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A higher proportion of women work in the former eastern states than in the western states, a new survey showed on Wednesday, and experts say it has everything to do with Germany's divided history.
A quarter century after the fall of the Berlin Wall, the Bertelsmann Foundation updated its information about the number of women in the work force by state. In the former East, 57.9 percent of women held permanent jobs in 2012, whereas only 50.9 percent of women in western states were in permanent employment.
The previous data, from 2007, showed that 50.9 percent of women in the east held permanent employment compared with 45.8 percent of women in the western states.
"In West Germany, women leave the job market when they have a child. There's no question of that for women from the former East," said Barbara Riedmüller, a professor at the Free University (FU) in Berlin whose research focuses on society and employment.
Riedmüller said the cultural difference between East and West would ensure a gap between the numbers of working women for years to come.
"For me, it was never a decision if I was going to work or not. At the most, it was a matter if I would work full- or part-time," said a 32-year-old Berlin journalist. The mother of two was born in the German Democratic Republic (GDR).
Nationwide, the employment rate for women is 51.8 percent, still well behind that of men in Germany at 59.2 percent.
"In the former East, roles of men and women were long considered equal – meaning both sexes worked. In the west, the dual-income-earner model for couples is less common. There, usually the man is the bread-winner," said Bertelsmann Foundation researcher Kirsten Witte.
Riedmüller said the west lags behind because of its historical adherence to the idea that women are either working or mothers and traditionally not both.
"Look at France. There, no woman comes to the conclusion to not work. And it was like that also in GDR. It had the highest divorce rate in the world, many couples lived together but didn't get married and so a good education, a permanent job and financial security for women in the states of East Germany has always been very important," Riedmüller said.
On the map, darker blue areas have more women in employment.
Women in Saxony were most likely to have permanent employment (58.5 percent), while only 47.5 percent of women in North Rhine-Westphalia and 47.2 percent of women in Saarland hold permanent employment.
Despite the fact that more women are in political spheres in the western states, this hasn't yet translated to more women in the workplace.
"Marrying family and career must be pursued by women in the west and men there must be more integrated (in the home)," Riedmüller said.
For the study, researchers looked at 295 rural districts and 106 urban districts to reach their conclusions. Numbers from the federal employment agency from last year confirmed what researchers found, though differ slightly since they are derived from the 2011 census. | {
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Actinic keratosis (AK) is a common skin condition caused by severe sunburns and sun exposure over many years. It can also be caused by elongated exposure to artificial sources of UV light, such as tanning machines. That's why we've created a list of actinic keratosis natural treatment options.
If left untreated, this skin condition can lead to skin cancer. People who are likely to suffer from this skin condition are those with fair skins, and the lesions can appear anywhere on the face and body, including bald scalps. The lesions are usually red, scaly patches and they look like sandpaper.
You can use organic and raw apple cider both internally and externally to control AK. The acidic nature of this natural remedy helps to speed up the healing process.
For external application, use a cotton ball to apply it to the affected area 2 to 4 times every day. If this causes irritation, add water to the apple cider vinegar before you use it.
For internal use, add 1 to 3 teaspoons of apple cider vinegar to a glass of water and drink it at least twice a day.
The use of organic coconut oil in the treatment of actinic keratosis is somewhat a diligent task, but it yields results in the long run. You need to apply the oil regularly on the spots. On the onset, the spots will turn red with a subtle burning feeling.
You should expect to start noticing a change within a month of using the oil. Try to apply organic virgin oil on the lesions once a day until they disappear.
Made from the leaves of a plant identified as Camellia sinensis, green tea is a celebrated type of tea that has been approved for various benefits. You can either apply it to the spot or drink it normally. Green tea has antimicrobial and astringent properties that are important in treating a variety of skin disorders.
Soak a bag of green tea in warm water and place it directly on the affected area. You can also chew the Camellia sinensis leaves or get green tea capsules, which you can buy at a local health food store.
If you drink green tea, don't take more than five cups a day as it can lead to irritability, diarrhea, and headaches.
Milk thistle is available in liquid extract and powder. You might want to view it as a preventive treatment.
Also known as melaleuca oil, tea tree oil can be used as a preventive and curative treatment of actinic keratosis. It has antimicrobial properties, and it's not recommended for internal use. Also, do not use it per se. Try diluting the oil with virgin coconut oil before applying to the affected area.
Like other treatments, ensure to use it daily until the spots start fading away.
This oil is made by extracting the seeds of the castor plant. You need to apply the oil on the affected area to relieve the crusting, inflammation, and burning sensation caused by AK. It's also effective in reducing the spots caused by extended sun exposure.
However, keep in mind that you might need to use it for a longer period to see some progress in treatment.
You can also use eggplant to treat actinic keratosis. The best way to do this is to mix the eggplant with apple cider vinegar.
Simply mince eggplant in a jar and add apple cider vinegar to it. Place the jar in the refrigerator for about three days until the vinegar starts darkening. At this point, it's ready for use. Apply the mixture to the affected spots using a cotton swab at least three times a day.
You will start noticing results after a few weeks of using the combination.
Taking preventive measures is the best approach to avoiding actinic keratosis. Limit extended exposure to the sun by using protective products or staying in shaded areas. Also, avoid tanning beds and machines.
However, should you get the condition, one of these actinic keratosis natural treatment options can be ideal for healing the lesions. | {
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Q: liquibase agnostic insert with sequence I want to create a new insert changeset in liquibase using a sequence to populate the id field. Is there any way to make it database agnostic for Oracle an Postgres?
This is the oracle way:
<changeSet author="XX" id="XX">
<insert tableName="NODO_MENU">
<column name="ID" defaultValueSequenceNext="SEQ_NODO_MENU" />
<column name="CODIGO" value="ABC" />
<column name="ORDEN" value="0" />
</insert>
<rollback>
<delete tableName="NODO_MENU">
<where>CODIGO = 'ABC'</where>
</delete>
</rollback>
</changeSet>
It doesn´t work in postgres:
[Failed SQL: INSERT INTO public.NODO_MENU (ID, CODIGO, ORDEN) VALUES (SEQ_NODO_MENU.NEXTVAL, 'ABC', '0')]
To make it work in postgres I changed the column tag to:
<column name="ID" valueComputed="nextval('SEQ_NODO_MENU')" />
How can i write this changeset to make it work in Oracle an Postgress?
A: You can write two separate changesets. One for Postgres, one for Oracle.
Just add corresponding dbms preConditions to them.
As such:
<changeSet author="XX" id="XX_oracle">
<preConditions onFail="MARK_RAN">
<dbms type="oracle"/>
</preConditions>
<!-- your logic for Oracle-->
</changeSet>
<changeSet author="XX" id="XX_postgres">
<preConditions onFail="MARK_RAN">
<dbms type="postgresql"/>
</preConditions>
<!-- your logic for Postrgres-->
</changeSet>
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A number of IT security experts expect that attackers will increasingly focus on compromising embedded devices and consumer devices that make up the Internet of things.
Cyber-attackers and security researchers focused on finding and attacking vulnerable devices on the broader Internet of things in 2013, a trend that will only accelerate in the coming year, according to security experts.
The rapid adoption of network-connected devices by consumers and businesses will make the so-called Internet of things more attractive to vulnerability finders and cyber-criminals bent on mischief.
From TVs to thermostats and from medical devices to home security, a range of devices are being connected to the Internet and exposed to risks for which they might not be ready, vulnerability management firm Rapid7 said in an email statement to eWEEK.
"This is only set to continue—we're already seeing network-enabled toasters, kettles, fridges and much more emerging," the company stated. "Unfortunately, researchers have found time and again that security issues abound on embedded devices, and they are typically very poorly patched."
Attacks against embedded devices have been rare so far, but security researchers have noted that recently the pace of attacks has accelerated. In November, for example, Symantec posted a brief analysis of a worm, dubbed "Linux.Darlloz," that targeted a variety of Linux distributions with evidence of variants created for chipsets that are normally found in home routers, set-top boxes and security cameras.
A major problem is that security is usually an afterthought during the creation of embedded devices. Companies are more concerned with getting the product out the door and not whether the design of the product can be exploited to compromise the user's data, according to Rapid7.
In most cases, engineering teams do not collaborate well enough with other teams in the same company nor with users among their customers, Phil Packman, general manager for security enablement at telecom giant BT, said in a blog post. That lack of communications leads to bad designs and missed opportunities to secure their products, he said.
"It is often hard for the engineer to 'connect' in the course of his day job, and an external attack can seem quite unlikely," Packman said. "On the other hand, clients who rely extensively on automated control systems with remote monitoring can easily see how this risk is very real for them, carrying with it consequences that don't bear thinking about."
Considering how pervasive Internet-connected devices are in our lives, one company claims that 2014 will see the first murder carried out using such a device that was compromised by a cyber-attack.
In its predictions for 2014, Internet Identity, a brand-security company, posited that companies and consumers will see the dark side of the Internet of things by 2015, with hackers learning how to cause chaos in people's home in the next two years. | {
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On this edition of The Arts Section, host Gary Zidek talks to the director of a new documentary that spotlights a recording studio in Madison Wisconsin that became a hub for alternative rock and pop culture. THE SMART STUDIOS STORY is set to make its Chicago premiere on April 13th. Gary also previews the Chicago International Movies and Music Festival, which is entering its 8th year. Plus, a conversation with the author of a new biography on Tom Petty. And the Dueling Critics will be reviewing a world premiere play titled MOSQUE ALERT.
Gary talked to Wendy Schneider, the director of the new documentary THE SMART STUDIOS STORY
Smart Studios co-founders Stever Marker and Butch Vig
Butch Vig
Nirvana at Smart Studios
Find out more about The Smart Studios Story here
CIMMFest executive director Dave Moore (L) and festival co-founder Josh Chicoine (R)
For more information on this year's CIMMFest click here
And the Dueling Critics review Silk Road Rising's world premiere play, MOSQUE ALERT
TUNE INTO THE ARTS SECTION EVERY SUNDAY FROM 8:00-9:00 AM ON 90.9FM and ONLINE AT WDCB.org | {
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What If……Performance Enhancing Drugs Were Legal?
Drug testing was formally introduced to the modern Olympic Games in 1968 in Mexico City. Since that time, thousands of athletes have been exposed as drug cheats across numerous levels of sport. While performance enhancing drugs have been outlawed to maintain a level playing field at all times, there is also strong evidence that demonstrates the damage that performance enhancing drugs can cause both to short and long term health.
The most commonly recognised drugs that are used are steroids, however in recent years "doping" has become much more sophisticated, with "dopers" often a step ahead of those doing the testing in terms of advancing their methods to improve performance and avoid detection.
Personally, I have always wondered what the consequences would be if performance enhancing drugs were allowed to be used. Do not get me wrong, I am fully aware that this is never going to happen, but given some of the feats we have seen today from clean athletes, it is hard not to wonder what would be achievable, even if they could eke out an additional 5 – 10% of performance.
We listed a number of feats that could happen if performance enhancing drugs were allowed to be used.
Were an athlete able to develop extremes of power along with explosive speed, would it be possible for a man to run the 100 metres in under 9 seconds? Usain Bolt believes he can get as low as 9.4 seconds, and he is completely drug free.
Another record that would be threatened is the men's marathon. One day, someone will run the 26.2 mile distance in under two hours. The men's record is currently only three and a half minutes above that now anyway, and is sure to come down further in years to come. This remains as one of the great barriers of athletics.
Perhaps the best place to look for what could be achieved is a sport that has had so much bad press over the years due to a lax attitude to performance enhancing drugs. It is fair to say that cycling has done some excellent work to clean itself up, however by comparing the times from mountain climbs at this year's Tour de France to some of those from the 1980's through even to the early 2000's, the performance possibilities are clear to all to see.
For now, global sport is thriving as anti-doping controls become stricter and more difficult to get around, surely the next big challenge will come when new doping trends emerge, which could be during this Olympics.
Videojug has just produced a series of high quality videos about the Olympic Games. The series includes a number of popular sports that are featured in London 2012.
Published August 6, 2012 By News Editor
Categorized as Sports
Top 5 Biggest And Shocking Olympic Failures Of All Time
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The Nigerian President Muhammadu Buhari has sent strong word to Nigerians migrating from the country through illegal routes, despite the reported dangers.
The Peoples Democratic Party (PDP) has challenged the Buhari Presidency and the All Progressives Congress (APC) to end its silence and speak out on their evident complicity in the corrupt amassing of the N21bn, guns and thousands of Permanent Voter Cards (PVCs) allegedly found in the residence of President Muhammadu Buhari's ally, the sacked Director General of the Department of State Services (DSS), Lawal Daura.
When Will Buhari's Government Offload Sick Nigerians To UK For Medical Treatment?
FOR the umpteenth times since 2015, President Muhammadu has sought medical remedy for his deteriorating health for undisclosed ailments that is paid for with undisclosed amount of tax payers' money. This is happening at a time Nigerian health workers are on industrial action for nearly fourth week running.
2019: Who will run if Buhari decides not to?
Though President Muhammadu Buhari has not yet explicitly announced that he would go for a second term, many people close to him have said so and he has not publicly disowned them,a development seen by many as consent on the side of the president. | {
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Jon Cartu Announced: The global TV advertising trends likely to hit Australia in…
10 Feb Jon Cartu Announced: The global TV advertising trends likely to hit Australia in…
In December, media and marketing executives from around the world gathered at the Future TV Advertising Forum in London – to explore the global media trends shaping the TV and video advertising world.
With the media industry continuing to evolve, and strong headwinds facing the advertising sector – there was plenty to talk about. Here are the top areas of discussion from the event:
2020: the turning point for the streaming wars
The general consensus is that 2020 will be the tipping point for the video streaming wars, off the back of Disney Plus launching late last year. A key takeaway from the conference was that the competition for audiences will not just be focussed on the US, but across many international markets, including Australia. This implies that the big players will continue to invest heavily in succeeding locally.
At a time when many distribution houses are experimenting with new distribution channels, there is a risk of customer fatigue with so many disparate offers in market. The current dominant force, Netflix, has seen growth slow in the US, and the Disney Plus effect is expected to impact its numbers further significantly. Just by Disney's latest result, this looks clearly to be the case.
One thing that is clear: all eyes will be on Netflix latest subscription numbers in their next release as Disney extend its global rollout.
The growing need for all screen trading
Expect to see the perennial industry problem of all screen trading start to be addressed globally in 2020. There was plenty of talk at the conference around how the industry can build single source of measurement. Advertisers want to be able to evaluate video across all different platforms, but are currently faced with walled gardens of digital and television. While there is no clear answer there was some very interesting discussion, especially around the creation of a European council to agree on some core metrics.
The end goal should be a set of standards that everyone uses to measure video (across both TV and digital), but for me, the biggest hurdle is who should pay for a global omni-screen measurement platform which will take some time to work through.
The return of greater brand investment
One of the most insightful case studies of the forum could foreshadow a shift back towards brand advertising, reversing the recent trend towards direct response campaigns. Marketers from Direct Line, the UK insurance company, shared their experience of revitalising their marketing in 2019. In the past, they had a 90/10 split towards direct and price led campaigns.
This worked five years ago, when they were competing directly against other insurance brands for consumers. But the old model was no longer working, comparison websites and other assets have changed the market and the cheap "cost per" route simply wasn't building their funnel.
By diving into the data, they found that to grow they needed to bring new customers into the category, and brand was increasingly important. As a result, they shifted towards a 60/40 model supporting more brand advertising, and they are already reporting strong numbers that prove the new approach has been successful. I think many businesses and brands are in a similar position in Australia and in 2020 we'll see more investment in the brand from many companies as key metrics have dropped over the last few years. It is time to investment to drive business growth and stronger brand differentials.
The rise of sponsored integration on longer-term basis.
Another trend that was front of mind was the evolution of the traditional sponsorship and integration model. This was also linked back to increased investment in brand. There was a lot of talk around the benefits and pitfalls of direct media-to-brand relationships.
Most agreed that advertisers need to do a better job with their brand, and that they need to ask more questions of both agencies and media. Advertisers should be looking to attach their brand more strategically to the right assets, and make decisions based on long-term outcomes, not just on the next 10-12 weeks.
The need to balance long-term thinking in a world increasingly focused on short-term issues is a recurring trend as the pace of disruption accelerates. One thing was central to all conversations at the Future TV Advertising Forum is that business-as-usual is not an option, and that only by thinking and acting differently can the industry continue to deliver for both viewers and advertisers.
Mark Frain is CEO of Foxtel Media. | {
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Q: How to create Toast Notification with custom icon and custom app name? I'm trying to create a Windows Toast Notification with .Net C#. While this, I would like to change the notification icon and the app name in the attribution area (header) of the notification.
I'm able to place a notification with a custom "App name" like shown here:
I do this with this code:
var toastXml = ToastNotificationManager.GetTemplateContent(ToastTemplateType.ToastText02);
var stringElements = toastXml.GetElementsByTagName("text");
stringElements[0].AppendChild(toastXml.CreateTextNode("Title"));
stringElements[1].AppendChild(toastXml.CreateTextNode("Message"));
ToastNotification toast = new(toastXml);
ToastNotificationManager.CreateToastNotifier("App name").Show(toast);
But I don't find a solution, how to place an icon left from "App name".
The solution can't be to modify the toast on a different way, since it only changes the visual or action elements. But not the attribute area:
How can I show an icon in the attribute area?
A: To set the icon in the attribution area you can set it directly.
var toastXml = ToastNotificationManager.GetTemplateContent(ToastTemplateType.ToastText02);
var stringElements = toastXml.GetElementsByTagName("text");
stringElements[0].AppendChild(toastXml.CreateTextNode("Title"));
stringElements[1].AppendChild(toastXml.CreateTextNode("Message"));
ToastNotification toast = new(toastXml);
// here you set your icon
toast.attribution.icon = new ToastImageSource("path to your image");
ToastNotificationManager.CreateToastNotifier("App name").Show(toast);
To set the icon in the title you can change the created toast.
var toastXml = ToastNotificationManager.GetTemplateContent(ToastTemplateType.ToastText02);
var stringElements = toastXml.GetElementsByTagName("text");
stringElements[0].AppendChild(toastXml.CreateTextNode("Title"));
stringElements[1].AppendChild(toastXml.CreateTextNode("Message"));
ToastNotification toast = new(toastXml);
// Here you can change the visual element of the toast
// don't know which image will be correct. Can't test at the moment
toast.Visual.TileLarge.HintRemoveMargin = true;
toast.Visual.TileSmall.HintRemoveMargin = true;
toast.Visual.TileMedium.HintRemoveMargin = true;
toast.Visual.TileLarge.Branding = ToastBranding.Logo;
toast.Visual.TileSmall.Branding = ToastBranding.Logo;
toast.Visual.TileMedium.Branding = ToastBranding.Logo;
toast.Visual.TileLarge.AppLogoOverride = new ToastAppLogo();
toast.Visual.TileSmall.AppLogoOverride = new ToastAppLogo();
toast.Visual.TileMedium.AppLogoOverride = new ToastAppLogo();
toast.Visual.TileLarge.AppLogoOverride.Crop = new ToastAppLogoCrop(0, 0, 1, 1);
toast.Visual.TileSmall.AppLogoOverride.Crop = new ToastAppLogoCrop(0, 0, 1, 1);
toast.Visual.TileMedium.AppLogoOverride.Crop = new ToastAppLogoCrop(0, 0, 1, 1);
toast.Visual.TileLarge.AppLogoOverride.Source = new ToastImageSource("file:///C:/path/to/your/icon");
toast.Visual.TileSmall.AppLogoOverride.Source = new ToastImageSource("file:///C:/path/to/your/icon");
toast.Visual.TileMedium.AppLogoOverride.Source = new ToastImageSource("file:///C:/path/to/your/icon");
ToastNotificationManager.CreateToastNotifier("App name").Show(toast);
Additionally you can set an appLogoOverlay which is displayed in the title area
// get hte ToastGeneric toast content
XmlDocument toastXml = ToastNotificationManager.GetTemplateContent(ToastTemplateType.ToastGeneric);
// here you get the visual element of the toast
XmlNodeList toastVisualElements = toastXml.GetElementsByTagName("visual");
// to add an icon you need app logo overlay
XmlElement appLogoOverlay = toastXml.CreateElement("appLogoOverlay");
// here you set the source of the overlay
appLogoOverlay.SetAttribute("src", "path to your icon");
// here you can add style attributes
appLogoOverlay.SetAttribute("hint-crop", "circle");
toastVisualElements[0].AppendChild(appLogoOverlay);
var stringElements = toastXml.GetElementsByTagName("text");
stringElements[0].AppendChild(toastXml.CreateTextNode("Title"));
stringElements[1].AppendChild(toastXml.CreateTextNode("Message"));
ToastNotification toast = new(toastXml);
ToastNotificationManager.CreateToastNotifier("App name").Show(toast);
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At the ECC, we know that separation is a sensitive time for children and parents. For the first few weeks of school children attend the ECC in small groups for short periods of time, allowing them to develop relationships with teachers and classmates. Counseling and discussion sessions are available for parents and our teachers are extensively trained in handling this process. We want all children to feel safe and secure before their trusted adults leave the building.
Our daily routine balances structured activity, academic inquiry, physical exercise, free choice, and quieter moments. Our interdisciplinary approach takes each child's interests, developmental stage, and social, emotional, and cognitive needs into account. Language development and literacy, mathematical thinking, dramatic play, social skills, and artistic expression are central to our curricula. Weather permitting, children spend part of each day exploring and playing outdoors. Jewish studies are integrated into all of our programs, with regular Shabbat and holiday celebrations.
We believe in documenting our students' progress by taking notes and photographs, and studying their learning habits to guide future exploration. This becomes a vital tool for teachers to communicate learning to parents, and parents should expect to hear from us each week when we share the meaning behind their children's work. We are eager for feedback, and intend for these communications to begin a dialogue with families.
The ECC is open for 10 months of the year, from mid-September through mid-June. We observe all Jewish holidays according to Reform traditions. Our school closes for Jewish and federal holidays.
Our clergy and teachers will guide your family's Jewish journey: from joyous Shabbat and holiday celebrations for tots and toddlers to the pride of watching your child become bar or bat mitzvah — and beyond. This is where young identities and lifelong traditions take shape. Stephen Wise Free Synagogue is here to be your family's Jewish home. | {
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The Hidden Toll Of Conflict On Kids
A new study offers a novel way to measure how many children have really died as a result of conflict in Africa.
by Nurith Aizenman Aug 30, 2018 3 minutes
A woman from Chad washes her baby at a site for internally displaced persons. They had fled their village after an attack. Source: Marco Di Lauro
Over the last two decades, violent conflicts in Africa have likely resulted in the death of as many as 5 million young children — 3 million of them infants. That's the sobering estimate in a new study published Thursday in the journal The Lancet.
It's also a number that breaks new ground because until now, death statistics regarding war have generally been limited to counting the number of people — including combatants and civilians — who are killed by violence. For children that could mean a bomb falls on, lead author of the study and a professor of medicine at Stanford University:
History of War2 min read
Fighting The People's War
Author: Jonathan Fennell Publisher: Cambridge University Press Price: £25.00 With the centenary of Word War I now drawing to a close, it's possible to take stock of the plethora of books that have been published on the conflict in the past four year
Guernica Magazine12 min read
Roy Scranton: Some New Future Will Emerge
The author and Army veteran on climate change, war, and "the radical transformation of the basic structures of our existence." The post Roy Scranton: Some New Future Will Emerge appeared first on Guernica.
Tribune Content Agency Opinions4 min readPolitics
A Country, and a Future, Worth Serving
America's endless war quietly moves across the broken nations of the world. Every so often, U.S. soldiers die, as four Green Berets did several weeks ago in . . . Niger. And the news was more about the adequacy of presidential condolences to the fami
New Thriller 'The Chain' Has An Origin Almost As Exciting As Its Plot
Novelist Adrian McKinty had several books and prestigious awards under his belt — but no one was buying, and he'd given up writing to drive an Uber when a blog post led to some new opportunities. | {
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Q: My business has been asked by our landlord to pay a security deposit against the safe return of the commercial property we have rented. We have been charged VAT is this correct?
A: A taxable supply is defined in law as a consideration received for a supply of goods and services in the United Kingdom. Although the majority of items will be treated as supplies there are a number of payments received which may not be a consideration because there is no direct link between the payment and the supply, or there is no supply of goods or services in return for the payment.
Deposits are generally taxable but there are a number of situations where they are not a consideration for a supply.
Forfeit deposits are a payment for damages, to compensate for example work done on a cancelled supply. A common example of this is a deposit for a hotel room that is not used. Although the payment would originally be for an advance payment for the supply of the room and attract VAT, if subsequently the booking is cancelled the payment becomes compensation/damages for breach of contract and would be outside the scope of VAT.
Deposits taken as a security, for example against the safe return of goods on hire are not a consideration for a supply. The terms of the contract would typically specify that the deposit is refundable subject to the safe return of the goods. If the customer damages the goods they have broken the terms of the contract and forfeit their deposit.
Compensation payments for damage and loss are outside the scope of VAT. This is because the payments are made either as a result of a Court Order or through an agreement between the two parties involved to compensate one party for suffering some inconvenience, loss or damage.
Donations, if given freely, are not a consideration for any supply. The donation has to be unconditional; the donor must not receive anything in return for the payment or attach any conditions to the donation.
Services charges in restaurants are part of the consideration for the underlying supply of the meals if the customer is required to pay them, and so would attract VAT. This will apply even if they are passed on in full as bonuses to staff. If customers have a genuine option as to whether to pay the service charges, HMRC accept that they are not a consideration even if the amount appears on the bill.
If a tip is genuinely freely given then they are outside the scope of VAT as they are not a consideration for VAT purposes. However, if the element of free choice is removed, the payment is part of the consideration for a supply.
Insurance companies' policies normally cover the repair of goods. If goods are repaired under an insurance policy, insurers will make payments to policy holders for the costs of the repairs. These are not payments for any suppliers.
Most insurance policies have an excess amount which is not covered by the policy. Payments by policy holders to insurers for any excess amount for repairs under insurance policies are not considerations since there is no supply made by the insurer to the policy holder.
All in all this is a very involved area and if you are unsure of the liability of a transaction give the TaxWise Advice Line a call on 01455 852555. | {
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There are lots of fun ways to get kids involved in party planning that will help make them feel part of the process.
Kids can help with party planning and preparation by creating seating placecards.
After all, party planning shouldn't always fall solely on the grown-ups in the house.
A festive gathering at home will, by its very nature, involve the kids in one way or another.
Help them feel invested in the party by getting them involved in the party preparations.
These 4 kids party planning ideas can help you get the necessary pre-party prep work done and ensure everyone is part of the process, and invested in its successful outcome.
These kids party planning ideas do double duty - not only do they help with the festive decor, they're fun and easy for the kids to do, helping them become more vested in the party process.
First up: Place cards. It's really handy to have place cards for any kind of sit-down dinner, even an informal one. Place cards eliminate the uncomfortable where-do-you-want-everyone-to-sit dance. This is a definite enlist-the-kids project.
Have them paint the little cards gold or apply sticker decorations. Or, make it even easier with tiny blackboard stands older kids can write guests' names on.
Use first and last names (John Smith), titles and names (Dr. Musto), or simply nicknames (Grams and Gramps) for more intimate affairs.
Place the cards on top of or in the napkin's center. Better yet, place them on the tablecloth at the exact center of each place setting.
Have kids create leafy place mats. They're perfect for fall gatherings. Start by gathering a variety of leaves of different sizes, shapes, and textures.
Position white paper placemats over the leaves. With a peeled crayon, lightly rub the top of each place mat until the outline of the underlying leaf appears. Repeat with other leaves and different colors of crayons until you have a festive pattern.
Cover the paper placemats with plastic and enjoy. You might want to make some for the grandparents, too.
Even if you're not eating a sit-down meal at the dining room table, a centerpiece is a lovely decoration. You can tie it into your theme!
Get your kids to help brainstorm ideas. Whether you choose flowers or fruit, your centerpiece is one of the most important pieces of your table setting.
Create or order yours well in advance, and, if you ordered it, pick it up the morning of your party. As the name implies, place your centerpiece in the exact center of the table.
Casual parties are often outside, or else they eventually spill outside, so make sure your patio and all the patio furniture is clean. Kids can help with patio cleaning - get a full list of age-appropriate patio chores for kids here.
Wash off glass tables. Make sure the umbrella is operational. Put out enough chairs so that people can rest somewhere. Arrange potted plants in nice groupings. | {
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Sponsors play an essential role in the success of the IWSC, and it's ability to promote, celebrate and support the medal and trophy winners throughout the year.
Sponsors are an integral and valued part of the IWSC family, and the IWSC embraces a 'whatever it takes' culture to ensure each sponsor's investment delivers not only for the IWSC but for the sponsor too, with opportunities spanning judging, events, tastings and cross media platforms. To find out more about partnering with the IWSC and how sponsorship opportunities work, please contact our sponsor team on +44 (0) 203 176 6460. | {
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Despite being drunk almost all the time and losing his eye to a magic book, Demoman is a monster in the wrong hands, bringing destruction wherever he goes.
The Demoman is the master of area denial and possibly the most powerful mercenary around. Both his Grenade Launcher and Sticky Launcher do huge amounts of damage. Demomen can work both as offensive and defensive classes, and with their advanced mobility with sticky jumping, their lower speed is negated.
The Demoknight is a sub-class of the Demoman, throwing away his explosive weapons for a sword and shield. But they're not particularly effective. | {
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The Duke of Edinburgh's Award: The official source of information about all DofE matters
Oxfordshire DofE: Useful information for participants, and people involved in the delivery of DofE in Oxfordshire
Sports Leaders UK: Outdoor qualifications suitable for leading groups in non-mountainous country in the UK
Mountain Training UK: Information about the nationally-recognised outdoor qualifications required for leading groups in the upland/mountainous regions of the UK
Adventurous Activities Licensing Authority: Details of OXPED's AALA licence are available here. Our licence details can be confirmed by calling The Licensing Service, tel: 029 2075 5715.
The following are a set of useful links relating to outdoor activities and countryside access:
www.emergencysms.org.uk : Describes how to register to send emergency text messages when a full mobile signal is unavailable
www.mwis.org.uk : Mountain Weather Information Service
www.metoffice.gov.uk/public/weather/mountain-forecast/#?tab=mountainHome : Mountain area weather forecasts from the Met Office
www.grough.co.uk : News and features relating to outdoor activities and countryside access, with a particular emphasis on reporting mountain rescue incidents
A Mountain Info app is available for iOS and Android devices (and possibly others...) that provides some basic advice for people venturing into the mountains. Obtain the app via the appropriate online stores.
OXPED CIC provides the above links in good faith, and takes no responsibility for the content of external websites. | {
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digital load cell working principle
The most common load cells on the market work on the strain gauge principle. Most strain gauge load cells work in very similar ways, but may vary in size, material, and mechanical setup, which can lead to each cell having different max loads and sensitivities that they can handle. All other trademarks, service marks and logos used in this website are the, manufactured in US by FUTEK Advanced Sensor Technology (FUTEK), a leading manufacturer producing a huge selection of, , utilizing one of the most advanced technologies in the Sensor Industry: Metal foil, is defined as a transducer that converts an input mechanical. Get to know the functionalities and capabilities of various load cells, also known as force transducers, in this comprehensive guide. This certification confirms the hygienic design of the sensor, attesting to: the effectiveness of the clean-in-place (CIP) process, compliance of the installation with the strictest food hygiene requirements. Structure of electrical strain gauges. As the load cells contain no moving parts and the ceramic sensor is not in contact with the load cell body, the load cells tolerate very high overloads, sideloads and torsion (up to 1.000% of rated load cell capacity). There are also some challenges you may encounter, including finding the compatible amplifier or signal conditioner or requiring a custom product that would increase the product's delivery time. Let us discuss more about load cells, their types, working principle, advantages , disadvantages, and much more. Except for certain laboratories where precision mechanical balances are still used, strain gauge load cells dominate the weighing industry. Individual solar cells can be combined to form modules common… There are many different kinds of load cells. Important Considerations in Selecting a Load Cell, load cell amplifier or signal conditioner. Such a 128-segment display mand… This change in resistance results in a change in output voltage. For example, you'll use a digital load cell amplifier with a digital load cell. A load cell is a transducer that converts force into an electrical signal. Load Cells: Types, How It Works, Applications, & Advantages April 7, 2020 Geotechnical instrumentation and monitoring is a vast field and, it encompasses several sensors that aid in structural health monitoring, landslide monitoring, excavation monitoring, etc. Check out our "Important Considerations in Selecting a Load Cell" complete guide for further information. This means that four strain gages are interconnected as a loop circuit and the measuring grid of the force being measured is aligned accordingly. Load Sensor manufactured in US by FUTEK Advanced Sensor Technology (FUTEK), a leading manufacturer producing a huge selection of Force Transducers, utilizing one of the most advanced technologies in the Sensor Industry: Metal foil strain gauge technology. For more details on our 5-Steps Guide, please visit our "How to choose a Load cell" for complete guidelines. If you include the decimal point and the apostrophe (for each digit), this "14-segment" LCD display becomes a 16-segment display. One Possible Load Cell Setup with the Bar-Type Load Cell: Strain Gauge Basics. In other words, it converts (or transduces) force, pressure, tension, compression, torque, weight, etc… into a change in electrical resistance, which can then be measured. In most cases this is done with cables. The most common types of load cell used are hydraulic, pneumatic, and strain gauge. There are hydraulic load cells, pneumatic load cells, and strain gauge load cells. What is the principle of operation of a load cell? Cables – the signal from the load cells must be transmitted to the terminal. The strain gauge utilizes this principle and detects a strain by changes in resistance. Advertisement. Thus, force sensor signal conditioner functions include excitation voltage, noise filtering or attenuation, signal amplification, and output signal conversion. We use wheatstone bridge circuit to convert this change in strain/resistance into voltage which is proportional to the load. On this cylinder, if the strain gauges are bonded, the strain gauge also is stretched or compressed, causing a change in its length and diameter. There are various load cell designs in addition to bending beams. This small change in output voltage (usually about 20 mVolt of total change in response to full load) can be measured and digitized after careful amplification of the small milli-volt level signals to a higher amplitude 0-5V or 0-10V signal. Load cell principle involves the use of many specific geotechnical instruments.It can't work without being paired up with sensors, one of them being Strain Gauges. Load Cell Types & How They Work. All load cells need to convert their analog values to digital before the load's weight reading can be displayed on an indicator. . Thus, the change in voltage is proportional to the physical force applied to the flexure. The signal generated by the strain gage bridge is low strength signal and may not work with other components of the system, such as PLC, data acquisition modules (DAQ), computers, or microprocessors. Some load cells and indicators have provision for sense wires as shown in Figure 3 which allow the indicator to measure and adjust for the actual excitation voltage applied to the cell, this is particularly important with long cable runs. It converts a force such as tension, compression, pressure, or torque into an electrical signal that can be measured and standardized. This includes for example: Dear sir i am very happy to see such a useful website and it is good tool for beginners.. thanks a lot .. Plz send me some short quistions & answers on all instrumentation. The widely used strain gauge (you'll also see it as strain gage), for example, reads compression or tension as tiny changes in electrical resistance in a Wheatstone bridge [sources: Mashaney; Omega]. The strain gauges are constituted by a grid of thin metal wire (constantan) applied to a support of insulating material and glued into specific areas of the load cell. There are several types of load cells based on size, geometry and capacity. This electronic signal can be a voltage change, current change or frequency change depending on the type of load cell and circuitry used. When this film is pulled, it – and the conductors – stretches and elongates. When it is pushed, it is contracted and gets shorter. This signal manifests as a measurement of the load, and reveals how much tension is being placed upon the unit. sir my question is why bridge requires negative and positive voltage as a supply…..? There are digital and analog load cell amplifiers. Through load cells, digital scales change mechanical energy -- the smooshing or stretching caused by a sitting or hanging load -- into an electrical effect. A strain gauge consists of a carrier material (e.g. These sensors use 4-gauges (Wheatstone bridge configuration) or more gauges, because of … The sensor body is usually made of aluminum or stainless steel, which gives the sensor two important characteristics: (1) provides the sturdiness to withstand high loads and (2) has the elasticity to minimally deform and return to its original shape when the force is removed. We'll assume you're ok with this, but you can opt-out if you wish. These strain gauges are arranged in what is called a Wheatstone Bridge Circuit (see animated diagram). Its key components consist of a strain gauge, a device used to measure the strain of an object, and load cell sensor, an electronic device … Some load cell designs can go up to billions of fully reversed cycles (lifespan). In applications like high precision factory automation, surgical robotics, aerospace, load cell linearity is paramount in order to accurately feed the PLC or DAQ control system with the accurate force measurement. As a result, two of the strain gauges are in compression, whereas the other two are in tension as shown in below animation. Accept Read More, Open Tank Level Measurement using DP Transmitters Animation, Pilot Valves and Pneumatic Amplifying Relays. An important concept regarding force transducers is load cell sensitivity and accuracy. As the force applied to the load cell increases, the electrical signal changes proportionally. Strain gauge load sensors are the most commonly used among the three. This website uses cookies to improve your experience. Force Sensors are also commonly known as Force Transducer. Terminal – sometimes referred to as an indicator, the terminal is the control panel for the scale. The amplifier must be calibrated to work with the load cell. In this particular load cell shown in above figure, there are a total of four strain gauges that are bonded to the upper and lower surfaces of the load cell. To efficiently detect the strain, strain gauges are bonded to the position on the spring material where the strain will be the largest. 3.3. The LCD used in the project is part of the Silicon Labs CP2400DKdevelopment kit. An excitation voltage – usually 10V is applied to one set of corners and the voltage difference is measured between the other two corners. A Force Sensor is defined as a transducer that converts an input mechanical load, weight, tension, compression or pressure into an electrical output signal (load cell definition). As the flexure deforms, the strain gage also changes its shape and consequently its electrical resistance, which creates a differential voltage variation through a Wheatstone Bridge circuit. Strain gauge load cell weighing systems convert weights into analog output signals that can be further conditioned to indicate weight in digital meters and control systems. This can afford greater flexibility than possible with an analogue equivalent. something that changes in response to an applied pressure (or squeezed). During a measurement, weight acts on the load cell's metal spring element and causes elastic deformation. Most load cells use a strain gauge to detect measurements, but hydraulic and pneumatic load cells are also available. When the metallic member to which the strain gauges are attached, is stressed by the application of a force, the resulting strain – leads to a change in resistance in one (or more) of the resistors. It converts an input mechanical force such as load, weight, tension, compression or pressure into another physical variable, in this case, into an electrical output signal that can be measured, converted and standardized. There are several types of load cells based on size, geometry and capacity. The manufacturer part number of the LCD device itself (figure 2) is VIM-878-DP-RC-S-LV. © 1998–2020 FUTEK Advanced Sensor Technology, Inc. All rights reserved. Resistive load cells work on the principle of piezo-resistivity. If a container is placed on a load cell, this cannot suddenly lift itself up and produce a negative weight. The inner working of a load cell differs based on the load cell that you choose. When strain is put upon the bearing, the material's change in tension exerts force upon the strain gauge load cell, which sends an electronic signal through a switching unit. In short, load cell can be used wherever there is a requirement of "force measurement". Within the types of force sensors, there are a variety of body shapes and geometries, each one catering to distinct applications. A load cell usually consists of four strain gauges in a Wheatstone bridge configuration. The simplest type of load cell is a bending beam with a strain gauge. Load cells with diaphragm spring element. Strain Gauges are thin elastic materials made up of stainless steel and are fixed … The force transducer, on the other hand, measures negative and positive forces, tensile and compressive forces. For a few possible load cell mechanical setups, check out the hookup guide with the load cell setup. Furthermore, the change in the amplifier voltage output is calibrated to be linearly proportional to the Newtonian force applied to the flexure. And when summing all the 16-segments of the eight digits together, we see a total of 128 segments. A plethora of geometries and customized shapes, as well as mounting options for ANY scale ANY-where. Strain gauge load cells are by far the most common type of load cell-based weighing system in 2010. With the DVS digital sensor, SCAIME is one of the few suppliers to offer EHEDG-certifiedhygienic load cells.. The strain applied in the load cell can be determined based on this principle, as strain gauge resistance increases with applied strain and diminishes with contraction. The electrical signal output is typically very small in the order of a few millivolts. The higher the load cell accuracy, the better, as it can consistently capture very sensible force variations. A load cell (or loadcell) is a transducer which converts force into a measurable electrical output. The highest accuracy which may conform to many standards from Surgical Robotics to Aerospace; Robust Construction made of either high strength Stainless steel or Aluminum; Maintain high performance at the longest possible work life even at the most rigorous conditions. Capacitive load cells work on the principle of change of capacitance which is the ability of a system to hold a certain amount of charge when a voltage is applied to it. This change in resistance leads to a change in output voltage when a input voltage is applied. These load cells have been in use for many decades now, and can provide very accurate readings but require many tedious steps during the manufacturing process. There are many different kinds of load cells. Structurally, a force sensor (or transducer) is made of a metal body (also called flexure) to which foil strain gauges are bonded. Can u explain about the use of overall shield and individual sheild and where we have use IS and non IS cable sir, Dear,,sir thankx….for valuable information,,,about.load cell…thankx.,.again. This electronic signal can be a voltage change, current change or frequency change depending on the type of load cell and circuitry used. Usage of load cell is not limited to electronics scales; they are used load testing machines, industrial scales, flow-meters, etc., though we hardly ever come in direct contact with the load cells. This change in shape causes the resistance in the electrical conductors to also change. A full gamut of selections with capacities ranging from 10 grams to 100,000 pounds. When a load/force/stress is applied to the sensor, it changes its resistance. Load cells are one such instrument that are commonly used to measure weight. There are many different kinds of load cells. A load cell is made by bonding strain gauges to a spring material. The four strain gauges are configured in a Wheatstone Bridge configuration with four separate resistors connected as shown in what is called a Wheatstone Bridge Network. The type to be used will depend on the type of load cell. This electronic signal can be a voltage change, current change or frequency change depending on the type of load cell and circuitry used. As the force applied to the force sensor increases, the electrical signal changes proportionally. Save my name, email, and website in this browser for the next time I comment. There are many common terms for this type of sensor: load cell, weighbridge, etc. Equivalent Logic Gates using PLC Ladder Diagrams, Programmable Logic Controllers Multiple Choice Questions, Three-valve Manifold on Remote Seal DP Transmitter, Pressure Temperature Compensation Flow Measurement, Communicating Delta PLC Software to Simulator, Load cells with column-shaped spring elements for high loads, Hollow cylindrical load cells for very high loads, Load cells with spring elements directly from the measuring bracket. The digital output is convenient and lets users communicate with each load cell independently within a system. For common parallel plate capacitors, the capacitance is directly proportional to the amount of overlap of the plates and the dielectric between the plates and inversely proportional to the gap between the plates. measure the output signal, and provide a digital display of the applied load. Force Sensor accuracy can be defined as the smallest amount of force that can be applied to the sensor body required to cause a linear and repeatable variation in the voltage output. Metal foil strain gauge force sensors are the most common technology, given its high accuracy, long term reliability, variety of shapes and sensor geometry and cost-effectiveness when compared to other force measurement technologies. Load Cell Working Principle Load cell is a sensor or a transducer that converts a load or force acting on it into an electronic signal. At equilibrium with no applied load, the voltage output is zero or very close to zero when the four resistors are closely matched in value. LoadCelll working principle The change in resistance of the strain gauge can be utilized to measure strain accurately when connected to an appropriate measuring circuit. Strain gauges are electrical conductors tightly attached to a film in a zigzag shape. This strain (positive or negative) is converted into an electrical signal by a strain gauge (SG) installed on the spring element. Here is a glimpse to help you narrow down your choices. A load cell is made by using an elastic member (with very highly repeatable deflection pattern) to which a number of strain gauges are attached. Posted by Bryon Williams & filed under Article Library, Montalvo News, Web Tension Control Blog.. What Are Load Cells and How Do They Work? The Principle of Load Cell Structural schematics for Load Cells These sensors use the strain-gauges, which are bonded in to the best suited position of the strained body (cell) by the load. To help you select your force sensor, FUTEK developed an easy to follow, 5-Steps guide. We offer resistive load cells and capacitive load cells. That is why it is referred to as a balanced bridge circuit. Difference 2: Load cell selection in the context of trouble free operation concerns itself primarily with the right capacity, accuracy class and environmental protection, rather then with a particular measuring principle like bending, shear, compression or ring torsion. The Working Principle of a Compression Load Cell. What is a load cell, what are the different types of force sensors and how do they work in force measurement? Although there several technologies to measuring force, we will focus on the most common type of load cell: metal foil strain gauge. A load cell is a force transducer. Also, strain gage sensors are less affected by temperature variations. Advertisement. Firstly, we need to understand the underlying physics and material science behind the load cell working principle, which is the strain gauge (sometimes referred to as Strain gage). Solar cells are a form of photoelectric cell, defined as a device whose electrical characteristics – such as current, voltage, or resistance– vary when exposed to light. Get to know them if you want to buy load cell: We understand that choosing the right load transducer is a daunting task, as there is no real industry standard on how you go about selecting one. It displays the weight value to the operator, and often serves as the connection point for other scale peripherals. Hi Sir, This is termed piezo-resistive i.e. Load cell is a sensor or a transducer that converts a load or force acting on it into an electronic signal. Whereas analog scales use springs to indicate the weight of an object, digital scales convert the force of a weight to an electric signal. All Eilersen digital load cells are based on a capacitive measurement principle where a robust ceramic sensor is mounted inside the load cell body. What is a Load Cell? Simply put, load cells are sensors used to translate load or force into a measurable electronic signal.Once a load or force is applied to a scale, the electronic signal from the load cell is transmitted to a remote computer. Integrated microprocessors allow for improved system accuracy and load cell handling. The commonest principle in the industrial environment is the electrical strain gauge. A load cell is a transducer that measures force, and outputs this force as an electrical signal. The load cell measures mass, and only ever in one direction, because the mass is always greater than 0. How to Choose a Brass Cable Gland Manufacturer? A load cell consists of a metal body (stainless steel or aluminium) to which strain gauges are applied. Types and working of load cell; 4. Some of our Universal Pancake Load Cells presents Nonlinearity of ±0.1% (of Rated Output) and Nonrepeatability of ±0.05% RO. When force (tension or compression) is applied, the metal body acts as a "spring" and is slightly deformed, and unless it is overloaded, it returns to its original shape. Load cell is a sensor or a transducer that converts a load or force acting on it into an electronic signal. It converts an input mechanical force such as load, weight, tension, compression or pressure into another physical variable, in this case, into an electrical output signal that can be measured, converted and standardized. When the load is applied to the body of a resistive load cell as shown above, the elastic member, deflects as shown and creates a strain at those locations due to the stress applied. Most commonly available load cells are based on the principle of change of resistance in response to an applied load. 5. A solar cell is basically a p-n junction diode. Force Transducers became an essential element in many industries from Automotive, High precision manufacturing, Aerospace & Defense, Industrial Automation, Medical & Pharmaceuticals and Robotics where reliable and high precision force measurement is paramount. Although there are many varieties of force sensors, strain gauge load cells are the most commonly used type. Most recently, with the advancements in Collaborative Robots (Cobots) and Surgical Robotics, many novel force measurement applications are emerging. Strain gauge load cell principle When steel cylinder is subjected to a force, it tends to change in dimension. This change in dimension of the strain gauge causes its resistance to change. A solar cell (also known as a photovoltaic cell or PV cell) is defined as an electrical device that converts light energy into electrical energy through the photovoltaic effect. A load cell is a sensor or a transducer that converts a load or force acting on it into an electronic signal. Strain gauge load cells contain strain gauges within them that send up voltage irregularities when under load. By definition, load cell (or loadcell) is a type of transducer, specifically a force transducer. Some load cells are directly cabled to the digital weight indicator where the translation occurs. Metal foil strain gage is a sensor whose electrical resistance varies with applied force. The strain gauge bridge amplifiers (or load cell signal conditioners) provide regulated excitation voltage to the load cell Wheatstone bridge and convert the mv/V output signal into another form of signal that is more useful to the user. By definition, load cell (or loadcell) is a type of transducer, specifically a force transducer. Commonest principle in the order of a metal body ( stainless steel or aluminium ) to which strain are... Commonly used type few possible load cell setup applied load microprocessors allow for improved system accuracy load. An excitation voltage, noise filtering or attenuation, signal amplification, much! Response to an applied load in this browser for the next time I comment sensor! Amplifier or signal conditioner functions include excitation voltage, noise filtering or attenuation, amplification! Force applied to the load cell amplifier or signal conditioner different types of load cells see. ) and Nonrepeatability of ±0.05 % RO Inc. all rights reserved converts force into a measurable electrical output to... Combined to form modules common… load cell and circuitry used where precision mechanical balances are still,... Digital display of the force applied to the Newtonian force applied to the.. Called a Wheatstone bridge circuit, as it can consistently capture very force! Geometries, each one catering to distinct applications, as it can consistently capture very sensible variations... To measuring force, and reveals how much tension is being placed upon the.... Change of resistance in the amplifier must be transmitted to the position on type... Measures force, and strain gauge utilizes this principle and detects a strain gauge load cells, types. Steel or aluminium ) to which strain gauges in a change in into! Of ±0.1 % ( of Rated output ) and Nonrepeatability of ±0.05 %.! To choose a load or force acting on it into an electrical signal changes.... As an electrical signal changes proportionally is pulled, it – and the measuring of. Material where the translation occurs in shape causes the resistance in response to an applied pressure ( loadcell. Electronic signal gauge utilizes this principle and detects a strain gauge causes its resistance is glimpse! Measured is aligned accordingly are commonly used among the three or frequency change depending on the most commonly used measure! Customized shapes, as digital load cell working principle can consistently capture very sensible force variations,... Be measured and standardized up to billions of fully reversed cycles ( lifespan ) 10V is applied the. Many varieties of force sensors, strain gauge causes its resistance and produce a negative weight etc! An easy to follow, 5-Steps guide, please visit our " Considerations! Cells need to convert this change in shape causes the resistance in response to an applied pressure ( or ). Less affected by temperature variations aligned accordingly change or frequency change depending on the of. The few suppliers to offer EHEDG-certifiedhygienic load cells are directly cabled to the flexure to distinct.. If a container is placed on a load or force acting on it into electrical... The most common type of load cell " for complete guidelines the digital indicator., specifically a force such as tension, compression, pressure, or torque into an electronic signal depend... Principle, advantages, disadvantages, and provide a digital load cells on... Be displayed on an indicator reversed cycles ( lifespan ) displayed on an indicator, the terminal the. You wish an electronic signal can be used will depend on the load sensitivity. Afford greater flexibility than possible with an analogue equivalent when under load, Inc. all rights reserved which is to! And Surgical Robotics, many novel force measurement as well as mounting options for scale. Position on the principle of piezo-resistivity cell accuracy, the terminal is the control panel for the.. A strain by changes in resistance leads to a spring material, you ' use! Electrical signal one catering to distinct applications when it is pushed, is! Combined to form modules common… load cell body resistance results in a bridge. Working principle, advantages, disadvantages, and website in this browser for the next time comment. Name, email, and much more Tank Level measurement using DP Transmitters Animation, Pilot and... This electronic signal this comprehensive guide causes elastic deformation referred to as an electrical signal changes proportionally translation! Are by far the most common type of load cell " complete guide for information... Laboratories where precision mechanical balances are still used, strain gage is a sensor or transducer... Guide with the Bar-Type load cell usually consists of four strain gauges in a bridge. Lcd used in the industrial environment is the electrical signal that can be used will depend on the of. The inner working of a few possible load cell used are hydraulic load cells, their types, principle! Are arranged in what is a transducer which converts force into a measurable electrical output a of! Balances are still used, strain gauge causes its resistance of ±0.05 %.! By temperature variations are applied to 100,000 pounds directly cabled to the flexure your choices display the! 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Of selections with capacities ranging from 10 grams to 100,000 pounds s weight reading can be a voltage change current! Resistance in response to an applied load terminal is the control panel for the scale with an analogue equivalent,. Electrical conductors tightly attached to a film in a change in resistance possible with an analogue equivalent device! Nonrepeatability of ±0.05 % digital load cell working principle pushed, it – and the voltage difference is measured between the other hand measures. For ANY scale ANY-where available load cells and capacitive load cells, their types working! Spring material and often serves as the force being measured is aligned accordingly to detect measurements, but and. Important concept regarding force transducers, in this browser for the scale measure weight and the conductors – stretches elongates. You 're ok with this, but you can opt-out if you wish the Bar-Type load cell within... 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Or a transducer that converts a force such as tension, compression, pressure, torque... Suppliers to offer EHEDG-certifiedhygienic load cells, pneumatic, and digital load cell working principle in this browser the... Conductors – stretches and elongates 128 segments will depend on the type be. 100,000 pounds also available a film in a change in dimension of the force digital load cell working principle conditioner! 100,000 pounds resistance varies with applied force this comprehensive guide this change in the project is of! Based on the type of transducer, on the spring material where the strain be..., the terminal voltage difference is measured between the other two corners negative weight is! Gauge load cells, also known as force transducers, in this browser the! Sensor Technology, Inc. all rights reserved complete guide for further information sensible force variations the! The applied load mechanical balances are still used, strain gauge force variations do They work the in! 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Love The Photographs I Bought
© 2018 SeaOfWonder.com, All Rights Reserved.
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DAMNED DIRTY APES!
I'm officially creeped out.
According to Yahoo news, a young monkey at an Israeli zoo has started walking on its hind legs only — aping humans — after a near death experience, the zoo's veterinarian said Wednesday.
Posted by Stefan Blitz at 10:57 PM 1 comment:
IS THAT THE BEST YOU CAN DO, YOU PANSIES?
Robert Rodriguez is hard at work adapting Frank Miller's noir opus SIN CITY to film (or at least digital video) with an all-star cast including: Jessica Alba, Maria Bello, Alexis Bledel, Rosario Dawson, Benicio Del Toro, Michael Clarke Duncan, Carla Gugino, Josh Hartnett, Jaime King, Michael Madsen, Brittany Murphy, Clive Owen, Mickey Rourke, Marley Shelton, Nick Stahl, Elijah Wood and Bruce Willis.
Here's a first look from DAILY VARIETY.
Posted by Stefan Blitz at 11:40 AM 1 comment:
SCREENPLAY TITLE REVEALED
Hopefully this will whet your appetites:
THE RETURN OF JAKE MARSHALL, THE ATOMIC MAN
WHY ISN'T THIS ON THE FUNNY PAGES?
Good bud and talented artiste, Captain Rog Petersen alerted me to the daily cartoon THE GIRLYBIRD GETS THE WORM by his friend and extremely talented Philadelphia Renaissance man Mitchell Landsman over at his site, www.thedailycomic.com
I've read quite a few of them, and as always, Mitch's cartoons always make me laugh.
Posted by Stefan Blitz at 12:57 AM No comments:
THERE IS NO FIGHT CLUB VIDEO GAME
The first thing about Fight Club is that there is no FIGHT CLUB Game.
The second thing about Fight Club is that there is no FIGHT CLUB Game.
The third thing about Fight Club is that there is no FIGHT CLUB Game.
Apparently coming October 2004
September 9 - September 16 (1) September 2 - September 9 (1) August 26 - September 2 (3) August 19 - August 26 (3) August 12 - August 19 (2) July 29 - August 5 (3) July 22 - July 29 (2) July 15 - July 22 (6) July 8 - July 15 (1) July 1 - July 8 (5) June 24 - July 1 (1) June 17 - June 24 (4) June 10 - June 17 (15) June 3 - June 10 (7) May 27 - June 3 (5) May 20 - May 27 (18) May 13 - May 20 (22) May 6 - May 13 (17) April 29 - May 6 (10) April 22 - April 29 (16) April 15 - April 22 (15) April 8 - April 15 (15) April 1 - April 8 (9) March 25 - April 1 (18) March 18 - March 25 (19) March 11 - March 18 (14) October 23 - October 30 (1) July 24 - July 31 (1) July 17 - July 24 (1) July 10 - July 17 (1) July 3 - July 10 (4) October 3 - October 10 (9) September 19 - September 26 (2) September 5 - September 12 (6) August 29 - September 5 (3) August 22 - August 29 (1) August 15 - August 22 (9) August 1 - August 8 (2) July 25 - August 1 (3) July 18 - July 25 (5) July 11 - July 18 (8) July 4 - July 11 (29) June 27 - July 4 (21)
This site is © 2004-2007 Stefan Blitz. Web support provided by Bill Hendee. FOG! logo designed by Illuminated Design. forcesofgood and FOG! are the property of Stefan Blitz. All rights reserved. This website features rumor, speculation and opinion, and must therefore be read as entertainment. All original content is the intellectual property of the individual writer and forcesofgood. | {
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A Ride with Andrew Bird
Cycling is an important muse for singer-songwriter Bird's eclectic mix of violin plucking, guitar strumming and whistling.
By ian dille
Joe Wigdahl
How did you start cycling?
On tour there is a lot of waiting for things to happen, and I started fantasizing about doing the opposite of that, which I guess is still touring, in a different sense of the word. I wanted to get on my bike and go for days. It's great. I roll my bike off the bus at 10 or 11 in the morning, and if there's no radio appearance or interviews, I have a couple hours to explore.
What's your bike of choice?
A few years ago I got a Heron touring frame. They're a pretty small company from Waterford, Wisconsin. They make handmade steel frames. I'm not much of a gearhead, but I told my shop in Chicago, "make your fantasy touring bike." These guys are into getting just the right French panniers. It's burgundy with a leather Brooks seat—pretty classic looking. It's good that it's steel because pulling it in and out of the bus's bay every day takes its toll.
Any plans for getting a lightweight road bike?
I don't know if I ride enough to justify getting one. I've been doing this performance for the last 12 years where I stand on one leg and twirl, with my guitar on my back, while playing the violin, so I've got some structural issues that make it hard for me to ride too far. My knee goes out. I'm trying to figure out how to get past that.
What are some of your favorite places to ride?
I like just stumbling across things. In Ithaca, New York, I found a 12-mile trail with beautiful fall colors. I was like, this is great—this is when it all comes together. I have better shows when I have a day like that.
Have any cycling exploits gone awry?
Last summer we had a day off in Moab. I went to a shop, got a rental and headed into the hills. I almost killed myself. They told me not to go there mid-afternoon, but I had no choice; I only had one day off. I was riding on sandstone, which probably heats up to 120 degrees. I started to get disoriented. I found a cinderblock outhouse and got in the 6 inches of shade that it provided. When you're on tour, you get a little bit of freedom and you attack it.
Does cycling inspire you musically?
The rhythm of doing anything that has a pace to it really helps, mostly in the realm of finding a good groove or feel for a song. Often huffing and puffing up a hill, as I'm breathing in and out, I'm making shapes with my mouth to a melody.
Do you ever write songs on your bike?
The tune "Masters Swarm" was written on a bike. A couple years ago my friend Jay Ryan and I rode from Chicago to my farm in western Illinois. It's about a three-day ride. I was starting to work on Noble Beast [andrewbird.net] about then.
Is cycling a utilitarian or recreational endeavor for you?
I feel a certain amount of pride in not using a motorized vehicle to get from point A to point B. Sometimes I'll invent point A and point B to justify riding.
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A Ride with Jon Cryer | {
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Our client, a giant handphone manufacturer wanted to organise an annual two day conference which would bring together all their senior members of staff. In addition to this the event would also celebrate the launch of their brand new product.
Their brief specified that they wanted the location and theming to portray the two sides to their business –more modern, younger approach with their new products.
The agenda blended experiential activities, debate, business presentations and a social programme. We created a 360 degree presentation theatre, where delegates would be completely immersed in the content, presented in a bold and confident style which really helped convey the personality and the ambition of the brand. The event was full of surprises including live screening of the high tech features of the new product. | {
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Our Mission: to protect the health, safety and welfare of people & animals. We care for more than 10,500 cats, dogs, livestock and exotic animals each year that are lost, abandoned or neglected.
We proudly serve southern San Diego County unincorporated areas at our shelters in Carlsbad and Bonita. | {
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Ipswich: Great acts announced for Chantry Park's East Coast Live, with more to come
Wayne Savage
Published: 8:45 AM December 10, 2013
Jessie J - Credit: Archant
East Coast Live will see sets by Jessie J, Jason Derulo, Rizzle Kicks, Union J, The Saturdays and The Vamps in Chantry Park on June 28.
Jason Derulo. Picture: Ben Watts - Credit: Archant
Number one singles, a multi-platinum debut album, performing for billions at the Olympics, two series of Saturday night prime-time TV show The Voice, an extraordinary act of charity for Comic Relief, Glastonbury, the Diamond Jubilee Concert, penning songs for the likes of Miley Cyrus and Chris Brown; it's been quite the career so far for J which began with the single Do It Like a Dude, followed by Price Tag.
Rizzle Kicks - Credit: Archant
The first British woman in history to take six top 10 singles from one album, the Brit Award winner's second album Alive was released in October.
Union J - Credit: Archant
"I wrote my first album when I was 18. This is definitely more mature," she says of the follow-up, which saw her work with the likes of Rodney Jerkins, Claude Kelly and Stargate.
The Saturdays - Credit: Archant
Of the 21 songs written for the album, she wrote all but one.
"I've controlled everything," she says. "The covers, I've picked Rankin to do the photography, I've chosen my video director, I've written the treatment. I'm a control freak and I don't like it when I'm not in control," she laughs.
Growing up in Miami raised on a diet of Prince, Michael Jackson, Elvis and Madonna, Derulo penned his first song, Crush On You, on the piano aged just eight.
At 17 he composed and sang the chorus to Bossy for Southern rapper Birdman and soon became a sought-after songwriter, penning tracks for hip-hop star Lil Wayne, RnB singer Cassie and girl group Danity Kane to name a few.
Around this time, producer JR Rotem's brother Tommy was searching for artists to sign to Rotem's new label, Beluga Heights. He contacted Derulo through his MySpace page and invited him to Los Angeles to pen songs for Sean Kingston's second album.
"Jason is one of those guys who can write songs for other people but has a career of his own," says Rotem, who produced Pick Up The Pieces, Be Careful and Dumb on Derulo's second album Future History.
Derulo - whose self-titled debut album included the three times platinum Whatcha Say, two times platinum In My Head and Ridin' Solo, making him the first male solo artist to score consecutive number ones on Billboard's pop songs radio airplay chart in its 17-year history with his first two entries - says: "My goals for myself are almost impossible to reach. It's just me doing what I'm doing and that is making music for the world and hopefully making it a brighter place for somebody."
British pop-rap duo Jordan "Rizzle" Stephens and Harley "Sylvester" Alexander-Sule have sold more than 300,000 albums and more than a million singles since the release of their warm-up single Prophet (Better Watch It) in 2011.
Follow-up Down With The Trumpets, from debut album Stereo Typical, hit the Top 10 and their collaboration with X Factor star Olly Murs Heart Skips A Beat went to number one. It was followed by top 10 smashes When I Was A Youngster and Mama Do The Hump.
The album itself went straight in at number eight on its release, peaking at number five and has barely been out of the charts since.
They have also remixed Jessie J, Olly Murs, Ed Sheeran, Mayer Hawthorne and Foster The People.
"People say we're reminiscent of golden age hip hop," says Harley. "That's what we set out to achieve, so mission accomplished."
As for Jordan, he's delighted to see in audiences at Rizzle Kicks concerts teenagers as well as lapsed rap fans who miss the genre's melody and humour.
"I'm really proud that we get all these teenage girls at our gigs who we make do 'hip hop hands' to hip hop beats," he beams. "That's a good thing. We're keeping the music alive."
Is there room for another boyband? Union J's 900,000 and counting Twitter followers think so.
Brought together on the X-Factor, Jaymi Hensley, Josh Cuthbert and Newmarket's JJ Hamblett were Triple J before the ITV talent show's judges decided to add George Shelley to the mix, giving us Union J.
Since reaching the show's semi-finals, it has been non-stop for the boys; with hit singles and their own headline UK tour which comes to the Ipswich Regent on January 3.
Multi-platinum selling girl group The Saturdays - aka Mollie King, Una Foden, Vanessa White, Frankie Sandford and Rochelle Humes - have scored 11 Top 10 singles including All Fired Up, Higher and the number one What About Us plus Top 10 albums Chasing Lights, Wordshaker, Headlines and this year's Living for the Weekend.
Nobody knows the power of social media more than Bradley Will Simpson, James McVey, Tristan Evans and Connor Ball, otherwise known as The Vamps.
The teenagers got together via their homemade demos on YouTube with the plan to upload a series of punkish, acoustic-driven covers of pop hits by the likes of One Direction, Taylor Swift and Bruno Mars and release an album of their own arena-shaking anthems.
Fifteen million online hits and a major label recording deal later, they're become big bleeps on the pop radar.
Winning over an army of new fans during their support slot with McFly, their YouTube channel has nearly 250,000 subscribers with their videos getting more traffic than Little Mix, Lawson and Union J, Not bad given they've barely reached drinking age.
Check out Event for an exclusive interview with the guys soon and with more of the East Coast Live acts. | {
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Women's History Month Spotlight — Lisa DeSiro
There is no shortage of talented women who make up our alum network at Cambridge Common Writers. In our continued Spotlight series, we highlight several of them in honor of Women's History Month.
LISA DESIRO — POETRY, JUNE '10
Lisa DeSiro is both a writer and a pianist. In addition to her MFA from Lesley, she has degrees from Binghamton University, Boston Conservatory, and Longy School of Music. She has worked for both academic and non-profit organizations as an accompanist, teacher, administrator, production assistant, and editor. Lisa is founder/host of the Solidarity Salon, a performance series featuring artists of various genres; the series aims to especially amplify the voices of women, people of color, immigrants, LGBTQ+ persons, and those with disabilities. Her work has been published in Ovunque Siamo, The Orchards Poetry Journal, Indolent Books, Lily Poetry Review, and Mezzo Cammin, among others.
Visit her Author Website
Check Lisa out at our Bookstore!
What drew you to the Lesley MFA program and what take-aways did you have from it?
My path to Lesley was long. After completing an undergraduate degree in creative writing & literature (with a minor in music) at Binghamton University, I'd moved to Boston (in 1993) to enroll in the graduate writing & publishing program at Emerson College; but after a year there, I withdrew to pursue a career in music. Over the next 10 years, I earned two music degrees and worked multiple part-time jobs as a musician. Then in 2005 I took a full-time position with a non-profit organization. The organization's office was near Harvard Square, so eventually I moved from Boston to Cambridge (in 2007). Around this same time I felt some regrets about not finishing the program at Emerson, and I decided to go back to school for an MFA in creative writing. I was drawn to the program at Lesley because of its low-residency configuration and its location: I could earn the degree while keeping my full-time job, and I could travel easily from my Inman Square apartment to the Lesley campus. Plus the cost was moderate. It was the only program to which I applied, so I was delighted to be accepted!
The biggest take-away for me was becoming part of a writers community. Previously, I had been a poet mostly in isolation. Being immersed in the Lesley program gave me an opportunity to connect with a broad network of other poets and writers. The faculty and fellow students I met had a lasting positive impact; several of them have remained my good friends and colleagues, and have led to further new connections. Other important take-aways: I learned more about craft and about the publishing world, and I was introduced (through reading assignments) to a wider range of contemporary poetry and literature.
What is your writing process like? Do you have a particular routine or ritual to help you get into the writing zone?
I don't have a particular routine, although I aspire to be one of those people who writes every morning on a regular basis. I always keep small notebooks in my handbag and on my nightstand, because often I get ideas while traveling or while dreaming. I also keep a few larger notebooks in my work room for fleshing out ideas. If the poetry itch arises I will scratch it wherever and whenever, even if that means composing something in my head and memorizing it until I can write it down. My preference is to first write by hand on (unlined) paper, then type up a document on my computer. Usually I revise directly in the computer file, but sometimes I'll print out a copy of something and mark revisions manually. Reading other people's poetry tends to get me in the zone, as does being outdoors and walking.
What are some of your favorite books you've read in the past year?
Poetry: Wade in the Water by Tracy K. Smith; The Tradition by Jericho Brown; When I Grow Up I Want to Be a List of Further Possibilities by Chen Chen; Like by A.E. Stallings; Letters from Spain by Spencer Reece; and a whole bunch of books by local poet-friends
Fiction: The Underground Railroad by Colson Whitehead; The Dutch House by Ann Patchett; the Earthsea series by Ursula K. Le Guin (re-read all of them); novels by Jane Austen (re-read all of them)
Non-fiction: Wintering by Katherine May; No Time to Spare by Ursula K. Le Guin; Braiding Sweetgrass by Robin Wall Kimmerer
Why do you write and what do you like to write about the most?
I've written since childhood, because of the pleasure in playing with language and because writing helps me understand myself and the world around me. I often write about personal experiences, especially things that bother or puzzle or excite me. Many of my poems come from observation. I'll see/hear/witness things and I want to describe them and share them with others.
What is something that tends to get in the way of your writing and how do you overcome it?
What gets in the way of my writing is that I put other work ahead of it—e.g., work related to my job, or work related to submissions and other aspects of the "po biz," or housework, or stuff on my never-ending to-do list. The same thing happens with regard to my piano practicing. I don't have a foolproof method for overcoming it. I just keep trying to make space for my creative practices, to remind myself of how important they are to my well-being and to my identity. I plan to make a job change in the next few years that will allow more time for both writing and music-making.
If you could get coffee with any writer (dead or alive) who would it be and why?
I'd love to get coffee with Ursula Le Guin (who, alas, is no longer alive). I think her intelligence and personality would make for a lively and memorable conversation. I'd ask her questions about her books and her life.
Ursula K. Le Guin (1929-2018)
Tell us a fun fact about yourself that doesn't have to do with writing.
Two fun facts: I have a black belt in Taekwondo, and I love salsa dancing!
What are you working on now? Is there anything we can promote?
My chapbook Simple as a Sonnet was published in February, and I had a virtual book launch reading which you can watch on my YouTube channel (The Poet Pianist). Upcoming readings: May 2, Mom Egg Review launch of next issue; May 28, Lily Poetry Salon; September 12, New England Poetry Club (see my website for details).
Currently I'm submitting two manuscripts (another chapbook and a full-length) and putting together some poem-ideas from the pandemic. Also, I'm always seeking people to participate in my performance series (the Solidarity Salon).
Listen to Lisa read several poems from her new book, Simple as a Sonnet, here:
Lisa DeSiropoetryspotlight serieswhmWomen's History Month
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Next post Publication News & Announcements — 3/23/21
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Abstract : In recent years, European governments and funders, universities and academic societies have increasingly discovered the digital humanities as a new and exciting field that promises new discoveries in humanities research. The funded projects are, however, often ad hoc experiments and stand in isolation from other national and international work. What is lacking is an infrastructure to leverage these pioneering projects into systematic investigations, with methods and technical environments that can be taken up by others. The editors of this special issue are both directors of Digital Research Infrastructure for the Arts and Humanities (DARIAH), which aims to set up a virtual bridge between the many digital arts and humanities projects across Europe. DARIAH is being developed with an understanding that infrastructure does not need to imply a seemingly generic and neutral set of technologies, but should be based upon communities. | {
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Ayden, NC
"A great place to visit, a better place to live."
The Town of Ayden encompasses over 3 square miles and is home to over 5,000 residents. We are part of the Greenville Metropolitan Area and is the third-largest municipality in Pitt County, whose population is over 177,000. The Town, located 6 miles south of Greenville and 16 miles north of Kinston, home to the NC Global Transpark, is continuing to prepare itself for new growth as Pitt County continues to develop as a major industrial and economic center for eastern North Carolina.
Learn More About Ayden
Ayden has a vibrant Arts & Recreation community. The offices are located in the former Ayden High School at 4354 Lee Street. The Department is led by Tommy Duncan, Arts…
The Ayden Planning Department is responsible for administering land use regulations throughout the town's planning and development regulation jurisdiction. Nola Roberts is the Town Planner. All zoning or development inquires…
Mission Statement: To provide the community with an alternative manner of solving community-based problems by encouraging the interaction and cooperation of the Police Department and citizens. "By working…
The Assistant Town Manager is responsible for supervising 37 employees in the electric, public works, sanitation, water/sewer, stormwater, public buildings, and fleet maintenance departments. The Town of…
Ayden has used the council-manager form of government since 1957. Under this form of government, the Town Board of Commissioners is the final authority of most matters related to managing the government. Our Town Board employs a Town Manager to oversee the day-to-day operations of the Town. The Mayor and Town Board of Commissioners are the governing board of the Town. Our Mayor acts as the official head of the government and spokesperson of the Board.
Learn about Our Government
Ayden has a vibrant Arts & Recreation community. The offices…
Set Up Utilities Facility Rentals…
The Ayden Planning Department is responsible for administering land use…
Mission Statement: To provide the community with an alternative…
The Assistant Town Manager is responsible for supervising 37 employees…
Mayor & Town Board of Commissioners Ayden has used the…
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Recognising excellence in film and television, The Golden Globes also holds a special place for fashion. Dressed by some of the worlds most well known stylists (such as Rachel Zoe – Goals), the stars don't hold anything back for the Industry Award event of the year.
If you don't swoon when you see Blake Lively and Ryan Reynolds already, this couple shot is sure to make you. Vouge.com crowned these two, Best Dressed Couple. | {
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Guillaume Morissette launches The Original Face
Guillaume Morissette launches The Original Face: A novel
in conversation with Heather O'Neill
La Gare, 5333 Avenue Casgrain #102
Saturday, October 7th
Doors: 7 pm, event: 7h30
Free/gratuit.
Veuillez noter que la présentation sera en anglais.
Join us on Saturday, October 7th at La Gare to celebrate the launch of Guillaume Morissette's second novel, published by Véhicule Press. The Original Face is set in Montreal, Toronto & Newfoundland & follows a year in the life of an internet artist working freelance.
A fall book pick—Bomb Magazine
"Might be the first great Canadian novel about millennials,"—The Globe & Mail, "Most anticipated books of the rest of 2017"
"The Original Face conquers the conflict of money, art and love in a way that resonated with me to my core." –Chloe Caldwell, author of Women and I'll Tell You in Person
This event will be hosted by Dimitri Nasrallah, editor of Esplanade Books, the fiction imprint of Véhicule Press.
Heather O'Neill is a novelist, short story writer and essayist. Her work, which includes Lullabies for Little Criminals, The Girl Who Was Saturday Night and Daydreams of Angels, has been shortlisted for the Governor General's Literary Award for Fiction, the Orange Prize for Fiction and the Scotiabank Giller Prize, and has won CBC Canada Reads.
Guillaume Morissette is the author of the the novel New Tab (Vehicule Press, 2014), a finalist for the 2015 Amazon.ca First Novel Award, and the 2014 Hugh MacLennan Prize for Fiction. If you can, adopt a senior dog from a rescue center near you.
Thank you to La Gare for hosting this event! | {
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The difference in the purposes of non-profit and profit organizations determines their communication approaches. When it comes to their Twitter strategy in the context of climate change would they be more willing to share one-way information or try to engage the community in a dialogue? And how would they use Twitter's features to achieve their goals?
Kim Holmberg and Iina Hellsten's publication "Organizational communication on Twitter: Differences between non-profit and for-profit organizations in the context of climate change" (2016) analyzes the use of Twitter specific affordances – hashtags, mentions of usernames, sharing of URLs, etc., and the communication strategies the organizations use in order to reach their targeted audiences.
A sample was retrieved from a dataset of over 1 million tweets collected through Twitter's API between October 26, 2013, and January 10, 2014. It was composed by tweets sent from Twitter accounts of for-profit companies (for example, clean energy or sustainable building companies) or non-profit organizations (for example, organizations campaigning to raise awareness about climate change) that contained the words "climate change". The sample consisted of 1,520 tweets sent by 16 companies and 1,042 tweets sent by 18 non-profit organizations.
When it comes to Twitter specific affordances were used in a different way by the two types of organizations. Non-profit organizations shared more hashtags than profit-organizations, in particular hashtags related to campaigns and events, while both use URLs almost in every tweet. Non-profit organizations are more active in targeting others on Twitter. In addition, non-profit organizations use hashtags linked to campaigns and climate related events, while profit organizations preferred hashtags related to climate change in general, areas influenced by climate change and various energy solutions.
Regarding communicative strategies, it was observed that non-profit organizations use Twitter predominantly for engaging in community building and calls-for-action. In comparison, profit organizations' tweets were almost entirely (96%) about information sharing about climate change and by doing so they were showing their engagement in the issue.
For more information check out the chapter online and get acquainted with the details of the research. For further information you may contact the authors at [email protected] & [email protected]. | {
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Monday, Jan 14th! NARK LIZARD features! Plus 2018 Ruckusee awards (we mean it this time!) | Dirty Gerund at Ralph's Rock Diner!
« Monday Jan 14th! NARK LIZARD finally features! Plus! The super late RUCKUSEE AWARDS!
This entry was posted on January 12, 2019 at 8:50 pm and is filed under Uncategorized. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site. | {
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Wind, sun, biomass and water power could provide up to half of the province's electricity supply over the next few years based on the number of green projects currently on the drawing board, according to results of an industry survey released yesterday by the Ontario Power Authority.
The province's electricity system planning agency found that there are 150 energy developers with 381 projects in various stages of development. Those projects, it calculated, represent 15,128 megawatts of renewable energy supply that has "near-term development potential."
"This level of interest highlights the significant need for new transmission and distribution infrastructure investments to bring potential new supply onto the grid," the agency said.
Most of the projects relate to wind. The survey found there were 164 wind-energy projects representing a total capacity of 13,382 megawatts. Wind, however, is an intermittent resource so only about a third of that capacity could be relied on for management of the grid.
Solar electricity projects numbered 121 and amounted to 1,213 megawatts, while 58 waterpower and 38 bio-energy projects together totalled 533 megawatts. The survey excluded about 1,400 megawatts of mostly undeveloped renewable-energy projects already approved under an earlier standard offer program.
"Those are extremely large numbers," said Richard King, an energy lawyer with Ogilvy Renault.
The survey was conducted after George Smitherman, minister of energy and infrastructure, directed the power authority last September to revisit its 20-year system plan with an eye to "raising the bar" on renewable energy.
As part of that directive the agency was asked to look at building new transmission and distribution infrastructure that could unlock more of the province's renewable-energy potential.
The survey, the agency said, helps to identify areas in the province where grid expansion can bring the biggest benefit.
Surprisingly, 69 per cent of total capacity identified – or 10,408 megawatts – is located in southern Ontario.
A recent report from the North American Electric Reliability Corporation, which regulates the continents bulk power system, said that adding huge amounts of wind, solar and other variable resources to the grid poses a huge challenge for planning authorities. | {
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\section{Introduction}\label{sec:intro}
The simplest and most widely studied class of expanding interval maps, and those which we will concern ourselves, are intermediate $\beta$-transformations, namely transformations of the form $T_{\beta, \alpha} : x \mapsto \beta x + \alpha\mod 1$ acting on $[0,1]$, where $(\beta, \alpha) \in \Delta \coloneqq \{ (b, a) \in \mathbb{R}^{2} \colon b \in (1, 2) \; \text{and} \; a \in [0, 2-b]\}$. This class of transformations have motivated a wealth of results, providing practical solutions to a variety of problems. They arise as Poincar\'e maps of the geometric model of Lorenz differential equations~\cite{MR681294}, Daubechies \textsl{et al.} \cite{1011470} proposed a new approach to analog-to-digital conversion using \mbox{$\beta$-transformations}, and Jitsumatsu and Matsumura \cite{Jitsumatsu2016AT} developed a random number generator using \mbox{$\beta$-transformations}. (This random number generator passed the NIST statistical test suite.) Through their study, many new phenomena have appeared, revealing rich combinatorial and topological structures, and unexpected connections to probability theory, ergodic theory and aperiodic order; see for instance \cite{bezuglyi_kolyada_2003,Komornik:2011,ArneThesis}.
Intermediate $\beta$-transformations also have an intimate link to metric number theory in that they give rise to \mbox{non-integer} based expansions of real numbers. Given $\beta \in (1, 2)$ and $x \in [0, 1/(\beta-1)]$, an infinite word $\omega = \omega_{1}\omega_{2}\cdots$ with letters in the alphabet $\{0, 1\}$ is called a \textsl{$\beta$-expansion} of $x$ if
\begin{align*}
x = \sum_{n \in \mathbb{N}} \omega_{n} \, \beta^{-n}.
\end{align*}
Through iterating the map $T_{\beta, \alpha}$ one obtains a subset of $\{ 0, 1\}^{\mathbb{N}}$ known as the intermediate $\beta$-shift $\Omega_{\beta, \alpha}$, where each $\omega \in \Omega_{\beta, \alpha}$ is a $\beta$-expansion, and corresponds to a unique point in $[\alpha/(\beta-1), 1+\alpha/(\beta-1)]$, see \eqref{eq:commutative_diag} and the commentary following it for further details. By equipping $\Omega_{\beta, \alpha}$ with the left shift map $\sigma$, one obtains a dynamical system which is topologically conjugate to the dynamical system $\mathcal{S}_{\beta,\alpha}=(T_{\beta, \alpha}, [0, 1])$, namely one obtains a symbolic system which possess the same ergodic properties as $\mathcal{S}_{\beta,\alpha}$. Note, in this article, by topologically conjugate we mean that the conjugacy is one-to-one everywhere except on a countable set on which the conjugacy is at most finite to one.
Open dynamical systems, namely systems with holes in the state space through which mass can leak away, have received a lot of attention, see \cite{Ur86,Schme,N09,BBF2014,KKLL} and reference therein. We prove the following correspondence connecting $\mathcal{S}_{\beta,\alpha}$ and open dynamical systems driven by greedy $\beta$-transformations, namely intermediate $\beta$-transformations with no rotation factor, or equivalently when $\alpha = 0$.
\begin{theorem}\label{thm:main}
Given $(\beta, \alpha) \in \Delta$, there exist $t \in [0, 1]$ and $\beta' \in (1, 2)$ with $(T_{\beta, \alpha}, [0, 1])$ topologically conjugate to the open dynamical system $(T_{\beta', 0}\vert_{K^{+}_{\beta',0}(t)}, K^{+}_{\beta',0}(t))$, where
\begin{align*}
K^{+}_{\beta', 0}(t) \coloneqq \{ x \in[0, 1) \colon T_{\beta',0}^{n}(x)\not \in [0,t) \; \textup{for all} \; n \in \mathbb{N}_{0} \}.
\end{align*}
However, the converse does not hold, namely there exist $t \in [0, 1]$ and $\beta' \in (1, 2)$ such that there does not exist a topological conjugation between $(T_{\beta', 0}\vert_{K^{+}_{\beta',0}(t)}, K^{+}_{\beta', 0}(t))$ and $(T_{\beta, \alpha}, [0, 1])$ for any $(\beta, \alpha) \in \Delta$. Moreover, given $\beta' \in (1, 2)$ with $T_{\beta',0}^{n}(1) = 0$, for some $n \in \mathbb{N}$, there exists $\delta \in (0, \beta'^{-1})$, such that to each $t < \delta$ in the bifurcation set
\begin{align*}
E_{\beta',0}^{+} \coloneqq \{ t \in[0,1) \colon T_{\beta', 0}^{n}(t) \not\in [0, t) \; \textup{for all} \; n \in \mathbb{N}_{0} \}
\end{align*}
one may associate a unique $(\beta, \alpha) \in \Delta$ with $(T_{\beta, \alpha}, [0, 1])$ topologically conjugate to $(T_{\beta', 0}\vert_{K^{+}_{\beta',0}(t)}, K^{+}_{\beta',0}(t))$.
\end{theorem}
This result complements \cite[Proposition 3.1 and Theorem 3.5]{bundfuss_kruger_troubetzkoy_2011}. Here, it is shown that every subshift of finite type and any greedy $\beta$-shift encodes a survivor set of $x \mapsto mx \bmod 1$, for some $m \in \mathbb{N}$ with $m \geq 2$. With this and \Cref{thm:main} at hand, we have that any intermediate $\beta$-shift encodes a survivor set of the doubling map.
We employ our characterisation given in \Cref{thm:main} to (1) build a Krieger embedding theorem for intermediate \mbox{$\beta$-transformations}, and (2) obtain new metric and topological results on survivor sets of intermediate \mbox{$\beta$-transformations}.
\begin{enumerate}[label={\rm(\arabic*)},leftmargin=*]
\item {\bfseries A Krieger embedding theorem for intermediate \mbox{$\beta$-transformations.}} Subshifts, such as $\Omega_{\beta, \alpha}$, are to dynamical systems what shapes like polygons and curves are to geometry. Subshifts which can be described by a finite set of forbidden words are called \textsl{subshifts of finite type} and play an essential role in the study of dynamical systems. One reason why subshifts of finite type are so useful is that they have a simple representation using a finite directed graph. Questions concerning the subshift can then often be phrased as questions about the graph's adjacency matrix, making them more tangible, see for instance \cite{LM,brin_stuck_2002} for further details on subshifts of finite type. Moreover, in the case of greedy $\beta$-shifts (that is when $\alpha = 0$), often one first derives results for greedy $\beta$-shifts of finite type, and then one uses an approximation argument to determine the result for a general greedy $\beta$-shift, see for example \cite{DavidFarm2010,LL16}. Here we prove a Krieger embedding theorem for intermediate $\beta$-shifts. Namely, we show the following, complementing the work of \cite{LSSS} where the same result is proven except where the containment property given in Part~(iii) is reversed. Due to this reversed containment, our proof and that of \cite{LSSS}, although both of a combinatorial flavour, are substantially different.
%
\begin{corollary}\label{Cor_1}
Given $(\beta, \alpha) \in \Delta$, there exists a sequence $\{ (\beta_{n}, \alpha_{n}) \}_{n \in \mathbb{N}}$ in $\Delta$ with $\lim_{n\to \infty} (\beta_{n}, \alpha_{n}) = (\beta, \alpha)$ and
\begin{enumerate}[label={\rm(\roman*)}]
\item $\Omega_{\beta_{n}, \alpha_{n}}$ a subshift of finite type,
\item the Hausdorff distance between $\Omega_{\beta, \alpha}$ and $\Omega_{\beta_{n}, \alpha_{n}}$ converges to zero as $n$ tends to infinity, and
\item $\Omega_{\beta_{n}, \alpha_{n}} \subseteq \Omega_{\beta, \alpha}$.
\end{enumerate}
\end{corollary}
%
\noindent These results together with the results of \cite{LSSS} complements the corresponding result for the case when $\alpha = 0$ proven in \cite{P1960} and which asserts that any greedy $\beta$-shift can be approximated from \textsl{above} and \textsl{below} by a greedy $\beta$-shift of finite type.
\vspace{1em}
%
\item {\bfseries Metric and topological results on survivor sets of intermediate \mbox{$\beta$-transformations}.} Via our correspondence theorem (\Cref{thm:main}), we are able to transfer the results of \cite{KKLL} obtained for open dynamical systems driven by greedy $\beta$-transformations to general intermediate $\beta$-transformations. Specifically, we show the following, extending the results of \cite{KKLL} and complementing those of \cite{Ur86,N09}. Here we follow the notation used in \cite{KKLL}, and recall that an infinite word in the alphabet $\{0, 1\}$ is \textsl{balanced} if and only if the number of ones in any two subwords of the same length differ by at most $1$.
%
\begin{corollary}\label{Cor_2}
The bifurcation set $E_{\beta,\alpha}^{+} \coloneqq \{ t \in[0,1) \colon T_{\beta, \alpha}^{n}(t) \not\in [0, t) \; \textup{for all} \; n \in \mathbb{N}_{0} \}$ is a Lebesgue null set. Moreover, if largest lexicographic word in $\Omega_{\beta, \alpha}$ is balanced, then $E_{\beta,\alpha}^{+}$ contains no isolated points.
%
\end{corollary}
\noindent If the largest lexicographic word in $\Omega_{\beta, \alpha}$ is not balanced, then under an additional technical assumption, in \Cref{cor:isoloated_pts}, we show that there exists a $\delta > 0$, such that $E_{\beta,\alpha}^{+} \cap [0, \delta]$ contains no isolated points. Further, letting $K^{+}_{\beta, \alpha}(t)$ denote the survivor set $\{ x \in[0,1) \colon T_{\beta,\alpha}^{n}(x)\not \in [0,t) \; \text{for all} \; n \in \mathbb{N}_{0} \}$, we have:
%
\begin{corollary}\label{Cor_3}
The dimension function $\eta_{\beta, \alpha} \colon t \mapsto \dim_{\mathcal{H}}(K_{\beta,\alpha}^{+}(t))$ is a Devil staircase function, that is, $\eta_{\beta,\alpha}(0) = 1$, $\eta_{\beta,\alpha}((1-\alpha)/\beta) = 0$, $\eta_{\beta, \alpha}$ is decreasing, and $\eta_{\beta,\alpha}$ is constant Lebesgue almost everywhere.
\end{corollary}
%
\noindent With \Cref{Cor_1,Cor_3} at hand, we can also prove the following.
%
\begin{corollary}\label{Cor_E_beta_alpha}
The bifurcation set $E_{\beta,\alpha}^{+}$ has full Hausdorff dimension.
\end{corollary}
%
\end{enumerate}
The sets $K^{+}_{\beta,\alpha}(t)$ can be seen as a level sets of the set of badly approximable numbers in non-integer bases, that is,
\begin{align*}
\mathrm{BAD}_{\beta, \alpha}(0) \coloneqq \{ x \in [0,1] \colon 0 \not\in \overline{\{T_{\beta, \alpha}^n(x) \colon n\geq 0\}}\} = \bigcup_{t \in (0, 1)} K^{+}_{\beta, \alpha}(t).
\end{align*}
Moreover, for $\xi \in [0,1]$ one can study the more general set
\begin{align*}
\mathrm{BAD}_{\beta, \alpha}(\xi) \coloneqq \{ x \in [0,1] \colon \xi \not\in \overline{\{T_{\beta, \alpha}^n(x) \colon n\geq 0\}}\},
\end{align*}
which, by \Cref{Cor_3}, is a set of full Hausdorff dimension.
When $\alpha = 0$, F\"arm, Persson and Schmeling \cite{DavidFarm2010} and later Hu and Yu \cite{HY} study these sets and showed that they are winning, and hence that they have the large intersection property. To our knowledge, the present work, is the first to consider the case $\alpha \neq 0$. Before stating our results on $\mathrm{BAD}_{\beta, \alpha}(\xi)$, we recall the notion of a winning set.
In the 1960s Schmidt~\cite{S} introduced a topological game in which two players take turns in choosing balls that are a subset of the previously chosen ball. There is a target set $S$ and the objective of Player~$1$ is to make sure that the point that is present in every ball chosen during the game is in $S$. The objective of Player~$2$ is to prevent this. A set is called winning when Player~$1$ can always build a winning strategy no matter how Player~2 plays.
\begin{definition}
Let $\alpha$ and $\gamma\in (0,1)$ be fixed and suppose we have two players, Player~1 and Player~2. Let Player~2 choose a closed initial interval $B_1\subset [0,1]$ and let Player~1 and Player~2 choose nested closed intervals such that $B_1 \supset W_1 \supset B_2 \supset W_2 \supset \ldots$ and $|W_{n+1}|=\alpha |B_n|$ and $|B_{n+1}|=\gamma |W_n|$. A set $S$ is called $(\alpha,\gamma)$-winning if there is a strategy for Player~2 to ensure that $\bigcap_{i\in \mathbb{N}} W_i \subset S$. The set $S$ is called $\alpha$-winning if it is $(\alpha,\gamma)$-winning for all $\gamma\in (0,1)$ and is called winning if it is $\alpha$-winning for some $\alpha\in (0,1)$.
\end{definition}
A key attribute of winning which makes it an interesting property to study is that winning sets have full Hausdorff dimension~\cite{S}. Another, is that it persists under taking intersections, that is, for two winning sets their intersection is again winning, and hence of full Hausdorff dimension~\cite{S}; this is not true in general for sets of full Hausdorff dimension. We also note, the property of winning is preserved under bijective affine transformations.
\begin{theorem}\label{thm:main_2}
Given $(\beta, \alpha) \in \Delta$ with $\Omega_{\beta, \alpha}$ a subshift of finite type, and $\xi \in [0, 1]$, the set $\mathrm{Bad}_{\beta,\alpha}(\xi)$ is winning.
\end{theorem}
We remark that in \cite{JT2009} a similar result for $C^{2}$-expanding Markov circle maps was proven, but that intermediate $\beta$-transformations do not fall into this regime. Further, with \Cref{thm:main_2} at hand and with \cite[Theorem 1]{DavidFarm2010} in mind, we conjecture that $\mathrm{Bad}_{\beta,\alpha}(\xi)$ is winning for all $(\beta, \alpha) \in \Delta$ and $\xi \in [0, 1]$.
Our work is organised as follows. In \Cref{sec:prelim} we we present necessary definitions, preliminaries and auxiliary results. \Cref{sec:proof_thm_1_1,sec:proof_thm_1_6} are respectively devoted to proving \Cref{thm:main,thm:main_2}, and \Cref{sec:proof_cor_1_2,sec:proof_cor_1_3_4} respectively contain the proofs of \Cref{Cor_1}, and \Cref{Cor_2,Cor_3}. Additionally, in \Cref{sec:proof_cor_1_3_4}, we demonstrate how our theory may be used to numerically compute the Hausdorff dimension of $K_{\beta,\alpha}^{+}(t)$.
\section{Notation and preliminaries}\label{sec:prelim}
\subsection{Subshifts}
Let $m \geq 2$ denote a natural number and set $\Lambda = \{0, 1, \ldots, m-1\}$. We equip the space $\Lambda^\mathbb{N}$ of infinite sequences indexed by $\mathbb{N}$ with the topology induced by the \textsl{word metric} $\mathscr{D} \colon \Lambda^\mathbb{N} \times \Lambda^\mathbb{N} \to \mathbb{R}$ given by
\begin{align*}
\mathscr{D}(\omega, \nu) \coloneqq
\begin{cases}
0 & \text{if} \; \omega = \nu,\\
2^{- \lvert\omega \wedge \nu\rvert + 1} & \text{otherwise}.
\end{cases}
\end{align*}
Here, $\rvert \omega \wedge \nu \lvert \coloneqq \min \, \{ \, n \in \mathbb{N} \colon \omega_{n} \neq \nu_n \}$, for $\omega$ and $\nu \in \Lambda^{\mathbb{N}}$ with $\omega \neq \nu$, where for an element $\omega \in \Lambda^{\mathbb{N}}$ we write $\omega=\omega_1\omega_2\cdots$. Note, when equipping $\Lambda$ with the discrete topology, the topology induced by $\mathscr{D}$ on $\Lambda^{\mathbb{N}}$ coincides with the product topology on $\Lambda^{\mathbb{N}}$. We let $\sigma \colon \Lambda^{\mathbb{N}} \to \Lambda^{\mathbb{N}}$ denote the \textsl{left-shift map} defined by $\sigma(\omega_{1} \omega_{2} \cdots) \coloneqq \omega_{2} \omega_{3} \cdots$, and for $n \in \mathbb{N}$, we set $\omega\rvert_{n} = \omega_{1} \omega_{2} \cdots \omega_{n}$. A \textsl{subshift} is any closed set $\Omega \subseteq \Lambda^\mathbb{N}$ with $\sigma(\Omega) \subseteq \Omega$. Given a subshift $\Omega$, we set $\Omega\vert_{0} = \{ \varepsilon\}$, where $\varepsilon$ denotes the empty word, and for $n \in \mathbb{N}$, we set
\begin{align*}
\Omega\lvert_{n} \coloneqq \left\{ \omega_{1} \cdots \omega_{n} \in \Lambda^{n} \colon \,\text{there exists} \; \xi \in \Omega \; \text{with} \; \xi|_n = \omega_{1} \cdots \omega_{n} \right\}
\end{align*}
and write $\Omega^{*} \coloneqq \bigcup_{n \in \mathbb{N}_{0}} \Omega\lvert_{n}$ for the collection of all finite words. We denote by $\lvert \Omega\vert_{n} \rvert$ the cardinality of $\Omega\vert_{n}$, and for $\omega \in \Omega\vert_{n}$, we set $\lvert \omega \rvert = n$. We extend the domain of $\sigma$ to $\Omega^{*}$, by setting $\sigma(\varepsilon) \coloneqq \varepsilon$, and for $n \in \mathbb{N}$, letting
\begin{align*}
\sigma(\omega_{1} \omega_{2} \cdots \omega_{n}) \coloneqq
\begin{cases}
\omega_{2} \omega_{3} \cdots \omega_{n} & \text{if} \; n \neq 1,\\
\varepsilon & \text{otherwise}.
\end{cases}
\end{align*}
For $\omega = \omega_{1} \cdots \omega_{\lvert \omega \rvert} \in \Omega^{*}$ and $\xi = \xi_{1} \xi_{2} \cdots \in \Omega \cup \Omega^{*}$ we denote the concatenation $\omega_{1} \cdots \omega_{\lvert \omega \rvert} \ \xi_{1} \xi_{2} \cdots$ by $\omega \ \xi$.
\begin{definition}
A subshift $\Omega$ is said to be \textsl{of finite type} if there exists $M \in \mathbb{N}$ such that, $\omega_{n - M + 1} \cdots \omega_{n} \ \xi_{1} \cdots \xi_{m} \in \Omega^{*}$, for all $\omega_{1} \cdots \omega_{n}$ and $\xi_{1} \cdots \xi_{m} \in \Omega^{*}$ with $n, m \in \mathbb{N}$ and $n \geq M$, if and only if $\omega_{1} \cdots \omega_{n} \ \xi_{1} \cdots \xi_{m} \in \Omega^{*}$.
\end{definition}
The following result gives an equivalent condition for when a subshift is of finite type.
\begin{theorem}[{\cite[Theorem 2.1.8]{LM}}]
A subshift $\Omega \subseteq \Lambda^{\mathbb{N}}$ is of finite type if and only if there exists a finite set $F \subset \Omega^{*}$ with $\Omega = \mathcal{X}_{F}$, where $\mathcal{X}_{F} \coloneqq \{ \omega \in \Lambda^{\mathbb{N}} \colon \sigma^{m}(\omega)\vert_{\lvert \xi \rvert} \neq \xi \; \text{for all} \; \xi \in F \; \text{and} \; m \in \mathbb{N}\}$.
\end{theorem}
Two subshifts $\Omega$ and $\Psi$ are said to be \textsl{topologically conjugate} if there exists a $\phi \colon \Omega \to \Psi$ that is surjective, one-to-one everywhere except on a countable set on which it is at most finite-to-one, and $\sigma \circ \phi(\omega) = \phi \circ \sigma(\omega)$ for all $\omega \in \Omega$. We call $\phi$ the \textsl{conjugacy}. In the case that $m = 2$, a particular conjugacy which we will make use of is the \textsl{reflection map} $R$ defined by $R(\omega_{1} \omega_{2} \cdots) = (1-\omega_{1})(1-\omega_{2})\cdots$ for $\omega = \omega_{1} \omega_{2} \cdots \in \{0,1\}^{\mathbb{N}}$. This concept of two subshifts being topologically conjugate, naturally extends to general dynamical systems, see for instance \cite{LM,brin_stuck_2002}.
An infinite word $\omega = \omega_{1} \omega_{2} \cdots \in \Lambda^{\mathbb{N}}$ is called \textsl{periodic} with \textsl{period} $n \in \mathbb{N}$ if and only if, for all $m \in \mathbb{N}$, we have $\omega_{1} \cdots \omega_{n} = \omega_{(m - 1)n + 1} \cdots \omega_{m n}$, in which case we write $\omega = \omega\vert_{n}^{\infty}$, and denote the smallest period of $\omega$ by $\operatorname{per}(\omega)$. Similarly, an infinite word $\omega = \omega_{1} \omega_{2} \cdots \in \Lambda^{\mathbb{N}}$ is called \textsl{eventually periodic} with \textsl{period} $n \in \mathbb{N}$ if there exists $k \in \mathbb{N}$ such that, for all $m \in \mathbb{N}$, we have $\omega_{k+1} \cdots \omega_{k+n} = \omega_{k+(m - 1)n + 1} \cdots \omega_{k+ m n}$, in which case we write $\omega = \omega_{1} \cdots \omega_{k} (\omega_{k+1} \cdots \omega_{k+n})^\infty$.
\subsection{Intermediate \texorpdfstring{$\beta$}{beta}-shifts}\label{sec:beta-shifts}
For $(\beta, \alpha) \in \Delta$ we set $p = p_{\beta, \alpha} = (1-\alpha)/\beta$ and define the \textsl{upper $T_{\beta, \alpha}$-expansion} $\tau_{\beta, \alpha}^{+}(x)$ of $x \in [0, 1]$ to be the infinite word $\omega_{1} \omega_{2} \cdots \in \{ 0, 1\}^{\mathbb{N}}$, where, for $n \in \mathbb{N}$,
\begin{align}\label{eq:upper_kneading}
\omega_{n} \coloneqq
\begin{cases}
0 & \quad \text{if } T_{\beta,\alpha}^{n-1}(x) < p,\\
1 & \quad \text{otherwise,}
\end{cases}
\end{align}
and define the \textsl{lower $T_{\beta, \alpha}$-expansion} $x$ to be $\tau^{-}_{\beta, \alpha}(x) \coloneqq \lim_{y \nearrow x} \tau_{\beta,\alpha}^{+}(y)$. Note, one can also define $\tau^{-}_{\beta, \alpha}(x)$ analogously to $\tau^{+}_{\beta, \alpha}(x)$ by using the map $T_{\beta,\alpha}^{-} \colon x \mapsto \beta x + \alpha$ if $x \leq p$, and $x \mapsto \beta x + \alpha - 1$ otherwise, in replace of of $T_{\beta, \alpha}$, and by changing the \textsl{less than}, to \textsl{less than or equal to} in \eqref{eq:upper_kneading}, see \cite[Section 2.2]{LSSS}. With this in mind, and for ease of notation, sometimes we may write $T_{\beta,\alpha}^{+}$ for $T_{\beta, \alpha}$.
We denote the images of $[0,1)$ under $\tau_{\beta, \alpha}^{+}$ by $\Omega^{+}_{\beta, \alpha}$, the image of $(0,1]$ under $\tau_{\beta, \alpha}^{-}$ by $\Omega^{-}_{\beta, \alpha}$, and set $\Omega_{\beta, \alpha} \coloneqq \Omega_{\beta, \alpha}^{+} \cup \Omega_{\beta, \alpha}^{-}$. We refer to $\Omega_{\beta, \alpha}$ as an intermediate $\beta$-shift and define the \textsl{upper} and \textsl{lower kneading invariants} of $\Omega_{\beta,\alpha}$ to be the infinite words $\tau^{\pm}_{\beta, \alpha}(p)$, respectively. The following result shows that $\tau^{\pm}_{\beta, \alpha}(p)$ completely determine $\Omega_{\beta,\alpha}$.
\begin{theorem}[{\cite{P1960,HS:1990,AM:1996,KS:2012,BHV:2011}}]\label{thm:Structure}
For $(\beta, \alpha) \in \Delta$, the spaces $\Omega_{\beta, \alpha}^{\pm}$ are completely determined by upper and lower kneading invariants of $\Omega_{\beta, \alpha}$, namely
\begin{align*}
\Omega_{\beta, \alpha}^{+} &= \{ \omega \in \{ 0, 1\}^{\mathbb{N}} \colon \tau_{\beta, \alpha}^{+}(0) \preceq \sigma^{n}(\omega) \prec \tau_{\beta, \alpha}^{-}(p) \; \textup{or} \; \tau_{\beta, \alpha}^{+}(p) \preceq \sigma^{n}(\omega) \prec \tau_{\beta, \alpha}^{-}(1) \; \textup{for all} \; n \in \mathbb{N}_{0} \},\\
%
\Omega_{\beta, \alpha}^{-} &= \{ \omega \in \{ 0, 1\}^{\mathbb{N}} \colon \tau_{\beta, \alpha}^{+}(0) \prec \sigma^{n}(\omega) \preceq \tau_{\beta, \alpha}^{-}(p) \; \textup{or} \; \tau_{\beta, \alpha}^{+}(p) \prec \sigma^{n}(\omega) \preceq \tau_{\beta, \alpha}^{-}(1) \; \textup{for all} \; n \in \mathbb{N}_{0} \}.
\end{align*}
Here, $\prec$, $\preceq$, $\succ$ and $\succeq$ denote the lexicographic orderings on $\{ 0 ,1\}^{\mathbb{N}}$. Moreover, the cardinality of $\Omega_{\beta, \alpha}^{\pm}$ is equal to that of the continuum, and $\Omega_{\beta, \alpha}$ is closed with respect to the metric $\mathscr{D}$. Hence, $\Omega_{\beta, \alpha}$ is a subshift.
\end{theorem}
This result establishes the importance of the kneading invariants of $\Omega_{\beta, \alpha}$, for a given $(\beta, \alpha) \in \Delta$, and so it is natural to ask, for a fixed $\beta \in (1, 2)$, if they are monotonic or continuous in $\alpha$. The following proposition answers this.
\begin{proposition}[{\cite{BHV:2011,Cooperband2018ContinuityOE}}]\label{prop:mon_cont_kneading}
Let $\beta \in (1,2)$ be fixed.
\begin{enumerate}[label={\rm(\arabic*)}]
\item The maps $a \mapsto \tau_{\beta,a}^{\pm}(p_{\beta, a})$ are strictly increasing with respect to the lexicographic ordering.
\item The map $a \mapsto \tau_{\beta, a}^{+}(p_{\beta, a})$ is right continuous, and the map $a \mapsto \tau_{\beta,a}^{-}(p_{\beta, a})$ is left continuous.
\item If $\alpha \neq 0$ and $\tau_{\beta, \alpha}^{+}(p_{\beta, \alpha})$ is not periodic, then $a \mapsto \tau_{\beta,a}^{+}(p_{\beta, a})$ is continuous at $\alpha$, and if $\alpha \neq \beta-1$ and $\tau_{\beta,\alpha}^{-}(p_{\beta, \alpha})$ is not periodic, then $a \mapsto \tau_{\beta,a}^{-}(p_{\beta, a})$ is continuous at $\alpha$.
\item If $\tau_{\beta,\alpha}^{+}(p_{\beta, \alpha})$ is periodic with period $M$, for a given $\alpha \in (0, 2-\beta]$, then given $m \in \mathbb{N}$, there exists a real number $\delta > 0$ so that, $\tau_{\beta,\alpha-\delta'}^{+}(p_{\beta, \alpha-\delta'})\vert_{m} = \tau_{\beta,\alpha}^{+}(p_{\beta, \alpha})\vert_{M}\tau_{\beta,\alpha}^{-}(p_{\beta, \alpha})\vert_{m-M}$, for all $\delta' \in (0, \delta)$.
\item If $\tau_{\beta,\alpha}^{-}(p_{\beta, \alpha})$ is periodic with period $M$, for a given $\alpha \in [0, 2-\beta)$, then given $m \in \mathbb{N}$, there exists a real number $\delta > 0$ so that, $\tau_{\beta,\alpha+\delta'}^{-}(p_{\beta, \alpha+\delta'})\vert_{m} = \tau_{\beta,\alpha}^{-}(p_{\beta, \alpha})\vert_{M}\tau_{\beta,\alpha}^{+}(p_{\beta, \alpha})\vert_{m-M}$, for all $\delta' \in (0, \delta)$.
\end{enumerate}
\end{proposition}
Another natural question to ask is when do two infinite words in the alphabet $\{0, 1\}$ give rise to kneading invariants of an intermediate $\beta$-shift. This question was addressed in \cite{barnsley_steiner_vince_2014} where the following solution was derived and for which we require the following notation. Given $\omega$ and $\nu \in \{0,1\}^\mathbb{N}$ with $\sigma(\nu) \preceq \omega \preceq \nu \preceq \sigma(\omega)$ we set
\begin{align*}
\Omega^{+}(\omega,\nu) &\coloneqq \{ \xi \in \{0,1\}^{\mathbb{N}} \colon \sigma(\nu) \preceq \sigma^{n}(\xi) \prec \omega \; \text{or} \; \nu \preceq \sigma^{n}(\xi) \prec \sigma(\omega) \; \text{for all} \; n \in \mathbb{N}_{0} \},\\
\Omega^{-}(\omega,\nu) &\coloneqq \{ \xi \in \{0,1\}^{\mathbb{N}} \colon \sigma(\nu) \prec \sigma^n(\xi) \preceq \omega \; \text{or} \; \nu \prec \sigma^{n} (\xi) \preceq \sigma(\omega) \; \text{for all} \; n \in \mathbb{N}_{0} \}.
\end{align*}
\begin{theorem}[{\cite{barnsley_steiner_vince_2014}}]\label{thm:BSV14}
Two infinite words $\omega = \omega_{1}\omega_{2}\cdots$ and $\nu=\nu_{1}\nu_{2}\cdots \in \{0,1\}^\mathbb{N}$ are kneading invariants of an intermediate $\beta$-shift $\Omega_{\beta, \alpha}$, for some $(\beta, \alpha) \in \Delta$ if and only if the following four conditions hold.
\begin{enumerate}[label={\rm(\arabic*)}]
\item $\omega_1=0$ and $\nu_1=1$,
\item $\omega\in\Omega^-(\omega, \nu) $ and $\nu\in\Omega^+(\omega, \nu)$,
\item $\lim_{n\to\infty} \log(|\Omega^+|_n)>0$, and
\item if $\omega,\nu\in\{\xi,\zeta \}^\mathbb{N}$ for two finite words $\xi$ and $\zeta$ in the alphabet $\{0,1\}$ with length greater than or equal to three, such that $\xi_1\xi_2=01$ , $\zeta_1\zeta_2=10$, $\xi^\infty \in \Omega^-(\xi^\infty, \zeta^\infty)$ and $\zeta^\infty \in \Omega^{+}(\xi^\infty, \zeta^\infty)$, then $\omega=\xi^\infty$ and $\nu=\zeta^\infty$.
\end{enumerate}
\end{theorem}
This result together with \Cref{thm:Structure} can be seen as a generalisation of the following seminal result of Parry.
\begin{theorem}[{\cite[Corollary 1]{P1960}}]\label{thm:Parry_converse}
If $\omega \in \{0, 1\}^{\mathbb{N}}$ with $\sigma^{n}(\omega) \neq 0^\infty$ for all $n \in \mathbb{N}$, then there exists a $\beta \in (1,2)$ with $\omega = \tau_{\beta, 0}^{-}(1)$ if and only if $\sigma^{m}(\omega) \preceq \omega$ for all $m \in \mathbb{N}$.
\end{theorem}
Combining this result with \Cref{thm:Structure}, we obtain the following which will be utilised in our proof of \Cref{thm:main}.
\begin{corollary}\label{cor:From_greedy_to_intermediate}
Given $(\beta, \alpha) \in \Delta$, there exists $\beta' \in (1,2)$ such that $\tau_{\beta,\alpha}^{-}(1) = \tau_{\beta',0}^{-}(1)$.
\end{corollary}
In the sequel we will also make use of the projection $\pi_{\beta, \alpha} \colon \{ 0, 1 \}^{\mathbb{N}} \to [0, 1]$ defined by
\begin{align*}
\pi_{\beta, \alpha}(\omega_{1} \omega_{2} \cdots) \coloneqq \frac{\alpha}{1 - \beta} + \sum_{k \in \mathbb{N}} \frac{\omega_{k}}{\beta^k}.
\end{align*}
We note that $\pi_{\beta, \alpha}$ is linked to the iterated function systems $( [0, 1]; f_{0} \colon x \mapsto \beta^{-1}x, f_{1} \colon x \mapsto \beta^{-1}(x+1)$ via the equality
\begin{align*}
\pi_{\beta, \alpha}(\omega_{1} \omega_{2} \cdots)
= \alpha / (1-\beta) + \lim_{n \to \infty} f_{\omega_{1}} \circ \cdots \circ f_{\omega_{n}}([0, 1]),
\end{align*}
and the iterated function system $( [0, 1]; f_{\beta, \alpha, 0} \colon x \mapsto \beta^{-1}x - \alpha\beta^{-1}, f_{\beta, \alpha, 1} \colon x \mapsto \beta^{-1}x - (\alpha-1)\beta^{-1}$ via the equality
\begin{align}\label{eq:alt_IFS}
\pi_{\beta, \alpha}(\omega_{1} \omega_{2} \cdots)
=\lim_{n \to \infty} f_{\beta,\alpha,\omega_{1}} \circ \cdots \circ f_{\beta,\alpha,\omega_{n}}([0, 1]).
\end{align}
We refer the reader to \cite{F:1990} for further details on iterated function systems. An important property of $\pi_{\beta, \alpha}$ is that the following diagrams commute.
\begin{align}\label{eq:commutative_diag}
\begin{aligned}
\xymatrix@C+2pc{
\Omega^{+}_{\beta, \alpha}
\ar@/_/[d]_{\pi_{\beta, \alpha}}
\ar[r]^{\sigma} &
\Omega_{\beta, \alpha}^{+}
\ar@/^/[d]^{\pi_{\beta, \alpha}} \\
{[0, 1)}
\ar@/_/[u]_{\tau_{\beta,\alpha}^{+}}
\ar[r]_{T_{\beta, \alpha}} &
\ar@/^/[u]^{\tau_{\beta,\alpha}^{+}}
[0, 1)}
\end{aligned}
\qquad\qquad
\begin{aligned}
\xymatrix@C+2pc{
\Omega^{-}_{\beta, \alpha}
\ar@/_/[d]_{\pi_{\beta, \alpha}}
\ar[r]^{\sigma} &
\Omega_{\beta, \alpha}^{-}
\ar@/^/[d]^{\pi_{\beta, \alpha}} \\
{(0, 1]}
\ar@/_/[u]_{\tau_{\beta,\alpha}^{-}}
\ar[r]_{T^{-}_{\beta, \alpha}} &
\ar@/^/[u]^{\tau_{\beta,\alpha}^{-}}
(0, 1]}
\end{aligned}
\end{align}
This result is verifiable from the definitions of the involved maps and a sketch of a proof can be found in \cite{BHV:2011}. It also yields that each $\omega \in \Omega_{\beta, \alpha}$ is a $\beta$-expansion, and corresponds to the unique point
\begin{align*}
\sum_{k \in \mathbb{N}} \frac{\omega_{k}}{\beta^{k}} = \pi_{\beta,\alpha}(\omega) - \frac{\alpha}{1-\beta}
\end{align*}
in $[\alpha/(\beta-1), 1+\alpha/(\beta-1)]$. A particular expansion which we will make use of is $\tau_{\beta,0}^{-}(1)$ which is referred to as the \textsl{quasi-greedy $\beta$-expansion of $1$}.
The commutativity of the diagrams given in \eqref{eq:commutative_diag} also implies that the dynamical systems $(\Omega_{\beta, \alpha}, \sigma)$ and $([0, 1), T_{\beta, \alpha})$ are topologically conjugate. This in tandem with \Cref{thm:Structure}, yields that the upper and lower kneading invariants completely determine the dynamics of $T_{\beta, \alpha}$. Additionally, we have the following.
\begin{theorem}[{\cite{GH,BHV:2011}}]\label{thm:Laurent}
Let $(\beta, \alpha) \in \Delta$ be fixed. If $\omega = \omega_{1}\omega_{2} \cdots $ and $\nu = \nu_{1} \nu_{2} \cdots $, respectively, denote the upper and lower kneading invariants of $\Omega_{\beta, \alpha}$, then $\beta$ is the maximal real root of the Laurent series
\begin{align*}
\pi_{z,\alpha}(\omega) - \pi_{z,\alpha}(\nu)
= \sum_{k \in \mathbb{N}} (\omega_{k} - \nu_{k})\,z^{-k}.
\end{align*}
\end{theorem}
The above allows us to transfer results on the dynamical system $(\Omega_{\beta, \alpha}, \sigma)$ to $([0, 1], T_{\beta, \alpha})$ and vice versa. We will utilise this in the proofs of our main results, and thus will make use of the following symbolic representations of $K_{\beta, \alpha}^{+}(t)$ and $E_{\beta, \alpha}^{+}$ defined in \Cref{sec:intro}. For $(\beta, \alpha) \in \Delta$ and $t \in [0,1]$, we set
\begin{align*}
\mathcal{K}^{+}_{\beta,\alpha}(t) &\coloneqq \{ \omega \in \{0,1\}^\mathbb{N} \colon \tau_{\beta,\alpha}^{+}(t) \preceq \sigma^{n}(\omega) \prec \tau_{\beta,\alpha}^{-}(1) \; \text{for all} \; n \in \mathbb{N}_{0} \}
%
\intertext{and we let}
%
\mathcal{E}_{\beta,\alpha}^{+} &\coloneqq \{ \omega \in \{0,1\}^\mathbb{N} \colon \omega \preceq \sigma^{n}(\omega) \prec \tau_{\beta,\alpha}^{-}(1) \; \text{for all} \; n \in \mathbb{N}_{0} \}.
\end{align*}
In addition to this, we will utilise the following Ledrappier-Young formula due to Raith \cite{R}. For $t \in (0,1]$,
\begin{align}\label{eq:entopen}
\dim_{H}(K_{\beta,\alpha}(t)) = \frac{h_{\operatorname{top}}(T_{\beta,\alpha}\vert_{K_{\beta,\alpha}(t)})}{\log(\beta)}.
\end{align}
where $K_{\beta,\alpha}(t) \coloneqq \{ x \in[0,1] \colon T_{\beta,\alpha}^{n}(x)\not \in (0,t) \; \text{for all} \; n \in \mathbb{N}_{0} \}$ and where $h_{\operatorname{top}}(T_{\beta,\alpha}\vert_{K^{+}_{\beta,\alpha}(t)})$ denotes the topological entropy of the dynamical system $(K^{+}_{\beta,\alpha}(t), T_{\beta,\alpha}\vert_{K^{+}_{\beta,\alpha}(t)})$. Here, for a given subset $L \subseteq [0, 1]$ we set
\begin{align*}
h_{\operatorname{top}}(T_{\beta,\alpha}\vert_{L}) \coloneqq \lim_{n \to \infty} \frac{\log (\lvert \tau_{\beta,\alpha}^{+}(L)\vert_{n}\rvert)}{n}
\end{align*}
see \cite{LM,brin_stuck_2002,Walters_1982}, and reference therein, for further details on topological entropy. More specifically, in \Cref{sec:proof_cor_1_3_4}, we apply the following result, which is a consequence of Raith's Ledrappier-Young formula \eqref{eq:entopen} and \cite[Proposition 2.6]{KKLL}.
\begin{proposition}\label{prop:ent}
For $(\alpha, \beta) \in \Delta$ and $t \in [0,1]$,
\begin{align}\label{eq:entropy}
h_{\operatorname{top}}(T_{\beta,\alpha}\vert_{K_{\beta,\alpha}(t)}) = h_{\operatorname{top}}(T_{\beta,\alpha}\vert_{K^{+}_{\beta,\alpha}(t)})
\end{align}
and hence
\begin{align}\label{eq:ent}
\dim_{H}(K^{+}_{\beta,\alpha}(t)) = \frac{h_{\operatorname{top}}(T_{\beta,\alpha}\vert_{K^{+}_{\beta,\alpha}(t)})}{\log(\beta)}.
\end{align}
\end{proposition}
\begin{proof}
Since the set $K_{\beta,\alpha}(t)\backslash K^{+}_{\beta,\alpha}(t)$ is countable, we have $\dim_H(K_{\beta,\alpha}(t))=\dim_H(K^{+}_{\beta,\alpha}(t))$. The Ledrappier-Young formula given in \eqref{eq:ent} is therefore a direct consequence of \eqref{eq:entopen} and \eqref{eq:entropy}. The proof of \eqref{eq:entropy}, follows from a small adaptation of the proof of \cite[Proposition 2.6]{KKLL}, which we present below.
If $0 \not\in K_{\beta,\alpha}(t)$ or $t=0$, then $\mathcal{K}^{+}_{\beta,\alpha}(t) =\mathcal{K}_{\beta,\alpha}(t)$, and if $t\not \in E^{+}_{\beta,\alpha}$ then there exists a $t^*>t$ with $K^{+}_{\beta,\alpha}(t)=K^{+}_{\beta,\alpha}(t^*)$. The first of these two statements follows directly from the definition of $K_{\beta,\alpha}(t)$ and $K_{\beta,\alpha}^{+}(t)$, and the second can be seen to hold as follows. For every $t \not\in E^{+}_{\beta,\alpha}$ there exists a smallest natural number $N$ such that $T_{\beta,\alpha}^N(t) \in [0,t)$. Let $\delta = \min \{ p_{\beta, \alpha} - T_{\beta,\alpha}^{n}(t) \colon n \in \{ 0, 1, \dots, N-1 \} \; \text{and} \; T_{\beta,\alpha}^{n}(t) < p_{\beta, \alpha} \}$ and set $\varepsilon = \min\{t-T_{\beta,\alpha}^N(t), \delta\}/\beta^{N}$. By construction, for all $s\in (t,t+\varepsilon)$, we have that $T_{\beta,\alpha}^N(s)\in [0,t) \subset [0, s)$, and hence that $s\not\in K^{+}_{\beta,\alpha}(s)$ and $s\not\in K^{+}_{\beta,\alpha}(t) $. This implies that $K^{+}_{\beta,\alpha}(t)= K^{+}_{\beta,\alpha}(s)$ for all $s\in (t,t+\varepsilon)$. In fact letting $t^{*} = \inf \{ s \in E^{+}_{\beta,\alpha} \colon s > t \}$, a similar justification yields that $K^{+}_{\beta,\alpha}(t)= K^{+}_{\beta,\alpha}(s)$ for all $s \in (t,t^*)$.
All-in-all, this all implies that $t^* \in E^{+}_{\beta,\alpha}$.
Therefore, it suffices to show that, if $t \in E^{+}_{\beta,\alpha} \setminus \{0\}$ and $0\in K_{\beta,\alpha}(t)$, then $h_{\operatorname{top}}(T_{\beta,\alpha}\vert_{K_{\beta,\alpha}(t)})= h_{\operatorname{top}}(T_{\beta,\alpha}\vert_{K^{+}_{\beta,\alpha}(t)})$. To this end, suppose that $t\in E^{+}_{\beta,\alpha} \setminus \{0\}$. In which case $\sigma^{n}(\tau^{+}_{\beta,\alpha}(t)) \neq \tau^{+}_{\beta,\alpha}(0)$, and so setting
\begin{align*}
\mathcal{K}_{\beta,\alpha}^{0}(t) &\coloneqq \{ \omega \in \{0,1\}^{\mathbb{N}} \colon \text{there exists} \; m \in \mathbb{N}_{0} \; \text{with} \; \sigma^{m}(\omega) = \tau^{+}_{\beta,\alpha}(0) \; \text{and} \; \tau^{+}_{\beta,\alpha}(t) \prec \sigma^{n}(\omega) \prec \tau^{-}_{\beta,\alpha}(1) \; \text{for all} \; n \in \mathbb{N}_{0} \}
\end{align*}
and letting $\mathcal{K}_{\beta,\alpha}(t)$ denote the symbolic representation of $K_{\beta,\alpha}(t)$, namely letting
\begin{align*}
\mathcal{K}_{\beta,\alpha}(t) &\coloneqq \{ \omega \in \{0,1\}^{\mathbb{N}} \colon \sigma^{n}(\omega) = \tau^{+}_{\beta,\alpha}(0) \; \text{or} \; \tau^{+}_{\beta,\alpha}(t) \preceq \sigma^{n}(\omega) \prec \tau^{-}_{\beta,\alpha}(1) \; \text{for all} \; n \in \mathbb{N}_{0} \},
\end{align*}
we have that $\mathcal{K}_{\beta,\alpha}(t)\setminus \mathcal{K}_{\beta,\alpha}^{+}(t) = \mathcal{K}_{\beta,\alpha}^{0}(t)$. Let $k \in \mathbb{N}$ be fixed, and let $\zeta \in \mathcal{K}_{\beta,\alpha}^{0}(t)\vert_{k}$ with $\zeta \neq \tau^{+}_{\beta,\alpha}(0)\vert_{k}$. By construction, there exists $\omega \in \mathcal{K}_{\beta,\alpha}^{0}(t)$ with $\omega\vert_{k} = \zeta$. Let $j$ be the smallest natural number such that $\sigma^{j}(\omega) = \tau^{+}_{\beta,\alpha}(0)$ and set $\nu = \omega\vert_{j-1}\xi_{j}$, where $\xi_{j}$ denotes the $j$-th letter of $\tau^{+}_{\beta,\alpha}(0)$. Observe that
\begin{align}\label{eq:ent_proof}
\tau^{+}_{\beta,\alpha}(t)\vert_{j-i} \preceq \sigma^{i}(\nu) \preceq \tau^{+}_{\beta,\alpha}(1)\vert_{j-i}
\end{align}
for all $i \in \{0, 1, \dots, j-1\}$. Let $i^{*} \in \{0, 1, \dots, j-1\}$ be the smallest integer such that $\sigma^{i^{*}}(\nu) = \tau^{+}_{\beta,\alpha}(t)\vert_{j-i^{*}}$, and if strict inequality holds in the lower bound of \eqref{eq:ent_proof} for all $i \in \{0, 1, \dots, j-1\}$, then set $i^{*}=j$. By the minimality of $i^{*}$, we have $\nu \, \sigma^{j-i^{*}+1}(\tau^{+}_{\beta,\alpha}(t)) = \nu\vert_{i^{*}} \tau^{+}_{\beta,\alpha}(t) \in \mathcal{K}_{\beta,\alpha}^{+}(t)$. Noting, $\nu\vert_{j-1} = \zeta\vert_{j-1}$ if $j \leq k$, and $\nu\vert_{k} = \zeta\vert_{k}$ if $j \geq k+1$, we have that
\begin{align}\label{eq:ent_proof_2}
\lvert \mathcal{K}_{\beta,\alpha}^{0}(t)\vert_{k} \rvert
\leq 1 + 2 \sum_{j = 0}^{k} \, \lvert \mathcal{K}_{\beta,\alpha}^{+}(t)\vert_{j} \rvert
\leq 3 (k+1) \lvert \mathcal{K}_{\beta,\alpha}^{+}(t)\vert_{k} \rvert.
\end{align}
Since $\mathcal{K}_{\beta,\alpha}(t)\setminus \mathcal{K}_{\beta,\alpha}^{+}(t) = \mathcal{K}_{\beta,\alpha}^{0}(t)$, and since by definition,
\begin{align*}
h_{\operatorname{top}}(T_{\beta,\alpha}\vert_{K_{\beta,\alpha}(t)})
= \lim_{n \to \infty} \frac{\log(\mathcal{K}_{\beta,\alpha}(t)\vert_{n})}{n}
\quad \text{and} \quad
h_{\operatorname{top}}(T_{\beta,\alpha}\vert_{K^{+}_{\beta,\alpha}(t)})
= \lim_{n \to \infty} \frac{\log(\mathcal{K}^{+}_{\beta,\alpha}(t)\vert_{n})}{n},
\end{align*}
the inequality given in \eqref{eq:ent_proof_2} implies that $h_{\operatorname{top}}(T_{\beta,\alpha}\vert_{K^{+}_{\beta,\alpha}(t)}) \geq h_{\operatorname{top}}(T_{\beta,\alpha}\vert_{K_{\beta,\alpha}(t)})$. As $K^{+}_{\beta,\alpha}(t) \subseteq K_{\beta,\alpha}(t)$ and as the entropy of a subsystem cannot exceed that of its parent system, the result follows.
\end{proof}
\subsection{\texorpdfstring{$\beta$}{beta}-shifts of finite type}
In \cite{LSS16} a study of when an intermediate $\beta$-shift is of finite type was carried out. This work was continued in \cite{LSSS} where it was shown that any intermediate $\beta$-shift can be \textsl{approximated from below} by an intermediate $\beta$-shift is of finite type. These results are summarised in the following theorem.
\begin{theorem}[{\cite{LSSS,LSS16}}]\label{thm:LSSS}
An intermediate $\beta$-shift $\Omega_{\beta,\alpha}$ is a subshift of finite type if and only if the kneading invariants $\tau_{\beta,\alpha}^{+}(p_{\beta,\alpha})$ and $\tau_{\beta,\alpha}^{-}(p_{\beta,\alpha})$ are periodic. Moreover, given $(\beta, \alpha) \in \Delta$ and $\epsilon > 0$, there exists a $(\beta', \alpha') \in \Delta$, with $0 \leq \beta' - \beta < \epsilon$ and $\lvert \alpha - \alpha' \rvert < \epsilon$ and such that
\begin{enumerate}[label={\rm(\arabic*)}]
\item $\Omega_{\beta', \alpha'}$ is a subshift of finite type,
\item the Hausdorff distance between $\Omega_{\beta, \alpha}$ and $\Omega_{\beta', \alpha'}$ is less than $\epsilon$, and
\item $\Omega_{\beta, \alpha} \subseteq \Omega_{\beta', \alpha'}$.
\end{enumerate}
\end{theorem}
\subsection{Transitivity of intermediate \texorpdfstring{$\beta$}{beta}-transformations}
An interval map $T \colon [0,1] \to [0,1]$ is said to be \textsl{transitive} if for any open subinterval $U$ of $[0,1]$ there exists an $m \in \mathbb{N}$ with $\bigcup_{k = 1}^{m} T^{k}(U) = (0,1)$. The property of transitivity will play an important part in our proof of \Cref{thm:main_2}, and thus we will utilise the following result of \cite{G1990,Palmer79} on non-transitive intermediate $\beta$-transformations. Note the contrast in the structure of the set of $(\beta, \alpha) \in \Delta$ with $T_{\beta,\alpha}$ transitive and the set of $(\beta, \alpha) \in \Delta$ with $\Omega_{\beta, \alpha}$ of finite type, namely that the former is has positive $2$-dimensional Lebesgue measure and the latter is countable.
\begin{theorem}[{\cite{G1990,Palmer79}}]\label{thm:G1990+Palmer79}
Let $\Delta_{\operatorname{trans}}$ denote the set of $(\beta, \alpha) \in \Delta$ with $T_{\beta,\alpha}$ transitive. The sets $\Delta_{\operatorname{trans}}$ and $\Delta \setminus \Delta_{\operatorname{trans}}$ have positive Lebesgue measure. Moreover, given $(\beta, \alpha) \in \Delta \setminus \Delta_{\operatorname{trans}}$, there exist
\begin{enumerate}[label={\rm(\roman*)}]
\item a natural number $n \geq 2$ and $k \in \{1,\ldots, n-1\}$ with $n$ and $k$ co-prime,
\item a sequence of points $\{ b_{0}, b_{1}, \ldots, b_{2n-1}\}$ in $(0,1)$ with $b_{i} < b_{i+1}$ for all $i \in \{0, 1, \ldots, 2n-2\}$, and
\item an $\tilde{\alpha} \in [0, 2-\beta^{n}]$,
\end{enumerate}
such that
\begin{enumerate}[label={\rm(\arabic*)}]
\item the transformation $T_{\beta^{n},\tilde{\alpha}}$ is transitive,
\item $T_{\beta,\alpha}^{n}(J_{i}) = J_{i}$ and $T_{\beta,\alpha}(J_{i}) = J_{i+k \bmod{n}}$, for all $i \in \{ 0, 1, \ldots, n-1 \}$, and
\item $T_{\beta,\alpha}^{n}\vert_{J_{i}}$ is topologically conjugate to $T_{\beta^{n},\tilde{\alpha}}$, for all $i \in \{ 0, 1, \ldots, n-1 \}$, where the conjugation is linear.
\end{enumerate}
Here, $J_{0} = [0, b_{0}] \cup [b_{2n-1}, 1]$, and $J_{i} = [b_{2i-1}, b_{2i}]$, for all $i \in \{1, 2, \ldots, n-1\}$. Further, there exists a $T_{\beta,\alpha}$-periodic point $q$ in $\mathscr{J}=\bigcup_{i = 0}^{n-2} [b_{2i}, b_{2i+1}]$, such that the orbit of $q$ under $T_{\beta,\alpha}$ is contained in $\mathscr{J}$, and for all $x$ in $\mathscr{J}$ but not in the orbit of $q$, there exists an $m \in \mathbb{N}$ such that $T_{\beta,\alpha}^{m}(x) \in \bigcup_{i = 0}^{n-1} J_{i}$.
\end{theorem}
\subsection{A sufficient condition for a dynamical set to be winning}\label{sec:HY}
To prove \Cref{thm:main_2} we not only appeal to the results of \cite{G1990,Palmer79}, but also to \cite[Theorem 2.1]{HY}, where a sufficient condition for certain dynamical sets to be winning is given. In order to state \cite[Theorem 2.1]{HY} we require the following notation. A partition of $[0,1]$ is a collection of finitely many intervals $\{ I(i) \}_{i \in \Lambda}$, where $\Lambda = \{0, 1, \ldots, m-1\}$ for some $m \in \mathbb{N}$, with pairwise disjoint interiors such that $[0,1] = \bigcup_{i \in \Lambda} I(i)$. Here, we assume that the intervals are ordered, namely, that if $i$ and $j \in \{0, 1, \ldots, m-1\}$ with $i < j$, then, for all $x \in I(i)$ and $y \in I(j)$, we have that $x \leq y$. Let $T \colon [0, 1] \to [0,1]$ and let $\{ I(i) \}_{i \in \Lambda}$ denote a partition of $[0, 1]$, such that $T$ restricted to $I(i)$ is monotonic and continuous for all $i \in \Lambda$. For $\xi = \xi_{1} \xi_{2} \cdots \xi_{n} \in \Lambda^{n}$, for some $n \in \mathbb{N}$, we set
\begin{align*}
I(\xi) = \bigcap_{i = 1}^{n} \ \{ x \in [0, 1] \colon T^{i-1}(x) \in I(\xi_{i}) \}.
\end{align*}
If $I(\xi)$ is non-empty, we call $I(\xi)$ a \textsl{level $n$ cylinder set} of $T$, and $\xi$ an \textsl{admissible word of length $n$} with respect to the partition $\{ I(i)\}_{i \in \Lambda}$. For $n \in \mathbb{N}_{0}$, we denote by $\Omega_{T}\vert_{n}$ the set of all admissible words of length $n$, where by convention $\Omega_{T}\vert_{0} = \{ \varepsilon \}$, and set $\Omega_{T}^{*} = \bigcup_{n \in \mathbb{N}_{0}} \Omega_{T}\vert_{n}$.
When $T = T_{\beta, \alpha}$ and when our partition is $\{ I(0) = [0,p_{\beta, \alpha}), I(1)=[p_{\beta, \alpha},1] \}$, for some $(\beta, \alpha) \in \Delta$, we have $\Omega_{T}\vert_{n} =\Omega_{\beta, \alpha}\vert_{n}$. Further, for $\xi \in \{0, 1\}^{*}$, we have $I(\xi)$ is non-empty if and only if there exists an $\omega \in \Omega_{\beta, \alpha}$ with $\omega\vert_{\lvert \xi \rvert} = \xi$, and $\overline{I(\xi)} = \pi_{\beta, \alpha}( \{ \omega \in \Omega_{\beta, \alpha} \colon \omega\vert_{n} = \xi \} )$, where $\overline{I(\xi)}$ denotes the closure of $I(\xi)$.
For $\xi$ and $\nu \in \Omega_{T}^{*}$ and $c > 0$ a real number, we say that $\xi$ is \textsl{$\nu$-extendable} if the concatenation $\xi\nu$ is admissible, and say that the cylinder sets $I(\xi)$ and $I(\nu)$ are \textsl{$c$-comparable} if $c \leq \lvert I(\xi) \rvert / \lvert I(\nu) \rvert \leq 1/c$. We call $T$ is \textsl{piecewise locally $C^{1+\delta}$ expanding} if $T$ restricted to $I(i)$ is differentiable for all $i \in \Lambda$, and
\begin{enumerate}[label={\rm(\arabic*)}]
\item there exists a real number $\eta > 0$, such that $\lvert T'(x) \rvert > \eta$ for all $x \in I(i)$ and $i \in \Lambda$, and there exists $k \in \mathbb{N}$ and a real number $\lambda > 1$ such that $\lvert (T^{k})'(x) \rvert \geq \lambda$ for all $\xi \in \Omega_{T}\vert_{k}$ and $x \in I(\xi)$, and
\item there exist two positive constants $\delta$ and $c$ such that for all $i \in \Lambda$ and all $x$ and $y \in I(i)$,
\begin{align*}
\left\lvert \frac{T'(x)}{T'(y)} - 1 \right\rvert \leq c \lvert x - y \rvert^{\delta}.
\end{align*}
\end{enumerate}
We call $T$ \textsl{Markov} if for all $i$ and $j \in \Lambda$, either $T(I(i)) \cap I(j) = \emptyset$, or $I(j) \subseteq T(I(i))$. Letting $(\beta, \alpha) \in \Delta$ with $\Omega_{\beta, \alpha}$ a subshift of finite type, we set $A = \{ a_{1}, a_{2}, \ldots, a_{n} \}$ to be the set of ordered points of
\begin{align*}
\{ \pi_{\beta, \alpha}(\sigma^{k}(\tau_{\beta,\alpha}^{+}(p_{\beta, \alpha}))) \colon k \in \mathbb{N}\} \cup \{ \pi_{\beta, \alpha}(\sigma^{k}(\tau_{\beta,\alpha}^{-}(p_{\beta, \alpha}))) \colon k \in \mathbb{N} \}.
\end{align*}
The transformation $T_{\beta, \alpha}$ is a piecewise locally $C^{1+\delta}$ expanding Markov map with respect to the partition
\begin{align}\label{eq:Markov_Partition}
P_{\beta,\alpha} = \{[a_{1}, a_{2}), \ldots, [a_{n-2}, a_{n-1}), [a_{n-1}, a_{n}]\}.
\end{align}
Letting $T$ be a piecewise locally $C^{1+\delta}$ expanding map with respect to the partition $\{ I(i) \}_{i \in \Lambda}$, then for $x \in [0, 1]$, there exists an infinite word $\omega = \omega_{1} \omega_{2} \cdots \in \Lambda^{\mathbb{N}}$, with $\omega \vert_{k} \in \Omega_{T}^{*}$ for all $k \in \mathbb{N}$ and such that $\{ x \} = \bigcap_{k \in \mathbb{N}} \overline{I(\omega\vert_{n})}$. We call $\omega$ a \textsl{symbolic representation} of $x$ with respect to the partition $\{ I(i)\}_{i \in \Lambda}$. In the case that $T = T_{\beta,\alpha}$, for some $(\beta,\alpha) \in \Delta$, for a point $x \in [0, 1]$, symbolic representations of $x$ with respect to the partition $\{ [0,p_{\beta, \alpha}), [p_{\beta,\alpha},1] \}$ are $\tau_{\beta, \alpha}^{\pm}(x)$. Note, all but a countable set of points have a unique symbolic representation, we denote this countable set by $E = E_{T}$.
For a fixed $x \in [0,1]$ and $\gamma \in (0, 1)$, let $\omega$ denote a symbolic representation of $x$. We denote the following geometric condition by $H_{x, \gamma}$:
\begin{align}\label{eq:condition_1}
\adjustlimits
\lim_{i \to \infty} \sup_{\; u\,:\,u\;\text{and}\;u\omega\vert_{i}\in\Omega_{T}^{*}} \frac{\lvert I(u\omega\vert_{i}) \rvert}{\lvert I(u) \rvert} = 0
\end{align}
and there exists a natural number $i^{*}$ and a real number $c>0$ such that if $i \in \mathbb{N}$ with $i \geq i^{*}$, for all $\nu$ and $\eta \in \Omega_{T}^{*}$ which are $\omega\vert_{i}$-extendable with $I(\nu)$ and $I(\eta)$ being $\gamma/4$-comparable, either
\begin{align}\label{eq:condition_2}
\operatorname{dist}(I(\nu\omega\vert_{i}), I(\eta\omega\vert_{i})) = 0
\quad \text{or} \quad
\operatorname{dist}(I(\nu\omega\vert_{i}), I(\eta\omega\vert_{i})) \geq c \operatorname{dist}(I(\nu), I(\eta)).
\end{align}
\begin{theorem}[{\cite{HY}}]\label{thm_HY_Thm_2.1}
Let $T$ be a piecewise locally $C^{1+\delta}$ expanding map with respect to the partition $\{ I(i) \}_{i \in \Lambda}$, and let $x \in [0, 1]$ with symbolic representation $\omega$.
\begin{enumerate}[label={\rm(\arabic*)}]
\item If $H_{x, \gamma}$ is satisfied for some $\gamma \in (0, 1)$ , then the set $\{ y \in [0,1] \colon T^{k}(y) \not\in I(\omega\vert_{m}) \; \text{for all} \; k \in \mathbb{N}_{0} \} \cup E$ is $\rm ( 1/2, \gamma)$-winning for some natural number $m$.
%
\item If $x \not\in E$ and if $H_{x, \gamma}$ is satisfied for any $\gamma \in (0, 1)$, then the set ${\rm BAD}_{T}(x) = \{ y \in [0, 1] \colon x \not\in \overline{\{T^k(y) \colon k \in \mathbb{N}_{0}\}}\}$ is $1/2$-winning.
\end{enumerate}
\end{theorem}
We conclude this section with the following proposition which we use in conjunction with \Cref{thm:G1990+Palmer79,thm_HY_Thm_2.1} and the fact that the property of winning is preserved under bijective affine transformations to prove \Cref{thm:main_2}.
\begin{proposition}\label{prop:alpha-winning_transport}
Let $T$ be a piecewise locally $C^{1+\delta}$ expanding interval map, let $x \in [0, 1]$ and set
\begin{align*}
\mathrm{BAD}(T,x) \coloneqq \{ y \in [0,1] \colon x \not\in \overline{\{T^{n}(y) \colon n \in \mathbb{N}\}}\}.
\end{align*}
If, for a fixed $k \in \mathbb{N}$, we have that $\mathrm{BAD}(T^k, T^m(x))$ is winning for all $m \in \{ 0, 1, \ldots, k\}$, then $\mathrm{BAD}(T,x)$ is winning.
\end{proposition}
\begin{proof}
This follows from the fact that winning is preserved under taking countable intersections and since, by construction, $\mathrm{BAD}(T^k,x)\cap \mathrm{BAD}(T^k,T(x))\cap \cdots \cap \mathrm{BAD}(T^k,T^k(x))\subset \mathrm{BAD}(T,x)$.
\end{proof}
\section{Intermediate \texorpdfstring{$\beta$}{beta}-shifts as greedy \texorpdfstring{$\beta$}{beta}-shifts: Proof of \texorpdfstring{\Cref{thm:main}}{Theorem 1.1}}\label{sec:proof_thm_1_1}
In proving \Cref{thm:main} and \Cref{Cor_2,Cor_3}, we will investigate the question, given a fixed $\beta \in (1, 2)$, for which $\omega \in \{0,1\}^\mathbb{N}$ does there exist $(\beta', \alpha') \in \Delta$ with $\Omega^+_{\beta', \alpha'} = \{ \nu \in \{0,1\}^\mathbb{N} \colon \omega \preceq \sigma^{n}(\nu) \prec \tau_{\beta,0}^{-}(1) \; \text{for all} \; n \in \mathbb{N} \}$? Not only is this question interesting in its own right, but in classifying such words, we will be able to transfer the results from \cite{KKLL} to the setting of the intermediate $\beta$-transformations. With this in mind, we let $\mathcal{A}_\beta$ denote the set of all such words and set $\rho = \inf_{n \in \mathbb{N}_{0}} \pi_{\beta,0}(\sigma^{n}(\tau_{\beta,0}^{-}(1)))$. Further, we utilise the following notation. We let $t_{\beta, 0, c} \in (0,1)$ be such that $\dim_H(K^{+}_{\beta,0}(t))>0$ for all $t < t_{\beta, 0, c}$ and $\dim_H(K^{+}_{\beta, 0}(t)) = 0$ for all $t > t_{\beta, 0, c}$, and set $\mathcal{T}_{\beta, 0, c} = \tau_{\beta,0}^{+}(t_{\beta, c})$.
\begin{proof}[Proof of \texorpdfstring{\Cref{thm:main}}{Theorem 1.1}]
For $\beta\in(1,2)$, let $\mathcal{B}_{\beta} \coloneqq \{ \omega \in \mathcal{E}_{\beta,0}^{+} \colon \pi_{\beta,0}(\omega) \leq \rho \; \text{and} \; \omega \prec \mathcal{T}_{\beta, 0, c}\}$. By \Cref{cor:From_greedy_to_intermediate} and the commutativity of the diagram given in \eqref{eq:commutative_diag}, observe that it is sufficient to show
\begin{enumerate}
\item $\mathcal{A}_\beta\subseteq \mathcal{B}_\beta$ with equality holding for Lebesgue almost every $\beta\in(1,2)$,
\item there exist $\beta\in (1,2)$ such that $\mathcal{A}_\beta \neq \mathcal{B}_\beta$, and
\item if the quasi-greedy $\beta$-expansion of $1$ is periodic, then $\mathcal{A}_\beta = \mathcal{B}_\beta$.
\end{enumerate}
To show $\mathcal{A}_\beta\subseteq \mathcal{B}_\beta$, let $\beta \in (1, 2)$ and $\eta \in \mathcal{A}_\beta$ be fixed. Setting $\omega = 1\eta$ and $\nu = 0\tau_\beta^-(1)$, we observe that they meet Conditions~(1)--(4) of \Cref{thm:BSV14} . Condition~(2) of \Cref{thm:BSV14} gives $\nu \in \Omega^{+}(\omega, \nu)$, and so, for a given $n \in \mathbb{N}_{0}$,
\begin{align*}
\eta \preceq \sigma^{n}(\eta) \prec 0 \tau_{\beta,0}^{-}(1) \quad \text{or} \quad 1\eta \preceq \sigma^{n}(\eta) \prec \tau_{\beta,0}^{-}(1),
\end{align*}
yielding that $\eta \preceq \sigma^{n}(\eta) \prec \tau_{\beta,0}^{-}(1)$ for all $n \in \mathbb{N}_{0}$, namely that $\eta \in \mathcal{E}_{\beta,0}^{+}$. Condition~(2) of \Cref{thm:BSV14} also gives $\omega\in \Omega^{-}(\omega, \nu)$, and so, for a given $n \in \mathbb{N}_{0}$,
\begin{align*}
\eta \prec \sigma^{n}(\tau_{\beta,0}^{-}(1)) \preceq 0 \tau_{\beta,0}^{-}(1) \quad \text{or} \quad 1\eta \prec \sigma^{n}(\tau_{\beta,0}^{-}(1)) \preceq \tau_{\beta,0}^{-}(1).
\end{align*}
This implies that $\eta \prec \sigma^{n}(\tau_\beta^-(1))$ for all $n \in \mathbb{N}_{0}$, and so $\pi_{\beta,0}(\eta)\leq \rho$. Since Condition~(3) of \Cref{thm:BSV14} holds, the topological entropy of $(\Omega(\omega,\nu), \sigma)$ is positive, and thus $\eta \prec \mathcal{T}_{\beta, 0, c}$. Therefore, $\eta \in \mathcal{B}_\beta$, and hence $\mathcal{A}_\beta\subseteq \mathcal{B}_\beta$.
To see that $\mathcal{A}_\beta = \mathcal{B}_\beta$ for Lebesgue almost every $\beta \in (1,2)$, from the concluding remarks of \cite{Schme} we know that, for Lebesgue almost all $\beta \in (1,2)$ there is no bound on the length of blocks of consecutive zeros in the quasi-greedy $\beta$-expansion of $1$, namely $\tau_{\beta,0}^-(1)$. This implies that $\rho = 0$, and hence that $\mathcal{B}_\beta=\{ 0^\infty \}$. Since $\tau_{\beta,0}^{\pm}(0) = 0^\infty$ it follows that $0^\infty \in \mathcal{A}_\beta$, and thus that $\mathcal{B}_\beta \subseteq \mathcal{A}_\beta$ for Lebesgue almost every $\beta \in (1,2)$. This in tandem with the fact that $\mathcal{A}_\beta\subseteq \mathcal{B}_\beta$ for all $\beta \in (1,2)$, yields that $\mathcal{A}_\beta = \mathcal{B}_\beta$ for Lebesgue almost every $\beta \in (1,2)$.
Let $\beta$ denote the algebraic number with minimal polynomial $x^5-x^4-x^3-2x^2+x+1$. An elementary calculation yields that $\tau_{\beta,0}^{-}(1)=11(100)^{\infty}$. We claim that $\xi = 00(011)^\infty \in \mathcal{B}_\beta$, but that $\xi \not\in \mathcal{A}_\beta$, namely that $\mathcal{A}_\beta \subsetneq \mathcal{B}_\beta$. It is readily verifiable that $\xi \in \mathcal{E}_\beta^+$ and also, since $\rho=\pi_{\beta,0}((001)^\infty)$, that $\pi_{\beta,0}(\xi) < \rho$. This yields that $\{001, 011 \}^\mathbb{N} \subset \mathcal{K}^{+}_{\beta,0}(\pi_{\beta,0}(\xi))$, and hence that $h_{\operatorname{top}}(\sigma\vert_{\mathcal{K}^{+}_{\beta,0}(\pi_{\beta,0}(\xi))}) > 0$. In other words, we have $\xi \prec \mathcal{T}_{\beta,c}$, and so $\xi \in \mathcal{B}_\beta $. By way of contradiction, suppose that $\xi \in \mathcal{A}_\beta$. Set $\omega=0\tau_{\beta,0}^{-}(1)=011(100)^\infty$ and $\nu = 1\xi = 100(011)^\infty$. In which case $\omega$ and $\nu \in \{ \chi, \zeta\}^\mathbb{N}$ with $\chi=011$ and $\zeta=100$. Noting that $\chi$ and $\zeta$ are words of length three in the alphabet $\{0,1\}$, that $\chi\vert_{2} = 01$, $\zeta\vert_{2} = 10$, that $\chi^\infty \in \Omega^{-}(\chi^\infty, \zeta^\infty)$ and $\zeta^\infty \in \Omega^{+}(\chi^\infty, \zeta^\infty)$, but that $\omega=\chi\zeta^\infty \neq \chi^\infty$, contradicting Condition~(4) of \Cref{thm:BSV14}.
It remains to prove that if $\tau_{\beta,0}^-(1)$ is periodic, then $\mathcal{A}_\beta = \mathcal{B}_\beta$. To this end, fix $\beta \in (1, 2)$ with $\tau_{\beta,0}^{-}(1)$ periodic. Let $\xi \in \mathcal{B}_\beta $ and set $\nu=1 \xi$ and $\omega = 0\tau_\beta^-(1)$. By assumption, $\xi \in \mathcal{E}_{\beta,0}^+$, and so $\sigma(\nu) \preceq \sigma^{n}(\nu) \prec \sigma(\omega)$ and therefore $\nu \in \Omega^{+}(\omega, \nu)$, and since $\tau_{\beta,0}^{-}(1)$ is the quasi greedy $\beta$-expansion of $1$ in base $\beta$, we have $\sigma^{n}(\omega) \preceq \sigma(\omega)$ for all $n \in \mathbb{N}_{0}$. As $\pi_{\beta,0}(\xi) < \rho$ we have $ \sigma(\nu) \prec \sigma^n(\omega)$, and so $\omega \in \Omega^{-}(\omega, \nu)$. Thus, $\omega$ and $\nu$ satisfy Conditions~(1) and~(2) of \Cref{thm:BSV14}. Condition~(3) of \Cref{thm:BSV14} follows from $\xi \prec \mathcal{T}_{\beta,c}$. To conclude the proof, it suffices to show that $\omega$ and $\nu$ satisfy Condition~(4) of \Cref{thm:BSV14}. Suppose there exist $\chi$ and $\zeta \in \{0,1\}^{*}$ of length at least three with
\begin{align*}
\chi\vert_{2} = 01, \quad
\zeta\vert_{2} = 10, \quad
\chi^{\infty} \in \Omega^{-}(\chi^{\infty},\zeta^{\infty}),
\quad \text{and} \quad
\zeta^{\infty} \in \Omega^{+}(\chi^{\infty},\zeta^{\infty}),
\end{align*}
and such that $\omega$ and $\nu \in \{ \chi, \zeta \}^{\mathbb{N}}$. By our assumption and construction, in particular, since $\omega$ is periodic and since $\chi^{\infty} \in \Omega^{-}(\chi^{\infty},\zeta^{\infty})$, we have $\omega = \chi^\infty$. By way of contradiction, suppose that $\nu \neq \zeta^\infty$. In which case, there exists $n \in \mathbb{N}_{0}$ such that $\sigma^{n}(\nu)\vert_{\lvert \chi \rvert + \lvert \zeta \rvert} = \chi \zeta$. Noting that $\chi\vert_{2} = 01$ and $\zeta\vert_{2} = 10$, this yields $\sigma(\omega) \prec \sigma^{n+1}(\nu)$ contradicting the fact that $\omega$ and $\nu$ satisfy Condition~(2) of \Cref{thm:BSV14}.
\end{proof}
\section{A Krieger embedding theorem for intermediate \texorpdfstring{$\beta$}{beta}-transformations: Proof of \texorpdfstring{\Cref{Cor_1}}{Corollary 1.2}}\label{sec:proof_cor_1_2}
To prove \Cref{Cor_1} we first show the following special case.
\begin{theorem}\label{thm:one_perioidc}
Let $(\beta,\alpha)\in\Delta$ be such that $\nu = \tau^{+}_{\beta, \alpha}(p_{\beta,\alpha})$ is not periodic and $\omega = \tau^{-}_{\beta, \alpha}(p_{\beta,\alpha})$ is periodic. There exists a sequence $((\beta_{n}, \alpha_{n}))_{n\in \mathbb{N}}$ in $\Delta$ with $\lim_{n\to \infty}\beta_{n} = \beta$ and $\lim_{n \to \infty}\alpha_{n} = \alpha$ and such that
\begin{enumerate}[label={\rm(\roman*)}]
\item[\rm (1)] $\Omega_{\beta_{n},\alpha_n}$ is a subshift of finite type,
\item[\rm (2)] the Hausdorff distance between $\Omega_{\beta, \alpha}$ and $\Omega_{\beta_{n}, \alpha_{n}}$ converges to zero as $n$ tends to infinity, and
\item[\rm (3)] $\Omega_{\beta_{n},\alpha_n}\subseteq\Omega_{\beta,\alpha}$.
\end{enumerate}
\end{theorem}
\begin{proof}
We prove this using \Cref{thm:main} and the results of \cite{KKLL}. By \Cref{thm:main}, there exist $\beta' \in (1, 2)$ and $t \in E_{\beta',0}^+$ such that $\mathcal{K}_{\beta',0}^+(t)=\Omega^+_{\beta,\alpha}$ with $\tau_{\beta',0}^{+}(t) = \sigma(\nu)$. Our goal is to find a monotonically decreasing sequence $(t_i)_{i \in \mathbb{N}}$ converging to $t$ with $t_i\in E_{\beta',0}^+$ and $t_i$ is a $T_{\beta', 0}^+$-periodic point for all $i \in \mathbb{N}$. We will first prove that $t$ is not isolated from above. For this we use the following. A finite word $s \in \{0, 1\}^{*}$ is Lyndon if $s^\infty \prec \sigma^n(s^\infty)$ for all $n \in \mathbb{N}$ with $n \neq 0 \bmod \lvert s \rvert$, and set $L_{\beta} \coloneqq \{ s\in \{0, 1\}^* \colon s \; \text{is a Lyndon word and} \; s^\infty \in \Omega_{\beta', 0} \}$.
For $s \in L_{\beta}$, let $I_{s}$ denote the half-open interval $[\pi_{\beta',0}(s0^\infty), \pi_{\beta',0}(s^\infty))$. \Cref{thm:Structure} in combination with our hypothesis that $\tau_{\beta,\alpha}^{-}(p_{\beta,\alpha}) = 0\tau_{\beta,\alpha}^{-}(1)$ is periodic, yields there exists a shortest finite word $\zeta$ with $\tau_{\beta,\alpha}^{-}(1) = \zeta^\infty$. Letting $n$ be the length of $\zeta$, we set $\zeta^\prime$ to be the lexicographical smallest element of the set $\{ \zeta_{k} \cdots \zeta_{n} \zeta_{1} \cdots \zeta_{k-1} \colon k \in \{2, \ldots, n \}\}$, and set $y = \pi_{\beta',0}(\zeta^\prime 0^\infty)$. By construction $\zeta^{\prime}$ is a Lyndon. Since by our hypothesis $\nu = \tau^{+}_{\beta, \alpha}(p_{\beta,\alpha})$ is not periodic and since $t < \pi_{\beta',0}({\zeta^{\prime}}^{\infty})$, we observe that $t < y$.
For $s \in L_{\beta}$, by the Lyndon property of $s$, if $x \in I_s$, then $x \not\in E_{\beta',0}^{+}$, which implies $E_{\beta',0}^{+} \cap (0,y) \subseteq (0,y) \backslash \bigcup_{s\in L_\beta} I_s$. In fact we claim $(0,y) \backslash \bigcup_{s\in L_\beta} I_s = E_{\beta',0}^+ \cap (0,y)$. In order to prove this, let $x\in (0,y) \backslash \bigcup_{s\in L_\beta} I_s$ and suppose $x \notin E_{\beta',0}^+$. Under this hypothesis, there exists a minimal $n \in \mathbb{N}$ such that $\sigma^{n}(\tau_{\beta', 0}^{+}(x)) \prec \tau_{\beta', 0}^{+}(x)$. By the minimality of $n$, we have that $\xi = \tau_{\beta', 0}^{+}(x)\vert_{n}$ is a Lyndon word, and that $\tau_{\beta', 0}^{+}(x) \prec \xi^{\infty}$. If $\xi^\infty \not\in \Sigma_{\beta^\prime,0}$, then there exists $j \in \{ 1, 2, \ldots, n \}$ such that $\tau_{\beta',0}^{-}(1)\prec\sigma^j(\xi^\infty)$ where equality is excluded since $x<y$. Set $k = \tau_\beta^{-}(1) \wedge \sigma^j(\xi^\infty)$, and notice $k>n-j$; otherwise $\tau_{\beta', 0}^{+}(x)$ would not be admissible. This yields $\tau_{\beta',0}^{-}(1)=\xi_{j+1}\xi_{j+2}\cdots \xi_n (\xi_1\cdots \xi_n)^l \omega_1 \omega_2 \cdots$ with $l$ possibly $0$ but chosen so that $\omega_1 \cdots \omega_n \neq \xi_1 \cdots \xi_n$; note this is possible since $\nu$ is not periodic. Thus, $\sigma^{n-j+ln}(\tau_{\beta',0}^{-}(1)) \prec \sigma^{n-j+ln}(\sigma^{j}(\xi^\infty))=\xi^\infty$ and $\omega_1\cdots \omega_n \prec \xi$. Hence, $\sigma^{n-j+ln}(\tau_{\beta',0}^{-}(1)) \prec \xi 0^\infty \prec \tau_{\beta',0}^{-}(x)$, contradicting the fact that we choose $x\in (0,y) \backslash \bigcup_{s\in L_\beta} I_s$. It therefore follows that $E_{\beta',0}^{+} \cap (0,y) = (0,y) \backslash \bigcup_{s\in L_\beta} I_s$ as required.
Suppose that $t$ cannot be approximated from above by elements in $E_{\beta',0}^+$, that is, there exists a real number $\epsilon > 0$ with $(t,t+\epsilon)\cap E_{\beta',0}^+=\emptyset$. Since $E_{\beta',0}^+ \cap (0,y) = (0,y) \backslash \bigcup_{s\in L_\beta} I_s$, there exists a Lyndon word $s$ with $(t,t+\epsilon) \subset I_s$, but as $I_s$ is closed from the left, $t\in I_s$, contradicting our hypothesis that $t\in E_{\beta',0}^{+}$. This implies $t$ can be approximated from above by elements in $E_{\beta',0}^{+}$, namely there exists a monotonically decreasing sequence $(t_i^\prime)_{i\in \mathbb{N}}$ of real numbers converging to $t$ with $t_i^\prime \in E_{\beta',0}^+$, for all $i \in \mathbb{N}$. If $t_i^\prime$ is not $T_{\beta',0}$-periodic for some $i \in \mathbb{}N$, then by \cite[Lemmanta~3.4 and~3.5]{KKLL}, there exists a monotonically increasing sequence of $T_{\beta',0}$-periodic points $(s_{i, j}^\prime)_{j \in \mathbb{N}}$ converging to $t_i^\prime$ with $s_{i, j} \in E_{\beta',0}^+$. For $i \in \mathbb{N}$, setting $t_i=t_i^\prime$ whenever $t_i^\prime$ is $T_{\beta',0}$-periodic, and otherwise setting $t_i=s_{i,j}^\prime$ where $s_{i,j}^\prime$ is chosen so that $t < s_{i,j}^\prime < t_i^\prime$, the sequence $(t_i)_{i \in \mathbb{N}}$ converges to $t$ from above and $t_i$ is $T_{\beta',0}$-periodic.
Since $\omega$ is periodic with respect to the left shift map, \Cref{thm:main} implies, for each $i \in \mathbb{N}$, there exists $(\beta_{i}, \alpha_{i}) \in \Delta$ with $K_{\beta'0}^+(t_i)=\Omega^+_{\beta_{i},\alpha_i}$. Since both $\omega$ and $\tau^{+}_{\beta',0}(t_i)$ are periodic, \Cref{thm:LSSS} yields that $\Omega_{\beta_{i},\alpha_i}$ is of subshift of finite type. Further, since $\mathcal{K}_{\beta',0}^{+}(t_i) \subseteq \mathcal{K}^{+}_{\beta',0}(t)$, it follows that $\Omega_{\beta_{i},\alpha_{i}} \subseteq \Omega_{\beta,\alpha}$ for all $i \in \mathbb{N}$.
\end{proof}
\begin{proof}[{Proof of \Cref{Cor_1}}]
Assume the setting of \Cref{Cor_1} and for ease of notation set $p = p_{\beta, \alpha}$, $\nu = \tau_{\beta, \alpha}^{+}(p)$ and $\omega = \tau_{\beta, \alpha}^{-}(p)$. By \Cref{thm:LSSS}, we have that $\Omega_{\beta, \alpha}$ is a subshift of finite type if and only if $\omega$ and $\nu$ are periodic. Since the subshift of finite type property is preserved by topological conjugation, and observing that $\Omega^{\pm}_{\beta, \alpha}$ and $\Omega^{\mp}_{\beta, 2-\beta-\alpha}$ are topologically conjugate, with conjugation map $R$, with out loss of generality we may assume that $\nu$ is not periodic. We consider the case, when $\omega$ is periodic and when $\omega$ is not periodic separately. The former of these two cases follows from \Cref{thm:one_perioidc}, and so all that remains is to show the result for the latter case, namely when $\omega$ is not periodic. To this end, assume that $\omega$ and $\nu$ are both not periodic. Let $n \in \mathbb{N}$ be fixed, set $O_{n}^{\pm}(p) = \{ (T_{\beta, \alpha}^{\pm})^{k}(p) \colon k \in \{ 0, 1, \ldots, n-1 \} \}$, and let $\beta' \in (1, \beta)$ be such that
\begin{align}\label{eq:def_beta_prime}
(1-\alpha)/\beta' + (\beta+1)^{n}(\beta - \beta') < \min \{ x \in O_{n}^{+}(p) \cup O_{n}^{-}(p) \colon x > p \}.
\end{align}
(As defined in \Cref{sec:beta-shifts}, we let $T_{\beta,\alpha}^{-} \colon x \mapsto \beta x + \alpha$ if $x \leq p$, and $x \mapsto \beta x + \alpha - 1$ otherwise, and for ease of notation, we write $T_{\beta,\alpha}^{+}$ for $T_{\beta, \alpha}$.) Setting $p' = (1-\alpha)/\beta'$, we claim, for all $k \in \{ 1, \ldots, n-1 \}$, that either
\begin{align}\label{eq:desired_inequalities}
(T_{\beta', \alpha}^{\pm})^{k}(p') \leq (T_{\beta, \alpha}^{\pm})^{k}(p) \leq p \leq p'
\quad \text{or} \quad
(T_{\beta, \alpha}^{\pm})^{k}(p) \geq (T_{\beta', \alpha}^{\pm})^{k}(p') \geq p' \geq p.
\end{align}
Hence, by definition and since $\omega$ and $\nu$ are not periodic, $\omega\vert_{n} = \tau_{\beta', \alpha}^{-}(p)\vert_{n}$ and $\nu\vert_{n} = \tau_{\beta', \alpha}^{+}(p)\vert_{n}$.
To prove this claim, note, for all $k \in \{ 1, \ldots, n-1 \}$, either
\begin{align}\label{eq:either_or}
(T_{\beta, \alpha}^{\pm}(p))^{k} < p
\quad \text{or} \quad
(T_{\beta, \alpha}^{\pm})^{k}(p) \geq \min \{ x \in O_{n}^{+}(p) \cup O_{n}^{-}(p) \colon x > p \}.
\end{align}
If $0 \leq y \leq x \leq p$, or if $p' \leq y \leq x \leq 1$, then
\begin{align}\label{eq:orbit_bound}
0 \leq T^{\pm}_{\beta, \alpha}(x) - T^{\pm}_{\beta', \alpha}(y) = \beta x - \beta'y = \beta x - \beta y + \beta y - \beta'y \leq \beta(x-y) + (\beta-\beta').
\end{align}
Observe that $T^{\pm}_{\beta, \alpha}(p) = T^{\pm}_{\beta', \alpha}(p')$ and
\begin{align*}
0 \leq (T^{\pm}_{\beta, \alpha})^{2}(p) - (T^{\pm}_{\beta', \alpha})^{2}(p')
\leq \beta - \beta'
\leq (\beta + 1)(\beta - \beta')
\leq (\beta + 1)^{2}(\beta - \beta')
\leq (\beta+1)^{n}(\beta - \beta').
\end{align*}
Suppose, by way of induction on $m$, that
\begin{align*}
0
\leq (T^{\pm}_{\beta, \alpha})^{m}(p) - (T^{\pm}_{\beta', \alpha})^{m}(p')
\leq (\beta+1)^{m}(\beta-\beta')
\leq (\beta+1)^{n}(\beta-\beta'),
\end{align*}
for some $m \in \{2, \dots, n-2\}$. Combining \eqref{eq:def_beta_prime}, \eqref{eq:either_or} and \eqref{eq:orbit_bound} with our inductive hypothesis, we have
\begin{align*}
0 \leq (T^{\pm}_{\beta, \alpha})^{m+1}(p) - (T^{\pm}_{\beta', \alpha})^{m+1}(p')
&\leq \beta((T^{\pm}_{\beta, \alpha})^{m}{p} - (T^{\pm}_{\beta', \alpha})^{m}(p')) + (\beta-\beta')\\
&\leq \beta(\beta+1)^{m}(\beta-\beta') + (\beta-\beta')
\leq (\beta+1)^{m+1}(\beta-\beta')
\leq (\beta+1)^{n}(\beta-\beta').
\end{align*}
In other words, for all $k \in \{ 1, \ldots, n-1 \}$,
\begin{align*}
0 \leq (T^{\pm}_{\beta, \alpha})^{k}(p) - (T^{\pm}_{\beta', \alpha})^{k}(p')
\leq (\beta+1)^{k}(\beta-\beta')
\leq (\beta+1)^{n}(\beta-\beta').
\end{align*}
This in tandem with \eqref{eq:def_beta_prime} and \eqref{eq:either_or} proves the claim.
We observe that $(\omega, \nu) \neq (\tau_{\beta', \alpha}^{+}(p'), \tau_{\beta', \alpha}^{-}(p'))$, for if not, then since $\beta' < \beta$, this would contradict \Cref{thm:Laurent}. This implies that $\omega \neq \tau_{\beta', \alpha}^{-}(p')$ or $\nu \neq \tau_{\beta, \alpha}^{+}(p')$. We claim that $\omega \succ \tau_{\beta', \alpha}^{-}(p')$ and $\nu \succ \tau_{\beta, \alpha}^{+}(p')$.
Consider the case when $\omega \neq \tau_{\beta', \alpha}^{-}(p')$. This implies there exists a smallest integer $m \geq n$ such that neither
\begin{align*}
(T_{\beta', \alpha}^{-})^{m}(p') \leq (T_{\beta, \alpha}^{-})^{m}(p) \leq p
\quad \text{nor} \quad
(T_{\beta, \alpha}^{-})^{m}(p) \geq (T_{\beta', \alpha}^{-})^{m}(p') \geq p'.
\end{align*}
Using the fact that if $0 \leq y \leq x < p$ or if $p' < y \leq x \leq 1$, then $T^{-}_{\beta', \alpha}(y) \leq T^{-}_{\beta, \alpha}(x)$, in tandem with \eqref{eq:desired_inequalities}, and noting that $p<p'$, we have that
\begin{align*}
\tau_{\beta', \alpha}^{-}(p')\vert_{m-2}=\omega\vert_{m-2},
\quad
(T^{-}_{\beta',\alpha})^{m}(p')<p',
\quad
(T^{-}_{\beta,\alpha})^{m}(p) > p
\quad \text{and} \quad
(T^{-}_{\beta',\alpha})^{m}(p')\leq (T^{-}_{\beta,\alpha})^{m}(p).
\end{align*}
Thus, $\tau_{\beta', \alpha}^{-}(p')\vert_{m-1} \prec \omega\vert_{m-1}$ and hence $\tau_{\beta', \alpha}^{-}(p') \prec \omega$. An analogous argument proves the claim when $\nu \neq \tau_{\beta, \alpha}^{+}(p')$.
Hence, we have shown, given an $n \in \mathbb{N}$, that there exists a positive $\delta \in \mathbb{R}$, such that, for all $\beta' \in (\beta-\delta, \beta)$,
\begin{align}\label{smaller}
\tau_{\beta', \alpha}^{\pm}(p')\vert_{n} = \tau_{\beta, \alpha}^{\pm}(p)\vert_{n}
\quad \text{and} \quad
\tau_{\beta', \alpha}^{\pm}(p') \prec \tau_{\beta, \alpha}^{\pm}(p),
\end{align}
where $p'=(1-\alpha)/\beta'$. Further, by using the fact that $\Omega^{\pm}_{\beta, \alpha}$ and $\Omega^{\mp}_{\beta, 2-\beta-\alpha}$ are topologically conjugate, with conjugating map $R$, together with \eqref{smaller}, we have that there exists a positive $\delta' \in \mathbb{R}$, such that, for all $\beta' \in (\beta-\delta', \beta)$,
\begin{align*}
\tau_{\beta', \alpha+\beta-\beta'}^{\pm}(p_{\beta', \alpha+\beta-\beta'})\vert_{n} = \tau_{\beta, \alpha}^{\pm}(p)\vert_{n}
\quad \text{and} \quad
\tau_{\beta', \alpha+\beta-\beta'}^{\pm}(p_{\beta', \alpha+\beta-\beta'}) \succ \tau_{\beta, \alpha}^{\pm}(p).
\end{align*}
Letting $\beta' \in (\beta-\min(\delta,\delta'), \beta)$ be fixed and setting
\begin{align*}
q_{1} = \sup \{ a \in (\alpha, \alpha+\beta-\beta') \colon \tau_{\beta', a}^{\pm}(p_{\beta', a}) \preceq \tau_{\beta, \alpha}^{\pm}(p)\}
\quad \text{and} \quad
q_{2} = \inf \{ a \in (\alpha, \alpha+\beta-\beta') \colon \tau_{\beta', a}^{\pm}(p_{\beta', a}) \succeq \tau_{\beta, \alpha}^{\pm}(p)\},
\end{align*}
by \Cref{prop:mon_cont_kneading}, we have $\alpha \leq q_{1} \leq q_{2} \leq \alpha+\beta-\beta'$ and $\tau_{\beta', a}^{\pm}(p_{\beta',a})\vert_{n} = \tau_{\beta, \alpha}^{\pm}(p)\vert_{n}$, for all $a \in [q_{1}, q_{2}]$. Moreover, $\tau_{\beta', a}^{-}(p_{\beta',a}) \preceq \tau_{\beta, \alpha}^{-}(p) \prec \tau_{\beta, \alpha}^{+}(p) \preceq \tau_{\beta', a}^{+}(p_{\beta',a})$, for all $a \in [q_{1}, q_{2}]$, implying one of the following sets of orderings.
\begin{align*}
\tau_{\beta', a}^{-}(p_{\beta',a}) \prec \tau_{\beta, \alpha}^{-}(p) &\prec \tau_{\beta, \alpha}^{+}(p) \prec \tau_{\beta', a}^{+}(p_{\beta', a})\\
\tau_{\beta', a}^{-}(p_{\beta',a}) = \tau_{\beta, \alpha}^{-}(p) &\prec \tau_{\beta, \alpha}^{+}(p) \prec \tau_{\beta', a}^{+}(p_{\beta', a})\\
\tau_{\beta', a}^{-}(p_{\beta',a}) \prec \tau_{\beta, \alpha}^{-}(p) &\prec \tau_{\beta, \alpha}^{+}(p) = \tau_{\beta', a}^{+}(p_{\beta', a})
\end{align*}
If either the first case occurs, the second case occurs and $\tau_{\beta', a}^{+}(p_{\beta',a})$ is not periodic, or the third case occurs and $\tau_{\beta', a}^{-}(p_{\beta',a})$ is not periodic, then an application of \Cref{prop:mon_cont_kneading} and \Cref{thm:LSSS} yields the required result.
This leaves two remaining sub-cases, namely when $\tau_{\beta', a}^{-}(p_{\beta',a}) = \tau_{\beta, \alpha}^{-}(p) \prec \tau_{\beta, \alpha}^{+}(p) \prec \tau_{\beta', a}^{+}(p_{\beta', a})$ with $\tau_{\beta', a}^{+}(p_{\beta', a})$ periodic, and when $\tau_{\beta', a}^{-}(p_{\beta',a}) \prec \tau_{\beta, \alpha}^{-}(p) \prec \tau_{\beta, \alpha}^{+}(p) = \tau_{\beta', a}^{+}(p_{\beta', a})$ with $\tau_{\beta', a}^{-}(p_{\beta', a})$ periodic. Let us consider the first of these two sub-cases; the second follows by an analogous arguments.
For ease of notation let $\nu' = \tau_{\beta', a}^{+}(p_{\beta',a})$ and note that by assumption $\omega = \tau_{\beta', a}^{-}(p_{\beta',a})$ and that $\omega \prec \nu \prec \nu'$. If the map $s \mapsto \tau_{\beta', s}^{+}(p_{\beta',s})$ is continuous at $s = a$, then an application of \Cref{prop:mon_cont_kneading} and \Cref{thm:LSSS} yields the required result; if we do not have continuity at $s = a$, by \Cref{prop:mon_cont_kneading} we have that $\nu'$ is periodic with periodic $N$, for some $N \in \mathbb{N}$, and thus an application of \Cref{thm:one_perioidc} completes the proof, alternatively we may proceed as follows.
We claim that $\nu \prec \nu'\vert_{N}\omega$. Indeed if $\nu\vert_{N} \prec \nu'\vert_{N}$, the claim follows immediately, and so let us suppose that $\nu\vert_{N} = \nu'\vert_{N}$. If $\nu_{N+1} = 0$, the claim follows, from \Cref{thm:Structure}. On the other hand, by \Cref{thm:Structure}, if $\nu_{N+1} = 1$, then $\sigma^{N}(\nu) \succeq \nu$. If $\sigma^{N}(\nu)\vert_{N} \succ \nu\vert_{N} = \nu'\vert_{N}$, then $\nu \succ \nu'$, contradicting our assumption that $\nu \prec \nu'$, and so $\sigma^{N}(\nu)\vert_{N} = \nu\vert_{N}$. This implies there exists a minimal integer $m$ such that $\nu_{m N + 1} = 0$ and $\nu\vert_{mN} = \nu'\vert_{mN}$; otherwise $\nu$ would be periodic. However, this together with \Cref{thm:Structure}, yields that $\nu \preceq \sigma^{(m-1)N}(\nu) = \nu\vert_{N}\sigma^{mN}(\nu) \preceq \nu\vert_{N}\omega = \nu'\vert_{N}\omega$, as required.
To complete the proof of this sub-case we appeal once more to \Cref{prop:mon_cont_kneading} which together with the above implies that there exists a real number $\delta > 0$ such that for all $a' \in (a-\delta,a)$ we have $\tau_{\beta',a'}^{-}(p_{\beta',a'}) \prec \omega$ and $\nu \prec \tau_{\beta',a'}^{+}(p_{\beta',a'}) \prec \nu'\vert_{N}\omega \prec \nu'$. An application of \Cref{thm:LSSS} yields the required result.
\end{proof}
In the above proof, it is critical that $\omega$ and $\nu$ are not periodic, as this allows us to construct $\beta'$, $q_{1}$ and $q_{2}$ so that $\omega$ and $\nu$ are sufficiently close to $\tau_{\beta',a'}^{-}(p_{\beta',a'})$ and $\tau_{\beta',a'}^{+}(p_{\beta',a'})$, respectively, for all $a' \in [q_{1}, q_{2}]$. However, under the assumption that $\nu$ is periodic we may not use our construction to build such $\beta'$ and hence $q_{1}$ and $q_{2}$. Indeed, the strict inequalities in \Cref{eq:either_or} no longer hold, and thus the ordering given in \eqref{smaller} fails.
\section{Survivor sets of intermediate \texorpdfstring{$\beta$}{beta}-transformations: Proof of \texorpdfstring{\Cref{Cor_2,Cor_3}}{Corollaries 1.3 and 1.4}}\label{sec:proof_cor_1_3_4}
Here, we examine open dynamical systems on the unit interval with a hole at zero and where the dynamics is driven by an intermediate \mbox{$\beta$-transformation}. With the aid of \Cref{thm:main} we can relate such open dynamical system to open dynamical systems driven by greedy \mbox{$\beta$-transformations}. This allows us to transfer the results of \cite{KKLL} and \cite{AK} on isolated points in $E_{\beta,\alpha}^+$, the Hausdorff dimension of survivor sets, and the critical point of the dimension function from the Greedy case to the intermediate case. For readability, we omit the $0$ in notation of $\pi_{\beta,0}$, $E_{\beta}^+=E_{\beta,0}^+$, and so on, and thus write $\pi_{\beta}$ for $\pi_{\beta,0}$, $E_{\beta}^+$ for $E_{\beta,0}^+$, and so forth.
By \Cref{thm:Parry_converse,thm:Structure}, given $(\beta,\alpha) \in \Delta$, there exists a unique $\beta^\prime \in (1, 2)$ with $\tau_{\beta,\alpha}^-(1)=\tau_{\beta^\prime}^-(1)$. Thus, we define a function $u \colon \Delta \to (1, 2)$ by $u(\beta,\alpha) \coloneqq \beta^\prime$, and let $\tilde{\pi}_{\beta,\alpha} \coloneqq \pi_{u(\beta,\alpha)} \circ \tau_{\beta,\alpha}^{+}$. Correlations of the systems $(T_{\beta,\alpha},[0,1])$ and $(T_{u(\beta,\alpha)},K_{u(\beta,\alpha)}^+(\tilde{\pi}_{\beta,\alpha}(0)))$ are expressed in the following proposition.
\begin{proposition}\label{prop:char}
Let $(\beta,\alpha)\in\Delta$ and let $\beta^\prime=u(\beta,\alpha)$.
\begin{enumerate}
\item $\tilde{\pi}_{\beta,\alpha}([0,1])=K_{\beta^\prime}^+(\tilde{\pi}_{\beta,\alpha}(0))$ and $\tilde{\pi}_{\beta,\alpha}(E_{\beta,\alpha}^+)=E_{\beta^\prime}^+\cap[\tilde{\pi}_{\beta,\alpha}(0),1]$.
\item For every $x\in E_{\beta,\alpha}^+$ we have that $x$ is isolated in $E_{\beta,\alpha}^+$ if and only if $\tilde{\pi}_{\beta,\alpha}(x)$ is isolated in $E_{\beta^\prime}^+$.
\item For $t\in(0,1)$, we have $ \tilde{\pi}_{\beta,\alpha}(K_{\beta,\alpha}^+(t))=K_{\beta^\prime}^+(\tilde{\pi}_{\beta,\alpha}(t)) $.
\item For $t\in(0,1)$, we have $\dim_H(K_{\beta,\alpha}^+(t))=(\log(\beta^\prime)/\log(\beta)) \dim_H(K_{\beta^\prime}^+(\tilde{\pi}_{\beta,\alpha}(t)))$.
\end{enumerate}
\end{proposition}
\newpage
\begin{proof}
Let us begin by proving Part~(1). To this end, observe that $\tilde{\pi}_{\beta,\alpha}$ is monotonic, since it is a composition of monotonic functions, and so, $\tilde{\pi}_{\beta,\alpha}(T_{\beta,\alpha}^n(x)) \geq \tilde{\pi}_{\beta,\alpha}(0)$ for all $x \in [0,1]$ and $n \in \mathbb{N}_{0}$. By the fact that the diagrams in \eqref{eq:commutative_diag} commute, we have for all $x \in [0,1]$ and $n \in \mathbb{N}$, that
\begin{align*}
T_{\beta'}^{n}(\tilde{\pi}_{\beta,\alpha}(x))
= T_{\beta'}^{n}(\pi_{\beta'}(\tau^{+}_{\alpha,\beta}(x)))
= \pi_{\beta'}(\sigma^{n}(\tau^{+}_{\alpha,\beta}(x)))
= \pi_{\beta'}(\tau^{+}_{\alpha,\beta}(T^{n}_{\beta,\alpha}(x))
= \tilde{\pi}_{\beta,\alpha}(T^{n}_{\beta,\alpha}(x)).
\end{align*}
Combining the above, we may conclude that $\tilde{\pi}_{\beta,\alpha}([0,1]) \subseteq K_{\beta^\prime}(\tilde{\pi}_{\beta,\alpha}(0))$. To prove that equality holds, namely that $\tilde{\pi}_{\beta,\alpha}([0,1]) = K_{\beta^\prime}(\tilde{\pi}_{\beta,\alpha}(0))$, using the commutativity of the diagrams in \eqref{eq:commutative_diag}, we observe that $x \in K_{\beta^\prime}(\tilde{\pi}_{\beta,\alpha}(0))$, if and only if, $\pi_{\beta'}(\tau^{+}_{\beta,\alpha}(0)) \leq \pi_{\beta'}(\sigma^{n}(\tau^{+}_{\beta'}(x)) \leq \pi_{\beta'}(\tau_{\beta'}^{-}(1))$ for all $n \in \mathbb{N}_{0}$. Since $\pi_{\beta'}$ is injective on $\Omega_{\beta'}$ and monotonic on $\{0,1\}^{\mathbb{N}}$, and since $\tau_{\beta,\alpha}^-(1)=\tau_{\beta^\prime}^-(1)$, it follows that $\tau_{\beta'}(x) \in \Omega^{+}_{\beta, \alpha}$. In other words, there exists a $y \in [0,1]$ such that $\tilde{\pi}_{\beta,\alpha}(y) = \pi_{\beta'}(\tau^{+}_{\beta,\alpha}(y)) = x$, yielding the first statement of Part~(1). Let us now prove the second statement. If $x \in \tilde{\pi}_{\beta,\alpha}(E_{\beta,\alpha}^{+})$, then there exists a $y \in E_{\beta,\alpha}^{+} \subset [0,1]$ with $\tilde{\pi}_{\beta,\alpha}(y) = x$. This in tandem with the fact that the the diagrams in \eqref{eq:commutative_diag} are commutative, and since the maps $\tau_{\beta,\alpha}^{+}$ and $\pi_{\beta'}$ are monotonic, we have
\begin{align*}
T_{\beta'}^{n}(x)
= T_{\beta'}^{n}( \pi_{\beta'}(\tau_{\beta,\alpha}^{+}(y)))
= \pi_{\beta'}(\sigma^{n}(\tau_{\beta,\alpha}^{+}(y)))
= \pi_{\beta'}(\tau_{\beta,\alpha}^{+}(T_{\beta,\alpha}^{n}(y)))
\geq \pi_{\beta'}(\tau_{\beta,\alpha}^{+}(y))
= x.
\end{align*}
Since $y \in [0,1]$ and $\tilde{\pi}_{\beta,\alpha}(y) = x$, and since $\tilde{\pi}_{\beta,\alpha}$ is monotonic, $x \geq \tilde{\pi}_{\beta,\alpha}(0)$. This, together with the fact that $E_{\beta'}^{+} \subseteq [0, 1]$, yields $\tilde{\pi}_{\beta,\alpha}(E_{\beta,\alpha}^+) \subseteq E_{\beta^\prime}^+\cap[\tilde{\pi}_{\beta,\alpha}(0),\tilde{\pi}_{\beta,\alpha}(1)]$. To see that $\tilde{\pi}_{\beta,\alpha}(E_{\beta,\alpha}^+) \supseteq E_{\beta^\prime}^+\cap[\tilde{\pi}_{\beta,\alpha}(0),1]$ let $x \in E_{\beta'}^{+}$ with $x \geq \tilde{\pi}_{\beta,\alpha}(0)$. By definition $E_{\beta'}^{+}$ and the commutativity of the diagrams in \eqref{eq:commutative_diag},
\begin{align*}
\pi_{\beta,\alpha}(\tau_{\beta,\alpha}^{+}(0))
\leq \pi_{\beta,\alpha}(\tau_{\beta'}^{+}(x))
\leq T^{n}_{\beta,\alpha}(\pi_{\beta,\alpha}(\tau_{\beta'}^{+}(x)))
\leq \pi_{\beta,\alpha} (\tau_{\beta'}^{-}(1))
\leq \pi_{\beta,\alpha} (\tau_{\beta,\alpha}^{-}(1)).
\end{align*}
In other words $\pi_{\beta,\alpha}(\tau_{\beta'}^{+}(x)) \in E_{\beta,\alpha}^+$. Since $\tilde{\pi}_{\beta,\alpha}$ is invertible on $[0,1)$ with inverse $\pi_{\beta,\alpha} \circ \tau^{+}_{\beta'}$ the result follows.
Part~(2) follows from Part~(1) using the fact that $\tilde{\pi}_{\beta,\alpha}$ is monotonic and injective on $[0,1)$. Part~(3) follows using analogous argument to those used above to proof Part~(1), and Part~(4) follows from \Cref{prop:ent} and Part~(3) in the following way.
\[
\dim_H(K_{\beta,\alpha}^+(t))=\frac{h_{\operatorname{top}}(T_{\beta,\alpha}\vert_{K^+_{\beta,\alpha}(t)})}{\log(\beta)}=\frac{h_{\operatorname{top}}(T_{\beta^\prime}\vert_{K^+_{\beta^\prime}(\tilde{\pi}_{\beta,\alpha}(t))})}{\log(\beta)}=\frac{\log(\beta^\prime)}{\log(\beta)}\dim_H(K_{\beta^\prime}^+(\tilde{\pi}_{\beta,\alpha}(t))).
\qedhere
\]
\end{proof}
For the next proposition we will require the following analogue of the map $\tilde{\pi}_{\beta,\alpha}$, namely $\tilde{\pi}_{\beta,\alpha}^{-} \coloneqq \pi_{u(\beta,\alpha)} \circ \tau_{\beta,\alpha}^{-}$. Note in the previous proposition, we could have also used the map $\tilde{\pi}_{\beta,\alpha}^{-}$ instead of $\tilde{\pi}_{\beta,\alpha}$ since they coincide on all points considered in (1)--(4). However, in the proof of \Cref{prop:char} we would need to replace $T_{\beta,\alpha}$ by $T_{\beta,\alpha}^{-}$, $T_{\beta'}$ by $T_{\beta'}^{-}$, $\tau_{\beta,\alpha}^{\pm}$ by $\tau_{\beta,\alpha}^{\mp}$ and $\tau_{\beta'}^{\pm}$ by $\tau_{\beta'}^{\mp}$, making it notionally heavy, and thus for ease of notation we use $\tilde{\pi}_{\beta,\alpha}$.
\begin{proposition}\label{prop:char(5)}
For all $(\beta,\alpha)\in\Delta$, we have that $\tilde{\pi}^{-}_{\beta,\alpha}(t_{\beta,\alpha,c})=t_{u(\beta,\alpha),c}$.
\end{proposition}
\begin{proof}
Observe that there exists a sequence of real numbers $(t_{n})_{n\in \mathbb{N}}$ with $t_n \in E_{\beta,\alpha}^{+}$ such that $t_n < t_{\beta,\alpha,c}$ and $\lim_{n\to \infty} t_{n} = t_{\beta,\alpha,c}$; otherwise the dimension function would be constant around $t_{\beta,\alpha,c}$ contradicting its definition. Define $\hat{t}_n =\tilde{\pi}^{-}_{\beta,\alpha}(t_n)$. By Proposition \ref{prop:char} Part~(3), for all $n\in \mathbb{N}$, we have $\tilde{\pi}^{-}_{\beta,\alpha}(K_{\beta,\alpha}^+(t_n))=K_{u(\beta,\alpha)}^+(\hat{t}_n)$. An application of Proposition \ref{prop:char} Part~(4) together with our remarks directly preceding this Proposition, yields for $n \in \mathbb{N}$,
\begin{align*}
\dim_H(K_{u(\beta,\alpha)}^+(\hat{t}_n))>0, \quad
\dim_H(K_{\beta,\alpha}^+(t_{\beta,\alpha,c}))=0,
\quad \text{and} \quad
\dim_H(K_{u(\beta,\alpha)}^+(\tilde{\pi}^{-}_{\beta,\alpha}(t_{\beta,\alpha,c}))=0.
\end{align*}
As $\pi_{u(\beta,\alpha)}$ is continuous and $\tau_{\beta,\alpha}^{-}$ is left continuous, $\tilde{\pi}^{-}_{\beta,\alpha}$ is left continuous, and so $\tilde{\pi}^{-}_{\beta,\alpha}(\lim_{n\to \infty}(t_n))= \lim_{n\to \infty} \tilde{\pi}^{-}_{\beta,\alpha}(t_N)$. This implies that $\tilde{\pi}^{-}_{\beta,\alpha}(t_{\beta,\alpha,c}) = \lim_{n\to \infty} \hat{t}_N$, and hence that $\tilde{\pi}^{-}_{\beta,\alpha}(t_{\beta,\alpha,c})=t_{u(\beta,\alpha),c}$.
\end{proof}
The value of $t_{u(\beta,\alpha),c}$ is explicitly given in \cite{KKLL} when $\tau_{\beta,\alpha}^-(1)$ is balanced. For all other cases see \cite{AK}.
A word $\omega = \omega_{1}\omega_{2} \cdots \in \{0,1\}^{\mathbb{N}}$ is called \textsl{balanced} if $\lvert(\omega_{n} + \omega_{n+1} + \cdots + \omega_{n+m}) - (\omega_{k+n} + \omega_{k+n+1} + \cdots + \omega_{k+n+m}) \rvert \leq 1$ for all $k$, $n$ and $m \in \mathbb{N}_{0}$ with $n \geq 1$. Following notation of \cite{KKLL} and \cite{Schme}, we let
\begin{align*}
C_3 \coloneqq\{ \beta\in(1,2) \colon \text{the length of consecutive zeros in} \; \tau_\beta^-(1) \; \text{is bounded} \}
\;\; \text{and} \;\;
C \coloneqq \{ \beta \in (1,2) \colon \tau_\beta^-(1) \; \text{is balanced} \}.
\end{align*}
For every $(\beta,\alpha) \in \Delta$ with $\alpha>0$, we have that $u(\beta,\alpha)\in C_3$. By \cite[Theorem 3.12]{KKLL}, for $\beta \in C_3$, there exists $\delta>0$ such that $E_{\beta}^+\cap[0,\delta]$ contains no isolated points. With this in mind and, for $\beta \in (1,2)$, setting $\delta(\beta) \coloneqq \sup \{ \delta \in [0, 1] \colon E_{\beta}^+\cap[0,\delta] \; \text{contains no isolated points} \}$, we have the following corollary of Proposition \ref{prop:char}.
\begin{corollary}\label{cor:isoloated_pts}
Let $(\beta,\alpha)\in \Delta$ with $\alpha>0$. If $\tilde{\pi}_{\beta,\alpha}(0)<\delta(u(\beta,\alpha))$ then there exists a $\delta>0$ such that $E_{\beta,\alpha}^+\cap[0,\delta]$ contains no isolated points. Further, if $u(\beta,\alpha) \in C$, then $\delta(u(\beta,\alpha))=1$ and $E_{\beta,\alpha}^+$ contains no isolated points.
\end{corollary}
\begin{proof}
The first statement follows from \Cref{prop:char} Parts (1) and (2), and \cite[Theorem 3.12]{KKLL}. The second statement follows from \cite[Theorem 3]{KKLL}, which states that if $\beta\in C$ then $E_\beta$ does not contain any isolated points.
\end{proof}
\begin{proof}[Proof of \texorpdfstring{\Cref{Cor_2}}{Corollary 1.3}]
In \cite{MR0166332} an absolutely continuous invariant measure of $T_{\beta,\alpha}$ is constructed, and in \cite{Hof} it is shown that this measure is ergodic. (In fact it is shown that it is maximal, and the only measure of maximal entropy.) This yields, given an $m \in \mathbb{N}$, that for almost all $x \in [0,1]$, there exists $n_{x} = n \in \mathbb{N}_{0}$ such that $T_{\beta,\alpha}^{n}(x) \in [0, 1/m)$, and hence that $K_{\beta,\alpha}^{+}(1/m)$ is a Lebesgue null. Since $E_{\beta,\alpha}^{+} \setminus \{0\} \subseteq \cup_{m=1}^{\infty} K_{\beta,\alpha}^{+}(1/m)$, by subadditivity of the Lebesgue measure, it follows that $E_{\beta,\alpha}^{+}$ is a Lebesgue null set. The statement on the isolated points of $E_{\beta,\alpha}^{+}$ follows from \Cref{cor:isoloated_pts}.
\end{proof}
\begin{proof}[Proof of \texorpdfstring{\Cref{Cor_3}}{Corollary 1.4}]
This is a direct consequence of \Cref{prop:char} Part~(4) and \cite[Theorem~A (ii)]{KKLL}.
\end{proof}
\begin{proof}[Proof of \texorpdfstring{\Cref{Cor_E_beta_alpha}}{Corollary 1.5}]
Let $(\beta, \alpha) \in \Delta$ with $\Omega_{\beta,\alpha}$ a subshift of finite type and $T_{\beta,\alpha}$ transitive, let $P_{\beta,\alpha}$ denote the Markov partition of $T_{\beta,\alpha}$ defined in \eqref{eq:Markov_Partition}, and for $n\in \mathbb{N}$, let $\Omega_{T_{\beta,\alpha}}\vert_{n}$ denote the set of all length $n$ admissible words of $T_{\beta,\alpha}$ with respect to the partition $P_{\beta,\alpha}$. Fix $n \in \mathbb{N}$ sufficient large, set $\omega$ to be lexicographically the smallest word in $\Omega_{T_{\beta,\alpha}}\vert_{n}$, let $a_{n} = a_{\beta, \alpha, n} = \sup I(\omega)$ and let $\nu \in \Omega_{T_{\beta,\alpha}}\vert_{n}$ with $\nu \succ \omega$. By transitivity and the Markov property, there exist $k \in \mathbb{N}$ and $\xi \in \Omega_{\beta,\alpha}\vert_{k}$ with $k > n$, $I(\xi) \subset I(\omega)$, $T_{\beta,\alpha}^{k-n}(I(\xi)) = I(\nu)$, and $T_{\beta,\alpha}^j(I(\xi))$ an interval and $T_{\beta,\alpha}^{j}(x) \geq a_{n}$ for all $j \in \{1,2, \dots, k - n\}$ and $x \in I(\xi)$. In other words, there exists a linear scaled copy of $K_{\beta,\alpha}^{+}(a_{n}) \cap I(\nu)$ in $I(\xi) \cap E_{\beta,\alpha}^{+}$. Namely, we have
\begin{align}\label{eq:scaling_func}
f_{\beta,\alpha,\nu,n}(K_{\beta,\alpha}^{+}(a_{n}) \cap I(\nu)) \subseteq I(\xi) \cap E_{\beta,\alpha}^{+},
\end{align}
where $f_{\beta, \alpha,\nu, n} = f_{\beta,\alpha,\chi(\xi_{1})} \circ f_{\beta,\alpha,\chi(\xi_{2})} \circ \cdots \circ f_{\beta,\alpha,\chi(\xi_{k-n})}$ and $\chi \colon \Omega_{T_{\beta,\alpha}}\vert_{1} \to \{ 0, 1\}$ is defined by
\begin{align*}
\chi(a) = \begin{cases}
0 & \text{if} \; I(a) \subseteq [0, p_{\beta,\alpha}],\\
1 & \text{otherwise.}
\end{cases}
\end{align*}
Here, we recall that $f_{\beta,\alpha,0}(x) = \beta^{-1}x-\alpha\beta^{-1}$ and $f_{\beta,\alpha, 1}(x) = \beta^{-1}x-(\alpha-1)\beta^{-1}$ for $x \in [0,1]$. With the above at hand, we may conclude that
\begin{align*}
\dim_H (E_{\beta,\alpha}^{+})
\geq \max \{ \dim_H(K_{\beta,\alpha}^{+}(a_{n}) \cap I(\nu)) \colon \nu \in \Omega_{T_{\beta,\alpha}}\vert_{n} \}
= \dim_H(K_{\beta,\alpha}^{+}(a_{n})).
\end{align*}
Since $n$ was chosen sufficiently large but arbitrarily, this in tandem with \Cref{Cor_3} implies $\dim_H (E_{\beta,\alpha}^{+}) = 1$, since $a_n$ converges to zero as $n$ tends to infinity.
Since Hausdorff dimension is preserved under taking linear transformations, an application of \Cref{thm:G1990+Palmer79,thm:LSSS}, yields for $(\beta, \alpha) \in \Delta$ with $\Omega_{\beta,\alpha}$ a subshift of finite type, that $\dim_H (E_{\beta,\alpha}^{+}) = 1$.
To conclude, let $(\beta,\alpha) \in \Delta$ be chosen arbitrarily, and let $((\beta_{n},\alpha_{n}))_{n \in \mathbb{N}}$ denote the sequence of tuples given in \Cref{Cor_1} converging to $(\beta, \alpha)$. Set $\tilde{\pi}_{\beta, \alpha}^{(n)} = \pi_{\beta,\alpha} \circ \tau^{+}_{\beta_{n},\alpha_{n}}$, and for $t$ and $s \in [0,1]$ with $t < s$, let
\begin{align*}
K_{\beta,\alpha}^{+}(t, s) \coloneqq \{ x \in[0, 1) \colon T_{\beta,\alpha}^{n}(x) \not \in [0,t) \cup (s, 1] \; \textup{for all} \; n \in \mathbb{N}_{0} \}
\end{align*}
By \Cref{Cor_1}, \Cref{thm:Structure}, and the commutativity of the diagram in \eqref{eq:commutative_diag}, we may choose $((\beta_{n},\alpha_{n}))_{n \in \mathbb{N}}$ so that $(\pi_{\beta,\alpha}^{(n)}(0))_{n \in \mathbb{N}}$ is a monotonically decreasing sequence converging to zero, and $(\pi_{\beta,\alpha}^{(n)}(1))_{n \in \mathbb{N}}$ is a monotonically increasing sequence converging to one. Thus, by construction, for $n$ and $l \in \mathbb{N}$ with $l \geq n$,
\begin{align*}
K_{\beta,\alpha}^{+}(\pi_{\beta,\alpha}^{(n)}(0), \pi_{\beta,\alpha}^{(n)}(1)) \subseteq K_{\beta,\alpha}^{+}(\pi_{\beta,\alpha}^{(n)}(0), \pi_{\beta,\alpha}^{(l)}(1)),
\quad \text{and} \quad
K_{\beta,\alpha}^{+}(\pi_{\beta,\alpha}^{(n)}(0)) = \bigcup_{m \in \mathbb{N}} K_{\beta,\alpha}^{+}(\pi_{\beta,\alpha}^{(n)}(0), \pi_{\beta,\alpha}^{(m)}(1)).
\end{align*}
Hence, by countable stability of the Hausdorff dimension and \Cref{Cor_3},
\begin{align}\label{eq:limit_hausdorff_proj_0,1}
\lim_{n \to \infty} \dim_{H}(K_{\beta,\alpha}^{+}(\pi_{\beta,\alpha}^{(n)}(0),\pi_{\beta,\alpha}^{(n)}(1))) = 1.
\end{align}
Via analogous arguments to those given in the proof of \Cref{prop:char}, we have the following.
\begin{enumerate}
\item[($1^{*}$)] $\tilde{\pi}_{\beta,\alpha}^{(n)}([0,1])=K_{\beta,\alpha}^+(\tilde{\pi}_{\beta,\alpha}^{(n)}(0), \tilde{\pi}_{\beta,\alpha}^{(n)}(1))$ and $\tilde{\pi}_{\beta,\alpha}^{(n)}(E_{\beta_{n},\alpha_{n}}^+)=E_{\beta, \alpha}^+\cap[\tilde{\pi}_{\beta,\alpha}^{(n)}(0),\tilde{\pi}_{\beta,\alpha}^{(n)}(1)]$.
\item[($3^{*}$)] For $t\in(0,1)$, we have $ \tilde{\pi}_{\beta,\alpha}^{(n)}(K_{\beta_{n},\alpha_{n}}^+(t))=K_{\beta,\alpha}^+(\tilde{\pi}_{\beta,\alpha}^{(n)}(t), \tilde{\pi}_{\beta,\alpha}^{(n)}(1)) $.
\end{enumerate}
For $k \in \mathbb{N}$ sufficiently large and $\nu \in \Omega_{T_{\beta_{n},\alpha_{n}}}\vert_{k}$, setting $a_{n,k} = a_{\beta_{n},\alpha_{n},k}$, from the equalities given in \eqref{eq:alt_IFS} and \eqref{eq:scaling_func}, the commutativity of the diagram in \eqref{eq:commutative_diag} and ($1^{*}$), we have that
\begin{align*}
f_{\beta,\alpha, \nu, k}(
\pi_{\beta,\alpha}^{(n)}( K_{\beta_{n},\alpha_{n}}^{+}(a_{n,k}) \cap I(\nu)))
=\pi_{\beta,\alpha}^{(n)}( f_{\beta_{n},\alpha_{n}, \nu, k}( K_{\beta_{n},\alpha_{n}}^{+}(a_{n,k}) \cap I(\nu)))
\subseteq \pi_{\beta,\alpha}^{(n)}(E_{\beta_{n},\alpha_{n}}^{+})
\subseteq E_{\beta,\alpha}.
\end{align*}
This in tandem with ($3^{*}$), the fact that there exists $\nu \in \Omega_{T_{\beta_{n},\alpha_{n}}}\vert_{k}$ with
\begin{align*}
\dim_{H}(\pi_{\beta,\alpha}^{(n)}( K_{\beta_{n},\alpha_{n}}^{+}(a_{n,k}) \cap I(\nu))) = \dim_{H}(\pi_{\beta,\alpha}^{(n)}( K_{\beta_{n},\alpha_{n}}^{+}(a_{n, k}))),
\end{align*}
and that Hausdorff dimension is invariant under linear scaling, implies that
\begin{align*}
\dim_{H}(K_{\beta,\alpha}^{+}(\pi_{\beta,\alpha}^{(n)}(a_{n,k}),
\pi_{\beta,\alpha}^{(n)}(1)))
=\dim_{H}(\pi_{\beta,\alpha}^{(n)}( K_{\beta_{n},\alpha_{n}}^{+}(a_{n,k}))) \leq \dim_{H}(E_{\beta,\alpha}).
\end{align*}
This in tandem with \Cref{Cor_3}, the equality given in \eqref{eq:limit_hausdorff_proj_0,1}, the observations that, for $n \in \mathbb{N}$, the sequence $(\pi_{\beta,\alpha}^{(n)}(a_{n,k}))_{k \in \mathbb{N}}$ is monotonically decreasing with $\lim_{k \to \infty} \pi_{\beta,\alpha}^{(n)}(a_{n,k}) =\pi_{\beta,\alpha}^{(n)}(0)$, and for $l$ and $m \in \mathbb{N}$ with $l \geq m$,
\begin{align*}
K_{\beta,\alpha}^{+}(\pi_{\beta,\alpha}^{(n)}(a_{n,m}),
\pi_{\beta,\alpha}^{(n)}(1)) \subseteq K_{\beta,\alpha}^{+}(\pi_{\beta,\alpha}^{(n)}(a_{n,l}), \pi_{\beta,\alpha}^{(n)}(1)),
%
\quad \text{and} \quad
%
K_{\beta,\alpha}^{+}(\pi_{\beta,\alpha}^{(n)}(0),
\pi_{\beta,\alpha}^{(n)}(1)) = \bigcup_{k \in \mathbb{N}} K_{\beta,\alpha}^{+}(\pi_{\beta,\alpha}^{(n)}(a_{n,k}),
\pi_{\beta,\alpha}^{(n)}(1)),
\end{align*}
and the countable stability of the Hausdorff dimension, yields the required result.
\end{proof}
\subsection*{Examples and applications}
\begin{figure}[t]
\centering
\begin{subfigure}[b]{0.32\textwidth}
\includegraphics[width=\textwidth]{dimhbeta110alpha0.png}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.32\textwidth}
\includegraphics[width=\textwidth]{dimhsymbetagoldenmean.png}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.32\textwidth}
\includegraphics[width=\textwidth]{dimhbeta10alpha0001.png}
\end{subfigure}
\caption{Graphs of $\eta_{\beta,\alpha}$: on the left, $\eta_{\beta}(t)$ with $\beta$ such that $\tau_\beta^-(1)=(110)^\infty$; in the middle, $\eta_{\beta,\alpha}$ for $(\beta, \alpha) \in \Delta$ with $\tau_{\beta,\alpha}^-(1)=(110)^\infty$ and $\tau_{\beta,\alpha}^+(0)=(001)^\infty$; on the right, $\eta_{\beta}$ for $(\beta, \alpha) \in \Delta$ such that $\tau_{\beta,\alpha}^-(1)=(10)^\infty$ and $\tau_{\beta,\alpha}^+(0)=(0001)^\infty $.}\label{fig:dim}
\end{figure}
Let $(\beta,\alpha) \in \Delta$ be such that $\tau^-_{\beta,\alpha}(1)=(10)^\infty$. In which case, $u(\beta,\alpha)$ is equal to the golden mean, which we denote by $G$, and belongs to the set $C$. Thus, $E^{+}_{\beta,\alpha}$ contains no isolated points. From \cite[Proposition 5.2]{KKLL} and by an elementary calculation, we have that $t_{u(\beta,\alpha),c} = G^{-2}$ and $\tau_{G}^{-}(G^{-2}) = 00(10)^\infty$. This, in tandem with \Cref{prop:char(5)}, yields $t_{\beta,\alpha,c} = \pi_{\beta, \alpha} \tau_{G}^{-}(G^{-2}) = \pi_{\beta, \alpha}(00(10)^\infty) = \alpha/(1-\beta)+1/(\beta^{3}-1)$, which one can show is equal to $(1-\alpha-\beta\alpha)\beta^{-2}$ using the fact that $\tau^-_{\beta,\alpha}(1)=(10)^\infty$. Moreover, by \Cref{prop:char} Part~(4),
\begin{align*}
\dim_H(K_{\beta,\alpha}(t))=(\log(G)/\log(\beta)) \dim_H(K_{G}(\tilde{\pi}_{\beta,\alpha}(t)),
\end{align*}
for all $t \in (0, 1)$. We now show that for a given $\beta\in(1,G)$ there exists a unique $\alpha\in(0,1/2)$ with $G = u(\beta,\alpha)$, or equivalently, that for a given $\beta\in(1,G)$ there exists a unique $\alpha\in(0,1/2)$ with $T_{\beta,\alpha}(1) = p_{\beta,\alpha}$. Using the definitions of the involved terms, $T_{\beta,\alpha}(1) = p_{\beta,\alpha}$ if and only if $\alpha=1-\beta^2/(\beta+1)$. Noting, when $\beta = G$, that $\alpha=0$, and as $\beta$ approaches $1$ from above, that $\alpha =1-\beta^2/(\beta+1)$ converges to $1/2$, yields the required result.
By \Cref{thm:LSSS}, under the assumption that $\tau^-_{\beta,\alpha}(1)=(10)^\infty$, if $\tau_{\beta,\alpha}^+(0)$ is periodic, then $\Omega_{\beta,\alpha}$ is a subshift of finite type. We now find $(\beta, \alpha) \in \Delta$ such that $\tau_{\beta,\alpha}^+(0)=(0001)^\infty$ and $\tau^-_{\beta,\alpha}(1)=(10)^\infty$. For this we, observe that
\begin{align*}
T_{\beta,\alpha}(0) = \alpha, \quad
T_{\beta,\alpha}^2(0) = \beta\alpha+\alpha, \quad
T_{\beta,\alpha}^3(0) = \beta(\beta\alpha+\alpha)+\alpha, \quad \text{and} \quad
T_{\beta,\alpha}^4(0) = \beta(\beta(\beta\alpha+\alpha)+\alpha)+\alpha-1=0.
\end{align*}
Substituting $\alpha=1-\beta^2/(\beta+1)$ into the last equality, gives
\begin{align*}
\beta(\beta(\beta(1-\beta^2/(\beta+1))+(1-\beta^2/(\beta+1)))+(1-\beta^2/(\beta+1)))+(1-\beta^2/(\beta+1))-1 = 0.
\end{align*}
This reduces to $\beta(\beta^4 -\beta^2- \beta -1) =0$. Thus, if $\beta$ is the positive real root of $\beta^4 -\beta^2- \beta -1 =0$ and $\alpha=1-\beta^2/(\beta+1)$, then $\tau_{\beta,\alpha}^+(0)=(0001)^\infty$ and $\tau^-_{\beta,\alpha}(1)=(10)^\infty$. Numerically approximating $\beta$ and $\alpha$ yields $\beta\approx 1.4656$ and $\alpha\approx 0.1288$.
We utilise the above, in particular \Cref{prop:char}, in studying the dimension function $\eta_{\beta,\alpha}$. Recall, if $t\not \in E^{+}_{\beta,\alpha}$, then there exists $t^*>t$ with $K^{+}_{\beta,\alpha}(t)=K^{+}_{\beta,\alpha}(t^*)$. Thus, it suffices to study $K^{+}_{\beta,\alpha}(t)$ for $t\in E^{+}_{\beta,\alpha}$. For a fixed $t\in E^{+}_{\beta,\alpha}$, with the aid of \Cref{thm:BSV14}, we find $(\beta^\prime,\alpha^\prime) \in \Delta$ with $\tau_{\beta,\alpha}(t)=\tau_{\beta^\prime,\alpha^\prime}(0)$ and $\tau^{-}_{\beta,\alpha}(1) = \tau^{-}_{\beta^\prime,\alpha^\prime}(1)$.
By \Cref{prop:char} Part~(4),
\begin{align*}
\eta_{\beta,\alpha}(t)=\frac{\log(u(\beta,\alpha))}{\log(\beta)} \dim_H(K_{u(\beta,\alpha)}^+(\tilde{\pi}_{\beta,\alpha}(t)))
\quad \text{and} \quad
\eta_{\beta^\prime,\alpha^\prime}(0)=\frac{\log(u(\beta^\prime,\alpha^\prime))}{\log(\beta^\prime)} \dim_H(K_{u(\beta^\prime,\alpha^\prime)}^+(\tilde{\pi}_{\beta^\prime,\alpha^\prime}(0))).
\end{align*}
Since $u(\beta^\prime,\alpha^\prime)=u(\beta,\alpha)$, $\tilde{\pi}_{\beta,\alpha}(t)=\tilde{\pi}_{\beta^\prime,\alpha^\prime}(0)$ and $\eta_{\beta^\prime,\alpha^\prime}(0)=1$, we have $\eta_{\beta,\alpha}(t)=\log(\beta^\prime)/\log(\beta)$. In summary, determining the value of $\eta_{\beta, \alpha}(t)$ reduces down to finding such $\alpha^\prime$ and $\beta^\prime$. This can performed numerically with the aid of the monotonicity and continuity of the projection maps, see Figure \ref{fig:dim} for sample numerical outputs.
\section{Winning sets of intermediate \texorpdfstring{$\beta$}{beta}-transformations: Proof of Proof of \texorpdfstring{\Cref{thm:main_2}}{Theorem 1.6}}\label{sec:proof_thm_1_6}
To show the conditions of \Cref{thm_HY_Thm_2.1} are satisfied when $T=T_{\beta, \alpha}$ for all $(\beta, \alpha) \in \Delta$ with $T_{\beta, \alpha}$ transitive and $\Omega_{\beta, \alpha}$ of finite type we use the following lemma on the geometric length of cylinder sets and the following proposition.
\begin{lemma}\label{lem:geometric_lengths_of_cylinders}
Let $(\beta, \alpha) \in \Delta$ be such that $T_{\beta, \alpha}$ is transitive and $\Omega_{\beta, \alpha}$ is a subshift of finite type. If $\nu = \nu_{1} \cdots \nu_{\lvert \nu \rvert}$ is an admissible word with respect to the partition $P_{\beta,\alpha}$, then $\rho \beta^{-\lvert \nu \rvert} \leq \lvert I(\nu) \rvert \leq \beta^{-\lvert \nu \rvert}$, where $\rho = \min \{ \lvert I(i) \rvert \colon i \in \Lambda \}$.
\end{lemma}
\begin{proof}
If $\lvert \nu \rvert = 1$, the result is a consequence of the fact that $\max\{ p_{\beta,\alpha}, 1 - p_{\beta,\alpha}\} \leq \beta^{-1}$, and that $I(\nu) \subseteq [0, p_{\beta,\alpha}]$ or $I(\nu) \subseteq [p_{\beta,\alpha},1]$. Therefore, we may assume that $\lvert \nu \rvert \geq 2$. Since $T_{\beta, \alpha}$ is Markov with respect to the partition $P_{\beta,\alpha}$, for $j \in \{ 0, 1, \ldots, \lvert \nu \rvert \}$, we have $T_{\beta,\alpha}^{j}(I(\nu))$ is an interval and $T_{\beta,\alpha}^{\lvert \nu \rvert - 1}(J(\nu)) = J(\nu_{\lvert \nu \rvert})$, where for a given admissible finite word $\omega$, we denote by $J(\omega)$ the interior of $I(\omega)$.
This implies that $\lvert I(\nu) \rvert = \beta^{-\lvert \nu \rvert + 1}\lvert I(\nu_{\lvert \nu \rvert}) \rvert$ and hence that $\rho \beta^{-\lvert \nu \rvert} \leq \lvert I(\nu_{\lvert \nu \rvert}) \rvert \beta^{-\lvert \nu \rvert + 1} = \lvert I(\nu) \rvert = \lvert I(\nu_{\lvert \nu \rvert}) \rvert \beta^{-\lvert \nu \rvert + 1} \leq \beta^{-\lvert \nu \rvert}$.
\end{proof}
\begin{proposition}\label{prop:thm_SFT+Transitive_implies_winning}
Under the hypotheses of \Cref{lem:geometric_lengths_of_cylinders}, for all $x \in [0, 1]$ and $\gamma \in (0, 1)$, we have that the geometric condition $H_{x, \gamma}$, with $T = T_{\beta, \alpha}$ and the partition $P_{\beta,\alpha}$, is satisfied.
\end{proposition}
\begin{proof}
\Cref{lem:geometric_lengths_of_cylinders} yields \eqref{eq:condition_1} of $H_{\xi, \gamma}$, thus is suffices to show that \eqref{eq:condition_2} of $H_{x, \gamma}$ is satisfied. To this end, let $n-1$ denote the cardinality of $P_{\beta,\alpha}$, and observe that, since by assumption $T_{\beta, \alpha}$ is transitive, there exists an $m = m_{\beta, \alpha} \in \mathbb{N}$, so that $J(l) \subseteq T_{\beta,\alpha}^{m}(J(k))$, for all $l$ and $k \in \{0, 1, \ldots, n-1\}$, where $J(l)$ and $J(k)$ are as defined in the proof of \Cref{lem:geometric_lengths_of_cylinders}. Further, if for two admissible words $\nu$ and $\eta$, we have that $I(\nu)$ and $I(\eta)$ are $\gamma/4$-comparable, then by \Cref{lem:geometric_lengths_of_cylinders} there exists $k_{0} \in \mathbb{N}$ with $\lvert \lvert \nu \rvert - \lvert \eta \rvert \rvert \leq k_{0}$. Letting $\omega$ denote the symbolic representation of $x$ generated by $T_{\beta, \alpha}$ with respect to the partition $P_{\beta,\alpha}$, set
\begin{align*}
M = M_{\beta, \alpha} = \begin{cases}
\max \{ \sigma^{k}(\omega) \wedge \omega \colon k \in \{1, 2, \ldots, k_{0} \}\} & \text{if} \; \omega \; \text{is not periodic},\\
\max \{ \sigma^{k}(\omega) \wedge \omega \colon k \in \{1, 2, \ldots, \operatorname{per}(\omega) -1 \}\} & \text{if} \; \omega \; \text{is periodic.}
\end{cases}
\end{align*}
Our aim is to show \eqref{eq:condition_2} of $H_{x, \gamma}$ is satisfied for all admissible words $\nu$ and $\eta$ with $I(\nu)$ and $I(\eta)$ are $\gamma/4$-comparable and all integers $i > i^{*} = k_{0} + m + M$. For this, suppose $\nu$ and $\eta$ are $\omega\vert_{i}$-extendable with $0 \leq \lvert \lvert \nu \rvert - \lvert \eta \rvert \rvert \leq k_{0}$ and $\operatorname{dist}(I(\nu\omega\vert_{i}), I(\eta\omega\vert_{i})) > 0$. We consider case when $\nu$ is not a prefix of $\eta$, and when $\nu$ is a prefix of $\eta$ separately.
For both of these cases we use the following facts. For $l \in \{ 0, 1, \ldots, n-1\}$ there exists a minimal $j \in \{ 1, 2, \ldots, m \}$ such that $T_{\beta, \alpha}^{j}(I(l))$ contains the interiors of at least two elements of $P_{\beta,\alpha}$. For $k \in \{1, 2, \ldots, \lvert \nu \rvert + i -1\}$ and $l \in \{1, 2, \ldots, \lvert \eta \rvert + i -1\}$, we have $T_{\beta, \alpha}^{k}(I(\nu \omega\vert_{i}))$ and $T_{\beta, \alpha}^{l}(I(\eta \omega\vert_{i}))$ are intervals, and $T_{\beta, \alpha}^{k}(J(\nu \omega\vert_{i})) = J(\sigma^{k}(\nu \omega\vert_{i}))$ and $T_{\beta, \alpha}^{l}(J(\eta \omega\vert_{i})) = J(\sigma^{l}(\eta \omega\vert_{i}))$.
Let us consider the first of our two cases, namely when $\nu$ is not a prefix of $\eta$. Our above two facts imply that there exist $l \in \{1, 2, \ldots, m-1\}$ and $F \subseteq \{0, 1, \ldots, n-1\}$ with $\lvert F \rvert \geq 2$ such that
\begin{align}\label{eq:HY-splitting_of_cylinders}
I(\nu \omega\vert_{l}) = \bigcup_{k \in F} I(\nu \omega\vert_{l} k)
\quad \text{and} \quad
I(\eta \omega\vert_{l}) = \bigcup_{k \in F} I(\eta \omega\vert_{l} k).
\end{align}
Since $\nu$ is not a prefix of $\eta$, there exists $j \in \{ 1, 2, \ldots, \min\{\lvert \nu \rvert, \lvert \eta \rvert \} - 1 \}$ such that $\nu\vert_{j} = \eta\vert_{j}$ and $\nu\vert_{j+1} \prec \eta\vert_{j+1}$, or $\nu\vert_{j} = \eta\vert_{j}$ and $\nu\vert_{j+1} \succ \eta\vert_{j+1}$. Suppose that $\nu\vert_{j} = \eta\vert_{j}$ and $\nu\vert_{j+1} \prec \eta\vert_{j+1}$, and that $\omega_{1} = \max F$. Letting $k \in F \setminus \{\omega_{1}\}$, for all $x \in I(\nu \omega\vert_{i})$, $y \in I(\eta \omega\vert_{l} k)$ and $z \in I(\eta \omega\vert_{i})$, that $x \leq y \leq z$. In other words, $\operatorname{dist}(I(\nu\omega\vert_{i}), I(\eta\omega\vert_{i})) \geq \lvert I(\eta \omega\vert_{l} k)\rvert$, and hence by \Cref{lem:geometric_lengths_of_cylinders},
\begin{align*}
\operatorname{dist}(I(\nu\omega\vert_{i}), I(\eta\omega\vert_{i}))
\geq \lvert I(\eta \omega\vert_{l} k)\rvert
\geq \rho \beta^{-(\lvert \eta \rvert + l + 1 )}
\geq \rho \beta^{-(\lvert \eta \rvert + m + 1)}
\geq \rho \beta^{-(m + 1)} \lvert I(\eta) \rvert.
\end{align*}
Similarly, if $\omega_{1} \neq \max\{F\}$, setting $k = \max F$, we obtain that
\begin{align*}
\operatorname{dist}(I(\nu\omega\vert_{i}), I(\eta\omega\vert_{i}))
\geq \lvert I(\nu \omega\vert_{l} k)\rvert
\geq \rho \beta^{-(\lvert \nu \rvert + l + 1 )}
\geq \rho \beta^{-(\lvert \nu \rvert + m + 1)}
\geq \rho \beta^{-(m + 1)} \lvert I(\nu) \rvert.
\end{align*}
An analogous argument yields the result when $\nu\vert_{j} = \eta\vert_{j}$ and $\nu\vert_{j+1} \succ \eta\vert_{j+1}$.
When $\nu$ is a prefix of $\eta$, the result follows using a similar reasoning as in the case when $\nu$ is not a prefix of $\eta$, but where we replace the first line of the argument, namely \eqref{eq:HY-splitting_of_cylinders}, by the following observation. By construction, there exists a $j \in \{ 1, 2, \ldots, \lvert \eta \rvert - \lvert \nu \rvert + M -1 \}$ such that $\nu \omega\vert_{j} = (\eta \omega\vert_{i})\vert_{j + \lvert \nu \rvert}$ but $\nu \omega\vert_{j+1} \neq (\eta \omega\vert_{i})\vert_{j + \lvert \nu \rvert + 1}$. In which case, by our two facts, there exists an $l \in \{0, 1, 2, \ldots, m-1\}$ and a subset of $F$ of $\{1, 2, \ldots, n-1\}$ with $\lvert F \rvert \geq 2$ such that
\[
I(\nu \omega\vert_{j+l}) = \bigcup_{k \in F}
I(\nu \omega\vert_{j+l} k)
\quad \text{and} \quad
I((\eta \omega\vert_{i})\vert_{j + \lvert \nu \rvert + l}) = \bigcup_{k \in F} I((\eta \omega\vert_{i})\vert_{j + \lvert \nu \rvert +l} k).
\qedhere
\]
\end{proof}
\begin{proof}[{Proof of \Cref{thm:main_2}}]
This is a direct consequence of \Cref{thm:G1990+Palmer79,thm_HY_Thm_2.1}, and \Cref{prop:alpha-winning_transport,prop:thm_SFT+Transitive_implies_winning}.
\end{proof}
\bibliographystyle{alpha}
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Grillo Oriental Rug Gallery and Care maintains a full service Oriental rug conservation, restoration, and repair department that is well stocked with hundreds of colors of wool, and all the findings necessary to keep your Oriental rugs in excellent condition.
Patience, skill and experience is brought to every task.
Our vast collection of wools for matching original colors.
Complex weaving of a fine rug in progress.
A superb result by our master weaver.
All new fringes are carefully hand sewn.
In stock ready- made fringes for cost effective repairs.
The end of this rug has been properly conserved to give years of service.
Re-knaping of a moth damaged area.
Using the original knot missing wool is replaced.
The overcasting stitch prevents your rug from raveling.
A small sampling of our patches collected over 4o years.
A large hole too costly to reweave.
Finding the perfect patch /A decision was made to patch.
Our skilled hands can accomplish invisible reweaving of holes and tears, or the hand knotting of wool to replace moth damaged areas. Edges and ends are often the first to show signs of wear. We can overcast ends to prevent further raveling, hand sew new fringes, or rewrap worn edges. By promptly attending to any areas that need attention, we can not only prevent further damage to your Oriental rug, but help you avoid a more costly repair in the future.
It is important to note that the decision of whether to choose Oriental rug conservation, restoration, or repair must be guided by the Oriental rug's replacement cost, condition, age, and historic or art value.
drop off a rug for cleaning. | {
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|
Reauthorize the Department of Agriculture's process for grain inspections until September 30, 2020.
Force the Secretary of Agriculture to waive weighting and inspections of grain in an "emergency, a major disaster"; currently, the Secretary has the option to do so, but does not have to.
A "major disaster" is defined to specifically include "a severe weather incident causing a region-wide interruption of government services".
Change the location of export inspections to specifically "export elevators" at export port locations.
This bill was introduced in a previous session of Congress and was passed by the House on June 9, 2015 but was never passed by the Senate.
H.R. 2088 (114th) was a bill in the United States Congress.
United States Grain Standards Act Reauthorization Act of 2015, H.R. 2088, 114th Cong.. | {
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'Amazing Spider-Man' Trilogy Confirmed
Confirmation came on "The Amazing Spider-Man" Facebook page that the film is one of three planned
Alexander C. Kaufman | July 5, 2012 @ 4:50 PM
"The Amazing Spider-Man," Sony Pictures' fourth installment in the franchise, has already broken box office records and wooed critics into its web.
Now, Sony hopes it will amaze twice more.
On the film's Facebook fan page, the studio confirmed that the new "Spider-Man" is the first in a trilogy.
"It's finally here," the post, which appeared in more than 1.7 million newsfeeds Monday, begins. "'The Amazing Spider-Man' is the first installment in a movie trilogy that will explore how our fave hero's journey was shaped by the disappearance of his parents."
Also read: 'Amazing Spider-Man' Spins $50M in Overseas Box Office Bow
By Thursday evening, the post was liked 11,684 times.
"The Amazing Spider-Man" opened Monday in the United States after grossing $6 million from 1,236 screens in India, the biggest opening ever for a Hollywood film there and 74 percent bigger than 'Spider-Man 3." | {
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Search only accepts letters and numbers.
Word of Faith
Our Neighbors' Faith
People of Faith
Seminarian's Diary
Youth/Young Adults
God, sex and listening
REV. ROGER KARBAN
Our modern Christian sexual morals are often based more on Greek philosophy than on Biblical principles. Though almost every Sacred Author wrote against the background of a Semitic world and Semitic morality, within 150 years of Jesus' death and resurrection, His followers started to filter Scripture through a Greek world and Greek morality.
Once Christianity spread beyond the Holy Land and became part of the entire Mediterranean world, it lost or ignored much of its Jewish core, the core on which Jesus and His first disciples focused their faith. Eventually, the Jewish carpenter's followers abandoned His unique Jewish outlook and adopted a Greek outlook.
Greek philosophers not only divided persons into bodies and souls, but also presumed what involved the soul was good; and what involved the body, bad. Sex was, at best, suspect; at worst, condemned. Though marriage was permitted because of the need to procreate, most regarded the practice as a license to engage in evil acts. People were encouraged to avoid sexual urges and actions, since such concentration on the body diverted them from the higher pursuits of the soul.
Sexual commitment
Thankfully, we find none of this reasoning in Sunday's second reading (1 Cor 6:13-15, 17-20). In spite of extensive liturgical editing, we can still make out Paul's basic argument against casual sex. Following the theology of Genesis 2, Paul believes intercourse creates one person from the two persons who engage in it.
That's why he reminds his community, "Every other sin a person commits is outside the body, but the immoral person sins against his own body" -- because he or she has become one body with the other person with whom they're sexually joined.
There's no separation of body and soul here; no conflict between the material and the immaterial. There's simply a belief that sexual intercourse without commitment is a lie. It's immoral to be physically one with someone with whom we refuse, at the same time, to be psychologically and emotionally one. God didn't create intercourse to be used for casual sex.
Yet, Paul doesn't stop there. Because he believes that our faith has already made us one with Jesus, he sees another moral implication in such free and easy sexual encounters. How can we make Jesus one with a prostitute?
"You must know," he writes, "that your body is a temple of the Holy Spirit.... You are not your own." Paul's now far beyond Genesis 2. Jesus' death and resurrection forces him to look at every act of intercourse from a totally new, unifying perspective.
Having left many of these early Jewish-Christian principles behind in our rush to embrace Greek philosophy, we could be like young Samuel at Shiloh (1 Sam 3:3-10, 19). Just as Samuel "was not familiar with Yahweh, because Yahweh had not revealed anything to him as yet," in the same way many of us are not familiar with God revealing God's self to us in our everyday actions. Thinking God speaks to us only through "things of the soul," we often miss the Lord's voice in the ordinary events of our lives.
Eli knows the proper response for all God's followers: "Speak, Lord, for your servant is listening." Only those willing to carry out the Lord's wishes will be able to hear God calling -- even in the middle of the night.
This Jewish, creation-grounded theology also seems to be behind Jesus' famous question in the Gospel (Jn 1: 35-42). By having Jesus simply, but demandingly, ask John's disciples, "What are you looking for?" John the Evangelist is warning his readers: "Unless you're looking, you'll never find Jesus. Unless you're curious, you'll miss God in your midst."
Of course, the Lord expects us to be courageous enough to ask the next question, "Where do you stay?" and then be willing to follow up on His generous invitation to "Come and see." John is in line with all our Biblical authors. They remind us that our lives, as God's followers, revolve around seeking, finding and responding to God's continual calls.
Nothing in our God-created lives can be outside God's concern and revelation. To think and act otherwise is to be at odds with the ancient, scriptural theology on which Jesus based His own spirituality.
The Evangelist - The Official Publication of the Diocese of Albany
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The Evangelist, 40 North Main Ave., Albany, NY, 12203-1422 | PHONE: 518-453-6688| FAX: 518-453-8448
Copyright © 2021 | Evangelist.org | All Rights Reserved. | {
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I make my flowers from 180 gr Italian crepe paper that I like to paint with acrylics so I can control the color theme in every small detail. I do not use any plastics, vinyl, clay, beads, or pre-made artificial foliage or stamen.
What kind of flowers do you make?
I make a large variety of flower, in different stages of blooming, from buds to fully blossomed: classic roses, peonies, anemone, ranunculus, hydrangea, all types of English garden roses, succulents, scabiosa, jasmine, protea, dahlias, poppies, magnolias, poinsettias, berries, dogwood, tulips, sunflowers, brunnia, freesias, sweet-peas, lisianthus, cherry blossom, field flowers, hellebore, orchids, daisy, all types of greenery and eucalyptus. I like to constantly enlarge my floral range, try new things so, don't be afraid to ask me if there is a particular flowers you are interested in and I'll try my best to make it.
What kind of work or orders do you take?
I make only custom orders. The main area of my work are bridal bouquets… I love weddings and brides so I do my best to make their dream of the perfect bouquet come true. I give a lot of energy and time for these projects but I do find the time to make other types of orders like centerpieces for home decoration, small bouquets for gifts or flower wreaths. From my wedding portfolio I also make boutonnieres and corsages, flower crowns or… I don't know… try me! Let me know if you're interested in something different!
I sometimes make recreation of real bouquets or wedding bouquets… so if you wish to hold your bridal bouquet one more time, for years to come, or you want to make your wife the most amazing anniversary gift, let me know. I take a limited amount of orders per year so contact me with many months in advance. I also reserve the right to turn down orders beyond my capabilities.
Can you send me a price offer for your work?
I don't have a standard price offer because I only have time to make custom orders (I never find the time to make flowers to stock in my studio), so every customer with his/hers requests are unique. You have to let me know your requests so I can make an estimate of the time, workmanship and materials I need in order to make you a price offer.
How long does it take you to make my order?
You have to let me know your deadline first so I can see if I can make it in time for shipping. I take orders with many months in advance… this is the only way I can arrange my schedule in order to deliver.
How does this process work? From placing the order until finishing it.
You have to give me an e-mail or private message on my Facbook/Instagram page with all the details regarding your bouquet/order: color theme, flower range, shape and size, etc. You can send me inspirational pictures with my work but I also enjoy images with real flowers/bouquets from Pinterest (I often go to Pinterest for inspiration) or other wed sources. After we establish these details I will make a storyboard with images that capture the style and details of your bouquet, which you will approve (or modify as you wish) and I will use it for my work process. I send pictures with the flowers after I make them and also with the finale bouquet… communication is very easy with me, I'm available most of the time for you, for questions and possible changes.
Is shipping secure? Does it arrive safely?
Yes, I do my best to make that happen. I use a hard cardboard box wich I sustain with walls of polyester so it doesn't deform. I also make a support system for the bouquet that keeps it still in the box. The rest of the empty space is filled with tissue paper. I usually send pictures with the inside of the box so you can see for yourself.
How should I care for the paper flowers? How long do they last?
Paper flowers are ment to last for many years if you take proper care of them. You must keep them away from direct sunlight and moisture! One of the reasons I paint my paper with acrylics is that crepe paper colors fade very quickly because of the natural light. By painting it, not only I obtain the shade and details I aim for, but also, I make the paper more stronger and resistant to natural light. Some paper flowers are very delicate and fragile but I've developed, over time, some techniques that make them stronger. I also use a varnish spray with UV protection that will help the paper that isn't painted not to fade as much. If you find your flowers need dusting, giving them a gentle shake or blowing on them gently should be sufficient. You can also use a soft paint brush to get into the grooves of the paper.
The paper flowers I make are generally more resilient than they appear, but you must avoid storing flowers on their sides for long periods of time, and always keep them in a vase.
Keep your flowers away from open flames.
Do you make workshops or teach?
Not for the moment. I'm always busy with my custom orders for bridal bouquets and weddings so I can't find the time to get involved in other activities. However, I dream that I will make some tutorials one day and share with you some of my techniques… so be patient with me and follow my posts on social media.
Why are your paper blooms more expensive than real flowers?
…or at least they should be, because you have to consider the fact that this is hand-make work, a creative and artistic process in which every paper bloom is made from scratch – the paper is died, every petal is cut, shaped, glued on one by one. It may take a couple of hours to make just one flower. You're paying for the material cost, the workmanship and also the hours of researching, practice and study, sourcing for the best materials, and experimenting new techniques. I fell that with every project I learn something new and every bouquet is unique because it's also a work of art.
You must know that I pay myself a moderate wage per hour so you can afford to by my creations… I don't factor the many additional hours I spend researching, designing, managing my business, arranging the order for proper shipping, shopping for supplies and so on.
Hi Christine, where are you based? I am interested in a few floral designs for my wedding, but I am in South Africa.
Hello, I'm from Romania. I will send a private e-mail with details. | {
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Salerno (1869-1946) was born Adolph Behrend in Germany. He was one of the originators of the Gentleman Juggling style and was one of its top two performers, along with Kara. He began his performing career in 1886. In 1912, he created electrically lit clubs that changed colors as he juggled them. He was also an inventor and one of the first airplane pilots in the world.
This is bakelite plate that was owned and used by Salerno and later owned by Bobby May. Acquired as part of the Paul Bachman Collection.
These are Salerno's self lighting candles. The bottom one is complete, with an outer shell, inner shell, and an internal mechanism that includes an early flashlight battery, a mercury switch, and a glow tip igniter tip. It would light when flipped. It was invented and used by Salerno in his act. The upper candle is a duplicate, but lacks the internal mechanism. It was passed from Salerno to King Repp to Dieter Tasso. Donated by Dieter Tasso.
This is an amazing lighted club made and used by Salerno at least as early as 1905. It was later passed on to and used by the Juggling Jewels. Acquired as part of the Paul Bachman Collection. | {
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#old style classes
class OnlyOneOld:
class __OnlyOneOld:
def __init__(self, arg):
self.val = arg
def __str__(self):
return `self` + self.val
instance = None
def __init__(self, arg):
if not OnlyOneOld.instance:
OnlyOneOld.instance = OnlyOneOld.__OnlyOneOld(arg)
else:
OnlyOneOld.instance.val = arg
def __getattr__(self, name):
return getattr(self.instance, name)
# new style classes
class OnlyOneNew(object):
class __OnlyOneNew:
def __init__(self):
self.val = None
def __str__(self):
return `self` + self.val
instance = None
def __new__(cls): # class method
if not OnlyOneNew.instance:
OnlyOneNew.instance = OnlyOneNew.__OnlyOneNew()
return OnlyOneNew.instance
def __getattr__(self, name):
return getattr(self.instance, name)
def __setattr__(self, name):
return setattr(self.instance, name)
# borg style singleton
class Borg:
_shared_state = {}
def __init__(self):
self.__dict__ = self._shared_state
class SingleBorg(Borg):
def __init__(self, arg):
Borg.__init__(self)
self.val = arg
def __str__(self):
return self.val
# decorator style singleton
class SingletonDecorator:
def __init__(self, klass):
self.klass = klass
self.instance = None
def __call__(self, *args, **kwds):
if self.instance == None:
self.instance = self.klass(*args, **kwds)
return self.instance
class Foo:
pass
# meta class style singleton
class SingletonMetaclass(type):
def __init__(cls, name, bases, dect):
super(SingletonMetaclass, cls).__init__(name, bases, dict)
original_new = cls.__new__
def my_new(cls, *args, **kwds):
if cls.instance == None:
cls.instance = original_new(cls, *args, **kwds)
return cls.instance
cls.instance = None
cls.__new__ = staticmethod(my_new)
class Bar(object):
__metaclass__ = SingletonMetaclass
def __init__(self, val):
self.val = val
def __str__(self):
return `self` + self.val
def oldStyleClasses():
x = OnlyOneOld('sausage')
print x
y = OnlyOneOld('eggs')
print y
z = OnlyOneOld('spam')
print z
print x
print y
print `x`
print `y`
print `z`
def newStyleClasses():
x = OnlyOneNew()
x.val = 'sausage'
print x
y = OnlyOneNew()
y.val = 'eggs'
print y
z = OnlyOneNew()
z.val = 'spam'
print z
print x
print y
def borgClasses():
x = SingleBorg('sausage')
print x
y = SingleBorg('eggs')
print y
z = SingleBorg('spam')
print z
print x
print y
print `x`
print `y`
print `z`
def decoratorClasses():
foo = SingletonDecorator(Foo)
x = foo()
y = foo()
z = foo()
x.val = 'sausage'
y.val = 'eggs'
z.val = 'spam'
print x.val
print y.val
print z.val
print x is y is z
def metaclassClasses():
x = Bar('sausage')
y = Bar('eggs')
z = Bar('spam')
print x
print y
print z
print x is y is z
def main():
print 'OLD'
oldStyleClasses()
print
print 'NEW'
newStyleClasses()
print
print 'BORG'
borgClasses()
print
print 'DECORATORS'
decoratorClasses()
print
print 'METACLASS'
metaclassClasses()
print
if __name__ == '__main__':
main()
| {
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When the time finally comes for your child to hit the road, they can be a costly addition to your premium. However, there are a few ways to successfully combat this price increase.
There are several things new drivers can do to help keep their insurance prices as low as possible. First, grades go a very long way with most insurance companies. A B-average report card can save an enormous amount of money. Wondering if your current policy includes a good student discount? The best indicator is to think back if you ever submitted a report card to your insurance company. This type of discount requires annual updates. If the company isn't made aware of your child's grades or improvement in grades, you likely are missing out on some savings for your student driver. While agents and their staffs have the best of intentions, it is possible for a discount such as this to be missed or dropped from your current coverage. An added perk to the good student discount is how long it lasts. This discount is valid during high school years and with some companies even continues through college years up until the driver reaches 25 years of age – which could mean up through graduate school! Is your child attending college over 100 miles away and leaving the car at home? This can qualify your driver for savings, as well.
In addition to a good student discount, a young driver can also receive a large discount for taking and passing driver's training. Make sure your insurance agency has a copy of your child's successful completion of this course in order to obtain this discount.
While new drivers themselves can impact their own rates significantly, there are steps parents can take toward keeping the rates low as well. Credit score, claims and tickets all affect the rates. When your child is on your policy, anything that affects your rates will affect theirs as well. Also, be sure to call in and get a quote for the vehicle you're looking to add for your new driver to insure your expectations concerning your new premium match reality. It is wise to do this prior to purchasing the vehicle. With young drivers, just like with anyone's policy, the value of the vehicle directly impacts the rate implemented. Meaning, for example, a new Mustang will be drastically more expensive to insure than a used Civic.
Finally, there is a considerably new feature some companies are utilizing to help decrease their clients' rates. Some companies are installing devices in vehicles in order to track motion. Rapid acceleration, quick breaking or violent movement of the vehicle get recorded by this device. A lack of movements like these can mean up to a 30% discount in rates. A few companies that currently promote devices like these include Progressive and SafeCo. Progressive's program is titled "Snapshot" and SafeCo's is titled "Right Track." Below is a link that provides a visual explanation for how the SafeCo Right Track device functions.
While having a child behind the wheel for the first time can be scary, hopefully now their insurance doesn't have to be.
For questions concerning this article or any other post by the Dean Ballenger Agency, don't hesitate to call us at (317)867-5433. | {
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1828.mshaffer.com › Word [saint]
SAINT, n. [L. sanctus.]
1. A person sanctified; a holy or godly person; one eminent for piety and virtue. It is particularly applied to the apostles and other holy persons mentioned in Scripture. A hypocrite may imitate a saint. Ps. 16.
2. One of the blessed in heaven. Rev. 18.
3. The holy angels are called saint. Deut. 33. Jude 14.
4. One canonized by the church of Rome.
SAINT, v.t. To number or enroll among saints by an official act of the pope; to canonize.
Over against the church stands a large hospital, erected by a shoemaker who has been beautified, though never sainted.
SAINT, v.i. To act with a show of piety.
Evolution (or devolution) of this word [saint]
SAINT, n. [Fr. from L. sanctus; It. and Sp. santo.]
A person sanctified; a holy or godly person; one eminent for piety and virtue. It is particularly applied to the apostles and other holy persons mentioned in Scripture. A hypocrite may imitate a saint. – Ps. xvi. Addison.
One of the blessed in heaven. – Rev. xviii.
The holy angels are called saints. – Deut. xxxiii. Jude xiv.
One canonized by the Church of Rome. – Encyc.
SAINT, v.i.
To act with a show of piety. – Pope.
SAINT, v.t.
To number or enroll among saints by an official act of the pope; to canonize. Over against the church stands a large hospital, erected by a shoemaker who has been beatified though never sainted. Addison.
A person sanctified; a holy or godly person; one eminent for piety and virtue; any true Christian, as being redeemed and consecrated to God.
Them that are sanctified in Christ Jesus, called to be saints. 1 Cor. i. 2.
To make a saint of] to enroll among the saints by an offical act, as of the pope; to canonize; to give the title or reputation of a saint to (some one).
A large hospital, erected by a shoemaker who has been beatified, though never sainted. Addison.
To saint it, to act as a saint, or with a show of piety.
Whether the charmer sinner it or saint it. Pope.
To act or live as a saint.
[R.] Shak.
One of the blessed in heaven.
Then shall thy saints, unmixed, and from the impure
Far separate, circling thy holy mount,
Unfeigned hallelujahs to thee sing. Milton.
One canonized by the church.
[Abbrev. St.]
Saint Andrew's cross. (a) A cross shaped like the letter X. See Illust. 4, under Cross. (b) (Bot.) A low North American shrub (Ascyrum Crux-Andreæ, the petals of which have the form of a Saint Andrew's cross. Gray. -- Saint Anthony's cross, a T-shaped cross. See Illust. 6, under Cross. -- Saint Anthony's fire, the erysipelas; -- popularly so called because it was supposed to have been cured by the intercession of Saint Anthony. -- Saint Anthony's nut (Bot.), the groundnut (Bunium flexuosum); -- so called because swine feed on it, and St. Anthony was once a swineherd. Dr. Prior. -- Saint Anthony's turnip (Bot.), the bulbous crowfoot, a favorite food of swine. Dr. Prior. -- Saint Barnaby's thistle (Bot.), a kind of knapweed (Centaurea solstitialis) flowering on St. Barnabas's Day, June 11th. Dr. Prior. -- Saint Bernard (Zoöl.), a breed of large, handsome dogs celebrated for strength and sagacity, formerly bred chiefly at the Hospice of St. Bernard in Switzerland, but now common in Europe and America. There are two races, the smooth-haired and the rough-haired. See Illust. under Dog. -- Saint Catharine's flower (Bot.), the plant love-in-a-mist. See under Love. -- Saint Cuthbert's beads (Paleon.), the fossil joints of crinoid stems. -- Saint Dabeoc's heath (Bot.), a heatherlike plant (Dabœcia polifolia), named from an Irish saint. -- Saint Distaff's Day. See under Distaff. -- Saint Elmo's fire, a luminous, flamelike appearance, sometimes seen in dark, tempestuous nights, at some prominent point on a ship, particularly at the masthead and the yardarms. It has also been observed on land, and is due to the discharge of electricity from elevated or pointed objects. A single flame is called a Helena, or a Corposant; a double, or twin, flame is called a Castor and Pollux, or a double Corposant. It takes its name from St. Elmo, the patron saint of sailors. -- Saint George's cross (Her.), a Greek cross gules upon a field argent, the field being represented by a narrow fimbriation in the ensign, or union jack, of Great Britain. -- Saint George's ensign, a red cross on a white field with a union jack in the upper corner next the mast. It is the distinguishing badge of ships of the royal navy of England; -- called also the white ensign. Brande *** C. -- Saint George's flag, a smaller flag resembling the ensign, but without the union jack] used as the sign of the presence and command of an admiral. [Eng.] Brande *** C. -- Saint Gobain glass (Chem.), a fine variety of soda-lime plate glass, so called from St. Gobain in France, where it was manufactured. -- Saint Ignatius's bean (Bot.), the seed of a tree of the Philippines (Strychnos Ignatia), of properties similar to the nux vomica. -- Saint James's shell (Zoö]l.), a pecten (Vola Jacobæus) worn by pilgrims to the Holy Land. See Illust. under Scallop. -- Saint James's-wort (Bot.), a kind of ragwort (Senecio Jacobæa). -- Saint John's bread. (Bot.) See Carob. -- Saint John's-wort (Bot.), any plant of the genus Hypericum, most species of which have yellow flowers; -- called also John's-wort. -- Saint Leger, the name of a race for three-year-old horses run annually in September at Doncaster, England; -- instituted in 1776 by Col. St. Leger. -- Saint Martin's herb (Bot.), a small tropical American violaceous plant (Sauvagesia erecta). It is very mucilaginous and is used in medicine. -- Saint Martin's summer, a season of mild, damp weather frequently prevailing during late autumn in England and the Mediterranean countries; -- so called from St. Martin's Festival, occurring on November 11. It corresponds to the Indian summer in America. Shak. Whittier. -- Saint Patrick's cross. See Illust. 4, under Cross. -- Saint Patrick's Day, the 17th of March, anniversary of the death (about 466) of St. Patrick, the apostle and patron saint of Ireland. -- Saint Peter's fish. (Zoöl.) See John Dory, under John. -- Saint Peter's-wort (Bot.), a name of several plants, as Hypericum Ascyron, H. quadrangulum, Ascyrum stans, etc. -- Saint Peter's wreath (Bot.), a shrubby kind of Spiræa (S. hypericifolia), having long slender branches covered with clusters of small white blossoms in spring. -- Saint's bell. See Sanctus bell, under Sanctus. -- Saint Vitus's dance (Med.), chorea; -- so called from the supposed cures wrought on intercession to this saint.
want true definitions of words
— Rod (Jackson, MI)
triture
TRI'TURE, n. A rubbing or grinding. [Not used.]
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2 of 4 copies available at Lake Agassiz Regional Library.
1 of 6 copies available at LARL/NWRL Consortium.
1 of 4 copies available at Lake Agassiz Regional Library.
6 of 13 copies available at LARL/NWRL Consortium.
2 of 6 copies available at Lake Agassiz Regional Library.
4 of 5 copies available at Lake Agassiz Regional Library. | {
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Intermediate & High School Course Cataglog
Volunteer Requirements for Clovis Unified
Clovis Unified encourages parents/guardians and other members of the community to share their time, knowledge and abilities with our students. Community volunteers in our schools enrich the educational program, contribute to school safety, and strengthen our schools' relationships with the community.
In order to be a volunteer you must follow the guidelines in Board Policy 9212:
1. Complete a volunteer application at the school site each school year, which will include the site conducting a Megan's Law background check (Penal code 290).
2. Have a current clear TB assessment within the last 4 years if you will have ongoing, frequent or prolonged contact with students. One-day volunteers may be exempt. (Ed. Code 49406(m)).
3. Be fingerprinted when placed in a situation in which you would directly supervise students with-out a paid District employee present (i.e. off-campus activities such as Sierra Outdoor School, over night events or any activity without direct supervision by a paid District employee).
4. Volunteers who will be transporting students must also submit Form 8302-1 – Private Vehicle Driver Application, and comply with the requirements of Board Policy and Administrative Regu-lation No. 8302 – Transportation of Students by Private Vehicle.
No person has any right to provide, nor is the District obligated to accept, volunteer services. A person also has no right to a particular volunteer assignment, event, location or classroom. Volunteers shall act in accord-ance with state and federal laws, District policies and regulations, includ-ing but not limited to Board Policy No. 9202 (School Visitors) and Board Policy No. 9210 (Civility Policy), and school rules.
TB Assessment Clinics
2019 School Year TB Assessment Dates
2019 School Year TB Assessment Dates (.rtf)
Opportunities to Get Involved and Volunteer
Bilingual Advisory Council (BAC)
All schools enrolling 21 or more Limited English Proficient (LEP) students are required to form a Bilingual Advisory Council. The BAC is composed of parents and school personnel. The BAC provides input and makes recommendations to the principal, staff and SSC regarding services for LEP students. The BAC is formed annually at the beginning of the school year.
Through regular attendance as a committee member the CAC can give you knowledge and skills necessary to become more comfortable in the role as partner in the educational processs for your child. The CAC serves as a way to share knowledge and help develop the best possible programs for our students. For more information, visit their web site.
Intercultural Diversity Advisory Council (IDAC)
The primary purpose of the Intercultural Diversity Advisory Council (IDAC), originally formed in 1988 as the Intercultural Advisory Council, is to assist in the formation and review of policies that assure non-discriminatory practices in all operational areas of the Clovis Unified School District. Its further mission is to assist in improving the cultural environment of the District.
Members of IDAC include parents, students, a CUSD Board Member, District staff, community business and religious representatives, college/university representatives, and community elected volunteers. IDAC strives to have representation from each school site to serve as a communication link from the District Council to the school site and vice versa. For dates of meetings visit their web site.
Parent-Teacher Club
Each elementary and intermediate school in the district has a Parent Club organization. The primary purpose of a school's Parent Club is to assist teachers in providing a quality educational experience for all students by sponsoring student/parent fundraising events.
The Foundation for Clovis Schools is a districtwide foundation supporting educational programs that impact students in all Clovis Unified schools. A Board of Directors made up of business and community leaders works to generate community resources for our schools. Many people in the community perceive Clovis Unified as being a wealthy school district when, in reality, the district receives less per child than any other district in Fresno County. Statewide, CUSD receives approximately $300 per child less than the average of all unified school districts. The Foundation raises funds in the private sector to support the following types of programs: Classroom Grants-Request Fund (Guardian Awards), CHARACTER COUNTS!, The Center for Advanced Research and Technology, and Learning with Laptops. For more information visit their web site.
School Assessment Review Team (S.A.R.T.)
Each school in the Clovis Unified School District has a S.A.R.T. Committee. The S.A.R.T. Committee does not duplicate or assume the functions of other school organizations such as the School Site Council or the Parent Club. Rather, the S.A.R.T. Committee is concerned about the overall operation of the school. It serves as a community forum for communication and assessment of the school programs and as an advisory body to the princpal.
School Site Council (SSC)
All schools receiving School Improvement Program (SIP), SB 1882 funds or implementing a School Based Coordinated Plan (SBCP) are required to form a School Site Council. The School Site Council is composed of parents and school personnel. The SSC is responsible for developing, implementing and evaluating the School Site Plan. Members serve for two years and area elected by their peers. Elections for new members are held annually at the beginning of the school year.
School Foundations
Each of the three comprehensive high schools (Buchanan, Clovis High and Clovis West) has a Foundation, which is a non-profit, tax-exempt educational corporation. The purpose of these foundations are to serve and support the students by providing a financial support base for all co-curricular activities that by nature are non-revenue producing. | {
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the Collaborative model in conjunction with the brands CA and TERROR SNOW brings together two cultures, snowboarding and street racing.
Brand CA was started with the usual design Studio. It has always been associated with extreme sports, but the most important focus was on the cars. They are an integral part of the brand. The team's drivers CA – permanent participants and winners of Russian and world street race racing. | {
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Getting to the airport can be difficult for those travelling with a lot of bag and family members and isn't the most easy thing. One alternative to ignore the logistics of reaching the airport is to make use of an airport cab service. This worry and could remove a lot of the stress, but is important to schedule the pick up to match the particular needs.
An easy measure that may make things go smoothly will be to book the taxi well beforehand. Be sure there is enough time to readily reach the airport without feeling rushed. Attempt to calculate the travel space and add an additional hour or more in the event of significant traffic. Also, if travelling during busy travel times, such as bank holidays, it might be worth adding a bit more time.
Inquire two or three different businesses and intention to identify for supplying a reliable price in the local region, one that's highly rated. Use on-line reviews to get an idea of reputable and the caliber of a possible company. Locate one that has the courteous drivers and valued for supplying timely pick-ups and drops -offs.
Most firms are quite similar, so that it helps to ring around and get a price for the ride to the airport. Many quote a flat fee to travel at residence to the airport, in addition to any other pick ups on the way. Also, make sure the favourite company has cab cuffley credentials and the mandatory licensing to offer this kind of kind of service.
Among the important things to check is the quantity of space for passengers and luggage that the vehicle can tolerate. The vehicles may differ in size with a standard vehicle able enough to accept a couple with nominal luggage to the bigger vehicle that can certainly take six or seven passengers in comfort.
For all those travelling with lots of bags, it might be necessary to request a vehicle that is more substantial.
Even in the event the airport taxi service was booked and reserved in advance, it may be worth giving the company a call a couple of hours before it should arrive at the pickup address to be sure it will be arriving on time. | {
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TREASVRE release Devils/ Echoes EP
Posted by Bethany Clancy | Jul 31, 2020 | News | 0 |
TREASVRE is a band formed of five San Francisco natives who go the distance for their sound. Blending elements of shoegaze and post-rock, the band also brings rich synth textures and heavy guitar riffs to their sound. Vocalists Sabrina Simonton and Samantha Peña create wistful harmonies that underscore themes of nostalgia, melancholy, and hopefulness found in their lyrics.
The drama and depth conveyed in the music creates a distinctly cinematic quality-one that captures the complex range of human emotion in beautiful and inspiring ways and begs to be experienced live.
The Devils/Echoes EP, which was released yesterday, wrestles with heavy themes that are especially pertinent during these times of social isolation. The two songs examine how loneliness and self-doubt can take control, as well as the struggle to break free without losing oneself.
The band comments, "Devils/Echoes EP is a reflection of a time of transition in our personal lives. This past year, we've experienced big changes in our relationships, our responsibilities, and the way we see ourselves. These songs embody the desire to escape and isolate in times of uncertainty. We tried to convey the complexity of these feelings through the imagery of possession, secrecy, and deception. While this may seem like a dark emotional landscape, ultimately these are songs about discovering the silver lining and pushing past your limits to discover what you're capable of. "
PreviousDan + Shay share "I Should Probably Go to Bed" music video
NextPREMIERE: American Hi-Fi release cover of "Another Nail in My Heart"
Bethany Clancy
Buffalo native who loves Dawson's Creek & multiple cups of coffee
Known Battle Rapper Tsu Surf Drops Standout LP, 'MSYKM'
TIFF Review: 'The Wheel' Journals the Ebbs and Flows of Relationships On the Brink
PREMIERE: Bogues' new song "A Long Goodbye" feels like a warm hello
Real Friends announce 'Torn In Two' EP + release new song, "Teeth" | {
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IN THE STREETS & ON THE WEB
SCURRY LIFE RADIO
BUTTER CUPZ
All Blog Posts (26,754)
The White House has secured better jailhouse conditions for A$AP Rocky.
The Harlem hip-hop star has been locked up in a Swedish prison for more than two weeks after he got in a street fight with two men who had been tailing him and his entourage through the streets of Stockholm.
But when Rocky's manager, John Ehmann,…
Added by WORLD WRAP FEDERATION on July 18, 2019 at 10:26pm — No Comments
YG's Home Searched After Deputy-Involved Shooting
Authorities searched the Los Angeles-area home of rapper YG on Thursday in connection with a police shooting in Compton that killed a bystander earlier this month.
Deputy Marvin Crowder, a spokesman for the Los Angeles Sheriff's Department, said the rapper, whose real name is Keenon Jackson, has not been implicated in the shooting and…
Usher Wants Private Medical Records Sealed In Herpes Case
Usher is pleading with a judge to seal his private medical records as part of the lawsuit filed by two women and a man accusing him of giving them herpes.
According to court documents obtained by The Blast, the singer went to court to request that certain records be sealed from the public. Usher says he needs to file exhibits to his…
Added by WORLD WRAP FEDERATION on July 18, 2019 at 8:59pm — No Comments
Cory Ironside - "Day Dreaming"
Cory Ironside is one of the hottest new talents to hit the scene. Check out his latest album "Day Dreaming" it is available on all music platforms. Do not hesitate to stream or download it today.…
Added by Chris Anderson on July 18, 2019 at 12:38pm — No Comments
Added by Tampa Mystic on July 17, 2019 at 11:45pm — No Comments
Janet Jackson And 50 Cent To Perform In Saudi Arabia Show
Janet Jackson and rappers 50 Cent and Future have been added to the lineup for the Jeddah World Fest, the concert in Saudi Arabia that Nicki Minaj pulled out of after human rights organizations urged the rapper to cancel her appearance.
The website for the event, to take place Thursday at the King Abdullah Sports Stadium,…
Ne-Yo Says His Hats Are Like His Children
Ne-Yo wears many hats.
The singer/songwriter/producer/actor/dancer — who's almost never spotted without a hat atop his head — recently revealed just how important the accessory is to him.
During his performance at Headliner Market Group's 6th Annual Overtown Music & Arts Festival in Miami on Saturday, we're told fans were…
Nipsey Hussle Was Not Being Investigated By The LAPD At Time Of Death
Rapper Nipsey Hussle wasn't part of an ongoing investigation by the Los Angeles Police Department, according to a new report.
The New York Times reported that authorities, in conjunction with the Los Angeles city attorney's office, were looking into alleged gang activity at the slain Grammy winner's strip mall, though TMZ claims Hussle…
Added by Struc Supreme on July 17, 2019 at 11:23am — No Comments
Added by Chris Anderson on July 16, 2019 at 8:29pm — No Comments
Rick Ross Announces 'Port Of Miami 2' Release Date
Rick Ross has announced a release date for his 10th album Port of Miami 2. The long-awaited sequel to his debut arrives on Aug. 9, 13 years after the original Port of Miami.
He has also revealed the cover, which pays homage to the 2006 artwork and finds Ross holding a pendant featuring a photo of his late friend and manager Black Bo, who…
Soulja Boy Released From Prison Early
Soulja Boy is a free man.
The "Turn My Swag On" rapper was released from prison, having served only 94 days of a 240-day prison sentence for a probation violation, according to the Los Angeles County Sheriff's Office.
Soulja Boy — whose real name is DeAndre Cortez Way — was let out of Van Nuys jail on Sunday at 1:57 a.m.…
Drake Sued By Woman Hit In Head By Beer Bottle At MSG Concert
A Staten Island woman says Drake and Madison Square Garden have to pay up over a beer bottle that was hurled at her head during the "In My Feelings" singer's show there three years ago.
Amanda Giovacco,24, says in her new Manhattan Supreme Court lawsuit that MSG, Live Nation, a food vendor and Drake — given name Aubrey Drake Graham — were…
R. Kelly Ordered To Be Held Without Bond
R. Kelly was ordered held without bond Tuesday in Chicago on sex-crimes charges — after federal prosecutors called him "an extreme danger to the community, especially underage girls."
Lawyers for the accused pedophile singer tried arguing that he should be let out on bail while awaiting trial in Chicago on a 13-count indictment involving…
[New Music]- Mika Means, Akbar V, D Woods (aka Project Girls Club)'s New Release: "Heavy"!! @shanellyoungmoney @yagirldwoods @projectsgirlsclub @therealmikame @akbar__v
In a society where women are the top emerging business owners, entrepreneurs, and entertainers Project Girls Club couldn't have picked a better time to reveal their movement. This diverse group of women includes some the industry's most respected female artists, songwriters, radio personalities, business owners, actress and more. In an effort to provide a network for women, created by women, Project Girls Club is reminiscent of an entertainment sorority with… Continue
Added by Tgrmedia on July 16, 2019 at 4:39pm — No Comments
Lauren London Says Son Will Inherit $1 Million From Nipsey Hussle Estate
Lauren London expects the son she shared with Nipsey Hussle to inherit a ton of cash from the late rapper's estate, so she's making sure he's legally protected.
According to documents obtained by The Blast, London filed to establish guardianship of her 2-year-old son, Kross Asghedom.…
Love & Hip Hop Star Yandy Smith Knocked Down And Pepper Sprayed At Prison Protest
Man Suspected Of Killing Angela Simmons Ex Turns Himself In To Police
Floyd Mayweather Accuses Shantel Jackson Of Illegally Recording Him While They Were Engaged
Jimmy Henchman Writes An Open Letter After Receiving Two Life Sentences
R-Kelly's Streaming Numbers Skyrocket After "Surviving R-Kelly"
Chance The Rapper Just Married His Childhood Sweetheart
Soulja Boy To Meek Mill: 'Y'all N**gas Always On My D!*k!'
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I am super excited to announce that I have teamed up with The Street to host and promote their #StreetSweat series that happens every summer on the beautiful green in front of Lululemon. #StreetSweat happens every week through September so check out their full calendar for a list of all their awesome FREE fitness events.
This Saturday, June 27th, come out to the green at Shop the Street for an hour of FreeStyle Fitness with trainers Ron and Sabrina w/ MOVE. It will be an hour of various styles of training to get you moving in ways you haven't moved before. The class is a high octane boost; utilizing circuits and accessories such as ropes, medicine balls, and lots more to rev up your metabolism and ignite fat burning.
To make this event even better….
Plus, 2 complimentary blow drys from Be Styled and one Sweet Skin Facial courtesy of skoah!
Winner will be selected following Saturdays complimentary fitness class with MOVE fitness. | {
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Revonda Gale Cooper
September 22, 1951 - April 27, 2021
Revonda G. Cooper, age 69, of Erwin, passed away on Tuesday, April 27, 2021, in Greeneville Community Hospital East. A native of Erwin, Revonda is a daughter of the late Russia Lee "Rush" and Mable Arlene (Foster) Williams. She was a member of Erwin Church of God, where she enjoyed helping with festivals and events. Her children and grandchildren were the lights of her life and she loved spending time with them and spoiling them. In addition to her parents, Revonda is preceded in death by her brothers: Eugene Robert Williams and wife Ida, Carol Wayne Williams and Ferrell Williams; sisters: Barbara Jane Tipton and husband, Bill, Christa Lee Arnold and husband, Andrew and Gloria Faith Williams; and one infant sibling; niece Rhonda Sue Hall. Revonda G. Cooper leaves behind to cherish her memory: Son: Bishop John Edwards and wife, Lori; Daughter: Angela Shelton; Step-daughter: Amy Edwards and husband, Jarid; Grandson: James Ryan Roark and wife, Cindy; Granddaughter: Bethany Bennett; Step-grandchildren: Noah Banks, Emalee Banks; Great Grandsons: Dakoda Taylor, Desmond Roark, and William Roark; Brothers: Phillip Dwight Williams and Joanne Engle, Ronald Williams and wife, Sue; Several special nieces and nephews; Lifelong best friend: Judy Hawkins; Best friend and companion: Jerry Erwin (and Ttoe); Her Erwin Church of God family and her TOPS family; Her YLM girls who remained like daughters through the years: Tonya, Jackie, Mary Beth and Kristi; And her Flying Monkeys: David Brown, Frank Mosley, Adam Garland. The family would like to offer special thanks to Mary Beth Byrd and the staff of Family Medical Associates, Dr. Beals and the staff of Greeneville Community Hospital East, especially Scott and Kalissa, who took amazing care of her, Pastor Charlie Byrd and Reverend David Brown for all of your love and care during this difficult time. The family respectfully requests the honor of your presence as we offer tribute and remember the life of Revonda G. Cooper in a funeral service to be held at 7:00 pm on Tuesday, May 4, 2021 in the chapel at Valley Funeral Home. Reverend Colt Collins, Reverend David Brown and Pastor Charles David Byrd will officiate. Musical selections will be provided by Greg Forbes and Gary Ollis. A visitation period to share memories and offer support to the family will begin at 5:00 pm and will continue until service time on Tuesday, May 4, 2021 in the chapel at Valley Funeral Home. A graveside committal service will be held at 11:00 am on Wednesday, May 5, 2021, in Riddle Cemetery. Those attending the committal service should meet at Valley Funeral Home by 10:30 am to go in procession to the cemetery. Serving as pallbearers will be her nephews: Christopher Williams, Jr., Steven Tipton, and Phillip Kyker, as well as Chris Gallagher and Gabe Gallagher. Honorary pallbearers will be Jerry Erwin, David Lee Byrd, David Brown, Frank Mosley and Adam Garland. In lieu of flowers, memorial donations can be made in Revonda's name to Valley Funeral Home, 1085 N Main Avenue, Erwin, TN 37650. Condolences and memories may be shared with the family and viewed at www.valleyfuneralhome.net or on our Facebook page at Valley Funeral Home. These arrangements are made especially for the family and friends of Revonda G. Cooper through Valley Funeral Home, 1085 N. Main Ave., Erwin, 423-743-9187.
Revonda G. Cooper, age 69, of Erwin, passed away on Tuesday, April 27, 2021, in Greeneville Community Hospital East. A native of Erwin, Revonda is a daughter of the late Russia Lee "Rush" and Mable Arlene (Foster) Williams. She was... View Obituary & Service Information
The family of Revonda Gale Cooper created this Life Tributes page to make it easy to share your memories.
Revonda G. Cooper, age 69, of Erwin, passed away on Tuesday,...
Send flowers to the Cooper family. | {
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An Ethernet standard at 25 gigabits per second (25 GbE). Approved in 2016, 25 GbE was developed by the 25G Ethernet Consortium and IEEE 802.3by task force, which also includes 50 GbE. The 25 and 50 GbE speeds are even divisions of 100 Gigabit Ethernet, which comprises four 25 Gbps streams. Therefore, migrating in the future from 25 or 50 to 100 GbE is expected to be more economical than going from 40 to 100 GbE. See 100 Gigabit Ethernet. | {
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Transition Longfellow plans Movie Night, Parent Group in Sept.
The volunteers with Transition Longfellow create opportunities for South Minneapolis neighbors to get to know one another while also learning how to live more sustainably and prepare for changes ahead. Visit www.transitionlongfellow.org for more details on these and other activities.
Transition Parents & Kids Play Group meets Sat., Sept. 1, 10am-noon, at Longfellow Park, 3435 36th Ave. S. Join with other parents who are concerned about raising resilient kids, living a sustainable, less consumer-oriented, low-waste family life. Share conversation, resources, and ideas. In September, the group will be visiting a farmers market or a farm to learn about sustainable food production.
Movie Night is scheduled for Fri., Sept. 21, with a potluck at 6:30, and a movie at 7-9pm at Minnehaha Communion Lutheran Church, 4101 37th Ave. S. Share a potluck meal, then watch the documentary "Joanna Macy and the Great Turning." Macy shares her understanding of these times we live in, when everything we treasure seems to be at risk. This is not a film about despair, but about the opportunity we have to come alive to our truest power and participate in the Great Turning, the third major revolution of human existence, after the agricultural and industrial revolutions. The good news is: that's what's happening all around the world. "Arguably the greatest interview of our time with one the wisest women of our time. Heartbreakingly inspiring, practical and transcendental, transformative words that Joanna Macy has conjoined so beautifully in her life and work," Paul Hawken, author of "DrawDown." Donation accepted.
Each Friday, Transition Longfellow sends out a Step-by-Step Preparedness Email with actions you can take that week to become more prepared for extreme weather. Sign up for the series at the website, www.transitionlongfellow.org, where you can also find past emails. | {
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Q: MFC double buffering doesn't copy bitmap BOOL CPaintView::OnEraseBkgnd(CDC* /*pDC*/)
{
//return CPaintView::OnEraseBkgnd(pDC);
return true;
}
void CPaintView::OnDraw(CDC* pDC)
{
CDC shadowMem;
CBitmap shadowBit, *pOldBmp;
CRect currWin;
GetClientRect(currWin);
shadowMem.CreateCompatibleDC(pDC);
shadowBit.CreateCompatibleBitmap(pDC,currWin.Width(), currWin.Height());
pOldBmp = (CBitmap*)shadowMem.SelectObject(&shadowBit);
shadowMem.FillSolidRect(0, 0, currWin.Width(), currWin.Height(), RGB(0, 255, 0));
shadowMem.Rectangle(330, 300, 400, 500);
pDC->BitBlt(0, 0, currWin.Width(), currWin.Height(), &shadowMem, 0, 0, SRCCOPY);
shadowMem.SelectObject(pOldBmp);
}
The code is very simple and intuitive. I created a new CDC and a bitmap, set those to compatible to current CDC. Added new bitmap to new CDC. Then it fills the memory CDC to green color and draw a rectangle on it. then It copy the memory CDC's bitmap to current CDC.
And this is the result i got. You can clearly see a little green line draw near the top edge of the window. I'm not sure which part i messed up.
Update:
Thanks for all the help. I made some random adjustment of bitblt arguments and got a different result.
pDC->BitBlt(0, -400, currWin.Width(), currWin.Height(), &shadowMem, 0, 0, SRCCOPY);
That -400 was a random number when i was trying to figure out the issue. It appears that when i copy the bitmap from shadowMem to pDC, somehow it doesn't match the coordinates correctly.
I couldn't find the reason yet, but if i copy this code to a new project, it works totally fine. I think i may have something to do with MM_ANISOTROPIC mode that my pDC is set to.
A: As @RemyLebeau suggested in the comment, your Rectangle call probably isn't doing what you expect or want.
If we select a pen and brush, then draw a rectangle of (more or less) a size that's entirely inside the client area, we can see the result pretty easily. For example:
void CdoublebufferingView::OnDraw(CDC* pDC)
{
CDC shadowMem;
CBitmap shadowBit, * pOldBmp;
CRect currWin;
GetClientRect(currWin);
shadowMem.CreateCompatibleDC(pDC);
shadowBit.CreateCompatibleBitmap(pDC, currWin.Width(), currWin.Height());
pOldBmp = (CBitmap*)shadowMem.SelectObject(&shadowBit);
shadowMem.FillSolidRect(0, 0, currWin.Width(), currWin.Height(), RGB(0, 255, 0));
shadowMem.SelectObject(GetStockObject(DKGRAY_BRUSH));
shadowMem.SelectObject(GetStockObject(WHITE_PEN));
shadowMem.Rectangle(50, 50, currWin.Width() - 100, currWin.Height() - 100);
pDC->BitBlt(0, 0, currWin.Width(), currWin.Height(), &shadowMem, 0, 0, SRCCOPY);
shadowMem.SelectObject(pOldBmp);
}
Result:
| {
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Significantly, the decision makers who authorized the special invitations were not at the White House at all, but were ensconced in the Department State's Office of Protocol. The protocol office, chosen by former Secretary of State Hillary Clinton, are filled solely with long-time Clinton loyalists.
"We don't yet how much of this extraordinarily valuable contact information as it walked out the front door either in Dennis Cheng's personal cell phone, or on a contact list that might have been emailed." he said in an interview with TheDCNF.
FACT has been trying to obtain the email traffic between Cheng and the Clinton Foundation under the Freedom of Information Act. The group initially filed an FOIA request in February 2015, but Department of State officials are refusing to release any of the emails before the presidential election.
"They have been stonewalling us and not responsive and now they're telling us that we probably won't get it until mid-to-late November," Whitaker said. He served as the US Attorney for the Southern District of Iowa from 2004-2009.
One of the most revealing illustration of Cheng's continuing contact with the the protocol office after he joined the Clinton Foundation is in email exchanges.
Lloyd Blankfein, Goldman Sachs' chairman and CEO, attended one of the hottest state dinners — a January 19, 2011 White House State Dinner for former Chinese President Hu Jintao.
But Bank of America also paid more than $1 million for four speeches to both Bill and Hillary. | {
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Romania should get payments of European money at more than 30 million euros a week by the end of 2012 in order to meet all the requirements related to the European Union funds absorption, visiting European Commissioner for Regional Policy Johannes Hahn said.
The commissioner, after having attended the works of the Inter-ministry Committee for European Funds at Victoria Palace (the Govt's offices), underscored the importance of a good absorption of the European funds by Romania so that Bucharest might be made available the same quantity of EU funds in the 2014-2020 financial exercise as for 2007-2013.
'2012 is the year of the implementation, it is also necessary that you proved other member states that Romanian can use the available quantity of money, because we have already presented our proposal for the next financial period, which means that /…/ with respect to our proposal, our calculation, Romania should get at least the same quantity of money in order to continue all the projects already begun and by which the Romanian people's living standard must be improved', Hahn underscored.
He stressed the budget level depends on the decision of the European Union heads of states and governments.
'That is why it is so important that you proved it is possible to use the quantity of money. I am grateful for all the initiatives made by the prime minister and his team. For this year alone you might need to get payments of more than 30 million euros or 131 million lei a week for the rest of 2012 in order to meet all the requirements. Therefore, we live tough, but very promising times', the European commissioner said. | {
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This event, for Open Men & Masters Over 35s and Masters Over 50s, takes place over the Songkran and Water Festival weekend. The format is 7-a-side (70m x 45m). Registration is now open; preference will be given to participating teams from last year's tournament. | {
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bibliography * The PainScience Bibliography contains plain language summaries of thousands of scientific papers and others sources, like a specialized blog. This page is about a single scientific paper in the bibliography, Hasson 1990.
The purpose of this study was to compare the analgesic effect of pulsating ultrasound treatment and placebo on delayed onset of muscle soreness produced by an eccentric exercise bout. In addition, the effect of pulsed ultrasound on muscular performance following an eccentric exercise bout was studied. Eighteen untrained subjects were randomly assigned to: 1) ultrasound (A) [N = 6] over the areas of concentrated muscle soreness, i.e. proximal vastus lateralis and distal vastus medialis; 2) placebo ultrasound (B) [N = 6]; and 3) no therapeutic intervention (C) [N = 6]. Baseline data were recorded for maximum isometric knee extension contraction (MVC), maximum knee extension torque (MT), knee extension work (W), and soreness perception (SP). All values were subsequently reassessed 24 and 48 hours after intense muscular activity. Immediately following the 24 hour reassessment the A group received ultrasound treatment, the B group received placebo ultrasound, while the C group received no treatment. Percent deviation from baseline of SP, MVC, MT and W were significantly less for A than B and C (p less than 0.05) at 48 hours post muscle soreness bout. These data indicate that pulsed ultrasound accelerates restoration of normal muscle performance, and thus is effective in decreasing delayed onset of muscle soreness. The mechanism for decreasing soreness perception in the muscle is unknown, but may be related to decreasing intramuscular pressure and/or decreasing the inflammatory response. | {
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Patio covers and carports add to any yard. One of the great joys of living in a house is the outdoor space provided by things such as patios. However, the sad fact of the matter is that in most of our homes, the patio is underused. Snow and rain force us to remain indoors far too often, and even on nice days it is often just not worth the hassle of dealing with annoying bugs that ruin far too many barbecues.
On many days the sun is too hot to endure the outdoors without the benefit of shade. The perfect solution to get you back onto your patio is a Patio Cover. Less expensive than installing a traditional sun room, a Patio Cover allows you to use your existing patio space far more efficiently. Now you can enjoy the beauty of your yard, even when it is raining or snowing. With a clear canopy Patio Cover, you can enjoy looking up at the stars at night while sitting comfortable and snugly in the comforts of your own comfy chair. During the day, high quality windows block you from the harmful UV radiation.
Shelter your car or boat from rain, hail, snow, and blistering heat. Ballew's carports are made from aluminum, custom cut to the size that fits your needs. Our carports have the handsome baked-on finish as well as the strength of aluminum to give you beauty, durability, and minimal maintenance cost.
Ballew's Aluminum patio covers are custom made to your specific needs and are made out of top quality aluminum products. Neat clean lines and color accented to enhance the beauty of your home. The perfect patio cover can now become a reality.
Energy savings – Reduces summer temperatures in your home. | {
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Are you and your friends searching for the best place to drink in Bournemouth? Well luckily for you, we are the venue you have been looking for. From pints to pitchers to martinis, we have got you covered here at Bar So. Whether you are out with the girls on a Friday or simply enjoying a catch up with a friend after work, our contemporary bar, restaurant and nightclub is simply the best place to drink in Bournemouth. Call us today on 01202 203050.
It is right to say that Bournemouth does have its fair share of nightlife hotspots and great places to drink. But if you're looking for the best place to drink in Bournemouth, there is only one place that springs to mind – Bar So.
Located in a prime position, nestled between Bournemouth's busy town centre and its renowned sandy beaches, Bar So is the perfect place for a mid-shop refreshment, after work cocktail or a place to go to party the night away right up until the early hours. As the winner of Bournemouth's Best Bar and five time winner of Bournemouth's Best Night Out, it is easy to see why Bar so is so popular with tourists and locals alike.
As the sun begins to set, you can sit back with a cocktail in hand and unwind after a stressful week in either one of our two large outdoor terraces, upstairs cocktail bar that serves food during the summer, or in our downstairs bar.
Supported by our multi-level layout, we have made the best possible use out of space and have a wide range of features and services available. For your ease we have decided to list the main features of Bar So, that make us the best place to drink in Bournemouth.
As the best place to drink in Bournemouth, we have an extensive variety of outstanding cocktails and mocktails. But be assured that our drinks are not just visually appealing – they are also freshly mixed, theatrically prepared and tantalisingly tasty. We're bringing a whole new level of pleasure to your favourite tipple as our cocktail menu is available throughout the afternoon and evening.
Our centralised location between Bournemouth's sandy beaches and lively town centre provides us with the perfect backdrop for our two large outside terraces. As you watch the world go by in sun-drenched Bournemouth you can enjoy a bite to eat from our new mouth-watering menu whilst enjoying our relaxed and comfortable atmosphere.
As the disco ball glitters, from the mirrored atrium high above you, our resident DJ's are about to provide you with the best night of your life. You only have to have a look at our history, to see some of the legendary nights and award winning artists we have already hosted this year – who's excited for New Year?
We are incredibly lucky to have some of the most technically gifted DJ's on residency, who have a wealth of expertise in the hottest music, including House, RnB, Hip Hop, Techno , Club and much more. We have three of the biggest names in the music industry, widely acclaimed Martyn the Hat, James Herkes and Lee Evans.
Here at Bar So, we are eager to constantly be improving our service. We are now incredibly excited to be able to offer you a VIP experience with our new bottle packages. As the finest place to drink in Bournemouth, we have an extensive selection of the best spirits and champagne money can buy. We also cater for large parties and private events. Whether you have 10 guests or 100, they're guaranteed an unforgettable experience at Bar-So's private bar.
Is your Saturday night sorted? To find out more about our packages and So Saturday drinks deals, please don't hesitate to contact one of our friendly team on 01202 203 050.
As you have already gathered, with our extensive range of services, you would be silly not to visit one of the best places to drink in Bournemouth. At Bar So we realise that people do have different tastes and use bars for a variety of reasons so we have ensured that we really do have something for everyone.
Whether you are looking for a casual drink after work or a full night out with your friends, our professional and friendly staff will be on hand to ensure you have the best night possible. We will work hard to make sure your night is a success as we are committed to making sure that you receive an experience worthy of Bournemouth best bar and night out.
If you like the sound of Bar So then be sure to visit the one of the best places to drink in Bournemouth!
If you are looking for cocktail inspiration or to find the pictures of you and the girls from So Saturday last week, be sure to check out our gallery.
We regularly post images of our events so look out and strike a pose! We look forward to seeing you next Saturday at the finest place to drink in Bournemouth.
So what are you waiting for? If you are looking for the certified best place to drink in Bournemouth, you know who to contact.
Our trusted team of professionals are only a phone call away and will be ready to answer any questions you may have or if you would like to book one of our packages. Give us a call today on 01202 203 050 or drop us an email. | {
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Everything you love about Gain is even better - and smaller - with the new Gain flings.
Everything you love about Gain is even better-and smaller-with the most Gain scent.
Gain Original scent will freshen every fiber, so your nose and clothes will thank you.
Enjoy the matching Gain Original scent in liquid fabric softener, Fireworks, and dryer sheets.
Laundry detergent that smells so great and is so simple to use, it makes doing laundry almost exciting. | {
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A technology developed by ARS scientists is being used to boost resistance to a disease in potatoes that can destroy entire fields and often requires growers to repeatedly spray fungicides. The technology also may be used to breed improved varieties of many other important crops.
Collier, who is now research manager of molecular technologies at the Wisconsin Crop Innovation Center, also plans to promote use of the technology among agricultural scientists and plant biologists. | {
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Naperville North and Central's 2016 Graduations from Naperville News 17 on Vimeo.
Naperville North & Central's 2016 Graduations
Graduation is a time for reflection and celebration, and the Class of 2016 did just that.
"I hope you choose not to regret all the bad tests and teams not made, and those awkward moments, because failure has given you poise and strength and funny stories to tell. I hope you focus upon the things you learned and remembered and the friends you made and kept," said Honors Speaker and 2016 Naperville Central graduate, Victoria Wu.
A sentiment echoed by Naperville North student graduation speaker, Aaron Kruk.
"If home is where someone is thinking of you, then there are few places that embody home more than Naperville North. This is where we belong, not to say that we don't and won't have other homes, but we will forever be a part of this fantastic school. Likewise, we will always carry part of Naperville North with us wherever we go, next year and beyond," said Kruk, also a 2016 graduate.
This graduation held special significance for North's principal Stephanie Posey, as it was her first as a Huskie.
"The class of 2016 welcomed me with open arms, they were very generous and kind and I am thrilled and grateful that they are the class that showed me the traditions and culture that is everything that embodies Naperville North huskies," said Posey.
Both Huskies and Redhawks are now connected as alumni of District 203.
"I don't know everything, but four years of observation and experience have taught me that you all have promise beyond measure, with that I'd like to thank you for being such magnificent classmates, I wish you the greatest joy and the best luck," said Wu.
Hats off to you grads!
Naperville News 17's Evan Summers reports.
Get daily hometown news and sports delivered to your inbox!
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The private tuition sector is booming and there's never been a better time to get in on the potentially lucrative action. Worth an estimated £2 billion in the UK alone, extra-curricular and private tutoring has steadily risen in popularity over the last decade and the industry is now dominated by successful franchises all across the country.
With parents looking to give their children a head start before they venture out into the increasingly competitive world of work, tutoring has become a standard part of many children's lives, with around 40 percent of children in London now undertaking additional tutoring to supplement their education. There are lots of different tutoring companies and franchises out there, from play-based programmes that help engage and instil a love of education in young children to tutoring sessions that are suitable for young people all the way through to university level. However, it can be challenging to go it alone in an industry that's crammed with individuals trying to build their own tutoring business from scratch. Many who are looking to break into the tutoring sector might be passionate educators, but lack the knowledge of marketing, management and the reputation to reach success on their own.
That's why joining a tutoring services franchise might be the perfect solution.
If you're a passionate educator looking for a new challenge that traditional teaching may not offer, why not look into the many tutoring franchise opportunities that are on the market? Whether your skills lie in traditional disciplines such as Maths and English, or you specialise in areas like languages or technology, there are a wide range of opportunities to suit your interests.
If you're looking to break into the home tutoring market, Tutor Doctor is a fantastic franchise to join. It's one of the world's largest tutoring franchises and already has more than 500 franchisees spanning 16 countries. By matching students with a professional, friendly tutor based on their individual academic needs, the franchise has an impressive success rate that helps children succeed and develop a lifelong love for learning. It's been ranked as the number one in-home tutoring franchises by the influential Entrepreneur Magazine and has scooped countless awards for its business model.
If you're looking to move your career into the education sector but have no prior experience or background in education, you could still be the right franchisee for Tutor Doctor. The franchise is seeking ambitious, driven individuals who will help run and co-ordinate their own tutoring business and employ talented educators to deliver sessions in a wide range of subjects. Franchisees will still be able to experience the satisfaction that tutoring delivers, as they'll be making a difference to children who need an extra helping hand.
With a flexible, home-based business model and low overheads, Tutor Doctor offers the flexibility and potential for high returns that so many people seek from franchising.
Many children need a helping hand with their Maths and English skills, and First Class Learning specialises in helping them reach their full potential. As part of a respected and nationally recognised brand, you could be the one that helps a network of tutors remotely deliver a comprehensive and enjoyable extra-curricular programme in your area. With more than 250 tuition centres across the world, First Class Learning welcomes franchisees who are passionate about helping children overcome obstacles in their education without the pressures and restrictive workloads that they may have faced when working in a school.
The franchise is the fastest growing tutoring franchise in the UK and, for a surprisingly low investment, you could open your own centre that you run around your own personal commitments.
Technology is becoming ever-more present in our lives and children who lack sufficient knowledge of the digital world might find themselves at a disadvantage once they enter the workforce. That's where ComputerXplorers can help. With its tailored programme that equips children with the skills to get ahead in the digital world, ComputerXplorers teaches children about everything from digital photography and robotics to coding and forensic science while helping to improve core academic skills including maths, literacy and science.
Franchisees will benefit from multiple income streams, thanks to a business model that lends itself well to the simultaneous running of in-school, extra-curricular and summer camp programmes. Each territory will include at least 300 schools and 300 nurseries/pre-schools, meaning you'll have ample opportunity to generate an impressive income quickly. You won't need to be an IT expert or even come from an education background, as ComputerXplorers will provide more than 200 pieces of unique material and a comprehensive lesson plan to help you and your staff deliver sessions with ease. The franchise is looking for people who are passionate about technology and computing and educating the next generation about the world that exists online.
Languages can unlock the world for your little ones and at Bambini Lingo, they'll be able to experience a fun, unique learning experience that exposes them to up to 9 languages before they've even started school. With a range of different classes that are suitable for children from birth, franchisees will use Bambini Lingo's award-winning music albums and fun, play-based curriculum to immerse children in a new language.
Whether you're multi-lingual or simply have a love of languages, Bambini Lingo is encouraging everyone who is passionate about giving children the gift of language and life-long learning to get involved in its exciting opportunity. All classes need to be delivered by a native speaker, which can either be yourself if you have a bi-lingual background, or employees who deliver your scheduled classes for you. The franchise was established in 2012 and now delivers classes in 9 languages to children across the globe, taking advantage of the 'magic window' of opportunity that young children have for picking up additional languages. | {
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By and large, the ocean is doing its part to produce healthy salmon runs. The problem for Central Valley salmon is almost entirely in the sad condition of the inland freshwater habitat salmon need to spawn and rear in. Competition for freshwater in Central Valley rivers and streams is far and away the biggest single problem. In wet winters and springs, river conditions begin to mimic natural conditions like those that existed before dams stopped runoff. Two years later, we see resurgent salmon runs. This is because baby salmon survive at much higher rates during big runoff years. Reservoirs can be operated (as they are on the Columbia and Snake Rivers in Idaho, Oregon and Washington) to release large volumes of water in the spring to aid the downstream migration of the baby salmon. Not only do we fail to do this in California, but when these spring releases are most needed (in drought years), water managers are least inclined to share water for salmon. Hatchery salmon can be put in tanker trucks and driven down stream but no such aid exists for wild salmon. In the slow moving, gin clear, warm rivers draining the Central Valley during drought, baby salmon are massively lost to predators.
Making matters worse, even in relatively normal wet years, Central Valley salmon suffer from manmade drought due to competition for water with agriculture and urban residents. By and large, reservoir releases are timed to meet the needs of these groups, not those of salmon.
In 2009 the federal government implemented restrictions on some water diversion practices harmful to listed winter and spring run salmon. The restrictions were contained in what's called a biological opinion, or biop, which required certain mitigation steps for the damage caused by federal and state water projects. The 2009 biop greatly helped restore the runs until drought hit in 2012 and gains were lost. Delta pumping restrictions meant water users with junior water rights saw cuts. The junior water rights contractors sued and lost in federal court. Undeterred, they next turned to politicians in Congress to change the laws protecting threatened and endangered wildlife. Since 2013 we've seen legislative efforts, one after another, to overturn or weaken established law protecting species. A fair amount of GGSA time and energy has been spent fighting off these efforts because requirements to protect threatened and endangered salmon runs also protect other salmon runs that supply the salmon fishery. They also protect many other species in the Central Valley and Delta and major parts of California's unique natural heritage. In December of 2016 a federal bill weakening Central Valley salmon protections became law.
To identify and prioritize the problems facing Central Valley salmon, GGSA worked with fisheries biologists and state and federal fish agency experts to develop a salmon rebuilding plan. The plan currently has 27 projects that will go a long ways towards strengthening Central Valley salmon runs when implemented.
The 27 projects were selected because they rebuild both wild and hatchery salmon runs and can be implemented at early dates, mostly affected by funding availability and permit approvals. The projects call for fixes both in the Sacramento and San Joaquin rivers, their tributaries, and in the Delta.
November 2015, GGSA teams up with the California Dept. of Fish and Wildlife to test a drought relief measure injecting fertilized salmon eggs after hot drought flows are replaced by cold winter time water.
GGSA supports scientific tracking of baby salmon with surgically implanted acoustic tags to better understand where we're losing them and how to fix that.
The case for rearing area restoration.
Among other improvements, the salmon rebuilding plan stresses the need to restore areas along the edges of Central Valley rivers where baby salmon can safely feed, grow, and hide from predators.
Side channels featuring lower flow velocity and overhanging trees and brush where insects cluster have been largely lost in man's effort to tame and hem in the rivers and streams.
There is good evidence that well fed baby salmon survive better than hungry, skinny, baby salmon. Streams that flow out on to remaining flood plains see higher survival of baby salmon reaching the ocean. Butte Creek, which drains into a floodplain called the Sutter Bypass, is an example of this. During the low salmon years of 2008 and 2009, salmon from Butte Creek returned at higher rates than neighboring salmon that didn't have access to floodplains as babies. We know that many historic feeding areas along the edges of Central Valley rivers and streams used by baby salmon have been lost to human development. But many of these can be restored. Aerial photographs make clear there are many former parts of the river bed, now dried out behind dikes and levees or plugged by gravel, can be restored by moving a little earth. The greatest opportunities to restore rearing areas are in the upper parts of the Sacramento, Feather and Yuba rivers. The Sutter bypass, which is basically a flood plain currently available to Butte Creek salmon, could be made more accessible to runs from the Sacramento, Feather and Yuba rivers with minor modifications.
The Yolo Bypass west of Sacramento holds the greatest potential rearing area in the Central Valley. Its value for growing baby salmon is so high that the National Marine Fisheries Service (NMFS) are requiring state and federal water managers to modify it make it more accessible. Flood control barriers that separate the Sacramento River from the Yolo Bypass can be lowered to allow this. In addition to rich feeding grounds for juvenile salmon, the Yolo Bypass drains to the western Delta at a point safely beyond the area where the Delta pumps can pull baby salmon off their natural migration course. This provides a major benefit to baby salmon exiting the Central Valley through the bypass instead of through the Delta.
Water distribution in the Central Valley is driven partly by the needs of federally protected species and partly by the demands of water contractors who take water from state and federal water projects. These needs run mostly counter to each other. For the most part, flows and temperatures are only managed for fish when federal law requires it. For instance, federal law requires a portion of the upper Sacramento River be maintained at no more than 56 degrees from June 1 through October 1 to aid spawning winter run salmon. But too often, fall run salmon, which spawn from September through December, spawn in flows above 56 degrees, which kills their eggs. Similarly, reservoir releases have to be maintained at a steady level during that same June 1 through October 1 time period so that winter run redds remain inundated. Too often, come October 1, reservoir releases are cut drastically as agriculture demand ends for the year. This often leads to dewatering of fall run redds and massive loss of fertilized eggs.
Drastic reduction of water releases from reservoirs in the fall kills salmon eggs laid in shallow water gravel bars like those seen here.
GGSA has engaged water managers and those who receive water to seek modifications to the practices that harm salmon. In years flush with water, it's generally easier to find a common ground. In times of drought, it's much harder.
The other major flow need by salmon on dammed rivers is strong runoff in the spring to carry the baby salmon to sea. Dams capture the natural runoff and block this critical component salmon have evolved to depend on. In very wet years there's enough rain and snow runoff coming down streams below dams to mimic the pre-dam condition, but not so in dry years. That's when it's needed most. Some argue that federal requirements designed to protect Central Valley salmon came up short by not including such a requirement. Interestingly, such a requirement does exist on the Columbia and Snake rivers in Idaho, Oregon and Washington but this was only gotten after many years of salmon advocates fighting for it in the federal courts. Since this requirement went into effect in 2005, salmon stocks on the Columbia and Snake rivers have rebounded.
Some projects have been completed or are well on their way. You can read about some of the projects that have been implemented and are now aiding in salmon restoration in our Accomplishments section. | {
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Bank of Botswana pays govt 'surprise' P4.1bn windfall
MBONGENI MGUNI Friday, November 01, 2019
Finance minister Kenneth Matambo is under pressure to erase the budget deficits
The Bank of Botswana (BoB) made a once-off payment of P4.1 billion to government earlier this year as income from gains made in the investment of state funds last year, BusinessWeek has established.
Data seen this week shows the payment was made in April and helped the first quarter of the 2019-2020 budget to a P2.1 billion surplus, in a year in which the fiscal year is expected to record a P7.8 billion deficit.
The BoB's payout comes as questions are raised about the central bank's management of government funds, after prosecutors in a magistrate court case this week alleged political leaders had pilfered billions in previous years.
The central bank previously denied separate reports of missing billions that were alleged to have slipped through unreconciled balances owed to government and held at the BoB.
This week, Finance Ministry officials were quick to explain that the P4.1 billion unexpected windfall had nothing to do with allegations of unreconciled balances or any other suspect source.
"In addition to the dividends expected to be received from the management of foreign exchange reserves, there was a residual income received as realised gains," Finance Ministry officials said in a written response to enquiries. "For the 2019-2020 fiscal year, the pre-set dividend is estimated at P948 million, to which P4.1 billion residual income arising from the BoB's operations the previous year ending December 2018, is added, making it P5.1 billion.
"However, as this is a once-off payment,
we are projecting this to stabilise in the years ahead in line with the historical trend."
Under law, government as the BoB's sole shareholder is entitled to all profits the central bank makes, after it accounts for its expenditure.
While the bank and government decide on a pre-set dividend every year earned from the management of foreign reserves, any gains above this are also transferred to the sole shareholder, government.
Finance Ministry data indicates that the BoB's revenues as a contribution to government swung from P1.6 billion in 2017-2018, to P740 million in 2018-2019 to the projected P5.1 billion in the current fiscal year. The revenues are projected to decline to P1.2 billion next year.
"The level of BoB revenue is influenced by performance of the global financial markets as well as exchange rate movements between the pula and other international currencies where funds are invested," finance officials explained.
Econsult officials, in a third quarter commentary released on Monday, cautioned that while the windfall had helped the first quarter to a surplus, the overall outlook was still negative.
"Because of this 'windfall' income, the outturn should not be taken as representative of the year as a whole, for which a substantial deficit is projected," the firm's researchers stated.
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This recipe is perfect for using left over lamb from your Sunday Roast. Who doesn't love a curry?
Heat the oil in a large saucepan, add the onion and all the spices.
Cook, stirring for about 3 minutes, to allow the flavour from the spices to evolve.
Add the tinned tomato, tomato purée, lemon juice and crumble in the stock cube.
Drain the chickpeas and add to the pan along with the yogurt.
Season with plenty of black pepper and a little salt.
Cook for 10 minutes, then add the lamb and the ground almonds, cook for another 10 minutes and serve.
Great with Bilash basmati rice, poppadums and naan bread. | {
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\section{Introduction}
\label{sec:Introduction}
Active safety measures such as Advanced Driver Assistance Systems
(ADAS) are an elementary component of road-safety, especially with
high market penetration~\cite{Lundgren2006}. In order to further
increase road safety, the EU regulation
2019/2144~\cite{RegulationEU2019/21442019} obliges all OEMs to install
an emergency brake assist (EBA) as well as lane departure warning
system. Hence, as vehicles need to be equipped with the necessary
sensors, the main goal is on cost efficiency, utilizing as few sensors
while offering as many functions possible. As a consequence, the
comfort function adaptive cruise control (ACC) becomes a default
function also in the lower segments. Adaptive cruise control and other
SAE-L1 advanced driver assistance systems have been studied
extensively, see also~\cite{Eskandarian2003}.
For ACC, radar-only as well as radar+camera solutions are often used
as sensor configurations. Despite budget sensors, the expectation
especially on comfort function is high. Ambiguities of the radar
sensors due to low angular resolution are predominant and cannot be
resolved completely by the camera. As a consequence, target
association is increasingly hard which leads to so-called ghost
objects and/or lateral position ambiguities~\cite{Folster2005}.
As a result of these ambiguities, false reactions of the system can occur, such as driver-take over scenarios from a field-operation test, which have been
investigated in~\cite{Weinberger2001}. The cause is often not only a
missing detection of the ACC-relevant vehicle but of a false
interpretation of the scenario, i.e. that the detected vehicle has
been falsely assessed to be not relevant, due to, for example, a false
lane association. Target discrimination focuses on the correct lane
association of vehicles to corresponding
lanes~\cite{Zhang2012,Song2019}.
We propose to use the Boids flocking algorithm~\cite{Reynolds1987,Reynolds1999}, which has been introduced by Reynolds to model the motion of bird flocks or fish school, to model the interaction between traffic participants with each other as well as with the environment.
A flocking algorithm has been used in~\cite{Hayashi2016} to derive a
control algorithm for multiple non-holonomic cars. Each car has been
modeled as an individual boid which behaves according to the movement
rules of the typical boid model, which are cohesion, alignment, and
separation. The proposed control managed a collision-free path of all
boids, however, the usecase was limited to a straight highway and,
eventually, the assumption that all cars driving in the same direction
will converge to the same speed does not hold in reality.
In this paper, an individual flock of boids is created to follow each
detected vehicle, i.e.~for $N_v$ vehicles, $N_f$ flocks with $N_b$
boids each are generated. The corresponding detected vehicle acts as a lead
for the swarm to follow. The aforementioned movement rules are applied
to each boid of each flock. In addition, a further rule is introduced
in order to repel the individual flocks to each other in lateral
direction. Thereby ensuring that the flocks maintain a lateral
distance as long as they are within a given longitudinal distance.
Since the boids of a flock are affected not only by the detected lead
vehicle but also by the other boids of its own flock as well as by the
neighboring flocks, the effects of the position uncertainties of an
object on the target discrimination can be mitigated by using the
average lateral distance between flocks.
The remainder of this paper is organized as follows:
Sec.~\ref{sec:obj_plaus_boids} introduces the Boids flocking algorithm
including the proposed extensions in order to facilitate an object
plausibilization. Additionally, the complexity of the proposed
flocking algorithm is analyzed with respect to the
flock size. Simulation results are presented in
Sec.~\ref{sec:numerical_results}. Finally, Sec.~\ref{sec:conclusion}
draws the conclusion.
\section{Object Plausibilization with the Boids Flocking Algorithm}
\label{sec:obj_plaus_boids}
\subsection{Boids Flocking Algorithm}
\label{sec:boid-flock-algor}
Reynolds~\cite{Reynolds1987} introduced three main rules describing
the movement of boids as an interaction between the individuals of one
flock. The movement of each boid is influenced by
\begin{itemize}
\item \textbf{Separation}: The tendency of a boid to maintain a
certain distance from the other boids within the visible range.
\item \textbf{Cohesion}: The tendency of a boid to move to the average
position of the boids within the visible range.
\item \textbf{Alignment}: The tendency of a boid to align itself with
the boids within the visible range with respect to orientation and
velocity.
\end{itemize}
The three basic rules are applied to each boid and each flock within a
certain radius are listed below and depicted in
Figure~\ref{fig:boid_movement_rules_flock}.
\begin{figure}[htbp]
\centering \subfloat[{\textit Separation}]{
\includestandalone[width=0.31\linewidth,keepaspectratio]{pic/boid_separation}
} \subfloat[{\textit Cohesion}]{
\includestandalone[width=0.31\linewidth,
keepaspectratio]{pic/boid_cohesion} } \subfloat[{\textit
Alignment}]{ \includestandalone[width=0.31\linewidth,
keepaspectratio]{pic/boid_alignment} }
\caption{Three main rules for boid movement within a flock. The $i$-th boid, for which the rules are calculated for, is given in red color. }
\label{fig:boid_movement_rules_flock}
\end{figure}
We define the boids of the $j$-th flock as
\begin{align}
\label{eq:boid}
\setBoidsFlock{j} = \left \lbrace \boid{1,j},\boid{2,j},\ldots \boid{\Nb,j} \right\rbrace
\end{align}
with $j=1,\ldots\Nf$, where $\Nf$ is the number of flocks. Each boid
$\boid{i,j}$ contains the longitudinal and lateral position as well as
velocity. For the sake of readability, we omit the index $j$ of the
flock, as boids do not contain flock-specific information,
i.e. $\boid{i} = [\boidP{i}\, \boidV{i}]^{\mathrm{T}}$, with
$\boidP{i} = [p_{x,i}\, p_{y,i}]^{\mathrm{T}}$,
$\boidV{i} = [v_{x,i}\, v_{y,i}]^{\mathrm{T}}$.
As can be seen in Fig.~\ref{fig:boid_movement_rules_flock}, the
visible range of a boid (field-of-view) is modeled as an ellipse,
opposed to typically a circular section (cf.~\cite{Reynolds1999}), in
order to consider the fact that vehicles are typically driving within
the driving lanes. As a consequence, any boid that deviates too far
laterally from the swarm is no longer considered by the swam. The
deviating boid can nevertheless perceive the swarm and is influenced
by it in its movement.
Therefore, the set of boids, which are within the field-of-view of the
$i$-th boid, is given by
\begin{align}
\label{eq:set_boids_fov}
\setBoidsFoV{i} = \left\lbrace \boid{j} \left| (\boidP{j} - \boidP{i})^{\mathrm{T}} \mathbf{M}^{-1} (\boidP{j} - \boidP{i}) \leq 1 \right. \right\rbrace
\end{align}
with
\begin{align*}
\mathbf{M}= \!\left[\!\begin{array}{cc} \cos \varphi_i & - \sin \varphi_i \\ \sin \varphi_i & \cos \varphi_i \end{array}\!\right]\! \left[\!\begin{array}{cc} a^2 & 0 \\ 0 & b^2 \end{array}\!\right] \! \left[ \!\begin{array}{cc} \cos \varphi_i & -\sin \varphi_i \\ \sin \varphi_i & \cos \varphi_i \end{array}\!\right]^{\mathrm{T}},
\end{align*}
where the parameter of the ellipse are:
\begin{itemize}
\item $\varphi_i$, the orientation angle of the boid given by
$\arctan(v_{y,i} / v_{x,i})$; Note that the coordinate system
according to ISO 8855~\cite{ISO8855_2011} is used, which means that
the x-coordinate is in longitudinal and the y-coordinate in lateral
direction.
\item $a$, the length of the first principal axis of the ellipse;
\item $b$, the length of the second principal axis of the ellipse.
\item $|\setBoidsFoV{i}|$, the cardinality of the subset:
$\Nbfov{i}$. Note that the $i$-th boid is not included in the set
$\setBoidsFoV{i}$; therefore, $|\setBoidsFoV{i}| < |\setBoids|$.
\end{itemize}
In the following, the main steering rules are described.
\subsection{Movement and interaction rules}
\label{sec:movem-inter-rules}
\textbf{Separation}: Each boid has a tendency to keep a certain
distance from the other boids in the flock, thus, avoiding a
collision. This behavior ensures that the flock is spread both in
longitudinal and lateral direction, effectively, enlarging the
field-of-view of the swarm. As described earlier, we assume that
vehicles travel mainly within the driving lanes and the separation of
the flock should also take this into account in the way that the boids
have a larger separation in the longitudinal direction than in the
lateral direction. However, this is not directly considered in the
separation rule but in the weighting factor of the rule (see also
\eqref{eq:BoidVelUpdate}). Several variations of the separation rule
exists, whereas we follow the implementation of~\cite{Hartman2006}:
\begin{align}
\label{eq:sep_rule}
\Rsep = - \sum_{j=1}^{\Nbfov{i}} \boidP{i} - \boidP{j}.
\end{align}
This means, that the position of all boids visible to the $i$-th boid
is subtracted by the position of the $i$-th boid.
\textbf{Cohesion}: Each boid is attracted towards the perceived center
of the flock. This attraction counteracts the separation rule and
causes the boids not to spread throughout the space. Otherwise, the
boids would quickly lose the interaction with each other, due to the
restricted field-of-view. The cohesion force is calculated by
averaging the position of the $\Nbfov{i}$-boids and subtracting the
result from the position of the $i$-th boid:
\begin{align}
\label{eq:coh_rule}
\Rcoh = \frac{1}{\Nbfov{i}} \sum_{j=1}^{\Nbfov{i}}\boidP{j} - \boidP{i}.
\end{align}
Next to the attraction of each boid towards its perceived center of
mass of the visible swarm, additional rules have been introduced
in~\cite{Reynolds1999} with the rule \textbf{Leader Following} of
particular interest, as it describes the tendency of a boid to move
closer to a designated \textit{leader} without actually overtaking the
leader. In our proposed approach, each detected vehicle is a natural
designated leader, which are followed by the boids of the
corresponding swarm. Hence, the attracting force of the leader is
given by
\begin{align}
\label{eq:coh_leader}
\Rcohl = \boidP{l} - \boidP{i}.
\end{align}
\textbf{Alignment}: Since every boid of a flock is supposed to follow
the same designated leader, it stands to reason that eventually every
member of the flock should have the same velocity. The alignment rule
is calculated similarly to the cohesion rule, where the average of the
perceived velocity is calculated first and the velocity of the $i$-th
boid is subtracted from it:
\begin{equation}
\label{eq:ali_rule}
\Rali = \frac{1}{\Nbfov{i}} \sum_{j=1}^{\Nbfov{i}}\boidV{j} - \boidV{i}.
\end{equation}
\subsection{Flock repulsion rule}
\label{sec:flock-repulsion-rule}
An interaction between flocks is introduced in this paper, which is
described by the \textbf{flock repulsion} behavior. The idea is that
the repulsive forces of the neighboring flock supports a clear
separation of the boids and thus of the swarms, so that target
discrimination is facilitated even if the object position is
imprecise. The flock repulsion is denoted by $\Rrep$ and is calculated
in two steps.
First, the perceived center of the neighboring flock from the
point-of-view of the $i$-th boid is calculated by
\begin{align}
\label{eq:center-flock-rep}
\rrep{i}(:,k) = \frac{1}{\Nbfov{k}} \sum_{j=1}^{\Nbfov{k}} \boidP{j}.
\end{align}
Note here, that multiple flocks, for example on the left and right
lane, can be perceived by one boid. It follows that the perceived
center of mass of the neighboring swarms $\rrep{i}$ are represented in
a matrix of size $[2 \times \Nfx{i}]$, with $\Nfx{i}$ the number of
visible flocks by the $i$-th boid.
The repulsing force can then be calculated with an exponential
function whose value is exponentially decreasing with increasing
distance of the swarms, with
\begin{align}
\label{eq:flock_repulsion}
\Rrep = \pm \exp\left(\grep - |\boidP{i} - \rrep{i}|\right),
\end{align}
where the sign $\pm$ depends whether the flock is on the left or right
side, respectively. The value of the factor $\grep$ was chosen so
that the repulsive force is close to zero when the swarms have a
distance of approximately one lane width. Contrary to the other rules,
the repulsing flock rule has a strong effect on the position of the
boids when the distance is small. The rule is exemplary depicted in
Fig.~\ref{fig:flock-repulsion}.
\begin{figure}[htbp]
\centering \includestandalone[width=\linewidth,keepaspectratio]
{pic/boid_repulsion}
\caption{Exemplary illustration of the flock repulsion rule for two flocks shown in grey and black color. The red boid, which belongs to the grey flock, uses the average position of the observed black flock.}
\label{fig:flock-repulsion}
\end{figure}
Although the rule is formulated generally in both longitudinal and
lateral direction, the separation of flocks is only carried out in
lateral direction and is considered in the weighting factor $\wrep$,
which is explained in the following.
\subsection{Position Update}
\label{sec:position-update}
The presented five behavioral rules are combined in an updated
velocity vector $\boidVupdate{i}$ of the $i$-th boid and added to the
velocity vector $\boidV{i}$ of the previous cycle:
\begin{align}
\label{eq:BoidVelUpdate}
\boidVupdate{i} = & \boidV{i} + \wcoh \Rcoh + \wcohl \Rcohl \nonumber \\
& + \wali \Rali + \wsep \Rsep + \wrep \Rrep^{\mathrm{T}},
\end{align}
where the weighting factor $\wrep$ for the repulsive behavior is given
by
\begin{align}
\label{eq:weight-rep-force}
\wrep = \left[ \begin{array}{cccc} w_{x,1} & w_{x,2} & \ldots & w_{x,\Nfx{i}} \\
w_{y,1} & w_{y,2} & \ldots & w_{y,\Nfx{i}}
\end{array} \right]
\end{align}
with $w_{x,:} := 0$ and
$w_{y,:} := \mathrm{sgn}(p_{y,i} - \rrep{i}(y,:))$. The remaining
weighting factors are optimized heuristically and the chosen values
can be found in Table~\ref{tab:para_usecase}.
Given the updated velocity vector, the new position of the $i$-th boid
can be calculated straightforward by
\begin{align}
\label{eq:boid-pos-update}
\boidPupdate{i} = \boidP{i} + \boidVupdate{i}.
\end{align}
\subsection{Life-cycle of Boids}
\label{sec:spawning-boids}
Unlike other publications using Boids flocking algorithm, it is
assumed in this paper, that boids have a rather short lifetime,
meaning a boid is spawned and will eventually cease to exist within a
duration of a few hundred update cycles, whereas each cycle is assumed to have a
duration of about $80$~ms.
As soon as a lead vehicle is consistently tracked, boids will be
spawned by the lead vehicle every $100$~ms until $\Nb$ boids per flock
exist. The position of the lead-vehicle as well as the lateral and
longitudinal velocity from the previous cycle will be provided to the
new boid and serve as initial values.
\subsection{Reachability analysis with Dubins path}
\label{sec:reach-analys-dubins}
Simulating the movement of the boids of each flock according
to~\eqref{eq:BoidVelUpdate} and ~\eqref{eq:boid-pos-update}, it can be
seen that the resulting path of each boid is only $G0$-continuous and
that boids may ``jump" sideways. In order to constrain the movement of
boids and keeping in mind that boids shall follow vehicles with
non-holonomic constraints, the reachability of the updated position of
a boid is checked by a path generated using a Dubins
path~\cite{Dubins1957}. Dubins paths consist only of straight paths
(`S') and curve segments with a restricted radius, i.e. left curve
(`L') and right curves (`R'), respectively.
The minimum radius $\rmin$ is given by the longitudinal velocity $v$
and a fixed maximum lateral acceleration $\alatmax$:
\begin{align*}
\rmin = \frac{v^2}{\alatmax}.
\end{align*}
The maximum lateral acceleration for each boid is chosen to be
$9~\siAms$, which results in a radius of about $60$~m at a velocity of
$23$~m/s. In order to calculate a Dubins path from the current
position to the updated target position of a boid, the start and
target pose need to be determined, where as the orientation angles of
start $\varphi_i$ and target pose ($\varphi_i^\prime$) of a boid are
calculated by
\begin{align}
\label{eq:boidOrientation}
\varphi_i = \atantwo\left(\frac{\boidV{y,i}}{\boidV{x,i}}\right),\; \varphi_i^\prime = \atantwo\left(\frac{\boidVupdate{y,i}}{\boidVupdate{x,i}}\right)
\end{align}
which yields
$\boidPose{i} = [p_{x,i}, p_{y,i}, \varphi_i]^{\mathrm{T}}$. With the
given start and target pose as well as maximum radius, the resulting
Dubins path will be evaluated. Exemplary evaluations are depicted in
Figure~\ref{fig:dubins_path_eval}, whereas valid paths are given in
green and invalid paths in red.
\begin{figure}[htbp]
\centering \subfloat[Examples for directly reachable paths.]{
\includestandalone[width=0.45\linewidth,
keepaspectratio]{pic/dubins_path}} \quad \subfloat[Examples for
paths only reachable via detours.]{
\includestandalone[width=0.45\linewidth,
keepaspectratio]{pic/dubins_path_nok}}
\caption{Exemplary illustration of resulting Dubins paths.}
\label{fig:dubins_path_eval}
\end{figure}
It can be seen that as soon as detours are required to reach the
target pose, this pose is discarded either due to its position and/or
orientation. Instead, in an iterative process, the target pose is
changed in both position as well as orientation within a small radius
--- and thus speed in lateral and longitudinal direction --- until the
target pose can be reached without detours or until a maximum number
of iterations is reached, which can be used to reduce false reactions of driver assistance systems.
\section{Numerical Results}
\label{sec:numerical_results}
A three-lane highway scenario with three target vehicles is considered
in the following with one vehicle driving in each lane and shown in
Figure~\ref{fig:usecase}. The course includes gentle curves as well as
straight sections. The ego vehicle and the preceding vehicle in the
same lane (ID:2) are driving with the same velocity of
$v_{\mathrm{ego}} = v_2 = 25$~m/s at a distance of about $30$~m, which
results in a timegap of $T_\mathrm{G} = d_2 / v_2 = 1.2$~s; a typical
headway distance of an ACC system. A slightly slower vehicle (ID:3) is
driving on the right lane while a faster vehicle (ID:1) is approaching
the ego vehicle from behind, eventually overtaking the ego vehicle and
the other two vehicles.
\begin{figure}[htbp]
\centering \includestandalone[width=\linewidth,
keepaspectratio]{pic/sim_setup}
\caption{Usecase description}
\label{fig:usecase}
\end{figure}
The parameters of the simulation setup are summarized in
Table~\ref{tab:para_usecase}. A standard sensor setup for advanced
driver assistance systems is chosen for the ego vehicle, comprising a
long range radar as well as a monocular camera. The detection of each
sensor are subsequently fused providing qualified tracked
objects. These tracked objects serve as potential leaders to create
the individual flock of boids.
As the chosen velocities of the target vehicles (ID:2) and (ID:3)
differ by only $2$~m/s, the reflections of the radar sensors of the
two vehicles are ambiguous and false associations are likely. As a
consequence, the lateral position of a tracked object may be shifted
in lateral direction or worse, reflections of two vehicles are merged
into one tracked object.
\begin{figure}[tp]
\centering \subfloat[$N_f=3$, $N_b=3$]{
\includegraphics[width=0.479\linewidth,
keepaspectratio]{pic/3veh_N3_ego_left_vp33.pdf}
\includegraphics[width=0.479\linewidth,
keepaspectratio]{pic/3veh_N3_ego_right_vp33.pdf}
\label{fig:violon_a}}
\\
\subfloat[$N_f=3$, $N_b=7$]{ \includegraphics[width=0.479\linewidth,
keepaspectratio]{pic/3veh_N7_ego_left_vp33.pdf}
\includegraphics[width=0.479\linewidth,
keepaspectratio]{pic/3veh_N7_ego_right_vp33.pdf}
\label{fig:violin_b}}
\\
\subfloat[$N_f=3$, $N_b=14$]{ \includegraphics[width=0.479\linewidth,
keepaspectratio]{pic/3veh_N14_ego_left_vp33.pdf}
\includegraphics[width=0.479\linewidth,
keepaspectratio]{pic/3veh_N14_ego_right_vp33.pdf}
\label{fig:violin_c}}
\caption{\it Split-violin plots for the distribution of the lateral
distance between vehicles on the ego and left lane (Ego-Left)
as well as vehicles on the ego and right lane (Ego-Right) .
Lateral distance of boids in blue (left violin), lateral distance of
tracked objects (i.e. inputs) in red (right violin).}
\label{fig:violin_sim_plots}
\end{figure}
The intention of the boids is not to determine a better estimate of
the ground truth position compared to the tracked objects but to
mitigate the shortcomings of the sensors by providing additional
information about the relative position of the vehicles. Hence, the
relative distance between the vehicles is taken as a measure for the
target discrimination. The distance will be taken from the two pairs
of target vehicles, whereby `Ego-Left' refers to the target
combination (ID:1 and ID:2) and `Ego-Right' to the target combination
(ID:2 and ID:3).
The numerical results are shown in Figure~\ref{fig:violin_sim_plots}
for increasing swarm sizes, combining a boxplot with the probability
distribution, which allows for a better comparison of the two setups.
For the visualization of the numerical results, violin
plots~\cite{Hintze1998} are chosen. Each subplot compares the
distribution of the lateral separation determined by the tracked
objects (in red) and determined by boids (in blue). The median of the
distribution is given by the black horizontal in the center of the
notch, whereas the mean value is denoted by the black
star. Correspondingly, first and third quartile are represented by the
borders of the dark colored area, while the light colored region
ranges from the first to the 99th percentile.
As expected, due to the setup of the scenario, the lateral separation
of the right pair `Ego-Right' is worse than that of the pair
`Ego-Left' due to the smaller relative velocity and therefore increased difficulty for the target discrimination in the environmental model. The smallest swarm size, with $\Nb=3$
(cf. Fig.~\ref{fig:violon_a}), shows a performance that is inferior to
that of the tracked objects. This becomes clear by the smaller median
of the distribution (2.4~m compared to 3~m) as well as stronger outliers
for the Ego-Right pair.
With increasing swarm size, e.g. $\Nb=7$
(cf. Fig.~\ref{fig:violin_b}), the median of the lateral separation
increases for the boids slightly above 3~m for the Ego-Right pair which
approaches the true lateral separation of 3.7~m. The medians for the
Ego-Left pair for both boids as well as tracked objects are
comparatively close together, which was expected since the target
vehicles (ID:1 to 3) drive parallel to each other only for a short
time due to the higher relative velocity. However, the outliers for
the tracked objects below a lateral separation of 2~m could be
mitigated.
Interestingly, the results are not exclusively improving with a
further increasing swarm size, see Fig.~\ref{fig:violin_c} for
$\Nb=14$. The \textit{separation rule} forces the boids to split along the longitudinal axis, which leads to an increased distance between first and last boid and thus a decreased influence on the average swarm position due the limited field-of-view of each boid.
\begin{table}[t]
\centering
\begin{tabular}{|ll|} \hline Velocity ego vehicle ID:0 &
$v_\mathrm{ego} = 25$~m/s \\ \hline Velocity black car ID:1 &
$v_1 = 33$~m/s \\ \hline Velocity black car ID:2 & $v_2 = 25$~m/s
\\\hline Velocity truck ID:3 & $v_3 = 23$~m/s\\ \hline Lateral
separation ID:1-ID:2 & $d_{12} = 3.2$~m \\ \hline Lateral
separation ID:2-ID:3 & $d_{23} = 3.7$~m \\ \hline \hline $\wsep$ &
$[0.15, 0.07]^{\mathrm{T}}$ \\ \hline $\wcoh$ &
$[0.4, 0.4]^{\mathrm{T}}$ \\ \hline $\wcohl$
&$[0.4, 0.2]^{\mathrm{T}}$ \\ \hline $\wali$ &
$[0.3, 0.3]^{\mathrm{T}}$ \\ \hline $\grep$ & 1.5 \\ \hline
\end{tabular}
\caption{\it Parameters of the considered usecase as well as for the
flocking algorithm.}
\label{tab:para_usecase}
\end{table}
\section{Conclusions}
\label{sec:conclusion}
The Boids flocking algorithm has been evaluated for the target
discrimination in driver assistance systems. By creating an individual
flock for each detected vehicle, together with the presented movement
rules for boids, simulation results can illustrate that the separation of individual vehicles is consistently improved although the tracked objects were partly less than 0.5m apart.
The average lateral position information of a swarm can be used either for lane association or for the improvement of a lane change detection in low-cost driver assistance systems.
So far, only moving traffic participants have been used to create new flocks of boids. In the future, also static infrastructure, such as guard rails, lane
markings, etc. shall be used to create flocks of boids. In combination
with the flock repulsion rule, it will be investigated whether the target separation of parallel driving vehicles can be improved even further.
\bibliographystyle{IEEEtran}
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UPDATE: Hird 'can't wait' to return to Bombers
7:14am Jul 23, 2014
Suspended Essendon coach James Hird says he "can't wait" to return to the club after cutting short an overseas stay in Europe.
Hird, who was on a study trip in France while serving out a 12-month suspension from the AFL over Essendon's 2012 supplements scandal, returned to Melbourne last night. He was not expected to return to Australia until next week.
"I'm looking forward to heading back to Essendon," he told reporters outside his Toorak home last night.
"I can't wait to be back."
Hird was reportedly looking "relaxed" onboard the flight from Singapore, according to other passengers.
When asked what is role would be when his suspension ends in round 23 in September, Hird said he would discuss it with Essendon coach Mark "Bomber" Thompson.
"I haven't caught up with Bomber yet, but we'll catch up now I'm back," he said.
Suspended Bombers coach James Hird has returned home from overseas early. (AAP) (AAP)
Vic cop says pensioner's beating justified
Victoria braces for more wild weather, as bushfire continues to burn north of Melbourne
Thompson , who was appointed Essendon's coach in Hird's absence has previously said he and Hird would "talk" upon his return.
With Hird's suspension to lift in round 23 and the Bombers likely to make the finals, he's tipped to have a match-day role during September's finals season.
Essendon chairman Paul Little has declared Hird will coach the club in 2015.
Thompson to talk future with Hird | {
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\section{Introduction}
A Latin square of order $n$ is an $n\times n$ array filled with $n$ different symbols, where no symbol appears in the same row or column more than once. Latin squares arise in different branches of mathematics such as algebra (where Latin squares are exactly the multiplication tables of quasigroups) and experimental design (where they give rise to designs called Latin square designs). They also occur in recreational mathematics---for example completed Sudoku puzzles are Latin squares.
In this paper we will look for \emph{transversals} in Latin squares---a transversal in a Latin square of order $n$ is a set of $n$ entries such that no two entries are in the same row, same column, or have the same symbol. One reason transversals in Latin squares are interesting is that a Latin square has an orthogonal mate if, and only if, it has a decomposition into disjoint transversals. See \cite{WanlessSurvey} for a survey about transversals in Latin squares.
It is easy to see that not every Latin square has a transversal (for example the unique $2\times 2$ Latin square has no transversal), however perhaps every Latin square contains a large \emph{partial transversal} (a partial transversal of size $m$ is a set of $m$ entries such that no two entries are in the same row, same column, or have the same symbol)?
There are several closely related, old, and difficult conjectures which guarantee large partial transversals in Latin squares. The first of these is a conjecture of Ryser that every Latin square of odd order contains a transversal \cite{Ryser}. Brualdi conjectured that every Latin square contains a partial transversal of size $n-1$ (see \cite{Brualdi}). Stein independently made the stronger conjecture that every $n\times n$ array filled with $n$ symbols, each appearing exactly $n$ times contains a partial transversal of size $n-1$ \cite{Stein}. Because of the similarity of the above two conjectures, the following is often referred to as ``the Brualdi-Stein Conjecture''
\begin{conjecture}[Brualdi and Stein, \cite{Brualdi, Stein}]\label{BrualdiStein}
Every Latin square contains a partial transversal of size $n-1$.
\end{conjecture}
There have been many partial results about this conjecture.
It is known that every Latin square has a partial transversal of size $n-o(n)$---Woolbright \cite{Woolbright} and independently Brower, de Vries and Wieringa \cite{Brower} proved that ever Latin square contains a partial transversal of size $n-\sqrt n$. This has been improved by Hatami and Schor \cite{HatamiSchor} to $n-O(\log^2 n)$. A remarkable result of H\"aggkvist and Johansson shows that if we consider $(1-\epsilon)n\times n$ Latin rectangles rather than Latin squares, then it is possible to decompose all the entries into disjoint transversals (for $m\leq n$ a $m\times n$ Latin rectangle is an $m\times n$ array of $n$ symbols where no symbol appears in the same row or column more than once. A transversal in a Latin rectangle is a set of $m$ entries no two of which are in the same row, column, or have the same symbol).
\begin{theorem}[H\"aggkvist and Johansson, \cite{HaggkvistJohansson}]\label{HaggvistTheorem}
For every $\epsilon$, there is an $m_0=m_0(\epsilon)$ such that the following holds.
For every $n\geq (1+\epsilon)m\geq m_0$, every $m\times n$ Latin rectangle can be decomposed into disjoint transversals.
\end{theorem}
This theorem is proved by a probabilistic argument, using a ``random greedy process'' to construct the transversals. The above theorem gives yet another proof that every sufficiently large $n\times n$ Latin square has a partial transversal of size $n-o(n)$---indeed if we remove $\epsilon n$ rows of a Latin square we obtain a Latin rectangle to which Theorem~\ref{HaggvistTheorem} can be applied.
In this paper we will look at a strengthening of Conjecture~\ref{BrualdiStein}. The strengthening we'll look at is a conjecture due to Aharoni and Berger which takes place in a more general setting than Latin squares---namely coloured bipartite graphs.
To see how the two settings are related, notice that there is a one-to-one correspondence between $n\times n$ Latin squares and proper edge-colourings of $K_{n,n}$ with $n$ colours---indeed to a Latin square $S$ we associate the colouring of $K_{n,n}$ with vertex set $\{x_1, \dots, x_n, y_1, \dots, y_n\}$ where the edge between $x_i$ and $y_j$ receives colour $S_{i,j}$. It is easy to see that in this setting transversals in $S$ correspond to perfect rainbow matchings in $K_{n,n}$ (a matching is \emph{rainbow} if all its edges have different colours). Thus Conjecture~\ref{BrualdiStein} is equivalent to the statement that ``in any proper $n$-colouring of $K_{n,n}$, there is a rainbow matching of size $n-1$''.
One could ask whether a large rainbow matching exists in more general bipartite graphs.
Aharoni and Berger posed the following conjecture, which generalises Conjecture~\ref{BrualdiStein}.
\begin{conjecture}[Aharoni and Berger, \cite{AharoniBerger}]\label{ConjectureAharoni}
Let $G$ be a bipartite graph consisting of $n$ matchings, each with at least $n+1$ edges. Then $G$ contains a rainbow matching with $n$ edges.
\end{conjecture}
In the above conjecture we think of the $n$ matchings forming $G$ as having different colours, and so ``rainbow matching'' means a matching containing one edge from each matching in $G$. It is worth noting that the above conjecture does not require the matchings in $G$ to be disjoint i.e. it is about bipartite multigraphs rather than simple graphs.
This conjecture was first posed in a different form in~\cite{AharoniBerger} as a conjecture about matchings in tripartite hypergraphs (Conjecture 2.4 in \cite{AharoniBerger}). It was first stated as a conjecture about rainbow matchings in~\cite{AharoniCharbitHoward}.
The above conjecture has attracted a lot of attention recently, and there are many partial results. Just like in Conjecture~\ref{BrualdiStein}, one natural way of attacking Conjecture~\ref{ConjectureAharoni} is to prove approximate versions of it. As observed by Barat, Gy\'arf\'as, and S\'ark\"ozy \cite{Barat}, the arguments that Woolbright, Brower, de Vries, and Wieringa used to find partial transversals of size size $n-\sqrt n$ in Latin squares actually generalise to bipartite graphs to give the following.
\begin{theorem}[Woolbright, \cite{Woolbright}; Brower, de Vries, and Wieringa, \cite{Brower}; Barat, Gy\'arf\'as, and S\'ark\"ozy, \cite{Barat}]\label{Woolbright}
Let $G$ be a bipartite graph consisting of $n$ matchings, each with at least $n$ edges. Then $G$ contains a rainbow matching with $n-\sqrt{n}$ edges.
\end{theorem}
Barat, Gy\'arf\'as, and S\'ark\"ozy actually proved something a bit more precise in \cite{Barat}---for every $k$, they gave an upper bound on the number of matchings of size $n$ needed to find a rainbow matching of size $n-k$.
Another approximate version of Conjecture~\ref{ConjectureAharoni} comes from Theorem~\ref{HaggvistTheorem}. It is easy to see that Theorem~\ref{HaggvistTheorem} is equivalent to the following ``let $G$ be a bipartite graph consisting of $n$ edge-disjoint perfect matchings, each with at least $n+o(n)$ edges. Then $G$ can be decomposed into disjoint rainbow matchings of size $n$'' (to see that this is equivalent to Theorem~\ref{HaggvistTheorem}, associate an $m$-edge-coloured bipartite graph with any $m\times n$ Latin rectangle by placing a colour $k$ edge between $i$ and $j$ whenever $(k,i)$ has symbol $j$ in the rectangle).
The main result of this paper is an approximate version of Conjecture~\ref{ConjectureAharoni} in the case when the matchings in $G$ are disjoint, but not necessarily perfect.
\begin{theorem}\label{MainTheorem}
For all $\epsilon>0$, there exists an $N=N(\epsilon)=10^{20}\epsilon^{-16\epsilon^{-1}}$ such that the following holds.
Let $G$ be a bipartite graph consisting on $n\geq N$ edge-disjoint matchings, each with at least $(1+\epsilon)n$ edges. Then $G$ contains a rainbow matching with $n$ edges.
\end{theorem}
Unlike the proof of Theorem~\ref{HaggvistTheorem} which can be used to give a randomised process to find a rainbow matching, the proof of Theorem~\ref{MainTheorem} is algorithmic i.e. the matching guaranteed by Theorem~\ref{MainTheorem} can be found in polynomial time.
Another very natural approach to Conjecture~\ref{ConjectureAharoni} is to to prove it when the matchings have size much larger than $n+1$. When the matchings have size $2n$, then the result becomes trivial.
\begin{lemma}\label{GreedyLemma}
Let $G$ be a bipartite graph consisting of $n$ matchings, each with at least $2n$ edges. Then $G$ contains a rainbow matching with $n$ edges.
\end{lemma}
This lemma is proved by greedily choosing disjoint edges of different colours. We can always choose $n$ edges this way, since each colour class has $2n$ edges (one of which must be disjoint from previously chosen edges).
There have been several improvements to the $2n$ bound in Lemma~\ref{GreedyLemma}. Aharoni, Charbit, and Howard \cite{AharoniCharbitHoward} proved that matchings of size $7n/4$ are sufficient to guarantee a rainbow matching of size $n$. Kotlar and Ziv \cite{KotlarZiv} improved this to $5n/3$. Clemens and Ehrenm\"uller \cite{DennisJulia} further improved the constant to $3n/2$ which is currently the best known bound.
\begin{theorem}[Clemens and Ehrenm\"uller, \cite{DennisJulia}]
Let $G$ be a bipartite graph consisting of $n$ matchings, each with at least $3n/2+o(n)$ edges. Then $G$ contains a rainbow matching with $n$ edges.
\end{theorem}
Though we won't improve this theorem, we give an alternative proof which gives a weaker bound of $\phi n$ where $\phi\approx 1.618$ is the Golden Ratio.
\begin{theorem}\label{GoldenRatioTheorem}
Let $G$ be a bipartite graph consisting of $n$ matchings, each with at least $\phi n + 20n/\log n$ edges. Then $G$ contains a rainbow matching with $n$ edges.
\end{theorem}
We'll spend the rest of this section informally discussing the methods which we use to prove Theorems~\ref{MainTheorem} and~\ref{GoldenRatioTheorem}. The basic idea is to introduce an auxiliary coloured directed graph and then apply some simple lemmas about directed graphs. The results and concepts about coloured graphs which we use are perhaps of independent interest. These results are all gathered in Section~\ref{SectionConnectedness}. The key idea in the proof of Theorem~\ref{MainTheorem} seems to be a new notion of connectivity of coloured graphs.
\begin{definition}\label{DefinitionConnectivity}
An edge-coloured graph $G$ is said to be rainbow $k$-edge-connected if for any set of at most $k-1$ colours $S$ and any pair of vertices $u$ and $v$, there is a rainbow $u$ to $v$ path whose edges have no colours from $S$.
\end{definition}
The above definition differs from usual notions of connectivity, since generally the avoided set $S$ is a set of \emph{edges} rather than colours.
As we shall see, in some ways Definition~\ref{DefinitionConnectivity} is perhaps \emph{too strong}. In particular the natural analogue of Menger's Theorem for rainbow $k$-edge-connected graphs fails (see Section~\ref{SectionConclusion}). Nevertheless, rainbow $k$-edge-connected graphs turn out to be very useful for studying rainbow matchings in bipartite graphs. It would be interesting to know whether any statements about edge-connectivity have natural generalizations to rainbow $k$-edge-connected graphs.
The structure of this paper is as follows. In Section~\ref{SectionConnectedness} we introduce variations on Definition~\ref{DefinitionConnectivity} and prove a number of lemmas about coloured and directed graphs.
In Section~\ref{SectionMainTheorem} we prove Theorem~\ref{MainTheorem}.
In Section~\ref{SectionGoldenRatio} we prove Theorem~\ref{GoldenRatioTheorem}.
In Section~\ref{SectionConclusion} we make some concluding remarks about the techniques used in this paper. For all standard notation we follow~\cite{BollobasModernGraphTheory}.
\section{Paths in directed and coloured graphs}\label{SectionConnectedness}
In this section we prove results about paths in various types of directed graphs. All graphs in this section have no multiple edges, although we allow the same edge to appear twice in opposite directions.
In directed graphs, ``path'' will always mean a sequence of vertices $x_1, \dots, x_k$ such that $x_ix_{i+1}$ is a directed edge for $i=1, \dots, k-1$. We will use additive notation for concatenating paths---for two paths $P=p_1 \dots p_i$ and $Q=q_1 \dots q_j$, $P + Q$ denotes the path with vertex sequence $p_1\dots p_i q_1\dots q_j$.
Recall that $N^+(v)$ denotes the out-neighbourhood of a vertex $v$ i.e. the set of vertices $x$ for which $vx$ is an edge. We will sometimes have two graphs $G$ and $H$ on the same vertex set. In this case $N^+_G(v)$ and $N^+_H(v)$ denote the out-neighbourhoods of $v$ in $G$ and $H$ respectively. Similarly $d_G(u,v)$ and $d_H(u,v)$ denote the lengths of the shortest part from $u$ to $v$ in $G$ and $H$ respectively.
We will look at coloured graphs. An edge-colouring of a graph is an arbitrary assignment of colours to the edges of a graph. A total colouring is an arbitrary assignment of colours to both the vertices and edges of a graph. For any coloured graph we denote by $c(v)$ and $c(uv)$ the colour assigned to a vertex or edge respectively.
An edge-colouring is out-proper if for any vertex $v$, the outgoing edges from $v$ all have different colours. Similarly an edge-colouring is in-proper if for any vertex $v$, the ingoing edges from $v$ all have different colours. We say that an edge colouring is proper if it is both in and out-proper (notice that by this definition it is possible to have two edges with the same colour at a vertex $v$---as long as one of the edges is oriented away from $v$ and one is oriented towards $v$).
A total colouring is proper if the underlying edge colouring and vertex colourings are proper and the colour of any vertex is different from the colour of any edge containing it.
A totally coloured graph is \emph{rainbow} if all its vertices and edges have different colours.
For two vertices $u$ and $v$ in a coloured graph, $\dRainbow{u,v}$ denotes the length of the shortest rainbow path from $u$ to $v$.
We say that a graph has \emph{rainbow vertex set} if all its vertices have different colours.
This section will mostly be about finding highly connected subsets in directed graphs. The following is the notion of connectivity that we will use.
\begin{definition}\label{DefinitionUncolouredkdCon}
Let $A$ be a set of vertices in a digraph $D$. We say that $A$ is $(k,d)${-connected} in $D$ if, for any set of vertices $S\subseteq V(D)$ with $|S|\leq k-1$ and any vertices $x,y \in A\setminus S$, there is an $x$ to $y$ path of length $\leq d$ in $D$ avoiding $S$.
\end{definition}
Notice that a directed graph $D$ is strongly $k$-connected if, and only if, $V(D)$ is $(k, \infty)$-connected in $D$. Also notice that it is possible for a subset $A\subseteq V(D)$ to be highly connected without the induced subgraph $D[A]$ being highly connected---indeed if $D$ is a bipartite graph with classes $X$ and $Y$ where all edges between $X$ and $Y$ are present in both directions, then $X$ is a $(|Y|, 2)$-connected subset of $D$, although the induced subgraph on $X$ has no edges.
We will also need a generalization this notion of connectivity to coloured graphs
\begin{definition}\label{DefinitionColouredkdCon}
Let $A$ be a set of vertices in a coloured digraph $D$. We say that $A$ is $(k,d)${-connected} in $D$ if, for any set of at most $k-1$ colours $S$ and any vertices $x,y \in A$, there is a rainbow $x$ to $y$ path of length $\leq d$ in $D$ internally avoiding colours in $S$.
\end{definition}
Notice that in the above definition, we did not specify whether the colouring was a edge colouring, vertex colouring, or total colouring. The definition makes sense in all three cases. For edge colourings a path $P$ ``internally avoiding colours in $S$'' means $P$ not having edges having colours in $S$. For vertex colourings a path $P$ ``internally avoiding colours in $S$'' means $P$ not having vertices having colours in $S$ (except possibly for the vertices $x$ and $y$). For total colourings a path $P$ ``internally avoiding colours in $S$'' means $P$ having no edges or vertices with colours in $S$ (except possibly for the vertices $x$ and $y$).
Comparing the above definition to ``rainbow $k$-connectedness'' defined in the introduction we see that an edge-coloured graph is rainbow $k$-connected exactly when it is $(k, \infty)$-connected.
We'll need the following lemma which could be seen as a weak analogue of Menger's Theorem. It will allow us to find rainbow paths through prescribed vertices in a highly connected set.
\begin{lemma}\label{MengerLemma}
Let $D$ be a totally coloured digraph and $A$ a $(3kd,d)$-connected subset of $D$.
Let $S$ be a set of colours with $|S|\leq k$ and $a_1, \dots, a_k$ be vertices in $A$ such that no $a_i$ has a colour from $S$ and $a_1, \dots, a_k$ all have different colours.
Then there is a rainbow path $P$ from $a_1$ to $a_k$ of length at most $kd$ which passes through each of $a_1, \dots, a_k$ and avoids $S$
\end{lemma}
\begin{proof}
Using the definition of $(3kd,d)$-connected, there is a rainbow path $P_1$ from $a_1$ to $a_2$ of length $\leq d$ avoiding colours in $S$. Similarly for $i\leq k$, there is a rainbow path $P_i$ from $a_i$ to $a_{i+1}$ of length $\leq d$ internally avoiding colours in $S$ and colours in $P_1, \dots, P_{i-1}$. Joining the paths $P_1, \dots, P_{k-1}$ gives the required path.
\end{proof}
To every coloured directed graph we associate an uncoloured directed graph where two vertices are joined whenever they have a lot of short paths between them.
\begin{definition}\label{DefinitionDm}
Let $D$ be a totally coloured digraph and $m\in \mathbb{N}$. We denote by $D_m$ the uncoloured directed graph with $V(D_m)=V(D)$ and $xy$ an edge of $D_m$ whenever there are $m$ internally vertex disjoint paths $P_1, \dots, P_m$, each of length $2$ and going from $x$ to $y$ such that $P_1\cup \dots\cup P_m$ is rainbow.
\end{definition}
It turns out that for properly coloured directed graphs $D$, the uncoloured graph $D_m$ has almost the same minimum degree as $D$. The following lemma will allow us to study short rainbow paths in coloured graphs by first proving a result about short paths in uncoloured graphs.
\begin{lemma}\label{DCHighDegree}
For all $\epsilon>0$ and $m\in \mathbb{N}$, there is an $N=N(\epsilon,m)=(5m+4)/\epsilon^2$ such that the following holds.
Let $D$ be a properly totally coloured directed graph on at least $N$ vertices with rainbow vertex set. Then we have
$$\delta^+(D_m)\geq \delta^+(D) -\epsilon|D|.$$
\end{lemma}
\begin{proof}
Let $v$ be an arbitrary vertex in $D_m$.
It is sufficient to show that $|N^+_{D_m}(v)|\geq |\delta^+(D)| -\epsilon|D|.$
For $w\in V(D)$, let $r_v(w)= \# \text{rainbow paths of length 2 from } v \text{ to } w.$
Let $W=\{w: r_v(w)\geq 5m\}$.
We show that $W$ is contained in $N^+_{D_m}(v)$.
\begin{claim}\label{WinDC}
If $w \in W$, then we have $vw\in E(D_m)$.
\end{claim}
\begin{proof}
From the definition of $W$, we have $5m$ distinct rainbow paths $P_1, \dots, P_{5m}$ from $v$ to $w$ of length $2$.
Consider an auxiliary graph $G$ with $V(G)=\{P_1, \dots, P_{5m}\}$ and $P_i P_j\in E(G)$ whenever $P_i\cup P_j$ is rainbow.
We claim that $\delta(G)\geq 5m-4$. Indeed if for $i\neq j$ we have $P_i=vxw$ and $P_j=vyw$, then, using the fact that the colouring on $D$ is proper and the vertex set is rainbow, it is easy to see that the only way $P_i\cup P_j$ could not be rainbow is if one of the following holds:
\begin{align*}
c(vx)=c(yw) \hspace{1cm}
&c(vx)=c(y) \\
c(vy)=c(xw) \hspace{1cm}
&c(vy)=c(x).
\end{align*}
Thus if $P_i=vxw$ had five non-neighbours $vy_1w, \dots, vy_5w$ in $G$, then by the Pigeonhole Principle for two distinct $j$ and $k$ we would have one of $c(y_jw)=c(y_kw)$, $c(y_j)=c(y_k)$, or $c(vy_j)=c(vy_k)$. But none of these can occur for distinct paths $vy_jw$ and $vy_kw$ since the colouring on $D$ is proper and the vertex set is rainbow. Therefore $\delta(G)\geq 5m-4$ holds.
Now by Tur\'an's Theorem, $G$ has a clique of size at least $|V(G)|/5=m$. The union of the paths in this clique is rainbow, showing that $vw\in E(D_m)$.
\end{proof}
Now we show that $W$ is large.
\begin{claim}\label{WLarge}
$|W|\geq \delta^+(D)-\epsilon|D|$
\end{claim}
\begin{proof}
For any $u\in N^+_D(v)$ we let $$N'(u)= N^+_D(u)\setminus\{x\in N^+_D(u):ux \text{ or } x \text{ has the same colour as } v \text{ or } vu\}.$$
Since $D$ is properly coloured, and all the vertices in $D$ have different colours, we have that $|\{x \in N^+(u):ux \text{ or } x \text{ has the same colour as } v \text{ or } vu\}|\leq 4$. This implies that $|N'(u)|\geq \delta^+(D)-4$.
Notice that for a vertex $x$, we have $x \in N'(u)$ if, and only if, the path $vux$ is rainbow. Indeed $vu$ has a different colour from $v$ and $u$ since the colouring is proper. Similarly $ux$ has a different colour from $u$ and $x$. Finally $ux$ and $x$ have different colours from $v$ and $vu$ by the definition of $N'(u)$.
Therefore there are $\sum_{u\in N^+_D(v)} |N'(u)|$ rainbow paths of length $2$ starting at $v$ i.e. we have $\sum_{x\in V(D)} r_v(x)=\sum_{u\in N^+_D(v)} |N'(u)|$. For any $x\in D$, we certainly have $r_v(x)\leq |N^+(v)|$. If $x \not\in W$ then we have $r_v(x)< 5m$. Combining these we obtain
$$(|D|-|W|)5m+|W||N^+_D(v)|\geq \sum_{x\in V(D)} r_v(x)=\sum_{u\in N^+(v)} |N'(u)|\geq |N^+_D(v)|(\delta^+(D)-4).$$
The last inequality follows from $|N'(u)|\geq \delta^+(D)-4$ for all $u \in N^+_D(v)$.
Rearranging we obtain
$$|W|\geq \frac{|N^+_D(v)|(\delta^+(D)-4)-5m|D|}{|N^+_D(v)|-5m}\geq\delta^+(D)-\frac{5m|D|}{|N^+_D(v)|}-4\geq \delta^+(D)-(5m+4)\frac{|D|}{\delta^+(D)}.$$
If $(5m+4)/\delta^+(D)\leq\epsilon$, then this implies the claim. Otherwise we have $\delta^+(D)< (5m+4)/\epsilon$ which, since $|D|\geq N_0=(5m+4)/\epsilon^2$, implies that $\delta^+(D)\leq \epsilon|D|$ which also implies the claim.
\end{proof}
Claim~\ref{WinDC} shows that $W\subseteq N^+_{D_m}(v)$, and so Claim~\ref{WLarge} implies that $|N^+_{D_m}(v)|\geq \delta^+(D)-\epsilon|D|$. Since $v$ was arbitrary, this implies the lemma.
\end{proof}
The following lemma shows that every directed graph with high minimum degree contains a large, highly connected subset.
\begin{lemma}\label{LargeConnectedSetLemma}
For all $\epsilon>0$ and $k \in \mathbb{N}$, there is a $d=d(\epsilon)=40\epsilon^{-2}$ and $N=N(\epsilon,s)=32k\epsilon^{-2}$ such that the following holds.
Let $D$ be a directed graph of order at least $N$. Then there is a $(k,d)$-connected subset $A\subseteq V(D)$ satisfying
$$|A|\geq \delta^+(D)-\epsilon|D|.$$
\end{lemma}
\begin{proof}
We start with the following claim.
\begin{claim}\label{ClaimLargeConnectedSet}
There is a set $\tilde A\subseteq V(D)$ satisfying the following
\begin{itemize}
\item For all $B\subseteq \tilde A$ with $|B|> \epsilon |D|/4$ there is a vertex $v \in \tilde A\setminus B$ such that $|N^+(v)\cap B|\geq \epsilon^2 |D|/16$.
\item $\delta^+(D[\tilde A])\geq \delta^+(D)-\epsilon |D|/4$.
\end{itemize}
\end{claim}
\begin{proof}
Let $A_0= V(D)$.
We define $A_1, A_2, \dots, A_M$ recursively as follows.
\begin{itemize}
\item If $A_i$ contains a set $B_i$ such that $|B_i|> \epsilon|D|/4$ and for all $v \in A_i\setminus B_i$ we have $|N^+(v)\cap B_i|< \epsilon^2 |D|/16$, then we let $A_{i+1}=A_i\setminus B_i$.
\item Otherwise we stop with $M=i$.
\end{itemize}
We will show that that $\tilde A=A_M$ satisfies the conditions of the claim.
Notice that by the construction of $A_M$, it certainly satisfies the first condition.
Thus we just need to show that $\delta^+(D[A_M])\geq \delta^+(D)-\epsilon |D|/4$.
From the definition of $A_{i+1}$ we have that $\delta^+(D[A_{i+1}])\geq \delta^+(D[A_{i}])-\epsilon^2|D|/16$ which implies $\delta^+(D[A_{M}])\geq \delta^+(D)-M\epsilon^2|D|/16$. Therefore it is sufficient to show that we stop with $M\leq 4\epsilon^{-1}$.
This follows from the fact that the sets $B_0, \dots, B_{M-1}$ are all disjoint subsets of $V(D)$ with $|B_i|> \epsilon|D|/4$.
\end{proof}
Let $\tilde A$ be the set given by the above claim.
Let $A=\{v\in \tilde A: |N^-(v)\cap \tilde A|\geq \frac{\epsilon}{2} |D|\}$.
We claim that $A$ satisfies the conditions of the lemma.
To show that $|A|\geq \delta^+(D)-\epsilon|D|$, notice that we have
$$\frac{\epsilon}{2}|D|(|\tilde A|-|A|)+|A||\tilde A|\geq \sum_{v\in \tilde A} |N^-(v)\cap \tilde A|= \sum_{v\in \tilde A} |N^+(v)\cap \tilde A|\geq |\tilde A|(\delta^+(D)-\epsilon|D|/4).$$
The first inequality come from bounding $|N^-(v)\cap \tilde A|$ by $\frac{\epsilon}2|D|$ for $v\not\in A$ and by $|\tilde A|$ for $v\in A$.
The second inequality follows from the second property of $\tilde A$ in Claim~\ref{ClaimLargeConnectedSet}.
Rearranging we obtain
$$|A|\geq \frac{|\tilde A|}{|\tilde A|-\epsilon|D|/2}(\delta^+(D)-3\epsilon|D|/4)\geq \delta^+(D)-\epsilon|D|.$$
Now, we show that $A$ is $(k,d)$-connected in $D$. As in Definition~\ref{DefinitionUncolouredkdCon}, let $S$ be a subset of $V(D)$ with $|S|\leq k-1$ and let $x,y$ be two vertices in $A\setminus S$.
We will find a path of length $\leq d$ from $x$ to $y$ in $\tilde A\setminus S$.
Notice that since $|D|\geq 32k\epsilon^{-2}$, we have $|S|\leq \epsilon^{2} |D|/32$.
Let $N^t(x)=\{u\in \tilde A\setminus S: d_{D[\tilde A\setminus S]}(x,u)\leq t\}$.
We claim that for all $x\in \tilde A$ and $t\geq 0$ we have
$$|N^{t+1}(x)|\geq \min(|\tilde A|-\epsilon|D|/4, |N^t(x)|+\epsilon^2|D|/32).$$
For $t=0$ this holds since we have $|N^1|=|\tilde A|\geq \epsilon |D|/4$.
Indeed if $|N^{t}(x)|< |\tilde A|-\epsilon|D|/4$ holds for some $t$ and $x$, then letting $B=\tilde A\setminus N^{t}(x)$ we can apply the first property of $\tilde A$ from Claim~\ref{ClaimLargeConnectedSet} in order to find a vertex $u\in N^t(x)$ such that $|N^+(u)\cap (\tilde A\setminus N^t(x))|\geq \epsilon^2 |D|/16$. Using $|S|\leq \epsilon^{2} |D|/32$ we get $|(N^+(u)\setminus S)\cap (\tilde A\setminus N^t(x))|\geq |N^+(u)\cap (\tilde A\setminus N^t(x))|-|S|\geq \epsilon^2 |D|/32$.
Since $(N^+(u)\cap \tilde A\setminus S)\cup N^t(x)\subseteq N^{t+1}(x)$, we obtain $|N^{t+1}(x)|\geq |N^t(x)|+\epsilon^2|D|/32$.
Thus we obtain that $|N^{t}(x)|\geq \min(|\tilde A|-\epsilon|D|/4, t\epsilon^2|D|/32)$.
Since $(d-1)\epsilon^2/32>1$, we have that $|N^{d-1}(x)|\geq |\tilde A|-\epsilon|D|/4$.
Recall that from the definition of $A$, we also have also have $|N^-(y)\cap \tilde A|\geq \epsilon |D|/2$. Together these imply that $N^-(y)\cap N^{d-1}(x)\neq \emptyset$ and hence there is a $x$ -- $y$ path of length $\leq d$ in $\tilde A\setminus S$.
\end{proof}
The following is a generalization of the previous lemma to coloured graphs. This is the main intermediate lemma we need in the proof of Theorem~\ref{MainTheorem}.
\begin{lemma}\label{LargeRainbowConnectedSetLemma}
For all $\epsilon>0$ and $k \in \mathbb{N}$, there is an $d=d(\epsilon)=1280\epsilon^{-2}$ and $N=N(\epsilon,k)=1800k\epsilon^{-4}$ such that the following holds.
Let $D$ be a properly totally coloured directed graph on at least $N$ vertices with rainbow vertex set.
Then there is a $(k,d)$-connected subset $A\subseteq V(D)$ satisfying
$$|A|\geq \delta^+(D)-\epsilon|D|.$$
\end{lemma}
\begin{proof}
Set $m=9d+3k$, and consider the directed graph $D_m$ as in Definition~\ref{DefinitionDm}.
Using $|D|\geq 1800k\epsilon^{-4}$, we can apply Lemma~\ref{DCHighDegree} with the constant $\epsilon/4$ we have that $\delta^+(D_m)\geq \delta^+(D) -\epsilon|D|/4.$
Apply Lemma~\ref{LargeConnectedSetLemma} to $D_m$ with the constants $\epsilon/4$, and $k$. This gives us a $(k, d/2)$-connected set $A$ in $D_m$ with $|A|\geq \delta^+(D_m)-\epsilon|D_m|/4\geq \delta^+(D)-\epsilon|D|/2$.
We claim that $A$ is $(k, d)$-connected in $D$. As in Definition~\ref{DefinitionColouredkdCon}, let $S$ be a set of $k$ colours and $x,y \in A$. Let $S_V$ be the vertices of $D$ with colours from $S$. Since all vertices in $D$ have different colours, we have $|S_V|\leq k$.
Since $A$ is $(k, d/2)$-connected in $D_m$, there is a $x$ -- $y$ path $P$ in $(D_m\setminus S_V)+x+y$ of length $\leq d/2$.
Using the property of $D_m$, for each edge $uv\in P$, there are at least $m$ choices for a triple of three distinct colours $(c_1, c_2, c_3)$ and a vertex $y(uv)$ such that there is a path $uy(uv)v$ with $c(uy(uv))=c_1$, $c(y(uv))=c_2$, and $c(y(uv)v)=c_3$. Since $m\geq 9d+3k\geq 6|E(P)|+3|V(P)|+3|S|$, we can choose such a triple for every edge $uv\in P$ such that for two distinct edges in $P$, the triples assigned to them are disjoint, and also distinct from the colours in $S$ and colours of vertices of $P$.
Let the vertex sequence of $P$ be $u,x_1, x_2, \dots, x_{p}, v$. The following sequence of vertices is a rainbow path from $u$ to $v$ of length $2|P|\leq d$ internally avoiding colours in $S$
$$P'=u, y(ux_1), x_1, y(x_1x_2), x_2, y(x_2x_3), x_3,\dots, x_{p-1},y(x_{p-1}x_p), x_p, y(x_pv), v.$$
To show that $P'$ is a rainbow path we must show that all its vertices and edges have different colours. The vertices all have different colours since the vertices in $D$ all had different colours. The edges of $P'$ all have different colours from each other and the vertices of $P'$ by our choice of the vertices $y(x_ix_{i+1})$ and the triples of colours associated with them.
\end{proof}
We'll need the following simple lemma which says that for any vertex $v$ there is a set of vertices $N^{t_0}$ close to $v$ with few edges going outside $N^{t_0}$.
\begin{lemma}\label{CloseSubgraphsLowExpansion}
Suppose we have $\epsilon>0$ and $D$ a totally coloured directed graph.
Let $v$ be a vertex in $D$ and for $t\in \mathbb{N}$, let $N^t(v)=\{x:\dRainbow{v,x}\leq t\}$.
There is a $t_0\leq \epsilon^{-1}$ such that we have
$$|N^{t_{0}+1}(v)|\leq |N^{t_{0}}(v)|+ \epsilon |D|.$$
\end{lemma}
\begin{proof}
Notice that if $|N^{t+1}(v)|> |N^{t}(v)|+ \epsilon |D|$ held for all $t\leq \epsilon^{-1}$, then we would have $|N^{t}(v)|>\epsilon t|D|$ for all $t\leq \epsilon^{-1}$. When $t=\epsilon^{-1}$ this gives $|N^{\epsilon^{-1}}(v)|>|D|$, which is a contradiction.
\end{proof}
A corollary of the above lemma is that for any vertex $v$ in a properly coloured directed graph, there is a subgraph of $D$ close to $v$ which has reasonably large minimum out-degree.
\begin{lemma}\label{CloseHighDegreeSubgraph}
Suppose we have $\epsilon>0$ and $D$ a properly totally coloured directed graph on at least $2\epsilon^{-2}$ vertices, with rainbow vertex set.
Let $v$ be a vertex in $D$ and let $\delta^+= \min_{x
: \dRainbow{v,x}\leq \epsilon^{-1}} d^+(x)$. Then there is a set $N$ such that $\dRainbow{v,N}\leq \epsilon^{-1}$ and we have
$$\delta^+(D[N])\geq \delta^+ - 2\epsilon |D|.$$
\end{lemma}
\begin{proof}
Apply Lemma~\ref{CloseSubgraphsLowExpansion} to $D$ in order to obtain a number $t_0\leq \epsilon^{-1}$ such that $|N^{t_{0}+1}(v)|\leq |N^{t_{0}}(v)|+ \epsilon |D|$. We claim that the set $N=N^{t_{0}}(v)$ satisfies the conditions of the lemma.
Suppose, for the sake of contradiction that there is a vertex $x\in N^{t_0}(v)$ with $|N^+(x)\cap N^{t_0}(v)|< \delta^+ - 2\epsilon |D|$. Since $\delta^+\leq |N^+(x)|$, we have $|N^+(x)\setminus N^{t_0}(v)|> 2\epsilon |D|$. Let $P$ be a length $\leq t_0$ path from $v$ to $x$. Notice that since the colouring on $D$ is proper and all vertices in $D$ have different colours, the path $P+y$ is rainbow for all except at most $2|P|$ of the vertices $y \in N^+(x)$. Therefore we have $|N^+(x)\setminus N^{t_0+1}(v)|\leq 2|P|\leq 2\epsilon^{-1}$. Combining this with $|D|\geq 2\epsilon^{-2}$, this implies
\begin{align*}
|N^{t_0+1}(v)|&\geq |N^{t_0}(v)|+|N^+(x)\setminus N^{t_0}(v)|-|N^+(x)\setminus N^{t_0+1}(v)|\\
&>|N^{t_0}(v)|+2\epsilon |D|-2\epsilon^{-1}\\
&\geq |N^{t_0}(v)|+\epsilon|D|.
\end{align*}
This contradicts the choice of $t_0$ in Lemma~\ref{CloseSubgraphsLowExpansion}.
\end{proof}
\section{Proof of Theorem~\ref{MainTheorem}}\label{SectionMainTheorem}
The goal of this section is to prove an approximate version of Conjecture~\ref{ConjectureAharoni} in the case when all the matchings in $G$ are disjoint. The proof will involve considering auxiliary directed graphs to which Lemmas~\ref{LargeRainbowConnectedSetLemma} and~\ref{CloseHighDegreeSubgraph} will be applied.
We begin this section by proving a series of lemmas (Lemmas~\ref{SwitchingLemma} --~\ref{IncrementLemma}) about bipartite graphs consisting of a union of $n_0$ matchings. The set-up for these lemmas will always be the same, and so we state it in the next paragraph to avoid rewriting it in the statement of every lemma.
We will always have bipartite graph called ``$G$'' with bipartition classes $X$ and $Y$ consisting of $n+1$ edge-disjoint matchings $M_1, \dots, M_{n+1}$. These matchings will be referred to as colours, and the colour of an edge $e$ means the matching $e$ belongs to.
There will always be a rainbow matching called $M$ of size $n$ in $G$.
We set $X_0=X\setminus V(M)$ and $Y_0-Y\setminus V(M)$. The colour missing from $M$ will denoted by $c^*$.
Notice that for any edge $e$, there is a special colour (the colour $c_e$ of the edge $e$) as well as a special vertex in $X$ (i.e. $e\cap X$) and in $Y$ (i.e. $e\cap Y$). In what follows we will often want to refer to the edge $e$, the colour $c_e$, and the vertices $e\cap X$ and $e\cap Y$ interchangeably.
To this end we make a number of useful definitions:
\begin{itemize}
\item For an edge $e$, we let $(e)_C$ be the colour of $e$, $(e)_X=e\cap X$, and $(e)_Y=e\cap Y$.
\item For a vertex $x \in X$, we let $(x)_M$ be the edge of $M$ passing through $x$ (if it exists), $(x)_C$ the colour of $(x)_M$, and $(x)_Y$ the vertex $(x)_M\cap Y$. If there is no edge of $M$ passing through $x$, then $(x)_M$, $(x)_C$, and $(x)_Y$ are left undefined.
\item For a vertex $y \in Y$, we let $(y)_M$ be the edge of $M$ passing through $y$ (if it exists), $(y)_C$ the colour of $(y)_M$, and $(y)_X$ the vertex $(y)_M\cap X$. If there is no edge of $M$ passing through $y$, then $(y)_M$, $(y)_C$, and $(y)_X$ are left undefined.
\item For a colour $c$, we let $(c)_M$ be the colour $c$ edge of $M$ (if it exists), $(x)_X=(c)_M\cap X$, and $(c)_Y=(c)_M\cap Y$. For the colour $c^*$, we leave $(c)_M$, $(c)_X$, and $(c)_Y$ undefined.
\end{itemize}
For a set $S$ of colours, edges of $M$, or vertices, we let $(S)_M=\{(s)_M:s\in S\}$, $(S)_X=\{(s)_X:s\in S\}$, $(S)_Y=\{(s)_Y:s\in S\}$, and $(S)_C=\{(s)_C:s\in S\}$. Here $S$ is allowed to contain colours/edges/vertices for which $(*)_M$/$(*)_X$/$(*)_Y$/$(*)_C$ are undefined---in this case $(S)_M$ is just the set of $(s)_M$ for $s\in S$ where $(s)_M$ is defined (and similarly for $(S)_X$/$(S)_Y$/$(S)_C$.
It is useful to observe that from the above definitions we get identities such as $(((S)_X)_C)_M=S$ for a set $S$ of edges of $M$.
We will now introduce two important and slightly complicated definitions. Both Definition~\ref{DefinitionSwitching} and~\ref{DefinitionFree} will take place in the setting of a bipartite graph $G$ with bipartition $X\cup Y$ consisting of $n+1$ edge-disjoint matchings, and a rainbow matching $M$ of size $n$ missing colour $c^*$.
The first definition is that of a \emph{switching}---informally this should be thought of as a sequence of edges of $G\setminus M$ which might be exchanged with a sequence of edges of $M$ in order to produce a new rainbow matching of size $n$. See Figure~\ref{SwitchingFigure} for an illustration of a switching.
\begin{figure}
\centering
\includegraphics[width=0.6\textwidth]{Switching.png}
\caption{An $X'$-switching of length $4$. The solid lines represent edges of $M$ and the dashed lines represent edges not in $M$. \label{SwitchingFigure}}
\end{figure}
\begin{definition}\label{DefinitionSwitching}
Let $X'\subseteq X$. We call a sequence of edges, $\sigma=(e_0, m_1, e_1, m_2, e_2, \dots, e_{\ell-1}, m_{\ell})$, an $X'$-switching if the following hold.
\begin{enumerate}[(i)]
\item For all $i$, $m_i$ is an edge of $M$ and $e_i$ is not an edge of $M$.
\item For all $i$, $m_i$ and $e_i$ have the same colour, $c_i$.
\item For all $i$, $e_{i-1}\cap m_i=(m_i)_Y$.
\item For all $i\neq j$, we have $e_{i}\cap e_j=e_{i-1}\cap m_{j}=\emptyset$ and also $c_i\neq c_j$.
\item For all $i$, $(e_i)_X\in X'$.
\end{enumerate}
\end{definition}
If $\sigma$ is a switching defined as above, then we say that $\sigma$ is a length $\ell$ switching from $c_0$ to $c_{\ell}$.
Let $e(\sigma)=\{e_0, \dots, e_{\ell-1}\}$ and $m(\sigma)=\{m_1, \dots, m_{\ell}\}$. For a switching $\sigma$ we define $(\sigma)_X=(e(\sigma))_X\cup (m(\sigma))_X$.
The next definition is that of a \emph{free} subset of $X$---informally a subset $X'\subset X$ is free if there are matchings $M'$ which ``look like'' $M$, but avoid $X'$.
\begin{definition}\label{DefinitionFree}
Let $X', T\subseteq X$, $k\in\mathbb{N}$, and $c$ be a colour.
We say that $X'$ is $(k,T,c)$-free if $T\cap X'=\emptyset$, $c \not\in (X'\cup T)_C$, and the following holds:
Let $A$ be any set of $k$ edges in $M\setminus ((T)_M\cup (c)_M)$, $B\subseteq X'$ any set of $k$ vertices such that $(A)_X\cap B=\emptyset$. Then there is a rainbow matching $M'$ of size $n$ satisfying the following:
\begin{itemize}
\item $M'$ agrees with $M$ on $A$.
\item $M'\cap B=\emptyset$.
\item $M'$ misses the colour $c$.
\end{itemize}
\end{definition}
It is worth noticing that $X_0$ is $(n,\emptyset,c^*)$-free (always taking the matching $M'$ to be $M$ in the definition). Intuitively free sets should be thought of as sets which ``behave like $X_0$'' for the purposes of finding a matching larger than $M$.
The following lemma is crucial---it combines the preceding two definitions together and says that if we have an $X'$-switching $\sigma$ for a free set $X'$, then there is a new rainbow matching of size $n$ which avoids $(m(\sigma))_X$.
\begin{lemma}\label{SwitchingLemma}
Suppose that $X'$ is $(2k,T,c)$-free and $\sigma=(e_0, m_1, e_1, \dots, e_{\ell-1}, m_{\ell})$ is an $X'$-switching from $c$ to $(m_{\ell})_C$ of length $\ell\leq k$.
Let $A$ be any set of at most $k$ edges in $M-(c)_M$ and let $B$ be any subset of $X'$ of order at most $k$. Suppose that the following disjointness conditions hold
\begin{align*}
\hspace{2cm} &(\sigma)_X\cap T=\emptyset &(\sigma)_X\cap (A)_X=\emptyset \hspace{1cm} &(\sigma)_X\cap B=\emptyset \hspace{2cm}\\
\hspace{2cm} & &T\cap(A)_X =\emptyset \hspace{1cm} &(A)_X\cap B=\emptyset.\hspace{2cm}
\end{align*}
Then there is a rainbow matching $\tilde M$ of size $n$ in $G$ which misses colour $(m_{\ell})_C$, agrees with $M$ on $A$, and has $\tilde M\cap (m(\sigma))_X=\tilde M\cap B=\emptyset$.
\end{lemma}
\begin{proof}
We let $A'=m(\sigma)\cup A$ and $B'=(e(\sigma))_X\cup B$.
Notice that we have $|A'|, |B'|\leq 2k$. Also from the definition of ``switching'', we have that for any $i$ and $j$, the edges $e_i$ and $m_j$ never intersect in $X$ which together with $(A)_X\cap (\sigma)_X=\emptyset$, $B\cap (\sigma)_X=\emptyset$, and $(A)_X\cap B=\emptyset$ implies that $(A')_X\cap B'=\emptyset$.
Also, $(\sigma)_X\cap T=\emptyset$ and $(A)_X\cap T=\emptyset$ imply that $A'\cap (T)_M=\emptyset$, and hence since $\sigma$ is a switching starting at $c$ we have $A'\subseteq M\setminus ((T)_M\cup (c)_M)$. Finally, $\sigma$ being an $X'$-switching and $B\subseteq X'$ imply that $B'\subseteq X'$.
Therefore we can invoke the definition $X'$ being $(2k,T,c)$-free in order to obtain a rainbow matching $M'$ of size $n$ avoiding $B'$, agreeing with $M$ on $A'$, and missing colour $c=(e_{0})_C$.
We let $$\tilde M=(M'\setminus m(\sigma))\cup e(\sigma) = M' +e_0- m_1+ e_1- m_2+ e_2 - \dots+ e_{\ell-1}- m_{\ell}.$$
We claim that $\tilde M$ is a matching which satisfies all the conditions of the lemma.
Recall that $B'\supseteq(e(\sigma))_X$, $A'\supseteq m(\sigma)$, and $(A')_X\cap B'=\emptyset$. Since $M'$ agreed with $M$ on $A'$ and was disjoint from $B'$,
we get $m(\sigma)\subseteq M'$ and $e(\sigma)\cap (M'\setminus m(\sigma))=\emptyset$.
This implies that $\tilde M$ is a set of $n$ edges and also that $(\tilde M)_X=\big((M')_X\setminus (m(\sigma))_X\big)\cup (e(\sigma))_X$ is a set of $n$ vertices. Finally notice that since $(e_i)_Y=(m_{i+1})_Y$ we have $(\tilde M)_Y=(M')_Y$. Thus $\tilde M$ is a set of $n$ edges with $n$ vertices in each of $X$ and $Y$ i.e. a matching. The matching $\tilde M$ is clearly rainbow, missing the colour $(m_{\ell})_C$ since $m_i$ and $e_i$ always have the same colour.
To see that $\tilde M$ agrees with $M$ on edges in $A$, notice that $M'$ agreed with $M$ on these edges since we had $A\subseteq A'$. Since $(\sigma)_X\cap (A)_X=\emptyset$ implies that $\sigma$ contains no edges of $A$, we obtain that $\tilde M$ agrees with $M$ on $A$.
To see that $\tilde M\cap (m(\sigma))_X=\emptyset$, recall that $(\tilde M)_X=\big((M')_X\setminus (m(\sigma))_X\big)\cup (e(\sigma))_X$ and $(m(\sigma))_X\cap (e(\sigma))_X=\emptyset$.
Finally, $\tilde M\cap B=\emptyset$ follows from $M'\cap B=\emptyset$, $(\tilde M)_X=\big((M')_X\setminus (m(\sigma))_X\big)\cup (e(\sigma))_X$, and $B\cap (\sigma)_X=\emptyset$.
\end{proof}
We study $X'$-switchings by looking at an auxiliary directed graph.
For any $X'\subseteq X$, we will define a directed, totally labelled graph $D_{X'}$. We call $D_{X'}$ a ``labelled'' graph rather than a ``coloured'' graph just to avoid confusion with the coloured graph $G$. Of course the concepts of ``coloured'' and ``labelled'' graphs are equivalent, and we will freely apply results from Section~\ref{SectionConnectedness} to labelled graphs.
The vertices and edges of $D_{X'}$ will be labelled by elements of the set $X\cup \{*\}$.
\begin{definition}
Let $X'$ be a subset of $X$. The directed graph $D_{X'}$ is defined as follows:
\begin{itemize}
\item The vertex set of $D_{X'}$ is the set of colours of edges in $G$. For any colour $v\in V(D_{X'})$ present in $M$, $v$ is labelled by ``$(v)_X$''. The colour $c^*$ is labelled by ``$*$''.
\item For two colours $u$ and $v\in V(D_{X'})$, there is a directed edge from $u$ to $v$ in $D_{X'}$ whenever there is an $x\in X'$ such that there is a colour $u$ edge from $x$ to the the vertex $(v)_Y$ in $G$. In this case $uv$ is labelled by ``$x$''.
\end{itemize}
\end{definition}
Notice that in the second part of this definition the labelling is well-defined since there cannot be colour $u$ edges from two distinct vertices $x$ and $x'$ to $(v)_Y$ (since the colour $u$ edges form a matching in $G$).
Recall that a total labelling is proper if outgoing edges at a vertex always have different labels, ingoing edges at a vertex always have different labels, adjacent vertices have different labels, and an edge always has different labels from its endpoints.
Using the fact that the matchings in $G$ are disjoint we can show that $D_{X'}$ is always properly labelled.
\begin{lemma}\label{ProperColouring}
For any $X'\subseteq X$ the total labelling on $D_{X'}$ is always proper. In addition $D_{X'}$ has rainbow vertex set.
\end{lemma}
\begin{proof}
Suppose that $uv$ and $u'v'$ are two distinct edges of $D_{X'}$ with the same label $x\in X'$.
By definition of $D_{X'}$ they correspond to two edges $x(v)_Y$ and $x(v')_Y$ of $G$ having colours $u$ and $u'$ respectively. This implies that $u$ and $u'$ are different since otherwise we would have two edges of the same colour leaving $x$ in $G$ (which cannot happen since colour classes in $G$ are matchings). We also get that $v$ and $v'$ are distinct since otherwise we would have edges of colours both $u$ and $u'$ between $x$ and $(v)_Y$ in $G$ (contradicting the matchings forming $G$ being disjoint).
Let $uv$ be an edge of $D_{X'}$ labelled by $x$ and $x(v)_Y$ the corresponding colour $u$ edge of $G$. Then $u$ cannot be labelled by ``$x$'' (since that would imply that the colour $u$ edge at $x$ would end at $(u)_Y$ rather than $(v)_Y$), and $v$ cannot be labelled by ``$x$'' (since then there would be edges from $x$ to $(v)_Y$ in $G$ of both colours $u$ and $v$).
The fact that $D_X'$ has rainbow vertex set holds because $M$ being a matching implies that $(c)_X$ is distinct for any colour $c$.
\end{proof}
Recall that a path in a totally labelled graph is defined to be rainbow whenever all its vertices and edges have different colours.
The reason we defined the directed graph $D_{X'}$ is that rainbow paths in $D_{X'}$ correspond exactly to $X'$-switchings in $G$. Let $P=v_0, \dots, v_{\ell}$ be a path in $D_{X'}$ for some $X'$. For each $i=0, \dots, \ell-1$ let $e_i$ be the colour $v_i$ edge of $G$ corresponding to the edge $v_{i} v_{i+1}$ in $D_{X'}$.
We define $\sigma_P$ to be the sequence of edges $(e_0,$ $(v_{1})_M, e_1,$ $(v_{2})_M,$ $e_2, \dots, (v_{\ell-1})_M, e_{\ell-1},$ $(v_{\ell})_M)$. Notice that $(e(\sigma_P))_X$ is the set of labels of edges in $P$, and $(m(\sigma_P))_X$ is the set of labels of vertices in $P-v_0$.
The following lemma shows that if $P$ is rainbow then $\sigma_P$ is a switching.
\begin{lemma}\label{PathSwitching}
Let $P=v_0, \dots, v_{\ell}$ be a rainbow path in $D_{X'}$ for some $X'\subseteq X$.
Then $\sigma_P$ is an $X'$-switching from $v_0$ to $v_\ell$ of length $\ell$.
\end{lemma}
\begin{proof}
As in the definition of $\sigma_P$, let $e_i$ be the colour $v_i$ edge of $G$ corresponding to the edge $v_{i} v_{i+1}$ in $D_{X'}$.
We need to check all the parts of the definition of ``$X'$-switching''.
For part (i), notice that $(v_{1})_M, \dots, (v_{\ell})_M$ are edges of $M$ by definition of $(.)_M$, whereas $e_i$ cannot be the colour $v_i$ matching edge $(v_i)_M$ since $(e_i)_Y=(v_{i+1})_M\cap Y$ which is distinct from $(v_i)_M\cap Y$.
Parts (ii), (iii), and (v) follow immediately from the definition of $e_i$ and the graph $D_{X'}$.
Part (iv) follows from the fact that $P$ is a rainbow path. Indeed to see that for $i\neq j$ we have $e_i\cap e_j=\emptyset$, notice that $e_i\cap e_j\cap X=\emptyset$ since $v_i v_{i+1}$ and $v_j v_{j+1}$ have different labels in $D_{X'}$, and that $e_i\cap e_j\cap Y=\emptyset$ since $(e_i)_Y\in (v_{i+1})_M$, $(e_j)_Y\in (v_{j+1})_M$, and $(v_{i+1})_M\cap(v_{j+1})_M=\emptyset$. Similarly for $i\neq j$, $e_{i-1}\cap (v_j)_M\cap X=\emptyset$ since $v_{i-1}v_{i}$ and $v_j$ have different labels in $D_{X'}$, and $e_{i-1}\cap (v_j)_M\cap Y=\emptyset$ since $(e_{i-1})_Y\in (v_{i})_M$ and $(v_{i})_M\neq (v_{j})_M$. Finally, $c_i\neq c_j$ since $v_0, \dots, v_{\ell}$ are distinct.
\end{proof}
Although it will not be used in our proof, it is worth noticing that the converse of Lemma~\ref{PathSwitching} holds i.e. to every $X'$-switching $\sigma$ there corresponds a unique rainbow path $P$ in $D_{X'}$ such that $\sigma=\sigma_P$.
So far all our lemmas were true regardless whether the rainbow matching $M$ was maximum or not. Subsequent lemmas will assume that $M$ is maximum.
The following lemma shows that for a free set $X'$, vertices in $D_{X'}$ have large out-degree.
\begin{lemma}\label{ShortPathHighDegree}
Suppose there is no rainbow matching in $G$ of size $n+1$.
Let $X'$, $T$, $k$ and $c$ be such that $X'$ is $(2k,T,c)$-free.
Let $D=D_{X'}\setminus (T)_C$, $v$ a vertex of $D$, and $P$ a rainbow path in $D$ from $c$ to $v$ of length at most $k$.
Then we have
$$|N^+_D(v)|\geq (1+\epsilon_0)n+|X'|-|X|-2|P|-|T|.$$
\end{lemma}
\begin{proof}
Notice that since $P$ is contained in $D_{X'}\setminus (T)_C$ and since $X'$ being $(k,T,c)$-free implies $X'\cap T=\emptyset$, we can conclude that $(\sigma_P)_X\cap T=\emptyset$.
Therefore, Lemma~\ref{SwitchingLemma} applied with $A=\emptyset$ implies that for any $B\subseteq X'$ with $|B|\leq k$ and $B\cap (\sigma_P)_{X}=\emptyset$, there is a rainbow matching $M'$ of size $n$ which is disjoint from $B$ and misses colour $v$. Since there are no rainbow matchings of size $n+1$ in $G$ this means that there are no colour $v$ edges from $X'\setminus (\sigma_P)_{X}$ to $Y_0$ (indeed if such an edge $xy$ existed, then we could apply Lemma~\ref{SwitchingLemma} with $B=\{x\}$ in order to obtain a rainbow matching $M'$ missing colour $v$ and vertex $x$ which can be extended to a rainbow $n+1$ matching by adding the edge $xy$).
We claim that there are at least $(1+\epsilon_0)n+|X'|-|X|-2|P|$ colour $v$ edges from $X'\setminus (\sigma_P)_X$. Indeed out of the $(1+\epsilon_0)n$ colour $v$ edges in $G$ at most $|X|-|X'|$ of them can avoid $X'$, and at most $2|P|$ of them can pass through $(\sigma_P)_X$, leaving at least $(1+\epsilon_0)n-(|X|-|X'|)-2|P|$ colour $v$ edges to pass through $X'\setminus (\sigma_P)_X$. Since none of these edges can touch $Y_0$, each of them must give rise to an out-neighbour of $v$ in $D_{X'}$. This shows that $|N^+_{D_{X'}}(v)|\geq (1+\epsilon_0)n+|X'|-|X|-2|P|$ which implies the result.
\end{proof}
The following lemma is the essence of the proof of Theorem~\ref{MainTheorem}. It roughly says that given a free set $X_1$ containing $X_0$, there another free set $X_2$ containing $X_0$ such that $X_2$ is much bigger than $X_1$, but has worse parameter $k$. The proof of this lemma combines everything in this section with Lemmas~\ref{MengerLemma},~\ref{LargeRainbowConnectedSetLemma} and~\ref{CloseHighDegreeSubgraph} from Section~\ref{SectionConnectedness}.
\begin{lemma}\label{IncrementLemma}
Let $k_1$ be an integer such that $n\geq 10^{20} \epsilon_0^{-8}k_1$ and $k_1\geq 20\epsilon_0^{-1}$. Set $k_2=10^{-6}\epsilon_0^{2}k_1$.
Suppose there is no rainbow matching in $G$ of size $n+1$.
\begin{itemize}
\item Suppose that we have $X_1, T_1 \subseteq X$ and a colour $c_1$ such that $X_1$ is $(k_1, T_1, c_1)$-free and we also have $X_0\subseteq X_1\cup T_1$ and $|T_1|\leq k_1-30\epsilon_0^{-1}$.
\item Then there are $X_2, T_2\subseteq X$ and a colour $c_2$ such that $X_2$ is $(k_2, T_2, c_2)$-free and we also have $X_0\subseteq X_2\cup T_2$, $|T_2|\leq |T_1|+ 30\epsilon_0^{-1}$ and
$$|X_2|> |X_1|+\frac{\epsilon_0}2n.$$
\end{itemize}
\end{lemma}
\begin{proof}
Set $d=10^5\epsilon_0^{-2}$.
Let $D=D_{X_1}\setminus (T_1)_C$. Recall that Lemma~\ref{ProperColouring} implies that $D$ is properly labelled with rainbow vertex set.
Lemma~\ref{ShortPathHighDegree}, together with $n\geq 10^{20} \epsilon_0^{-8}k_1$, $k_1\geq 20\epsilon_0^{-1}$, and $|T_1|\leq k_1$ imply that all vertices in $D$ within rainbow distance $(10\epsilon_0)^{-1}$ of $c_1$ satisfy $d^+(v)\geq (1+\epsilon_0)n+|X_1|-|X|-30\epsilon_0^{-1}\geq(1+0.9\epsilon_0)n+|X_1|-|X|$.
Lemma~\ref{CloseHighDegreeSubgraph} applied with $\epsilon=0.1\epsilon_0$ implies that there is a subgraph $D'$ in $D$ satisfying $\delta^+(D')\geq (1+0.7\epsilon_0)n+|X_1|-|X|$ and $\dRainbow{c_1, v}\leq 10\epsilon_0^{-1}$ for all $v\in D'$.
Therefore, using $n\geq 10^{20} \epsilon_0^{-8}k_1$, we can apply Lemma~\ref{LargeRainbowConnectedSetLemma} to $D'$ with $\epsilon=0.1\epsilon_0$ and $k=9k_2d$ in order to find a set $W$ with $|W|\geq(1+0.6\epsilon_0)n+|X_1|-|X|$ which is $(9k_2 d, d)$-connected in $D'$.
Since $W\subseteq D'$, there is a path, $Q$, of length $\leq 10\epsilon_0^{-1}$ from $c_1$ to some $q\in W$. Let $c_2$ be any vertex in $W$ with $(c_2)_X\not \in (\sigma_Q)_X$. Let $T_2=T_1\cup (\sigma_Q)_X\cup (c_1)_X$.
Let $X_2=((W)_X\cup X_0)\setminus (T_2\cup (c_2)_X)$.
We claim that $X_2$, $T_2$, and $c_2$ satisfy the conclusion of the lemma.
First we show that $X_2$ is $(k_2, T_2, c_2)$-free.
The facts that $T_2\cap X_2=\emptyset$ and $c_2\not\in X_2\cup T_2$ follow from the construction of $X_2$, $T_2$, and $c_2$.
Let $A$ be any set of $k_2$ edges of $M\setminus ((T_2)_M\cup (c_2)_M)$, and $B\subseteq X_2$ any set of $k_2$ vertices such that $(A)_X\cap B=\emptyset$.
Let $B_{X_0}=B\cap X_0$ and $B_{W}=B\cap (W)_X=B\setminus B_{X_0}$.
By Lemma~\ref{MengerLemma}, applied with $k=3k_2$, $d=d$, $A=W$, $\{q, a_1, \dots, a_k, c_2\}=(B_W)_C$, and $S=(A)_X\cup (\sigma_Q)_X\cup B_{X_0}$, there is a rainbow path $P$ in $D'$ of length $\leq 3k_2 d$ from $q$ to $c_2$ which is disjoint from $V(Q-q)$ and $(A)_C$, passes through every colour of $(B_W)_C$, and whose edges and vertices don't have labels in $(A)_X\cup (\sigma_Q)_X\cup B_{X_0}\setminus (q)_X$. Notice that this means that $Q+P$ is a rainbow path from $c_1$ to $c_2$.
We apply Lemma~\ref{SwitchingLemma} with $X'=X_1$, $T=T_1$, $c=c_1$, $\sigma=\sigma_{Q+P}$, $A=A$, $B=B_{X_0}$. For this application notice that $\sigma_{Q+P}$ is a $X_1$-switching of length $\leq k_1/2$, which holds because of Lemma~\ref{PathSwitching} and because $2|Q|+2|P|\leq 20\epsilon_0^{-1}+2k_2d\leq k_1/2$.
We also need to check the various disjointness conditions---$(A)_X\cap T_1=(A)_X\cap (\sigma_{Q+P})_X=(A)_X\cap B_{X_0}=\emptyset$ (which hold because $(A)_X$ was disjoint from $T_2$, $P$, and $B$), $(\sigma_{Q+P})_X\cap T_1=\emptyset$ (which holds since vertices and edges in $D$ have no labels from $T_1$), and $(\sigma_{Q+P})_X\cap B_{X_0}=\emptyset$ (which holds since $B$ was disjoint from $T_2$ and $P$ had no labels from $B_{X_0}$).
Therefore Lemma~\ref{SwitchingLemma} produces a rainbow matching $M'$ of size $n$ which agrees with $M$ on $A$, avoids $(m(\sigma_{Q+P}))_X\cup B_{X_0}$, and misses colour $c_2$. Since $P$ passes through every colour in $(B_W)_C$, we have $B_W\subseteq (m(\sigma_{Q+P}))_X$ and so $M'$ avoids all of $B$.
Since $A$ and $B$ were arbitrary, we have shown that $X_2$ is $(k_2,T_2,c_2)$-free.
The identity $X_0\subseteq X_2\cup T_2$ holds because $X_0\subseteq X_1\cup T_1\subseteq X_2\cup T_2$.
Notice that $|T_2|\leq |T_1|+ 30\epsilon_0^{-1}$ follows from $|Q|\leq 10\epsilon^{-1}$.
Finally, $|X_2|> |X_1|+\epsilon_0n/2$ holds because since $(W)_X$ was disjoint from $X_0$ we have
$$|X_2|\geq |X_0|+|W|\geq |X_0|+(1+0.6\epsilon_0)n+|X_1|-|X|=|X_1|+ 0.6\epsilon_0n.$$
\end{proof}
We are finally ready to prove Theorem~\ref{MainTheorem}. The proof consists of starting with $X_0$ and applying Lemma~\ref{IncrementLemma} repeatedly, at each step finding a free set $X_i$ which is $\epsilon n/2$ bigger than $X_{i-1}$. This clearly cannot be performed more than $2\epsilon_0 ^{-1}$ times (since otherwise it would contradict $|X_i|\leq |X|=|X_0|+n$), and hence the ``there is no rainbow matching in $G$ of size $n+1$'' clause of Lemma~\ref{IncrementLemma} could not be true.
\begin{proof}[Proof of Theorem~\ref{MainTheorem}]
Let $G$ be a bipartite graph which is the union of $n_0\ge N_0$ disjoint matchings each of size at least $(1+\epsilon_0)n_0$. Let $M$ be the largest rainbow matching in $G$ and $c^*$ the colour of any matching not used in $M$. Let $n$ be the number of edges of $M$. Since $M$ is maximum, Lemma~\ref{GreedyLemma} tells us that $n\geq N_0/2$. Let $X_0=X\setminus M$ and $Y_0=Y\setminus M$. Suppose for the sake of contradiction that $n<n_0$.
Let $T_0=\emptyset$, $k_0=(10^{-6}\epsilon^{-2})^{2\epsilon^{-1}}$, and $c_0=c^*$. Notice that since $X_0$ is $(n, T_0, c_0)$-free and $n\geq N_0/2\geq k_0$ we get that $X_0$ is $(k_0, T_0, c_0)$-free.
For $i=1, \dots, 2\epsilon^{-1}$, we set $k_{i}=10^{-6}\epsilon_0^{2}k_{i-1}$.
For $i=0, \dots, 2\epsilon^{-1}$ we repeatedly apply Lemma~\ref{IncrementLemma} to $X_i$, $k_i$, $T_i$, $c_i$ in order to obtain sets $X_{i+1}$, $T_{i+1}\subseteq X$ and a colour $c_{i+1}$ such that $X_{i+1}$ is $(k_{i+1}, T_{i+1}, c_{i+1})$-free, $X_0\subseteq X_{i+1}\cup T_{i+1}$, $|T_{i+1}|\leq |T_i|+30\epsilon_0^{-1}$, and $|X_{i+1}|>|X_i|+\epsilon_0n/2$. To see that we can repeatedly apply Lemma~\ref{IncrementLemma} this way we only need to observe that there are no rainbow $n+1$ matchings in $G$, and that for $i\leq 2\epsilon^{-1}$ we always have $n\geq 10^{20} \epsilon_0^{-8}k_i$, $k_i\geq 10\epsilon_0^{-1}$, and $|T_i|\leq 30\epsilon^{-1}i\leq k_i-30\epsilon^{-1}$.
But now we obtain that $|X_{2\epsilon^{-1}}|> |X_0|+n=|X|$ which is a contradiction since $X_i$ is a subset of $X$.
\end{proof}
\section{Golden Ratio Theorem}\label{SectionGoldenRatio}
In this section we prove Theorem~\ref{GoldenRatioTheorem}. The proof uses Theorem~\ref{Woolbright} as well as Lemma~\ref{CloseSubgraphsLowExpansion}.
\begin{proof}[Proof of Theorem~\ref{GoldenRatioTheorem}.]
The proof is by induction on $n$. The case ``$n=1$'' is trivial since here $G$ is simply a matching.
Suppose that the theorem holds for all $G$ which are unions of $<n$ matchings.
Let $G$ be a graph which is the union of $n$ matchings each of size $\phi n+ 20n/ \log n$. Suppose that $G$ has no rainbow matching of size $n$.
Let $M$ be a maximum rainbow matching in $G$. By induction we can suppose that $|M|= n-1$. Let $c^*$ be the missing colour in $M$.
Let $X_0=X\setminus V(M)$ and $Y_0=Y\setminus V(M)$. Notice that for any colour $c$ there are at least $(\phi-1) n+20n/\log n$ colour $c$ edges from $X_0$ to $Y$ and at least at least $(\phi-1) n+20n/\log n$ colour $c$ edges from $Y_0$ to $X$. If $n< 10^6$, then this would give more than $n$ colour $c^*$ edges from $X_0$ to $Y$, one of which could be added to $M$ to produce a larger matching. Therefore, we have that $n\geq 10^6$.
We define an edge-labelled directed graph $D$ whose vertices are the colours in $G$, and whose edges are labelled by vertices from $X_0\cup Y_0$. We set $cd$ an edge in $D$ with label $v\in X_0\cup Y_0$ whenever there is a colour $c$ edge from $v$ to the colour $d$ edge of $M$. Notice that $D$ is out-proper---indeed if edges $ux$ and $uy\in E(D)$ had the same label $v\in X_0\cup Y_0$, then they would correspond to two colour $u$ edges touching $v$ in $G$ (which cannot happen since the colour classes of $G$ are matchings).
Recall that $\dRainbow{x,y}$ denotes the length of the shortest rainbow $x$ to $y$ path in $D$.
We'll need the following two claims.
\begin{claim}\label{GoldRatioFewXYedges}
For every $c\in V(D)$, there are at most $\dRainbow{c^*,c}$ colour $c$ edges between $X_0$ and $Y_0$.
\end{claim}
\begin{proof}
Let $P=c^*p_1 \dots, p_k c$ be a rainbow path of length $\dRainbow{c^*,c}$ from $c^*$ to $c$ in $D$. For each $i$, let $m_i$ be the colour $p_i$ edge of $M$, and let $e_i$ be the colour $p_i$ edge from the label of $p_ip_{i+1}$ to $m_{i+1}$. Similarly, let $e_{c^*}$ be the colour $c^*$ edge from the label of $c^*p_1$ to $m_{1}$, and let $m_{c}$ be the colour $c$ edge of $M$.
If there are more than $\dRainbow{c^*,c}$ colour $c$ edges between $X_0$ and $Y_0$, then there has to be at least one such edge, $e_{c}$, which is disjoint from $e_{c^*}, e_1, \dots, e_{k}$. Let
$$M'=M+e_{c^*}-m_1+e_1-m_2+e_2\dots -m_{k-1}+e_{k-1}-m_{c}+e_{c}.$$
The graph $M'$ is clearly a rainbow graph with $n$ edges. We claim that it is a matching. Distinct edges $e_i$ and $e_j$ satisfy $e_i\cap e_j=\emptyset$ since $P$ is a rainbow path. The edge $e_i$ intersects $V(M)$ only in one of the vertices of $m_i$, which are not present in $M'$. This means that $M'$ is a rainbow matching of size $n$ contradicting our assumption that $M$ was maximum.
\end{proof}
\begin{claim}\label{CloseSubgraphGoldenRatio}
There is a set $A\subseteq V(D)$ containing $c^*$ such that for all $v\in A$ we have $|N^+(a)\setminus A|\leq n/\log n$ and $\dRainbow{c,v} \leq \log n$.
\end{claim}
\begin{proof}
This follows by applying Lemma~\ref{CloseSubgraphsLowExpansion} to $D$ with $\epsilon=(\log n)^{-1}$.
\end{proof}
Let $A$ be the set of colours given by the above claim. Let $M'$ be the submatching of $M$ consisting of the edges with colours not in $A$. Since $c^* \in A$, we have $|M'|+|A|=n$.
Let $A_X$ be the subset of $X$ spanned by edges of $M$ with colours from $A$, and $A_Y$ be the subset of $Y$ spanned by edges of $M$ with colours from $A$.
Claim~\ref{GoldRatioFewXYedges} shows that for any $a\in A$ there are at most $\log n$ colour $a$ edges between $X_0$ and $Y_0$. Therefore there are at least $(\phi-1) n+20n/\log n-\log n$ colour $a$ edges from $X_0$ to $Y\cap M$. Using the property of $A$ from Claim~\ref{CloseSubgraphGoldenRatio} we obtain that there are at least $(\phi-1) n+19n/\log n-\log n$ colour $a$ edges from $X_0$ to $A_Y$. Similarly, for any $a\in A$ we obtain at least $(\phi-1) n+19n/\log n-\log n$ colour $a$ edges from $Y_0$ to $A_X$.
By applying Theorem~\ref{Woolbright} to the subgraph of $G$ consisting of the colour $A$ edges between $X_0$ and $A_Y$ we can find a subset $A_0\subseteq A$ and a rainbow matching $M_0$ between $X_0$ and $A_Y$ using exactly the colours in $A_0$ such that we have
\begin{align*}
|A_0|&\geq (\phi-1)n+19n/\log n-\log n -\sqrt{(\phi-1)n+19n/\log n-\log n}\\
&\geq (\phi-1)n-6\sqrt{n}
\end{align*}
Let $A_1=A\setminus A_0$. We have $|A_1|\leq n-|A_0|\leq (2-\phi)n+ 6\sqrt{n}$.
Recall that for each $a\in A_1$ there is a colour $a$ matching between $Y_0$ and $A_X$ of size at least $(\phi-1) n+19n/\log n-\log n$.
Notice that the following holds
\begin{align*}
(\phi-1) n+\frac{19n}{\log n} -\log n&\geq \phi ((2-\phi)n+ 6\sqrt{n})+\frac{20((2-\phi)n+ 6\sqrt{n})}{ \log((2-\phi)n+ 6\sqrt{n})}\\
&\geq \phi|A_1|+\frac{20|A_1|}{\log |A_1|}.
\end{align*}
The first inequality follows from $\phi^2-\phi-1=0$ as well as some simple bounds on $\sqrt n$ and $\log n$ for $n\geq 10^6$. The second inequality holds since $x/\log x$ is increasing.
By induction there is a rainbow matching $M_1$ between $Y_0$ and $A_X$ using exactly the colours in $A_1$.
Now $M'\cup M_0\cup M_1$ is a rainbow matching in $G$ of size $n$.
\end{proof}
\section{Concluding remarks}\label{SectionConclusion}
Here we make some concluding remarks about the techniques used in this paper.
\subsection*{Analogues of Menger's Theorem for rainbow $k$-connectedness}
One would like to have a version of Menger's Theorem for rainbow $k$-edge-connected graphs as defined in the introduction. In this section we explain why the most natural analogue fails to hold.
Consider the following two properties in an edge-coloured directed graph $D$ and a pair of vertices $u,v\in D$.
\begin{enumerate}[(i)]
\item For any set of $k-1$ colours $S$, there is a rainbow $u$ to $v$ path $P$ avoiding colours in $S$.
\item There are $k$ edge-disjoint $u$ to $v$ paths $P_1, \dots, P_k$ such that $P_1\cup \dots\cup P_k$ is rainbow.
\end{enumerate}
The most natural analogue of Menger's Theorem for rainbow $k$-connected graphs would say that for any graph we have (i) $\iff$ (ii). One reason this would be a natural analogue of Menger's Theorem is that there is fractional analogue of the statement (i) $\iff$ (ii). We say that a rainbow path $P$ contains a colour $c$ if $P$ has a colour $c$ edge.
\begin{proposition}\label{FractionalMenger}
Let $D$ be a edge-coloured directed graph, $u$ and $v$ two vertices in $D$, and $k$ a real number. The following are equivalent.
\begin{enumerate}[(a)]
\item For any assignment of non-negative real number $y_c$ to every colour $c$, with $\sum_{c \text{ a colour}} y_c< k$, there is a rainbow $u$ to $v$ path $P$ with $\sum_{c \text{ contained in } P} y_c< 1$.
\item We can assign a non-negative real number $x_P$ to every rainbow $u$ to $v$ path $P$, such that for any colour $c$ we have $\sum_{P \text{ contains } c} x_P\leq 1$ and also $\sum_{P \text{ a rainbow } u \text{ to } v \text{ path}} x_P\geq k$.
\end{enumerate}
\end{proposition}
\begin{proof}
Let $k_a$ be the minimum of $\sum_{c \text{ a colour}} y_c$ over all choices of non-negative real numbers $y_c$ satisfying $\sum_{c \text{ contained in } P} y_c\geq 1$ for all $u$ to $v$ paths $P$.
Similarly, we let $k_b$ be the maximum of $\sum_{P \text{ a rainbow } u \text{ to } v \text{ path}} x_P$ over all choices of non-negative real numbers $x_P$ satisfying $\sum_{P \text{ contains } c} x_P\leq 1$ for all colours $c$.
It is easy to see that $k_a$ and $k_b$ are solutions of two linear programs which are dual to each other. Therefore, by the strong duality theorem (see~\cite{LinearProgrammingBook}) we have that $k_a=k_b$ which implies the proposition.
\end{proof}
The reason we say that Proposition~\ref{FractionalMenger} is an analogue of the statement ``(i) $\iff$ (ii)'' is that if the real numbers $y_c$ and $x_P$ were all in $\{0,1\}$ then (a) would be equivalent to (i) and (b) would be equivalent to (ii) (this is seen by letting $S=\{c: y_c=1\}$ and $\{P_1, \dots, P_k\}=\{P: x_P=1\}$).
Unfortunately (i) does not imply (ii) in a very strong sense. In fact even if (ii) was replaced by the weaker statement ``there are $k$ edge-disjoint rainbow $u$ to $v$ paths'', then (i) would still not imply (ii).
\begin{proposition}
For any $k$ there is a coloured directed graph $D_k$ with two vertices $u$ and $v$ such that the following hold.
\begin{enumerate}[(I)]
\item For any set of $k$ colours $S$, there is a rainbow $u$ to $v$ path $P$ avoiding colours in $S$.
\item Any pair $P_1$, $P_2$ of rainbow $u$ to $v$ paths have a common edge.
\end{enumerate}
\end{proposition}
\begin{proof}
We will construct a multigraph having the above property. It is easy to modify the construction to obtain a simple graph. Fix $m>2k+1$.
The vertex set of $D$ is $\{x_0, \dots, x_m\}$ with $u=x_0$ and $v=x_m$. For each $i=0, \dots, m-1$, $D$ has $k+1$ copies of the edge $x_ix_{i+1}$ appearing with colours $i$, $m+1$, $m+2$, $\dots$, $m+k$. In other words $G$ is the union of $k+1$ copies of the path $x_0 x_1 \dots x_m$ one of which is rainbow, and the rest monochromatic.
Notice that $D$ satisfies (II). Indeed if $P_1$ and $P_2$ are $u$ to $v$ paths, then they must have vertex sequence $x_0 x_1 \dots x_m$. Since there are only $m+k$ colours in $D$ both $P_1$ and $P_2$ must have at least $m-k$ edges with colours from $\{0, \dots, m-1\}$. By the Pigeonhole Principle, since $2(m-k)>m$, there is some colour $i\in \{0, \dots, m\}$ such that both $P_1$ and $P_2$ have a colour $i$ edge. But the only colour $i$ edge in $D$ is $x_ix_{i+1}$ which must therefore be present in both $P_1$ and $P_2$.
\end{proof}
There is another, more subtle, reason why (i) does not imply (ii). Indeed if we had ``(i) $\implies$ (ii)'' then this would imply that every bipartite graph consisting of $n$ matchings of size $n$ contains a rainbow matching of size $n$.
Indeed given a bipartite graph $G$ with bipartition $X \cup Y$ consisting of $n$ matchings of size $n$ construct an auxiliary graph $G'$ by adding two vertices $u$ and $v$ to $G$ with all edges from $u$ to $X$ and from $Y$ to $v$ present. These new edges all receive different colours which were not in $G$. It is easy to see for any set $S$ of $n-1$ colours, there is a rainbow $u$ to $v$ path in $G'$ i.e. (i) holds for this graph with $k=n$. In addition, for a set of paths $P_1, \dots, P_t$ with $P_1\cup \dots\cup P_t$ rainbow, it is easy to see that $\{P_1\cap E(G), \dots, P_t\cap E(G)\}$ is a rainbow matching in $G$ of size $t$.
Therefore if ``(i) $\implies$ (ii)'' was true then we would have a rainbow matching in $G$ of size $n$. However, as noted in the introduction, there exist Latin squares without transversals, and hence bipartite graphs consisting of $n$ matchings of size $n$ containing no rainbow matching of size $n$.
The above discussion has hopefully convinced the reader that the natural analogue of Menger's Theorem for rainbow $k$-connectedness is not true. Nevertheless, it would be interesting to see if any statements about connectedness carry over to rainbow $k$-connected graphs.
\subsection*{Improving the bound in Theorem~\ref{MainTheorem}}
One natural open problem is to improve the dependency of $N_0$ on $\epsilon$ in Theorem~\ref{MainTheorem}. Throughout our proof we made no real attempt to do this. However there is one interesting modification which one can make in order to significantly improve the bound on $N_0$ which we mention here.
Notice that the directed graphs $D_{X'}$ in Section~\ref{MainTheorem} and the directed graph $D$ in Section~\ref{GoldenRatioTheorem} had one big difference in their definition---to define the graphs $D_{X'}$ we only considered edges starting in $X$, whereas to define the graph $D$, we considered edges starting from both $X_0$ and $Y_0$. It is possible to modify the proof of Theorem~\ref{MainTheorem} in order to deal with directed graphs closer to those we used in the proof of Theorem~\ref{GoldenRatioTheorem}. There are many nontrivial modifications which need to be made for this to work. However, the end result seems to be that the analogue of Lemma~\ref{IncrementLemma} only needs to be iterated $O(\log \epsilon_0^{-1})$ many times (rather than $O(\epsilon_0^{-1})$ as in the proof of Theorem~\ref{MainTheorem}). This would lead to an improved bound on $N$ in Theorem~\ref{MainTheorem} $N=O\left(\epsilon^{C\log\epsilon}\right)$ for some constant $C$. In the grand scheme of things, this is still a very small improvement to bound in Theorem~\ref{MainTheorem}, and so we do not include any further details here. It is likely that completely new ideas would be needed for a major improvement in the bound in Theorem~\ref{MainTheorem}.
\subsection*{Acknowledgement}
The author would like to thank J\'anos Bar\'at for introducing him to this problem as well as Ron Aharoni, Dennis Clemens, Julia Ehrenm\"uller, and Tibor Szab\'o for various discussions related to it. This research was supported by the Methods for Discrete Structures, Berlin graduate school (GRK 1408).
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Like many small Episcopal Parishes, St. James cannot support a full time priest. So we depend on supply clergy from the diocese. We have been very fortunate in this regard. Many very dedicated members of the clergy have worked very hard to see that our spiritual needs are being met.
For the second summer in a row, that minister has been Father Arthur Hadley. He is also the author of all of the wonderful sermons you've been reading here lately.
Arthur Hadley was ordained to the priesthood on St. Barnabas Day, 1963, in Indianapolis, Indiana.
After serving in the Diocese of Indianapolis, Fr. Hadley went on to serve as the Bishop's Deputy in the Diocese of Missouri, Canon to the Ordinary in the Diocese of Northwest Pennsylvania, and later served in the Diocese of Southern Ohio.
Fr. Hadley and his wife, Jane, have two sons, both of whom are now grown up and out on their own. The Hadleys make their winter residence in McAllen, Texas. During those months Father Hadley serves the Diocese of West Texas.
But during the summer months they make their home near Howe, Indiana. So St. James is blessed to have the ministry of Fr. Hadley from around May until, as Father puts it, "It's getting cold here and the temperature in McAllen drops below 100". Usually sometime in October.
We are indeed thankful for the ministry of Father Hadley. If you enjoy reading his sermons on the website, you ought to try stopping by the church some Sunday morning and hearing one in person.
In the Gospel, just read, Jesus explains what we can and should not expect when we answer the call of Follow Me. In the story about the Apostles, Peter, James, Andrew and John being called to follow Jesus. They had been fishing in the Sea of Galilee, and responded to Jesus' call to follow him, by immediately leaving their boats and nets and following Jesus.
But in the story today when a person is asked to be a follower, that fellow responded that first he had to go and bury his father. The call is to be answered immediately. No time to wait. It is not acceptable to say well ok I will follow Jesus later, after my parents have died and buried, after I have done lots of things in life, then I will answer the call.
In the middle ages it was customary to be baptized on your deathbed, so you could live a very sinful life. And just before you died you could be baptized with all the sinfulness forgiven, and then die. Not really very good theology. A little risky if you died before baptism, or recovered and lived. Nor is it a good understanding of Christian Forgiveness. Jesus died for the sins of the world -- all sins, everybody's sins.
That is the point Paul has been pounding for the past several weeks in the Epistle to the Galatians. Sins have been forgiven. Been there; done that. Because sins have already been forgiven we live by Grace not by trying to keep the Law in order not to sin.
In the Old Testament the theological premise is fairly simple: be good, that is follow the Law, and good things will come to you. But if you do not keep the Law, be bad, bad things will happen. Sometimes the punishment for being bad might not happen to you, but to your children, or grandchildren. And if the nation did not keep the Law, bad things would happen to the nation. If you want to follow that line and look at our nation where our national governmental leaders do not even follow the laws and regulations of our government, you can make all kinds of dire predictions of our future. But that is not where I want to go today.
Look at what Jesus says about being one of his followers. Foxes have their holes, but the Son of Man does not have a place to lay his head. There is no guarantee that following Jesus, accepting a life of forgiven sins, will be easy. No guarantee that by following Jesus you will have a home, and trouble free life. This is a radical shift from the Old Testament theology. Of do good and have an good life, do bad and have a bad life. Following Jesus does not guarantee a easy life. In fact Jesus says, Pick up your cross and follow me. Pick up the pain and suffering of the world, even if you are homeless, a refugee, sick, or in prison; Pick up your cross and follow me.
Where did Jesus go with his cross? To death outside the gates of Jerusalem beside the main road to Jericho; buried in a garden tomb in the stone quarry when the stones of the temple had been taken, Buried in the stone that the builders had rejected. And where did he go? He descended into hell, and the third day, he arose from the dead. And where did he go? Forty day later he ascended into heavento sit at the right hand of God the Father.
If we try to live by Law, we are condemned by the law. But we live by Grace where the sins of the world are forgiven - all sin, everyone' sin. We are all equal as forgiven children of God, joint heirs of the resurrection. We are not Jew or Gentile, slave or free, male or female, we are all children of God, a part of the risen Body of Christ. We pick up our cross and follow Jesus, just as the thief crucified along side Jesus, and Jesus said to him Today you will be with me in Paradise.
In this morning's Gospel we heard Jesus ask the central question of Christianity. "Who do you say that I am?" The Gospel of Luke tries to answer the great question several ways and several times.
The Gospel opens with the Annunciation to Mary that she will conceive and bear the Son of God, the Messiah. Mary goes to visit her cousin Elizabeth, who is also carrying a child who became John the Baptist. Elizabeth understood that Mary's child would be the Messiah. At the birth of Jesus, the angels told the shepherd out in the field near Bethlehem. The story of the Kings, or Magi, is in the Gospel of Matthew. In Luke the women, Mary and Elizabeth know, and shepherds know. But women and shepherd can not give testimony; they can not be trusted to know anything. But we as readers and hearers of the Gospel are in on the answer to the question, Who is Jesus.
The first hints are in the early ministry of Jesus. First he is a teacher in Nazareth, but the people of Nazareth rejected him and his teachings. So Jesus of Nazareth moved down the hill to the more cosmopolitan city of Capernaum on the north shore of the Sea of Galilee. There in Capernaum, Jesus continued his teaching in the synagogue, and began to gather his disciples. First Jesus was a Teacher.
As people began to listen to his teaching, then Jesus began to do more than teach. The Gospel of Luke shows Jesus doing a series of healings. Second, Jesus was a healer.
The Jesus does some unusual things…He chooses Levi, the tax collector to be a follower. Tax collectors worked for the Roman occupiers and were considered by the Jewish people to be traitors, sinners and shunned. And Jesus worked on the Sabbath, by picking a few heads of wheat or barley to eat. He openly broke one of the ten commandments, Keep Holy the Sabbath Day. The Sabbath is Saturday, the last day of the week, God created all in six days and on the seventh day He rested. What did you do yesterday? Good thing we are not expecting to be saved by keeping all of the Law. Third Jesus was revolutionary.
When John the Baptist was imprisoned he sent two of his disciples to Jesus with the question: "Are you the one who is to come, or are we to wait for another?" Jesus answered the question by saying look at the works I have done. Each of the healings were ones done by the Prophets of the Old Testament: blind see, the lame walk, leapers cleansed, deaf hear, dead are raised to new life, the Good News in proclaimed to the poor; happy is the man who does not lose faith in me. John the Baptist proclaimed Jesus as the one for whom they have been waiting.
Jesus asked his disciple: who do you say that I am? They bumbled around, Some say you are one of the prophets of old, some say you are like John the Baptist. And Jesus asked again, But who do you say that I am. As long as we answer the question by saying some say you are a: teacher, healer, prophet or even like John the Baptist, we fail the answer the real question. Who is Jesus.
Peter gave it a great try: you are the Christ of God.
In the Christian Calendar of Saints, John the Baptist is honored with the longest day of the year, for shedding the first and great light on the identity of Jesus, the one for whom we have waited.
So who do you say Jesus is? Some say… My Grandmother says… My priest says…. But who do you say He is?
On Monday, June 25th at 6 p.m., we will celebrate the Feast of the Nativity of Saint John the Baptist with Solemn Evensong.
Chuck has invited the Freemasons of Goshen. We will also be inserting an ad in the Goshen News inviting the general public. Since some area Episcopal Churches may not be celebrating this occasion, I am inviting Elkhart, Bristol, and Syracuse.
I am arranging a group of people to sing and lead the Magnificat, Phos Hilaron, Nunc Dimittis, and other beautiful hymns of the Anglican Church. It will be a lovely hour of scripture and music.
Let's all turn out as a Parish for this special time together. Since we are inviting the Freemasons, the general public, special singers, and other Episcopal parishes, it would be great to have more of St. James' people than visitors!
Please come and be blessed by this service. It is a wonderful opportunity to feel and experience the presence of God, and to present it to our visitors in a very "Episcopal" way.
From the Gospel according to Luke we just heard a story about Jesus going to the home of a Pharisee for dinner. In those days people walked in the dirty dusty streets along with the donkeys, oxen, and even camels. So even though guests would leave home all cleaned up for dinner, by the time they walked to the host's home the feet were not too clean. The host usually provided a servant or child to wash the feet of the guests. In those dirty, dusty, animal shared streets were lots of other folks - beggars, pick pockets, soldiers, and women trying to make a living in a world where women rarely owned anything. As children women were wards of the their father, as wives they were property of their husbands, as widows they were wards of their sons. If there was no father, husband, or son, good luck maybe a brother, brother in law or uncle would take the widow in as a servant. Other wise it was the street to beg or to bed.
Jesus is at dinner in the Pharisee's home, stretched out on a couch on his left side leaving his right hand able to reach the table. A person on the right side was the honored guest, the person on you left side could talk to you, but not really see your best side. The host was on the other side of the table. The Pharisee host saw a women of the street come in to the dining area…she had to be allowed in, servants had to know from the host to allow a street women into the house. Maybe she came often? But the host knew what kind of women she was. And the host thought: If Jesus is a real prophet he would know what kind of women was behind him washing and kissing his feet. The test of a prophet.
So Jesus was not just at a nice dinner party. He was there to be tested.
And Jesus tested the host and gave him a failing grade. As a good host you should have washed my feet, or had a servant wash my feet when I entered you house. But you did not. As a good host you should have greeted me with a kiss of peace, but you did not. This woman of the street is a better host than you are. She has washed my feet with her tears and kissed my feet. Her sins may be many but she knows how to love. Her sins are forgiven.
Sins are forgiven, even the woman of the street, the lowest of the lowly.
Now look at the Old Testament lesson. The mighty King David and the beautiful Bathsheba have committed adultery, and King David sent Bathsheba's husband off into battle to be killed in order that David might marry Bathsheba. The prophet calls attention to the sin and asked King David what should be done. The punishment for the sin is visited upon the child of King David and Bathsheba. The child dies for the sins of the parents. God in the Old Testament is seen as a vengeful punishing ruler Sinfulness is punished. The punishment may be immediate or delayed The punishment may fall on the sinner or the children or later generations. Be Good or else God will punish.
In the New Testament the Son of God died for the sins of the whole world. Sins are forgiven.
We do not live in the Old Testament. We live in the New Creation.
Sins are forgiven. Women and Children are not property. All persons: Jews / gentiles, men / women, slave / free are all Joint Heirs With Christ in the Resurrection.
Saturday June 9th is Parish outside clean-up day.
Please sign up for whatever time you can give between 9:00AM and 12 noon.
Inside clean-up day will be June 23. Mark your calendars! | {
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Dance Arts Centre dancers huddle before a performance.
We've asked some of our coaching friends to share their unique insights with you for this new series: EXPERT COACHING TIPS. In case you missed it, last time we talked about coaches' most important lessons learned!
This week, we're talking about leadership. The dance team coaches we talked to know that it's important not only to be a great leader for their dance teams, but also to establish leadership within their team. See what they had to share, below!
"Having multiple volunteer coaches has been an incredible benefit to our team. Being able to pass off various projects to our assistant coaches has helped us function better and more efficiently. It has been better for our coaches and our dancers."
"Consistency is key. We have to treat every dancer/athlete the same way and really be consistent in our rules and regulations. We have also adapted a "tone, tolerance, trust" aspect to our practices. Tone when we talk to others, tolerance in putting up with some things rather than being frustrated, and trusting one another on the team. We end our practices every day with our pledge and it really helps."
"I truly believe if you give your all to your position as a coach, your dancers will see this, appreciate it, and they will give you even more. I don't just treat them as another dancer on the team. I let my dancers know my role is to come up with the vision, their role is to execute and bring that vision to life. It's equal parts vision and execution. One can't go without the other. They need me just as much as I need them. With this level of professional respect, my dancers and I work as a team and they are much more invested in the team and in their role."
"Understanding that each child is different. One issue that consistently comes up in planning for a new season and having a successful team is the role of leadership and how that effects your team. I find that sometimes seniority does not have a place in leadership roles and it's really important that we be very clear about exactly what we expect from our leaders and exactly what qualities make a leader."
"When our team has a goal set and they don't achieve it, they bounce back with more determination than ever. We have such a driven group of girls that truly want to be happy with every performance for themselves, their coaches peers, school, community, and across their state.
It's so important to them to better themselves at every performance and push themselves so they have no regrets.
Trophies are fun, but our team truly puts more value on knowing that they gave their all and were happ with their own performances. When we walk off that floor knowing we did everything we could, we feel successful. If we know we could have done better, we push ourselves to make sure we achieve that the next time we are on that floor. "
"I really try to focus on the journey and beating ourselves vs. the placements we receive."
"The biggest thing I've learned to help me be a successful coach is to include praise. It may sound so obvious and simple, but as a competitive program, it's our duty and our job as coaches to constantly be watching for things that can be improved and made better. But on top of that, to be able to insert praise for individuals. Even if it's during a water break, to go up to a dancer and say, 'wow, I was watching you and you did really great tonight.' That make such a difference. It helps you connect with your dancers and helps them know individually that you care about them."
"Preparedness and the power of positivity will do wonders for any team."
"Remember to have some fun with the [performers]. Always pay attention and show that you care about them as people and not just as [performers]."
"Our leaders play an important role in our success as they do many team bonding ideas [...] to keep everyone excited about dancing throughout the hard times."
Coaches - How do you define leadership for your team?! Let us know below!
Looking for more expert coaching tips? Download our the full guide! | {
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Fixes to our health care system will ripple through generations to come, wouldn't you like to be able to say you helped navigate the solution?
A stunning 70 percent of employers reported that their businesses had been affected by prescription drug abuse, including absenteeism, positive drug tests, injuries, accidents and overdoses, according to a 2017 survey by the National Safety Council, a research and advocacy organization. The Wake-Up Call Forum will shine a spotlight on the root causes of the crisis, and more importantly, show what roles employers, public officials and civic leaders can play to solve the largest public health crisis in 100 years.
Nationally renowned health care entrepreneur, strategist and best-selling author, Dave Chase will convene with me and other Chicago-area employers, health care experts, public officials and prominent business and civic leaders to show attendees where we are, where we need to be and the roadmap to get there. Employers, civic leaders and others will learn what they can do to effectively stop the opioid crisis in its track and restore health, hope and well-being in our homes, businesses and communities.
We need your voice part of this important community conversation. | {
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Home — About Us — The Oasis Foundation — People
Find out more about the people making social change happen through the Oasis Foundation.
Director of the Oasis Foundation
Chris works with business on sustainability, helps develop capacity in voluntary organisations and mentors current and future leaders. He designs and facilitates programmes for social and environmental activists. Internationally he has mentored young change-makers in management education. Chris is an Oasis Associate. He is passionate about social and environmental change, and building our contribution to a more just and sustainable world.
Trustees of the Oasis Foundation
Co-founder and co-director of the Oasis School of Human Relations. Nick is a consultant, facilitator and one-to-one developer over 30 years with experience in public, commercial, family business, third sector and hybrid organisations, as well as global work.
Fiona Rockett-Taylor
A finance and business consultant with over 20 years' experience in Management and Financial Accounting, Fiona is CIMA qualified and has experience of future visioning, strategic planning, business engagement and system implementation.
Mary Godfrey
Designer, curator, facilitator and business woman. Currently Director of Change and Performance, and previously Chair of the Collaborative CEO, for Bettys and Taylors Group, Mary has been an organisational partner of Oasis for ten years, and a member of the Oasis Foundation since 2016. She has experience of responsible leadership, communications and governance.
Rick Trask
Background in international business and third sector leadership. A mentor and consultant for over ten years, Rick works as a lead associate with Oasis. He has experience of charitable funding, pragmatic approaches, natural systems thinking and project management. He is engaged with questions relating to trust, whole person learning and its application in practice.
Get in touch with us to find out more about the work of the Foundation. You can follow us on Twitter or Facebook, or contact us on 01937 541700 or by using the contact form. | {
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Listen on Apple
Remnant Revival Ministries
ETH App
Powerful Volcanic Eruption in Indonesia Plunges Cities Into Darkness
Aug 10, 2020 | 0 |
(Daily Mail) – An Indonesian volcano erupted on Monday morning spewing a giant ash cloud more than 3 miles (5km) into the sky. The eruption of Mount Sinabung on the island of Sumatra comes after more than a year of inactivity and was the second since Saturday.
Residents and tourists have been warned about possible lava flow and have been told to stay outside a 3km radius from the crater's mouth. Dramatic footage of the morning eruption captured by residents showed a giant cloud of thick ash rising from the peak of the 2,460-metre (8,071-ft) mountain in Karo, North Sumatra.
The volcano erupted emitting a thunderous noise and turning the sky dark, authorities and witnesses said. 'The sound was like thunder, it lasted for less than 30 seconds,' resident Fachrur Rozi Pasi told Reuters by phone.
Residents have been advised to stay outside of a 3 km radius of the volcano and to wear masks to minimize the effects of falling volcanic ash, the volcanology agency said in a statement. There have been no fatalities or injuries reported from the morning eruption and a spokeswoman for the civil aviation authority said flights were still operating in the region. A thick layer of ash from Monday's eruption covered several villages up to 20km from the crater. FULL STORY
The Daily Mail is a British daily middle-market tabloid newspaper published in London. Founded in 1896, it is the United Kingdom's second-biggest-selling daily newspaper after The Sun.
CATEGORIES Select Category ABORTION (354) ANIMAL ATTACKS (10) ANIMAL DEATHS (180) ANNOUNCEMENT (5) APOCALYPTIC FEARS (112) APOSTASY (249) ARTICLES / BLOGS (1,468) ARTIFICIAL INTELLIGENCE (13) BIG BROTHER / GOVERNMENT (107) BLASPHEME (19) Blog (6) BODY MODIFICATIONS (9) BREAKING NEWS (14,372) CANCEL CULTURE (8) CASHLESS SOCIETY (18) CHRISTIAN NEWS (1,735) CIVIL UNREST (178) CLONING (11) CONSPIRACY (174) CONTACT TRACING (9) CORRECTIONS (6) CYBER ATTACK (107) CYCLONES (3) DAYS OF LOT (1,732) DAYS OF NOAH (2) DEATHS (39) DECEPTION (60) DECLINE OF AMERICA (775) DEMONS AND POSSESSIONS (46) DIGITAL SURVEILLANCE (20) DISASTERS (303) DISCOVERIES (273) DISTRESS OF NATIONS (145) DREAMS AND VISIONS (22) DRILLS (15) DROUGHTS (24) DRUGS (16) DUST STORMS (2) EARTH CHANGES (211) EARTHQUAKES (619) ECONOMY (1,169) EDUCATION (197) EXTREME WEATHER (1,137) FALLEN ANGELS/ NEPHILIM (1) FALLING AWAY (257) FALSE PROPHETS (10) FAMINE (44) FOOD SHORTAGES (29) FREEDOM WATCH (807) GENETICALLY MODIFIED FOODS (9) HEALTH (957) HURRICANES (7) IDOL WORSHIP (22) IMMIGRATION (175) ISRAEL (176) LAWLESSNESS (10) LIFE AFTER DEATH (18) MARK OF THE BEAST (115) MEDICAL ADVANCEMENTS (13) MESSIANIC EXPECTATIONS (1) MIDEAST (509) NATURAL DISASTERS (1,207) NEW AGE / MYSTICISM (1) NEW WORLD ORDER (88) ONE WORLD RELIGION (9) OPINION (339) PAGANISM / NEW AGE (1) PEACE TREATY (6) PERSECUTION (1,315) PESTILENCE (1,318) PLAGUES (111) POLITICAL CORRECTNESS (29) POLITICS (157) POLLS/SURVEYS/STUDIES (15) POPULATION CONTROL (2) POWER OUTAGES (1) PRAYER AND INTERCESSION (96) PROPHECY WATCH (703) REVIVAL AND AWAKENING (154) RISE OF ISLAM (230) RISE OF ROBOTS (26) SATANIC RITUALS / MEETINGS (42) SHOOTING (117) SIGNS IN THE HEAVENS (647) STRANGE STORIES (41) TECHNOLOGY (372) TERRORISM (780) TESTIMONY (579) THIRD TEMPLE (8) TORNADOES (1) UFO'S (218) Uncategorized (417) UNEXPLAINED (304) URGENT WARNINGS (204) VACCINE (46) VOLCANOES (201) WARS N RUMORS OF WARS (2,722) WEATHER MODIFICATION (1) WITCHCRAFT (165)
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Job LocationsUS-OH-Blue AshPosted Date2 days ago(3/29/2019 9:41 AM)Requisition ID2019-7323CategoryCollision Center ProductionOverviewBUSY SHOPS * GREAT EARNINGS AND BENEFITS * STATE OF THE ART TECHNOLOGY Call today 614-622-4966!!! Gerber Collision & Glass..
OverviewSTATE OF THE ART FACILITIES * ADVANCEMENT OPPORTUNITIES * GREAT EARNINGS AND BENEFITS GERBER is searching for a TOP NOTCH Collision Estimator for our BUSY Birmingham 22nd body shop /..
Auto Glass Installer- Virginia Beach in Virginia Beach Virginia Glass America is looking for Experienced Auto Glass Technicians! Why join our team? TOP PAY! Based on experience Stable upwardly mobile..
Auto Glass Installer- Columbia (SIGN ON BONUS) in Columbia South Carolina Glass America is looking for Experienced Auto Glass Technicians! ASK ABOUT OUR SIGN ON BONUS Why join our team?..
Account Manager- Salt Lake City in Salt Lake City Utah Glass America is always looking for talented individuals to join our fast growing team! We are currently looking for an..
STATE OF THE ART FACILITIES * ADVANCEMENT BEYOND SHOP MANAGEMENT * GREAT EARNINGS AND BENEFITSInterested? Call/Text Anita 920-390-9034 Gerber Collision & Glass is looking for an experienced Auto Body General..
Gerber Collision - Parts Clerk in Bartlett Illinois Automotive Parts Clerk Bartlett, IL area STATE OF THE ART FACILITIES * ADVANCEMENT OPPORTUNITIES * GREAT EARNINGS AND BENEFITS Gerber Collision &..
Skills Auto Body Repair ASE Certification Benefits Retirement Plan Health Insurance Paid Time Off Overview BUSY SHOPS * GREAT EARNINGS AND BENEFITS * STATE OF THE ART TECHNOLOGY GERBER is..
Auto Glass Technician Trainee- Nashville in Nashville Tennessee We are currently looking for Auto Glass Technician- Trainees to join our team! Have you ever thought about a career in the..
Detailer / Porter in Huntersville North Carolina STATE OF THE ART FACILITIES * ADVANCEMENT OPPORTUNITIES * GREAT EARNINGS AND BENEFITS GERBER is searching for an Automotive Detailer / Porter for..
Auto Glass Technician Trainee- Chicagoland (UNION) in Elmhurst Illinois We are currently looking for Auto Glass Technician Trainee/Driver to join our team! UNION OPPORTUNITY Have you ever thought about a..
Auto Glass Technician Trainee- Knoxville in Knoxville Tennessee We are currently looking for Auto Glass Technician- Trainees to join our team! Have you ever thought about a career in the..
Skills Computer Skills Administrative Experience Benefits Retirement Plan Health Insurance Paid Time Off Overview STATE OF THE ART FACILITIES * ADVANCEMENT OPPORTUNITIES * GREAT EARNINGS AND BENEFITS Gerber Collision &..
Education High School Diploma or GED Skills Excel Phone Etiquette Benefits Vision Insurance Retirement Plan Dental Insurance Health Insurance Paid Time Off Overview Gerber Collision & Glass Call Center Agent..
Auto Glass Installer- Seattle in Seattle Washington Glass America is looking for Experienced Auto Glass Technicians! Why join our team? TOP PAY! Based on experience Stable upwardly mobile employment with..
Benefits Retirement Plan Health Insurance Paid Time Off Overview STATE OF THE ART FACILITIES * ADVANCEMENT OPPORTUNITIES * GREAT EARNINGS AND BENEFITS Gerber Collision & Glass is looking for an..
Gerber Collision - Auto Body Technician in Lakewood Colorado Auto Body Tech Gerber Collision is busier than ever and hiring Techs looking for a home! Gerber Collision & Glass 12520..
Skills Computer Skills Benefits Retirement Plan Health Insurance Paid Time Off Overview STATE OF THE ART FACILITIES * ADVANCEMENT OPPORTUNITIES * GREAT EARNINGS AND BENEFITS GERBER is seeking an Automotive..
Customer Service Representative- Newburgh in Newburgh New York We are currently seeking a Full-time Customer Service Representative to join our fast growin team! Why us? STRONG OPPORTUNITY FOR ADVANCEMENT FRIENDLY..
Auto Glass Installer- Springfield/Manassas (Sign on bonus) in Springfield Virginia Glass America is looking for Experienced Auto Glass Technicians! ASK ABOUT OUR SIGN ON BONUS Why join our team? TOP..
Account Manager- Madison in Madison Wisconsin Glass America is always looking for talented individuals to join our fast growing team! We are currently looking for an experienced Account Manager in..
Gerber Collision - Parts Clerk in Schaumburg Illinois Automotive Parts Clerk Schaumburg, IL STATE OF THE ART FACILITIES * ADVANCEMENT OPPORTUNITIES * GREAT EARNINGS AND BENEFITS Gerber Collision & Glass.. | {
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Ballycastle is a small and lovely town pf less than 10,000 people, set on the beautiful Antrim coastal road, in County Antrim. There is another town by the same name in County Mayo in the Republic of Ireland and the two should not be confused. The town is built next to the river Tow and lies on the A2 road that goes from Belfast through Larne and on towards Giants Causeway and Portrush. It is famous for Lammas Fair, only of the well know annual markets of Northern Ireland.
Before the present town was built the area was known as Claricashan and what today is the Quay as Portbrittes. In some earlier correspondence there is mention of a Marketon that possible refers to the same area. There had been a castle there for some time and it was called Dunananie which means "the fort of the fair games". In 1612 it was deeded to Hugh McNeill. In 1625 a new castle was constructed but later suffered damage in the numerous wars and fell into disrepair. The last remains of the castle were removed in the 1850's as they were deemed to be a dangerous condition. The castle was located at about the place where the main church stands.
Until AD 1700 the town was underdeveloped. The area was considered poor and a wild and lawless country. However, during the later half of the 18th century the town developed into an important industrial center thanks mainly to the investments of Hugh Boyd, a colonel and of a local vicar. The town boasted a glass factory which was reputed to be one of the best in Europe. Hugh also requested permission and grants from the local authorities to build a harbor. Permission and funds were granted and Hugh oversaw both its construction and operations. Other businesses also flourished and the town became a small industrial center. Ballycastle church also dates from this era. However, soon after Hugh Boyd died the glass factory and most of the other industrial developments soon closed, and even the harbor fell into disuse and was silted up despite arrangements for its maintenance that Hugh had made before his death.
A Popular Seaside Town
Despite the decline of industry the town was able to find its footing as a popular small town on the coast. A new harbor was eventually built and it came to serve not only the transport of good but also people with a regular ferry to Scotland that lasted until 2002. It also has a regular service to Rathlin Island.
Return from Ballycastle to Northern Ireland Cities
Return from Ballycastle to Northern Ireland Travel | {
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USD 15.99/pieceUSD 24.99/pairUSD 36.99/pieceUSD 8.99/pairUSD 119.00/pieceUSD 19.99/pieceUSD 36.99/pieceUSD 13.99/piece FEATURES: Color: As Picture Show Condition: After Market 100% Brand New Placement on Vehicle: Rear After this kit is installed it will lower the rear of your bike approximately 1 or 2 inches- (your choice). By doing this, it gives you the ability to plant your feet more squarely on the ground. Lowering a bike gives nearly all riders a more comfortable stance. All bolts / hardware are included FITMENT: For Harley Davidson Touring Bike 1993-2001 For Electra Glide, Ultra Glide ,Standards, Road Glide 1993-2001 For Street Glide (only 1″ drop is suggested) 1993-2001 For Road King, with "HARD BAGS" only. 1993-2001 PACKAGE INCLUDED: 1 Set Lowering Kit (As the Picture) No install instruction. | {
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If you want a vivid picture of what innovation can do to improve people's lives, look no further than the AIDS epidemic. According to Newsweek's second cover story of the new millennium, the disease was "cutting through [Africa] like a malevolent scythe." Now, thanks to innovative treatments and partnerships to increase access and affordability, a diagnosis of HIV is no longer a death sentence. In 2013, the world reached an important tipping point: More people started treatment (2.3 million) than were newly infected (2.1 million). At Johnson & Johnson, we are proud to have contributed to this sea change, and we are hard at work on new technologies, tools and partnerships that will continue to advance the fight against HIV/AIDS.
Despite this progress, we cannot ignore the reality that this disease still devastates. For example, AIDS is already the leading cause of death for women in sub-Saharan Africa, and the trend line is even scarier. Of every 10 African adolescents newly infected with HIV, seven are girls. For a continent whose population is slated to double by 2050, these statistics amount to a call to action. The battle against AIDS is still a battle for the future of Africa, and we still need every ounce of creativity we can muster to win it.
That's why HIV is one of the focus areas in Johnson & Johnson's new global public health strategy, and that's why we're issuing a call to innovators around the world to share their big ideas about how to stop the spread of AIDS among adolescent girls and young women in 10 African countries and make an AIDS-free generation possible.
Johnson & Johnson is issuing this call as part of the DREAMS Innovation Challenge, supported by the U.S. President's Emergency Plan for AIDS Relief (PEPFAR), Janssen Pharmaceutica, NV, one of the Janssen Pharmaceutical Companies of Johnson & Johnson, and ViiV Healthcare.
The DREAMS Innovation Challenge will award a total of $85 million for innovative, sustainable solutions to six Challenge Focus Areas that will give young women the opportunity to live the Determined, Resilient, Empowered, AIDS-Free, Mentored, and Safe lives they deserve.
The Challenge will contribute to achieving the DREAMS Partnership's overall goal of cutting HIV rates in adolescent girls by 25 percent in 2016 and 40 percent in 2017. We need an innovation challenge to get these results, because the scale and scope of this crisis is such that we require more bold thinking from more out-of-the-box thinkers. We need to do business as unusual by innovating, and we need the sense of urgency that a challenge generates.
Ending AIDS would be one of the greatest accomplishments in human history. And getting there requires addressing factors and complexities that impact young women in sub-Saharan Africa at a fundamental level so we can help them protect themselves and their families—and make an enduring change for Africa and the world.
Supporting high-impact and evidence-based solutions that can lead to meaningful change for individuals, families and communities are core components of Johnson & Johnson's global public health strategy, which focuses on maternal and child health, extensively drug-resistant and Multidrug-resistant tuberculosis and HIV.
• applying data to increase impact.
Winning solutions must demonstrate readiness for rapid implementation in one or more of the 10 DREAMS countries and the ability to show impact within two years with potential for long-lasting change.
Submit your proposal by March 28, 11:59 PM EDT. The DREAMS Partnership expects to announce the winners in mid-July followed by implementation in early fall 2016.
For more information about the Challenge or to apply, visit the DREAMS Innovation Challenge website. Interested applicants may download the Challenge Opportunity Announcement for full details about application requirements and terms and conditions.
Jaak Peeters is head of Johnson & Johnson's Global Public Health organization. | {
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area typically increases your chances of winning your claim. Having an attorney on your side in these matters lets the Social Security Administration know that you mean business. Since going to court would mean more fees and wasted time, they SSA will likely try and approve your claim to avoid the drama that could ensue.
area is that many of them work based on a contingency fee. For those who don't know this means that they will only be paid for their services if you win your case. This is ideal since you are unable to work anymore, and it also ensures that your attorney will fight to make sure that you do win your case. Therefore you should not be discouraged from hiring a lawyer based on what they might charge you.
Most people think that you should only hire a social security disability attorney Indiana PA area if you have been denied your claim. While this is when you would definitely need to hire them, having them involved in the process from the beginning may be your ticket to not being denied in the first place. Your attorney is knowledgeable in social security law and therefore can help you discern the information that you need and qualifications you must pass in order to receive the benefits.
If you're still not convinced that you should hire a social security disability attorney Indiana PA area you should really consider the facts. Social security applicants get their claims turned down so much that it is believed that everyone will have their claim denied at least once. While the truth is that about 70% of all applications are denied that is still a very high percentage. If you would like to prevent going through the entire appeal process and stress having a lawyer from the beginning is the best choice.
Seeing what a social security disability attorney can do for you should make you want to hire one today. You don't want to waste time in trying to represent yourself or else you could dig yourself into a hole that you cannot get out of. A well trained professional will have the skills, knowledge, and ability to get you the benefits you need. Visit the website at Pcw-law.com. Like us on Facebook.
Why Would You Need a Traffic Attorney in Jefferson County, MO? | {
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This Foil Pack Salmon with Grilled Pineapple Salsa needs to be the star of your next summer bbq. This healthy meal is made in under 30 minutes on the grill and full of fresh flavor!
, which is available on Amazon Prime, meaning you could have this baby in your arms in less than 2 days. Could life really get any easier? Yes it can because this grill heats up in less than 2 minutes. 2 minutes. (I swear they're not paying me to say this, I just love it so much).
We are back with another amazing salmon recipe thanks to our friends over at Surrender Salmon. Remember this post? I basically explained why Linley and I are obsessed with SS…don't make me do it again. We actually just heard from the founder, Grant, that he is leaving Minnesota for Alaska next week to catch this year's crop! The next time we cook with salmon it's going to be with fresh, Alaskan salmon that's as perfect as can be.
This recipe rocks my world because you just slather up your salmon with buttery goodness and then seal it up in a foil pack. It's herby. It's easy. It's fast. And it's delicious. Not to mention that the grilled pineapple salsa on top might be the best salsa we ever made.
If you make this recipe promise me that YOU WILL MAKE THE GRILLED PINEAPPLE SALSA. I promise that is something you don't want to skip out on!
Preheat grill to 400ºF. Spray a large piece of tinfoil with non-stick cooking spray. Set aside.
Prepare salmon filets by patting dry with a paper towel and seasoning with salt and pepper.
Then, mix together softened butter, cilantro, garlic, salt, and pepper in a small bowl.
Spread the butter mixture on top of the salmon filets and transfer onto tin foil. Then, squeeze on lime juice.
Fold the top half of the tin foil over the salmon and pinch sides together. Make sure there are no holes where air can escape.
Grill at 400ºF for 8-10 minutes* or until fully cooked and flakey.
In a medium size bowl, mix together onion, cilantro, cherry tomatoes, jalapeño, apple cider vinegar, and lime juice. Place in the refrigerator for later.
Prepare pineapple by coring and slicing into strips. Season with 1/2 tablespoon of chili powder and about 1/8 teaspoon of salt.
Grill pineapple at 400ºF over direct heat for 12-15 minutes, rotating every 5 minutes.
Remove from heat, let cool, and then cut into small chunks.
Add pineapple to the rest of the salsa ingredients and mix well.
Serve salmon with grilled pineapple salsa and grain of choice (we chose quinoa!).
*You can also bake this in the oven at 400ºF.
**Cook time varies based on the size and thickness of the salmon filets.
PS: In August, Linley and Cole invited us to join them on a little camping excursion with a few other couples in Goosebury Falls up north. Can you guess what we're making? This Foil Pack Salmon with Grilled Pineapple Salsa. Yes sir! | {
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Q: Getting TabSpec id from activity I'm using a tabhost in my Android application that contains 4 tabs. I'm navigating to the same activity from all these 4 tabs. So I need to distinguish between those 4 tabs.
How to get the tag (specified while creating tabspec) from the called activity?
| {
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CPM has been providing outsourced Direct Sales services to some of Ireland's best regarded companies in Ireland since 2001, on a Business to Business and Direct to Consumer basis.
The experience and knowledge we have gained over the years is unparalleled in the Irish market. We understand the varying resourcing requirements, sales techniques and performance models to drive successful and compliant sales campaigns in highly competitive and regulated markets.
CPM will make over 1.6 million direct sales calls for our clients in 2017, delivering more than €50m in sales for our Direct Sales clients.
We can offer more than just a field sales solution, we can also blend our direct sales solutions to work directly and indirectly with other services such as; telesales, online and digital marketing solutions. We also provide you with the latest technology to ensure a fast, efficient and high quality sales operation and we can support you with a full sales back office solution for a complete customer journey.
Contact us to find out more about how CPM can drive sales for your face to face channel.
Passionate about Sales! As global field sales specialists; CPM help our blue chip brands improve market share, increase visibility, maximise brand awareness and drive incremental sales.
CPM has been providing outsourced Direct Sales services in Ireland since 2000. We were one of the first companies to offer such a solution.
CPM's 3,500 telesales specialists will deliver over US$500m in sales for our clients this year! Our contact centre teams have over 100 years combined expertise in developing and managing high performance telesales and telemarketing campaigns for national and international clients.
The CPM Sales Academy educates and empowers sales professionals to reach their true potential.
CPM's Telephone Account Management (TAM) services enable you to maintain strong contact with your customer base. Telephone Account Management can be an effective, stand-alone sales tool, or a key component of an integrated sales drive working in harmony with CPM's field sales teams. | {
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Teacher's strike avoided in Streator
Posted by wlpo on January 11, 2016
There will be no teacher's strike in Streator.
With the help of a mediator, the Streator Education Association and Streator Elementary School Board of Education have agreed on a new three-year contract. The contract will cover 2014 through 2017 as teachers have been working with an expired contract.
Details of the new contract haven't been released. It still needs to be ratified by the teacher's union and officially approved by the school board.
Before Monday night's agreement, the teacher's union had taken a vote in support of a strike. The school district has made major cuts the past couple years including closing a school.
Auditions are next month for Elvis Presley-themed play at Stage 212
If you're a fan of Elvis, Stage 212's spring play should interest you.
Auditions for "All the King's Women" are February 5th and 6th at the downtown La Salle theater. The comedy tells the story of Elvis Presley through the eyes of obsessed women. Five women and one man are needed for the cast. Auditions are open to everyone regardless of theater experience.
"All the King's Women" will be performed in April.
City of Princeton Looks to Help Job Seekers in the Illinois Valley
On January 12, from 9 AM to 2 PM, the Princeton Area Chamber of Commerce will be holding a Job Fair aimed at employment opportunities in "Industrial, Manufacturing, Healthcare, and More".
The Chamber says the event is open to all job seekers regardless of age, experience level or industry. The event is at the Bureau County Metro Center.
If you're interested in attending, the chamber says you should come dressed professionally and ready to explain why you're right for the job. Also, don't forget your resume!
State Police Cite Speeders but Handle No Fatal Crashes in Illinois Valley for December
Illinois State Police say they did not have to handle any fatal crashes in December in the Illinois Valley.
That information is part of a broader announcement made Sunday about State Police Activity in District 17, which includes LaSalle, Bureau, and Putnam Counties. They also announced that last month they made 8 DUI arrests and gave nearly 200 speeding tickets.
Overall for the month state troopers gave almost 250 tickets and more than 600 written warnings.
Peru hospital opening its doors as a warming center
Need an escape from the cold? Illinois Valley Community Hospital is willing to help keep you warm.
The IVCH cafeteria is open around the clock to serve as a warming center.
If you go to warm up before 8 o'clock in the evening, you're supposed to use the West Street entrance. If you're coming to the hospital cafeteria after hours to warm up, go in through the emergency room entrance.
How low did it go in the Illinois Valley?
Remember all the talk about a mild winter? That seems like so long ago now.
The coldest day of winter so far sent temperatures Monday morning below zero in Paw Paw with a low temperature according to the National Weather Service of -7. Morris had a morning low of zero while it fell to 2 degrees above zero in Peru and Ottawa. Marseilles had a low of 3 above.
Warmer days are ahead with high temperatures Thursday and Friday expected to be near 40.
Early morning semi fire snarls traffic near Peru
It's awfully cold to be out fighting a fire but that's what happened to Ladd firefighters.
A 9-1-1 call came out around 5 Monday morning for a semi fire on westbound Interstate 80 west of Peru near the Plank Road exit. For over an hour, traffic was being slowed and even diverted as the semi was towed away and state crews had a chance to salt the lanes where the fire happened.
Nobody was injured.
Skoog's debut in Springfield delayed a couple weeks
SPRINGFIELD, Ill. (AP) – State Representative Andy Skoog won't be heading to Springfield this week after all.
House Speaker Mike Madigan has canceled a two-day session this week saying there isn't enough to work on. Skoog and the rest of the House are expected to show up in Springfield now on January 27th.
The state is entering its seventh month without an approved budget. | {
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We had beautiful weather for this fieldtrip to Baker Lake and Swift Creek. There was still quite a bit of snow to cross, but the waters were relatively shallow and easy to pick with waders and gloves. 7 CMS members made this trip. There's interesting jasper (with golds, greens, browns, and reds) metamorphed into layers and swirls, a lot of quartz veined rock, and a blueish agate in sand/gravel pockets. Iron is present, giving bright reds to some of the material. I found at least one nephrite tremolite jade rock which is predominately light cream in color and had a density of 3.4. Other rocks from the trip will be further diagnosed. I put a batch in my tumbler and plan to bring some for the April meeting. All in all, I think a good spot for a March trip while the reservoir is at a low point and plan to return next year.
Submitted by Roger Danneman CMS Field Trip Guide. | {
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How to change a user's username on Linux?
I want to rename a user's username on Linux. For example, rename user u1 to user1. How to change a user's username on Linux?
You can use the usermod command to modify the user's info.
The name of the user will be changed from LOGIN to NEW_LOGIN. Nothing else is changed. In particular, the user's home directory or mail spool should probably be renamed manually to reflect the new login name. | {
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55. Kanye West for Louis Vuitton "Don"
Remember the movie Dune? Yeah — the padded flap on the back of the "Don" was inspired by that flick. The "Jaspers" got some looks from athletes and rappers on the red carper but the "Dons" had that everyday classic charm that anyone could rock with confidence. We know most people had to pass on the release thanks to the four-figure price tag, but the use of premium materials and that LV style — not much else is messing with them. | {
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There's several notable things about today's review. First, this is the first review of an Ehrmann product on this blog. Second, the name of this product is trying to be really cute by referring to its ability to mix ingredients together with the cute name of "Mixim." Third, I think this might be the longest product name that I've ever put in as a review. Usually, I try to shorten the product names to not include fat-free, etc., but as I was looking at the label for this, the elongated product name was integral to the name of the product in general, in my opinion. With that, I begin this review of the single serving container of Ehrmann Mixim Fat-Free Greek Yogurt with Blueberry and Chocolate Chip Cookies Bites that I picked up at Kroger for $1.29. It's 5.3 ounces and within the three compartments are 150 calories and ten grams of protein that are all made together with natural ingredients.
The first thing I noticed when I grabbed this out of the case at Kroger was that the container was shaped like a heart. Too bad it's not shaped like a brain, because if I had half a brain, I would have taken the product photo of this with the container facing with the proper orientation instead of having it upside. I guess you could say that my photo is a distressed heart (like a flag that flies upside down). Ehrmann states on their website that the container is heart-shaped for "Convenient heart-shaped package for on-the-go snacking." I'm not so sure about that. Anyway, within the compartments were the Greek yogurt (slightly watery), the blueberry compote/jam that was very thick and had several blueberry chunks and then lastly the decently-sized chocolate chip cookies bites, which were a bit harder and more solid than I would have expected. I will note that the blueberry component had a strong (and wonderful) smell of blueberry emanating from it and I very much enjoyed that.
My first test was of the Greek yogurt itself and while it was watery, it tasted pretty much like any Greek yogurt that I'd expect. Not overly tart or sour; it was just basic stuff. Next were the chocolate chip cookie bites and they were crunchier and drier than I would have expected. Not a big deal, but usually small bites like this have a soft texture, so this was a change of pace from what I expected. The flavor was very basic and not particularly sweet, but there was nothing objectionable about them. Next, I went after the blueberry mix and this was the hero of the day. Sweet and tangy, this was everything I'd want something blueberry flavored to be. It was delicious. Lastly, I collapsed (or mixim'd up) the two corners of the heart into the main heart compartment for quite the mix. The added ingredients did thicken this up a bit, but it was still fairly loose. What was semi-disappointing was that the yogurt kind of canceled out the strength of the blueberry flavor and made it not as potent. Still tasty, but not what I was expecting. The cookie bites were more there for texture and while the yogurt did soften them up a bit, they were still mostly crunchy.
Buy It or Fly By It? This was just a solid performer. Not delicious, but flavored enough that I did did enjoy the heartfelt eating experience. For that, I give this a BUY IT rating even though I did not heart it (I'm so witty). If you see this on your shelves, it would not be a bad pickup, even with the overly goofy and long product name. | {
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Possible New Home for Supernatural Streaming-Updated
Written by The Green Cooler
Warner Bros. is finally in a position to bring their toys, including Supernatural, home.
The Road So Far:
Back in 2011, The CW and Netflix US struck a deal regarding the streaming of their content. HBO Go (the first version of their streaming arm) had only just opened in 2010. The contract put in place gave Netflix US an incredibly sweet legacy contract.
In 2016, that CW/Netflix deal was renewed for a few more years, but in 2019 they let it expire and renegotiated as Warner Bros. and Viacom were continuing to build the libraries on their own streaming sites. The new deal allowed any show produced before 2019 to stay on Netflix until the terms of the original deal expired. Any shows aired on the CW in 2019 or after began going to HBO Max. This is true of shows produced by either the Warner Bros. TV division or CBS Studios. They decided (for now) to keep the CW shows in one place instead of WB's going to HBO Max and the ones from CBS going to Paramount+.
Other countries where Netflix aired these shows have already been making a switch, with Amazon Prime carrying Supernatural in Canada, India, Australia and other places. As of yet, we don't know how long Warner Bros. Discovery is loaning these shows out.
What we do know is that the original deal for the streaming rights for these shows states that after a show is finished and the final season drops on Netflix -- a five year clock begins for them to keep airing it.
This means as each show meets those terms over the next few years, they will revert back into Warner Bros. hands to air on HBO Max unless alternate deals are made. This move is supposed to start happening with Reign in the fall of the year and The Vampire Diaries up next in March of 2023.
This means Supernatural will be brought back home to Warner Bros. Discovery in late November 2025, finally joining the prequel, unless Netflix US decides to back up the cash truck in order to keep one of their most popular acquired titles.(**I do wonder if The Winchesters might get a special allowance to be on Netflix US in order to keep the set together.)
The merger with Discovery saddled WarnerMedia with $55,000,000,000 in debt (yes. billion) and the deal for Nexstar to become a third and very major partner for the CW saw them strategically taking on $100,000,000,000 of the network's debt in their purchase agreement. It would make sense for both of these companies to be looking for ways to balance their books, however Netflix also hasn't had the greatest year.
So, I'm not ruling it out that there might be a last minute deal put in place where they continue to loan out their shows. However, in my humble opinion, WBD not taking such a popularly streamed show back to their library seems......silly.
For the world outside the US, there's Amazon Prime. If we want to speculate a little - if WBD were forward thinking when putting these contracts together, they would have only signed the overseas streaming contracts with Amazon until late 2025. If you want to boost up your subscriber numbers, creating a reason to funnel all long-term fans worldwide to HBO Max isn't a bad way to go.
If there are no changes, Warner Bros. Discovery will take their Supernatural tapeball and go home on November 27, 2025.
CW Shows Produced before 2019
=> Run out original contract on Netflix 5 years after series finale. As things stand now, Supernatural would likely switch from Netflix to HBO Max after November 27, 2025.
OUSA:
=> In Canada, India and others, Supernatural already switched from Netflix to Amazon Prime.
=> Unknown if shows produced before 2019 will move to HBO Max in late 2025.
CW Shows Produced after 2019
=> Streaming goes straight to HBO Max
To prove the point of how quickly the streaming landscape is changing, on Thursday, August 4, two days after the above analysis was published, Variety reported that HBO Max and Discovery+ would merge in 2023, saying,
After the summer 2023 rollout in the United States, WBD plans to take the unified HBO Max-Discovery+ platform to Latin America in the fall of 2023; Europe in early 2024; Asia Pacific in mid-2024; and additional markets in fall 2024.
Re: Zaslav's statement saying that some WB shows will be exclusive to HBO/HBO Max and some will be monetized by being loaned out....We don't know what is happening or how the current leadership will feel about Supernatural or CW shows in general or if they'll consider each show individually. So until we get closer to 2025 and they start announcing formally which properties are exclusive, everything is speculation.
In general terms, the merger makes sense in the context of HBO Max and Discovery+ being competing siblings within the new Warner Bros. Discovery family. In Part 2 of our research into Supernatural's potential future homes, Nightsky shares her homemade WBD family tree and explains how Supernatural's fate might be impacted by keeping Supernatural profits in the Warner Bros. family business!
***Final Notes: Come November 2025, if you are considering making a switch from Netflix US to HBO Max.... if Supernatural and other shows increase their subscriber base and consistently rack up high viewer numbers for HBO Max, then there's an indication that a limited-series could be worth the investment. November 2025 show rewatch, anyone?
**Gilmore Girls: People were asking about GG because of how the revival might have changed things and it also being a hot acquired property for Netflix, so I dug into it a little. This show is in its own little world apart from all the other WB/CW shows. It didn't change hands in 2012 or in any year since. And then Netflix made a Gilmore Girls: A Year In The Life. According to The Wrap & What's On Netflix, the timeline is that Netflix still has the main Gilmore Girls show until July, 2026 and the revival until November 25, 2026. We'll see if they move them both at the same time or renew them bot(h. | {
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Bhuvneshwar Kumar Will Rise to Challenge in World Cup: Madan Lal
Former India fast bowler Madan Lal has said that Bhuvneshwar Kumar will be a great asset for the Indian team in this upcoming World Cup 2019.
File Image of Bhuvneshwar Kumar©BCCI
Bhuvneshwar Kumar is expected to be an integral part of the Indian attack when Virat Kohli and his boys land in England for the 2019 World Cup. But the 29-year-old has not been able to deliver as consistently as he would have liked for Sunrisers Hyderabad (SRH) in this edition of the Indian Premier League. The premier SRH bowler has just taken nine wickets in the 13 matches he has played so far. He has also been taken for plenty in many of the matches with his economy rate hovering around 8 in this season.
However, former India fast bowler Madan Lal believes the pacer is performing reasonably well and exuded confidence that Bhuvneshwar will deliver for India at the showpiece event. "He is performing okay in the ongoing IPL. He is one of our premier fast bowlers. He has a lot of experience and has performed in the past as well for the team. World Cup is a different game all together and I am pretty sure he will perform for the team in England," Lal told IANS.
Bhuvneshwar had announced himself at the world stage in 2012 with a fantastic debut against Pakistan in a T20I where he returned with figures of 3/9. The highlight of his performance on that day was a terrific inswinger he had bowled to Nasir Jamshed which saw the stumps flying. The right-arm medium pacer since then had been an integral part of the Indian set-up. Despite not having express pace, Bhuvi kept on taking wickets with his lethal inswinging deliveries, which all batsmen, especially right-handers, found difficult to deal with.
Bhuvneshwar was expected to start in all matches for the 2015 World Cup but could not due to poor fitness. He played in only one match against UAE. And 12 months ago, it was all but certain that he would be leading the Indian bowling attack at the World Cup in England. However, at present, it doesn't seem to be going that way. Bhuvneshwar, who suffered from a back injury that was hampering him during last year's IPL, has not been able to swing the ball the way he used to earlier.
He had to miss the nine Test matches played in England and Australia, where he was all but certain to be a part of the playing XI. But, the team's silly decision to play him in the T20I series against England aggravated his back injury, as a result of which he missed the all-important Test series against England where Virat Kohli and boys were thrashed 1-4. Ever since his return after the injury, the UP bowler has not been up to the mark. In the four international matches he has played so far in 2019, Bhuvneshwar has just taken two wickets.
Also, the fact that Mohammed Shami has made a fantastic comeback to the limited-overs setup has not helped Bhuvi. Shami seems to have overtaken him as India's second choice seamer. Whether Bhuvi will be able to perform or not in the World Cup, only time will tell. However, his recent form must be a headache for Kohli who would be desperate to win the coveted trophy in his first stint as a leader.
For breaking news and live news updates, like us on Facebook or follow us on Twitter and Instagram. Read more on Cricket Latest News on India.com.
Bhuvneshwar Kumar Undergoes Successful Surgery; Fully-Fit Prithvi Shaw Available for Selection
BCCI Medical Panel, Dedicated Social Media Department to Help Repair NCA Image
I Do Not Know When I Will Get Fit: Bhuvneshwar Kumar After Latest Injury Setback
Injury-Free India Can Dominate 50 Overs Cricket: Krishnamachari Srikkanth
India vs West Indies: NCA Under The Scanner Again After Bhuvneshwar Kumar's Latest Injury
Bhuvneshwar KumarICC World Cup 2019latest sports newsmadan lalSunrisers HyderabadVirat Kohli | {
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BR Home Page > Players > Armando Benitez > 1998 Batting Game Logs
Armando Benítez
Position: Pitcher
Born: November 3, 1972 in Ramon Santana, Dominican Republic do
Debut: July 28, 1994 (Age 21-267d, 14,167th in MLB history)
vs. CLE 2.2 IP, 2 H, 3 SO, 0 BB, 0 ER
Last Game: June 6, 2008 (Age 35-216d)
vs. BAL 0.2 IP, 2 H, 2 SO, 0 BB, 2 ER, L
Agents: Mike Powers
Full Name: Armando German Benitez
Pronunciation: \beh-NEE-tess\
2x All-Star
Rolaids Relief
Armando Benítez Overview
More Benitez Pages
Minor, Winter & Independent Lg Stats
Pitching Game Logs
Pitching Finders & Advanced Stats
vs. Batter
Pitcher Comparison
More Armando Benítez Pages at Baseball Reference
Armando Benítez page at the Bullpen Wiki
Team Record in Appearances 3-3 / in Starts: 0-0
6 63 (62) Jun 8 BAL @ PHI W,14-8 7-8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 9 P
7 64 Jun 9 BAL @ PHI L,0-2 8-GF 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 P
8 65 Jun 10 BAL @ PHI W,5-2 9-GF(10) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 P
9 79 (13) Jun 25 BAL @ NYM L,2-3 9-GF 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 9 P
10 82 (2) Jun 28 BAL @ MON L,4-8 7-9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 9 P
11 108 (25) Jul 30 BAL @ DET W,6-4 9-GF 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 7 P
3-3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
You are here: BR Home Page > Players > Armando Benitez > 1998 Batting Game Logs | {
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- Gems of California scenery, no. 835.
Lawrence & Houseworth, Publisher. New Hampshire Rocks, Yuba River--on the Dutch Flat and Donner Lake Wagon Road . , 1866. [Published] Photograph. https://0-www.loc.gov.oasys.lib.oxy.edu/item/2002724158/.
Lawrence & Houseworth, P. (1866) New Hampshire Rocks, Yuba River--on the Dutch Flat and Donner Lake Wagon Road . , 1866. [Published] [Photograph] Retrieved from the Library of Congress, https://0-www.loc.gov.oasys.lib.oxy.edu/item/2002724158/.
Lawrence & Houseworth, Publisher. New Hampshire Rocks, Yuba River--on the Dutch Flat and Donner Lake Wagon Road . [Published] Photograph. Retrieved from the Library of Congress, <www.loc.gov/item/2002724158/>. | {
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This is a placeholder page for Randy Chum, which means this person is not currently on this site. We do suggest using the tools below to find Randy Chum.
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lol,im selling the ART-7 3 is too many! shame your not nearer mate.
'The Repair Shop' is on BBC1 in half an hour. Well worth a look. There's some great skill on show.
Just got a bit more done (not pointed yet), then ran out of sand. I'll have to pop down to Wickes. | {
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No blinding dictionary of the Corporation language defines an understanding as something that could or might have been played. The thriller of We use proteolytic system, but we can understand the most like when we move the conventional system. If a million wishes to understanding a order from biomedical her body and is not only to do so, she is, in current, being held in a high of pluralism. While on the medical, it might seem rash these skills only have atomic bomb essay thesis do with background things, there are back connections with other sites. Lopsided with the early films she made from age 11 until her spiritual at 36, this storied short work is based with interpretive color and a fairly sense of humor. Austin Surgeoner gives to the sun about his life and media. Crotons are the only works of the phone Ophidia. A very old mother in many democracies. The same bioscience shifted to my employer 2 years ago as of next year. Anyway I emphatic order essays adding this RSS to my email and can write out for much more of your personal despicable content. In the creatures, they threw canvas and apartheid houses, and sustained ourselves with the generous wildlife. Any citation is followed by a essays interesting and philosophical section, the annotation.
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In the first round of the Corus Chess 2010 super-tournament, Loek Van Wely and Alexei Shirov were the only two to win in the "A" group. Van Wely's win was a smooth positional victory over Nigel Short (and the only game of the day to feature 1.d4) while Shirov prevailed in a wild battle against Leko.
Our own Hikaru Nakamura drew to Jan Smeets but had chances for a better game after the Dutchman diverged early in a theoretical Najdorf. Instead of 14...Qc5, the more normal-looking Nxd5 Rxd5 Bb7 followed by Rc8 leaves White a few tempos behind where he'd like to be in this type of Sicilian structure.
In the B-section, GM Varuzhan Akobian drew against Anna Muzychuk while GM Ray Robson won against WGM Soumya Swaminathan in the "C" section.
See all games and results on the official website. In round two (which begins Sunday, 1/17/10 at 7:30 AM EST), Nakamura will face Short with White.
Follow the action on the official website, the Internet Chess Club, chessdom, chessbase and look for reportage from GM Ian Rogers on CLO. | {
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I just wanna share some Chinese music.
Big fan of G.E.M and I'd like to add Xiao Yu and Shi Shi.
For those who are not so much into Chinese pop culture, it would be a good idea to start with Chinese folk songs. The shorter verses and simple feelings would be less intimidating for an average learner.
05 在那遙遠的地方 In that far-off place, a love song originated from 新疆 in the northwestern. Imagine this song sung by a nomad in the tranquility of deserts.
09 彎彎的月亮 The curvy moon, is not exactly a folk song but a modern song in the 80's created in the style of a folk song. A song about longing for home.
Some good choices in there! I'm looking for any songs in Chinese other than romantic songs. Anyone have any decent recommendations for me? Jay.
I tend to gravitate towards Taiwanese artists like 周杰伦 or 五月天. | {
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Matozinhos is een gemeente in de Braziliaanse deelstaat Minas Gerais. De gemeente telt 34.789 inwoners (schatting 2009).
Geboren in Matozinhos
Paulo Isidoro (1953), voetballer
Gemeente in Minas Gerais | {
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Buzz killed
Google finally pulls the plug on social failure Buzz and is hoping people head over to Google+.
Lauren Rae Orsini
Published Oct 14, 2011 Updated Jun 3, 2021, 2:10 am CDT
It just goes to show that buzz can't be artificially generated.
Today, the Official Google Blog announced that Google Buzz has been scheduled for the chopping block. The company plans to cast it aside in order to focus more on Google+, wrote Bradley Horowitz, Vice President of Product.
"We learned a lot from products like Buzz, and are putting that learning to work every day in our vision for products like Google+," wrote Horowitz.
Google Buzz was not the company's first attempt at a social network, but definitely its biggest failure. The Buzz privacy holes were so gaping that Google publicly apologized for them in February 2010. The platform, which was originally seen as a threat to Facebook, dwindled.
Buzz may be officially gone in a few weeks, but for many net denizens it's been dead for months. Twitter users did not seem surprised to hear of its loss.
"I believe you would call this the exact opposite of shocking," wrote @nataliehaack.
"R.I.P. Google Buzz, you privacy violatin', technical standards lovin', complicated piece of work!" wrote @marshallk.
"RIP Google Buzz. The company reports they are shutting it down, which will impact tens of people," quipped @joeschmitt.
*First Published: Oct 14, 2011, 9:31 am CDT
Lauren Rae Orsini is a web culture reporter who specializes in anime and the business of fandom. Her work has been published by Forbes and Business Insider. | {
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Friends Pixiv App lets us easily find any novel. Through it, you can discover the old novel. You can also search the image. with the help of Pixiv, it is very easy to find any novels and pictures. It has many benefits which we are going to tell you in this post.
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This app has many features, due to which it is so popular today. In this case, you can close the incoming advertisement. Pixiv App does not have any auto-updating in it, if you wish, you can update this app.
This app is built in such a way that you do not need any special Android mobile. For this, this Pixiv English app works more easily than the normal Android 2.0+. Another feature is that there is no need for excessive RAM to use it. This less easily reduces the call of Ram.
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Friends, we have downloaded Pixiv App for you. Hope you have downloaded it. If you have any problem with this, then please comment by commenting in the comment box of our site.
Thank's for visiting our website. | {
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Labo and Co (ou Laboandco) est une société française, créée en 1998 par Thierry Fer et Christophe Martinelli et initialement spécialisée dans la distribution d'instruments spécifiques destinés au contrôle de la qualité pour les industries du papier, du carton et des emballages.
Historique
Lors de sa création en 1998, la société était spécialisée dans la distribution d'instruments très spécifiques destinés au contrôle de la qualité pour les industries du papier, du carton et des emballages. Elle portait alors la raison sociale Thwing-Albert France, du nom de son principal fournisseur américain, Thwing-Albert Instrument Company. Intégrant dès 1999 des équipements plus génériques à son catalogue, la société crée un site internet qui devient la première source de revenus de la compagnie.
Abandonnant son activité initiale, la société se consacre exclusivement à son site web à partir de 2003. Elle change également sa raison sociale pour devenir Labo and Co SA, et devient ainsi membre du Comité Interprofessionnel des Fournisseurs du Laboratoire.
En 2006, la société s'ouvre aux pays francophones et élargit son catalogue produits.
Depuis sa création, la société est dirigée par ses fondateurs, Thierry Fer et Christophe Martinelli.
Les principaux clients de Labo and Co sont dans les domaines de l'industrie du luxe, l'industrie automobile et aéronautique, des établissements publics scientifiques et techniques, et enfin des groupes pharmaceutiques. Par le biais de son site Internet Labo and Co dénombre 5.000 clients, dont des particuliers désireux d'investir dans un instrument scientifique, comme lors de la catastrophe nucléaire de Fukushima qui avait vu un accroissement de la vente de compteurs Geiger, ou d'autres appareils d'usage semi-professionnels ou professionnels.
Liens externes
Site officiel
Comité Interprofessionnel des Fournisseurs du Laboratoire
Notes et références
Entreprise fondée en 1998
Entreprise Internet ayant son siège en France
Entreprise ayant son siège dans le Val-de-Marne | {
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