Ein Abenteuerspiel der etwas anderen Art kann der Spieler im Spiel Icy Tower erwarten. Und jetzt viel Spaß und Icy Tower kostenlos spielen auf argentera.eu . Das Onlinespiel Icy Tower ist ein kostenloses online Flash Game aus der Kategorie Koordination, welches du im Browser ohne Download oder Anmeldung. Ziel ist es bis nach ganz oben im Turm zu kommen. Um einen Kommentar zu schreiben kostenlos anmelden. Harold steht am unteren Ende des Turms und er möchte doch so gerne nach oben, an die Spitze gelangen. Immer höher, immer schwieriger! Habt ihr bestimmte Stunts geschafft, steigt ihr einen Rang auf. Deine zuletzt gespielten Spiele. Die besten Onlinespiele auf einer Seite! Icy Tower kostenlos spielen Harold steht am unteren Ödegaard zu bayern des Turms und er möchte doch so atlantis spiel nach oben, an die Beste innenverteidiger gelangen.
Because the tower is of infinite height, it is impossible to reach the top floor. Scores are recorded in several categories including highest score, highest floor, and longest combo.
Players may also publish their scores online high score lists. Online high-score lists may require players to provide game replays to verify their results.
The game follows a sequence-based software versioning scheme. The current version is 1. Changes in the minor number denote the entire package of the game being changed, including its sound and graphic resources.
Changes in the release number denote a release of a patch , i. Patches for Icy Tower are distributed inside and outside of official game packages.
Lesser changes include an audiovisual overhaul of the game 1. Four new combo rewards were also introduced: On March 14, , a press release was published on the Free Lunch Design website announcing a partnership between FLD and mobile game developer Xendex Entertainment for the porting of Icy Tower to the mobile environment.
Due to the then-current limitations of mobile platforms, the game lacked several features in its PC counterpart, such as recording or sharing replays, and changing game characters.
The producer estimates that on November 23, , the number downloads of Icy Tower 2 downloads surpassed one million.
Internal structure of Icy Tower allows for the use of skins through downloading and installation repackaged resource files. Since the version 1.
Making a new character only requires the player to create an image file of a certain structure. It is possible, however, to also create more complex ones, using which will cause the game to modify its soundtrack  and colour palette [ citation needed ].
As of [update] , no final version of ITFB has been released. Readme file named " Icy Tower v1. Archived from the original on July 3, Archived from the original on February 11, Make special jumps to activate combo mode and score even more points!
Adjust game screen size. Icy Tower Adjust Screen Maximize. Add to your favorites Remove from favorites Add to your hearts Remove from hearted.
Icy Tower Walkthrough Stuck? See how to complete the game. Add this game to your web page! Open embedded menu Add this game to your web page!
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Copyright by kostenlosspielen. Das Spiel wird in 20 Sekunden beginnen! Kostenlose Spiele Willkommen auf spiele. Einige der beliebtesten Kämpfer Spiele sind jene, die auf karate und boxen Motto werden. Earn To Die Part 2. Online Spiele Ein Online Spiel ist ein videospiel, über irgendeine form von computer-netzwerk gespielt, mit einem Personal Computer, Videospiel-Konsole oder handheld-konsole.
The character can also bounce off the vertical walls on the left and right edges of the tower, allowing them to switch direction while maintaining momentum.
If a player builds enough momentum, he can climb several floors in a single jump. Extended sequences of such multi-floor jumps are referred to as combos.
A combo ends when a player makes a jump which covers only one floor, falls off a floor and lands on a lower floor, or fails to make a jump within a certain time frame about 3 seconds.
When the player reaches floor 5, the floors begin to move slowly downward. It also ends if the player is standing on a floor that moves off the bottom of the screen.
The goal is to reach the highest possible floor and accumulate the greatest number of points. Because the tower is of infinite height, it is impossible to reach the top floor.
Scores are recorded in several categories including highest score, highest floor, and longest combo. Players may also publish their scores online high score lists.
Online high-score lists may require players to provide game replays to verify their results. The game follows a sequence-based software versioning scheme.
The current version is 1. Changes in the minor number denote the entire package of the game being changed, including its sound and graphic resources.
The producer estimates that on November 23, , the number downloads of Icy Tower 2 downloads surpassed one million. Online Icy Tower communities allow their members to exchange high scores, replays and comments about the game and its gameplay.
In every round, contestants are required to record one or more replays with a result in a specific category of the game. Internal structure of Icy Tower allows for the use of skins through downloading and installation repackaged resource files.
Since the version 1. Making a new character only requires the player to create an image file of a certain structure. It is possible, however, to also create more complex ones, using which will cause the game to modify its soundtrack  and colour palette [ citation needed ].
Xjump , the game which inspired the makers of Icy Tower. Modifications and utilities for Icy Tower. Since Icy Tower 1. Also known as ITR Checker.
Retrieved September 8, This measure against slowdown was subsequently abandoned in favour of the SDbuster program and, later, inbuilt features of the game since Icy Tower 1.
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These swaddle blankets make a great gift for any newborn or mom to be. Excellent for swaddling, covering up while nursing, tummy time, and makes an excellent prop for pictures. Ideal for a little baby. Blanket measures a nice 80cm x 100cm which also makes it ideal to use as a bassinet or stroller blanket. Before purchasing please make yourself familiar with the rest of the Joba Baby range of affordable luxury essentials.


February 1, 2010, Hangzhou Hi-tech products interactive whiteboard (Interactivewhiteboard) get TUVAT (CE testing the world's most authoritative body) issued by the CE certification, certified products cover the full range of GOAKE Whiteboard, performance in line with EU standards: EN60065-2006LVD instructions: 2006/95/EC. In recent years, Hangzhou High-tech business to keep doubling of international trade, particularly in Cologne, Germany in 2009 after the branch into the EU market Strong market demand, has great potential. CE mark obtained access to the tech whiteboard smooth access to the EU and the European Union Free Trade Association countries, the market permits, for the European market development laid the foundation. CE is a French ConformitéEuropéenne (European Conformity Assessment) abbreviation. CE mark attached to their products that meet the standards of the European Directive and the European coordination of the basic requirements, namely, public safety, health, environmental protection and the Consumer protection. CE mark (CEMark) implemented by the EU of a product logo. It is the elimination of technical barriers to trade to establish a single market and promote market liquidity, one of the means of economic development of member countries. CE certification mandatory To the EU market of products, must be attached to CE safety certification and CE mark. CE marking is a safety certification mark, any mark affixed to products with CE can be sold in the EU in all member states, not required to meet Requirements of each member in ord er to achieve the goods within the EU member states the freedom of circulation.


Was assigned to a ship in the north Atlantic whose mission was mine sweeping but was there for only two months.
Went to Miami anti-sub school to learn how to use sonar for two months.
Then was moved to Key West to instruct other officers on submarine warfare if the attack came to the States.
Next he was stationed to Columbia University to teach Midshipmen about Navigation for a year and did not enjoy it.
He tried to apply for submarine warfare in the North Atlantic but was assigned to Okinawa as a Yard Patrol (YP).
He was the CO of a single screw wooden hull ship with three 50. Cal. and was only a Lieutenant Junior Grade. While he was there he practiced using depth charges for anti submarine warfare but also carried provisioned foods to marine bases around the area.
During Christmas his ship broke down in the China sea where the Japanese had just laid down mines in the north so he had to steer his ship though a mine field while shooting a few that got in their way and drifted towards Shang Hi, were he was found and towed back to port and stuck there for six months.
July 4, 1946 after four years in the Navy he was discharged and joined the reserves for another two years in a command group.
During one of his liberty leaves in China he said he was on his way back to the ship when two Chinese guys in a car pulled up and offered him a ride. Two blocks before the dock they stopped and they pointed a 45 at him and robbed him and then let him go.


1. How has the Maker's Mark Secretariat Center set the standard for Thoroughbred Aftercare?
Over the last 5 years, the awareness of the issue of aftercare has grown. The industry is doing so much more to support Thoroughbreds once their racing careers are over, and many new non-profit and for-profit organizations dedicated to the rehab, reschooling, and rehoming of off-track Thoroughbreds have blossomed. Within this movement, the MMSC has chosen to focus on quality and depth of training, as well as a dedication and exigence to matching its horses with the ideal potential adopter. Our focus is horse-centric, and we pride ourselves on addressing each animal's needs, however long that may take.
2. Thoroughbreds often have very short careers and it is important to have a backup plan when they cannot race anymore. What do you believe is the best way to help these Thoroughbreds continue to have a purpose and fulfilling life?
This is a complicated issue and needs to be addressed on several fronts. First, education. All Thoroughbred owners and breeders, especially those who are new to the industry, need to understand the responsibility and necessity of caring for a horse if and when it no longer has a career on the track. This awareness must occur at the onset and prior to undertaking ownership of a racehorse. It must be understood that it comes with a moral commitment to aftercare. In this way it is similar to having and raising a child. Moreover, trainers must be educated about the need and value of retiring horses from their racing careers before they are utterly broken down. In this regard, the Retired Racehorse Project's directory of the numerous aftercare organizations in existence in the United States today is a tool that should be readily available and indeed distributed at every track in the United States.
Second, proper funding. It is no surprise that horses are expensive at best and even at the very least require care and nurturing for subsistence. This is a price that could and should be born not only by the owners and breeders, but by everyone within the racing industry, including those who wager and those who enjoy going to the racetrack for pleasure. The industry needs to be commended for the initiatives it has already undertaken, such as the Jockey Club check-off program, the Thoroughbred Incentive Program, the allocations of a small portion of the handle that certain tracks are making to various organizations such as the TAA, etc. While all of this is encouraging, this movement needs to build and as awareness and public pressure grow, no doubt funding will grow too.
3. How is the MMSC constantly improving its Aftercare program?
With the rise of numerous reschooling organizations and efforts going on around the country, it is imperative for the MMSC to amend its program continuously so that it meets the needs of the horses. Given that our location at the Kentucky Horse Park provides us with a unique and an international stage from which to broadcast the versatility and athleticism of the Thoroughbred in a second career, the MMSC can expand its educational and promotional efforts to help everyone who is involved in this noble pursuit.
4. What are the long term goals for the Maker's Mark Secretariat Center in 2016?
The long-term goals for the MMSC are as follows: 1) To procure additional land where horses coming directly off the track can have a period of rest to recuperate before transitioning to a new career. 2) A covered or indoor arena, so that reschooling and training efforts can take place year-round. 3) To acquire a secure base of funding that will cover the needs and costs of the horses and ensure the longevity of the MMSC and all of its programs.
5. How can an interested adopter learn more about the Maker's Mark Secretariat Center?
An interested adopter can go to our website, our Facebook page, our blog, call us directly, and/or visit. We are always happy to answer questions, to talk about our program, and to show off our facility!


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Mussel Soup
This is a delicious, healthy soup which is equally at home on a cold winter's night or as an appetiser for a summer dinner party. For the recipe for my Mussel Soup.
Fawlty Towers - Fav Food Quotes - A Touch of Class
Series No. 1, Episode No. 1, Title "A Touch of Class"
Sign reads FAWLTY TOWERS (Crooked S)
Synopsis: As Basil tries to raise the tone of the hotel, the aristocratic Lord Melbury comes to stay at the hotel. Basil fawns over him at every opportunity, causing himself to neglect or annoy other guests, until Polly discovers Melbury is actually a confidence trickster. Meanwhile, Sybil orders Basil to hang a picture.
"Manuel, there is too much butter on those trays."
"Que."
"There is too much butter on those trays."
"No, no no señor. Not, not on those trays, no sir. Un, dos tres."
"Are you ready to order?" "Ah yeah, what's a, a grail frit?"
"Grape fruit."
"Oh and a cream pot, pot rouge."
"Portuguese, tomato soup.
"I'll have the grail frit."
"And the palm carousel, lamb?"
"Lamb casserole."
"Does it come with a smile?"
"It comes with sprouts and carrots."
Review of The Botanical
Saturday, 23 February 2013, Dinner
169 Domain Road South Yarra, 03 9820 7888, www.thebotanical.com.au
Myrna received a voucher for a night at a city hotel for her birthday in January. The night was scheduled for 23 February and the hotel was Art Series - The Blackman Hotel at 452 St Kilda Road, Melbourne.
Leaves of Grass 20/40 Edition by Charles Blackman
We are both Blackman fans with our own limited edition lithograph of "Leaves of Grass" gracing my study wall.
It was a Studio Suite Balcony and was a very pleasant room with views of Faulkner Park to the East and Albert Park to the West.
Faulkner Park Edged With St Kilda Road High-rise
With many restaurants on offer in the vicinity, we chose The Botanical for a couple of reasons, firstly it was awarded Australia's Best Wine List of the Year 2012, Best Pub Restaurant of the Year 2012 as published in "Wine" magazine by Gourmet Traveller and secondly, we'd been told by a trusted source that it does a steak that is second to none. Furthermore, it's a nice little stroll from The Blackman to sharpen the appetite.
The Botanical is located adjacent to Melbourne's famous Botanic Gardens and is a multifaceted venue complete with formal restaurant, informal wine store and café, a bar and lounge and the alfresco-inspired "Archer Room' for casual dining and private events. The restaurant is open for lunch and dinner seven days, the Wine Store from 7am till late seven days and the Bar from 12noon till late seven days.
It was Saturday night and the place was abuzz with the sounds of happy conversation, laughter and the merriment of young people (and not so young) meeting at a social watering hole. As we watched the happy drinkers and even happier diners coming and going, we ate crusty sour dough with homemade butter and sipped tall glasses of well-chilled 4 Pines beer. I had the Keller Door ($9.00) and Myrna the Pale Ale ($7.50). Check it out here: http://www.4pinesbeer.com.au/beers.
Predictably for entrée, Myrna went beetroot. It was a pretty little thing entitled Organic Beetroot Salad with hazelnuts, whipped goat's curd and watercress ($15). I could tell by the look on Myrna's face it wasn't perfectly to her liking. I had a taste and thought it was great, light and refreshing, well-balanced and attractive but her concerns were the radish taste outweighed the beetroot taste, there was insufficient beetroot and the hazelnuts were missing. Apart from the hazelnuts missing, it was purely a matter of personal preference so we agreed to disagree.
I thought I'd opt for the calamari which I hadn't had for some time. My dish was entitled Crispy Fried Calamari served with Szechuan salt and squid-ink aioli ($21). It was delicious to eat but I could have done with 50% more.
Organic Beetroot Salad
Crispy Fried Calamari
Other entrées included Roast pumpkin and Cumin Soup ($14), Smoked Salmon Terrine ($15), Yellow fin Tataki ($18) and The Botanical Charcuterie Board for two ($28).
There were two specials on offer. One was a 300g hanger steak ($46) and the other was a fresh 500g flounder ($48) served with Spring Bay mussels, Crystal Bay prawns and cream cider sauce. Add to the choices a John Dory Fillet, Devon Duck, Crumbed Pork Cotoletta and a Wagyu Beef Burger and we were facing a dilemma of epic proportions.
Wikipedia - A hanger steak is a cut of beef steak prized for its flavour. Derived from the diaphragm of a steer or heifer, it typically weighs between 450g and 675g. In the past, it was sometimes known as "butcher's steak" because butchers allegedly kept it for themselves rather than offer it for sale. Hanger steak resembles flank steak in texture and flavour. It is a vaguely V-shaped pair of muscles with a long, inedible membrane down the middle. The hanger steak is not particularly tender and is best marinated and cooked quickly over high heat and served rare or medium rare, to avoid toughness.
As you probably guessed, I went with the hanger steak and I'm very glad I did, it was superb and cooked exactly as described above in the quote from Wikipedia. It was served with char-grilled red onions which tasted beautifully sweet and a perfect accompaniment to the medium rare steak and pepper sauce with a pepper-béarnaise sauce. Wow, what a feed.
Hanger Steak
Myrna almost decided on the hanger stead as well but at the last minuted was worried out of it by the 300g size. She opted instead for the John Dory fillet with fennel puree, piedmontese peppers and fennel salad ($35). While the John Dory was delightful, I think we both knew who had taken the prize for best main.
With dinner I enjoyed a glass or two of 2010 VillaCampa Tempranillo from Ribera del Duero, Spain ($9.50 per glass) and Myrna enjoyed a glass of 2011 Fraser Gallop Chardonnay from the Margaret River ($11).
Desserts were also very pleasant with a reasonable selection around $15 mark. I chose the Raspberries with Fromage Blanc with Breton shortbread and Szechuan pepper sorbet and Myrna won the dessert choice with Ultra Creamy Dark and Milk Valrhona Chocolate Macaroon with Bailey's ice-cream and salted caramel.
Raspberries with Fromage Blanc
Ultra Creamy Dark and Milk Valrhona Chocolate Macaroon
Overall Positives
Vibrant, lively atmosphere
Something a little different e.g. hanger steak
Good spacing between courses
Knowledgeable waiting staff and great service
Accepts Entertainment Gold Card for a good discount
Fascinating wine list and "Wall of Wine"
Overall Areas To Improve
Cost of on-tap beer expensive
Website not as good as some
Restaurant Rating
Faulty Towers - Food Stories
Who remembers the tele-series known as "Faulty Towers" and the misspelt signs at the introduction of each episode "Farty Towels", "Flay Otters" or "Watery Fowls" for example. There are so many memorable moments in the mere 12 episodes. Interestingly, some of the best lines are related to culinary "stuff-ups".
Review of Immerse Restaurant (Yarra Valley)
Saturday, 16 February 2013, Lunch
1548 Melba Highway Dixons Creek, 03 5965 2444, www.immerse.com.au
We've been to Immerse many times but not since starting Let It be Food. I was looking forward to a return visit and writing a review. I'd been thinking about it a bit lately.
Sitting at my desk on Saturday morning, gastronomically satisfied after enjoying a feast of Myrna's Raspberry and Blueberry Pancakes, I was deep in thought while preparing a blog post about them. The phone rang, it was son Daniel.
"Do you have anything planned for today?"
"Nothing specific, we were thinking of going to a pub for lunch."
I could tell by the reaction he had something a little more upmarket in mind. Being the opportunist I said
"How about Immerse."
"Done!"
Immerse it was. Funny how thinking about something can sometimes be enough to make it happen.
An hour later we were turning off the Melba Highway and heading through the gates that lead to the restaurant car park. It was hot, dry and dusty but the thought of what was to come was exciting. Immerse is a winery, a restaurant and a reception centre with onsite accommodation for up to 20 people. We even stayed there ourselves on the night of Dan's and Roz' wedding just over three years ago.
Walking through the front door to the restaurant is quite a contrast between a rural, rustic outside and a modern, simple, semi-glamorous inside. Just outside the front door is a small outside eating area shaded by leafy trees with sparrows hopping about in a profitable search of crumbs. Hard left through the entry is the cellar door. Wine tastings are popular and one always witnesses a number of Yarra Valley winery tours bringing guests for a tasting and a meal. Tastings are free, well-structured and well-presented and cover off about half a dozen of Immerse's staple wines.
After sitting at the table for a minute or so, we realise just how hot it was outside and we drank some cold water. The waitress refilled the glasses and we drank again. We then sipped glasses of Immerse Reserve Chardonnay 2010 while we studied the menu. We consciously aimed to maximise the breadth of coverage of the menu with three different entrees, four mains and four desserts between us. The prices are about fair and reasonable for a winery restaurant of this calibre with entrees, mains and desserts weighing in at $18, $36 and $15 respectively.
My entree was called Sticky Pork and it was a stunner. It was a Thai basil salad with coconut, tomato, peanut, cucumber and inch sized cubes of pork belly cooked to perfection and giving the salad both a rich, sweet taste with the brilliant texture of well-cooked pork. Of course everyone wanted to exchange a taste of their entree for a taste of mine. Well, that's half the fun. Myrna and Roz had calamari with smoky eggplant, spiced chickpea crunch, lemon yoghurt, and fennel and mint salad while Dan had pan-fried gnocchi, prawn and roast pumpkin. I've never had an entree at Immerse that hasn't made me groan with pleasure. The Sticky Pork left my taste buds vibrating and looking forward to the next course.
Sticky Pork
Pan-fried gnocchi
I would have been happy with any of the mains on the menu. I settled for beef sirloin with chorizo croquette, smoked onion and chimichurri. [Chimichurri originated in Argentina and is used as a meat sauce. It's made from finely chopped parsley, minced garlic, olive oil, oregano and wine vinegar.] As predictably as me and the steak, Dan picked pork belly with potato and fennel cream, apple salad and, cider and mustard jus. Roz picked the salmon fillet with zucchini, caper, raisin, pine nuts, potato and Persian fetta and Myrna picked the Yeringberg lamb rump with spice roasted cauliflower, almond cream and pomegranate. My only complaint in all my visits to Immerse would be that this time my beef was slightly over-cooked. I ordered medium-rare and got medium. It was still lovely and tender, char-grilled to a fabulous crust on the outside. The chorizo croquette was amazing with its smooth, creamy potato filling and delightful crispy crust. I enjoyed a glass of Immerse Shiraz Viognier 2008 which is one of my Immerse favourites.
Yeringberg lamb rump
Beef sirloin
Again, we got good coverage with the desserts. I had the passion-fruit cream caramel with black sesame ice-cream, popcorn macaroon and morello cherry while Myrna had fried custard, Pedro Ximenez ice cream, citrus salad and almond brittle, Roz had warm dark-chocolate fondant, salted caramel ice cream and raspberry and Dan had lemon verbena pannacotta, pistachio cake, raspberry sorbet and strawberry. Once more we did the shared tasting and each dessert was in its own right a masterpiece. There was a lot to like about my passion-fruit cream caramel, from the subtlety of the black sesame ice-cream to the piquancy of the fruit in the cream caramel and delicious passion-fruit sauce. I think though that the popcorn macaroon was the standout, crisp outside chewy inside and a beautiful sweetness that finished the meal with a bang.
Fried custard
Lemon verbena pannacotta
Warm dark-chocolate fondant
Passionfruit cream caramel
Great ambiance (especially in cooler months when the fireplace is alive)
Great menu
Great wine
Pleasant surrounds
Watch the order for steak matches the served dish
Join the Entertainment Book
Pumpkin and Sweetcorn Soup
This soup has a certain wholesomeness. It's filling, it's delicious, it's easy, it's quick and it's inexpensive. Notwithstanding its delicious taste, this soup is very healthy with a sweet creaminess from the pumpkin and a bulky consistency from the corn and carrot. Click here for the recipe of our Pumpkin and Sweetcorn soup.
This is a delicious, healthy and inexpensive dish that is easy to prepare and cook. The refrigeration of the cooked vegetables can add to the time taken to prepare but if you get organised you can use the time to walk the dog or even write for your blog. Please follow the link for the Vegetable Croquette recipe and for a brilliant accompaniment of Beetroot and Mint Salad.
This is a delicious way to kick off a weekend. Sweet, tasty, filling and not too expensive. It's healthy but maybe a little high in calories. I prefer it on the weekend so I can walk off the calories whilst taking Louie and Chloe for a nice, long walk. Click for the recipe for Raspberry and Blueberry Pancakes.
Who doesn't like a good bowl of spag-bol? Kids love, adults love and I hope it doesn't sound off putting but even dogs love it. I always save a little for a treat for Louis and Chloe just before we wash up the dishes (and BTW there aren't many dishes to wash with this dish). It's fun holding strands of Bolognese covered spaghetti for the dogs to jump at, as it inscribes a horizontal circle. Snap! Snap! Sometimes it disappears down their throat and other times its circular momentum wraps the spaghetti around their muzzle like a rubber band. Whether you have dogs, cats or canaries, I think you'll enjoy our take on spag-bol so click for the Spaghetti Bolognese recipe.
This recipe is quite simple but it does take some time to prepare. You'd need to set aside about 2 hours including cooking time but it's arguably worth the wait. Click for the recipe for Sun-dried Tomato and Roasted Capsicum Quiche.

redpajama_set_name: RedPajamaCommonCrawl


text:
hendrixprofile.com
Fact From Fiction: Your Guide To Spotting Fake News
In early December of last year, a man walked into a Washington, D.C. pizzeria and threatened employees and customers with a rifle, firing several shots inside the restaurant. He came to this restaurant believing that it was the center of a child sex ring run by Hillary Clinton. Yes, you read that right. The so-called "Pizzagate" conspiracy theory is an extreme example of how stories that are completely ludicrous can not only be taken to be true but can have real consequences. The 2016 election saw the spread of so-called "fake news" – stories that are passed around as legitimate when in fact they are stories that are fabricated or heavily skewed. A particularly popular one was that Pope Francis had endorsed Donald Trump for President of the United States in the closing days of the campaign (his Holiness did not, in fact, endorse Trump).
These stories are able to spread rapidly via social media, with people sharing them among their friends. When you see a news story on your friends' Facebook or Twitter feeds, how do you know if it's legitimate or not? It's a common problem. According to the Pew Research Center, 84 percent of 18-to-29-year-olds get at least some of their news from social media. Although Pew's surveys have shown that people are more wary of news stories they encounter online, their survey data also shows that around a third of Americans have shared a fake news story online. The rise of fictitious stories being promoted and shared around social media can contribute to a less-informed public. A common example is from the initial debate around the Affordable Care Act, when several sites on the Internet picked up and spread the false notion that the healthcare reform law included "death panels" that would decide if sick people would receive life-saving care or not. Despite this claim being PolitiFact's "Lie of the Year" in 2009, the notion of "death panels" negatively shaped the public's initial perception of the ACA.
The problem of determining which news stories and sources are trustworthy or not is especially relevant for Hendrix students who, like most college students, tend to get their news from Twitter and other non-conventional platforms. In a survey of students from Hendrix Today, the overwhelming majority of students surveyed said they received their news primarily from online sources. Only 3 of the 44 students surveyed indicated that their primarily news platforms were newspapers or TV broadcasts.
One student who likes using social media as a news outlet in particular is Guneev Sharma '17. Sharma says he likes using Twitter for news as it allows him to get a steady stream of updates as stories unfold. He says he follows the major news networks such as CNN and Fox as well as more politically focused outlets such as Politico and The Hill.
"If most of the sources are tweeting the same facts," Sharma said, "I'll deem it to be true."
He says that he always moves beyond the tweet's headline and checks out the stories being reported online as a way to learn more. When it comes to Facebook, Guneev says he always Googles the stories that pop up to confirm that they're true.
"I'll determine if it's true if there's no other news organization refuting the facts and if all sources are credible and not just on Breitbart or Occupy Democrats," Sharma said.
Greer Veon '17 tries to rely on more conventional platforms such as broadcast news, which she makes a point of trying to watch in the morning while she gets ready for her day. Veon likes to stick to more established news organizations like CBS, where she interned this past summer. This experience gave her a more inside perspective of what good journalism is supposed to look like, with advice from CBS Evening News anchor Scott Pelley serving as particularly relevant: always ask yourself, is it right, is it fair, and is it honest? This advice can be helpful for both journalists as well as critical news consumers.
"I notice when reading stories created by younger brands (Snapchat shows a lot from Buzzfeed, for example), there's a definite bias within the writing," Veon said. "This is why the writing is a key aspect in what I look at when deciding what's trustworthy or not," Veon says, with the wording used for the headlines of news reports in particular being critical.
Her tips for watching news broadcasts center on the questions being asked by anchors and correspondents: "Are they asking questions for both sides of an issue? Are they attacking a side they don't agree with?" Greer's tips for filtering out biased and fake news from the real, objective thing have much in common with tips on how to spot fake news from the Annenberg Public Policy Center's FactCheck.org. The tips are fairly simple, from reading beyond the headline to checking to see the evidence cited in support of the story actually back up what's being reported.
Seeking out news organizations that are trustworthy and that consistently produce high-quality journalism is just as important as avoiding dubious sources of information. There are several outlets in the United States and in Great Britain that have been recognized for their work in the journalism field. These include wire services such as the Associated Press and Reuters, which supply syndicated news stories to local newspapers, newspapers considered to be more liberal in their editorial work, such as the New York Times, and newspapers considered to be more conservative, such as The Wall Street Journal.
Once one knows how to avoid falling for fake news stories, the next step is to approach news spread online, and on social media in particular, with a critical lens. If there's a story going around Twitter or popping up on several friend's Facebook pages, take a second to examine the story closer before sharing and passing it along to your network. A better-informed discourse with our relatives and our old high-school classmates on Facebook and other social media platforms starts with all sides being able to recognize the (non-alternative) facts.
This story orginally appeared in the April Print Edition of the Profile
Photo credit: Aditya Katke
April 15, 2017 Barrett GoodwinFake News, Greer Veon, Guneev Sharma, Hendrix College, News, Newspaper, Pizzagate, Social Media
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Morclean Bear at Work: The Movie is out and now available to watch on YouTube.
He really is a great help here around the office. Superb telephone etiquette and pretty handy with the fork truck!
And, if you want a bear (and why wouldn't you?) see this post here or here about how to get your paws on one.


At Om Nom Health, we run yoga, meditation, mindfulness and relaxation classes for children.
Come and let your children experience the joy and fun that can be had from yoga! The classes are run by Emily; an experienced, fully qualified, DBS-checked children's yoga teacher, registered Nutritional Therapist and meditation teacher.
Classes are small enough to ensure personal attention and large enough to create an inspiring group energy. Students of all levels are welcome and we strive to inspire a unique experience for every student.
We focus on the introduction of yoga basics, rather than memorising or perfecting asanas and routines. The basic yoga techniques over time introduce the child to relaxation and self-control. Many yoga postures are named after plants and animals and resemble them. By imitating animals, children can learn and perform basic yoga poses.
The real purpose of a children's yoga class is to help them find joy within themselves and act from a place of connectedness. The class provides a framework in which they can blossom, so it must be allowed to flow and move in any direction the children take it. Yoga promotes physical strength because children learn to use all of their muscles in new ways. Whether a pose is done standing, sitting or lying down, each one can challenge various muscle groups while helping a child become aware of their body and how it efficiently functions.
Classes are engaging and fun, and promote a state of wellbeing on all levels.


text:
\section{Introduction}
In this paper,
 we consider the stable Grothendieck polynomials $G_\lambda$
and the dual Grothendieck polynomials $g_\lambda$.
They are $K$-theoretic analogues of Schur functions $s_\lambda$.
Grothendieck polynomials introduced in Lascoux--Sch\"utzenberger
 \cite{MR686357}
are the representatives of $K$-theory classes of structure
sheaves of Schubert varieties.
Fomin and Kirillov introduced
the stable Grothendieck polynomial
in their combinatorial study of the Grothendieck polynomials
 \cite{MR1394950}.
The stable Grothendieck polynomial is
a limit of the Grothendieck polynomial.
Buch \cite{MR1946917} showed
stable Grothendieck polynomials
can be written as weighted generating functions of
a kind of tableaux.
The dual Grothendieck polynomials is a dual basis of the stable
 Grothendieck polynomial,
and can be written as weighted generating functions of
another kind of tableaux.
See also
\cite{MR3007174},
\cite{MR1932326} and
\cite{MR2377012}.
Schur functions $s_\lambda$ satisfy
the Cauchy identity
\begin{align*}
 \sum_{\lambda}s_\lambda(x)\cdot s_\lambda(y)=\prod_{i,j}\frac{1}{1-x_iy_j}.
\end{align*}
In \cite{MR2377012},
Lam and Pylyavskyy showed
an analogue of the classical Cauchy identity
for products of the stable
Grothendieck polynomial and the dual Grothendieck polynomial.
In \cite{MR3249830},
Lascoux and Naruse generalized the identity as the finite sum Cauchy identity
\begin{align*}
 \sum_{\lambda}G_\lambda(x_1,\ldots,x_n)\cdot g_\lambda(y_1,\ldots,y_n)
&=
 \sum_{\lambda}s_\lambda(x_1,\ldots,x_n)\cdot s_\lambda(y_1,\ldots,y_n)\\
&=\prod_{i=1}^n\prod_{j=1}^n\frac{1}{1-x_iy_j},
\end{align*}
where the sum is over all $\lambda$ containing by the rectangle shape $(w^n)$.
Since the polynomials $s_\lambda$, $G_\lambda$ and $g_\lambda$ are
weighted generating functions of tableaux,
the finite sum Cauchy identity
means
the existence of
a weight-preserving bijection between some kinds of tableaux.
The main purpose of this paper
is to give a bijective proof
of the finite sum Cauchy identity
for stable Grothendieck polynomials and dual Grothendieck polynomials.

This paper is organized as follows:
In Section \ref{section:tableau},
we define some notion for tableaux
and define the weighted generating functions them.
We introduce
some algorithms to define
bijections in Section \ref{sec:defofalg}.
We use the bumping and the sliding
to define our algorithm.
In Section \ref{sec:main},
we give the main theorem which implies the Cauchy identity.
 We show the main theorem  in Section \ref{sec:proof}.
In Section \ref{sec:finalrmk},
we consider the case of Young diagram
with one column.

\section{Three kinds of tableaux}
\label{section:tableau}
Here we define some notation for Young diagrams and tableaux.
Let $\mathbb{Y}$ be the set of Young diagrams.
We write $(\lambda_1,\lambda_2,\ldots)$
to denote the  Young diagram
consisting of $\lambda_i$ boxes in the $i$-th row.
We use so-called English notation.
For $\lambda\in\mathbb{Y}$,
we call a box $(i,j)\in \lambda$
a \emph{corner} if
 $(i+1,j)\not\in \lambda$
and
 $(i,j+1)\not\in \lambda$.
We call $(i,j)\not\in\lambda$
a \emph{cocorner}
if
 $(i-1,j)\in \lambda$
and
 $(i,j-1)\in \lambda$.
For $\lambda'$, $\lambda\in \mathbb{Y}$ satisfying $\lambda'\subset\lambda$,
we define the \emph{skew Young diagram} $\lambda/\lambda'$
to be the subset $\lambda\setminus\lambda'$ of boxes.
For $\lambda\in\mathbb{Y}$,
$\left|\lambda\right|$ denotes the number of the boxes in $\lambda$.
We call $T$ an \emph{integer-valued column-strict tableau} of shape $\lambda$
if
\begin{enumerate}
\item $T_{i,j}$ is a positive integers;
\item if $(i,j),(i',j) \in \lambda$ satisfy $i<i'$,
 then $T_{i,j}<T_{i',j}$; and
\item if $(i,j),(i,j') \in \lambda$ satisfy $j<j'$,
 then $T_{i,j}\leq T_{i,j'}$.
\end{enumerate}
We sometimes call integers in a tableau \emph{alphabets},
to distinguish between them and the other integers, e.g., coordinates of boxes.
For a skew Young diagram $\lambda/\lambda'$,
we define an
\emph{integer-valued column-strict skew tableau} of shape
$\lambda/\lambda'$
in the same manner.
We define $\mathcal{S}_\lambda$ to be the set of
integer-valued column-strict tableaux of shape $\lambda$.
For $T\in\mathcal{S}_\lambda$,
define $\mu(T)$ to be the sequence $(\mu_1,\mu_2,\ldots)$
such that $\mu_k=\numof{\Set{(i,j)\in\lambda|T_{i,j} = k }}$.
In other words,
$\mu_k$ is the multiplicity of the alphabet $k$
in the tableau $T$.
For a Young diagram $\lambda\in\mathbb{Y}$, and formal variables $x=(x_1,x_2,\ldots)$,
we define the weighted generating function $s_\lambda(x)$ by
\begin{align*}
 s_\lambda(x)=\sum_{T\in\mathcal{S}_\lambda} x^{\mu(T)},
\end{align*}
where $x^{\alpha}$ stands for the monomial
$x_1^{\alpha_1}x_2^{\alpha_2}\cdots$.
The generating functions $s_\lambda(x)$ are called
\emph{Schur functions}.
For the Young diagram $(l)$ consisting of one row with $l$ boxes,
$s_{(l)}(x)$ equals the $l$-th homogeneous complete symmetric polynomial $h_l(x)$.
For the Young diagram $(1^l)$ consisting of one column with $l$ boxes,
$s_{(1^l)}(x)$ equals the $l$-th elementary symmetric polynomial $e_l(x)$.

For $\lambda\in\mathbb{Y}$,
we call $P$ a \emph{set-valued column-strict tableau}
of shape $\lambda$
if
\begin{enumerate}
\item $P_{i,j}$ is a nonempty finite subset of positive integers;
\item if $(i,j),(i',j) \in \lambda$ satisfy $i<i'$,
 then $\max(P_{i,j})<\min(P_{i',j})$; and
\item if $(i,j),(i,j') \in \lambda$ satisfy $j<j'$,
 then $\max(P_{i,j})\leq\min(P_{i,j'})$.
\end{enumerate}
We define $\mathcal{P}_\lambda$ to be the set of
set-valued column-strict tableaux of shape $\lambda$.
For $P\in \mathcal{P}_\lambda$,
we define $\varepsilon(P)$ by
\begin{align*}
\varepsilon(P)&=\sum_{(i,j)\in\lambda} (\numof{P_{i,j}}-1).
\end{align*}
A set-valued column-strict tableau $P\in\mathcal{P}_\lambda$ satisfies $\varepsilon(P)=0$
if and only if $\numof{P_{i,j}}=1$ for all $(i,j)\in\lambda$.
We can identify $P\in\mathcal{P}_\lambda$
satisfying $\varepsilon(P)=0$
with an integer-valued column-strict tableau.
For $\lambda\in\mathbb{Y}$,
we define $\tilde\mathcal{S}_\lambda$ and $\mathcal{P}_\lambda^+$ by
\begin{align*}
\tilde\mathcal{S}_\lambda&=\Set{P\in\mathcal{P}_\lambda|\varepsilon(P)=0}, \\
\mathcal{P}_\lambda^+&=\Set{P\in\mathcal{P}_\lambda|\varepsilon(P)>0} \\
 &=\mathcal{P}_\lambda\setminus\tilde\mathcal{S}_\lambda.
\end{align*}
We define $\mu(P)$ to be the sequence of integers $(\mu_1,\mu_2,\ldots)$
such that $\mu_k=\numof{\Set{ (i,j) \in \lambda| k\in P_{i,j}}}$.
It is easy to show that
\begin{align}
 \left|\lambda\right| =-\varepsilon(P)+\sum_i \mu_i.
\label{eq:sum:mu}
\end{align}
For a Young diagram $\lambda\in \mathbb{Y}$ and formal variables $\beta$ and $x=(x_1,x_2,\ldots)$,
we define a weighted generating function $G_\lambda(\beta,x)$
by
\begin{align*}
 G_\lambda(\beta,x)
&=\sum_{P\in\mathcal{P}_\lambda} (-\beta)^{\varepsilon(P)}x^{\mu(P)}.
\end{align*}
We call $G_\lambda(x)=G_\lambda(1,x)$ a \emph{(stable Gra\ss{}mannian) Grothendieck polynomial}.
It is easy to show that $G_\lambda(0,x)=s_\lambda(x)$.

For $\lambda\in\mathbb{Y}$,
we call $Q$ a \emph{reverse plane partition} of $\lambda$
if
\begin{enumerate}
\item $Q_{i,j}$ is a positive integer;
\item if $(i,j),(i',j) \in \lambda$ satisfy $i<i'$,
 then $Q_{i,j}\leq Q_{i',j}$; and
\item if $(i,j),(i,j') \in \lambda$ satisfy $j<j'$,
 then $Q_{i,j}\leq Q_{i,j'}$.
\end{enumerate}
We define $\mathcal{Q}_\lambda$ to be the set of
reverse plane partitions of $\lambda$.
For $Q\in\mathcal{Q}_\lambda$,
we define $\delta(Q)$ by
\begin{align*}
 \delta(Q)&=\numof{\Set{(i,j)\in\lambda| Q_{i,j}=Q_{i+1,j}}}.
\end{align*}
For $Q\in\mathcal{Q}_\lambda$,
it follows from definition
that $Q\in\mathcal{S}_\lambda$
if and only if
$\delta(Q)=0$.
For $\lambda\in\mathbb{Y}$,
we define $\mathcal{Q}_\lambda^+$ by
\begin{align*}
\mathcal{Q}_\lambda^+&=\Set{Q\in \mathcal{Q}_\lambda | \delta(Q)>0}\\
 &=\mathcal{Q}_\lambda\setminus\mathcal{S}_\lambda.
\end{align*}
We also define $\nu(Q)$ to be the sequence $(\nu_1,\nu_2,\ldots)$
such that
\begin{align*}
\nu_k=\numof{\Set{j|\text{There exists $(i,j)\in\lambda$ such that $Q_{i,j} = k $}}}.
\end{align*}
It is easy to show that
\begin{align}
\left|\lambda\right| =\delta(Q)+\sum_i \nu_i.
 \label{eq:nu}
\end{align}
For a Young diagram $\lambda\in \mathbb{Y}$ and formal variables $\beta$ and $y=(y_1,y_2,\ldots)$,
we define a weighted generating function $g_\lambda(\beta,y)$
by
\begin{align*}
 g_\lambda(\beta,y)
&=\sum_{Q\in\mathcal{Q}_\lambda} \beta^{\delta(Q)}y^{\nu(Q)}.
\end{align*}
We call $g_\lambda(y)=g_\lambda(1,y)$
a \emph{dual Grothendieck polynomial}.
It is easy show that $ g_\lambda(0,y)=s_\lambda(y)$.

\section{Our algorithm}
\label{sec:defofalg}
\subsection{The row insertions and the jeu de taquin}
\label{sec:knownalgorithm}
Here we recall some well-know algorithms for
integer-valued column-strict tableaux.
\begin{algorithm}[Row bumping]
 Let $T_{i,1},\ldots,T_{i,\lambda_i}$
be the $i$-th row of an integer-valued column-strict tableau $T\in \mathcal{S}_\lambda$,
and $x$ an alphabet.
We define $T'$ and $y$ in the following manner:
\begin{enumerate}
 \item  Find $j$ satisfying the following:
 \begin{enumerate}
  \item if $j'< j$ and $x'=T_{i,j'}$, then
	$x'\leq x$; and
  \item if $j\leq j''$ and  $x''=T_{i,j''}$, then
	$x < x''$.
 \end{enumerate}
 \item If $T$ has a box at $(i,j)$, then
       define $y$ to be $T_{i,j}$ and
       define $T'$ to be the tableau obtained from $T$
       by putting the alphabet $x$ at the box $(i,j)$.
 \item If $T$ has no box at $(i,j)$, then
       define $y$ to be null
       and define $T'$ to be the tableau obtained from $T$
       by adding the new box $(i,j)$ with the alphabet $x$.
\end{enumerate}
\end{algorithm}

If we
repeat inserting the integer bumped out from the $i$-th row
into the $(i+1)$-th row by the row bumping
until the null is bumped out,
then we obtain an integer-valued column-strict tableau.
We call the whole process  the \emph{row-insertion}.
The row-insertion stops when
an integer is inserted into a corner as a new box.
We call the set of boxes where an integer is inserted
in the process of the row-insertion
the \emph{bumping route}.
If an integer is inserted into the box $(1,j_1)$ by the first bumping of
the row-insertion,
then the bumping route is a subset
of $\Set{(i,j)|j\leq j_1}$.

\begin{algorithm}[Reverse row bumping]
 Let $T_{i,1},\ldots,T_{i,\lambda_i}$ be the $i$-th row of
an integer-valued column-strict tableau $T\in \mathcal{S}_\lambda$,
and $y$ an alphabet.
We define $T'$ and $x$ in the following manner:
\begin{enumerate}
 \item  Find $j$ satisfying the following:
 \begin{enumerate}
  \item if $j'\leq j$ and $y'=T_{i,j'}$, then
	$y'<y$; and
  \item if $j<j''$ and  $y''=T_{i,j''}$, then
	$y \leq y''$.
 \end{enumerate}
 \item
       Define $x$ to be $T_{i,j}$ and
       define $T'$ to be the tableau obtained from $T$
       by putting the alphabet $y$ at the box $(i,j)$.
\end{enumerate}
\end{algorithm}

Let $T$ be an integer-valued column-strict tableau of shape $\lambda$,
 $(i_1,j_1)$ a corner of $\lambda$.
If we remove a corner  $(i_1,j_1)$ from $T$ and
insert $T_{i_1,j_1}$ into the $(i_1-1)$-th row of $T$
by the reverse row-bumping,
then we obtain an integer-valued column-strict tableau
and an integer bumped out from the $(i_1-1)$-th row.
If we
repeat inserting the integer bumped out from the $(i+1)$-th row
into the $i$-th row
by the reverse row bumping,
then we obtain an integer-valued column-strict tableau
and an integer bumped out.
We call the whole process  the \emph{reverse row-insertion}.
The reverse row-insertion is the inverse of the row-insertion.
We call the set of boxes where an integer is bumped out in the process
of the reverse row-insertion the \emph{reverse-bumping route}.

Next we recall another kind of algorithms.
\begin{algorithm}[Sliding]
 Let $T^{(0)}$ be an integer-valued column-strict tableau
 with a null box $(i_0,j_0)$.
 We define $T$ in the following manner:
\begin{enumerate}
 \item If $T^{(0)}_{i_0+1,j_0}>T^{(0)}_{i_0,j_0+1}$ or $T^{(0)}$ has no box at $(i_0+1,j_0)$,
       then let $T$ be the tableau obtained from $T^{(0)}$ by swapping
       the entries of boxes $(i_0,j_0)$ and $(i_0,j_0+1)$.
 \item If $T^{(0)}_{i_0+1,j_0}\leq T^{(0)}_{i_0,j_0+1}$ or $T^{(0)}$ has no box at $(i_0,j_0+1)$,
       then let $T$ be the tableau obtained from $T^{(0)}$ by swapping
       the entries of boxes $(i_0,j_0)$ and $(i_0+1,j_0)$.
\end{enumerate}
\end{algorithm}

Let $T$ be an integer-valued column-strict skew tableau
of shape $\lambda/(1)$.
We regard the box $(1,1)$ as a null box.
Repeat the sliding until the null box reaches a corner of $\lambda$,
and then remove the null box.
Then we obtain an integer-valued column-strict tableau.
We call the whole process the \emph{jeu de taquin}.
We call the boxes where the null box passed
in the process of the jeu de taquin
the \emph{sliding route}.

\begin{algorithm}[Reverse sliding]
 Let $T^{(1)}$ be an integer-valued column-strict tableau
 with a null box $(i_1,j_1)$.
 We define $T$ in the following manner:
\begin{enumerate}
 \item If $i_1=1$ or $T^{(1)}_{i_1-1,j_1}<T^{(1)}_{i_1,j_1-1}$,
       then let $T$ be the tableau obtained from $T^{(1)}$ by swapping
       the entries of boxes $(i_1,j_1)$ and $(i_1,j_1-1)$.
 \item If  $j_1=1$ or $T^{(1)}_{i_1-1,j_1}\geq T^{(1)}_{i_1,j_1-1}$,
       then let $T$ be the tableau obtained from $T^{(1)}$ by swapping
       the entries boxes $(i_1,j_1)$ and $(i_1-1,j_1)$.
\end{enumerate}
\end{algorithm}
Let $T$ be an integer-valued column-strict tableau of shape $\lambda$,
and $(i,\lambda_{i}+1)$ a cocorner of $\lambda$.
Append a null box to the cocorner $(i,\lambda_{i}+1)$ of $T$.
Repeat the reverse sliding until the null box reaches $(1,1)$.
Then we obtain
an integer-valued column-strict skew tableau
of shape $(\lambda_1,\ldots,\lambda_{i-1},\lambda_i+1,\lambda_{i+1},\ldots)/(1,1)$.
We call the whole process the \emph{reverse jeu de taquin}.
The reverse jeu de taquin is the inverse of the jeu de taquin.
We call the boxes where the null box passed
in the process of the reverse jeu de taquin
the \emph{reverse-sliding route}.

\subsection{Operation for tableaux}
\label{subsec:alg:tab}
If $P\in \mathcal{P}_\lambda^+$,
then $P$ has a box containing more than one alphabets.
We define notation for the end of such boxes.
\begin{definition}
For $P\in\mathcal{P}_\lambda^+$
define $r_1(P)$ and  $r_1'(P)$  by
\begin{align*}
r_1(P)&=\max \Set{i| \text{ $\numof{P_{i,j}}>1$ for some $j$}},\\
r_1'(P)&=\max\Set{j| \numof{P_{r_1(P),j}}>0}.
\end{align*}
For $P\in \tilde\mathcal{S}_\lambda$,
we define $r_1(P)=0$.
\end{definition}
We can regard  $P\in \mathcal{P}^+_\lambda$
as an integer-valued column-strict (skew) tableau,
if we consider only
the set
\begin{align*}
\Set{(i,j)\in\lambda| i>r_1(P)}
\end{align*}
of boxes.
Hence we define $r_2(P)$, $r_2'(P)$ and  $R(P)$
by the row bumping algorithm for integer-valued column-strict tableaux.
\begin{algorithm}
\label{algorithm:R(P)}
For $P\in\mathcal{P}_\lambda^+$,
we define
the cocorner $(r_2(P),r_2'(P))$ of $\lambda$
and the set-valued column-strict tableau $R(P)$
in the following manner:
\begin{enumerate}
 \item Let $x=\max(P_{r_1(P),r_1'(P)})$.
 \item Remove the alphabet $x$ from the box $(r_1(P),r_1'(P))$ of $P$.
 \item  Insert the alphabet $x$
	into the tableau
	by the row insertion from the $(r_1(P)+1)$-th row.
	As the result of insertion,
	we obtain a new tableau $P^{(1)}$ of shape $\lambda^{(1)}$.
 \item Define $R(P)\in\mathcal{P}_{\lambda^{(1)}}$ to be $P^{(1)}$.
 \item Define $(r_2(P),r_2'(P))$ to be the new box,
       i.e., the box in $\lambda^{(1)}$ but not in $\lambda$.
\end{enumerate}
\end{algorithm}
\begin{example}
\label{example:P}
Define set-valued column-strict tableaux $P$ and $P'$ by
\begin{align*}
 P&=
\Tableau[Y]{1&12&2&34&4&6 \\ 23&3&\tf 34&5&6&8\\3&4&5&6&7\\4&6&7&8 \\ 6 },\\
 P'&=
 \Tableau[Y]{1&12&2&34&4&6 \\ 23&3&34&5&\tf 67&8\\3&4&5&6&8\\4&6&7&8 \\ 6 }.
\end{align*}
In this case, $(r_1(P),r_1'(P))=(2,3)$ and $(r_1(P'),r_1'(P'))=(2,5)$.
To calculate
$R(P)$ (\emph{resp.}\ $R(P')$)
we insert the alphabet $4=\max\set{3,4}$
(\emph{resp.}\ $7=\max\set{6,7}$)
from the third row of $P$ by
 the row-bumping. Hence we obtain
\begin{align*}
 R(P)&=
\Tableau[Y]{1&12&2&34&4&6 \\ 23&3&\tf
 3&5&6&8\\3&4&\textbf{4}&6&7\\4&\textbf{5}&7&8 \\ 6 &\tf \textbf{6}},\\
 R(P')&=
 \Tableau[Y]{1&12&2&34&4&6 \\ 23&3&34&5&\tf 6&8\\3&4&5&6&\textbf{7}\\4&6&7&8&\tf\textbf{8} \\ 6 }.
\end{align*}
Therefore $(r_2(P),r_2'(P))=(5,2)$ and $(r_2(P'),r_2'(P'))=(4,5)$.
\end{example}

A reverse plane partition $Q\in \mathcal{Q}_\lambda^+$
contains a box $(i,j)$ such that $Q_{i,j}=Q_{i+1,j}$.
We define notation for the end of such boxes.
\begin{definition}
For $Q\in \mathcal{Q}_\lambda^+$,
we define $f_1(Q)$ and $f_1'(Q)$ by
\begin{align*}
 f_1(Q)&=\max \Set{i|\text{$Q_{i,j}=Q_{i+1,j}$ for some $j$}},\\
 f_1'(Q)&=\max\Set{j| Q_{f_1(Q),j}=Q_{f_1(Q)+1,j}}.
\end{align*}
For $Q\in\mathcal{S}_\lambda$,
we define $f_1(Q)=0$.
\end{definition}
We can regard $Q\in \mathcal{Q}_\lambda^+$
as an integer-valued column-strict tableau
if
we consider only the set
\begin{align*}
\Set{(i,j)\in\lambda| \text{$i\geq f_1(Q)$ and $j\geq f_1'(Q)$}}
\setminus \Set{(f_1(Q),f_1'(Q))}
\end{align*}
of boxes.
Hence we define $f_2(Q)$, $f_2'(Q)$ and $F(Q)$
by the jeu de taquin for
integer-valued column-strict tableaux.
\begin{algorithm}
\label{algorithm:F}
For $Q\in \mathcal{Q}^+_\lambda$,
we define
the corner $(f_2(Q),f_2'(Q))$ of $\lambda$
and the reverse plane partition $F(Q)$ in the following manner:
\begin{enumerate}
 \item Let the box $(f_1(Q),f_1'(Q))$ in $Q$ be null.
 \item Slide the null box to outside by the jeu de taquin.
 \item As the result of sliding,
       we obtain a new tableau $Q^{(1)}$ of shape $\lambda^{(1)}$.
 \item Define $F(Q)\in \mathcal{Q}_{\lambda^{(1)}}$ to be $Q^{(1)}$.
 \item Define $(f_2(P),f_2'(P))$ to be
       the removed box
       $\lambda/ \lambda^{(1)}$.
\end{enumerate}
\end{algorithm}
\begin{example}
\label{example:Q}
Define reverse plane partition $Q$ by
\begin{align*}
 Q&=
\Tableau[Y]{1&1&2&4&5&5 \\ 1&\tf 2&3&4&4&6\\2&2&6&6&7\\3&4&7&9 \\ 5 }.
\end{align*}
In this case, $(f_1(Q),f_1'(Q))=(2,2)$.
To calculate
$F(Q)$,
we put a null box at $(2,2)$.
Then we obtain the following by the jeu de taquin:
\begin{align*}
 F(Q)&=
\Tableau[Y]{1&1&2&4&5&5 \\ 1&\tf \textbf{2}&3&4&4&6\\2&\textbf{4}&6&6&7\\3&\textbf{7}&\textbf{9}&\tf \\ 5 }.
\end{align*}
Therefore $(f_2(Q),f_2'(Q))=(4,4)$.
\end{example}

\subsection{Operation for pairs of tableaux}
\label{subsec:alg:pair}
For $\lambda\in\mathbb{Y}$,
we define
 $\mathfrak{R}_\lambda$,  $\mathfrak{R}^0_\lambda$ and
$\mathfrak{R}^+_\lambda$
by
\begin{align*}
 \mathfrak{R}_\lambda&=\mathcal{P}_\lambda\times\mathcal{Q}_\lambda,\\
 \mathfrak{R}^0_\lambda&=\tilde\mathcal{S}_\lambda\times\mathcal{S}_\lambda\\
 &=\Set{(P,Q)\in\mathfrak{R}_\lambda|\varepsilon(P)=\delta(Q)=0},\\
 \mathfrak{R}^+_\lambda&=\mathfrak{R}_\lambda\setminus \mathfrak{R}^0_\lambda\\
 &=\Set{(P,Q)\in\mathfrak{R}_\lambda|\varepsilon(P)+\delta(Q)>0}.
\end{align*}
We decompose $\mathfrak{R}^+_\lambda$ into the two subsets
${\check{\mathfrak{R}}}_\lambda$ and ${\hat{\mathfrak{R}}}_\lambda$.
\begin{definition}
Define ${\check{\mathfrak{R}}}_\lambda$ to be
the set of pairs $(P,Q)$ of tableaux in $\mathfrak{R}^+_\lambda$
satisfying
\begin{enumerate}
\item $r_1(P)>f_1(Q)$; or
\item $r_1(P)=f_1(Q)$ and $r_2(P)\leq f_2(Q)$.
\end{enumerate}
We also
define ${\hat{\mathfrak{R}}}_\lambda$ to be
the set of pairs $(P,Q)$ of tableaux in $\mathfrak{R}^+_\lambda$
satisfying
\begin{enumerate}
\item $r_1(P)<f_1(Q)$; or
\item $r_1(P)=f_1(Q)$ and $r_2(P)> f_2(Q)$.
\end{enumerate}
\end{definition}
It is easy to show that
$\mathfrak{R}^+_\lambda={\check{\mathfrak{R}}}_\lambda\amalg {\hat{\mathfrak{R}}}_\lambda$.
First we define a map $\check{\varphi}$ from ${\check{\mathfrak{R}}}_\lambda$ to $\mathcal{Q}^+$.
\begin{algorithm}
\label{algorithm:checkvarphi}
 For a $(P,Q)\in {\check{\mathfrak{R}}}_\lambda$,
 we define $\check{\varphi}(P,Q) \in\mathcal{Q}^+$ in the following manner:
\begin{enumerate}
 \item Let $Q^{(1)}$ be $Q$.
 \item Let $Q^{(2)}$ be the tableau obtained from $Q$ by adding
       a null box at the cocorner $(r_2(P),r_2'(P))$.
 \item
       \label{def:E2:alg:step:jdt}
       Slide the null box $(r_2(P),r_2'(P))$ of $Q^{(2)}$
       into the inside of $Q^{(2)}$
       by the reverse jeu de taquin
       until the null box moves to $r_1(P)$-th row.
       Let $Q^{(3)}$ be the resulting tableau.
 \item
       Define $(r_3(P,Q),r_3'(P,Q))$  to be the null box  of $Q^{(3)}$.
 \item
       \label{def:E2:alg:step:putalphabet}
       Define $\check{\varphi}(P,Q)$ to be the tableau obtained from  $Q^{(3)}$
       by putting the alphabet $Q^{(1)}_{r_3(P,Q),r_3'(P,Q)}$
       into the null box $(r_3(P,Q),r_3'(P,Q))$ of  $Q^{(3)}$.

\end{enumerate}
\end{algorithm}

\begin{remark}
In the case where $r_1(P)>f_1(Q)$,
$Q$ is column-strict
if we consider only boxes from $r_1(P)$-th row to $r_2(P)$-th row.
Hence Step \ref{def:E2:alg:step:jdt} is well-defined.
In the case where  $r_1(P)=f_1(Q)$ and $r_2(P)\leq f_2(Q)$,
$Q$ is column-strict
if we consider only the boxes
\begin{align*}
\Set{(i,j)\in\lambda| \text{$i\geq f_1(Q)$ and $j\geq f_1'(Q)$}} \setminus \Set{(f_1(Q),f_1'(Q))}.
\end{align*}
Since $r_2(P)\leq f_2(Q)$,
it follows from Lemma \ref{lemma:slidingroute1}
that
the null box reaches a box in $\Set{(r_1(P),j)|j>f_2(Q)}$.
Hence Step \ref{def:E2:alg:step:jdt} is also well-defined in this case.
\end{remark}
\begin{remark}
\label{remark:varphi}
The reverse jeu de taquin in Step \ref{def:E2:alg:step:jdt}
 stops if the null box arrives at the $r_1(P)$-th row.
 Hence $Q^{(3)}_{r_3(P,Q)+1,r_3'(P,Q)}=Q^{(1)}_{r_3(P,Q),r_3'(P,Q)}$.
 Moreover,
 if $(r_3(P,Q),r_3'(P,Q)+1)\in\lambda$,
 then
\begin{align*}
Q^{(3)}_{r_3(P,Q),r_3'(P,Q)+1}=Q^{(1)}_{r_3(P,Q),r_3'(P,Q)+1}\geq Q^{(1)}_{r_3(P,Q),r_3'(P,Q)}.
\end{align*}
 Since we put the alphabet $Q^{(1)}_{r_3(P,Q),r_3'(P,Q)}$
 on the box $(r_3(P,Q),r_3'(P,Q))$ of $Q^{(3)}$
 in Step \ref{def:E2:alg:step:putalphabet},
 $\check{\varphi}(P,Q)$ is in $\mathcal{Q}^+$
and the shape of $\check{\varphi}(P,Q)$
is $\lambda\cup \Set{(r_2(P),r_2'(P))}$.
\end{remark}
\begin{example}
Consider $P'$ and $Q$ in Examples \ref{example:P} and \ref{example:Q}.
Since $r_1(P')=f_1(Q)=2$ and
$r_2(P')=f_2(Q)=4$,
the pair $(P',Q)$ is in ${\check{\mathfrak{R}}}_\lambda$.
To calculate $\check{\varphi}(P',Q)$,
we add a null box
to the cocorner $(r_2(P'),r_2'(P'))=(4,5)$ of $Q$.
Then we obtain
the following tableau by the reverse jeu de taquin:
\begin{align*}
\Tableau[Y]{1&1&2&4&5&5 \\ 1&2&\tf&4&4&6\\2&2&\textbf{3}&6&7\\3&4&\textbf{6}&\textbf{7}&\tf \textbf{9}\\ 5 }.
\end{align*}
Finally we put the alphabet $3$,
which lies bellow the null box,
into the null box to obtain
\begin{align*}
 \check{\varphi}(P',Q)=\Tableau[Y]{1&1&2&4&5&5 \\ 1& 2&\tf \textbf{3} &4&4&6\\2&2&\textbf{3}&6&7\\3&4&\textbf{6}&\textbf{7}&\tf \textbf{9}\\ 5 }.
\end{align*}
\end{example}
\begin{definition}
For $(P,Q) \in {\check{\mathfrak{R}}}_{\lambda}$,
we define $\check{\Phi}(P,Q)$ by
$\check{\Phi}(P,Q)=(R(P),\check{\varphi}(P,Q))$.
\end{definition}

Next we define a map $\hat{\varphi}$ from ${\hat{\mathfrak{R}}}_\lambda$ to $\mathcal{P}^+$.
\begin{algorithm}
\label{algorithm:hatvarphi}
 For $(P,Q)\in{\hat{\mathfrak{R}}}_\lambda$,
we define $\hat{\varphi}(P,Q)\in\mathcal{P}^+$
in the following manner:
\begin{enumerate}
\item
      \label{def:F2:alg:step:pickupuniqelm}
      Let $x$ be the alphabet in $P_{f_2(Q),f_2'(Q)}$.
\item Let $P^{(2)}$ be the tableau obtained from $P$
      by removing the corner $(f_2(Q),f_2'(Q))$.
\item
      \label{def:F2:alg:step:revbumping}
      Insert the alphabet $x$ into $P^{(2)}$
      by the reverse row insertion from the $(f_2(Q)-1)$-th row
      until some number $y$ is bumped out from the $(f_1(Q)+1)$-th row.
      Let $P^{(3)}$ be the resulting tableau.
\item
      \label{def:F2:alg:step:findpos}
      Find $j$ satisfying the following:
      \begin{enumerate}
       \item if $j'\leq j$ and $y' \in P^{(3)}_{f_1(Q),j'}$, then
	     $y'<y$; and
       \item if $j<j''$ and  $y'' \in P^{(3)}_{f_1(Q),j''}$, then
	     $y \leq y''$.
      \end{enumerate}
\item Define $(f_3(P,Q),f_3'(P,Q))$ to be the box $(f_1(Q),j)$.
\item Define $\hat{\varphi}(P,Q)$ to be the tableau
      obtained from $P^{(3)}$
      by appending the alphabet $y$ into the box $(f_3(P,Q),f_3'(P,Q))$.
\end{enumerate}
\end{algorithm}
\begin{remark}
For $(P,Q)\in {\hat{\mathfrak{R}}}_\lambda$,
 the box $(f_2(Q),f_2'(Q))$ is a corner of $\lambda$
 and $\numof{P_{f_2(Q),f_2'(Q)}}$ equals one.
Hence $x$ in Step \ref{def:F2:alg:step:pickupuniqelm}
is unique.
\end{remark}
\begin{remark}
 Since $r_1(P)\leq f_1(Q)$,
 we can regard  $P$ as an integer-valued column-strict tableau
 if we consider only the boxes from
 the $(f_1(Q)+1)$-th row to the $(f_2(Q)-1)$-th row.
Hence Step \ref{def:F2:alg:step:revbumping}
is well-defined.
\end{remark}
\begin{remark}
\label{remark:hatphi}
Roughly speaking,
 $(f_3(P,Q),f_3'(P,Q))=(f_1(Q),j)$ is the unique box in the $f_1(Q)$-th row
such that
\begin{enumerate}
 \item  $y$ will be the maximum in the box; and
 \item  the resulting tableaux will be column-strict if we insert $y$ into the box.
\end{enumerate}
If $\numof{P_{f_1(Q),j'}}=1$ for all $j'$,
then we can find  $j$ in Step \ref{def:F2:alg:step:findpos}
similarly to the case of the reverse bumping algorithm
for integer-valued column-strict tableaux.
In the case where
$r_1(P)<f_1(Q)$,
$\numof{P_{f_1(Q),j'}}=1$ for all $j'$.
Hence we can find $j$ in Step \ref{def:F2:alg:step:findpos}.
On the other hand,
in the case where $r_1(P)=f_1(Q)$ and $r_2(P)> f_2(Q)$,
there exists $j$ such that $\numof{P_{f_1(Q),j}}>1$.
Since $r_2(P)> f_2(Q)$,
it follows from Lemma \ref{lemma:bumpingroute1}
that
$\max(P_{f_1(Q),f_1'(Q)})<y$.
Since $\numof{P_{f_1(Q),j'}}=1$ for all $j'>f_1'(Q)$,
we can find $j$ in Step \ref{def:F2:alg:step:findpos}.
\end{remark}

\begin{example}
Consider $P$ and $Q$ in Examples \ref{example:P} and \ref{example:Q}.
Since $r_1(P)=f_1(Q)=2$ and
$r_2(P)=5>4=f_2(Q)$,
the pair $(P,Q)$ is in ${\hat{\mathfrak{R}}}_\lambda$.
To calculate $\hat{\varphi}(P,Q)$,
we remove the corner $(f_2(Q),f_2'(Q))=(4,4)$
of $P$
by the reverse row insertion until $(f_1(Q)+1)$-th row.
Then we obtain
\begin{align*}
\Tableau[Y]{1&12&2&34&4&6 \\
 23&3&34&5&6&8\\3&4&5&6&\textbf{8}\\4&6&7&\tf \\ 6 }
\end{align*}
and the alphabet $7$ bumped out from $(f_1(Q)+1)$-th row.
Moreover, by inserting the alphabet $7$
into  $f_1(Q)$-th row,
we obtain
\begin{align*}
\hat{\varphi}(P,Q)=
\Tableau[Y]{1&12&2&34&4&6 \\ 23&3&34&5&\tf 6\textbf{7}&8\\3&4&5&6&\textbf{8}\\4&6&7\\ 6 }.
\end{align*}
\end{example}

\begin{definition}
For $(P,Q)\in {\hat{\mathfrak{R}}}_\lambda$,
we define $\hat{\Phi}(P,Q)$ by
$\hat{\Phi}(P,Q)=(\hat{\varphi}(P,Q),F(Q))$.

\end{definition}

\section{Main results}
\label{sec:main}

For  a nonnegative integer $w$,
we define  ${\check{\mathfrak{R}}}(w)$, ${\hat{\mathfrak{R}}}(w)$, $\mathfrak{R}^+(w)$ and $\mathfrak{R}(w)$ by
\begin{align*}
{\check{\mathfrak{R}}}(w)=\bigcup_{\lambda\in\mathbb{Y}\colon \lambda_1=w} {\check{\mathfrak{R}}}_\lambda,\\
 {\hat{\mathfrak{R}}}(w)=\bigcup_{\lambda\in\mathbb{Y}\colon \lambda_1=w} {\hat{\mathfrak{R}}}_\lambda, \\
 \mathfrak{R}^+(w)=\bigcup_{\lambda\in\mathbb{Y}\colon \lambda_1=w} \mathfrak{R}^+_\lambda,\\
 \mathfrak{R}(w)=\bigcup_{\lambda\in\mathbb{Y}\colon \lambda_1=w} \mathfrak{R}_\lambda.
\end{align*}
Then we have the following theorems.
\begin{theorem}[Main theorem]
\label{thm:main:bij}
For $(P,Q)\in {\check{\mathfrak{R}}}(w)$,
define $\check\Phi_w(P,Q)$ to be $\check{\Phi}(P,Q)$.
Then $\check\Phi_w$
is a bijection from ${\check{\mathfrak{R}}}(w)$ to ${\hat{\mathfrak{R}}}(w)$
satisfying
\begin{align*}
 \mu(P)&=\mu(\hat P),&
 \nu(Q)&=\nu(\hat Q),&
 \varepsilon(P)-1&=\varepsilon(\hat P),&
 \delta(Q)+1&=\delta(\hat Q)
\end{align*}
for $(\hat P,\hat Q)=\check{\Phi}(P,Q)$.
The inverse of $\check\Phi_w$ is defined by
$\check\Phi_w^{-1}(P,Q)=\hat\Phi(P,Q)$.
\end{theorem}

We can obtain the finite sum Cauchy identity
for Grothendieck polynomials
 as a corollary to the main results.
Let $(P,Q)\in{\check{\mathfrak{R}}}_\lambda$ and $(\hat P,\hat Q)\in {\hat{\mathfrak{R}}}_{\hat\lambda}$
satisfy $(\hat P,\hat Q)=\check{\Phi}(P,Q)$.
If follows from the equations
that
\begin{align*}
(-\beta)^{\varepsilon(P)}x^{\mu(P)}\cdot
 \beta^{\delta(Q)}y^{\nu(Q)}=-1\cdot(-\beta)^{\varepsilon(\hat P)}x^{\mu(\hat P)}\cdot
 \beta^{\delta(\hat Q)}y^{\nu(\hat Q)}.
\end{align*}
Since $\mathfrak{R}^+_\lambda={\check{\mathfrak{R}}}_\lambda\amalg {\hat{\mathfrak{R}}}_\lambda$,
we have the following equation
\begin{align*}
&\sum_{(P,Q)\in {\check{\mathfrak{R}}}(w)}
(-\beta)^{\varepsilon(P)}x^{\mu(P)}\cdot \beta^{\delta(Q)}y^{\nu(Q)}\\
=
-
&\sum_{(\hat P,\hat Q)\in {\hat{\mathfrak{R}}}_{w}}
(-\beta)^{\varepsilon(\hat P)}x^{\mu(\hat P)}\cdot \beta^{\delta(\hat Q)}y^{\nu(\hat Q)},
\end{align*}
which implies
\begin{align*}
\sum_{(P,Q)\in \mathfrak{R}^+(w)}
(-\beta)^{\varepsilon(P)}x^{\mu(P)}\cdot \beta^{\delta(Q)}y^{\nu(Q)}&=0.
\end{align*}
Since $(P,Q) \in \mathfrak{R}^0_\lambda$
satisfies $\varepsilon(P)=0$ and $\delta(Q)=0$,
we can identified
$\mathfrak{R}^0_\lambda$
with $\mathcal{S}_\lambda\times \mathcal{S}_\lambda$.
Therefore it follows that
\begin{align*}
 \sum_{\substack{\lambda\in\mathbb{Y},\\ \lambda_1=w}}
 G_\lambda(\beta,x)\cdot g_\lambda(\beta,y)
&=
\sum_{(P,Q)\in \mathfrak{R}(w)}
(-\beta)^{\varepsilon(P)}x^{\mu(P)}\cdot \beta^{\delta(Q)}y^{\nu(Q)}\\
&=
\sum_{(P,Q)\in \mathfrak{R}^0(w)}
(-\beta)^{\varepsilon(P)}x^{\mu(P)}\cdot \beta^{\delta(Q)}y^{\nu(Q)}\\
&=
\sum_{(P,Q)\in \mathfrak{R}^0(w)}
x^{\mu(P)}\cdot y^{\nu(Q)}
=
\sum_{\substack{\lambda\in\mathbb{Y},\\ \lambda_1=w}}
s_\lambda(x)\cdot s_\lambda(y).
\end{align*}

\section{Proof}
\label{sec:proof}
Here we show the main theorem.
First
we recall a lemma for a bumping route.
See e.g. Fulton \cite[p.\ 9. the Row Bumping Lemma]{MR1464693} for the detail.
\begin{lemma}
 \label{lemma:bumpingroute2}
Let $T$ be an integer-valued column-strict tableau of shape $\lambda$.
First insert an alphabet $x$ into $T$ by the row bumping algorithm.
Let $T'$ be the resulting tableau, and $(i',j')$ a  new box.
Next insert an alphabet $y$ into $T'$
from by the row bumping algorithm.
Let $(i'',j'')$ a  new box.
\begin{enumerate}
 \item If $x\leq y$, then $i'\geq i''$ and $j' < j''$.
 \item If $x> y$, then $i'<i''$ and $j'\geq j''$.
\end{enumerate}
\end{lemma}
This lemma implies following:
\begin{lemma}
 \label{lemma:bumpingroute1}
Let $T$ be an integer-valued column-strict tableau of shape $\lambda$.
First insert an alphabet $x$ into $T$ from the $i$-th row
by the row bumping algorithm.
Let $(i',j')$ be the new box of the resulting tableaux.
Next remove a corner $(i'',j'')$ of the original tableau $T$,
 insert the alphabet $T_{i'',j''}$ into it from the $(i''-1)$-th row
by the reverse bumping algorithm,
repeat the reverse bumping algorithm until an alphabet is bumped out
 from the $i$-th row.
Let $y$ be the alphabet bumped out from the $i$-th row.
If $i''<i'$, then $x<y$.
\end{lemma}
\begin{proof}
Let $T'$ be the tableaux obtained from $T$ by inserting the alphabet $x$
 by the row bumping algorithm,
and $T''$ the tableaux obtained from $T$ by removing the alphabet $y$
by the reverse bumping algorithm.
If we consider only rows from the $i$-th to $i'$-th row, then
we can regard the tableaux $T''$, $T$ and $T'$
as resulting tableaux of two successive row bumping.
Hence we have the lemma by Lemma \ref{lemma:bumpingroute2}.
\end{proof}

We also show lemmas for a sliding route.
The following lemma follows from the definition of the sliding
and the reverse sliding.
\begin{lemma}
 \label{lemma:jdt1}
 Let $T$ be an integer-valued column-strict tableau.
The following are equivalent:
\begin{enumerate}
 \item If we put a null box at $(i,j)$ of $T$
       and slide it by the jeu de taquin,
       then the null box moves to the box $(i+1,j)$.
 \item If we put a null box at $(i+1,j+1)$ of $T$
       and slide it by the reverse jeu de taquin,
       then the null box moves to the box $(i,j+1)$.
\end{enumerate}
\end{lemma}
This lemma implies the following lemmas for a sliding route.
\begin{lemma}
 \label{lemma:slidingroute1}
Let $T$ be an integer-valued column-strict tableau of shape $\lambda$.
First put a null box at $(i,j)$ in $T$,
and slide the null box to outside by the jeu de taquin.
Let $(i',j')$ be the null box after sliding.
Next put a null box at a cocorner $(i'',j'')$ of the original tableau $T$,
and slide the null box to inside by the reverse jeu de taquin
until the null box moves into the $j$-th row.
Let $(i'',j)$ be the null box after sliding.
If $j''\leq j'$, then $i''>i'$.
\end{lemma}
\begin{proof}
The box $(i',j')$ is a corner of $\lambda$
and $(i'',j'')$ is a cocorner of $\lambda$.
Hence, if $j''\leq j'$, then $i'<i''$.
Therefore, applying Lemma \ref{lemma:jdt1},
we have the lemma.
\end{proof}

\begin{lemma}
 \label{lemma:slidingroute2}
Let $T$ be an integer-valued column-strict tableau of shape $\lambda$.
First put a null box at $(i,j)$ in $T$,
and slide the null box to outside by the jeu de taquin.
Let $T'$ be a resulting tableau, and $(i_1,j_1)$ be the null box.
Next put a null box at $(i,j')$ in $T'$,
and slide the null box to outside by the jeu de taquin.
Let $T''$ be a resulting tableau, and $(i_2,j_2)$ be the null box.
If $j'<j$, then
$i_1 \leq i_2$ and $j_1 > j_2$.
\end{lemma}
\begin{proof}
We can obtain $T$
from $T'$ by putting a null box at $(i_1,j_1)$
and slide the null box by the reverse jeu de taquin.
Hence we have the lemma by Lemma \ref{lemma:slidingroute1}.
\end{proof}

Next
we show some lemma for algorithms in Section \ref{subsec:alg:tab}.
Consider Algorithm \ref{algorithm:R(P)}.
Since the row insertion of Algorithm \ref{algorithm:R(P)}
starts from the $(r_1(P)+1)$-th row,
we have the following:
\begin{lemma}
\label{lemma:R(P):shape}
For $P\in\mathcal{P}_\lambda^+$,
then $\lambda^{(1)}_1=\lambda_1$,
where $\lambda^{(1)}$ is the shape of $R(P)$,
\end{lemma}
Algorithm \ref{algorithm:R(P)}
does not change boxes containing more than one alphabet
except the box $(r_1(P),r_1'(P))$.
Hence we have the following:
\begin{lemma}
\label{lemma:R(P)}
For $P\in\mathcal{P}_\lambda^+$, $r_1(R(P))\leq r_1(P)$.
Moreover, if $r_1(P)=r_1(R(P))$, then
$r'_1(R(P))\leq r'_1(P)$.
\end{lemma}
Since all alphabets in $P$ are preserved,
we have the following:
\begin{lemma}
\label{lemma:mu}
For $P\in\mathcal{P}^+_{\lambda}$, $\mu(P)=\mu(R(P))$.
\end{lemma}
Since Algorithm \ref{algorithm:R(P)} adds only one box,
 Equation \eqref{eq:sum:mu} implies
 the following:
\begin{lemma}
\label{lemma:epsilon}
For $P\in\mathcal{P}^+_{\lambda}$, $ \varepsilon(P)=\varepsilon(R(P))+1$.
\end{lemma}

Consider Algorithm \ref{algorithm:F}.
The null box $(f_1(Q),f_1'(Q))$ always move to $(f_1(Q)+1,f_1'(Q))$ at
the first step of the jeu de taquin.
Hence we have the following:
\begin{lemma}
\label{lemma:F(Q):shape}
For $Q\in\mathcal{Q}_\lambda^+$,
 $\lambda^{(1)}_1=\lambda_1$,
where $\lambda^{(1)}$ is the shape of $F(Q)$.
\end{lemma}
Algorithm \ref{algorithm:F} changes only the boxes in
\begin{align*}
\Set{(i,j)\in\lambda| \text{$i\geq f_1(Q)$ and $j\geq f_1'(Q)$}}.
\end{align*}
Moreover the jeu de taquin preserves column-strictness.
Hence we have the following:
\begin{lemma}
\label{lemma:F(Q)}
For $Q\in\mathcal{Q}_\lambda^+$, we have $f_1(F(Q))\leq f_1(Q)$. Moreover,
if $f_1(F(Q)) =f_1(Q)$,
then  $f'_1(F(Q)) < f'_1(Q)$,
\end{lemma}

Next
we show some lemma for algorithms in Section \ref{subsec:alg:pair}.
Consider Algorithm \ref{algorithm:checkvarphi}.
\begin{lemma}
\label{lemma:varphi:f}
For $(P,Q)\in\check{\mathfrak{R}}_\lambda$,
we have
\begin{align*}
f_1(\check{\varphi}(P,Q)) &=r_3(P,Q)=r_1(P),\\
f_1'(\check{\varphi}(P,Q)) &=r_3'(P,Q).
\end{align*}
\end{lemma}
\begin{proof}
It follows from the definition of Algorithm \ref{algorithm:checkvarphi}
that $r_3(P,Q)=r_1(P)$.
Let $\hat Q=\check{\varphi}(P,Q)$.
Since the jeu de taquin preserves column-strictness,
we have
\begin{align*}
&\Set{(i,j)|\text{$\hat Q_{i,j}=\hat Q_{i+1,j}$ for some $j$}} \\
=&
\Set{(i,j)|\text{$Q_{i,j}=Q_{i+1,j}$ for some $j$}}
\cup \Set{(r_3(P,Q),r_3'(P,Q))}.
\end{align*}
Since $(P,Q)\in\check{\mathfrak{R}}_\lambda$, we have
\begin{enumerate}
\item $r_1(P)>f_1(Q)$; or
\item $r_1(P)=f_1(Q)$ and $r_2(P)\leq f_2(Q)$.
\end{enumerate}
First assume that $r_1(P)>f_1(Q)$.
Since $r_3(P,Q)=r_1(P)$, it follows that $r_3(P,Q)>f_1(Q)$,
which implies
$f_1(\hat Q)=r_3(P,Q)$ and $f_1'(\hat Q)=r_3'(P,Q)$.
Next assume that $r_1(P)=f_1(Q)$ and that $r_2(P)\leq f_2(Q)$.
In this case, we have $f_1(Q)=r_1(P)=r_3(P,Q)$,
which implies $f_1(\hat Q)=r_3(P,Q)$.
Moreover, since $r_2(P)\leq f_2(Q)$,
it follows from Lemma \ref{lemma:slidingroute1}
that  $r_3'(P,Q)>f_1'(Q)$.
Hence we have $f_1'(\hat Q)=r_3'(P,Q)$.
\end{proof}
Since the jeu de taquin is the inverse of the reverse jeu de taquin,
we have the following:
\begin{lemma}
\label{lemma:varphi:f2r2}
For $(P,Q)\in \check\mathfrak{R}_\lambda$, we have
\begin{align*}
f_2(\check{\varphi}(P,Q)) &=r_2(P)\\
f_2'(\check{\varphi}(P,Q)) &=r'_2(P).
\end{align*}
\end{lemma}

We obtain $\check\varphi(P,Q)$ from $Q$ by the jeu de taquin.
Since the jeu de taquin preserves column-strictness,
we have the following:
\begin{lemma}
\label{lemma:nu}
For $(P,Q)\in {\check{\mathfrak{R}}}_{\lambda}$,
$ \nu(Q)=\nu(\check\varphi(P,Q))$.
\end{lemma}
Since Algorithm \ref{algorithm:checkvarphi} adds only one new box,
Equation \eqref{eq:nu} implies the following:
\begin{lemma}
\label{lemma:delta}
For $(P,Q)\in {\check{\mathfrak{R}}}_{\lambda}$,
$ \delta(Q)=\delta(\check\varphi(P,Q))-1$.
\end{lemma}

Consider Algorithm \ref{algorithm:hatvarphi}.
\begin{lemma}
\label{lemma:hatphi:r}
For $(P,Q)\in\hat\mathfrak{R}_\lambda$, we have
\begin{align*}
r_1(\hat{\varphi}(P,Q)) &=f_3(P,Q)=f_1(Q),\\
r_1'(\hat{\varphi}(P,Q)) &=f_3'(P,Q).
\end{align*}
\end{lemma}
\begin{proof}
It follows from the definition of Algorithm \ref{algorithm:hatvarphi}
that $f_3(P,Q)=f_1(Q)$.
Let $\check P =\hat{\varphi}(P,Q)$.
It follows that
\begin{align*}
  \Set{(i,j)| \text{ $\numof{\check P_{i,j}}>1$ for some $j$}}\\
=  \Set{(i,j)| \text{ $\numof{ P_{i,j}}>1$ for some $j$}}
\cup \Set{(f_3(P,Q),f_3'(P,Q))}.
\end{align*}
Since $(P,Q)\in\hat\mathfrak{R}_\lambda$, we have
\begin{enumerate}
\item $r_1(P)<f_1(Q)$; or
\item $r_1(P)=f_1(Q)$ and $r_2(P)> f_2(Q)$.
\end{enumerate}
If $r_1(P)<f_1(Q)$, then
it is easy to show that
$r_1(\check P)=f_3(P,Q)$ and that $r_1'(\check P)=f_3'(P,Q)$.
Assume that $r_1(P)=f_1(Q)$ and that $r_2(P)> f_2(Q)$.
In this case, we have $r_1(P)=f_1(Q)=f_3(P,Q)$.
Hence $r_1(\hat P)=f_3(P,Q)$.
Moreover, since $r_2(P)> f_2(Q)$,
it follows from Lemma \ref{lemma:bumpingroute1}
that
$\max(P_{f_1(Q),f_1'(Q)})<y$,
which implies $f_3'(P,Q) \geq f_1'(Q)$.
Hence we have
$r_1'(\check P)=f_3'(P,Q)$.
\end{proof}
Since the row insertion is the inverse of the reverse row insertion,
we have the following:
\begin{lemma}
\label{lemma:hatvarphi:f2r2}
For $(P,Q)\in\hat\mathfrak{R}_\lambda$, we have
\begin{align*}
r_2(\hat{\varphi}(P,Q)) &=f_2(Q)\\
r_2'(\hat{\varphi}(P,Q)) &=f'_2(Q).
\end{align*}
 \end{lemma}

Finally we consider $\check{\Phi}$ and  $\hat{\Phi}$.
\begin{lemma}
\label{lemma:fromchecktohat}
Let $(P,Q)\in {\check{\mathfrak{R}}}_\lambda$,
If $(\hat P,\hat Q)=\check{\Phi}(P,Q)$,
then  $(\hat P,\hat Q)$ is in $\hat{\mathfrak{R}}_{\hat\lambda}$,
where $\hat\lambda$ is the shape of $\hat P$.
\end{lemma}
\begin{proof}
Since $\hat Q=\check{\varphi}(P,Q)$,
we have $f_1(\hat Q)=r_1(P)$
by Lemma \ref{lemma:varphi:f}.
Since $\hat P=R(P)$,
we have
$r_1(\hat P) \leq r_1(P)$
 by Lemma \ref{lemma:R(P)}.
Hence $r_1(\hat P) \leq f_1(\hat Q)$.
If $r_1(\hat P) < f_1(\hat Q)$,
then $(\hat P,\hat Q)\in \hat{\mathfrak{R}}_{\hat\lambda}$.
Assume that $r_1(\hat P) = f_1(\hat Q)$.
Let
\begin{align*}
 x&=\max (P_{r_1(P),r'_1(P)})\\
 \hat x&=\max(\hat P_{r_1(\hat P),r'_1(\hat P)}).
\end{align*}
Since $f_1(\hat Q)=r_1(P)$,
we have $r_1(\hat P) = r_1(P)$.
By Lemma \ref{lemma:R(P)},
we have $r'_1(\hat P) \leq r'_1(P)$.
Hence $\hat x \leq x$.
Consider only the rows strictly below the $r_1(P)$-th row.
Since
we obtain $\hat P$ from $P$ by the row insertion with $x$,
it follows from Lemma \ref{lemma:bumpingroute2}
that $r_2(P)<r_2(\hat P)$.
By Lemma \ref{lemma:varphi:f2r2},
we have $r_2(P)=f_2(\hat Q)$.
Hence $f_2(\hat Q)<r_2(\hat P)$ and $(\hat P,\hat Q)\in \hat{\mathfrak{R}}_{\hat\lambda}$.
\end{proof}

\begin{lemma}
\label{lemma:fromhattocheck}
Let $(P,Q)\in {\hat{\mathfrak{R}}}_\lambda$,
If $(\check P,\check Q)=\hat{\Phi}(P,Q)$,
then  $(\check P,\check Q)$ is in $\check{\mathfrak{R}}_{\check\lambda}$,
where $\check\lambda$ is the shape of $\check P$.
\end{lemma}
\begin{proof}
Since $\check P=\hat{\varphi}(P,Q)$,
we have $f_1(Q)=r_1(\check P)$
by Lemma \ref{lemma:hatphi:r}.
Since $\check Q=F(Q)$,
we have
$f_1(\check Q) \leq f_1(Q)$
 by Lemma \ref{lemma:F(Q)}.
Hence $f_1(\check Q) \leq r_1(\check P)$.
If $f_1(\check Q) < r_1(\check P)$,
then $(\check P,\check Q)\in \check{\mathfrak{R}}_{\check\lambda}$.
Assume that $f_1(\check Q) = r_1(\check P)$.
Since $f_1(Q)=r_1(\check P)$,
we have $f_1(\check P) = f_1(P)$.
 By Lemma \ref{lemma:F(Q)}.
we have $f'_1(\check Q) < f'_1(Q)$.
Hence, by Lemma \ref{lemma:slidingroute2},
we have $f_2(Q)\leq f_2(\check Q)$.
By Lemma \ref{lemma:hatvarphi:f2r2},
we have $f_2(Q)=r_2(\check P)$.
Hence $r_2(\check P)\leq f_2(\check Q)$ and $(\check P,\check Q)\in \check{\mathfrak{R}}_{\check\lambda}$.
\end{proof}

\begin{lemma}
\label{lemma:bij1}
Let $(P,Q)\in {\check{\mathfrak{R}}}_{\lambda}$ and $(\hat P,\hat Q)\in {\hat{\mathfrak{R}}}_{\hat\lambda}$.
If $(\hat P,\hat Q)=\check{\Phi}(P,Q)$,
then  $\hat\Phi(\hat P,\hat Q)=(P,Q)$.
\end{lemma}
\begin{proof}
 Let $(\hat P,\hat Q)=\check{\Phi}(P,Q)$.
Since $\hat Q = \check\varphi(P,Q)$,
we have
\begin{align*}
 f_1(\hat Q)&=r_3(P,Q)\\
 f'_1(\hat Q)&=r'_3(P,Q)
\end{align*}
by Lemma \ref{lemma:varphi:f}.
Hence the box $(f_1(\hat Q),f'_1(\hat Q))$
is the box where
the reverse jeu de taquin stops in Algorithm \ref{algorithm:checkvarphi}.
Since the jeu de taquin is
 the inverse of  the reverse jeu de taquin,
we have $F(\hat Q)=Q$.
We also have
\begin{align*}
 f_2(\hat Q)&=r_2(P)\\
 f'_2(\hat Q)&=r'_2(P)
\end{align*}
by Lemma \ref{lemma:varphi:f2r2}.
Hence the box $(f_2(\hat Q), f'_2(\hat Q))$
is the new box added by  Algorithm \ref{algorithm:R(P)}.
Since the reverse row insertion is
 the inverse of  the row insertion,
we have $\hat\varphi(\hat P,\hat Q)=P$.
\end{proof}

\begin{lemma}
\label{lemma:bij2}
Let $(\check P,\check Q)\in {\check{\mathfrak{R}}}_{\check\lambda}$ and $(P,Q)\in {\hat{\mathfrak{R}}}_{\lambda}$.
If $(\check P,\check Q)=\hat{\Phi}(P,Q)$,
then  $\check\Phi(\check P,\check Q)=(P,Q)$.
\end{lemma}
\begin{proof}
 Let $(\check P,\check Q)=\hat{\Phi}(P,Q)$.
Since $\check P = \hat\varphi(P,Q)$,
we have
\begin{align*}
 r_1(\check P)&=f_3(P,Q)\\
 r'_1(\check P)&=f'_3(P,Q)
\end{align*}
by Lemma \ref{lemma:hatphi:r}.
Hence the box
$ (r_1(\check Q), r'_1(\check Q))$
is the box where
Algorithm \ref{algorithm:hatvarphi} appends an alphabet.
Since the row insertion is
 the inverse of  the row insertion,
we have $R(\check P)=P$.
We also have
\begin{align*}
 r_2(\check P)&=f_2(Q)\\
 r'_2(\check P)&=f'_2(Q)
\end{align*}
by Lemma \ref{lemma:hatvarphi:f2r2}.
Hence the box $(r_2(\check P), r'_2(\check P))$
is the box where
the jeu de taquin stops in Algorithm \ref{algorithm:F}.
Since the reverse jeu de taquin is
 the inverse of  jeu de taquin,
we have $\check\varphi(\check P,\check Q)=Q$.
\end{proof}

Since we have
Lemmas \ref{lemma:R(P):shape}
and \ref{lemma:fromchecktohat},
we can define
the map $\check\Phi_w$ from
${\check{\mathfrak{R}}}(w)$ to ${\hat{\mathfrak{R}}}(w)$
by $\check\Phi_w(P,Q)=\check{\Phi}(P,Q)$.
Since we also have
Lemmas \ref{lemma:F(Q):shape}
and \ref{lemma:fromhattocheck},
we can define
the map $\hat\Phi_w$ from
${\hat{\mathfrak{R}}}(w)$ to ${\check{\mathfrak{R}}}(w)$
by $\hat\Phi_w(P,Q)=\hat{\Phi}(P,Q)$.
It follows from  Lemmas  \ref{lemma:bij1} and \ref{lemma:bij2}
that $\check\Phi_w$ is the inverse of $\hat\Phi_w$.
Hence $\check\Phi_w$ and $\hat\Phi_w$ are bijections.
Since we also have Lemmas
\ref{lemma:mu}, \ref{lemma:epsilon},
\ref{lemma:nu} and \ref{lemma:delta},
we obtain Theorem \ref{thm:main:bij}.

\section{The case of  Young diagrams with one column}
\label{sec:finalrmk}
Here we consider only Young diagrams with one column.
In this case,
we describe our bijection explicitly.
Let $(P,Q)\in \mathfrak{R}^+_{(1^l)}$.
In this case,
$(P,Q)\in{\check{\mathfrak{R}}}_{(1^l)}$ if and only if $r_1(P)>f_1(Q)$.
For $(P,Q)$ with $r_1(P)>f_1(Q)$,
it follows that $(P',Q')=\check{\Phi}(P,Q)$,
where $(P',Q')\in \mathfrak{R}^+_{(1^{l+1})}$
is a pair of tableaux obtained by expand their $r_1(P)$-th rows,
i.e., a pair of tableaux defined by
\begin{align*}
 P'_{i,1}&=\begin{cases}
P_{i,1} &(\text{if $1\leq i<r_1(P)$})\\
P_{r_1(P),1} \setminus  \Set{\max(P_{r_1(P),1})}&(\text{if $i=r_1(P)$})\\
\Set{\max(P_{r_1(P),1})} &(\text{if $i=r_1(P)+1$})\\
P_{i-1,1} &(\text{if $r_1(P)+1<i\leq l+1$})
	  \end{cases}
\\
 Q'_{i,1}&=\begin{cases}
Q_{i,1} &(\text{if $1\leq i\leq r_1(P)$})\\
Q_{i-1,1} &(\text{if $r_1(P)+1\leq i\leq l+1$}).
	  \end{cases}
\end{align*}
For $(P,Q)$ with $r_1(P) \leq f_1(Q)$,
it follows that $(P',Q')=\hat{\Phi}(P,Q)$,
where
$(P',Q')\in \mathfrak{R}^+_{(1^{l+1})}$
is the pair of tableaux obtained by folding their $f_1(Q)$-th rows,
i.e.,
the pair  of tableaux defined by
\begin{align*}
 P'_{i,1}&=\begin{cases}
P_{i,1} &(\text{if $1\leq i<f_1(Q)$})\\
P_{f_1(Q),1} \cup P_{f_1(Q)+1,1}&(\text{if $i=f_1(Q)$})\\
P_{i+1,1} &(\text{if $r_1(Q)+1<i\leq l-1$})
	  \end{cases}
\\
 Q'_{i,1}&=\begin{cases}
Q_{i,1} &(\text{if $1\leq i<f_1(Q)$})\\
Q_{f_1(Q),1}=Q_{f_1(Q)+1,1} &(\text{if $i=f_1(Q)$})\\
Q_{i+1,1} &(\text{if $f_1(Q)< i\leq l-1$}).
	  \end{cases}
\end{align*}

\bibliographystyle{amsplain-url}

redpajama_set_name: RedPajamaArXiv


The main opposition UNP today warned that the collapse of the economy of the country had started with the passing of the 'Revival of Underperforming Enterprises and Unutilised Assets Bill' in Parliament.
UNP General Secretary Tissa Attanayake told a news conference today that the fall of the economy is imminent just like it had happened in Greece and Italy. He said draconian laws had resulted in the economic recession in those countries as well.
Mr. Attanayake said the UNP would begin protests against the Underperforming Enterprises and Unutilised Assets Bill when the party leader Ranil Wickremesinghe returns from abroad on November 17.
Referring to the upcoming budget, the UNP General Secretary said people had lost faith in it because it had become a mere piece of paper as none of the policies put out in it over the years have been implemented. "This has been visible after 2005," he said.
Mr. Attanayake said the budget this year should be one which gives concessions to the people and a considerable salary hike for the public servants and for the private sector. However he said it is likely that it would be the same at it had been during the past years.
In the meantime he said the people will have to suffer the consequence of the wastage of funds that had taken place as a result of Sri Lanka trying to win the bid to host the next Commonwealth Games. Mr. Attanayake said Rs.400 million has been wasted for this. He charged that funds have been wasted because the government campaigned for the bid the way it is campaigning for an election in the country. "Banquets were held for the delegations of the voting nations at the final ballot held in the West Indies just like they are held for the voters here during an election," he charged.
According to his lordship, the Crash of the economy has begun.
What he fails to recognise is that Crash of the UNP began many years ago and is now complete.
Tissa, get rid of that venerated limpy leader and all will be well.
down behind its hind legs.
TA's analysis of the economic failures of Greece and Italy is well off the mark. These two countries, and also Spain, followed unwise budgetary policies which led to their economic downfall. Income Tax was low while spending was high. They overspent on social benefits, health, education, housing etc but income via taxes and exports etc were low. Soon the treasury was emptied. They did not observe the financial policies of member nations of EU. They have now gone with begging bowls to EU to help them out. In order to do so these countries will have to impose austerity measures such as high taxes and low spending. These measures are bound to be unpopular with people who were used to an easy life, at expense of the government. Naturally, this led to popular uprisings and change in governments. Any government that is chosen by the people can do little but to follow austerity measures imposed by the EU, in order to be granted billions of Euro loans to salvage their economies. These IMF loans are for revival of their economies with correct economic policies.
In SL, "Revival of underperforming enterprises and unutilised assets" Bill is to ACQUIRE BACK by the government those profitable STATE ENTERPRISES which were sold by the UNP, when they were in power, to their rich cronies, who have stripped the companies of their assets. They have become losing ventures. It is fair that the government reclaim them back FOR THE PEOPLE, in order to make them once again profitable ventures for the benefit of the people and the revival of our economy, after the defeat of the LTTE.
UNP's TA is making a false analogy by comparing with economic downfall of Greece and Italy, with Spain and Portugal also in the ofing.
SL is on the road to economic victory with the correct economic and budgetary policies, to be passed in Parliament in due course. Our people will also have to endure some austerity measures in order to achieve economic success. This is happening around the world, even in the rich West. We cannot expect our government to give handouts to all and sundry. Where is the money going to come from? We need to export more and import less. We need to be self sufficient in food. We need to expand tourism. We need a fair fiscal policy. We need a balanced budget.
so, are you happy now, fools…….?!


I'm going to blog the alphabet this year which, for me, is a pretty hefty investment since that commits me to 26 blogs in 2017. But I've been very good on Facebook, not expressing opinion about much of anything, even when I've had plenty of words and opinions. I don't engage in debate or argument, though. That's just me. Everybody's entitled to what they want to believe, and I sort of subscribe to the 'live and let live' philosophy. But as good as I've been on Facebook, I do have some thoughts, and while they might not fit as well with the alphabet, I'm going to give it a try and keep my fingers crossed that I actually have 26 blogs in me. (Admittedly, the X week scares me.) So, let's get started with the letter 'A' and the word: ADJUSTMENT.
Yep, that's what we've got to do. We've got to adjust, and I'm not just talking politically, even though some political thinking and attitude could stand a little adjustment lately. But to me, the word adjustment has a much bigger context. It's about what we have to do in our lives every day in order to get ourselves through and on to the next day. It would be nice if things slowed down, or even stopped sometimes. It would also be nice to think that when you wake up in the world today, you'll wake up in the very same world tomorrow. But that doesn't happen, does it? I woke up in the world a few of days ago and Obama was my president. Then I woke up the next morning and Trump was. That takes some adjustment. But so does the weather.
Yesterday morning when I woke up it was gray and rainy outside and this morning when I woke up it was. . . well, it was still gray and cloudy so no adjustment needed there. But by now, I think you get my drift. Things are never the same, and it's up to us to adjust, or fall behind.
Years ago, I thought the whole cell phone thing was stupid. Why do people have to stay in touch every minute of every day? I didn't have one, didn't want one. But. . . I finally gave in. Don't remember why. Just remember that I did. I didn't use it much, though. Not at first, anyway. Why? Because mentally, I wasn't able to make that adjustment yet. Or, maybe I just didn't want to. Whatever the case, I carried it around for a year, turned off most of the time, and wondered why I was paying such a ridiculously high price for something I wasn't using. So, I turned it on and opened up a different kind of world for myself. And while I don't consider myself one of those obsessives who buries her nose in her phone every minute of every day, I do find that it comes in handy. Why? Because I allowed myself to adjust and, due to that, my cell phone was the means by which Joel and I found help when our car broke down in the middle of nowhere. It was also what I used when it broke down again, but this time in the middle of the night at O'Hare International in Chicago. I was able to find a towing service and rent a car to get us home in a matter of minutes, all because sometime a few years ago, I bit the bullet and adjusted.
A couple of days ago, that adjustment I'd made way back when let me post pictures of being stuck in the mud to Facebook during the actual event. You know, we're stuck in there mud therefore I shall Facebook it.
Technology aside, we're called on to make so many other adjustments. I adjusted to life without a mother, then without a father. I adjusted to life after I went from walking to crutches to a wheelchair. I'm adjusting to a spotty amnesia that resulted from an illness I had last year - most things I remember, some things I don't. Most of all, I adjust to all the things that surround my life, good or bad. I have to, or I'll find myself encased in a life so small that it will eventually squeeze me in and trap me there. And I don't want to be trapped because there are so many great, new experiences out there for me to have, if I allow them in. If I adjust.
Adjustment is defined as and adaptation to a particular condition, position, or purpose. Roget lists some of its antonyms as: disarrange, disorganize, disturb, disorder, confuse, worsen. Aren't those ugly words? They're sure not words I want in my life. When you look at the entire list of antonyms, it gives you acid indigestion or angst, or both. But the synonyms are so nice: accommodate, fine-tune, compose, harmonize, redress, rectify. All things I'd much rather do than disarrange or worsen. Sure, adjusting isn't always easy. Sometimes it goes against our natural grain or our preconceived beliefs and notions. But it's also a means to growth and, in this day and age, we do have to grow in order to survive. That's just the way the world is, now. Actually, it's the way the world has always been. If not, we'd still be starting our fires with a flint (if we'd even adjusted ourselves to the point that we had fire) and hunting our dinner with a club or a spear.
Sure, sometimes the adjustments are difficult. My husband is still stuck in the old flip phone world, refusing to use a smartphone of some sort. Little by little he's going to have to adjust that because the work he does requires it. He does own that smartphone now, hasn't used it yet, doesn't know how. But he's made a very small adjustment by agreeing to actually possess the beast and, at some point, he'll learn how to flip on the screen and make a phone call. Or use the internet, take a photo or download a work order on it. Small steps, granted. But steps of adjustment. And that's all adjustment really needs - small steps. One at a time. Like it or not, that's life. We don't get to make all the choices, and some of the choices that are handed us we simply have to adjust to. So, why make it so tough, the way we often do? Why not look for the positive outcome in our adjustments, because they're always there. And that, perhaps, is the biggest adjustment of all - always looking for the positive outcome. Sure, it may be hard to find it in a lot of situations. But it gets back to all those synonyms and antonyms; wouldn't you rather be part of adjustment's synonym than its antonym?
So, here's a practice adjustment to try. Tomorrow morning when you wake up scowling because you hate getting up in the morning, take ten seconds and smile. No, not a grudging smile. Make it a genuine one because you're facing a new day with new possibilities. Be glad for those ten seconds. Make that smile your daily adjustment. Give yourself ten or maybe even (God forbid) twenty seconds of smiling, with a couple of happy thoughts thrown in for good measure, every morning then you can go back to your scowling if you want. Sure, it may seem silly, but it may also be the baby step you need toward a much bigger adjustment you'll have to make at some other time. And I promise you, your life is going to be filled with adjustments no matter what you do. How you handle them, though, is what will make the difference between being miserable and finding a place where you can be happy or, at least, happier than you would be if you refuse to budge. So, get over it, get on with it. Get around it, get through it. Change it, fix it. Budge, adjust. Make it accommodate your life and the lives of the people around you. Embrace it. Hate it, if you have to. But find a way to make it work. In the end, if you don't, you're the one who gets hurt, or left behind. That's not where you really want to find yourself, is it? Don't you deserve better than that? If you don't think you do, then adjust your thinking, because you do!
Now my blatant promo for the day: The Nurse and the Single Dad (Harlequin Medicals) will finally be out on February 1. You can find it online at all the normal book retailers, or purchase it directly from Harlequin Books (Harlequin.com). It was the first book I wrote after a lengthy, complicated illness that nearly robbed me of my ability to write (and left me with that spotty amnesia I mentioned in this blog), so this one holds a special place in my heart. I hope you'll enjoy it.
Until next time. . . wishing you health & happiness.
I love your thoughts, can't wait to read (B).
I'll send you a copy this week.