text:
Lecturer (Assistant Professor) Senior Lecturer/Reader (Associate Professor) - Applied Clinical Psychology
Job Title: Lecturer (Assistant Professor) Senior Lecturer/Reader (Associate Professor) - Applied Clinical Psychology
Location: Bath
Salary: Starting from £42,149, rising to £50,296 (Lecturer) or £51,799, rising to £65,573 (Senior Lecture/Reader)
Reference: CH8830
Dept: Psychology
Starting from £42,149, rising to £50,296 (Lecturer) or £51,799, rising to £65,573 (Senior Lecture/Reader)
Ref: CH8830
The Department of Psychology at the University of Bath is seeking a scientist with expertise relating to applied clinical psychology, potentially integrated with other areas of focus (e.g., cognitive science, developmental, health, or social psychology).
You will contribute to research and teaching in the Department, and the Masters in Applied Clinical Psychology in particular. Research excellence will be a key consideration for this vacancy. The Department is seeking applicants with proven research quality, with international reach, and the ability to contribute to teaching in our undergraduate and postgraduate programmes, while advancing our research and teaching.
Informal enquiries can be directed to the programme's incoming Director of Studies, Dr Melissa Atkinson (ma2293@bath.ac.uk; +44 (0)1225 386197) or to the Head of Department, Professor Greg Maio (g.r.maio@bath.ac.uk; +44 (0)1225 384647).
Research and teaching in the Department of Psychology occurs across five clusters: social and environmental psychology, clinical psychology, cognitive science, development and psychopathology, and health psychology. We are based in a new £30m purpose-built facility, containing two floors of state-of-the-art labs for interview, group observation, EEG/ERP, virtual reality, psychobiology, eye-tracking, pain, cross-modal and computerised testing.
In the 2014 Research Excellence Framework (REF) exercise, staff in the department were returned in different units of assessment, the largest of which (Allied Health Professions) obtained the highest REF2014 rating across the whole University. The University was ranked 12th overall in the UK in REF2014. The Department was recently ranked 2nd in the 2020 Complete University Guide. More information about the Department can be found at http://www.bath.ac.uk/psychology/
Closing date: Sunday 28 November 2021.
What we can offer you:
Our campus is surrounded by beautiful countryside, and the City of Bath with its historic charm and social activity only a short distance away creates an excellent environment in which to work.
a family friendly workplace
open to exploring flexible working opportunities which benefit you and the University.
recognising of the talents of our diverse workforce through an excellent reward package
encouraging of your personal and professional development
Find out from our staff what makes the University of Bath a great place to work. Follow us @UniofBath and @UniofBathJobs on Twitter for more information.
We aim to be an inclusive university, where difference is celebrated, respected and encouraged. We truly believe that diversity of experience, perspectives, and backgrounds leads to a better environment for our employees and students, creating a learning environment and organisational culture that enhances health and wellbeing across our community. We are very proud to have recently received Ambassadors for Autism certification and are an accredited Disability Confident Leader; committed to building disability confidence and supporting disabled staff.
The University of Bath is an equal opportunities employer and has an excellent international reputation with staff from over 60 different nations. We have made a positive commitment towards gender equality and intersectionality receiving a Silver Athena SWAN award. We are a family-friendly University, with an increasingly agile workforce, we are open to flexible working arrangements. We're also proud to be a disability confident employer and are happy to discuss any reasonable adjustments you may require.


/ What are the three toughest questions you can ever answer?
What are the three toughest questions that you can ever ask yourself?
I was recently at a high level workshop for experts and teachers and there were 60 of us there. We were all working on creating our next level product and service to help our clients. I spoke with many other experts in the field of health, nutrition, fitness, anti-aging, DNA restructuring, grief & anxiety counselling to name a few. We are all talking about a lot of great concepts designed to help people like you live a stronger and healthier life.
As I spoke with the other experts, I noticed that they all struggled with answering one of the three toughest questions I will give you in a minute. I'm curious to see if you currently struggle with one of these three questions. Are you ready to test yourself?
First question: Who are you?
The forest person is a person that is born and walks a specific life pathway. It's like walking in a forest of a specific area of study or industry. While the path may turn and twist and they can go from left to right, they never leave the forest. For example, musicians are in the forest of music. Movie actors are in the forest of acting. Chartered Accountants stay in the forest of accounting whether they are in public or private practice.
These people serve one industry their whole life. You are a forest person if you have a passion for one, life consuming area of expertise or interest.
The second type of person in life is a goal person. This is a person that has never really been beholden to one particular life purpose or forest. Instead, they go after project by project. If you're a goal person, you set your mind on a project and launch out to pursue it.
You want to become something or you want to do get something specific accomplished. You go get the project done. If you're a goal person you are always moving onto the next goal or project. It could be a similar goal to the one you last achieved or could be something completely a 180° different change of direction. This is good. The world needs accomplished goal getters who connect ideas with fanatical implementation.
The choice is really yours in knowing whether you're a forest person or goal person. Knowing who you are will really make a difference for you in how you accomplish your life mission and goals.
The next thing that you need to do in order to answer the question of who are you, is to devise out your own personal mission statement. Try to develop your mission statement that can be very eternal. Imagine it being written on your tombstone when you leave this earth.
Okay, so that's knowing who you are and knowing about your purpose or your life pathway.
The next question is: What do you want from life?
What do you want tomorrow?
What do you want a month from? Three months from now? Six months or in six years from now, what do you want?
What are you striving so hard for? What are you working for?
This question often really stops a lot of my clients from achieving their deepest desires and goals. How would you answer this question today?
I bet that if I were to work with you one-on-one, we might uncover a different answer than what you wrote down five minutes ago. It's natural to want one thing because of an external motivation, but then really long for an internally driven need or desire to accomplish a specific thing.
The next and third toughest question is: Why do you want what you want?
Why do you want to do that one particular thing that you're striving for?
At this point, I urge you to really think about this.
Sometimes you go after finding training and education to become something only to realize that it did make you happy. I can't tell you how many clients and friends who went through their first undergraduate degree to become a pharmacist or lawyer and then only realize after a few years in the profession that this was not what they imagined. One day they finally came clean with themselves.
The truth they later discovered is that being a pharmacist or lawyer was never their dream. It was their mother's or some other well-meaning person's dream for them. Some people bear down and feel that they have to prove to their father that they could accomplish something meaningful.
Before you commit and invest the time into something, it's really critical to ask "why do I want this" and if your motive is pure because it's something you feel would be worthy of your time, energy and resources, then I say congratulations. Make a plan and go for it.
So there you have it. Those are the three toughest questions that anybody can ask of themselves. If you can answer all three of them today, you are well off to being in the top 5% of all people in the world. It really does put you in the upper echelon of people set out to accomplish amazing things. You should be proud of this.
Then I say, rejoice, because you are at a place where you are struggling a little bit. That's okay and you get to choose the course of your life. You need to get comfortable with choosing what's right for you and struggling now on these three questions. The process will save you a lot of heart ache, pain and agony later on because you will have the clarity to advance forward with that calmness and confidence. Clarity of who you are, what you want and why you want it is what will save you a lot of time, money, aggravation and regret.
Congratulations if you're either certain of the answer to the three toughest questions or you're still figuring it out.
We have something that we're going to be launching here in the next little while. It's actually a movement starter.
I welcome your feedback in the comments below. I look forward to having a discussion with you. If you feel your questions or comments might be of a very personal nature. Feel free to send me an email at Shane[at]samuraiinnovation.com.


Wholesome nutrition, based upon feeding functional foods, is the key to maintaining a healthy immune system and resistance to disease. Discussion will focus on: ? Nutrigenomics, an emerging science that studies the molecular relationships between nutrition and the response of genes in promoting health. ? Different diets elicit different patterns of gene and protein expression as well as metabolite production; these are termed "molecular dietary signatures". ? Basic functional food ingredients and immune-balancing nutrients maintain health and reduce inflammation and disease. ? Promote healthy endocrine and immune function for longevity, well-being and cognitive awareness. ? Especially important to help maintain thyroid function and normal behavior in dogs with heritable and familial thyroid disorders.


text:
Darius Kasparaitis 1998-99 Road SEt 1 Jersey
NUMBER: 11 BIRTHDATE: October 16th 1972
HEIGHT: 5'11" BIRTHPLACE: Elektrenai, USSR
WEIGHT: 215 DRAFTED: NYI 1992 Entry Draft
SHOOTS: Left ROUND: 1st (5th Overall)
Darius began his career on Long Island in 1992-93 where he and the Islanders made a strong playoff push during which time, they defeated the 2-time defending Stanley Cup champion Pittsburgh Penguins. Darius quickly became a fan favorite, possessing a nasty edge to his game and getting known as one of the league's most gifted agitators. In 863 NHL games, he amassed 1379 well-earned PIMS. Following his time on the Island, he served with the Penguins for 6 seasons, and then with the Avalanche and Rangers. He has represented Russia in the 1996 World Cup of Hockey and the Olympic games.
 Darius Kasparaitis wore this road set 1 jersey for fifteen games. He picked up two assists while holding a plus four rating and logging fourteen penalty minutes.

redpajama_set_name: RedPajamaCommonCrawl


text:
We're committed to keeping our communities and our people safe. For the latest information related to COVID-19 click here
Minto.com
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Minto Group announces three exciting new appointments to its Board of Directors
OTTAWA, ON. August 31, 2020 – Minto Group announced the appointment of three new members to its Board of Directors this month. Floriana Cipollone, Gabriel Greenberg and Rachel Greenberg have joined the Board, bringing a depth of experience, leadership and perspective in finance, auditing, directorship, governance and philanthropy.
"With a Board in place for over 15 years, we have a strong history of governance within a privately-held company," said Michael Waters, CEO, Minto Group. "Our new members will complement the expertise within our long-standing Board—providing oversight on good governance policies and practices, and making key decisions to meet organizational and shareholder needs, aligned with our strategic vision."
Floriana Cipollone is the Chief Financial Officer of MCAN Mortgage Corporation (MKP-T) and previously CFO of Plaza REIT (PLZ.UN). Floriana brings over 30 years of robust financial experience to the table. Her knowledge of mergers and acquisitions, capital markets, corporate governance and risk, and investor relations will be a great asset to the Board.
Gabriel Greenberg is a Corporate Director of the Greenberg Family Enterprise, fully dedicated to the vision and success of the governance strategy. Before joining Minto's Board, he spent his career in the banking and real estate management industries in both Canada and the USA.
Rachel Greenberg continues to work for the Greenberg Family Enterprise, most notably driving the creation of a philanthropy strategy for the third generation of the family. In addition to joining Minto's Board, she works and volunteers for animal welfare clinics while training to be a veterinary nurse.
A complete list of Board Members can be found at minto.com.
About Minto Group:
Established in 1955, the Minto Group is a fully integrated real estate company offering new homes and condos, apartment rentals, furnished suites, commercial space, property and investment management. The company is a two-time winner of the Building Industry and Land Development Association's Green Builder of the Year award, four-time winner of EnerQuality's Ontario Green Builder of the Year award and two-time winner of the Ontario Home Builder Association's Home Builder of the Year award. The Minto Group was named one of Canada's Greenest Employers for 2019.
Minto Apartments wins Environmental Excellence and Certified Rental Building Program Award of Excellence at 2020 FRPO Awards
Minto Communities GTA wins six Ontario Home Builders' Association Awards
Minto Communities GTA is awarded prestigious EnerQuality Award for the first high-rise air leakage testing pilot program in Ontario
Minto Communities unveils the 20th Minto Dream Home for CHEO's Dream of a Lifetime Lottery
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The Minto Group is a fully integrated real estate company offering new homes and condos, apartment rentals, furnished suites, commercial space, property and investment management. The Minto Group refers collectively to Minto Holdings Inc. and its affiliated companies.
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Property has Great Potential for residential or commercial and is located on .56 of an acre in downtown area of Ashville, Al.! Home on property needs work and is not livable at this time. Property is located on Hwy 411 with 110 ft /- road frontage! Property is 309 ft. /- right side, 489 ft /- left side and 138 ft /- across the back. Property can't be financed because of condition of home and will be sold "As Is". This is a "Cash Only" sale, but Seller is negotiable on price.


Rodney Brown is a talented photographer from Hayward, Ca who currently based in Tracy, Ca, USA. He shoot with Polaroid SX-70.
Tony Viramontes was born in 1956 in Los Angeles, USA, of Spanish and Mexican parents.


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text:
import Debug.Trace
import Text.ParserCombinators.Parsec as PS
import System.IO
import System.Environment
import Data.List
import qualified Data.ByteString.Lazy as BL
import Data.Binary.Put
--import Control.Applicative as A
bin_size :: Int
bin_size = 100

-- the order of the automaton.  This counts the backbone as a degenerate case of skip nodes, you might call it the 0th order skip nodes.  So an automaton with a backbone and skipnodes that sum two consecutive nodes in the backbone for each skipnode would have a value of 2.
max_skipnode :: Int
max_skipnode = 3

desorption_thresh :: Float
desorption_thresh = 1.0

valouevFile :: GenParser Char st [(String, String, String, [Float])]
valouevFile =
            do  result <- PS.many record
                eof
                return result

record :: GenParser Char st (String, String, String, [Float])
record = do map_name <- fieldContent ; _ <- eol
            enz_name <- parseEnzyme ; _ <- char '\t'; enz_acr_name <- fieldContent; _ <- char '\t'; frags <- sepBy fieldContent (char '\t'); _ <- eol
            _ <- eol
            return (map_name, enz_name, enz_acr_name, fmap read frags)

parseEnzyme :: GenParser Char st String
parseEnzyme = do (char '\t' >> fieldContent)
                 PS.<|>
                 fieldContent

fieldContent :: GenParser Char st String
fieldContent = PS.many (noneOf "\t\n")

-- The end of line character is \n
eol :: GenParser Char st Char
eol = char '\n'

parseOM :: [Char] -> Either ParseError [(String, String, String, [Float])]
parseOM input = parse valouevFile "(unknown)" input

extract_frags :: (String, String, String, [Float]) -> [Float]
extract_frags (_, _, _, frags) = frags

frag_delim :: [Float]
frag_delim = [100000.0]

dumpnode :: (Int, Int) -> Put
dumpnode (lab, value) = do if lab == 0
                           then trace (show value) $ putWord32host $ fromIntegral lab
                           else  putWord32host $ fromIntegral lab
                           putWord32host $ fromIntegral value

enumerate :: [a] -> [(a, Int)]
enumerate list = (zip list [1..])

dumpOrderList :: [Int] -> Put
dumpOrderList order_list = do
  let enumerated = enumerate order_list
  let actions = fmap dumpnode enumerated   -- broken, need to do full list, not just first element
  sequence_ actions

make_skipnode n = if even n
                  then n + 1
                  else n

make_backbone n = if odd n
                  then n - 1
                  else n

process_backbone_node :: Float -> Int
process_backbone_node lab = if retval == 0
                        then trace ("proc_backbone: " ++ (show lab) )  retval
                        else retval
                            where retval :: Int
                                  retval =  fromIntegral $ make_backbone $  round (lab * 1000)

process_backbone_nodes :: [Float] -> [Int]
process_backbone_nodes labs = fmap process_backbone_node labs

process_actual_skipnode :: Float -> Int
process_actual_skipnode = (\lab -> fromIntegral $ make_skipnode $ round (lab * 1000))
process_actual_skipnodes :: [Float] -> [Int]
process_actual_skipnodes labs = fmap process_actual_skipnode labs

--dumpNodes :: [[Float]] -> Put
dumpNodes skip_nodes nodes = do
  let num_nodes = fromIntegral $ sum $ fmap length skip_nodes
  putWord32host $ fromIntegral $ num_nodes + 2 -- for initial and final
  --initnode :: (Int, Int)
  let initnode = (2100000000, 0) -- fixme, should be 2**32 - 1 but not sure if unsigned clean

  --finalnode :: (Int, Int)
  let finalnode = (process_backbone_node 0.0, (length nodes) + 1)

  dumpnode initnode
  let actual_skipnodes = init  skip_nodes
  let backbone = [last skip_nodes]
  let actions = (fmap dumpOrderList (fmap process_actual_skipnodes actual_skipnodes)) ++ (fmap dumpOrderList (fmap process_backbone_nodes backbone))
  sequence_ actions
  dumpnode finalnode
  --return ()

 -- this attaches numbers in the same order as they are written to disk, FIXME: figure out a way to assign these numbers at the time they are written
enumerateOrderList :: (Int, [a]) -> [(a, Int)]
enumerateOrderList (start, list) = zip list [start..]

enumerateNodes :: [[a]] -> [[(a, Int)]]
enumerateNodes skip_nodes = let lengths = fmap length skip_nodes
                                prefixes = scanl (+) 1 lengths
                                prefix_list_pairs = zip prefixes skip_nodes in
                            fmap enumerateOrderList prefix_list_pairs

quantize :: Int -> Int
quantize val = if (val `mod` bin_size) < (bin_size `quot` 2)
               then val - (val `mod` bin_size)
               else val - (val `mod` bin_size) + bin_size

float_quantize :: Float -> Float
float_quantize val = (fromIntegral (quantize (round (1000.0 * val)))) / 1000.0

--FIXME Edge and Node should probably be types
dumpEdgePiece :: (Float, Int) -> Put
dumpEdgePiece (_, pos) = do putWord32host $ fromIntegral pos

dumpEdge :: [(Float, Int)] -> Put
dumpEdge enum_nodes = do let actions = fmap dumpEdgePiece enum_nodes
                         sequence_ actions
                         --return ()

dumpEdges :: [[(Float, Int)]] -> Put
dumpEdges edge_list = do putWord32host $ fromIntegral $ length edge_list
                         sequence_ $ fmap dumpEdge edge_list
                         --return ()

--dumpGraph :: [[Float]] -> [[(Float, Int)]] -> Put
dumpGraph all_skipnodes edges nodes = do dumpNodes all_skipnodes nodes
                                         dumpEdges edges -- FIXME: TODO

-- FIXME: enable this function
add_reversed :: [[a]] -> [[a]]
add_reversed [] = []
add_reversed a = (head a) : (reverse (head a)) : (add_reversed (tail a))

-- FIXME: trim tips of rmaps

main :: IO ()
main = do
  args <- getArgs
  let fname = (head args)
  let ofname = (last args)
  hdl <- openFile fname ReadMode
  ohdl <- openFile ofname WriteMode
  contents <- hGetContents hdl
  -- case parseOM contents of
  --  Left x -> print $ show $ x
  --  Right x -> print $ show $ intercalate frag_delim $ fmap extract_frags x
  case parseOM contents of
   Left x -> print $ show $ x
   Right x -> sequence_ [show_stats,  dump_file ]
              where  show_stats = print $ "quantization bin size:" ++ (show bin_size) ++  "nodes: " ++ (show num_all_nodes) ++ " edges: " ++ (show (length edges)) ++ " skip edge bundles: " ++ (show (length skip_edge_list))
                     dump_file = BL.hPut ohdl $ runPut $ dumpGraph all_skipnodes edges nodes
                     nodes = {-filter (\a -> a > 0.001) $-} fmap float_quantize $ intercalate frag_delim $ {- add_reversed $ -}  fmap extract_frags x
                     skipnode_list n = take ((length nodes) - (n - 1)) $ nth_skipnodes n nodes
                     all_skipnodes = fmap skipnode_list $ reverse [1..max_skipnode]
                     all_enumerated_skipnodes = enumerateNodes all_skipnodes -- this enumeration is file position, not backbone position
                     num_all_nodes = length $ concat all_enumerated_skipnodes
                     enumerated_initnode = (0.0, 0)
                     enumerated_finalnode = (0.0, num_all_nodes + 1)

                     all_rights :: [[(Float, Int)]]
                     all_rights = [[enumerated_initnode]] ++ (rights all_enumerated_skipnodes)

                     all_lefts :: [[(Float, Int)]]
                     all_lefts = (lefts all_enumerated_skipnodes) ++ [[enumerated_finalnode]]

                     junction_nodes_pair :: [     ([(Float, Int)], [(Float, Int)])    ]
                     junction_nodes_pair = zip all_rights all_lefts

                     product ::  ([(Float, Int)], [(Float, Int)])   -> [[(Float, Int)]]
                     product (left, right) =  sequence [left, right]

                     --products a = concat $ map product a
                     edge_lists :: [[[(Float, Int)]]]
                     edge_lists = fmap product  junction_nodes_pair
                     edge_skipable :: [([(Float, Int)], [(Float, Int)])]
                     edge_skipable = zip (all_rights ++ [[enumerated_finalnode]]) ([[enumerated_initnode]] ++ all_lefts)
                     actual_skip = skip_edge_pairs edge_skipable
                     skip_edge_list = fmap product $ fmap productable actual_skip
                     edges :: [[(Float, Int)]]
                     edges =  (concat  edge_lists) ++ (concat skip_edge_list)

  hClose ohdl

productable :: (([(Float, Int)], [(Float, Int)]), ([(Float, Int)], [(Float, Int)])) -> ([(Float, Int)], [(Float, Int)])
productable ((a,b), (c,d)) = (a,d)

skip_edge_pairs :: [([(Float, Int)], [(Float, Int)])] -> [(([(Float, Int)], [(Float, Int)]), ([(Float, Int)], [(Float, Int)]))]
skip_edge_pairs (x:y:[]) = []
skip_edge_pairs (x:xs) = if small $ head xs
                         then skip_edge_run x xs
                         else skip_edge_pairs xs

skip_edge_run :: ([(Float, Int)], [(Float, Int)]) -> [([(Float, Int)], [(Float, Int)])] -> [(([(Float, Int)], [(Float, Int)]), ([(Float, Int)], [(Float, Int)]))]
skip_edge_run x (y:[]) = []
skip_edge_run x xs = if small $ head xs
                     then (x, head (tail xs)) : skip_edge_run x (tail xs)
                     else skip_edge_pairs xs

small :: ([(Float, Int)], [(Float, Int)]) -> Bool
small (rights, lefts) =  case (last rights) of (label, pos) ->  label < desorption_thresh

-- can probably generate the middle portion of the left and right ends of edges using: take, drop, and transpose
-- 'sequence' will form the cartesian product of lists of lists

-- takes a list of skip lists and returns a list of heads; assumes shortest list first and decreasing size : FIXME: check for or eliminate this assumption
-- lefts means all the skipnodes have their left edge aligned, which actually means they form the right side of a junction
lefts :: [[a]] -> [[a]]
lefts [] = []
lefts [[]] = []
lefts ([]:skip_lists) =  fmap head skip_lists : lefts ( fmap tail skip_lists )
lefts skip_lists =  fmap head skip_lists : lefts ( fmap tail skip_lists )

-- FIXME: rights is inefficient, as it forces the construction and retension of intermediate lists, this could be done lazily and allow the used heads to be GC'd
rights :: [[a]] -> [[a]]
rights skip_lists = reverse $ lefts $ fmap reverse skip_lists
-- thanks to http://stackoverflow.com/questions/27726739/implementing-an-efficient-sliding-window-algorithm-in-haskell
windows :: Int -> [a] -> [[a]]
windows m = transpose . take m . tails

nth_skipnodes :: Num b => Int -> [b] -> [b]
nth_skipnodes n nodes  = fmap sum $ windows n nodes

--TODO: mark backbone specially

redpajama_set_name: RedPajamaGithub


Søbakkehus Danois is the 11th foal from Søbakkehus Mai-Tai, Gold Medal and Elite Danish Warmblood mare. Mai-Tai is presently the only mare that has been graded with 4 times 10 in her DWB mare test … and the only mare that is placed in top 5 of the DWB performance index. Impressive since she is listed together with one of her daughters, namely Søbakkehus Mia-Bella who in 2011 was ranked no. 1 with Mai-Tai as no. 4.
This absolutely fabulous mover, Danois – the full brother of last years filly Maggie – promises to become just outstanding and a stallion prospect.


As we head into winter weather and driving conditions, the safety and reliability of our vehicles becomes a priority, which is made apparent by the amount of activity taking place at Symons Tire and Automotive Centre.
This family run Mission business is no stranger to busy days and hard work, starting from humble beginnings back in 1973, when Larry Symons first opened his doors on the site of the old Mission Auction Barn (now the location of Prospect Equipment). Not only managing the day to day running of a new business, Larry and wife Sharon were also bringing up their family and dealing with unforeseen circumstances like the flooding of the property, due to high river levels.
By 1976 they had built up the business enough to purchase their own premises in their current location on Glasgow Avenue. Over the years they extended both the building and number of staff, and now have 10 bays and a team of 16 employees. A very different story from the early days with their one and only employee Daryl Alstad, who still works for them 44 years on! Over the years the team have worked together developing from a tire service, to brakes and mufflers, suspension, as well as general automotive repairs – basically a one-stop-shop for your vehicle needs, from cars to RV's and commercial trucks.
Today Larry and Sharon are "taking it easy" compared to the early days, however, their idea of easing into retirement is still spending the majority of their time at the shop with son Jason taking over. Having grown up with the business, he follows in their footsteps and continues to build the reputation the business has developed for exceptional customer service. He's also a firm believer of the importance of giving back to the community and among other interests has been involved in the local soccer club, Search and Rescue team and has been a volunteer firefighter since the age of 19.
The truest testament to their customer's loyalty is that they have continued to support them over the years, with their client's children and grandchildren now also using Symons Tire and Automotive Centre.
In the last of our special 125th Anniversary editorials, recognizing Mission's long serving businesses, one common theme has emerged, without the support of local residents they would not be the thriving, successful businesses they are today.
Recognizing one long serving business each month as we celebrate our 125th Anniversary.


Some of the worst jobs you'll ever work will try to tell you that "You've gotta be tough to work here!" They'll make you think you're too weak if you feel like you have to quit for your mental health. This is very rarely true.
If you wanted to be physically strong, working in a gym would likely inspire you to build muscle. Being surrounded by people lifting weights all day, may motivate you to get fitter. And if the gym offered access to personal trainers and opportunities to use the equipment whenever you wanted, you'd probably find it easy to build muscle.
If, however, you worked 14 hour days in an office setting, it would be tougher to get buff. It'd be especially tough if the workplace cafeteria only served junk food and the boss expected you to do hours of work from home each night. The workaholic mentality would lead to fewer opportunities to lift weights and you'd probably find it much harder to get physically strong in that environment, compared to a job working in a gym. Well, mental strength is much like physical strength. The environment plays a huge role in your ability to build mental muscle.
Of course, it's easy to feel like a job with regular stress is toxic some days. You shouldn't quit your job just because you have a bad day. However, if your work is trying to tell you that you're not adequate just because you can't meet unreasonable demands—even if other people do it anyway—you don't have to accept that.
I Don't Like My Job. Does Quitting Mean I'm Not Mentally Strong Enough to Hack It? | Inc.


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DemDaily: The Latest Ads!
Posted May 26, 2016 | by KimS | in DemDaily, Resource, Uncategorized
With the final "super Tuesday" 6-state primaries and caucuses June 7th, Sanders and Clinton have unleashed a wave of campaign ads.
Tuesday, Sanders announced the launch of a new $1.5 million ad buy in California.
Not to be outdone, the following day the Clinton campaign announced a "six-figure" buy that starts this Friday in major California media markets, with ads featuring actor Morgan Freeman and civil rights activist Dolores Huerta. It also includes targeted ads to Latino and Asian American voters.
For those not viewing in a primary or battleground state, a look at what is out on the airwaves (just click on pic to watch):
The buys are a reflection of the increasingly tight race between Clinton and Sanders as they enter the final stretch of the primary season.
Recent polling shows that the race has closed in California. Clinton's 48-41 lead over Sanders in March narrowed to 46-44 in May. --the Public Policy Institute of California
California, with the largest # of delegates at 475, will either clinch the nomination for Clinton, or provide Sanders with a substantial win and enough delegates to continue the battle.
One of the better anti-Trump ads: "Speak" by PrioritiesUSA Action
For the much-needed comic relief outside of the live media circus, the latest from Saturday Night Live.
DemList is a free daily newsletter, calendar and resource for Democrats and allies nationwide. Please help us keep you informed with a contribution today!
Check out the DemList Convention Calendar, the free central source for all events and updates at the Democratic Convention July 25-28 in Philadelphia.
Source: Politico
DemDaily: The Power and Perplexity of White Women Voters
March 12, 2019 White women make up one of the largest voting blocks in the United States and are a critical target constituency in the 2020 presidential election. According to national exit...
DemDaily: The Composition of Congress
July 18, 2018 With less than four months until the end of the 115th Congress, and with a record number of members departing the halls of the Capitol, the timing is right for a quick look at the...
Convention Central: Sexy Liberal Convention Kick-Off!
The talent is lining up for the Democratic Convention, kicking-off Sunday, July 24th, with comedianStephanie Miller and her "Sexy Liberal Comedy Tour" at Philadelphia's Walnut...
DemDaily: Voices4Hillary!
The presidential election has inspired the formation of hundreds of outside groups – dedicated to supporting their candidates' platform and campaign through independent media, policy...


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PES President Sergei Stanishev regrets Mario Centeno's resignation
PES President (L) with outgoing Eurogroup President Mario Centeno
Portuguese Finance Minister Mario Centeno has announced he will be stepping down from government, as well as from his position as Eurogroup President.
During Mário Centeno's tenure, Portugal achieved sustained economic growth, consecutive years of economic convergence with the Euro Area, the first budget surplus in several decades and a significant decrease of public debt. It was also during his oversight that the Eurogroup pushed forward with fundamental tools for the stability and success of the Euro Area, particularly the budgetary capacity for convergence and competitiveness.
PES President Sergei Stanishev, commenting on the news, said:
"Mario, during his tenure in the government of Antonio Costa, has been instrumental in bringing Portugal's economy back to the path of economic recovery. He has also been successfully presiding over the Eurogroup for over two years, including helping develop a comprehensive European response to this pandemic. Mario has been a leading figure in promoting wellbeing in his country, stability in the eurozone and sustainability in Europe. It is therefore with great regret that I received the news of his resignation. I wish him all the best for the future."
In the past few months, Mário Centeno led the discussions between European finance ministers regarding our common response to the Covid-19 socioeconomic impact. The ambition of the proposals on the table is also a reflection of his work and dedication.


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U.K. to launch new digital identity document verification technology (IDVT) that enables data sharing between public bodies and businesses for the purpose of identity verification
The UK government is pushing ahead with its nationwide digital ID plans, despite half of the responses to its public consultation on digital identity opposing the idea.
On April 6, 2022, new digital identity document verification technology (IDVT) that enables data sharing between public bodies and businesses for the purpose of identity verification will be introduced. It will be made available to UK employers, landlords, and letting agents who can use it to digitally carry out pre-employment criminal record checks, right to work checks, and right to rent checks.
The introduction of this digital IDVT is part of the government's far-reaching digital ID plans which were announced in March. The government has framed these digital ID plans as a way for UK citizens to "easily and quickly prove their identity using digital methods instead of having to rely on traditional physical documents."
Under these digital ID plans, UK citizens will be able to "create a digital identity with a trusted organisation" which can be used "in-person or online" and "via a phone app or website." These trusted organizations will then be given a "legal gateway" to "carry out verification checks against official data held by public bodies to help validate a person's identity." The government will also allow the "trust" generated by a single successful digital identity check to be passed to other organizations "where appropriate."
The trusted organizations that provide these digital identity solutions will need to get accredited and certified under legislation that the government plans to introduce. Once accredited and certified, they'll be "given a trust mark to demonstrate their compliance and will be defined as being a trust-marked organisation."
A new interim governing body, the Office for Digital Identities and Attributes (ODIA), will be set up in the Department for Digital, Culture, Media, and Sport (DCMS) and it will have the power to issue these trust marks. The ODIA will also publish a publicly viewable list of trust-marked organizations.
Other companies that rely on the digital identity solutions provided by trust-marked organizations won't need to be certified but may be subject to "flow-down conditions" such as agreements to not share the information they receive.
Before announcing these digital ID plans, the government sought views and feedback on its proposed approach to digital identity via a public consultation.
50% of the responses to this consultation were "against digital identity in principle" but the government didn't include these responses in its statistical analysis of responses to the consultation because they "did not engage with the questions." However, the government insisted that "outside the context of producing the statistical analysis, we have taken these responses into account as part of this consultation exercise."
The government also admitted that some respondents feared that "digital identities are going to be made mandatory for all people" but dismissed these concerns as "false" and said it will seek feedback on how to "encourage more inclusive digital identities."
"As set out in the consultation, there are no plans to make digital identities mandatory, but we recognise they are an emerging technology and people may not be fully aware of the privacy and security benefits," the government said. "Therefore we will take steps to increase understanding amongst potential users and engage with civil society groups to receive their expert feedback on how to increase inclusion, now and into the future."
The government added that it's "committed to ensuring" that "people will still be able to use available paper documentation."
The government's digital ID framework has completed alpha testing. The next steps are a beta publication followed by beta testing before the framework is formalized in legislation.
The government cited "positive feedback received about the ability to conduct right to work and right to rent checks remotely during the COVID-19 pandemic" as one of its reasons for initiating its review of digital ID technology.
During the COVID-19 pandemic, the UK government embraced vaccine passports – a technology that shares many similarities with digital ID by requiring citizens to use a digital pass.
These vaccine passports were used to scoop up large amounts of data from UK citizens, some of which was shared with private companies.
Related: 🛡 How Big Government and Big Tech used COVID to accelerate the adoption of digital ID
Vaccine passports are one of many examples of the UK government using or proposing the use of technology to surveil its citizens. Other examples include it secretly surveilling millions of COVID vaccine recipients via their phones, proposing a social credit style app to encourage healthy eating, and proposing the increased use of surveillance drones to "protect" women.
Despite its history of surveillance, the government insists that this digital ID technology will have "strong security and privacy standards."
The government's digital ID plans were announced in the same month that the UK government's Online Safety Bill began its legislative journey. This bill mandates the implementation of identity and age verification technology on many large online platforms.
This current attempt to introduce digital ID comes almost a decade after the UK government launched its 2013 digital ID project "Verify" which was blasted by the National Audit Office and internal Parliamentary committees for "failing the public" and missing all of its performance targets.
Article: UK to launch new digital ID technology next month as part of nationwide digital ID push
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Next Next post: Indian law proposed to allow biometric data collection on arrest or detention


The juice is going down more quickly. It only took an hour today. On a side note, I actually had a movement this morning. Those were refreshingly missing yesterday!
I didn't juice as much Valencia as I wanted, due to time, so I will probably head to Urban Juicer to make up the difference.
My arms are sore from juicing!
My cycle started tonight. In hindsight, I realized I didn't have any cramps during the cleanse, and the mood dips were barely noticeable.


how many modifications a spaceship has and how fast the spaceship is. For every modification the spaceship costs 500 Zeni, and for every 100 words it knocks off travel time it's worth 500 additional Zeni.
Name: What is the name of the spaceship?
Appearance: What does your ship look like?
Modifications: Does your ship include any special things?
Travel Time: How many words are required to travel?
Cost: How much zeni does the ship cost?


(Keywords: Canola Oil Production, Canola Oil Plant, Canola Oil Processing, How Canola Oil Is Made, Canola Oil Production Line, Extracting Canola Oil, Canola Oil Production Line, Canola Oil Mill, Canola Oil Mill Project), Preparation of rapeseed oil use pre-pressing - leaching process mostly.
Rapeseed raw materials using vibrating screen or flat rotary screen screening impurity removal. As rapeseed contains more "side by side mud", also need to use vertical circular screen to the side of the mud to break, sieve out and remove with suction. After cleaning the rapeseed impurities should be less than 0.5%. The newly harvested rapeseed generally do not need to soften, but low water content of rapeseed to be used in softened layer of water in 9%, temperature 50-60 ℃ under softening conditions 12min. Then the use of vertical rolling mill or roller mill to roll the rapeseed into a 0.35mm thickness of the embryo, rapeseed into the rolling mill before the need to use the permanent magnet drum to remove the material contained in the iron. Rapeseed raw embryo by auxiliary steaming wok and pressing steamed frying pan after steaming into the pre-pressing machine mature embryo moisture content of 4-5%, temperature 110 ℃. The pre-squeezed oil is filtered and pre-extracted rapeseed oil is sent for refining. The pre-pressed cake was sent to the leaching plant for leaching. The solvent ratio was 1: (0.8-1). The mixed oil was pretreated, evaporated and stripped to obtain refined rapeseed oil. Leaching wet meal using high-layer steaming machine or vertical high-layer of steam stripping off, the rapeseed meal.
Challenge Maximum uptime while constantly improving your process.
Downtime is a costly business. To keep your operation up and running, a professional partner with the know-how to service your equipment and systems plus a reliable supply of spare parts are essential.
Dingzhou Yongsheng Part & Service staff helps optimize and keep your operations on-line throughout the life-time of the equipment.We are trying the shortest delivery time for spare parts ensure .
We take care of your service needs at our workshops or at your mill. Contact us 24/7 for technical assistance or a visit by a field service engineer. Our service include system trouble shooting, retrofits, repairs, hands-on staff training and site audits.


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Global Health in Medicine
Global Health in Medicine Program
More Within Global Health in Medicine Global Health in Medicine
Global Health Residency Elective & Track
Medicine Global Health Faculty
About Our Global Health Partnerships
Our global health partnerships guarantee optimal educational experiences for residents while serving local communities. Our main collaboration and training site is in Botswana, though residents can seek other international experience at other sites on a case-by-case basis. Also, many residents participate in local, regional and national global health rotations in underserved and marginalized communities through BIDMC's affiliated community health centers and the Indian Health Service.
Our main international partnership is with Scottish Livingstone Hospital (SLH) in Botswana. Through the Botswana Global Health Program, the Department of Medicine supports a full-time attending physician based at SLH and, and we offer robust residency and fellowship opportunities there. Learn more about our long-term partnership with SLH.
Beth Israel Deaconess Medical Center believes that community health centers (CHC) are in a unique position to provide accessible primary care and specialty services to medically underserved inner city and rural communities. One of the most successful "demonstration projects" of President Lyndon Johnson's War on Poverty, the community health center movement began in Boston in 1965. Beth Israel Deaconess was involved in community health from the earliest days, helping in 1969 to lead the conversion of Roxbury's New England Hospital for Women and Children into the Dimock Community Health Center. Our strong commitment and support of CHCs continues today as health centers are a vibrant and critical part of BIDMC's care delivery system.
A network of seven community health centers are affiliated with BIDMC, collectively providing care to 90,000 patients in 14 sites throughout Greater Boston and Cape Cod. In 1997, BIDMC was instrumental in helping these centers form a new network called Community Care Alliance (CCA). By collaborating together on clinical and administrative issues, CCA facilitates its members' delivery of high quality, cost-effective care by collectively contracting for services and funds, as well as sharing resources and expertise for the benefit of patients and communities.
Indian Health Service
The Indian Health Service (IHS), an agency within the Department of Health and Human Services, is responsible for providing federal health services to American Indians and Alaska Natives. Several BIDMC residents have participated in IHS rotations in Nome, Alaska and Shiprock, New Mexico.
Nome, Alaska is a town of about 3,000 people. Its hospital, Norton Sound Health Corporation, is also the referral center for about 14 other surrounding villages. Most of the physicians are either Family Medicine or Emergency Medicine trained. Participating residents will need to apply for an Alaska license prior to starting their rotation.
Shiprock, New Mexico is part of Navajo Nation, a Native American-governed territory covering almost 27,500 square miles in the Southwestern USA. Of the IHS locations available to BIDMC residents, Shiprock has been described as more "modern" with greater access to amenities. The local hospital, the Northern Navajo Medical Center, is also a larger hospital than Norton Sound, and there is clear division of departments (e.g. Internal Medicine, OB/GYN, and Surgery).
If you have more questions or are interested in pursuing a rotation through IHS, please email Dr. Anita Vanka, Associate Program Director of the Internal Medicine Residency Program. Visit the Indian Health Service website to learn more.
Other Global Health Projects and Partnerships
Many members of the Department of Medicine are involved in projects and partnerships in countries around the world, including:
Nephrologist Dr. Robert Brown is the current President of the volunteer organization TORCH (The Organization for Renal Care in Haiti), a consortium committed to bringing better kidney care to Haiti. TORCH works with the Partners in Health facility Hôpital Universitaire de Mirebalais in Mirebalais, Haiti.
Department Chair Dr. Mark Zeidel and other Medicine faculty have partnered with the Rambam Medical Center in Haifa, Israel, offering quality improvement and education consultation.
Dr. Eli Gelfand of the BIDMC CardioVascular Institute has partnered with global initiative Hearts Around the World since 2011. Dr. Gelfand and several members of the BIDMC community have traveled to Nairobi, Kenya to provide cardiovascular care and training at the Kenyatta National Hospital.
Dr. Tony Kaldany and the Division of Clinical Informatics are currently working to expand their international partnership to the University of Malaya in Kuala Lumpur, Malaysia.
Department Chair Dr. Mark Zeidel and other Medicine faculty members have partnered with the National University Health System (NUHS) in Singapore to create an exchange of education, research, and quality improvement resources, and to offer consultation on the NUHS residency and medical student curricula.
Dr. Lowell Schnipper of the Division of Hematology/Oncology partners with St. Albert's Mission Hospital in Harare, Zimbabwe, researching and implementing cervical cancer screening and prevention efforts. Dr. Schnipper's work centers around providing inexpensive, effective, and sustainable care for African women at risk for cervical cancer.


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"The Father of Locks" by Andrew Killeen (reviewed ...
Spotlight on September Books
Winners of the Light of the Burning Shadow Contest...
"Night Runner" by Max Turner (Mini-Review by Rober...
"The Choir Boats" by Daniel Rabuzzi (Reviewed by L...
Interview With Gary Gibson (Interviewed by Mark Ch...
News Flash Reminder: "The Quiet War" by Paul McAul...
"The Fall of Ossard" Book One in the Ossard Trilog...
"Prospero Lost" by L. Jagi Lamplighter (Reviewed b...
Memory, Physics and Identity: "The Einstein Girl"...
"The Light of Burning Shadows" by Chris Evans (Rev...
"Burning Skies" by David Williams (Reviewed by Mih...
"Water Keep: Far World Book 1" by J. Scott Savage ...
Interview with Adrian Tchaikovsky (Interviewed by ...
Sharing a World, Part II
2009 Booker Prize Nominee "The Children's Book" by...
"Traitors' Gate" by Kate Elliott with Bonus Q/A wi...
The Trojan War - A Reinterpretation: "The Troy Tri...
Spotlight Review: Man Booker Nominated Novel "Broo...
One More Superb Small Press Debut: "Angelglass" by...
"Warbreaker" by Brandon Sanderson (Reviewed by Mih...
The Hugo 2009 Finalists, Part 1 - The Graveyard Bo...
"Eyes Like Stars" Act One Theatre Illuminata by Li...
The Guardian Not The Booker Prize Stage 2: Longlis...
Masterpiece Debut: "Desideria" by Nicole Kornher-S...
The 2009 Hugo Awards - The Winners
"Blood of the Mantis" by Adrian Tchaikovsky (Revie...
The Legions of Rome: novelizations - "Soldier of...
"Soul Stealer: Blood and Rain" by Michael Easton &...
"Shiver" by Maggie Stiefvater (Reviewed by Cindy H...
Anticipation - Keep an eye on it via Convention Re...
Interview with Jennifer Fallon (Interviewed by Mih...
"Hitler's War" by Harry Turtledove (Reviewed by Li...
GIVEAWAY: "The Winds of Dune" Cosplay Contest!!!
"The Shadow Pavilion" by Liz Williams (Reviewed by...
"Land of the Dead" by Thomas Harlan (Reviewed by L...
"The Manhattan Prophet" by Jake Packard (Reviewed ...
Spotlight on August 2009 Books
"Prospero Lost" by L. Jagi Lamplighter (Reviewed by Liviu Suciu)
Official L. Jagi Lamplighter Website
Order "Prospero Lost" HERE or read an excerpt using the "look inside feature"
INTRODUCTION: I heard of "Prospero Lost" for a while but I never thought I will read it since it did not seem a book of great interest for me; in consequence I did not request an arc but when I saw it in stores, I opened it just to "do my duty" so to speak since I tend to open all new sff releases in bookstores just to get an idea of what's what. It hooked me from the first page so I bought it that day and it went to my top of the reading pile, while some 350 pages later I was still entranced and the announced sequel "Prospero in Hell" became an asap book for (hopefully) 2010.
OVERVIEW: Since escaping its island imprisonment described by Shakespeare - though that is just the official version approved by the Order of Solomon that keeps a lid on all things magical (!) - the sorcerer Prospero well, prospered; though his beneficent rule of Milan ended in flight in the 1500's and Miranda got jilted at the altar by Ferdinand, Prospero married several times across the centuries, got 6 more children beside Miranda, gave all them powerful magical staffs, founded Prospero, Inc which allowed science and technology to flourish leading to our current 21st century world by smoothing the dealings of the powerful supernatural beings and harnessing the elements so "science rather than chaos" works...
The company, Prospero, Inc is not perfect as our bloody and full of misery history shows, but the alternatives are too dreadful to contemplate. When the sorcerer disappears and leaves a message of danger and peril to Miranda who is his only child that remained active in the family business and lives in the Oregon main US mansion of the family, she needs to investigate and make sure that the "Three Shadowed Ones" do not wreak havoc on her family and on humanity at large.
Oh and Miranda is the "Lady's" handmaiden which allows her access to the "water of life"; this substance is what kept Prospero and his family immortal and is desired by all and sundry, at least the ones in the know, but it is very hard to get involving a long journey to "world's end" as well as being impossible to store except in special vials that only Miranda has.
And of course all depends on Miranda remaining a handmaiden with all that name implies, but if she can get to be a Sybill which is the supreme mortal rank in the "Lady's service", she will be free to marry, have children and so on...
"Prospero Lost" stands at about 350 pages and is narrated by Miranda. Mab the "incarnated wind" - aerie - chief detective of the family who resents deeply his "slavery" but is compelled to obey by Miranda's flute and Mephisto, her ne'er-do-well brother, moonlighting as a homeless lute player on the Chicago streets at the beginning of the novel are the other main characters. The ending is superb with a twist that will be explored in the next volume, though of course nothing is solved here so we need to wait for "Prospero in Hell".
ANALYSIS: "Prospero Lost" is another book of the "immerse in beautiful writing, expansive imagination and enjoy the ride" type which I so love when the novel in question engages me. Jasmyn, White is for Witching, Mortal Coils, and two upcoming releases that will be reviewed soon, The Choir Boats and Her Fearful Symmetry are similar books and in all, the one crucial factor is sustaining the suspension of disbelief. You engage with the novel and "live" in its world, you will love it, but do not look for consistent or detailed world building, deep psychological insights or non-stop action since then you will be disappointed and start nitpicking all of these kind of missing the point.
"Prospero Lost" is also quite whimsical and all creatures from myths, legends and fairy-tales make some appearance or are mentioned in its pages - or at least it seems so... While you can read an extended excerpt through the Amazon link above, the one line below represents the book well:
...Mephisto, and I had gone to Switzerland to meet with a yeti and discuss avalanches...
The novel is also strongly dependent on its narrator's voice, Miranda the 500 year old daughter of Prospero who has to solve the mystery of her father disappearance, keep civilization going by smoothing the dealings of supernatural beings which has been Prospero Inc mission for a while, warn her brothers and sister about the Shadowed Ones that Prospero belatedly let Miranda know about and just maybe finally become a Sybil so she can marry...
Mab and later Mephisto form a great counterpoint to Miranda's voice, Mab the "noir cynical detective" with a twist, and Mephisto, the clinically insane - or is he? - brother who somehow manages to see things clearer many times.
And there is Santa Claus first seen in a Mall of all places, the mixture of magical and mundane, hints of the big picture, snippets of past events with cameos of historical personages and events just to mention some more of the few goodies in the novel. And of course, not to forget, one of my favorite mythical beast, a Roc bird makes an appearance just in time to deal with a troublesome dragon that was attacking Miranda's plane, while another favorite, the Phoenix has an indirect role in the book so far.
A delight from page one to the end, Prospero Lost was another big positive surprise of 2009.
This may be an excellent book, but thanks to the way she comports herself online and in person at conventions, I'll be giving this one (and every one from her) a miss.
This sounds like an interesting book. Immortal Shakespaerean characters involved in a top-secret worldwide organization to hold society together--I have to admit, I've never quite read the like of it!
Rachel Heston Davis
www.rachelhestondavis.wordpress.com
I have this on hold at the library and I'm curious to see how I enjoy it.
As far as how the author conducts herself, I really do try to take personal views and lay them to the side.
Danielle Ackley-McPhail said...
I must say that I am saddened by the cancerous campaign being waged against L. Jagi Lamplighter. Having known her personally and professionally for over eight years, it is wholy undeserved and the result of one misconstrued comment that was not meant in the manner it was taken. I say with full confidence that she is the last person in the world to be bigoted or prejudice. So one word, "girl"--which she applies regularly to herself as well--has spawned and internet-wide anonymous slander campaign. I find this a very sad thing both for her, who has not earned such malevolence, and for the perpetuators, who do not believe in their own "righteousness" indignation enough to attach their name to their hateful comments and will not thus take the time to know a lovely and truly virtuous woman who has more understanding and compassion in her than I do.
Anonymous, you should all be ashamed of yourselves. You are doing the same thing to her that you accuse her of doing to others, only less honestly.


Ethara api Organization, organizing a forum to celebrate the opening of the Branch of the Organization of Greece in Greece.
Since independence in 1948, Sri Lanka had an average annual economic growth rate of 4.2 per cent and currently it is 2.6.This growth and Sri Lanka's market-oriented economic policy had a diverse impact on different sectors of society.
The blood donation program hosted by Qatar ethara api Branch was successfully held at the Sri Lankan Stafford School in Doha on 22 September. as the Chief invitees, Principal of the Sri Lankan Stafford School,Doha and Mrs.
Six sided Ethera api Challenge Cup, organized by the Rome Ethara Api Branch , was held in Campo Sportivo, Rome, on 24.09.29, in the glittering Lago Penestina. 18 cricket teams, including a Sri Lankan youth team from the napoli and 17 cricket teams from Rome, participated in the event.


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Author: Lee, D.A.
You may also wish to search for items by Lee, D. and Lee.
Counsell, J.F.; Lee, D.A.; Martin, J.F., Thermodynamic properties of organic oxygen compounds. Part XXVI. Diethyl ether, J. Chem. Soc. A, 1971, 313. [all data]
Counsell, J.F.; Lee, D.A.; Martin, J.F., Thermodynamic properties of organic oxygen compounds. Part XXVI. Diethyl ether, 1971, J. [all data]
Andon, R.J.L.; Counsell, J.F.; Lee, D.A.; Martin, J.F., Thermodynamic properties of aliphatic halogen compounds. Part 2. Heat capacity of 1,1,1-trichloroethane, J. Chem. Soc. Faraday Trans., 1973, I 69, 1721-1726. [all data]
Andon, R.J.L.; Counsell, J.F.; Lee, D.A.; Martin, J.F., Thermodynamic properties of organic oxygen compounds. Part 36. Heat capacity of isopropyl ether, J. Chem. Soc. Faraday Trans., 1974, I 70, 1914-1917. [all data]
Andon, R.J.L.; Counsell, J.F.; Lee, D.A.; Martin, J.F., Thermodynamic properties of organic oxygen compounds. 39. Heat capacity of n-propyl ether, J. Chem. Thermodynam., 1975, 7, 587-592. [all data]
Counsell, J.F.; Lee, D.A., Thermodynamic properties of organic oxygen compounds 30. Vapour heat capacity and enthalpy of vaporization of propanal, The Journal of Chemical Thermodynamics, 1972, 4, 6, 915-917, https://doi.org/10.1016/0021-9614(72)90013-4 . [all data]
Connett, J.E.; Counsell, J.F.; Lee, D.A., Thermodynamic properties of organic oxygen compounds XLIV. Vapour heat capacities and enthalpies of vaporization of methyl acetate, ethyl acetate, and propyl acetate, The Journal of Chemical Thermodynamics, 1976, 8, 12, 1199-1203, https://doi.org/10.1016/0021-9614(76)90129-4 . [all data]
Ambrose, D.; Ellender, J.H.; Gundry, H.A.; Lee, D.A.; Townsend, R., Thermodynamic properties of organic oxygen compounds LI. The vapour pressures of some esters and fatty acids, The Journal of Chemical Thermodynamics, 1981, 13, 8, 795-802, https://doi.org/10.1016/0021-9614(81)90069-0 . [all data]
Counsell, J.F.; Lee, D.A., Thermodynamic properties of organic oxygen compounds 31. Vapour heat capacity and enthalpy of vaporization of methanol, The Journal of Chemical Thermodynamics, 1973, 5, 4, 583-589, https://doi.org/10.1016/S0021-9614(73)80107-7 . [all data]
Andon, R.J.L.; Counsell, J.F.; Lee, D.A.; Martin, J.F., Thermodynamic properties of aliphatic halogen compounds. Part 2.---Heat capacity of 1,1,1-trichloroethane, J. Chem. Soc., Faraday Trans. 1, 1973, 69, 0, 1721, https://doi.org/10.1039/f19736901721 . [all data]
Counsell, J.F.; Lee, D.A.; Martin, J.F., J. Chem. Soc., Sect. A, 1971. [all data]
Ambrose, D.; Ellender, J.H.; Gundry, H.A.; Lee, D.A.; Townsend, R., Thermodynamic properties of organic oxygen compounds. LI. The vapour pressures of some esters and fatty acids, J. Chem. Thermodyn., 1981, 13, 795. [all data]
Ambrose, D.; Hall, D.J.; Lee, D.A.; Lewis, G.B., Vapor pressure of chlorine, J. Chem. Thermodyn., 1979, 11, 1089. [all data]
Lawrenson, I.J.; Lee, D.A., Thermodynamic properties of organic oxygen compounds XLVIII. The critical temperature of some alkanols, J. Chem. Thermodyn., 1978, 10, 1111. [all data]
Connett, J.E.; Counsell, J.F.; Lee, D.A., Thermodynamic Properties of Organic Oxygen Compounds. XLIV. Vapor Heat Capacities and Enthalpies of Vaporization of Methyl Acetate, Ethyl Acetate, and Propyl Acetate, J. Chem. Thermodyn., 1976, 8, 1199. [all data]
Counsell, J.F.; Lee, D.A., Thermodynamic Properties of Organic Oxygen Compounds 31. Vapor Heat Capacity and Enthalpy of Vaporization of Methanol, J. Chem. Thermodyn., 1973, 5, 583. [all data]
Counsell, J.F.; Lee, D.A., Thermodynamic Properties of Organic Oxygen Compounds 30. Vapor Heat Capacity and Enthalpy of Vaporization of Propanal, J. Chem. Thermodyn., 1972, 4, 915. [all data]
Counsell, J.F.; Lee, D.A.; Martin, J.F., Thermodynamic properties of organic oxygen compounds: xxvi diethyl ether, J. Chem. Soc. A, 1971, 1971, 313-6. [all data]
Jones, G.P.; Lee, D.A., The Enthalpy of Dilution of Concentrated Orthophosphoric Acid, J. Chem. Thermodyn., 1970, 2, 760-1. [all data]


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text:
Q: SUDOKU solver in c++ I need help to make a Sudoku solver in C++.
Requirements of the program:

*

*Input Sudoku:[Done]

*Read Sudoku: [Done]

*Validate function: (validates the Sudoku:) [Done]

*Display function [done]

*solve function [need help]

My plan:

*

*Make a function to validate the sudoku [done]
and assign possibility of 1-9 for every blank colume. and perform a brute-force
method to solve the puzzle. It will call the function validate each time
while doing brute force. It consumes a lot of time and I understood that it is
impossible.

I need a better algorithm for the solve function, please help me.
I have commented some of the code just to analyses the code: Apologize if that is the wrong way.
Here is what I have done so far:
/*  Program: To solve a SU-DO-KU:
    Date:23-sep-2013 @ 5.30:
*/

#include<iostream>
using namespace std;

int su_in[9][9]={
                 0,0,7   ,0,0,0    ,4,0,6,
                 8,0,0   ,4,0,0    ,1,7,0,
                 0,0,0   ,3,0,0    ,9,0,5,

                 0,0,0   ,7,0,5    ,0,0,8,
                 0,0,0   ,0,0,0    ,0,0,0,
                 4,0,0   ,2,0,8    ,0,0,0,

                 7,0,4   ,0,0,3    ,0,0,0,
                 0,5,2   ,0,0,1    ,0,0,9,
                 1,0,8   ,0,0,0    ,6,0,0
                };
int su_out[9][9];/*={

                 5,3,7   ,9,1,2    ,4,8,6,
                 8,2,9   ,4,5,6    ,1,7,3,
                 6,4,1   ,3,8,7    ,9,2,5,

                 9,1,3   ,7,4,5    ,2,6,8,
                 2,8,6   ,1,3,9    ,7,5,4,
                 4,7,5   ,2,6,8    ,3,9,1,

                 7,6,4   ,8,9,3    ,5,1,2,
                 3,5,2   ,6,7,1    ,8,4,9,
                 1,9,8   ,5,2,4    ,6,3,7
                };*/

struct chance{

    int poss[9];

};

int box_validate(int,int,int);
void read(int);
void display(int);
int validate(int);
int solve();

int main()
{
    int i,j,row,get_res=2;

    enum {input=1,output};

    cout<<"enter the sudoku: use 0 if the colum is empty:\n";

//  for(i=0; i<9; i++){
//  read(i);
//  }

    cout<<"\n\t\tTHE ENTERED SU-DO-CU IS:\n\n";
    display(input);

    get_res=validate(input);

    if(get_res==0)
        cout<<"valid input!\n";
        else if(get_res==1)
           cout<<"invalid input!\n";

//  display(input);
    for(i=0; i<9; i++)
    for(j=0; j<9; j++)
    su_out[i][j]=su_in[i][j];

//  display(output);
    get_res=validate(output);

    if(get_res==0){
        cout<<"valid solution!\n"; display(output);}
    //  else if(get_res==1)
    //     cout<<"invalid sudoku!\n";

    if(solve()==0) display(output);
    else cout<<"not solved buddy!!\n";

}

void read(int row) // function to read the SU-DO-CU
{
    unsigned num,i;
    cout<<"enter the row:'"<<row+1<<"'\n";

    for(i=0; i<9;   i++)
    {
    cin>>num;
    if(num<0||num>9)
    cout<<"error! invalid input: enter a number < or = 9 and > or = 0 \n";
    else
    su_in[row][i]=num;
    }
}

void display(int sudoku) // function to display the SU-DO-CU
{
    unsigned i,j;

        for(i=0;i<9; i++)
        {
            if(i%3==0)
            cout<<"\t\t-------------------------\n";

            cout<<"\t\t";
            for(j=0; j<9; j++)
                {
                    if(j%3==0)
                    cout<<"| ";

            //      if(su[i][j]==0)
            //      cout<<"_ ";
            //      else
                    if(sudoku==1)
                    cout<<su_in[i][j]<<" ";

                    else if(sudoku==2)
                    cout<<su_out[i][j]<<" ";

                    if(j==8)
                    cout<<"|";

                }

            cout<<"\n";
            if(i==8)
            cout<<"\t\t-------------------------\n";

        }

}

int validate(int sudoku) // function to validate the input SU-DO-CU
{
    unsigned i,j,k,n,count=0;
//..........................row validation
    for(i=0; i<9; i++)
        for(j=0; j<9; j++)
            for(k=0;k<9;k++)
                {
                    if(sudoku==1 && k!=j && su_in[i][j]!=0)
                       {
                        if(su_in[i][j]==su_in[i][k])
                        return 1;
                       }
                     else if(sudoku==2 && k!=j )
                       {
                        if(su_out[i][j]==su_out[i][k])
                        return 1;
                       }

                }
//..................................colume validation
    for(i=0;  i<9;  i++)
        for(j=0;  j<9;  j++)
            for(k=0;  k<9;  k++)
                {
                    if(sudoku==1 && k!=j && su_in[j][i]!=0)
                       {
                        if(su_in[j][i]==su_in[k][i])
                        return 1;
                       }
                     else if(sudoku==2 && k!=i )
                       {
                        if(su_out[j][i]==su_out[j][k])

                        return 1;
                       }
                }
    // each box validating.......................
    for(i=0;  i<=6;  i=i+3)
        for(j=0; j<=6; j=j+3)
        {
            if(box_validate(i,j,sudoku)==1)
            return 1;
        }

    // sum validation for output....

    if(sudoku==2)
    {
        for(i=0; i<9; i++)
        for(j=0; j<9; j++)
            count=count+su_out[i][j];

    if(count!=405) return 1;
    }

        return 0;

}

int box_validate(int i,int j,int sudoku)
{
    unsigned k=j,n=i;
    int temp_i=i,temp_j=j,temp_k=k, temp_n=n;
        for(i=temp_i;  i<temp_i+3;  i++)
        for(j=temp_j;  j<temp_j+3;  j++)
            for(k=temp_k;  k<temp_k+3;  k++)
                {
                  if(sudoku==1 && k!=j && su_in[i][j]!=0)
                 {
                    if(su_in[i][j]==su_in[i][k])
                     return 1;

                        for(n=temp_n;  n<temp_n+3;  n++)
                        {
                             if(sudoku==1 &&  su_in[i][j]!=0)
                              if(k==j && n==i)
                                ;else
                                  if(su_in[i][j]==su_in[n][k])
                                    return 1;

                        }
                }

                if(sudoku==2 && k!=j )
                {
                    if(su_out[i][j]==su_out[i][k])
                     return 1;

                        for(n=temp_n;  n<temp_n+3;  n++)
                        {
                             if(sudoku==1 )
                              if(k==j && n==i)
                                ;else
                                  if(su_out[i][j]==su_out[n][k])
                                    return 1;

                          }
                }
            }
    return 0;

}

int solve()
{
    unsigned i,j,k,m,n,x;

struct chance cell[9][9];

    for(i=0; i<9; i++)
        for(j=0; j<9; j++)
            if(su_in[i][j]==0)
                for(k=0;k<9; k++)
                cell[i][j].poss[k]=k+1;

                /*
    for(i=0; i<9; i++)
        for(j=0; j<9; j++)
            if(su_in[i][j]==0)
                for(k=0;k<9; k++)
                {   su_out[i][j]=cell[i][j].poss[k]=k+1;
                        su_out[i][j]=cell[i][j].poss[k];
                        for(m=0; m<9; m++)
                          for(n=0; n<9; n++)
                          for(x=1; x<=9; x++)
                          { if(x!=k+1)
                            cell[m][n].poss[x]==x;
                            if(validate(2)==0)
                               return 0;
                       }
            }*/

        for(i=0; i<9; i++)
        for(j=0; j<9; j++)
            if(su_in[i][j]==0)
            while (validate(2)==0)
            {

                for(k=0;k<9; k++)
                {   su_out[i][j]=k+1;
                        for(m=i+1; m <9 ; m++)
                          for(n=0; n <9 ; n++)
                            for(x=1; x<=9; x++)
                          { su_out[m][n]=x;
                            if(validate(2)==0)
                               return 0;
                       }
            }
          }
}

A: #include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
#define UNASSIGNED 0
#define N 9

bool FindUnassignedLocation(int grid[N][N], int &row, int &col);
bool isSafe(int grid[N][N], int row, int col, int num);

/* assign values to all unassigned locations for Sudoku solution
 */
bool SolveSudoku(int grid[N][N])
{
    int row, col;
    if (!FindUnassignedLocation(grid, row, col))
       return true;
    for (int num = 1; num <= 9; num++)
    {
    if (isSafe(grid, row, col, num))
    {
        grid[row][col] = num;
        if (SolveSudoku(grid))
            return true;
        grid[row][col] = UNASSIGNED;
    }
    }
    return false;
}

/* Searches the grid to find an entry that is still unassigned. */
bool FindUnassignedLocation(int grid[N][N], int &row, int &col)
{
    for (row = 0; row < N; row++)
        for (col = 0; col < N; col++)
            if (grid[row][col] == UNASSIGNED)
                return true;
    return false;
}

/* Returns whether any assigned entry n the specified row matches
   the given number. */
bool UsedInRow(int grid[N][N], int row, int num)
{
    for (int col = 0; col < N; col++)
    if (grid[row][col] == num)
        return true;
    return false;
}

/* Returns whether any assigned entry in the specified column matches
   the given number. */
bool UsedInCol(int grid[N][N], int col, int num)
{
    for (int row = 0; row < N; row++)
        if (grid[row][col] == num)
            return true;
    return false;
}

/* Returns whether any assigned entry within the specified 3x3 box matches
   the given number. */
bool UsedInBox(int grid[N][N], int boxStartRow, int boxStartCol, int num)
{
    for (int row = 0; row < 3; row++)
        for (int col = 0; col < 3; col++)
            if (grid[row+boxStartRow][col+boxStartCol] == num)
                return true;
    return false;
}

/* Returns whether it will be legal to assign num to the given row,col location.
 */
bool isSafe(int grid[N][N], int row, int col, int num)
{
    return !UsedInRow(grid, row, num) && !UsedInCol(grid, col, num) &&
       !UsedInBox(grid, row - row % 3 , col - col % 3, num);
}

void printGrid(int grid[N][N])
{
    for (int row = 0; row < N; row++)
    {
        for (int col = 0; col < N; col++)
            cout<<grid[row][col]<<"  ";
        cout<<endl;
    }
}

int main()
{
    int grid[N][N] = {{3, 0, 6, 5, 0, 8, 4, 0, 0},
                      {5, 2, 0, 0, 0, 0, 0, 0, 0},
                      {0, 8, 7, 0, 0, 0, 0, 3, 1},
                      {0, 0, 3, 0, 1, 0, 0, 8, 0},
                      {9, 0, 0, 8, 6, 3, 0, 0, 5},
                      {0, 5, 0, 0, 9, 0, 6, 0, 0},
                      {1, 3, 0, 0, 0, 0, 2, 5, 0},
                      {0, 0, 0, 0, 0, 0, 0, 7, 4},
                      {0, 0, 5, 2, 0, 6, 3, 0, 0}};
    if (SolveSudoku(grid) == true)
        printGrid(grid);
    else
        cout<<"No solution exists"<<endl;
    return 0;
}

A: Here's my code:
I think it's the best.
/*
 *  File: Sudoku-Solver.cc
 *
 *  Created On: 12, April 2016
 *      Author: Shlomo Gottlieb
 *
 * Description: This program holds A sudoku board (size: 9*9),
 *              it allows the user to input values that he choices
 *              and finds the first rational solution.
 *              The program also prints the sudoku whether A solution
 *              was found or not.
 *
 */

//========= Includes =========//
#include <iostream>

//========= Using ==========//
using namespace std;

//========= Constants =========//
const int N = 3;            // The size of 1 square in the sudoku.
const int EMPTY = 0;        // A sign for an empty cell.
const int STOP_INPUT = -1;  // A sign for stop get input.

//====== Function Declaration ======//
void input_sud(int sud[][N*N]);
bool fill_sud(int sud[][N*N], int row, int col);
void print_sud(const int sud[][N*N]);
bool is_legal(const int sud[][N*N], int row, int col, int val);
bool is_row_ok(const int row[], int col, int val);
bool is_col_ok(const int sud[][N*N], int row, int col, int val);
bool is_sqr_ok(const int sud[][N*N], int row, int col, int val);

//========== Main ===========//
int main()
{
    int sud[N*N][N*N] = { { EMPTY } };  // The sudoku board.

    input_sud(sud);
    fill_sud(sud, 0, 0);
    print_sud(sud);

    return 0;
}

//======== Input Sudoku ========//
// Gets the input for the sudoku,
// 0 means an empty cell.
void input_sud(int sud[][N*N])
{
    for(int i = 0; i < N*N; i++)
        for(int j = 0; j < N*N; j++)
            cin >> sud[i][j];
}

//======== Fill Sudoku =========//
// Tries to fill-in the given sudoku board
// according to the sudoku rules.
// Returns whether it was possible to solve it or not.
bool fill_sud(int sud[][N*N], int row, int col)
{
    // Points to the row number of the next cell.
    int next_row = (col == N*N - 1) ? row + 1 : row;

    // Points to the column number of the next cell.
    int next_col = (col + 1) % (N*N);

    // If we get here, it means we succeed to solve the sudoku.
    if(row == N*N)
        return true;

    // Checks if we are allowed to change the value of the current cell.
    // If we're not, then we're moving to the next one.
    if(sud[row][col] != EMPTY)
        return fill_sud(sud, next_row, next_col);

    // We're about to try and find the legal and appropriate value
    // to put in the current cell.
    for(int value = 1; value <= N*N; value++)
    {
        sud[row][col] = value;

        // Checks if 'value' can stay in the current cell,
        // and returns true if it does.
        if(is_legal(sud, row, col, value) && fill_sud(sud, next_row, next_col))
            return true;

        // Trial failed!
        sud[row][col] = EMPTY;
    }

    // None of the values solved the sudoku.
    return false;
}

//======== Print Sudoku ========//
// Prints the sudoku Graphically.
void print_sud(const int sud[][N*N])
{
    for(int i = 0; i < N*N; i++)
    {
        for(int j = 0; j < N*N; j++)
            cout << sud[i][j] << ' ';
        cout << endl;
    }
}

//========== Is Legal ==========//
// Checks and returns whether it's legal
// to put 'val' in A specific cell.
bool is_legal(const int sud[][N*N], int row, int col, int val)
{
    return (is_row_ok(sud[row], col, val) &&
            is_col_ok(sud, row, col, val) &&
            is_sqr_ok(sud, row, col, val));
}

//========= Is Row OK =========//
// Checks and returns whether it's legal
// to put 'val' in A specific row.
bool is_row_ok(const int row[], int col, int val)
{
    for(int i = 0; i < N*N; i++)
        if(i != col && row[i] == val)
            return false;       // Found the same value again!

    return true;
}

//========= Is Column OK =========//
// Checks and returns whether it's legal
// to put 'val' in A specific column.
bool is_col_ok(const int sud[][N*N], int row, int col, int val)
{
    for(int i = 0; i < N*N; i++)
        if(i != row && sud[i][col] == val)
            return false;       // Found the same value again!

    return true;
}

//========= Is Square OK =========//
// Checks and returns whether it's legal
// to put 'val' in A specific square.
bool is_sqr_ok(const int sud[][N*N], int row, int col, int val)
{
    int row_corner = (row / N) * N;
    // Holds the row number of the current square corner cell.

    int col_corner = (col / N) * N;
    // Holds the column number of the current square corner cell.

    for(int i = row_corner; i < (row_corner + N); i++)
        for(int j = col_corner; j < (col_corner + N); j++)
            if((i != row || j != col) && sud[i][j] == val)
                return false;       // Found the same value again!

    return true;
}

A: //here's a pseudo sudoku solver:

#include <iostream>
#include <cmath>
#include "time.h"
//for(int i=0; i< ++i){}
using namespace std;

int main(){
 int a[9][9]={{0,7,0,0,2,0,1,0,0},
             {6,0,3,0,0,0,0,0,0},
             {2,0,0,3,0,0,5,0,0},
             {0,0,0,0,3,0,0,6,0},
             {0,6,4,7,0,0,0,8,0},
             {0,5,0,0,9,0,0,4,0},
      /*54*/ {0,4,0,0,7,0,9,0,0},
             {0,2,0,0,0,8,0,5,0},
             {0,0,0,0,0,0,0,0,0}};/*this is the grid. I need to use two so that I can use one as a constant reference*/

  int a2[9][9]={{0,7,0,0,2,0,1,0,0},
             {6,0,3,0,0,0,0,0,0},
             {2,0,0,3,0,0,5,0,0},
             {0,0,0,0,3,0,0,6,0},
             {0,6,4,7,0,0,0,8,0},
             {0,5,0,0,9,0,0,4,0},
             {0,4,0,0,7,0,9,0,0},
             {0,2,0,0,0,8,0,5,0},
             {0,0,0,0,0,0,0,0,0}};

int x=0,y=0,z=0;
int oo=0,o=0,ooo=0,oooo=0;
 for(int i=0; i<9; ++i)/*this is the double for loop.  to keep track of  the cells*/
 {
 for(int j=0; j<9; ++j)
 {
 if(!a2[i][j])
 {a://label that if something doesn't work, to restart here.
 ++a[i][j];//increment the number in the cell by 1.
  if(a[i][j]>9)/*if no valid number is found/the number is out of range, execute the if statement.  **This is the backtracking algorithm.***/
{
             a[i][j]=0;//erase the invalid number.
             --j;if(j<0){--i;j=8;} /*if j equals < 0, j is reset to 8, as the first 9 numbers are easy to find, and if j needs to be reset, there will always be a number in the j equals 8 cell.  The "i" will never be less than zero, if it is a valid puzzle, so the j=8 will only happen after i is > 0.*/
    while(a2[i][j]){--j;if(j<0){--i;j=8;}}/*if there was a number present from the beginning, skip it until a variable cell space is found.*/

  goto a;//jump to a: and increment the current number in the cell by 1.
  }
 if(j<3){o=0;oo=3;}if(j>2 && j<6){o=3;oo=6;}/*locate the 3X9 grid j is in.*/
 if(j>5){o=6;oo=9;}

 if(i<3){ooo=0;oooo=3;}if(i>2 && i<6){ooo=3;oooo=6;}/*locate the 9X3 grid i is in.*/
 if(i>5){ooo=6;oooo=9;}

 for(int u=ooo; u<oooo; ++u)/*intersect the 3X9 and 9X3 grids to create a searchable 3X3 grid.*/
 {
 for(int p=o; p<oo; ++p)
 {

     if(a[u][p]==a[i][j]){++x;if(x>1){x=0;goto a;}}/*if the number is not valid, go to a: and increment the number in the cell by 1.*/

  }

  }
 x=0;
  for(int n=0; n<9; ++n)
  {
  if(a[i][j]==a[i][n]){++y; if(y>1){y=0;goto a;}}/*if no valid number is found, goto a: and increment the number in the cell by 1.*/
  }y=0;

  for(int m=0; m<9; ++m)
  {
   if(a[i][j]==a[m][j]){++y; if(y>1){y=0;goto a;}}//""
  } y=0;

  }

  }
  }

for(int i=0; i<9; ++i)//print the completed puzzle.
{
  for(int j=0; j<9; ++j)
    {
     cout<<a[i][j]<<" ";
    }
    cout<<endl;//print on the next line after 9 digits are printed.
}

return 0;
}

//if no valid puzzle is found, the screen hangs.  -  Harjinder

/*random sudokus' can be generated and solved in one program by copying the following code into an editor. hit enter to generate and solve a new puzzle:

//the following works the same way as above except that it is 2 solvers in a continuous while loop

//the first solver, to generate the random puzzle:

#include <iostream>

using namespace std;

int main(){
 while(1){
 int a[9][9]={0}, a2[9][9]={0};

 int w=0, x=0,y=0,z=0, yy=0, oo=0,o=0,ooo=0,oooo=0;
 for(int i=0; i<9; ++i)
 {
 for(int j=0; j<9; ++j)
 {
   if(!a2[i][j])
   {srand(time(0));if(w<5){++w;a[i][j]=rand()%9;}a:++a[i][j];//generate a random number from 0 to 9 and and then increment by 1.

   if(a[i][j]>9)
  {
             a[i][j]=0;
             --j;if(j<0){--i;j=8;}
   while(a2[i][j]){--j;if(j<0){--i;j=8;}}

  goto a;
  }
 if(j<3){o=0;oo=3;}if(j>2 && j<6){o=3;oo=6;}
 if(j>5){o=6;oo=9;}

 if(i<3){ooo=0;oooo=3;}if(i>2 && i<6){ooo=3;oooo=6;}
 if(i>5){ooo=6;oooo=9;}

 for(int u=ooo; u<oooo; ++u)
 {
 for(int p=o; p<oo; ++p)
 {
  if(a[u][p]==a[i][j]){++x;if(x>1){x=0;goto a;}}
  }}
x=0;
 for(int n=0; n<9; ++n)
 {
 if(a[i][j]==a[i][n]){++y; if(y>1){y=0;goto a;}}
 }y=0;

 for(int m=0; m<9; ++m)
 {
 if(a[i][j]==a[m][j]){++y; if(y>1){y=0;goto a;}}
 } y=0;}//if
 }
  }

//the second part below is to add zeros' to the puzzle, and copy the puzzle into the second solver arrays.
int n=40;//adjusts the amount of empty spaces. works only if landed on non-zero, or else the zero will be counted.
x=0;int xx=0;
o=0;int pp=0,ppp=0;
while(o<n)
{

pp=rand()%9;
ppp=rand()%9;
 a[pp][ppp]=0;
  pp=rand()%9;
 ppp=rand()%9;
a[ppp][pp]=0;

 ++o;
}
for(int i=0; i<9; ++i)
{
for(int j=0; j<9; ++j)
    {
     cout<<a[i][j]<<" ";

    }
     cout<<endl;
}
x=0,xx=0;

cout<<endl<<endl;
int a5[9][9]={0}, a6[9][9]={0};x=-1;
for(int i=0; i<9; ++i)
{
for(int j=0; j<9; ++j)
    {
     ++x;
     a5[i][j]=a[i][j];//copy first puzzle with added zeros into the new array.

    }

}
for(int i=0; i<9; ++i)
{
for(int j=0; j<9; ++j)
    {
     a6[i][j]=a5[i][j];//copy the updated puzzle with zeros, into the constant reference array.

    }

}

x=0;
////////////////////////////////////////////////////////////////////////
//second solver to solve the zero updated array:
cout<<endl<<endl;
o=0;
y=0,z=0;
oo=0,ooo=0,oooo=0;
for(int i=0; i<9; ++i)
{
for(int j=0; j<9; ++j)
{
if(!a6[i][j])
{c5:
 ++a5[i][j];
  if(a5[i][j]>9)
 {
             a5[i][j]=0;
             --j;if(j<0){--i;j=8;}
    while(a6[i][j]){--j;if(j<0){--i;j=8;}}

  goto c5;
  }
 if(j<3){o=0;oo=3;}if(j>2 && j<6){o=3;oo=6;}
 if(j>5){o=6;oo=9;}

 if(i<3){ooo=0;oooo=3;}if(i>2 && i<6){ooo=3;oooo=6;}
 if(i>5){ooo=6;oooo=9;}

 for(int u=ooo; u<oooo; ++u)
 {
 for(int p=o; p<oo; ++p)
 {
  if(a5[u][p]==a5[i][j]){++x;if(x>1){x=0;goto c5;}}
 }}
x=0;
for(int n=0; n<9; ++n)
{
if(a5[i][j]==a5[i][n]){++y; if(y>1){y=0;goto c5;}}
 }y=0;

for(int m=0; m<9; ++m)
{
 if(a5[i][j]==a5[m][j]){++y; if(y>1){y=0;goto c5;}}
 } y=0;} }
 }

for(int i=0; i<9; ++i)
{
for(int j=0; j<9; ++j)
    {
     cout<<a5[i][j]<<" ";
    }
    cout<<endl;
}

cout<<endl;
system("pause");//hit enter to load a new puzzle
}

return 0;
} */

A: The answer given above is similar to what a lot of people have posted at several websites like https://www.geeksforgeeks.org/sudoku-backtracking-7/. TBH, those solutions are not intuitive to me when they don't initialize row and col and call it in the FindUnassignedLocation function. I wanted to give it a shot myself with slight help from another answer I found on LeetCode. If it helps anyone, here you go.
// https://www.pramp.com/challenge/O5PGrqGEyKtq9wpgw6XP

#include <iostream>
#include <vector>

using namespace std;

// Print the sudoku
void print(vector<vector<char>> board){
    for(int i = 0; i < board.size(); i++){
        for(int j = 0; j < board[i].size(); j++){
            cout << board[i][j] << " ";
        }
        cout << endl;
    }
    cout << endl;
}

// Check if sudoku is all filled up
bool checkIfFilled(vector<vector<char>> board){
    for(int i = 0; i < 9; i++){
        for(int j = 0; j < 9; j++){
            if(board[i][j] == '.')
                return false;
        }
    }
    return true;
}

// Check if a given number can be inserted at a given position
bool checkIfInsertOkay(vector<vector<char>> board, int x, int y, char num){
    // Check row if num already exists
    for(int j = 0; j < 9; j++){
        if(board[x][j] == num)
            return false;
    }
    // Check column if num already exists
    for(int i = 0; i < 9; i++){
        if(board[i][y] == num)
            return false;
    }
    // Check 3x3 gird if num already exists
    // Find the corners
    // Find i
    if(x < 3)
        x = 0;
    else if(x < 6)
        x = 3;
    else
        x = 6;
    // Find j
    if(y < 3)
        y = 0;
    else if(y < 6)
        y = 3;
    else
        y = 6;
    // Check the 3x3 box
    for(int i = x; i < x+3; i++){
        for(int j = y; j < y+3; j++){
            if(board[i][j] == num)
                return false;
        }
    }
    return true;
}

// Helper function because of const issues
bool sudokuSolveHelper(vector<vector<char>> &board){
    // Base condition - if sudoku gets completely filled
    if(checkIfFilled(board))
        return true;

    // Iterate through the sudoku
    for(int i = 0; i < 9; i++){
        for(int j = 0; j < 9; j++){
            // If empty space
            if(board[i][j] == '.'){
                for(int k = 1; k <= 9; k++){
                    // Check if char(k) can be inserted at this empty location
                    char ch = '0' + k;
                    if(checkIfInsertOkay(board, i, j, ch)){
                        // Put k in this empty location and check
                        board[i][j] = ch;
                        // Check if done
                        bool flag = sudokuSolveHelper(board);
                        if(flag)
                            return true;
                        else
                            // Else, backtrack by making it empty again
                            board[i][j] = '.';
                    }
                }
                return false;
            }
        }
    }
    return true;
}

// Return true if a correct sudoku can be formed
// Apply backtracking
// Time complexity = O(9^empty spaces)
bool sudokuSolve(vector<vector<char>> &board){
    return sudokuSolveHelper(board);
}

int main() {

    vector<vector<char>> board = {
        {'.','.','.','7','.','.','3','.','1'},
        {'3','.','.','9','.','.','.','.','.'},
        {'.','4','.','3','1','.','2','.','.'},
        {'.','6','.','4','.','.','5','.','.'},
        {'.','.','.','.','.','.','.','.','.'},
        {'.','.','1','.','.','8','.','4','.'},
        {'.','.','6','.','2','1','.','5','.'},
        {'.','.','.','.','.','9','.','.','8'},
        {'8','.','5','.','.','4','.','.','.'}
    };
    print(board);
    bool flag = sudokuSolve(board);
    if(flag){
        cout << "A solution exists as below\n";
        print(board);
    }
    else
        cout << "No solution exists!\n";
    return 0;
}

A: //the following works like the one above accept there is a while loop instead of goto statements:

#include <iostream>
#include <cmath>
#include "time.h"
//for(int i=0; i< ++i){}
using namespace std;

int main(){
int a[9][9]={{2,0,0,0,0,0,0,0,1},
             {0,0,0,9,0,6,0,0,0},
             {0,0,0,8,0,1,7,2,0},
             {9,0,0,3,0,0,0,0,0},
             {0,0,8,0,0,0,2,0,4},
             {0,0,0,0,0,0,0,1,3},
      /*54*/ {1,0,3,0,0,5,0,0,9},
             {0,0,0,0,0,0,0,0,0},
             {0,4,6,2,0,0,0,0,0}};

 int a2[9][9]={{2,0,0,0,0,0,0,0,1},
             {0,0,0,9,0,6,0,0,0},
             {0,0,0,8,0,1,7,2,0},
             {9,0,0,3,0,0,0,0,0},
             {0,0,8,0,0,0,2,0,4},
             {0,0,0,0,0,0,0,1,3},
             {1,0,3,0,0,5,0,0,9},
             {0,0,0,0,0,0,0,0,0},
             {0,4,6,2,0,0,0,0,0}};

 int x=0,y=0,z=0;
 int colPlusThree=0,col=0,row=0,rowPlusThree=0;
  for(int i=0; i<9; ++i)
  {
  for(int j=0; j<9; ++j)
  {z=0;
  if(!a2[i][j])
  {while(z==0){ //the master while loop. loops until a valid number is found.
  ++a[i][j];
  if(a[i][j]>9)
 {
             a[i][j]=0;
             --j;if(j<0){--i;j=8;}
    while(a2[i][j]){--j;if(j<0){--i;j=8;}}

 ++a[i][j];if(a[i][j]>9){--j;}
 }

if(j<3){col=0;colPlusThree=3;}if(j>2 && j<6){col=3;colPlusThree=6;}
if(j>5){col=6;colPlusThree=9;}

if(i<3){row=0;rowPlusThree=3;}if(i>2 && i<6){row=3;rowPlusThree=6;}
if(i>5){row=6;rowPlusThree=9;}

 for(int u=row; u<rowPlusThree; ++u)
 {
 for(int p=col; p<colPlusThree; ++p)
 {
  if(a[u][p]==a[i][j]){++x;if(x>1){u=rowPlusThree;break;}}
 }}if(x<2){z=1;}else{z=0;}x=0;

for(int n=0; n<9; ++n)
{
if(z==0){y=2;break;}
if(a[i][j]==a[i][n]){++y; if(y>1){break;}}
}if(y<2){z=1;}else{z=0;}y=0;

for(int m=0; m<9; ++m)
{
 if(z==0){y=2;break;}
 if(a[i][j]==a[m][j]){++y; if(y>1){break;}}
} if(y<2){z=1;}else{z=0;}y=0;}}}}

for(int i=0; i<9; ++i)
{
for(int j=0; j<9; ++j)
    {
     cout<<a[i][j]<<" ";
    }
    cout<<endl;
}

return 0;
}//  -  Harjinder

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text:
Home Violent leftists dox conservative student, light door on fire — Report
Published: March 23, 2019 By The Editor For The Christian Journal
Violent leftists dox conservative student, light door on fire — Report
Updated: March 23, 2019 at 6:35 pm EST See Comments
Share on GAB
It seems that leftists won't be happy until the blood of conservatives is literally running in the streets. On Saturday, the Young Americans for Liberty (YAL) said that police were called to the Tulane University dorm room of Peyton Lofton, a student who is a chapter officer in the local organization.
According to the report, Lofton was recently doxxed by an anti-YAL social media account that exposed him for being a part of YAL. But it didn't stop there. The report said there was an attempt to light Lofton's dorm room door on fire.
"It's a sad day in America when radical activists are lighting dorm room doors on fire because they disagree with you politically. I'm proud of the work I'm doing on and off campus with YAL and TPUSA to bring students to the principles of the Constitution and individual liberty. This only encourages me to continue the battle to reach my classmates with the message of freedom," Peyton said in a statement.
"YAL stands behind Peyton and every other American college student that's want to have real dialogue and debate on campus. If a socialist student is your neighbor, the answer is to
The remainder of this article is available in its entirety at Conservative Firing Line
The views expressed in this news alert by the author do not directly represent that of The Christian Journal or its editors


text:
Q: EF code migration in reference dll coming through nuget We have a situation. Our company has Framework build on .net which is used by products developed using that Framework. Now this Framework is using EF code first. We want to implement EF Code Migration though generated migrations definitions because our entities are there in one of the Framework project. So I have implemented it. But the situation is this Framework we are distributing to other products as Nuget package (internal).
Now the situation is lets say ProductA is consuming the package which has xyz.dll which has migrations enabled in it. Now the developers of "ProductA" wants to upgrade the database created by Framework EF code first using that xyz.dll migrations, but this dll is only added as referenced dll. So running command in package manager console like Update-Database is not working because project is not in the current solution and its in reference dll
How do I solve this?

A: This is actually pretty easy to solve, the project which you add your nuget package to also needs to reference EF.
You can actually enforce this inside your nuget packages with a dependency eg:
    <dependencies>
        <dependency id="EntityFramework" version="4.2.0.0" />
    </dependencies>

in your nuspec (obviously update the version with the one you are using)
See: http://docs.nuget.org/docs/reference/nuspec-reference#Specifying_Dependencies
and
http://docs.nuget.org/docs/reference/versioning#Specifying_Version_Ranges_in_.nuspec_Files
for some more details around how the dependancy syntax works