text:
November 26, 2020 to current date : 1
July 18, 2017 to November 25, 2020 : 3
June 10, 2013 to June 11, 2017 : 6
Transportation and Infrastructure : 10
Government Operations : 8
Transportation : 1
Williams Lake : 10
Freshet response underway in Cariboo
A co-ordinated effort to ensure people have safe, reliable roads in the region is underway as work to repair flooding and damage continues at over 100 road sites in the Cariboo.
Repairs underway on flood-damaged roads southwest of Williams Lake
Wednesday, July 10, 2019 1:00 PM
The Chilcotin River and associated tributaries have experienced a one-in-200-year flood event, following more than 90 millimetres of rainfall in the past few days.
Highway 97 construction area shutting down for winter
Wednesday, October 31, 2018 8:15 AM
Construction on Highway 97 from Williams Lake Indian Reserve No. 1 to Lexington Road will soon be concluding for the season, resulting in a speed limit increase from the current 50 km/h construction zone to 70 km/h.
Increasing safety and capacity of Highway 97 in Williams Lake
Wednesday, September 13, 2017 10:00 AM
A section of the Highway 97 arterial through the City of Williams Lake will be widened from three to four lanes, improving safety within the city limits for local traffic, and increasing the capacity and efficiency for north and southbound vehicles on the highway.
Funding supports Anahim Lake and Williams Lake airport infrastructure projects
Wednesday, March 29, 2017 12:00 PM
Anahim Lake and Williams Lake airports will receive $78,750 in combined funding from the B.C. Air Access Program this year.
Jobs, growth boost rural communities as Cariboo Connector P3 moves ahead
The Province of British Columbia is building on its continued commitment to increase capacity and improve efficiency of the Cariboo Connector with a further $200 million of provincial funding.
Final project of Phase 2 Cariboo Connector goes out to tender
Improving the safety and reliability of the Cariboo Connector between Cache Creek and Prince George remains a priority for the Province of British Columbia, with the final section of four-laning and intersection improvements of Phase 2 going out to tender.
Williams Lake rest area improvements completed
Friday, December 16, 2016 10:00 AM
Northbound motorists on Highway 97 will experience improved conditions at the northbound pullout, approximately 10 kilometres south of Williams Lake, as work to improve the local amenities at this rest area is now complete.
Another section of four-laning on the way for Highway 97
Wednesday, August 31, 2016 1:30 PM
Construction is set to begin on widening and four-laning a section of Highway 97 south of Williams Lake, which will result in shorter travel times, less driver frustration and a safer highway.
Partnership with Williams Lake Indian Band to four-lane Highway 97
Thursday, June 25, 2015 11:30 AM
A tentative agreement between the B.C. government and the Williams Lake Indian Band to move forward on four-laning Highway 97 through the Williams Lake Indian Reserve #1 from 150 Mile House to Lexington Road, was announced by Minister of Transportation and Infrastructure Todd Stone.


Improvements to mixed use, commercial and industrial structures are eligible for tax exemption for a period of five (5) years commencing with the completion of an improvement .
Through these efforts, the Borough will increase ratables to stabilize taxes and encourage economic vitality.
Construction of a new restaurant along Crabbe Road, in addition to an improved park with a waterfront River Walk and non-motorized boat ramp will provide residents to enjoy beautified, protected landscape. These improvements will compliment improvements to Route 166, and provide the necessary waterfront overhaul while protecting our Green Acres.
1. To revitalize the Borough of South Toms River's Neighborhood Commercial and Multi-Use zoned areas.
3. To improve pedestrian safety along Route 166 and make the project area more pedestrian-friendly.


Title: Investigating participation in online policy discussion forums over time: Does network structure matter?
Citation: Phang, C.W.,Chua, Z.,Kankanhalli, A.,Goh, K.Y. (2007). Investigating participation in online policy discussion forums over time: Does network structure matter?. ICIS 2007 Proceedings - Twenty Eighth International Conference on Information Systems. ScholarBank@NUS Repository.
Abstract: Online policy discussion forums can be construed as a social network where citizens interact and collectively deliberate on policy issues. It is believed that citizen participation in such forums can help in the formulation of well-deliberated policy solutions. A key concern for governments is to promote participation in such forums. Social network literature reveals that the evolving interaction patterns between network participants can influence their collective behaviors. Thus, this study aims to investigate the longitudinal effects of network structural properties on forum participation using time series analysis. Our findings reveal that network structure can have both positive and negative effects on participation in terms of the number of postings and unique ideas contributed. This theoretically grounded empirical study on e-participation, employing ARIMA modeling with transfer function analysis on social network measures, contributes to the diversity of theories and methods in IS research. Implications for practice are also discussed.


text:
Home Page/Forums/HRIS/WkbjOOfGYyS
WkbjOOfGYyS
5 years, 7 months ago Keneth
Can I call you back? ordering viagra overnight delivery The U.S. Army's top general said it was significantlyharming day-to-day operations, and intelligence leaders say itis undermining their ability to monitor threats. A FederalReserve official said it could delay the central bank's abilityto assess whether its monetary stimulus efforts are stillneeded.
generic viagra australia reviews Add 475ml / 16 fl oz of water and use your hands to mix it into the cornflour. If you want to make coloured slime, add a couple of drops of food dye to the water first. You can wear disposable plastic gloves to avoid staining your hands.
best viagra online australia The Los Angeles County District Attorney's office cited that possibility along with insufficient evidence about the woman's level of intoxication when it formally rejected a felony rape charge against the "Forget You" crooner.
generic viagra 100mg indian "He's sore today, probably because he hasn't been running," Coughlin said. "He's not ready to come out two days in a row and do what he did (Thursday), but I'm encouraged by what he did."
cheap viagra jelly uk So what does it mean for you? Pencil in the eye doctor. Even if you're blessed with 20/20 vision, retinal imaging (a fancy term for the photo eye docs take of your eyes) does far more than test vision: It could be the easiest way yet to check in on your brain. It's also a good way to keep track of changes if you're at high risk for a disease like cardiovascular disease, Shalev says. Being able to compare images over time could help ID changes in midlife that hint towards problems. Otherwise, these changes could go unnoticed as they may not show symptoms until much later, he says.
buy generic viagra 100mg Aside from Mack, Ohio State's biggest problem was the heat and humidity. Both Miller and linebacker Ryan Shazier missed time with cramps. Buffalo didn't seem to have as many issues as the Buckeyes did.
50mg viagra no prescription Whitaker, 52, is no stranger to that tumultuous era. In an interview with Rolling Stone movie critic Peter Travers, the Oscar winner discussed growing up in Texas during the 1960s, knowing Black Panthers and being bused to their free breakfasts.
buy female viagra australia Yet while school dinners had improved significantly since celebrity chef Jamie Oliver's campaign against unhealthy meals like the 'Turkey Twizzler', six in 10 school children were eating packed lunches instead ahead of hot meals.
asda pharmacy viagra prices And when the United States began sending tens and then hundreds of thousands of young soldiers, many hardly more than teenagers, like myself, to a far off place called Vietnam in South East Asia, in the name of protecting our "Democracy", that was when I finally realized and indeed was forced to admit to myself, that everything I had ever been told and that had been force fed down my tender young throat, was a set of lies.
price viagra cialis levitra The researchers speculate that the disparities in life expectancy within the U.S. likely have a variety of causes. Socioeconomic factors such as poverty and education are known to play a role in longevity. Some places might have less access to medical facilities, fewer quality health care options available to residents or more people who lack health insurance. In other places, more people engage in riskier behaviors such as smoking, a poor diet or a lack of exercise. And, of course, healthy individuals might simply move from counties with a low life expectancy into counties with higher life expectancies.
why no generic viagra Phillips told the Daily News that during her Barneys checkout, the cashier fielded a brief phone call. Phillips left Barneys, walked three blocks to the Lexington Ave. subway station and was about to go through the turnstile when she was stopped by four plainclothes cops. Two approached her from behind, pushing her against a wall, as two others blocked the turnstile.

redpajama_set_name: RedPajamaCommonCrawl


Home Media You must have content. Period.
You must have content. Period.
Simply put, it doesn't matter how great a piece of technology you have invented, or how innovative a distribution platform you have created. You must have content.


text:
Get Updates From NCMEC
MISSING & EXPLOITED
Saving Zoe: AMBER Alert to the Rescue
AMBER Alert to the Rescue
By Patricia Davis
It was raining when Danielle Brown got to the restaurant, so she decided it would be best to leave her 10-month-old daughter Zoe right where she was – sleeping in her car seat, all snug in her pink onesie and tiny black jacket.
The new mom left the car on to keep Zoe warm and dashed into D'bo's Wings n' Things to pick up her take-out order. As she hurried out minutes later, she stared in disbelief at her car, which she'd parked close to the front door. It was backing out of the parking space.
"Stop!" she screamed at the strange man behind the wheel and grabbed the passenger door, yanking it open. "My daughter's in the car!"
But the man in the red hoodie slammed the door and took off, and Brown, her heart pounding, chased the car with the yellow "BABY ON BOARD" sign before it disappeared into the night. After the March 16, 2018 abduction, police issued an AMBER Alert with the car's description, and it wasn't long before motorists all over Memphis were looking for the 2016 black Honda Civic four-door sedan, tag number X30 00S – and Zoe.
When the car and Zoe weren't recovered in the first hour, Memphis Police Director Michael Rallings mobilized his entire police force. Officers, some off duty, worked through the night chasing leads all over the city, united in their mission: finding Zoe Jordan.
Memphis Police Director Michael Rallings
As the night wore on, Brown became increasingly distraught and blamed herself for being a "terrible mom" for leaving her infant daughter in the car. She had assumed, incorrectly, that since she had taken her car keys with her into the restaurant, Zoe would be safe for a few minutes. But the car was running; the carjacker didn't need her keys.
The next morning, a Saturday, Memphis retiree George Nicholes, who served 32 years in the military, saw a story on the news about the AMBER Alert. Concerned about the missing child, Nicholes punched the car's tag number into his contacts on his cellphone on the off chance he spotted the vehicle on the way to his daughter's, not far from the restaurant.
It had been nearly 13 hours since the abduction, but Nicholes checked all the tags on Hondas that he could see from his vantage point in his truck. Suddenly, his heart quickened. One said X30 00S.
The car was parked at an odd angle on a residential street. He checked the tag again to make sure he was right, then immediately called 911. He wasn't able to see inside the car and was told to sit tight in his truck.
Police arrived quickly, and Nicholes watched as an officer reached into the Honda Civic and pulled Zoe out of her car seat, touching off a spontaneous celebration of joy and tears. Nicholes was declared a hero, but he would have none of it. It was God who put him in the right place, at the right time, he said.
George Nicholes spotted the vehicle and called 911.
"I couldn't wait to hold her again and kiss her and then tell her I'm sorry," Brown said of going to the hospital where Zoe was taken as a precaution. "I've felt super guilty and just super emotional and just a lot, a lot of emotions."
Rallings, the police director, said the community can't thank Nicholes enough for taking the AMBER Alert seriously and getting involved. He said Zoe's case should serve as a cautionary tale: Never leave a child unattended in a car.
Here at the National Center for Missing & Exploited Children our analysis of nearly 554 non-family infant abductions showed that 11 percent occurred in the course of carjackings. In most of these cases, the carjacker didn't realize there was a child in the backseat. They often abandoned the children once they realized they were not alone in the car. They wanted a car – not a baby.
Rallings equates AMBER Alerts to seatbelts, which many years ago were seen by the public as annoying. But, Rallings said, seatbelts save lives.
"So the AMBER Alert, you know, the notification may be annoying to some, but AMBER Alerts save lives," Rallings said. "If you want to help save a life of an endangered child, then when you hear of an AMBER Alert going out, you need to stop what you're doing and get involved. That's what the AMBER Alert is all about. Law enforcement and the community working together to bring about a positive resolution of a very tense and uncertain situation."
The idea for AMBER Alerts came after the tragic 1996 abduction of 9-year-old Amber Hagerman, who was abducted while riding her bike with her brother in Arlington, Texas. As she kicked and screamed, Amber was pushed into the cab of the man's pickup. Her body was found five days later in a creek four miles away. Her killer has never been found.
Since AMBER Alerts were first used – 24 years ago this month – they have saved 977 abducted children, at least 61 of them from alerts over wireless phones. "National AMBER Alert Day" is observed each Jan. 13, the day Amber was abducted.
Nicholes has become close to Zoe and her mom and gave Zoe a present for her second birthday. He keeps photos of her in his cellphone. Brown is now a huge advocate for AMBER Alerts and pays close attention to every one she hears, going to the website to make sure she is armed with all available information.
"Normally, I'll get an AMBER Alert, and I'll be like, oh wow, and I just kinda, you know, push it to the side," said Brown. "And since then, I'm looking at every AMBER Alert, and I'm making sure I pay attention….I know this helped me, and I want to be able to help somebody else."
[Coming soon: A video interview with Zoe's mom, Danielle Brown, by our own Angeline Hartmann.]
Sign Up! ► Get the latest updates from NCMEC
Legal T&C
Donor Privacy
Copyright © National Center for Missing & Exploited Children. All rights reserved.
This Web site is funded, in part, through a grant from the Office of Juvenile Justice and Delinquency Prevention, Office of Justice Programs, U.S. Department of Justice. Neither the U.S. Department of Justice nor any of its components operate, control, are responsible for, or necessarily endorse, this Web site (including, without limitation, its content, technical infrastructure, and policies, and any services or tools provided).
3.2.21.P
Subscription Errors
Thank you. Your request has been submitted
Enter code or click refresh to get a new code.


After a challenging opening round for the eleventh season of Radical Australia Cup on Mount Panorama, the teams return to Sydney Motorsport Park this weekend for three races and more valuable championship points.
Bathurst saw some outstanding drives right throughout the field, but by the close of the second 50-minute race it was former champion Neale Muston who stood tall at the top of the points on his return to Radical SR3 competition, whilst for former rival and fellow three-time champion Peter Paddon the weekend didn't quite go to plan, a small technical issue whilst comfortably leading the opening race having a big impact on his points position.
Likewise title rival Chris Perini also suffered misfortune with his own problems in race two whilst comfortably holding a podium finish, the two of them though expected to reignite their Bathurst rivalry this weekend in Sydney.
Historically Paddon is the most successful Radical driver at Sydney across the last four seasons with five wins from eight starts, whilst Perini top qualified in both races that concluded the 2018 Radical Australia Cup season in Sydney, taking a win against Paddon in the process.
They'll both have a close eye on their GWR team-mate Kosta Pohurokov who returns after a frustrating run at Bathurst which saw contact with the barriers during practice, contact which brought his weekend to a premature end. With a stunning new livery on his Tilton Racing SR3RSX, Kosta is expected to be right in the fight at the front of the field, but he won't be the only local driver to make his presence felt.
Brad and Mitch Neilson were the standout team in the opening race at Bathurst, using strategy to perfection to dominate the first race after an early Safety Car intervention, the former 'Rookies of the Year' also expected to be right in the mix fighting for the podium.
Tony Haggarty returns to active competition this weekend alongside son Simon in the second of the HRT entries, having subbed his SR3RSX to rising West Australian star Aaron Love at Bathurst, the experienced Radical campaigners always in the mix in Sydney, whilst Phil Anseline – who was also strong at Bathurst in his new RSX – will likely make his presence felt towards the front of the field.
Peter Clare is another who is expected to feature, the Shared Runway driver having been on a steep learning curve at Bathurst alongside Australian superstar Joey Mawson, and the 2018 'Most Improved Driver of the Year' will be keen to implement that knowledge this weekend.
As the home-base of Radical Australia, the Sydney Motorsport Park layout is well known to the bulk of the field with drivers like Rowan Ross, Michael Whiting and Bill Medland enjoying intimate knowledge of the 3.93-kilometre circuit, whilst for RA Motorsport's Chris Medland, round two will present his first opportunity to share the driving duties with son Max who makes his Cup debut in Sydney.
Sydney also welcomes the debut of Texan Jim Hernandez, the now-Queensland based businessman competing in his first Radical event after months of testing in his SR3RSX, the motorsport tragic looking forward to his first event and an opportunity to apply the experience he has gained through the mentoring process developed by Radical Australia over the last 11 years.
Live streaming will be in operation on both Saturday and Sunday – details are available through through the Australian Production Car website and the Radical Australia Facebook page.


text:
Ramadan (n.)
ninth month of the Muslim year, period of the annual thirty-days' fast, 1590s, earlier Ramazan (c. 1500), from Arabic Ramadan (Turkish and Persian ramazan), originally "the hot month," from ramida "be burnt, scorched" (compare Mishnaic Hebrew remetz "hot ashes, embers"). In the Islamic lunar calendar, it passes through all seasons in a cycle of about 33 years, but evidently originally it was a summer month.
Share Ramadan
https://www.etymonline.com/word/Ramadan
Etymology of Ramadan by etymonline
Harper, D. (n.d.). Etymology of Ramadan. Online Etymology Dictionary. Retrieved $(datetime), from https://www.etymonline.com/word/Ramadan
Harper Douglas, "Etymology of Ramadan," Online Etymology Dictionary, accessed $(datetime), https://www.etymonline.com/word/Ramadan.
Harper, Douglas. "Etymology of Ramadan." Online Etymology Dictionary, https://www.etymonline.com/word/Ramadan. Accessed $(datetimeMla).
D. Harper. "Etymology of Ramadan." Online Etymology Dictionary. https://www.etymonline.com/word/Ramadan (accessed $(datetime)).
Dictionary entries near Ramadan
rambunctious


Businesses across the United States rely on Innovative Informative Technologies to enhance the potentials of their respective applications and products. At Innovative, we recognize that most information technology divisions face constant pressure to deliver more for less. In order to assist our clients in meeting this challenge, we provide project based and staff augmentation services.
Our project based solution service includes a combination of onsite, offsite, and onshore and offshore support. Through a range of services, including but not limited to, Custom Application Development, Testing and Maintenance through Deployment, Upgrades and Consulting, we can help you to achieve efficient investment in core business activities, while simultaneously maximizing returns from limited IT budgets.
Innovative's Application Development and systems integration service provides your business with consulting and delivery expertise in both end-to-end systems integration and also in custom application development. Our services can help you make improvements to existing services and products and can also support your expansion into new markets, all while reducing costs.
Our recognized testing approach utilizes a wide range of industry standard testing tools and techniques that leverage proven methodologies for Planning, Process Control, Performance Metrics, Test Automation, Test Execution, Defect Identification & Tracking and Reporting. The end result is superior software quality and enhanced business functionality that gives your business a competitive edge.


text:
vehicle lineup
luxury / sport
Accent from $15,364.00
Accent Sedan from $15,014.00
Elantra from $17,564.00
ELANTRA GT from $20,014.00
Elantra Coupe from $20,614.00
Veloster from $21,514.00
Veloster Turbo from $28,414.00
Sonata from $25,764.00
Sonata Hybrid from $24,764.00
Sonata 2.0T from $33,564.00
Genesis Coupe from $28,264.00
Genesis from $41,874.00
Equus from $66,674.00
Tucson from $23,374.00
Santa Fe Sport from $28,474.00
Santa Fe XL from $32,874.00
Certified Pre-owned Pre-owned Both
Find yours now
Model Accent Azera Elantra Elantra GT Elantra Touring Entourage Equus Genesis Genesis Coupe Santa Fe Santa Fe XL Sonata Sonata Hybrid Tiburon Tucson Veloster Veracruz XG300 XG350
2020 2019 2018 2017 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2004 2003 2002 2001 2000 1999 1998 1997 1996 1995 1994 1993 1992 1991 1990 1989 1988 1987 1986 1985 1984 1983 1982 1981 1980 1979 1978 1977 1976 1975 1974 1973 1972 1971 to 2020 2019 2018 2017 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2004 2003 2002 2001 2000 1999 1998 1997 1996 1995 1994 1993 1992 1991 1990 1989 1988 1987 1986 1985 1984 1983 1982 1981 1980 1979 1978 1977 1976 1975 1974 1973 1972 1971
Postal Code Distance:
25 km 100 km 200 km 500 km 1000 km Unlimited
Province Transmission
No Preference Alberta British Columbia Manitoba New Brunswick Newfoundland Northwest Territories Nova Scotia Nunavut Ontario Prince Edward Island Quebec Saskatchewan Yukon No preference Manual Automatic
$75,000 $74,000 $73,000 $72,000 $71,000 $70,000 $69,000 $68,000 $67,000 $66,000 $65,000 $64,000 $63,000 $62,000 $61,000 $60,000 $59,000 $58,000 $57,000 $56,000 $55,000 $54,000 $53,000 $52,000 $51,000 $50,000 $49,000 $48,000 $47,000 $46,000 $45,000 $44,000 $43,000 $42,000 $41,000 $40,000 $39,000 $38,000 $37,000 $36,000 $35,000 $34,000 $33,000 $32,000 $31,000 $30,000 $29,000 $28,000 $27,000 $26,000 $25,000 $24,000 $23,000 $22,000 $21,000 $20,000 $19,000 $18,000 $17,000 $16,000 $15,000 $14,000 $13,000 $12,000 $11,000 $10,000 $9,000 $8,000 $7,000 $6,000 $5,000 $4,000 $3,000 $2,000 $1,000 $0 to $75,000 $74,000 $73,000 $72,000 $71,000 $70,000 $69,000 $68,000 $67,000 $66,000 $65,000 $64,000 $63,000 $62,000 $61,000 $60,000 $59,000 $58,000 $57,000 $56,000 $55,000 $54,000 $53,000 $52,000 $51,000 $50,000 $49,000 $48,000 $47,000 $46,000 $45,000 $44,000 $43,000 $42,000 $41,000 $40,000 $39,000 $38,000 $37,000 $36,000 $35,000 $34,000 $33,000 $32,000 $31,000 $30,000 $29,000 $28,000 $27,000 $26,000 $25,000 $24,000 $23,000 $22,000 $21,000 $20,000 $19,000 $18,000 $17,000 $16,000 $15,000 $14,000 $13,000 $12,000 $11,000 $10,000 $9,000 $8,000 $7,000 $6,000 $5,000 $4,000 $3,000 $2,000 $1,000 $0
0 km 10,000 km 20,000 km 30,000 km 40,000 km 50,000 km 60,000 km 70,000 km 80,000 km 90,000 km 100,000 km 110,000 km 120,000 km 130,000 km 140,000 km 150,000 km 160,000 km 170,000 km 180,000 km 190,000 km to 10,000 km 20,000 km 30,000 km 40,000 km 50,000 km 60,000 km 70,000 km 80,000 km 90,000 km 100,000 km 110,000 km 120,000 km 130,000 km 140,000 km 150,000 km 160,000 km 170,000 km 180,000 km 190,000 km
Used Hyundai Cars, Trucks and SUVs in Canada
Sort By: Distance Price High - Low Price Low - High
Visit the hyundaicertified.ca website regularly to find the used Hyundai for sale that you're looking for at a Hyundai dealer in Canada.
A pre-owned Hyundai vehicle is your smart choice if you're looking for a reliable and economical car or SUV at a highly competitive price. In fact, Hyundai vehicles meet the highest quality standards, in addition to offering car owners a host of exclusive advantages.
What's more, our site is specially designed to help you find a second-hand Hyundai for sale in your region that meets your purchase criteria in terms of price, mileage, model-year and transmission type.
You can also locate a Hyundai dealership to easily find the model you need among a vast selection of pre-owned Hyundai models for sale in your area in Canada.
Hyundai vehicles available by province:
hyundai certified full inventory
hyundai pre-owned full inventory
view hyundai certified inventory
view hyundai pre-owned inventory
schedule a test-drive
parts and service special offers
about hyundai
TMThe Hyundai names, logo, product names, feature names, images and slogans are trademarks owned by Hyundai Auto Canada Corp. †Hyundai's Powertrain Limited Warranty coverage covers most vehicle components against defects in workmanship under normal use and maintenance conditions. ††Any warranty claims made during the vehicle's original 5-year Limited Comprehensive Warranty period have no deductible. Each separate warranty claim made during the additional warranty period, after the vehicle's original 5-year Limited Powertrain Warranty period has expired, is subject to a $100 deductible. *Some restrictions do apply, see dealer for details. ©2005-2019 CarProof. All Rights Reserved.


New light weight exhaust system for the 2008-14 Nytro 4-stroke that is 7 ½ lbs lighter than stock. It is ceramic coated and boasts a 3-4 hp gain with only 1-2 db louder with excellent sound quality. This exhaust system is designed to install using all factory mounts, plus the exhaust outlet port is turned to keep the exhaust off your storage rack and tunnel.


text:
"The tea comes on gleaming silver trays carried in from the kitchen."
You are here: Home / Current Events
The Republican Response
By Rita Welty Bourke on September 10, 2012 in Current Events 2
On January 20, 2009, the day Obama was inaugurated, twelve Republican lawmakers gathered in a private dining room of an exclusive Washington restaurant. The meeting lasted for four hours. Before they parted, each of the Senators and Representatives
Secular Law Trumps Religious Law
By Rita Welty Bourke on March 4, 2012 in Current Events 0
"Congress shall make no law respecting the establishment of religion or prohibiting the free exercise thereof…" unless religion goes off a cliff. When that happens, secular law trumps religious law. History is full of examples
Abortion in America
By Rita Welty Bourke on February 14, 2011 in Current Events 1
During the 2009 health care debate, fundamentalist congressmen inserted language that eliminated public funding for abortions. They are now proposing legislation that would make it impossible for a woman to buy private insurance with her own money, if that insurance includes coverage for abortion
Hacker Republic: Fact or Fiction
By Rita Welty Bourke on January 5, 2011 in Current Events 5
When WikiLeak hackers took down MasterCharge, Visa, and PayPal after the arrest of Julian Assange, I thought of Stieg Larsson's book, "The Girl Who Kicked the Hornet's Nest." It's a crime thriller. Fiction. Made up stuff, some would say. Yet the similarities between what happens in this last of the Millenium series and the headline-grabbing […]
Avandia Still on the Market: Shame on You, FDA
By Rita Welty Bourke on July 23, 2010 in Current Events 2
Three years ago I was on a shuttle bus headed into downtown London from Heathrow airport. I felt awful. When you've spent the night in a cramped seat on an overnight flight from the U.S., you'd kill for a few hours sleep. But you know you have to stay awake for the next 12 hours; […]
War on Drugs or War Against the Citizenry?
By Rita Welty Bourke on May 26, 2010 in Current Events 0
Dressed in camouflage and bristling with high-powered weapons, a police SWAT team, looking for illegal drugs, broke into an apartment in Columbia, Missouri. The family dog got up from his spot on the rug and began to bark. They shot him dead. While the man, his wife, and small son watched, the police shot a […]
By Rita Welty Bourke on April 29, 2010 in Current Events 3
No one wants to serve on a jury. You especially do not want to serve if it's a murder case. But that's where I found myself last week, sitting on a jury in a criminal case against a man charged with eight felonies, one of them felony murder. By the time I was called, both […]
By Rita Welty Bourke on April 9, 2010 in Current Events 5
Why, I wonder, have there been no acts of violence against the pedophile priests who have molested children? Why have the fathers of these children not gone looking for their abusers, and when they found them, proceeded to beat them to within an inch of their lives? I read a story once of a family […]
Publications by Rita Welty Bourke
Kylie's Ark: The Making of a Veterinarian published April 2016 on Amazon, iBooks, and RitaWeltyBourke.com
"Synchronous Fireflies" appears online in The Copperfield Review
"Milk House Water" appeared in the January 2014 Chaffin Review
"The Burn Hole" appeared in The Southwest Review (2013)
"The Larry Brown Discovery Tour" appeared in the Chattahoochee Review and online at The Smart Set
"Tobacco Road Revisited" appears online in the Steel Toe Review


Early in the dark hours of Wednesday, May 11th, I will go to SeaTac and board a plane for Denver. I don't like air travel, but there is something about being at an airport that screams "adventure!" at me, and it will mean some much-needed time away from reality, so I'm actually looking forward to getting up that early even though I will complain (loudly, on Twitter probably). Now, the Rockies were not my first choice; originally, this trip was supposed to be for Chicago and Milwaukee; a friend of mine is a huge Cubs fan, and we were going to go see both the Cubs and White Sox, and take a road trip to see the Brewers; but for now, that has to be put off a little longer while I pay off my last remaining veterinary bill for JJs surgery last year. I am hoping to be able to afford it by next summer.
The Rockies came into existence around 1993, and played both their '93 and '94 seasons at a place called Mile High Stadium. Then the Coors company bought naming rights to the new stadium, which opened in 1995, and that is where the Blake Street Bombers have lived ever since. I will be attending two games vs the Mets on Friday the 13th and then Saturday the 14th, sitting in two different areas of the stadium. I originally figured I'd only go to one game, but I'm there, I might as well go to two. Plus, with a stadium that was reportedly designed by the same people who did Safeco, I figured that there would be a lot of cool stuff to make sure I got photos of.
If you look off to the side of the top of the lower picture, you can sort of see a light purple line that runs around the seating in the upper deck. And section 331 is pretty much the same place as it is in Safeco.
An approximation of the view from 331. The outcropping in center field is an area called the Rockpile, and when the Rockpile seats existed in the old stadium, they were a buck apiece. Now? Around $4-6. That is correct; $6 for a major league ball game. I've been a fan long enough to remember when Safeco's center field bleacher tickets sold for $10. Those days are long gone. Kudos to the Rockies for keeping their prices down.
During game broadcasts throughout the season, the ROOT Sports team is reliable for at least two things; a visit to Gates BBQ in Kansas City, and focusing on the Coors Field water feature. The water feature sits in the batter's eye, and prior to actually reading up on it, I thought it was something that you could walk through; a public destination like maybe the little kids ballpark outside of Kauffman Stadium. Unfortunately, it is only for decoration, but it is really close to the bullpens, so conceivably, a baseball could get lost in the woods during warmups.
So now we've got some of the big stadium stuff out of the way (I am sure that the art direction there will also be fun to discover, and I've got two days to do it), the tickets are bought, the rivalry is set, and the important part must be planned out; what to eat and drink while I'm there. You may scoff at my prepare-ahead attitude on food, but ballparks have so many choices these days that I have to plan what I'm doing so that 1) I don't spend valuable picture-taking or trinket-shopping time running around the ballpark to find stuff to eat, and 2) I have a general idea of where everything is. Before I went to Citizens Bank, I knew that Tony Luke's was in center field, and Yuengling was everywhere. That gave me time to feel comfortable just walking around the park and taking pictures, getting a feel for the place; the water ice upstairs that I didn't know was there was an added bonus. As was the bacon on a stick I ran into at Camden. So what are we looking at in Denver? Well, I've been told a Helton burger must be part of my trip. Andrew Martin at Purple Row reviewed it a few years back, and while the $14 price tag from 2011 has probably gone up a bit, I can deal with a nice burger for dinner on Friday. Also, the onion rings look fantastic. Saturday night, though, it's all about Extreme Dogs. Check out this menu!
Coors Field also has The Sandlot, where Blue Moon makes their marvelous light wheat beers, and The Rooftop, which looks kind of like The Pen, but upstairs instead of under the park. I won't be going to The Rooftop, as unlike The Pen, tickets are required to get in there specifically. But I will gladly nurse a beer at the Sandlot on Friday/Helton Burger night. This is the bit where I address the elevation. I've been to Denver before, enough times to know that one good beer (ie; not weak, watery commercial beer) is enough if I want to keep my head on. My foreign ballpark limit is two, but I think I might have to cut that in half for this trip. I'd like to post some photos here that don't look like I took them from a moving car on a dirt road. So I'm going to try and find something that I can't get here at home; I could not find a beer list on the Rockies' MLB.com site, but I am guessing that a place called Coors Field in a town like Denver will probably have a lot of options.
The gates at Coors Field open two hours before game time. This doesn't feel like enough time to see stuff to me, but it totally will be. I am headed for either the A or E gates so I have as much time as possible to take everything in. There are multiple lots around the stadium to park in, all $15, and again I have to give props to the Rockies for keeping parking reasonably affordable. I have my secret spots in SoDo for the Mariners, but independent lots there to get pretty out of control. Even Camden had $10 lots all around the ballpark.
Bonus prize if you're in the area; The National Ballpark Museum on Blake Street, less than a block or so from the stadium. We do plan on going to this one as well; like a tiny Cooperstown in the mile high city! I am sure I have probably missed a ton of things, but that's what makes the visit more fun; a basic knowledge of the park and the area to get your bearings, and all the other things discovered during the trip fill in the gaps. Just two and a half weeks away. I can hardly wait!
Pictures used without permission, credit Purple Row and The Denver Channel.
This entry was posted in 1 and tagged 2016, Coors Field, Stadium preview, Stadium Tour. Bookmark the permalink.
Just a heads up, you don't actually need a ticket to get into the rooftop area, unless it has changed this season.
Ooo! Good to know. I was looking on their site and it did say "Rooftop tickets", but maybe I misread it. That would be great if so, since if my geography of the area is correct, the Rooftop would allow for a good view …west? -ish?


KARACHI: The biggest settlement of Christians in Sindh imperilled by a group of local extremists as they continue to harass the local Christians.
According to reports, on 26th of June a group of extremists allegedly led by a man known as Muhammad Akhtar Ali bulldozed the security walls of Street number 3 of Essa Nagri by force spreading terror among the residents. These walls were built in 2012 when this same group allegedly killed five youngsters from the local Christian community.
As reported this extremist group of forcibly enters Essa Nagri and demand money from local Christian shopkeepers and various other local Christians. If someone is reluctant to pay them, they would persecute that person publicly oftentimes shooting them dead.
The distressed local Christians reached an NGO for help thus approached the local police station to launch an FIR against the extremists and their alleged leader. As the NGO claims the SHO of the police station was reluctant to take action and paid no attention to their request. Later on the incident was brought before the Deputy Superintendent of Police and Deputy Commissioner of Karachi East but thus far no one has taken notice of this incident to take due legal action against the nominated culprits.
Seeing no hope, the locals in company with the NGO officials approached the Wing Commander Rangers on 27th of June and requested him to provide protection from continues assaults by the extremist group led by Muhammad Akhtar Ali as they stated: he was continuously threatening to burn down Churches in Essa Nagri.
Karachi city is under operation clean up against criminals and terrorists conducted by Sindh Rangers and Karachi Police but protection of Christians is not important for law enforcement agencies when five Christians have been already murdered in Essa Nagri and Christians were feeling safe after construction of security walls. A representative of the NGO said: We also nominate Terrorist Akhtar Ali because he threats our community on the behalf of terrorists. Muhammad Akhtar Ali says that they will burn the all Christian community of Essa Nagri, because he has hordes of arms and ammunition and no any force can protect Christian from him. Muhammad Akhtar Ali also claims that they have power and good relations with higher authorities.
Later on 27th of June, local Christians and activists from a Human Rights organization staged a protest in front of the Karachi Press Club holding banners and raising slogans while demanding the Chief Minister of Sindh to provide them protection from the extremist group. They also urged upon the Director General Sindh Rangers and Inspector General of Sindh police to take action against the alleged leader of the extremist group arresting him immediately. The protestors demanded that the security walls be built as son as possible so as to curb possible entry of these extremists in the Christian colony in future.
Previous articleLive And Let Live !!


I work with paper quite a bit. And the cutting mat is my best friend. While I love how useful it is, it doesn't look pretty. Thus far, I've always just left it on my craft table because I loath having to search for tools when my DIY bug bites. After rummaging through boxes and drawers and getting hot and bothered, my motivation nose dives and I'd rather be sipping a latte instead. When tools are accessible, it just makes it so much easier to start, don't you think? A place for everything, as Momma always said.
The idea was to turn the mat into wall art. It's flat. I don't need to use the back side of the mat. Perfect! The question was how to hang it. Enter the BUMERANG Pants Hanger. I chose the pants hanger instead of skirt hanger as I didn't want little teeth marks on my board.
Once that was sorted, getting it up was easy peasy.
With a wet cloth, wipe clean the back of the cutting mat. Let dry.
On the wall, mark where you want to hang the mat and stick on the 3M hook. Do not hang anything on it for an hour.
Spray or spread glue all over the back of the mat. (I used latex glue as I ran out of the spray stuff.) With a stiff cardboard, spread out the glue if you are not using spray glue.
Carefully stick on the wrapping paper, starting from one end and smoothing it down with a stiff cardboard. Make sure that all the edges are stuck on well.
I have a smaller mat too ... still thinking what to do with it.
Clip on the BUMERANG Pants Hanger, centralising it.
Hang it on the hook and you're done!
If you do find your mat slipping, clip on a piece of anti-slip grip pad between the back side of the mat and the hanger. It'll hold. Mine is hanging pretty well.
Wow! I love that idea! I'm going to do it myself.. Thanks for the tip!
Beautiful!!! Thank you for sharing :) I'll try to do it.
Do try it Dulce. It's super easy.
What a wonderful idea! Any ideas on what to do with a large one?
This was done on the large one. I will probably make a chalkboard with the smaller one.
Would it be possible to just clip the smaller one behind the large one? Just a thought.
LOVE this idea. i am going to buy the mat through amazon when available. THANK YOU for your blog and Ikea hackers site!


How To Install Wiring Harness For Led Light Bar: Offroad light bar accessories wire up led. Car light bars auto wiring harness kit led hid bar. .
General guide on how to wire relay harness w off switch. Led light bar install on a t r toyota runner forum. Offroad light bar accessories wire up led. Car light bars auto wiring harness kit led hid bar. Installing instructions. Led light bar on off switch relay wiring diagram. Wiring diagram for led lights on light bar readingrat with. Led light bar wiring diagram moesappaloosas. How to install a light bar super bright leds. Led light bar wiring harness diagram and. Led light wiring harness with switch and relay single. Rigid industries wiring harness. Off road lights as high beams tacoma world. Off road led light bars installation guide super bright leds. How to install led light bar. Light bar wire diagram wiring and schematic. Led light bar install on gr wrx subaru forum. Mictuning amp relay fuse laser blue driving.


text:
Chelonus bipicturatus är en stekelart som beskrevs av Vladimir Ivanovich Tobias 2000. Chelonus bipicturatus ingår i släktet Chelonus och familjen bracksteklar. Inga underarter finns listade i Catalogue of Life.

Källor

Bracksteklar
bipicturatus

redpajama_set_name: RedPajamaWikipedia


text:
A Pristine Roman Mosaic Is Discovered Deep Beneath a Vineyard. Was It from an 1,800-Year-Old Winery?
An extraordinary find in incredible condition has excited archaeologists in northeastern Italy. But local winemakers told us why it's a special moment for them too.
Among the fragments found are some representations of gladiators with their armor and weapons.
For more than a century, the residents of the little comune of Negrar in northeastern Italy's Veneto region knew there was something special in the terroir that yielded its acclaimed Valpolicella wines. Evidence of ruins from a vast ancient Roman villa first surfaced in 1887, but most of the site was reburied under bureaucratic woes and then dirt; for decades, the town couldn't locate some of the old finds and hadn't sought new ones. But excavations in the past few weeks have turned up ornate mosaics as vibrant as they looked 1,800 years ago—right under the storied vines of today.
As the town's government announced May 26, "After countless decades of failed attempts, part of the pavement and foundations of the Roman villa located north of the capital, discovered by scholars over a century ago, has finally been brought to light."
Lead archaeologist Gianni De Zuccato and his Società Archeologica dug more than 13 feet in some places to find the incredibly intact Roman mosaic, the villa's floor, dated to the 2nd or 3rd century A.D. The geometric and floral designs add color to the vineyard rows now, but what's more, Unfiltered has learned that the villa likely had a direct connection to winemaking going all the way back.
The lost Roman wine villa always turns up in the last place you looked: A researcher unearths. (Sap Archeologia srl and Comune di Negrar di Valpolicella)
"The Roman villa itself has a structure that highlights a pars rustica for the processing and conservation of agricultural products, in particular the 'Rhaetian wines' so famous in the Roman world," a rep from the office of Daniele Accordini, general director of the town's wine cooperative Cantina Valpolicella Negrar, told us via email. (The pars rustica was the part of a Roman country house devoted to agricultural work; Rhaetian wine is what Romans called the bounty of what is now the Veneto up to the Danube). "Studies also highlight other evidence that the villa area [where the mosaic was found] was particularly suited to the cultivation of vines."
It's early days—just a few trenches have been dug—but we had to wonder: What happens to the 2020 A.D. vineyard up top? Winemaker Manuele Bronzo, who oversees one of the plots, owned by his uncle, told us he's as excited by the find as anyone right now. "It's a wonderful finding," Bronzo said Monday. "I mean, in Verona there are a lot of those sites all over the city …. Finding something that nicely preserved inside our vineyard was a little bit shocking, but it was good, it was cool." (As it happens, the vineyard provided grapes for a cuvée from the co-op a few years ago, featuring a label with one of the mosaics found earlier.)
There's always the possibility that vines will have to be removed for the excavation to continue. "If they want to uncover all of it, we have to take off some of the vines. But it's just a small site, about probably 50 plants. Not much," said Bronzo. And the discovery has been a thrill for the family and the community in the sleepy area. "We worked with the archaeologists in the digging procedure, and it was very exciting," said Bronzo.
Detail of the mosaic (Sap Archeologia srl and Comune di Negrar di Valpolicella)
Negrar mayor Roberto Grison and the cantina's president are already discussing the best way to display the mosaic, and the local government is working with vineyard owners to see what can be done, though nothing is certain at this point, said Bronzo; a museum is being considered. As the co-op rep told us, "the extraordinary news of the archaeological discovery has increased the international visibility of our small town and its great wines": The ancient masterpiece has already become a source of modern pride.
Why Does Copper Turn Green Over Time? 10 Secret Facts About Statue Of Liberty
Metal Spoon Eaten By Very Strong Scid


text:
\section{Introduction}
The idea of studying the interplay between ring-theoretic properties of a ring $R$ and graph-theoretic properties of a graph defined after it, is quite recent. It was first introduced for commutative rings by Beck in 1988 \cite{be}. In Beck's definition the vertices of the graph are the elements of the ring and two distinct  vertices $x$ and $y$ are adjacent if and only if $xy=0$. Later, Anderson and Livingston \cite{an} slightly modified this idea, considering only the non-zero zero divisors of $R$ as vertices of the graph with the same adjacency condition. Redmond \cite{re} extended this notion of zero-divisor graph to noncommutative rings.

Given a noncommutative ring $R$, we define two different graphs associated to $R$. The directed zero divisor graph, $\Gamma(R)$, and the undirected zero divisor graph, $\overline\Gamma(R)$. Both graphs share the same vertex set, namely, the set $Z(R)^{*}$ of non-zero zero divisors of $R$. In $\Gamma(R)$, given two distinct vertices $x$ and $y$, there is a directed edge of the form $x\rightarrow y$ if and only if $xy=0$. On the other hand, two distinct vertices $x$ and $y$ of $\overline\Gamma(R)$ are connected by an edge if and only if either $xy=0$ or $yx=0$.
Several properties of zero divisor graphs of different general classes of rings are studied in \cite{ak, ak1, ku, re, wu}.

Recall that, if $n>1$ is a rational integer and $\langle n\rangle$ is the ideal in the Gaussian integers generated by $n$, then the factor ring $\mathbb{Z}[i]/\langle n\rangle$ is isomorphic to the Gaussian integers modulo $n$
\begin{equation*}\mathbb{Z}_n[i]:=\{a+bi : a, b\in\mathbb Z_n\}.\end{equation*}
The zero divisor graph of the ring of Gaussian integers modulo $n$ has recently received great attention \cite{ab, ab1, na}.

The algebraic construction defined above for the Gaussian integers can be easily extended to the ring $\mathbb{Z}[i,j,k]$ of Lipschitz integer quaternions. Indeed, let
again be $n>1$ a rational integer and denote by $\langle n\rangle$ the principal ideal in $\mathds Z[i, j, k]$ generated by $n$. Then, the factor ring $\mathds Z[i, j, k]/\langle n\rangle$ is isomorphic to
\begin{equation*}\mathds Z_n[i, j, k]:=\{a+ib+cj+dk:a, b, c, d\in\mathbb{Z}_n\},\end{equation*}
which is called the ring of Lipschitz quaternions modulo $n$.

The aim of this paper is to study the zero divisor graphs of the ring of Lipschitz quaternions modulo $n$, both the directed $\Gamma(\mathbb Z_n[i, j, k])$ and the undirected $\overline\Gamma(\mathbb Z_n[i, j, k])$.

If $n=p_1^{r_1}\ldots p_k^{r_k}$, is the prime power factorization of $n$, the Chinese remainder theorem induces a natural isomorphism
\begin{equation}\label{decom}\mathbb Z_n[i, j, k]\cong\mathbb Z_{p_1^{r_1}}[i, j, k]\oplus\ldots\oplus\mathbb Z_{p_k^{r_k}}[i, j, k].\end{equation}
Therefore, in order to study the structure of the rings $\mathbb Z_n[i, j, k]$ we can restrict ourselves to the prime power case.

If $p$ is an odd prime and $l$ is a positive integer, then  $\mathbb Z_{p^l}[i, j, k]$ is isomorphic to the full matrix ring $M_2(\mathbb Z_{p^l})$ \cite{gmo,vi}. Consequently, for an odd positive integer $n$ the ring $\mathbb Z_n[i, j, k]$ is isomorphic to the matrix ring $M_2(\mathbb Z_{n})$. Hence, in this case we can use known results about the zero divisor graph of matrix rings over commutative rings.

Unfortunately, if $n$ is even it is no longer true that $\mathbb Z_n[i, j, k]$ is isomorphic to the matrix ring $M_2(\mathbb Z_n)$. In fact, note that an element
$z=z_0+z_1i+z_2j+z_3k\in\mathbb Z_n[i, j, k]$ is a unit if and only if its norm $\|z\|=z_0^2+z_1^2+z_2^2+z_3^2$ is a unit in $\mathbb Z_n$. Since
\begin{equation*}\label{E1}\|z+w\|=\|z\|+\|w\|+2\textrm{Re}(z\overline{w}),\end{equation*}
the sum of two units of $\mathbb Z_n[i, j, k]$ is never a unit. This fact is clearly false in $M_2(\mathbb Z_n)$ and the claim holds.

\section{The number of vertices}

Recall that both graphs $\Gamma(\mathbb Z_n[i, j, k])$ and $\overline\Gamma(\mathbb Z_n[i, j, k])$ share the same set of vertices. Namely, the non-zero zero divisors of $\mathbb Z_n[i, j, k]$. Due to the isomorphism (\ref{decom}) we can focus on the case when $n$ is a prime power.

For an odd prime power $p^\alpha$, we have the isomorphism $\mathbb Z_{p^\alpha}[i, j, k]\cong M_2(\mathbb Z_{p^\alpha})$. Thus, it is enough to determine the number of non-zero zero divisors in the matrix ring $M_2(\mathbb Z_{p^{\alpha}})$. We do so in the following proposition.

\begin{Proposition}
Let $p$ be an odd prime number and $\alpha\geq 1$. Then, the
number of zero divisors in $M_2(\mathbb{Z}_{p^\alpha})$ is
$$p^{4\alpha-1}+p^{4\alpha-2}-p^{4\alpha-3}.$$
\end{Proposition}
\begin{proof}
Let us denote by $\mathcal{N}(M_2(\mathbb{Z}_{p^{\alpha}}))$ the nilradical of $M_2(\mathbb{Z}_{p^{\alpha}})$. It is known \cite[p.125]{ke} that $\mathcal{N}(M_2(\mathbb{Z}_{p^{\alpha}}))=M_2(\mathcal{N}(\mathbb{Z}_{p^{\alpha}}))$ and recall that $\mathcal{N}(\mathbb{Z}_{p^{\alpha}})=\{a\in\mathbb{Z}_{p^{\alpha}}:\gcd(a,p)>1\}$ is precisely the unique maximal ideal of the local ring $\mathbb{Z}_{p^{\alpha}}$.
This implies that the factor ring $M_2(\mathbb{Z}_{p^\alpha})/\mathcal{N}(M_2(\mathbb{Z}_{p^\alpha}))$ is simple, because every ideal of the full matrix ring $M_n(R)$ over a unital commutative ring $R$ is of the form $M_n(L)$ for some ideal $L$ of $R$.

Now, since every artinian simple ring is isomorphic to the full matrix ring over a division ring \cite[p.39]{lam} and by the well-known Wedderburn's little theorem a finite division ring is a field, we conclude that
$M_2(\mathbb{Z}_{p^\alpha})/\mathcal{N}(M_2(\mathbb{Z}_{p^\alpha}))$ is isomorphic to a full matrix ring over a finite field.

Moreover, since
\begin{equation*}\left|M_2(\mathbb{Z}_{p^\alpha})/\mathcal{N}(M_2(\mathbb{Z}_{p^\alpha}))\right|=\frac{p^{4\alpha}}{p^{4(\alpha-1)}}=p^4,\end{equation*}
it follows that $M_2(\mathbb{Z}_{p^\alpha})/\mathcal{N}(M_2(\mathbb{Z}_{p^\alpha}))$ is isomorphic to the full matrix ring $M_2(\mathbb{Z}_p)$.

Finally, since every element of $\mathcal{N}(M_2(\mathbb{Z}_{p^\alpha}))$ is a zero divisor and
there are exactly $p^3+p^2-p$ zero divisors in $M_2(\mathbb{Z}_p)$, it follows that the number of zero divisors in
$M_2(\mathbb{Z}_{p^\alpha})$ is
$$p^{4(\alpha-1)}(p^3+p^2-p)=p^{4\alpha-1}+p^{4\alpha-2}-p^{4\alpha-3},$$
as claimed.
\end{proof}

\begin{rem}
\label{remuniimp}
Since in a finite unital ring every element is either a zero divisor or a unit, it follows that the number of units in $M_2(\mathbb Z_{p^\alpha})$ for an odd prime $p$ is
\begin{equation*}\label{uniimpar}p^{4\alpha}-p^{4\alpha-1}-p^{4\alpha-2}+p^{4\alpha-3}.\end{equation*}
\end{rem}

If $p=2$ the isomorphism $\mathbb Z_{2^t}[i, j, k]\cong M_2(\mathbb Z_{2^t})$ is no longer true. Consequently, we must use a different approach in order to compute the number of non-zero zero divisors of $\mathbb Z_{2^t}[i, j, k]$.

\begin{Proposition}
\label{t1}
The number of vertices of the graph $\overline\Gamma(\mathbb Z_{2^t}[i, j , k])$ is  $2^{4t-1}-1$.
\end{Proposition}
\begin{proof}
Note that an element $z=x_0+x_1i+x_2j+x_3k\in\mathbb Z_{2^t}[i, j , k]$ is a unit if and only if its norm
 $\|x\|=x_0^2+x_1^2+x_2^2+x_3^2$ is a unit in $\mathbb Z_{2^t}$. On the other hand, an element is a unit in $\mathbb Z_{2^t}$ if its reduction modulo $2$ is a unit. So consider the composition of homomorphisms
 \begin{equation*}\mathbb Z_{2^t}[i, j , k] \xrightarrow{\|.\|}\mathbb Z_{2^t}\xrightarrow{mod\; 2}\mathbb Z_2,\end{equation*}
 the kernel of this composition if the set of zero divisors in $\mathbb Z_{2^t}[i, j , k]$. Thus,
 \begin{equation*}
\mathbb Z_{2^t}[i, j , k]/ Z(\mathbb Z_{2^t}[i, j , k])\cong \mathbb Z_2,
\end{equation*}
 where  $Z(\mathbb Z_{2^t}[i, j , k])$ denotes the set of zero divisors of $\mathbb Z_{2^t}[i, j , k]$. Therefore $|\mathcal U(\mathbb Z_{2^t}[i, j , k])|=| Z(\mathbb Z_{2^t}[i, j , k])|=2^{4t-1}.$ Hence, there are exactly $2^{4t-1}-1$ non-zero zero divisors in $\mathbb Z_{2^t}[i, j , k]$, which is the number of vertices of the graph $\overline\Gamma(\mathbb Z_{2^t}[i, j , k])$ as claimed.
\end{proof}

\begin{rem}
\label{remuni2}
Observe that,  as in Remark \ref{remuniimp},  the number of units in $\mathbb Z_{2^t}[i, j , k]$ is $2^{4t-1}$.
\end{rem}

Recall that, given a direct sum of rings $R=R_1\oplus\ldots\oplus R_k$, an element $r\in R$ is a unit if and only if every projection of $r$ in $R_i$ is a unit in $R_i$. Hence, if $n=2^tp_1^{\alpha_1}\cdots p_k^{\alpha_k}$ is the prime power decomposition of $n$, isomorphism (\ref{decom}) together with Remarks \ref{remuniimp} and \ref{remuni2} lead to:
\begin{equation*}
|\mathcal U(\mathbb Z_{n}[i, j , k])|=\begin{cases}
\prod\limits_{i=1}^k\left(p_i^{4\alpha_i}-p_i^{4\alpha_i-1}-p_i^{4\alpha_i-2}+p_i^{4\alpha_i-3}\right), & \textrm{if $t=0$};\\ 2^{4t-1}\prod\limits_{i=1}^k(p_i^{4\alpha_i}-p_i^{4\alpha_i-1}-p_i^{4\alpha_i-2}+p_i^{4\alpha_i-3}), & \textrm{if $t>0$}.
\end{cases}
\end{equation*}

As a consequence of the previous work we have the main result of this section.

\begin{Theorem}
Let $n=2^tp_1^{\alpha_1}\cdots p_k^{\alpha_k}$ be the prime power decomposition of $n$. Then, the number of vertices in
the graph $\Gamma(\mathbb Z_n[i, j, k])$ or $\overline\Gamma(\mathbb Z_n[i, j, k])$ is:
$$
|V(\Gamma(\mathbb Z_n[i, j, k]))|=\begin{cases}
n^4-\prod\limits_{i=1}^k\left(p_i^{4\alpha_i}-p_i^{4\alpha_i-1}-p_i^{4\alpha_i-2}+p_i^{4\alpha_i-3}\right)-1, & \textrm{if $t=0$};\\
n^4-2^{4t-1}\prod\limits_{i=1}^k(p_i^{4\alpha_i}-p_i^{4\alpha_i-1}-p_i^{4\alpha_i-2}+p_i^{4\alpha_i-3})-1, & \textrm{if $t>0$}.
\end{cases}
$$
\end{Theorem}
\begin{proof}
Just apply the previous observation recalling that the non-zero elements of a finite ring are either units or zero-divisors.
\end{proof}

\section{The diameter}

We recall that the {\it distance} between two distinct vertices $a$ and $b$ of a graph, denoted by $\textrm{d}(a,b)$, is the length of the shortest path connecting them (the distance being infinity if no such path exists). The {\it diameter} of a graph $G$, denoted by $\textrm{diam}(G)$, is given by
\begin{equation*}\textrm{diam}(G)=sup\{\textrm{d}(a, b) : \textrm{$a, b$ distinct vertices of $G$}\}.\end{equation*}

Our objective in this section is to find the diameter of the directed zero divisor graph $\Gamma(\mathbb Z_n[i, j, k])$ and of the undirected zero divisor graph $\overline\Gamma(\mathbb Z_{n}[i, j , k])$.

Recall that $Z_L(R)$ and $Z_R(R)$ denote, respectively, the set of left and right zero divisors of $R$. The following result was proved in \cite{re}.

\begin{Theorem}
\label{rediam}
Let $R$ be a noncommutative ring, with $Z^{*}(R)\neq\emptyset$. Then $\Gamma(R)$ is connected if and only if $Z_L(R)=Z_R(R)$. If $\Gamma(R)$ is connected, then $\rm{diam}(\Gamma(R))\leq 3$.
\end{Theorem}

Note that in any finite ring $R$ we have $Z_L(R)=Z_R(R)$. Hence, the previous theorem implies that the directed zero divisor graph $\Gamma(\mathbb Z_n[i, j, k])$ is connected and $diam(\Gamma(\mathbb Z_n[i, j, k]))\leq 3$. We will now see that, in many cases, equality holds. To do so, we first need a technical result involving the direct sum of finite unital noncommutative rings. A commutative version was established in \cite{ax}.

\begin{Lemma}
\label{l3}

Let $R=R_1\oplus R_2$, where $R_1$ and $R_2$ are finite unital noncommutative rings.  Then. ${\rm diam}(\Gamma(R))=3$.
\end{Lemma}
\begin{proof}
First note that, since  $R_1$ and $R_2$ are finite and noncommutaive if follows by the Wedderburn's little theorem that both $R_1$ and $R_2$ have nonzero zero divisors.

On the other hand, since $R_1$ and $R_2$ are rings with identity we can choose a unit $u_1$ from $R_1$ and a unit $u_2$ from $R_2$. Let
$x\in Z(R_1)^\star$ and $y\in Z(R_2)$ and consider the elements $(x, u_2), (u_1, y)\in Z(R)^\star$. We will prove that the distance between the vertices $(x, u_2)$ and  $(u_1, y)$ is 3. Indeed $(x, u_2)(u_1, y)=(xu_1, u_2y)\neq (0, 0)$. Hence $d((x, u_2),(u_1, y))>1$. On the other hand if $(a, b)\in Z(R)^\star$ satisfies
\begin{equation*}(x, u_2)(a, b)=(a, b)(u_1, y)=(0, 0),\end{equation*}
then we have $u_2b=0$ and $au_1=0$ implying $a=b=0$, a contradiction. Therefore, $d((x, u_2),(u_1, y))>2$. Finally, using theorem \ref{rediam} we get the result.$\Box$
\end{proof}

As a consequence, we have the following result.

\begin{Proposition}
Let $n$ be an integer divisible by at least two primes. Then,
$$\textrm{diam}(\Gamma(\mathbb{Z}_n[i,j,k]))= 3.$$
\end{Proposition}
\begin{proof}
It is enough to apply isomorphism (\ref{decom}) together with Lemma \ref{l3}.
\end{proof}

\begin{rem}
It is clear that $\textrm{diam}(\overline\Gamma(\mathbb{Z}_n[i,j,k]))\leq\textrm{diam}(\Gamma(\mathbb{Z}_n[i,j,k]))$. Now, if $n$ is divisible by at least two primes, there exist vertices in $\overline\Gamma(\mathbb{Z}_{n}[i,j,k])$ that are not at distance 2. Hence, we obtain the equality $\textrm{diam}(\overline\Gamma(\mathbb{Z}_n[i,j,k]))=\textrm{diam}(\Gamma(\mathbb{Z}_n[i,j,k]))$ in this case.
\end{rem}

Now, we must focus on the prime power case. We first look at the odd case, where the following technical lemma is useful \cite[Lem. 4.2; Cor. 4.1]{bo}.

\begin{Lemma}
\label{l1}
Let $R$ be a commutative ring. If every finite set of zero divisors from $R$ has a non-zero annihilator, then ${\rm diam}(\Gamma(M_n(R))=2$. In particular, if $F$ is a field, then ${\rm diam}(\Gamma(M_n(F))=2$.
\end{Lemma}

\begin{Proposition}
Let $t\geq 1$ and let $p$ be an odd prime. Then,
$$\textrm{diam}(\Gamma(\mathbb{Z}_{p^t}[i,j,k]))=2.$$
\end{Proposition}
\begin{proof}
If $t=1$, $\mathbb Z_p$ is a field and the result follows from the second part Lemma \ref{l1}. Now assume that $t>1$. In this case the maximal ideal of the local ring $\mathbb Z_{p^t}$ is the principal ideal generated by $p$. This maximal ideal is nilpotent with index of nilpotence $t$. Therefore, the element $p^{t-1}$ belongs to the annihilator of every zero divisor in $\mathbb Z_{p^l}$ and Lemma \ref{l1} applies again.
\end{proof}

\begin{rem}
Again, $\textrm{diam}(\overline\Gamma(\mathbb{Z}_{p^t}[i,j,k]))\leq\textrm{diam}(\Gamma(\mathbb{Z}_{p^t}[i,j,k]))$. Now, if $p$ is an odd prime, there exist vertices in $\overline\Gamma(\mathbb{Z}_{p^t}[i,j,k])$ that are not at distance 1. Hence, we obtain the equality $\textrm{diam}(\overline\Gamma(\mathbb{Z}_{p^t}[i,j,k]))=\textrm{diam}(\Gamma(\mathbb{Z}_{p^t}[i,j,k]))$ in this case.
\end{rem}

Finally, we turn to the $p=2$ case. The following result computes the diameter of the undirected zero divisor graph.

\begin{Proposition}
\label{t2}
Let $t\geq 1$. Then $\textrm{diam}(\overline\Gamma(\mathbb Z_{2^t}[i, j , k]))=2$.
\end{Proposition}
\begin{proof}
The graph $(\overline\Gamma(\mathbb Z_{2^t}[i, j , k])$ is not complete since the vertices $1+i$ and $1+j$  are not adjacent.
On the other hand, from equation (\ref{E1}) if follows that the sum of two zero divisors in $\mathbb{Z}_{p^t}[i,j,k]$ is a zero divisor. Hence, the set of all zero divisors is an ideal. Clearly, this ideal equals the nilradical $\mathcal N(\mathbb Z_{2^t}[i, j , k])$.
 Since the nilradical is nilpotent, it follows that there is a vertex of $\overline\Gamma(\mathbb Z_{2^t}[i, j , k])$ adjacent to all others, and thus $\textrm{diam}(\overline\Gamma(\mathbb Z_{2^t}[i, j , k]))=2$.
\end{proof}

Recall that a directed graph $G$ is called symmetric if, for every directed edge $x\rightarrow y$ that belongs to $G$, the corresponding reversed edge $y\rightarrow x$ also belongs to $G$. We are going to prove that the directed zero divisor graph $\Gamma(\mathbb Z_{2^t}[i,j,k])$ is symmetric. As a consequence, it will follow that $\textrm{diam}(\overline\Gamma(\mathbb{Z}_{2^t}[i,j,k]))=\textrm{diam}(\Gamma(\mathbb{Z}_{2^t}[i,j,k]))$.

A ring $R$ is called reversible \cite{co} if, for every $a,b\in R$, $ab=0$ implies that $ba=0$. Clearly, a ring $R$ is reversible if and only if its directed zero divisor graph $\Gamma(R)$ is symmetric. Thus, to prove that $\Gamma(\mathbb Z_{2^t}[i, j , k])$ is symmetric we will prove that $\mathbb{Z}_{2^t}[i,j,k]$ is reversible. To do so, we need a series of technical lemmata.

\begin{Lemma}
\label{L4}
Let $w\in \mathbb{Z}[i, j, k]$. If $\|w\|\equiv 0 \pmod{4}$, then either all the components of $w$ are even or all of them are odd.
\end{Lemma}
\begin{proof}
Put $w=a_1+a_2i+a_3j+a_4k$ and denote by $n_w:=\textrm{card} \{i : a_i\ \textrm{is even}\}$. Since $a_i^2\equiv 0,1\pmod{4}$, it follows that $0\equiv||w||\equiv n_w\pmod{4}$ and hence the result.
\end{proof}

\begin{Lemma}
\label{L8}
Let $w\in \mathbb{Z}[i,j,k]$. If $\|w\|\equiv 0 \pmod {8}$, then all the components of $w$ are even.
\end{Lemma}
\begin{proof}
Put $w=a_1+a_2i+a_3j+a_4k$. Since $\|w\|=a_1^2+a_2^2+a_3^2+a_4^2\equiv 0\pmod{4}$ the previous lemma implies that either all the components of $w$ are even or all of them are odd. Assume that all of them are odd and put $a_i=2a_i'+1$ for every $i$. Then, $\|w\|=a_1^2+a_2^2+a_3^2+a_4^2=4((a_1')^2+a_1'+(a_2')^2+a_2'+(a_3')^2+a_3'+(a_4')^2+a_4)+4$. Hence, $((a_1')^2+a_1'+(a_2')^2+a_2'+(a_3')^2+a_3'+(a_4')^2+a_4')+1\equiv 0\pmod{2}$. This is clearly a contradiction and the result follows.
\end{proof}

\begin{Proposition}
\label{rev}
Let $w,z \in \mathbb{Z}[i,j,k]$. If $wz \equiv 0 \pmod {2^s} $, then $zw \equiv 0 \pmod {2^s}$. In other words, the ring $\mathbb Z_{2^t}[i, j,k]$ is reversible.
\end{Proposition}
\begin{proof}
We will proceed by induction on $s$.

The case $s=1$ is obvious since $\mathbb{Z}[i, j, k]/ 2 \mathbb{Z}[i, j,k]$ is trivially commutative.

Let us consider $s=2$ and assume that $w z \equiv 0 \pmod {4}$. Hence, $\|w\|\|z\|=\|wz\|\equiv 0 \pmod {16}$. If $\|w\|\equiv 0\mod{8}$ we can apply Lemma \ref{L8} to conclude that $w=2w'$ for some $w'\in\mathbb{L}$. Hence (using the case $s=1$) we have,
$$w z \equiv 0 \pmod {4} \Leftrightarrow w'z\equiv 0\pmod{2} \Leftrightarrow zw'\equiv 0\pmod{2} \Leftrightarrow zw\equiv 0\pmod{4}.$$
the same holds if $\|z\|\equiv 0 \pmod{8}$. Finally, if both $\|w\|,\|z\|\equiv 0\pmod{4}$ we apply Lemma \ref{L4} to conclude that all the components of $w$ and $z$ are odd. But in this case it can be easily seen that $zw-wz\in 4\mathbb{Z}[i, j,k]$ and the result follows.

Now, assume that $s>2$ and that $w z \equiv 0 \pmod {2^s}$. In this case $\|w\|\|z\|=\|wz\|\equiv 0 \pmod {2^{2s}}$ and, since $s>2$ it follows that either $\|w\|\equiv 0 \pmod{8}$ or $\|z\|\equiv 0 \pmod{8}$. If, for instance, $\|w\|\equiv 0 \pmod{8}$ we apply Lemma \ref{L8} again to conclude that $w=2w'$ for some $w'\in\mathbb{Z}[i, j ,k]$ and we can proceed like in the previous paragraph. The same holds if $\|z\|\equiv 0\pmod{8}$ and the proof is complete.
\end{proof}

\begin{rem}
The concept of {\em symmetric} ring was defined by Lambek in \cite{SIM}: a ring $R$ is symmetric if, for every $a,b,c \in R$, $abc=0$ implies that also $acb=0$. It has sometimes been erroneously asserted (and even ``proved'') that reversible and symmetric are equivalent conditions. If a unital ring is symmetric, then it is also reversible. But this is no longer true for non-unital rings, as illustrated by an example of Birkenmeier \cite{BIR}. In the case of unital rings, the smallest known reversible non-symmetric ring was given in \cite{REV2}. Namely, it is the group algebra $\mathbb{F}_2 Q_8$ where $Q_8$is the quaternion group. In \cite{GUT} it was proved that this is in fact the smallest reversible group algebra over a field which is not symmetric. In \cite{LI} it was also confirmed that $\mathbb{F}_2 Q_8$ is indeed the smallest reversible group ring which is not symmetric. Note that $\mathbb{Z}_4[i, j, k]$ is a reversible ring due to Proposition \ref{rev} which is trivially non-symmetric. In fact, it is the smallest known ring with characteristic different from 2 with this property, having the same number of elements (256) as the aforementioned example $\mathbb{F}_2 Q_8$.
\end{rem}

The reversibility of $\mathbb Z_{2^t}[i, j,k]$ implies that $\textrm{diam}(\overline\Gamma(\mathbb{Z}_{2^t}[i,j,k]))=\textrm{diam}(\Gamma(\mathbb{Z}_{2^t}[i,j,k]))$, so from all the previous work we obtain the following result.

\begin{Theorem}
\label{diam}
Let $n$ be any integer. Then
$${\rm diam}(\overline{\Gamma}(\mathbb Z_n[i, j, k]))={\rm diam}(\Gamma(\mathbb Z_n[i, j, k]))=\begin{cases} 2, & \textrm{if $n$ is a prime power};\\ 3, & \textrm{otherwise}.\end{cases}$$
\end{Theorem}

Recall that a graph $G$ is {\emph complete} provided every pair of distinct vertices is connected by a unique edge. In \cite[Theorem 15]{ab1} it was proved that the undirected zero divisor graph for the ring of Gaussian integers modulo $n$, $\overline{\Gamma}(\mathbb{Z}_n[i])$, is complete if and only if $n=q^2$, where $q$ is a rational prime such $q\equiv 3\pmod{4}$. In our case we have the following.

\begin{Corollary}
The graph $\overline{\Gamma}(\mathbb Z_n[i, j, k])$ is never complete.
\end{Corollary}
\begin{proof}
The diameter of a complete graph is 1. Since this is not possible due to Theorem \ref{diam}, the result follows.
\end{proof}

\section{The girth}
A {\it cycle} in a graph is a path starting and ending at the same vertex. The {\it girth} of $G$, denoted by $\textrm{g}(G)$, is the length of the shortest cycle contained in $G$. If the graph does not contain any cycle, its girth is defined to be infinity. All the previous concepts can be defined for directed graphs just considering directed paths.

Let us consider the directed zero divisor graph $\Gamma(\mathbb{Z}_{n}[i,j,k])$. If $n$ is even then $g(\Gamma(\mathbb Z_n[i, j, k]))=2$, since
\begin{equation*}\left[\begin{array}{cc}1&0\\0&0\end{array}\right]\left[\begin{array}{cc}0&0\\0&1
\end{array}\right]=\left[\begin{array}{cc}0&0\\0&1\end{array}\right]\left[\begin{array}{cc}1&0\\0&0\end{array}\right]=
\left[\begin{array}{cc}0&0\\0&0\end{array}\right].\end{equation*}
For $n=2^t$ the ring $\mathbb Z_{2^t}[i, j, k]$ is reversible and consequently $g(\Gamma(\mathbb Z_n[i, j, k]))=2$.

Now, we turn to the undirected case. It is clear that $\overline\Gamma(\mathbb{Z}_{n}[i,j,k])$ is a simple graph; i.e., it does not contain loops and two vertices are not connected by more that one edge. Thus, it follows that $g(\overline\Gamma(\mathbb Z_{2^t}[i, j , k]))\geq3$. We will now see that equality holds.

To compute the girth of $\overline\Gamma(\mathbb Z_n[i, j, k])$ for odd $n$, we recall the following result \cite[Prop. 3.2]{bo}

\begin{Proposition}
Let $R$ be a commutative ring. Then $\textrm{g}(\overline\Gamma(M_n(R)))=3$.
\end{Proposition}

This proposition clearly implies that ${\rm g}(\overline\Gamma(\mathbb Z_{n}[i, j, k])=3$ because, for odd $n$, we have that $\mathbb Z_{n}[i, j, k]\cong M_2(\mathbb{Z}_n)$.

The case $n=2^t$ is analyzed in the following result.

\begin{Proposition}
Let $t\geq 1$. Then ${\rm g}(\overline\Gamma(\mathbb Z_{2^t}[i, j , k]))=3$.
\end{Proposition}
\begin{proof}
If $t=1$, we have the cycle (see the previous remark) $(1+i)$---$(j+k)$---$(1+i+j+k)$---$(1+i)$, for instance.
If $t=2$, we have the cycle $2$---$2i$---$(2+2i)$---$2$, for instance. Finally, if $t>2$ we can consider the cycle $2^{t-1}$---$2$---$2^{t-1}i$---$2^{t-1}$. This proves the result.
\end{proof}

Finally, if we recall isomorphism (\ref{decom}), the previous discussion leads to the following.

\begin{Theorem}
\label{gir}
For every integer $n$, ${\rm g}(\overline{\Gamma}(\mathbb Z_n[i, j, k]))=3$.
\end{Theorem}

A graph $G$ is {\emph complete bipartite} if its vertices can be partitioned into two subsets such that no edge has both endpoints in the same subset, and every possible edge that could connect vertices in different subsets is part of the graph. In \cite[Theorem 17]{ab1} it was proved that the undirected zero divisor graph for the ring of Gaussian integers modulo $n$, $\overline{\Gamma}(\mathbb{Z}_n[i])$, is complete bipartite if and only if $n=p^2$, where $p$ is a rational prime such $p\equiv 1\pmod{4}$ or $n=q_1q_2$, with $q_1,q_2$ rational primes such that $q_1\equiv q_2\equiv 3\pmod{4}$. In our case we have the following.

\begin{Corollary}
The graph $\overline{\Gamma}(\mathbb Z_n[i, j, k])$ is never complete bipartite.
\end{Corollary}
\begin{proof}
The girth of a complete bipartite graph is 4. Since this is not possible due to Theorem \ref{gir}, the result follows.
\end{proof}

\section{The domination number}

A {\emph dominating set} for a graph $G$ is a subset of vertices $D$, such that every vertex not in $D$ is adjacent to at least one member of $D$. The {\em domination number}, denoted by $\gamma(G)$, is the number of vertices in a minimal dominating set.

The problem of determining the dominating number of an arbitrary graph is  NP-complete \cite{ga}. Nevertheless, particular cases have been recently studied. In \cite{ab1}, for instance, the dominating number of the zero divisor graph of the ring of Gaussian integers modulo $n$ was studied. In particular, the authors characterized the values of $n$ for which the domination number of $\Gamma(\mathbb Z_n[i])$ is $1$ or $2$.

This section is devoted to study the domination number of the undirected zero divisor graph $\overline\Gamma(\mathbb Z_{n}[i, j , k])$. The easiest case arises when $n$ is a power of 2.

\begin{Theorem}  \label{dompot2}
The domination number of the undirected zero divisor graph $\overline\Gamma(\mathbb Z_{2^t}[i, j , k])$ is $1$ for every $t\geq 1$.
\end{Theorem}
\begin{proof}
Just observe that $\{2^{t-1}(1+i+j+k)\}$ is a dominating set of the graph $\overline\Gamma(\mathbb Z_{2^s}[i, j , k])$ because $2^{s-1}(1+i+j+k)z=0$ for every non-zero zero divisor $z$.
\end{proof}

The rest of the section will be devoted to study de case when $n$ is an odd prime. In particular we will prove that, for an odd prime number $p$, the domination number of $\overline\Gamma(\mathbb Z_p[i, j , k])$  is $p+1$.

Let $\mathcal{U}(M_2(\mathbb Z_p))$ be the group of units in the matrix ring $M_2(\mathbb Z_p)$. We can consider a natural left group action on the vertices of $\overline\Gamma(M_2(\mathbb Z_p))$ defined by $(U, X)\rightarrow UX$ from $\mathcal{U}(M_2(\mathbb Z_p))\times Z(M_2(\mathbb Z_p))^*$ to $Z(M_2(\mathbb Z_p))^*$.

In the following lemma we characterize the structure of the orbits under the natural left action.

\begin{Lemma}\label{lso}
The distinct orbits of  the regular left action on $Z^*(M_2(\mathbb Z_p))$  are
\begin{equation*}o\left(\left[\begin{array}{cc}1&a\\0&0\end{array}\right]\right),\;\;\;for\;\;some\;\;a\in\mathbb Z_p,
\;\;and \;\;o\left(\left[\begin{array}{cc}0&0\\0&1\end{array}\right]\right).\end{equation*}
\end{Lemma}
\begin{proof}
Let $A$ be an element of $ Z(M_2(\mathbb Z_p))^*$. As is well-known from linear algebra performing a row operation on a matrix, with entries in a field,  is equivalent to multiplying on the left by a suitable invertible matrix. Since by row operations we can reduce any matrix to an upper triangular matrix it follows that there is an invertible matrix $U\in M_2(\mathbb Z_p)$ suct that
\begin{equation*}UA=\left[\begin{array}{cc}\alpha&\beta\\0&\gamma\end{array}\right],\end{equation*}
for some $\alpha, \beta, \gamma\in\mathbb Z_p$.

Assume first that $\alpha\neq 0$. Since $A$ is singular it follows that $\gamma =0$. So, we have
\begin{equation*}\left[\begin{array}{cc}\alpha^{-1}&0\\0&\alpha^{-1}\end{array}\right]\left[\begin{array}{cc}\alpha&\beta\\0&0\end{array}\right]=
\left[\begin{array}{cc}1&\alpha^{-1}\beta\\0&0\end{array}\right],\end{equation*}
and we get the result.

Now assume that  $\alpha=0$.  If $\gamma\neq 0$, then we have
\begin{equation*}\left[\begin{array}{cc}\gamma^{-1}&-\gamma^{-1}\beta\\0&\gamma^{-1}\end{array}\right]\left[\begin{array}{cc}0&\beta\\0&\gamma\end{array}\right]=
\left[\begin{array}{cc}0&0\\0&1\end{array}\right],\end{equation*}
as desired. If $\gamma=0$, since $A$ is not the nul matrix it follows that  $\beta\neq 0$. Hence we have
\begin{equation*}\left[\begin{array}{cc}\beta^{-1}\gamma&\beta^{-1}\\ \beta^{-1}&0\end{array}\right]\left[\begin{array}{cc}0&\beta\\0&\gamma\end{array}\right]=
\left[\begin{array}{cc}0&0\\0&1\end{array}\right],\end{equation*}
and the proof is completed.
\end{proof}

Similarly, if we consider the right regular action of $\mathcal{U}(M_2(\mathbb Z_p))$ on the vertices of $\overline\Gamma(M_2(\mathbb Z_p))$ defined by $(U, X)\rightarrow XU$, we have that the
distinct orbits are
\begin{equation*}o\left(\left[\begin{array}{cc}1&0\\ b&0\end{array}\right]\right),\;\;\;for\;\;some\;\;b\in\mathbb Z_p,
\;\;and \;\;o\left(\left[\begin{array}{cc}0&0\\0&1\end{array}\right]\right).\end{equation*}

On the other hand, for any ring $R$, if $u\in R$ is a unit and $x\in R$ is any element of $R$ we have $ann_r(ux)=ann_r(x)$ and $ann_l(xu)=ann_l(x)$. Consequently, the distinct right
annihilators of a single element in the ring $M_2(\mathbb Z_p)$ are
\begin{equation}\label{eq1} ann_r\left(\left[\begin{array}{cc}1&a\\0&0\end{array}\right]\right)=\left\{\left[\begin{array}{cc}-a z&-a w\\z&w\end{array}\right]\;\;:\;\;z,w\in\mathbb Z_p\right\}\end{equation}
and
\begin{equation}\label{eq2} ann_r\left(\left[\begin{array}{cc}0&0\\0&1\end{array}\right]\right)=\left\{\left[\begin{array}{cc}z&w\\0&0\end{array}\right]\;\;:\;\;z,w\in\mathbb Z_p\right\}\end{equation}

Since any vertex in the graph $\overline\Gamma(M_2(\mathbb Z_p))$ belongs to a right annihilator, if follows that
\begin{equation*}D=\left\{\left[\begin{array}{cc}1&a \\0&0\end{array}\right]\;\;:\;\;a\in\mathbb Z_p\right\}\cup\left\{\left[\begin{array}{cc}0&0\\0&1\end{array}\right]\right\}\end{equation*}
is a dominating set for the graph $\overline\Gamma(M_2(\mathbb Z_p))$. Therefore, $p+1$ is an upper bound for the domination number of the graph
$\overline\Gamma(M_2(\mathbb Z_p))$. We will prove that this upper bound is the exact domination number of $\overline\Gamma(M_2(\mathbb Z_p))$.

Similarly, the distinct left annihilators are
\begin{equation}\label{eq3} ann_l\left(\left[\begin{array}{cc}1&0\\ b&0\end{array}\right]\right)=\left\{\left[\begin{array}{cc}-b x&x \\-b y&y\end{array}\right]\;\;:\;\;x,y\in\mathbb Z_p\right\}\end{equation}
and
\begin{equation}\label{eq4} ann_l\left(\left[\begin{array}{cc}0&0\\0&1\end{array}\right]\right)=\left\{\left[\begin{array}{cc}x&0\\y&0\end{array}\right]\;\;:\;\;x,y\in\mathbb Z_p\right\}\end{equation}

Note that the intersection of two distinct right annihilators (or two left annihilators) contains only the zero element.

\begin{Lemma}\label{li}
The intersection of a right annihilator of type (\ref{eq1}) or (\ref{eq2})  with a left annihilator of type (\ref{eq3}) or (\ref{eq4})   is a set with $p$ elements. \end{Lemma}
\begin{proof}
First note that both a left annihilator and a right annihilator are  subgroups, of order $p^2$, of the additive group $M_2(\mathbb Z_p)$.  Since no right annihilator of type (\ref{eq1}) equals a left annihilator of type (\ref{eq3}) it follows that their intersection is subgroup of order $p$ or else  the trivial subgroup.

The result is clearly true if the right annihilator is of type (\ref{eq2}) or the left annihilator is of type (\ref{eq4}). So, assume that the right annihilator is of type (\ref{eq1}) and the left annihilator is of type (\ref{eq3}).
Note that, the cardinality of the intersection
\begin{equation*}ann_r\left(\left[\begin{array}{cc}1&a\\0&0\end{array}\right]\right)\cap
ann_l\left(\left[\begin{array}{cc}1&0\\ b&0\end{array}\right]\right),\end{equation*}
equals the cardinality of the solution set of the homogeneous system of linear equation
\begin{equation*}a z-b x=0;\;\;x+a w=0;\;\;z+b y=0;\;\;w=y.\end{equation*}
Since the coeficient matrix of this system is singular we get the result.
\end{proof}

\begin{Theorem} \label{domop}
The domination number of the zero divisor graph $\overline\Gamma(\mathbb Z_p[i, j , k])$, where $p$ is an odd prime number, is $p+1$.\end{Theorem}
\begin{proof}
Suppose the theorem were false. Then, we could find a dominating set $D=\{D_1, \ldots, D_k\}$, where $k<p+1$. Hence, there is a right annihilator $ann_r(X)$ of type (\ref{eq1}) or (\ref{eq2}) that is not among the $k$ right annihilators $ann_r(D_1), \ldots, ann_r(D_k)$ and a left annihilator $ann_l(Y)$ that is not among the $k$ left annihilators $ann_l(D_1), \ldots, ann_l(D_k)$. Since all the elements in the right annihilator $ann_r(X)$ are vertices of the graph $\overline\Gamma(\mathbb Z_p[i, j , k])$, it follows that
\begin{equation*}ann_r(X)\subseteq \cap_{i=1}^k ann_l(D_i).\end{equation*}
Therefore, $ann_r(X)\cap ann_l(Y)=\{0\}$, which contradicts lemma \ref{li}.
\end{proof}

As a consequence of the previous result we can easily compute the domination number of $\overline\Gamma(\mathbb Z_n[i, j , k])$ when $n$ is an odd square-free integer.

\begin{Theorem} \label{domosf}
Let $n= p_1\cdots p_k$ with $p_i$ prime for every $i$. Then, the domination number of the zero divisor graph $\overline\Gamma(\mathbb Z_n[i, j , k])$ is $ k+p_1+...+p_k$.
\end{Theorem}
\begin{proof}
Let $C_i :=\{M_{i,1},...,M_{i, 1+p_i}\} $ be the dominating set for the graph $\overline\Gamma(\mathbb Z_{p_i}[i, j , k])$ given by Theore \ref{domop}. Now, it is easy to see that the set
$$   \bigcup_{i=1}^k \{(0,0,...,M_{i,1},...0,0..),...,(0,0,...,M_{i,1+p_i},...0,0..)\} $$
is a minimal dominating set of $\overline\Gamma(\mathbb Z_n[i, j , k])$ and hence the result.
\end{proof}

In a similar way, we can proof the following result.

\begin{Theorem}
Let $n=2^s p_1\cdots p_k$ with $p_i$ is prime for every $i$ and $s>0$. Then, the domination number of the zero divisor graph $\overline\Gamma(\mathbb Z_n[i, j , k])$ is $1+k+p_1+...+p_k$.
\end{Theorem}

We end this section presenting two open problems:
\begin{enumerate}
\item For an odd prime number $p$ and a positive integer $t$, what is the domination number of $\overline\Gamma(\mathbb Z_{p^t}[i, j , k])$?
\item Let $t$ be a positive integer and denote by $\mathbb F_{q}$ the finite field of $q=p^t$ elements. For a positive integer $n>2$, what is the domination number of the zero divisor graph $\overline\Gamma(M_n(\mathbb F_q))$?
\end{enumerate}

redpajama_set_name: RedPajamaArXiv


It's the afternoon of November 17, 2017. The sky is grey, yet friendly; the house is quiet; and my NaNoWriMo journal sits on the dining room table, waiting. It's been sorely neglected today, despite my best intentions.
November is a little more than halfway over, my journal is a little more than halfway full, and I'm closing in on 25,000 words.
When I think about it, having written close to 25,000 words in just over two weeks sounds like an amazing thing. But I'm behind. I should have been at that point on Wednesday. I had hoped to reach 30,000 words by Sunday night, and I'm not sure if I will make it!
But I must strive for it. Because I want to be able to say at the end of the day on November 30, 2017 that I did it. That, for the second time in my life, I won NaNoWriMo!
Even though I'm a little behind, I'm not discouraged. If I keep plugging along, I'll get there. Maybe I'll catch up, then get ahead, and finish a day early!
As long as I don't get any surprise out-of-town guests for Thanksgiving next week, I think that is doable.


IDC Retail Insights: Worldwide Retail Intelligent Product Merchandising and Marketing Strategies advisory service examines best practice, trends, market conditions, retailers' concerns, and vendors' offerings to provide authoritative advice on investment, life-cycle management, and use of analytics in core retail processes — merchandising, marketing, and customer management. It examines early-stage technologies such as machine learning and other artificial intelligence (AI) technologies to guide early adopters as well as later-stage technologies to support application portfolio rationalization. Our efforts are focused on helping retailers spot and prioritize opportunities, save time, minimize cost, and mitigate risk. Specific attention is given to the drivers, benefits, processes, uses cases, and best practices related to omni-experience customer engagement, digital supply chain optimization, and curated merchandise life-cycle management.
IDC Retail Insights: Worldwide Retail Intelligent Product Merchandising and Marketing Strategies global analyst team keeps in constant touch with retail CIOs and their staff, line-of-business executives, and product strategy executives of leading technology vendors. We develop deep and comprehensive perspectives to provide actionable recommendations. Our methods include in-depth interviews, quantitative and qualitative market research, and secondary research from conferences, publications, blogs, online communities, and traditional news sources.
How should I approach improving my company's analytics competencies?
What are some of the best approaches to customer-centric omni-channel assortment planning?
How will the convergence of personal technologies and retailers' systems of engagement change retail?
What is ahead for planning for retail — financial, assortment, buying, space, pricing, and promotions?
How can analytics help us better understand our customers?
Who are the startling start-ups that might challenge complacent incumbents?
How will analytics change how retailers facilitate each consumer's path to purchase?
What capabilities are emerging to orchestrate localized and omni-channel assortment strategies?
How can retailers best take advantage of new marketing technologies and related services?
IDC Retail Insights: Worldwide Retail Intelligent Product Merchandising and Marketing Strategies research advisory service is geared toward helping merchants, planners, brand and digital marketers, pricing analysts, stores organizations, and roles dependent on bringing predictive analytical insights into their decision management frameworks. In equal measure, our research helps IT executives align their application portfolio priorities to their business partners' needs, reduce the cost and risks in managing these portfolios, and develop midterm and longer-term business-led IT strategies. Our research addresses the different needs of general merchandise, apparel, footwear, food, drug, and other hard-line and soft-line specialty retailers.


« Change is coming… like it or not.
It all began in my bar, the Bahia. A debate led to a soccer match unprecedented. The expatriate world against an Hydriot squad of sporting enthusiasts.
I am going to call the sport soccer, as opposed to football, which a bunch of fans call it. It should be called football as it entails kicking, and the use of hands is not permitted. Football for some obscure reasons is a North American sport which uses hands to get the ball around a field, and only rarely is it kicked. Even then a specialist is brought on to boot a field goal through a set of posts that resemble those used in rugby.
Again, to clear up any confusion, while the shape of the ball, ovalish, and the goal posts require one to kick the ball twixt posts and at a level some meters above ground, the American sport has nothing in common with the more internationally recognised sport called rugby.
Americans are allowed to use body armour and crash helmets and may take frequent breaks to catch their breath so that sponsors may ply their wares and pretty, leggy girls can wave pom-poms and high kick in mini skirts to keep the fans occupied, lest boredom set in.
Our sport has rules too, in case one may be under the misapprehension that on this side of the puddle, we play a game for wussies. For attempting to garrott an enemy player, one earns the ignoble decision of ten minutes in the sin bin. If a player is bleeding too profusely, he is entitled to ten minutes' stitching in the blood bin.
Okay so lets get back to our international match, which, by the way, did accumulate trips to the blood bin due to the coarse nature of the gravel pitch and the lack of perpendicular participation of contestants, aided no doubt by the regular half-time alcohol-laced refreshments.
Pan was our man on the other side, employed to recruit a team of Hydriots brave enough to take on the rest of the planet. With the exception of Pan and Janni Drakopolos, the average age on their side was nineteen; on ours, it was about 40 or 30 or 60.
Ouzo Jimmy was the ref and doubled as refreshment dispenser, which meant he situated himself behind the goal line, next to the porta-fridge, and conducted the match behind a dangling fag of smoke.
Somewhere during the game, the girls decided that they couldn't kick as far as the gents and allowed themselves to throw the ball (I believe that is how rugby was invented). And so the match progressed. Every time we were losing too much, Pan switched sides and became an American, generally right in front of the goal and evened the game up.
Pan, bless him, was a dual citizen. Having supposedly been caught jumping ship in New York, he was offered deportation or US citizenship if he went on a vacation to Vietnam. Which he did with distinction.
At the end of the tournament, because how many goals had been scored could not be determined, we took the discussion down to the 'office,' the Liako taverna. It was agreed that the score was an even 22-22 draw, in itself another Guinness book record.


Sometimes people do things and we really just don't know what to say. This is Yolanda "Lana" Beeler. Lana has been a wonderful supporter of the Gwendolyn Strong Foundation because of the Gooden family (she and TJ Gooden worked together) and their darling little girl, Nora.


"Here is my proposal for the John Meadows Brand. I have a mix of that old school iron and something that's professional. After reading a little about him. It's a two part logo. The mark and name can be used together or separate as part of the brand. The top part is a monogram of his initials that have that gritty old school look and incorporate part of his existing brand the mountain. I think this takes a piece of what he already has and gives it a fresh look. I think that the abstract mountain shape also signifies the climb to the top. He had setbacks with health problems and continued to train hard and recover. The bottom is a mix between legible text and his signature in the background."


text:
Samsung prepares DexBook, a portable screen to transform your smartphone into a PC
Samsung DexBook is working on a portable screen called DexBook. Specially designed to connect to smartphones, it is intended for users of the DeX functionality who wish to maintain total mobility while using their smartphone as a central unit.
The DEX feature has come a long way since its launch in 2017. The idea of Samsung was to allow users of high-end Galaxy to experience the environment of Android with the comfort of a PC screen, a physical keyboard, and a mouse. Since then, it has been possible to use it without going through a dedicated dock and Samsung is also feeding the project for a fully wireless DeX.
The firm wishes to push the concept even further and is developing a new type of accessory. According to the Korean ETNews site, which is often very well informed about the manufacturer, the latter works on a portable screen that can offer complete mobility to DeX users. The slab which is a diagonal of 14.1 inches would weigh less than a kilogram and is announced with a 10,000 mAh battery that will allow it to be used on the fly, without the need to connect it to a source of food.
Samsung DexBook: what interest?
We hardly perceive the interest of this new type of screen especially since it would cost between 400,000 and 500,000 won, or between 310 and 388 euros. We imagine it to be of little use if not to replace laptops on a daily basis. We already use our Android smartphones to perform many tasks while their increasingly comfortable power allows us to run very resource-intensive applications.
But this bet is far from being won for Korean. The ETNews report does not tell us much about the features of the device, much less about its design. The use of a flexible screen would be very relevant in order to make it less bulky and even more mobile. But this characteristic would not make it more of an incentive for most people.
LG for its part would work on the same type of device. Its screen would also have a diagonal of 14.1 inches for a weight of 600 g, much lighter than the DexBook. But the capacity of the battery would be two times less: 5000 mAh. The two devices are rumored to be on display by the end of the year. The portable screen of LG would have a price located in the same bracket as that of Samsung.
Black Friday Deals 2021 : Ticwris Max 4G Smart Watch Deals
HONOR 20 Lite Dual SIM Smartphone Deals


text:
Polling Iran
We've seen in earlier polling that the public has no problem with Barack Obama's idea of high-level meetings with the leaders of "bad guy" nations. Now comes along this set of polling from Public Agenda (via David Shorr) that likewise shows specific support for a liberal approach to Iran -- diplomacy aimed at improving relations between Washington and Teheran is by far the most popular option, and it's popularity has only grown.
Lurking elsewhere in the polling, a strong 65 percent correctly say that it's not realistic to expect the government to be able to deliver cheap oil. At the same time, this poll and every other poll I've seen shows that people view the high cost of energy as a major problem in their lives. It's almost as if the public might be drawn to a serious proposal to develop alternatives to massive oil consumption.
Matthew Yglesias is a former writer and editor at The Atlantic.