was our first summer league game........ it was like the end of the first half and umm he um ..... collapsed and I still don't know why he passed away, It was crazy. OVP: How did you feel when it happened? MT: I was scared, confused and still to this day don't know what happened and can't understand it. It's just crazy.
OVP: How did you cope with it? MT: I talked, it was tough because one of my coping mechanisms is basketball. I didn't want to play it because of my loss, so I talked, I talked a lot and took pictures. I started to value my photography more and I started to realize how precious every moment is. I have a lot of pictures with him and that's all I have of him, those pictures, those memories. After people pass away you are not going to have that physical being, so you have to have another way to remember how they look. I started looking at the pictures we had together and that made me realize why I really like photography. Now I understand why parents take so many pictures of their children, once they grow up you will never be able to capture those childhood moments again.
MT: That's how I coped with it. Even though it was a tough experience, it was a great learning process. OVP: Was that your first time experiencing death? MT: Actually it's funny that you asked that. My grandfather died my junior year of high school and that was extremely tough for me. Even though I wasn't really close to him we were getting close and hanging out a lot. After he passed away my mother told me how similar we were, I started sharing interest in things that he really liked. He was big into photography, was in a printing business for most of his life printing pictures, taking pictures, it's crazy that even though I didn't know him that well, we were so similar in so many ways. That part of loosing him and not really knowing him and later finding out we were so similar is tough. It sucks because I can't do anything about it, I'm never going to see him again, it's just crazy to think about.
OVP: How do you feel about living your own life? What are your thoughts about when you go? What would you want to leave behind? MT: I thought about this a lot after Joey died. I don't want to be famous, I want to touch peoples lives. I want to start telling stories with my photography.
OVP: What are the three most important things you have learned in your life? MT: Umm one, cherish every moment you have. Two, make sure you let the people you love know that you love them. After losing Joey, I think I didn't express how much I cared about him and how much he was a good friend to me and now I can never tell him. OVP: How does that make you feel? MT: It sucks, because of that experience I tell people at least once a day, how much I care about them. MY third one is be happy, just make sure you are happy everyday and be around people that make you happy.
OVP: What does that tattoo mean to you? MT:This one is one of my first tattoos I got. It says family first, inside of the word family has the names of my family members. If you look at it closely you wouldn't see my fathers name because he hasn't really been in my life.
OVP: How is your relationship with your father MT: It is what it is. I lived with him when I was under 10 years old. Honestly I don't hate the guy he wasn't around for my life, he chose a path that didn't involve me. My sister tells me to reach out to him but it's not worth my time, he lives all the way in Florida. OVP: How do you feel not having a father in your life? MT: When I was younger um it was really was tough for me I was always asking where was my dad, then after a while I started finding other father figures. Good friends fathers became father figures to me. I started hanging out with older guys who taught me things I wouldn't have learned from my peers. I wouldn't know how living with a father would be, because I never had one. Not having a father doesn't bother me, my mother took over the role and to this day is there for me.
OVP: What advice would you give to those who are having a hard time dealing with the loss of a love one? MT: Don't bottle it up, the more you talk about it the easier it will get to accept it.
talk projectAnthony Adamick June 30, 2013 death, family, friends, love, photography, canon, 5d, people, talk, photos, boston, boston commons, portraits, bw, black and white
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Syria protest 9/9/13
Anthony Adamick September 9, 2013
A Talk With Spencer Braithwaite
peopleAnthony Adamick June 17, 2013 basketball, people, nike, canon, canon5d, 5dclassic, mom, dad, brother, family, love, color, flash, strobe, life, boston, nba, talent


Andy Oliver, global client engagement officer at PR agency Lewis, is to leave the business after 20 years, with Oliver saying he's in discussion with both agencies and brands about his next move.
Oliver joined Lewis PR, as it was then known, in 1996 as an account manager when it had around a dozen employees at its office in London's Covent Garden.
He established and headed the agency's US presence aged 26 with its first office in the States, in San Diego, and later set up Lewis' office in Los Angeles before moving to Singapore as senior vice president, APAC.
Oliver stayed in Singapore for five years, before moving to Hong Kong with the same role to oversee the acquisition of EBA Communications China. Last October he moved back to London to take his current position.
Oliver said: "I've had a fantastic round-the-world journey with the agency; proud that as part of the core team we grew the agency from startup to put it on the global map.
"I've been lucky to have had the opportunity to build the business in the US, Asia and EU. But after 20 years I think a change is as good as a rest. I want to apply my global experience into a different organisation, have new opportunities to learn in a different environment."
Regarding his next move, Oliver said: "I am in conversation with both agencies and brands. I'll take my time to decide what the best fit and opportunity is. I love challenges and I'm looking for my toughest assignment yet."
Lewis is the 28th biggest agency in the world by revenue, according to PRWeek's Global Agency Business Report 2015. It generated $68.6m (£51.6m) last year, a rise of 12 per cent on 2014.


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

A transitive group of permutations $G$ of a set $\Omega$ is called {\it primitive} if the only $G$-invariant equivalence relations on $\Omega$ are the universal and trivial relations.

Often, to say something useful about an infinite structure, one must first impose some finiteness condition upon it. In this paper we concern ourselves with infinite primitive groups whose point stabilisers have only orbits of finite length in $\Omega$. These orbits are called the {\it suborbits} of a group, and the cardinalities of these orbits are called the {\it subdegrees}. If the subdegrees of a group are all finite, the group is called {\it subdegree finite}. Thus, in this paper we examine infinite subdegree finite groups; such groups have been the subject of much research (see \cite{adeleke&neumann}, \cite{praeger} and \cite{me:SubdegreeGrowth} for example).

Associated with any permutation group $G$ there are directed graphs, called the {\it orbital digraphs} of $G$, which represent $G$ and its action on $\Omega$ in a very natural way. If $\alpha, \beta \in \Omega$ the set $(\alpha, \beta)^G = \{(\alpha^g, \beta^g) \mid g \in G\}$
is called an {\it orbital} of $G$. It is {\it diagonal} if $\alpha = \beta$. An {\it orbital digraph} of $G$ is a digraph whose vertex set is $\Omega$ and whose set of directed edges (or {\it arcs} as they are usually called) is an orbital of $G$.
Orbital digraphs of subdegree finite primitive groups are {\em locally finite}; that is, every vertex is adjacent to only finitely many vertices. Our digraphs are without loops and multiple edges, and unless otherwise stated the terms {\it path} and {\it distance} are used instead of the cumbersome {\it undirected path} and {\it undirected distance}. A digraph $\Gamma$ is {\it connected} if there is a path in $\Gamma$ between any two distinct vertices. A connected component of $\Gamma$ is a maximally connected subdigraph. Because the connected components of a digraph $\Gamma$ induce an equivalence relation on the set of vertices of $\Gamma$, an equivalence relation that is invariant under automorphisms of $\Gamma$, the non-diagonal orbital digraphs of a primitive group $G$ are connected.

For locally finite graphs and digraphs, an {\it end} is an equivalence class of {\it rays} (one-way infinite cycle-free paths), where two rays are equivalent if and only if they have infinitely many vertices in common with a third ray. Intuitively, one may think of them as the ``points of infinity'' of a graph or digraph. An end is {\it thin} if any pairwise disjoint set of rays in it is finite; otherwise it is {\it thick}. It was shown in \cite{me:OrbGraphs} that the ends of any two non-diagonal orbital digraphs of the same subdegree finite primitive group are essentially the same (they are homeomorphic as topological spaces). This is a special case of a deep property of compactly generated totally disconnected locally compact groups, detailed in \cite{kron_moller_analogues_of_Cayley_graphs}.

Our main result, presented in Section~\ref{section:orbits_on_pends}, shows that the structure of an infinite subdegree finite primitive group is determined by the length of any end orbit.

\begin{citethm}{\ref{thm:action_on_ends}} Let $G$ be a subdegree finite primitive group of permutations of an infinite set $\Omega$ that is closed in the natural complete topology of $\sym(\Omega)$. If $\epsilon$ is an end of any connected orbital digraph of $G$ then precisely one of the following holds,
\begin{enumerate}
\item $|\epsilon^G| = 1$ and $G = G_\epsilon$;
\item $|\epsilon^G| = \aleph_0$ and $G \cong G_{\alpha} \ast_{G_{\alpha, \epsilon}} G_\epsilon$;
\item $|\epsilon^G| = 2^{\aleph_0}$ and $G \cong G_{\alpha} \ast_{H_\alpha} H$, for some group $H$ satisfying $G_{\alpha, \epsilon} < H < G$.
\end{enumerate} \end{citethm}

In \cite{kron_moller_analogues_of_Cayley_graphs}  B. Kr{\"o}n and R. G. M{\"o}ller extend the notion of a Cayley graph of a finitely generated group to compactly generated totally disconnected locally compact groups, and they call these graphs {\em rough Cayley graphs}. The ends of any two Cayley graphs of a finitely generated group are the same, and this is also true of rough Cayley graphs. A discussion of their impressive paper is beyond the scope of this short note; it suffices to observe that an infinite primitive permutation group $G$ whose subdegrees are all finite has a rough Cayley graph. In fact, any non-diagonal orbital digraph of $G$ is rough Cayley graph of $G$. Consequently, the ends of any non-diagonal orbital digraph of $G$ are the rough ends of $G$. In Section~\ref{section:rough_ends} we briefly discuss rough Cayley graphs and their ends, and restate our theorem in terms of a primitive group and its action on its rough ends.

\section{Preliminaries}

It is well-known (see for example \cite[Remark 29.8]{adeleke&neumann}) that if $G$ is a subdegree finite primitive group of permutations of an infinite set $\Omega$, then $\Omega$ is countably infinite. And since $\Omega$ is countable, there is a natural complete topology on the symmetric group $\sym (\Omega)$, that of {\em pointwise convergence}. If we enumerate the set $\Omega$ as $\{\gamma_1, \gamma_2, \gamma_3, \ldots\}$, then a sequence of permutations $(g_n)$ tends to the limit $g$ if and only if, for any $k \geq 1$, we have $\gamma_k^{g_n} = \gamma_k^g$ and $\gamma_k^{g_n^{-1}} = \gamma_k^{g^{-1}}$ for all sufficiently large $n$. A basis for the open sets in this topology consists of all cosets of pointwise stabilisers of finite sets. If $G \leq \sym (\Omega)$, the {\em closure of $G$} is the intersection of all closed subgroups of $\sym (\Omega)$ that contain $G$, and is denoted by $\overline{G}$.

When examining actions on finite subsets of $\Omega$, there is no difference between a group $G$ and its closure $\overline{G}$.

\begin{thm} {\normalfont(\cite[Proposition 2.6]{cameron:oligomorphic})} If $G \leq \sym (\Omega)$ then $G$ and its closure $\overline{G}$ have the same orbits on $n$-tuples of $\Omega$ for all $n \geq 1$. \qed \end{thm}

The following result due to D. M. Evans in \cite{evans:a_note_on_automorphisms} will be fundamental to our argument; here we give the statement of this result as given in \cite{cameron:oligomorphic}.

\begin{thm} {\normalfont(\cite[Theorem 2.8]{cameron:oligomorphic})} \label{thm:evans_oligomorphic} If $G$ and $H$ are closed subgroups of $\sym(\Omega)$, where $\Omega$ is a countable set and $H \leq G$, then either $|G:H| \leq \aleph_0$ or $|G:H| = 2^{\aleph_0}$. The former holds if and only if $H$ contains the pointwise stabiliser in $G$ of some finite subset of $\Omega$. \qed \end{thm}

We will denote the stabiliser of any point $\alpha \in \Omega$ in $G$ by $G_\alpha$, and the setwise and pointwise stabilisers of a subset $\Lambda \subseteq \Omega$ by $G_{\{\Lambda\}}$ and $G_{(\Lambda)}$ respectively. This notation will be extended to ends, with $G_\epsilon$ used to denote the stabiliser in $G$ of the end $\epsilon$.

It was shown in \cite[Theorem 3.11]{me:OrbGraphs} that if $G$ is an infinite subdegree finite primitive group with an orbital digraph with more than one end, then there exists a canonical orbital digraph $\Gamma$ of $G$ that has a very simple, tree-like structure.

Like a tree, every vertex in $\Gamma$ is a {\em cut vertex}; that is, removing any vertex from $\Gamma$ results in a disconnected digraph (such digraphs are said to have {\em connectivity one}). But, unlike a tree,  $\Gamma$ contains non-trivial {\em lobes}: subdigraphs $\Lambda$ that are maximal subject to the condition that no vertex in $\Lambda$ is a cut vertex of $\Lambda$. The lobes of $\Gamma$ are all pairwise isomorphic, have at least three vertices and at most one end. Each lobe of $\Gamma$ is itself a primitive digraph, and any two vertices of $\Gamma$ lie in the same number $m$ of lobes. It was shown in \cite[Theorem 4.5]{me:OrbGraphs} that this canonical orbital digraph of $G$ is essentially unique, and as such can be denoted by $\Gamma(m, \Lambda)$.

Intuitively, one may think of $\Gamma(m, \Lambda)$ as the digraph obtained by glueing together infinitely many copies of $\Lambda$ such that no two copies share more than one vertex, and every vertex of $\Gamma(m, \Lambda)$ lies in $m$ copies of $\Lambda$.

In \cite{jung:watkins} H. Jung and M. Watkins characterise primitive graphs with connectivity one; this was extended to primitive digraphs with connectivity one in \cite{me:InfPrimDigraphs}. These graphs and digraphs all share this tree-like structure.

\begin{thm} {\normalfont(\cite[Theorem 3.3 \& Theorem 3.5]{me:InfPrimDigraphs})}
\label{thm:directed_connectivity_one_digraphs} If $\Gamma$ is a
vertex-transitive digraph with connectivity one, then it is
primitive if and only if the lobes of $\Gamma$ are primitive, pairwise isomorphic digraphs that are not directed cycles of odd prime length, and each lobe has at
least three vertices. \end{thm}

Thus, any locally finite primitive digraph with connectivity one may be written as $\Gamma(m, \Lambda)$, where $m \geq 2$ and $\Lambda$ is some primitive digraph that is not a directed cycle of odd prime length. In particular, note that such graphs have infinitely many ends.

We associate with any connectivity-one digraph $\Gamma$ a bipartite digraph $T$, whose vertices are coloured violet or blue. The violet vertices are the vertices of $\Gamma$, and the blue vertices are the lobes of $\Gamma$. A violet vertex $v$ is adjacent to a blue vertex $b$ in $T$ if and only if $v$ lies in $b$ in $\Gamma$. Adjacent vertices in $T$ are joined by two arcs, one in each direction. This construction yields a tree, which is known as the {\em block-cut-vertex tree} of $\Gamma$.

Applying the Bass-Serre theory of groups acting on trees (\cite{serre:trees}), one obtains the following.

 \begin{thm}{\normalfont(\cite[Theorem 5.2]{me:OrbGraphs})} \label{thm:G_amalgamated_free_product} If $G$ is an subdegree finite primitive group of permutations with an orbital digraph with more than one end, then $G$ has a canonical orbital digraph
$\Gamma(m, \Lambda)$ and
\[G \cong G_{\alpha} \ast_{G_{\alpha, \{\Lambda\}}} G_{\{\Lambda\}},\]
where $\alpha \in V\Lambda$, and $G_{\alpha, \{\Lambda\}}$ is a
non-trivial maximal proper subgroup of $G_{\{\Lambda\}}$ that
fixes no element in $V\Lambda \setminus \{\alpha\}$. \end{thm}

Of course not every group acting transitively on a digraph of the form $\Gamma(m, \Lambda)$ is primitive, and often one needs a quick test for imprimitivity.

\begin{thm} {\normalfont \cite[Theorem 2.5]{me:InfPrimDigraphs}}
\label{thm:Gya_eq_Gyb_not_prim} Let $G$ be a vertex-transitive group
of automorphisms of a connectivity-one digraph $\Gamma$ whose
lobes have at least three vertices, and let $T$ be the
block-cut-vertex tree of $\Gamma$. If there exist distinct
vertices $\alpha, \beta \in V\Gamma$ such that, for some vertex
$x \in (\alpha, \beta)_T$,
\[ G_{\alpha, x} = G_{\beta, x}, \]
then $G$ does not act primitively on $V\Gamma$. \qed \end{thm}

Associated with each orbital $\Delta = (\alpha, \beta)^G$ is its pair $\Delta^* := (\beta, \alpha)^G$, and associated with each suborbit $\Delta(\alpha) := \{\gamma \mid (\alpha, \gamma) \in \Delta\}$ is its pair $\Delta^*(\alpha) := \{\gamma \mid (\alpha, \gamma) \in \Delta^*\}$. In the orbital digraph $(\Omega, \Delta)$, the suborbit $\Delta(\alpha)$ is the set of all vertices $\gamma$ that are adjacent to $\alpha$ with an arc from $\alpha$ to $\gamma$. Similarly, the suborbit $\Delta^*(\alpha)$ is the set of all vertices $\gamma$ that are adjacent to $\alpha$ with an arc from $\gamma$ to $\alpha$. Thus, the out-valency of $\alpha$ in $(\Omega, \Delta)$ is $|\Delta(\alpha)|$, and the in-valency of $\alpha$ in $(\Omega, \Delta)$ is $|\Delta^*(\alpha)|$.

In a locally finite orbital digraph of an infinite primitive group the in-valency and the out-valency of the digraph must be equal. This is a consequence of a lovely theorem by C. E. Praeger.

\begin{thm} {\normalfont(\cite{praeger})} \label{thm:praeger} Let $\Gamma'$ be an infinite connected vertex- and edge-transitive directed graph with finite but unequal in-valency and out-valency. Then there is an epimorphism $\varphi$ from the vertex set of $\Gamma'$ to the set of integers $\mathbb{Z}$ such that $(\alpha, \beta)$ is an edge of $\Gamma'$ only if $\varphi(\beta) = \varphi(\alpha)+1$. \qed \end{thm}

\begin{cor} \label{cor:paired_suborbits_have_same_size} Suppose $G'$ is a primitive group of permutations of an infinite set $\Omega$, and every suborbit of $G'$ is finite. If $\alpha \in \Omega$ and $\Delta(\alpha)$ and $\Delta^*(\alpha)$ are paired $\alpha$-suborbits then $|\Delta(\alpha)| = |\Delta^*(\alpha)|$. \qed \end{cor}

\section{Orbits on ends}
\label{section:orbits_on_pends}

Suppose $G$ is an infinite primitive subdegree finite group of permutations with canonical orbital digraph $\Gamma(m, \Lambda)$, and let $T$ be the block-cut-vertex tree of $\Gamma$. Finally, assume that $G$ is closed, and recall that a group and its closure have the same orbits on $n$-tuples of $\Omega$, for all $n\geq 1$.

Using the following three lemmas, it is possible to scrutinise the action of $G$ on the thin ends of any connected orbital digraph of $G$.

\begin{lemma} \label{lemma:Ga_leq_closure Gepsilon_then_Ga_leq_Gepsilon} For any thin end $\epsilon$ of $\Gamma$, if $\underline{\alpha}$ is an $n$-tuple of vertices with $G_{\underline{\alpha}} \leq \overline{G_\epsilon}$, then $G_{\underline{\alpha}} \leq G_\epsilon$. \end{lemma}

\begin{proof} Enumerate the set $V\Gamma$ as $\{\gamma_1, \gamma_2, \gamma_3, \ldots \}$, and recall that an element $g \in G$ is the limit of a sequence of permutations $(g_n)$ of $V\Gamma$ if and only if, given $k \geq 1$, there exists an integer $N_k$ such that, for all $n \geq N_k$ we have $\gamma_i^{g_n} = \gamma_i^g$ and $\gamma_i^{g_n^{-1}} = \gamma_i^{g^{-1}}$ whenever $1 \leq i \leq k$.

Suppose $G_{\underline{\alpha}} \leq \overline{G_\epsilon}$ but $G_{\underline{\alpha}} \not \leq G_\epsilon$, and fix
$g \in G_{\underline{\alpha}} \setminus G_\epsilon.$
Since $g \in \overline{G_\epsilon}$, there exists a sequence $(g_n)$ of permutations in $G_\epsilon$ such that
$g_n \rightarrow g.$
If $\underline{\alpha} = (\alpha_1, \ldots, \alpha_n)$ then choose $k \geq 1$ such that $\{\alpha_1, \ldots, \alpha_n\} \subseteq \{\gamma_1, \ldots, \gamma_k\}$. Since $g_n \rightarrow g$, there exists an integer $N_k$ such that, for all $n \geq N_k$ we have $\gamma_i^{g_n} = \gamma_i^g$ whenever $1 \leq i \leq k$. Hence, for all $n \geq N_k$,
\[\underline{\alpha}^{g_n} = \underline{\alpha}.\]

Recall $T$ is the block-cut-vertex tree of $\Gamma$, and let $[\alpha_1, \epsilon)_T$ be the unique half-line of $T$ rooted at $\alpha_1$ lying in the end $\epsilon$. For all $n \geq N_k$ we have $g_n \in G_{\alpha_1, \epsilon}$, therefore $g_n$ fixes every vertex on $[\alpha_1, \epsilon)_T$. Since $g$ lies in $G_{\alpha_1}$ but not $G_{\epsilon}$, there exists $x \in [\alpha_1, \epsilon)_T$ such that $x^g \not = x$; furthermore, if $y \in [x, \epsilon)_T$ then $y^g \not = y$. Therefore, we may choose $\beta \in [x, \epsilon)_T \cap V\Gamma$ such that $\beta^g \not = \beta$. Since $\beta \in V\Gamma$, there is an integer $k^\prime$ such that $\beta = \gamma_{k^\prime}$. And because $g_n \rightarrow g$, there exists $N_{k^\prime} \geq 1$ such that for all $n \geq N_{k^\prime}$ we have $\gamma_{k^\prime}^{g_n} = \gamma_{k^\prime}^g$. Fix $N:= \max (N_k, N_{k^\prime})$. For all $n \geq N$, we have $g_n \in G_{\underline{\alpha}}$ but $\beta^{g_n} = \beta^g \not = \beta$, so $g_n \not \in G_\epsilon$, a contradiction. \end{proof}

\begin{lemma} \label{lemma:if_Ga_leq_Gepsilon} If there is an end $\epsilon$ of $\Gamma$ and an $n$-tuple $\underline{\alpha} = (\alpha_1, \ldots, \alpha_n)$ of vertices of $\Gamma$ such that $G_{\underline{\alpha}} \leq G_\epsilon$, then, for $1 \leq i \leq n$, the orbit $\epsilon^{G_{\alpha_i}}$ is finite. \end{lemma}

\begin{proof} Suppose $G_{\underline{\alpha}} \leq G_\epsilon$. Then, for all $i$ satisfying $1 \leq i \leq n$ we have $G_{\underline{\alpha}} \leq G_{\alpha_i, \epsilon}$. Therefore
\[|G_{\alpha_i} : G_{\underline{\alpha}}|
    = |G_{\alpha_i} : G_{\alpha_i, \epsilon}| |G_{\alpha_i, \epsilon} :
        G_{\underline{\alpha}}|
    = |\epsilon^{G_{\alpha_i}}| |G_{\alpha_i, \epsilon} :
        G_{\underline{\alpha}}|.\]
Observe
\[|G_{\alpha_1} : G_{\underline{\alpha}}|
    =  \prod_{m=1}^{n-1} |G_{\alpha_{1} \ldots \alpha_{m}} :
        G_{\alpha_{1} \ldots \alpha_{m+1}}|
    =  \prod_{m=1}^{n-1} |\alpha_{m+1}^{G_{\alpha_{1} \ldots
        \alpha_{m}}}|.\]

Since $G$ is subdegree finite, each orbit $|\alpha_{m+1}^{G_{\alpha_{1} \ldots \alpha_{m}}}|$ is finite, so $|G_{\alpha_1} : G_{\underline{\alpha}}|$ is finite. A similar argument shows $|G_{\alpha_i}:G_{\underline{\alpha}}|$ is finite for each $i$ satisfying $1 \leq i \leq n$. We have already seen that
\[|G_{\alpha_i} : G_{\underline{\alpha}}| = |\epsilon^{G_{\alpha_i}}| |G_{\alpha_i, \epsilon} : G_{\underline{\alpha}}|,\]
so one may deduce that $|\epsilon^{G_{\alpha_i}}|$ is finite. \end{proof}

\begin{lemma} \label{lemma:thin_end_in_small_Ga_orbit_for_all_a} If $\epsilon$ is a thin end of $\Gamma$ with $|\epsilon^G| < 2^{\aleph_0}$ then, for all $\alpha \in V\Gamma$, the orbit $\epsilon^{G_{\alpha}}$ is finite. \end{lemma}

\begin{proof} Let $\epsilon$ be a thin end of $\Gamma$, and let $H := \overline{G_\epsilon}$. Now
\[|\epsilon^G| = \ |G:G_\epsilon|
= \ |G:H| |H:G_\epsilon|.\]

If $|\epsilon^G| < 2^{\aleph_0}$, then $|G:H|<2^{\aleph_0}$, and, since $G$ and $H$ are both closed, we may apply Theorem~\ref{thm:evans_oligomorphic} to deduce that there exists an $n$-tuple of vertices $\underline{\alpha}$ such that $G_{\underline{\alpha}} \leq H$, and $|G:H| \leq \aleph_0$.

Suppose $|\epsilon^G| < 2^{\aleph_0}$ and $\alpha = (\alpha_1, \ldots, \alpha_n)$ is an $n$-tuple of vertices of $\Gamma$ such that $G_{\underline{\alpha}} \leq \overline{G_\epsilon}$. By Lemma~\ref{lemma:Ga_leq_closure Gepsilon_then_Ga_leq_Gepsilon} we have $G_{\underline{\alpha}} \leq G_\epsilon.$
Fix $\alpha \in V\Gamma$ and let $\underline{\alpha}^\prime:=(\alpha, \alpha_1, \ldots, \alpha_n)$; then $G_{\underline{\alpha}^\prime} \leq G_{\underline{\alpha}} \leq G_\epsilon$, and so, from Lemma~\ref{lemma:if_Ga_leq_Gepsilon}, the orbit $\epsilon^{G_\alpha}$ is finite. \end{proof}

We now prove our main result.

\begin{thm} \label{thm:action_on_ends}  Let $G$ be a subdegree finite primitive group of permutations of an infinite set $\Omega$ that is closed in the natural complete topology of $\sym(\Omega)$. If $\epsilon$ is an end of any connected orbital digraph of $G$ then precisely one of the following holds,
\begin{enumerate}
\item $|\epsilon^G| = 1$ and $G = G_\epsilon$;
\item $|\epsilon^G| = \aleph_0$ and $G \cong G_{\alpha} \ast_{G_{\alpha, \epsilon}} G_\epsilon$;
\item $|\epsilon^G| = 2^{\aleph_0}$ and $G \cong G_{\alpha} \ast_{H_\alpha} H$, for some group $H$ satisfying $G_{\alpha, \epsilon} < H < G$.
\end{enumerate} \end{thm}

\begin{proof} If an orbital digraph of $G$ has precisely one end, then all non-diagonal orbital digraphs of $G$ have just one end and (i) holds. So, suppose $G$ has an orbital digraph with more than one end. By Theorem~\ref{thm:G_amalgamated_free_product}, $G$ has a canonical orbital graph $\Gamma = \Gamma(m, \Lambda)$, and we may write
$G \cong G_{\alpha} \ast_{G_{\alpha, \{\Lambda\}}} G_{\{\Lambda\}}$. Furthermore, since the ends of any two connected orbital digraphs of $G$ are the same, we may assume $\epsilon$ is an end of $\Gamma$.

If $\epsilon$ is a thick end of $\Gamma$, then $\epsilon$ must be the end of some lobe $\Lambda'$ of $\Gamma$. Since $G$ permutes the lobes of $\Gamma$ transitively (because $G$ acts transitively on the arc set of $\Gamma$), the subgroups $G_{\{\Lambda\}}$ and $G_{\{\Lambda'\}}$ are isomorphic; every lobe contains precisely one end (which is thick); and $G_\epsilon \cong G_{\{\Lambda\}}$. There are countably many lobes in $\Gamma$, so $|\epsilon^G| = \aleph_0$ and (ii) holds.

Finally, suppose $\epsilon$ is a thin end of $\Gamma$, and $|\epsilon^G| < 2^{\aleph_0}$. Let $T$ be the block-cut-vertex tree of $\Gamma$. Since $\epsilon$ is a thin end of $\Gamma$, it is an end of $T$. By Lemma~\ref{lemma:thin_end_in_small_Ga_orbit_for_all_a}, if $|\epsilon^G| < 2^{\aleph_0}$ then for all $\alpha \in V\Gamma$, the orbit $\epsilon^{G_{\alpha}}$ is finite. Fix $\alpha \in V\Gamma$. Since $\epsilon^{G_{\alpha}}$ is finite, there exists a vertex $\alpha' \in V\Gamma$ that lies in $[\alpha, \epsilon)_T$ such that $G_{\alpha, \alpha'} \leq G_\epsilon$. Put $m:=|\alpha^{\prime \ G_\alpha}|$, and note $m = |\epsilon^{G_\alpha}|$. By Corollary~\ref{cor:paired_suborbits_have_same_size}, $|\alpha^{G_{\alpha'}}| = m$.

Now choose $\beta \in (\alpha', \epsilon)_T$. A similar argument shows there exists $\beta' \in (\beta, \epsilon)_T$ such that $G_{\beta, \beta'} \leq G_\epsilon$.

By choosing $\beta$ and $\beta'$ in this way, we ensure that $G_{\alpha, \beta'} \leq G_{\beta, \beta'}$ and $|\beta^{\prime G_\alpha}| = |\epsilon^{G_\alpha}| = m$. We again apply Corollary~\ref{cor:paired_suborbits_have_same_size} to deduce $|\alpha^{G_{\beta'}}| = m$. Hence
\begin{align*} m    =& \ |\alpha^{G_{\beta'}}|\\
                    =& \ |G_{\beta'}:G_{\alpha, \beta'}|\\
                    =& \ |G_{\beta'}:G_{\beta, \beta'}| |G_{\beta, \beta'}:G_{\beta' \alpha}|. \end{align*}

If $|G_{\beta, \beta'}:G_{\beta' \alpha}| = 1$ then $G_{\beta, \beta'} \leq G_\alpha$ and so $G_{\alpha, \alpha'} \leq G_{\beta, \beta'}$ and $G_{\beta, \beta'} \leq G_{\alpha, \alpha'}$, with $\alpha', \beta \in (\alpha, \beta')_T$ and $[\alpha, \alpha']_T \cap (\beta, \beta']_T = \emptyset$. Applying Theorem~\ref{thm:Gya_eq_Gyb_not_prim} we see $G$ is not primitive, which contradicts our original assumption; thus, we must have
$|G_{\beta, \beta'}:G_{\beta' \alpha}| > 1$. This implies $|G_{\beta'}:G_{\beta, \beta'}| < m$; that is, $|\beta^{G_{\beta'}}| < m$. However, $|\beta^{G_{\beta'}}| = |\beta^{\prime G_{\beta}}|$ by Corollary~\ref{cor:paired_suborbits_have_same_size}; furthermore, $\beta$ and $\beta'$ we chosen so that $G_{\beta, \beta'} \leq G_\epsilon$. Hence $|\epsilon^{G_\beta}| < m$.

Since $\alpha$ was chosen arbitrarily, we may now set $\alpha:=\beta$ and repeat the above argument. Eventually, we find a vertex $\alpha$ such that $|\epsilon^{G_\alpha}| = 1$. However, this is also a contradiction. Indeed, let $x$ be the vertex of $T$ that is adjacent to $\alpha$ in $[\alpha, \epsilon)_T$. This vertex corresponds to some lobe $\Lambda$ of $\Gamma$ that contains $\alpha$. By Theorem~\ref{thm:G_amalgamated_free_product}, the group $G_{\alpha, \{\Lambda\}}$ fixes no vertex in $V\Lambda \setminus \{\alpha\}$, and therefore cannot fix the half-line $[\alpha, \epsilon)_T$. Whence, we must have $|\epsilon^{G}| = 2^{\aleph_0}$.

Furthermore, since $x$ is the vertex of $T$ adjacent to $\alpha$ in $[\alpha, \epsilon)_T$,
\[G_{\alpha, \epsilon} \leq G_x.\]
The vertex $x$ corresponds to some lobe of $\Gamma$, and so $G_x \cong G_{\{\Lambda\}}$. Finally, we note that $G_{\alpha, \epsilon} \not = G_x$ because $|\epsilon^{G}| = 2^{\aleph_0}$. Taking $H = G_x$ we have $G \cong G_{\alpha} \ast_{H_\alpha} H$, with $G_{\alpha, \epsilon} < H < G$ by Theorem~\ref{thm:G_amalgamated_free_product}.
\end{proof}

As the following three examples show, this theorem cannot be improved upon. Indeed, the result fails if one does not require $G$ to be closed; it also fails if one does not require $G$ to be primitive. Furthermore, there are infinite imprimitive subdegree finite closed permutation groups that satisfy the conclusions of the theorem.

\begin{example}
Given any group $G$ of permutations of a countable set $\Omega$, one can find a subgroup $H$ of $G$ that is countable with the same orbits on tuples of $\Omega$ as $G$, using a method described in \cite{cameron:oligomorphic}.

Since $\Omega$ is countable, the set of all tuples of $\Omega$ is a countable union of countable sets, and is therefore countable. Enumerate pairs $(\underline{\alpha_n}, \underline{\beta_n})$ of tuples for which $\underline{\alpha_n}$ and $\underline{\beta_n}$ lie in the same orbit of $G$. For each natural number $n$ choose $g_n \in G$ such that $\underline{\alpha_n}^{g_n} = \underline{\beta_n}$, and let $H$ be the subgroup of $G$ generated by the elements $g_n$. This group is countable, and has the same orbits on tuples as $G$.

If $G$ is taken to be a closed arc-transitive primitive group of automorphisms of a locally finite orbital digraph $\Gamma$ with more than one end, and $H$ is constructed as described, then the orbit of $G$ on any thin end $\epsilon$ of $\Gamma$ with have length $2^{\aleph_0}$ by Theorem~\ref{thm:action_on_ends}, while the length of the orbit of $H$ on $\epsilon$ will be at most $\aleph_0$. Furthermore, because $G$ and $H$ have the same orbits on tuples, their orbital digraphs must be the same. In particular, all non-diagonal orbital digraphs of $H$ must be connected, and so $H$ is primitive by D. G. Higman's famous result (\cite{higman}). The group $H$ fails to satisfy the conclusions of Theorem~\ref{thm:action_on_ends} because $H$ is not closed.
\end{example}

\begin{example}
Let $S_2$ be the symmetric group on two letters, and let $S_3$ be the symmetric group on three letters, and let $G = S_2 \ast S_3$ be the free product of $S_2$ and $S_3$. This group acts without inversion on a bivalent tree $T$ in which one part of the natural bipartition of $T$ has valency two and the other has valency three, and $G$ acts transitively on each part of the bipartition. Colour the vertices with valency two violet, and the vertices with valency three blue. Let $V$ be the set of violet vertices of $T$, and let $\Gamma$ be the orbital digraph $(V, (\alpha, \beta)^G)$, where $\alpha, \beta \in V$ are at distance two in $T$.

If $x \in T$ is the vertex lying between $\alpha$ and $\beta$, then $G_x$ acts like $S_3$ on the vertices in $T$ adjacent to $x$. Whence, $\Gamma$ has connectivity one, and the lobes of $\Gamma$ are isomorphic to $K_3$, the complete digraph on three vertices, with each vertex in $\Gamma$ lying in precisely two lobes. Thus, $\Gamma = \Gamma(2, K_3)$, with block-cut-vertex tree $T$.

Now $\Gamma(2, K_3)$ is an orbital digraph of $G$ acting on $V$. The stabiliser $G_\alpha$ is isomorphic to $S_3$ and is therefore finite. Whence $G_\alpha$ is closed in the permutation topology, so we now replace $G$ with its closure and assume henceforth that $G$ is closed.

Again since $G_\alpha$ is finite, by Theorem~\ref{thm:Gya_eq_Gyb_not_prim}, the group $G$ does not act primitively on $\Gamma(2, K_3)$, although the full automorphism group of $\Gamma(2, K_3)$ is primitive by Theorem~\ref{thm:directed_connectivity_one_digraphs}. If $\epsilon$ is any thin end of $\Gamma$, then the end orbit $\epsilon^G$ has at most countably infinite length.

The closed group $G$ fails to satisfy the conclusions of Theorem~\ref{thm:action_on_ends} because $G$ is not primitive.
\end{example}

\begin{example}
Let $C_5$ denote a directed cycle on  five vertices, and let $\Gamma$ be the digraph $\Gamma(3, C_5)$. The group $\aut \Gamma$ acts imprimitively on $V\Gamma$ by Theorem~\ref{thm:directed_connectivity_one_digraphs}, but satisfies the conclusions of Theorem~\ref{thm:action_on_ends} (in fact the group $\aut \Gamma$ is of type (iii)).
\end{example}

\section{rough ends}
\label{section:rough_ends}

In this section we relate Theorem~\ref{thm:action_on_ends} to the exciting work of B. Kr{\"o}n and R. G. M{\"o}ller (\cite{kron_moller_analogues_of_Cayley_graphs}).

Recall that any permutation group $G$ of a countable set can be considered a topological group by imposing upon it the permutation topology. A basis for the open sets of this topology are the cosets of pointwise stabilisers of finite sets. It is a simple exercise to check that if $G$ is closed then the pointwise stabiliser of a single element in $\Omega$ is also closed.

If $G$ is a topological group, then a connected graph $\Gamma$ is said to be a {\em rough Cayley graph} of $G$ if $G$ acts transitively on $\Gamma$ and the stabilisers of vertices are compact open subgroups of $G$.

If $G$ has a rough Cayley graph then the ends of any two rough Cayley graphs are homeomorphic and can therefore be considered to be the same (\cite[Theorem 2.7 and Section 3.1.1 ]{kron_moller_analogues_of_Cayley_graphs}). The {\em rough ends} of $G$ are defined to be the ends of a rough Cayley graph of $G$ (\cite[Definition 3.12]{kron_moller_analogues_of_Cayley_graphs}).

\begin{thm} {\normalfont(\cite[Corollary 1]{moller:FC})}
If $G$ acts as a group of automorphisms of a locally finite connected graph $\Gamma$ such that $G$ is transitive on the vertex set of $\Gamma$ and the stabilisers of the vertices in $\Gamma$ are compact and open, then $G$ is compactly generated.
\end{thm}

\begin{lemma}  {\normalfont(\cite[Lemma 2]{{woess:topological_groups_and_infinite_graphs}})}
If $\Gamma$ is a vertex transitive infinite connected locally finite graph then the closure (in the permutation topology) of any subset $U \subseteq \aut(\Gamma)$ is compact if and only if the orbit $\alpha^U$ is finite for all $\alpha \in V\Gamma$.
\end{lemma}

Thus, if $G$ is a closed primitive group of permutations of a countable set, and the subdegrees of $G$ are all finite, then $G$ acts transitively on a non-diagonal orbital digraph $\Gamma$, which is necessarily connected, and the stabilisers of vertices in $\Gamma$ are open and compact. Whence $\Gamma$ is a rough Cayley graph of $G$, and the rough ends of $G$ are the ends of $\Gamma$.

We conclude by restating our main result as a theorem about a primitive group and its rough ends.

\begin{thm} \label{thm:action_on_ends}  Let $G$ be a subdegree finite primitive group of permutations of an infinite set $\Omega$ that is closed in the natural complete topology of $\sym(\Omega)$. If $\epsilon$ is a rough end of $G$ then precisely one of the following holds.
\begin{enumerate}
\item $|\epsilon^G| = 1$ and $G = G_\epsilon$;
\item $|\epsilon^G| = \aleph_0$ and $G \cong G_{\alpha} \ast_{G_{\alpha, \epsilon}} G_\epsilon$;
\item $|\epsilon^G| = 2^{\aleph_0}$ and $G \cong G_{\alpha} \ast_{H_\alpha} H$, for some group $H$ satisfying $G_{\alpha, \epsilon} < H \leq G$.
\end{enumerate} \end{thm}

\\

{\em \noindent Acknowledgements.} Some of the results in this paper are taken from the author's DPhil thesis, completed
at the University of Oxford, under the supervision of Peter Neumann and funded by the EPSRC; the remainder was completed at Syracuse University as a Philip T Church Postdoctoral Fellow. The author would
like to thank Dr Neumann and the EPSRC.

redpajama_set_name: RedPajamaArXiv


text:
CMR Approves Marine Turtle Study Resolution
News & Updates / August 19, 2011 August 13, 2019
11-88-CWS | August 19, 2011
BILOXI, Miss. – The Mississippi Commission on Marine Resources (CMR) voted unanimously to approve a Marine Turtle Study Resolution, which was presented to them for approval Tuesday, at the Aug. 16 CMR meeting.
The approved resolution was submitted to the U.S. Department of Commerce National Oceanic and Atmospheric Administration (NOAA), and requested a comprehensive Gulf of Mexico sea turtle population and distribution study to provide vital baseline data on the status of current sea turtle populations and furthermore, to more fully investigate all possible causes of sea turtle mortality.
The Marine Turtle Study Resolution acknowledges that increased turtle strandings are occurring in the northern Gulf of Mexico states, and there is a need to explore all options because timing of increased strandings does not correlate with spatial distribution of shrimping effort. The resolution also acknowledges the numerous proactive precautionary measures that have been taken by the Mississippi Department of Marine Resources (MDMR) to address fisheries interactions, which may affect sea turtles in state waters.
Additionally, the Marine Turtle Study Resolution states that the CMR's management of marine sea turtles is essential, and recognizes the need to know the status and distribution of the listed sea turtle species in the state and all Gulf of Mexico waters before enacting regulations, which may severely impact shrimp fisheries and dependent communities within the Mississippi Coast region.
Attached: Marine Turtle Study Resolution
Contact: Jennifer Goldman Leirer