Luckily, the solution is easy to implement and has the additional side-effect of making your code even more explicit.
A data register, from which we can drive the hardware pin high or low.
To help us out some kind soul has written a Register class that abstracts memory-mapped hardware access.
The implementation details of the Register class are unimportant for this article; the thing to note is that the Register class supports copying and moving explicitly (with the caveat that this is a contrivance for the purposes of this article).
We can now construct a GPIO class that encapsulates the three hardware registers we have to control and provide an abstract interface for clients.
Once again, implementation details for the GPIO class are unimportant.
Notice, however, that we haven't implemented any copy or move semantics. This is the Rule of Zero in action – we don't have to, since the Register class provides its own copy and move semantics.
I'm deliberately using push_back() (and, yes, deliberately not using emplacement) here to force the copy-construction of the GPIO objects into the vector.
Even though we haven't written a copy constructor for the GPIO class the compiler will automatically provide one for us that performs a (shallow) copy of the members of GPIO. This will invoke the copy constructor on the (composite) Register objects.
As vectors now support move semantics as well, we could 'improve' our design by moving the GPIO objects into the vector.
As before, since we haven't provided a move constructor for the GPIO class the compiler generates one for us, automatically moving the members of the class. This will invoke the move-constructor on the Register objects.
After a design review (you do those, right?) it is decided that our GPIO class should leave the hardware in a 'safe' state – for example, with any enabled GPIO functions turned off, or all pins set to inputs, etc. Quite reasonably, the destructor of the GPIO class is the best place to do this.
Revisiting our vector of GPIO objects, adding them by copy works exactly as before.
All the move-constructor calls have gone, replaced by copying!
OK, the code still works and it might not be a critical overhead in this particular case, but in another situation this could seriously impact the performance of your code.
The answer, in fact, lies with your GPIO class – or, more specifically, the rules the compiler uses when generating move and copy operations for your class.
Under normal circumstances the compiler will, indeed, generate a default move-constructor and move-assignment operator. However, as we've now added a destructor to the GPIO class the rules change.
Usually, the only reason to write a destructor is to perform some level of clean-up when an object goes out of scope. Of course, that's just what we've done in this case. The compiler, though, sees it as "I am doing resource management". If you've written a destructor the compiler assumes that the default 'shallow-move' semantics it can provide are no longer adequate. Therefore, it does not generate a move-constructor or move-assignment.
Why does my code still compile?
std::move doesn't actually do any moving. It simply casts the supplied object to an r-value reference. This then binds against the r-value overload of the vector's push_back() method. The r-value overload of push_back() attempts to move-construct the supplied object into the vector, thus calling the object's move-constructor.
If your class doesn't have a move-constructor the compiler searches for the next most-compatible constructor which, in this case, is the copy-constructor. Why? Because a reference-to-const can always be bound to an r-value object; just as it has always done in C++.
Thus, attempting to move an object that doesn't support move semantics always results in copying; and backward compatibility with pre-C++11 code is maintained.
Our GPIO class does need its destructor but it doesn't need deep-copy semantics. Does this mean we have to implement our own move-constructor and move-assignment and do the work the compiler used to do?
The =default syntax tells the compiler to generate the default implementation for these functions. You don't have to provide any body for them.
Hang on, wasn't this code just working? What's going on?
In order to use the default copy-constructor and copy-assignment operator you must explicitly declare them, as with the move-constructor and move-assignment.
The compiler will provide default implementations for copy-constructor and copy-assignment as required; you may provide your own implementations of either, letting the compiler provide the other.
If you provide a destructor the compiler will no longer generate default implementations for the move-constructor and move-assignment operators.
If you provide an implementation for either a move-constructor or move-assignment operator the compiler will no longer generate an implementation for the other.
If you provide an implementation for either a move-constructor or move-assignment operator the compiler will no longer generate an implementation for the copy-constructor or copy-assignment.
Where possible, adhere to the Rule of Zero – use resource-management classes rather than performing your own; and don't write destructor, copy/move operations.
If you are going to deviate from the Rule of Zero then be explicit about your entire move/copy policy.
This entry was posted in C/C++ Programming and tagged C++0x, C++11, C++1y, copy assignment, copy constructor, copy policy, copy semantics, Modern C++, move assignment, move constructor, move policy, move semantics. Bookmark the permalink.
Nice article, particularly the helpful table.
One thing, could you please rename Register to RegisterRef? It isn't a register, just a reference to one - ie the copy constructor doesn't make _another_ register, it makes another reference to the same register. And that's semantically important.
Overall a great article. However, it is nonsensical to say that "after moving, input and output are x-value objects"; xvalue is a value category, a static (compile-time) property of expressions. Rather, input and output become moved-from, a dynamic (runtime) property of objects. As you observe later in the article, simply calling std::move on an object does not cause it to become moved-from; the resulting xvalue expression must be passed or forwarded to a move-constructor or move-assignment operator.
I was being lazy in my writing. I've updated the article to to reflect your comments.
In the summary section you advocate being explicit about copy/move all the time, but the rule of zero is to rely on the compiler defaults as much as possible, by factoring out copy/move into classes dedicated to that function. You may argue for always being explicit, but that's not the rule of zero.
Nevertheless, you point out an interesting case where following the rule of zero is difficult, because you need a destructor but can't easily factor that into its own class, so I enjoyed the article from that perspective.
1. should explicitly declare copy/move constructor/assignment operator as default or delete.
2. should NEVER use a raw pointer to manage a resource.
combining the rule of big four and the rule of zero.
@Yindong: What I'm saying is: that has nothing to do with the rule of zero, other than being counter to it.
Rule of zero: Declare zero special members as often as possible.
Always declare move/copy is the opposite. It's more like The Rule of Always 4 or 5.


text:
Premiere Pro 15.0 crashing when opening larger pro...
Premiere Pro 15.0 crashing when opening larger projects
KazuTa
/t5/premiere-pro-discussions/premiere-pro-15-0-crashing-when-opening-larger-projects/td-p/13502275 Jan 18, 2023 Jan 18, 2023
I'm running Premiere Pro 15.0 on a MacBook Pro (Early 2015), OS 10.14.6.
I will be getting a new MacBook soon with a new OS, etc., but for a few more weeks, I need to keep using my current system.
For the past several weeks, when I try to open one of my larger projects, Premiere tries to load all the media, then crashes when there are 3 media files left. I've tried using a Production and breaking the project into three smaller projects, but the crash still occurs.
The only workaround I've found is to create a new project, then import the previous project into it. That seems to work. But it's an onerous workaround.
I tried to do a clean reinstall, but when I uninstalled v15 (I had a later one than 15.0), I wasn't able to find an installer for v15. Adobe sent me an installer but they only have v15.0.
Anyone have any ideas?
Have something to add?

redpajama_set_name: RedPajamaCommonCrawl


text:
Shift/Perspective
Meditative Nature
primaryPRISM
Exhibit Catalogs
Gallery Tempo is taking a break right now. Please be in touch if you are interested in collaborating on future projects!
Raw Emotion
Vignettes of Motherhood
Raw Emotion celebrates the power of community among artist mothers working tomake space in their own lives—and in the wider world—for their art. Each work featured tellsits own powerful story, but together, this exhibit proclaims, we are greater than the sum of ourparts.
This exhibition is part of the Taking. Up. Space. Initiative facilitated by the Artist/Mother and Thrive Networks. Taking Up Space or T.U.S. is a Grassroots Female Focused Community Global Art Exhibition empowering artists to take up more space and exhibit their art in the month of May, 2022. There are over 80 exhibitions taking place this year and Raw Emotion is one of them.
Gathering Studios, in Albuquerque, NM, is pleased to announce a group exhibition highlighting artists from two seemingly disparate cultures: skateboarders and mothers. While these two groups may seem like polar opposites, both have been historically marginalized, and seek authenticity, individuality, and freedom through creativity.This exhibition aims to celebrate the voices of these two vibrant cultures while bringing them into dialogue with one another.
Counterpoints is part of the TUS initiative
What is Gallery Tempo?
Gallery Tempo is a Boston based art gallery, founded in 2019 by Alison Judd. Providing visibility to emerging, mid career and established contemporary artists, Gallery Tempo pops up in vacant spaces and transforms them into short-term gallery spaces, bringing great art into neighborhoods.
Sign up for the newsletter to find out about our upcoming shows.
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Curry Chicken was much better than I was expecting, I definitely will order it again.
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text:
Hull & East Yorkshire News
Paula Williamson reveals truth about marriage to Charles Bronson after being dumped by criminal
Paula Williamson, who married the notorious criminal behind bars, says she wanted to take her own life
Geraldine McKelvie
Sign up to the Hull Live newsletter for daily updates and breaking news
The heartbroken wife of Britain's most notorious prisoner has told the Mirror how the bizarre relationship drove her to the brink of suicide.
Charles Bronson wants to divorce former soap actress Paula Williamson, 37, after pictures of a man burying his head in her chest during a wild night went viral.
Paula has now spoken out about the break up and said: "This relationship has driven me to the brink and I have been so low that I have even thought about taking my own life.
"No relationship should make you feel that way.
"It is time for people to know the truth of what happened between us. I have been used and controlled from behind bars as he has poked fun at my weight and told me I am an embarrassment.
Photos of Paula clubbing upset her prisoner husband (Image: Roland Leon/Sunday Mirror)
"I sat across from him in prison a couple of months ago when I hit rock bottom over the way he was squandering his chances of parole and I showed him scars on my arms from where I had tried to hurt myself.
"I feel like he has chewed me up and spat me out just like he did with the last woman who fell for him."
Who is Charles Bronson? The life and crimes of Britain's most notorious prisoner
The wedding of Paula and Bronson, now known as Charles Salvador, was at high security HMP Wakefield, last November, in a ceremony where only four friends attended.
Her horrified parents boycotted the nuptials – but Paula insisted the marriage would last for ever, and vowed to secure freedom for her partner who has spent nearly three decades behind bars.
Paula says she has been driven to the brink of suicide (Image: Julian Hamilton/Sunday Mirror)
Bronson was transferred to HMP Frankland in Co Durham just weeks following the ceremony after an argument with the deputy governor over the wedding pictures.
Britain's most notorious criminals behind bars in Hull and East Yorkshire
Paula, of Stoke-on-Trent, has been told by Bronson's lawyers that he is preparing to file for divorce over the images of her with a man in Tenerife.
Paula previously said she planned to have a baby with Charles (Image: SWNS.com)
The snaps also showed Paula kiss another reveller while she partied on the holiday island.
The bride says she was so distraught by the episode that she started to self-harm.
"If anyone should be filing for divorce it's me but I love him too much," said Paula.
Married Hull mum shares her journey into the world of bondage and domination
"Love and marriage are supposed to make you happy. We haven't even been married a year and I have reached points where I have been so low I didn't want to carry on.
"But I did carry on because I loved Charlie and I wanted to help him get his freedom back. But now I don't even know whether that was what he wanted.
"I can never say I regretted marrying him because I still feel that love. I defended that man to the hilt. I told people he had changed – that he wasn't the evil person everyone thinks he is. He has repaid me by breaking my heart."
Paula says she has been told Charles wants a divorce (Image: Adam Gerrard/Daily Mirror)
Paula, an actress who had roles in Coronation Street and Emmerdale, quit acting so she could concentrate on campaigning for his release after he proposed on Valentine's Day last year.
They became close when she visited him in jail after striking up a bond as penpals.
Bronson, 65, was first jailed in 1974 for armed robbery and has spent most of his life in Britain's toughest jails.
Wife reveals truth about marriage
Bronson is getting divorced
What Loose Women said about Bronson
He 'punched prison boss' after wedding
He spent time at East Yorkshire's Full Sutton high security dispersal prison, and notoriously smashed up a workshop in HMP Hull in 1975 after an altercation with a prison officer.
The prisoner also attacked a fellow inmate with a glass jug while there, and nine months were added to his ever-increasing sentence. He last tasted freedom in 1993.
He has stopped her visiting (Image: Adam Gerrard/Daily Mirror)
Wife Paula claims she was celibate for nearly two years – but accuses Bronson of writing to other women from his cell.
She said: "I fell in love with Charlie and he has treated me like this. He was even writing to other women while we were married. I can't believe I put up with that.
"I gave up everything – my job, my life, my family, my chance to have a physical relationship and children, my anonymity. I have hurt so many people around me because I fell in love.
"He made a vow to be with me forever and now over some stupid photographs in a gay bar, he has cut me off.
Charles Bronson's wife and ex-wife told on Loose Women 'he should never be let out of prison'
"He has banned me from even visiting him and he is still not allowed phone calls because of his last prison attack so I can't talk to him. I feel so helpless. I have been left with nothing and the worst thing is that I know everyone is thinking, 'I told you so'.
"The thing that hurts the most is that he knows how much I love him and he knows how much this is hurting me and he doesn't care."
Since the pictures were published in a newspaper earlier this week, Paula, who suffers from depression, has been in floods of tears.
It is claimed she challenged a group of six men to guess if her boobs were real – and one of the revellers took up the challenge.
Insiders say Bronson's pride has been dented by Paula's public betrayal but she says he could not care less about the split.
She originally insisted her hardman hubby would see the funny side and dismissed them as "banter", saying: "He knows I'm a flirt and he loves me for it."
Charles Bronson 'punched prison boss after being banned from his own wedding reception'
But on Friday she was told by Bronson's lawyers that the marriage was over.
She said: "Some friends went to visit him earlier this week and said he was laughing, joking and singing in his cell.
"Meanwhile he is getting his solicitor to ring me and tell me to expect divorce paperwork from them.
"He told me he loved me and would look after me no matter what, now he is taking the first chance to get out of this wedding."
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It's completely FREE and features all the latest news, sport and what's on information. You can download it for iPhone and iPad from Apple's App Store, or get the Android version from Google Play.
BusinessSecond homeowners set for £191m windfall from Omicron grantsAnalysis comes after Levelling Up minister Michael Gove announced plans to crack down on tax breaks for second homeowners from April next year
Boris Johnson 'attended another party in the run-up to Christmas 2020'
UK & World NewsThe Prime Minister is reported to have attended and gave a speech at the leaving do of his outgoing defence adviser


Celebrate the magic of falling in love! This adorable cartridge is filled with fun and whimsical shapes featuring dates, the proposal, the wedding, and much more. With creative features such as Card, Envelope, and Tag, you will be ready for every romantic occasion.
Layered images with shadows and blackout.


Founded in 1979, Allfloors Carpet One is operated by owners Michael & Vicki Henry. In 1986 we joined a young cooperative called "Carpet One", which grew into the largest buying group in our industry, with over 1,000 stores in North America and gross sales of over $1 billion per year. Being part of Carpet One enables us to pass along great prices, and also allows us to offer our exclusive "Beautiful Guarantee™". With everything that we install we provide a Lifetime Installation Warranty. We're locally owned, and family operated and here to make sure you're happy with your new floors or bathroom renovations.
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What was the goal for White Box Linux? To provide an unencumbered RPM-based Linux distribution that retains enough compatibility with Red Hat Linux to allow easy upgrades and to retain compatibility with their errata SRPMs. Being based off of Red Hat Enterprise Linux (RHEL) 3.0 means that a machine should be able to avoid the upgrade treadmill until October 2008 since RHEL promises errata availability for 5 years from date of initial release. Or more briefly, to fill the gap between Fedora and RHEL. Why was White Box Linux created? Its initial creation was sponsored by the Beauregard Parish Public Library in DeRidder, USA out of self interest. We have several servers and over 50 workstations running Red Hat Linux and were left high and dry by Red Hat's recent shift in business plan. Our choices were a difficult migration to another distribution or paying Red Hat an annual fee greater than the amortized value of our hardware. So we chose a third path, made possible by the power of open source.... White Box Linux.
Rate this project Eric Turgeon has announced that the first beta release of the upcoming GhostBSD 10.3, a desktop-oriented operating system based on FreeBSD, is now ready for download and testing: "GhostBSD 10.3-BETA1 is ready for testing. This first beta development release is ready for testing and debugging new features in GhostBSD 10.3; MATE and Xfce editions are available on SourceForge for the i386, amd64 and amd64-uefi architectures. What has been fixed? Localtime time has been fixed; installer support for UEFI; fix boot partition for GPT to supports freebsd-boot, bios-boot and EFI. We encourage you to use our issue system to report bug. The more issues are reported, the more of them will be fixed before the final release." Here is the brief release announcement with information on how to write the ISO images onto USB storage devices. Download the live images from SourceForge (pkglist): GhostBSD10.3-BETA1-mate-amd64.iso (1,860MB, SHA256), GhostBSD10.3-BETA1-xfce-amd64.iso (1,720MB, SHA256).


text:
Q: Chrome not recomputing element width when it should I need to size an element's width with respect to its height and thought I had worked out a pretty clever solution using the transparent img trick:
http://jsfiddle.net/6yc8a2yj/
The pink colored div correctly renders at a 1:2 ratio; however, it is supposed to resize when the viewport does (since it has height 100% of a 100vh element), but it doesn't.
If you resize the viewport, the img element increases width if the viewport height increases, but the container element does not grow in width to accommodate it, which is violating the rules for what display: inline-block means (if you check and uncheck the display style on .container in Chrome Console, it will resize properly to fit its child contents). I assume the problem is element dimension calculations are parent to child and thus when the img element resizes, Chrome does not recompute it's parents dimensions. Is there a way to trick it into doing so?

A: Chrome minimizes relayout cycles. And that's good in general.
Instead of .viewport I would suggest to set height:50vh on image itself and use width:max-content; on .container.
See: http://jsfiddle.net/0jgo5894/1/
And about vh units in general: http://sciter.com/css-vwvhvminvmax-units-are-quite-dangerous/

redpajama_set_name: RedPajamaStackExchange


When you are considering your metabolic surgery, you need to be aware that there will be an out of pocket cost for your treatment. Your costs will vary depending on your type of procedure and also your level of health insurance.
Our commitment to you, is to achieve the best possible outcome from your surgery. As a result, we have a team approach to your surgery. Your surgery costs will include Dr Baxter's fee, a surgical assistant fee & the anaesthetist fee. Additionally, there are consultation fees payable together with the dietetics package and a fee for your psychology assessment (pre-surgery).
Our team at The Sunshine Coast Medical Weight Loss Centre are dedicated to providing you with comprehensive and ongoing support.
There will be an out of pocket expense, after you have received your Medicare & private health insurance rebate for your weight loss surgery.
Will my private health insurance cover my surgery?
Most health funds will charge a hospital excess (a fee payable on your hospital admission) as well as applying a 12 month waiting period prior to being able to use your insurance for your weight loss surgery.
Can I still have weight loss surgery if I have no private health insurance?
Yes, however we recommend that you have health insurance prior to considering weight loss surgery. Without health insurance, you will have an out of pocket expense of approximately $16,000 for your hospital stay. Your total cost (hospital, surgery, assistant, anaesthetist, dietetics) is approximately $22,000. Additionally, if you have an unexpected prolonged hospital stay, you can expect significantly higher hospital costs.
Can I access my superannuation to assist with the cost of weight loss surgery?
We accept patients who wish to access early release of their superannuation to pay for their surgery & hospital stay. You will need to download the application form and bring them to your consultation. Additionally, your General Practitioner will be required to sign a form in support of your surgery. You can download the form here and all applications need to be submitted to the Australian Taxation office online via your MyGov account.


Historically, EMS professionals relied on the vital signs, specifically blood pressure, in conjunction with other physical findings to determine if a patient was in hypovolemic shock. Shock is a state of inadequate tissue perfusion. However, it has become clearer that blood pressure and heart rate may not be a good early indicator of a hypovolemic shock state and may actually mislead the EMS practitioner when considering a differential diagnosis.
Alteration in vital signs primarily results from both a reduction in blood volume and a cascade of neural and hormonal responses in an attempt to increase the blood pressure and conserve body fluid. We have always looked for profound changes in the blood pressure to assist in making a differential diagnosis of shock.
For example, a drop in the systolic blood pressure to 90 mm Hg is an indication that the shock state deteriorated from a compensatory stage to a decompensatory, or progressive stage. This dramatic drop, which is a clear but late finding, represents approximately a 30 percent blood loss in a healthy individual. The literature suggests that a patient could be in a true shock state and not initially present with a dramatic decrease in blood pressure or increase in heart rate [1,2]. Therefore, it is imperative to understand what the blood pressure is indicating and that the signs of poor perfusion can be assessed to identify early indicators of shock.
Blood pressure is determined by the cardiac output (CO) and peripheral vascular resistance (PVR). The equation BP = CO × PVR represents the interaction of the two variables. Cardiac output is the amount of blood ejected from the left ventricle in one minute. Peripheral vascular resistance is the resistance in the peripheral arteries and arterioles determined by the vessel size. A decrease in the vessel lumen will increase the resistance; whereas, a decrease in the vessel size will decrease the peripheral vascular resistance. An increase in the cardiac output or peripheral vascular resistance will lead to an increase in the blood pressure; whereas, a decrease will cause a decrease in blood pressure.
Cardiac output is an interaction of heart rate (HR) and stroke volume (SV), which is reflected in the equation CO = HR × SV. The stroke volume is defined as the amount of blood ejected from the left ventricle with each contraction and is determined by the preload, myocardial contractility and afterload. In general, an increase in heart rate or stroke volume will lead to an increase in cardiac output. Inversely, a reduction in heart rate or stroke volume will lead to a decrease in cardiac output.
The loss of blood associated with hypovolemic shock causes a reduction in the venous volume, which in turn diminishes the preload, stroke volume and cardiac output. A drop in cardiac output, which is reflected by a falling systolic blood pressure, results in a decrease in pressure in the carotid bodies and aortic arch, and triggers the baroreceptors (inhibitory stretch-sensitive receptors that constantly measure arterial pressure). When the baroreceptors sense a decrease in the arterial pressure, the sympathetic nervous system is prompted to initiate a cascade of neural and hormonal responses in an attempt to restore the pressure back to a normal state.
The direct neural stimulation and hormonal influence will increase the heart rate, increase myocardial contractility and increase peripheral vascular resistance through systemic vasoconstriction. The diastolic blood pressure is an indirect measure of peripheral vascular resistance; thus, as the vessels constrict and vascular resistance increases, the diastolic blood pressure is maintained or increases.
Even though the patient is losing blood and the venous volume and pressure is decreasing, the blood pressure will look relatively stable as the heart rate, myocardial contractility and peripheral vascular resistance increase as a means to compensate. This could produce a blood pressure that is deceiving and may lead the EMS practitioner into a false sense of patient stability.
For example, a blood pressure of 102/88 mm Hg surely falls within a normal limit; however, it could also be a clear sign of hypovolemia when assessed closer. It is important to not only look at the overall blood pressure, but also the pulse pressure, which can provide valuable information about the hemodynamic state. The pulse pressure is the difference between the systolic blood pressure and the diastolic blood pressure.
For example, using the pressure discussed previously, the pulse pressure is calculated at 14 mm Hg (102 – 88 = 14 mm Hg). If the difference is less than 25 percent of the systolic blood pressure, the pulse pressure is considered to be narrow. A wide pulse pressure is considered to be greater than 50 percent of the systolic blood pressure.
A narrow pulse pressure in a hypovolemic shock patient indicates a decreasing cardiac output and an increasing peripheral vascular resistance. The decreasing venous volume from blood loss and the sympathetic nervous system attempt to increase or maintain the falling blood pressure through systemic vasoconstriction. This increase in heart rate and myocardial contractility is reflected in the decreasing systolic BP, the increasing diastolic BP and the narrowing pulse pressure. Thus, a blood pressure of 102/88 mm Hg no longer appears to be "normal" and requires further assessment of heart rate, respiratory rate and other signs of perfusion, such as the skin color, temperature, condition and the patient's mental status.
Be careful when assigning a blood pressure assessment as "normal." The pulse pressure may provide more valuable and important information than the actual blood pressure itself. Blood pressure should be considered in the "whole" assessment of the patient and not purely as an independent finding.
1. Edelman D.A., White M.T., Tyburski J.G., et al: Post-traumatic hypotension: should systolic blood pressure of 90–109 mm Hg be included?. Shock 27. (2): 134-138.2007.
2. Victorino G.P., Battistella F.D., Wisner D.H.: Does tachycardia correlate with hypotension after trauma?. J Am Coll Surg 196. (5): 679-684.2003.


text:
Sea bream
Red Sea Bream
P 3.0 concept
PERSONAL DATA POLICY (GDPR)
Choose your country AUSTRALIA CHILE COSTA RICA DENMARK ECUADOR FINLAND FRANCE GERMANY ITALY NORWAY POLAND RUSSIA SPAIN SWEDEN TURKEY UNITED KINGDOM VIETNAM
Choose language EN EL-GR
BioMar icon Niels Alsted retires after 45 years in aquaculture
Niels Alsted, Executive Vice President of Business Relations in BioMar, nicknamed 'Mr Aquaculture' by his colleagues has retired after 45 years of service in the aquaculture industry.
"Niels has been one of the most important people forming not only BioMar but also the industry. His dedication to developing a sustainable and professional aquaculture has led to industry standards and the high-end feed ranges we see in the market. Furthermore, he has been one of the most important people forming the culture in BioMar: A culture built upon trust, relationships, professionalism and a desire to pioneer the future of the industry", said Carlos Diaz, CEO BioMar Group.
There are not many in the aquaculture industry who can claim 45 years of service. Niels' career started in 1974 on a small trout farm in Denmark before beginning his studies in aquaculture research at the University of Tromsoe, Institute of Fisheries in 1977 in Norway. He stayed in academic research for several years and was an associate professor at Aalborg University when he accepted to undertake his commercial PhD with BioMar in 1987.
Over the last 32 years Niels has held various positions in BioMar from R&D, sourcing, food safety and business relations and has been part of the executive management team in BioMar Group where he contributed to opening new markets like Chile and China. Niels is valued for his broad and deep technical knowledge and while at BioMar has published several papers on nutrition and sustainability. He was instrumental in the creation of the first ever environmentally friendly aquaculture feed product, Ecoline and is known for his scholar approach to feed product development.
Niels has represented the aquaculture industry and BioMar at numerous NGO events and been a member of various industry committees most recently chairing the Board of FEFAC and as a valued member of both the GSI feed task force and IFFO RS.
"I am very grateful to BioMar, I have seen most of the world meeting fantastic people and really enjoying my work in the aquaculture industry. I simply could not ask for more", concluded Mr. Alsted who now looks forward to spending more time with the family, but has not closed the door on potentially doing more aquaculture related projects.
Niels Alsted and delegation at Dae-Han feed factory in South Korea in 1989
Niels Alsted in 1987
BioMar Hellenic SA
2nd Industrial Zone of Volos
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GR-37500 Velestino
Phone +30 24250 61500
info@biomar.gr


text:
Headlines Opinion
The Raptors Should Have One Message for Kawhi Leonard: Good Riddance
Raptors fans feared the worst when Kawhi Leonard left Toronto for LA in the summer. But today? Things couldn't be better.
Kawhi Leonard who? | Source: Sean M. Haffey/Getty Images/AFP
Kawhi Leonard won an NBA with the Toronto Raptors before leaving for the Los Angeles Clippers.
The Raptors' victory over the Timberwolves took their unbeaten streak to 15 games – a franchise-best.
At the season's unofficial halfway point, it seems the Canadian outfit is looking better than ever – without Kawhi.
Toronto Raptors fans everywhere assumed that July 10, 2019, had ushered in the beginning of the end. That's the date when Kawhi Leonard signed with the Los Angeles Clippers. And the NBA dream in Toronto started to come apart.
Or so we thought.
We're now almost at the season's unofficial halfway point, and the Raptors' title dreams are far from crushed.
In fact, they're dreaming more than ever.
Maybe the Raptors Didn't Need Kawhi Leonard After All
That's not to say Kawhi Leonard wasn't an integral part of last year's title run.
Not only did the forward lead Toronto to its first NBA championship, but he also won the Finals MVP award. He's the first person ever to win that award with a team from both conferences.
Kawhi Leonard was – and will always be – a legend in the Great White North.
That's why it's understandable that Raptors fans freaked out at the prospect of enduring life without Kawhi.
Nobody could have guessed that all that hand-wringing would be unnecessary. Because the Raptors are even better at this stage of the season than during their championship run last year.
Does that mean they're favorites to go on and retain their title? Not really – many factors that come into play when determining who goes home with the championship.
But it's hard to dispute that things are looking pretty good for the reigning champs. Despite losing Kawhi Leonard.
The NBA Winning Streak No One Is Talking About
As the Raptors notched another win over the Timberwolves and extended their winning streak to an all-time franchise best 15 games, you have to wonder if the Canadian team is getting the respect they deserve.
We still remember the noises made about the 16-game streak the Golden State Warriors put together in 2014, and the Lakers have been heralded for a similar run in the late 1990s.
Where's the love for the Raptors?
Their title victory was "celebrated" by the media in a similarly muted fashion. Is that how it goes? If it doesn't happen in Boston, New York, or LA, it doesn't matter?
With games against the Nets, the Pacers, and the Suns upcoming before they face the high-flying Bucks, there's a good chance the Raptors could increase their streak to an incredible 18 wins.
But no one seems to care.
The Raptors Still Aren't Getting Any Love
A quick look at the betting favorites to win the NBA championship this year shows that the likes of the Bucks along with the Clippers and Lakers are leading the line.
Once again, the Raptors aren't even in the conversation.
And why would they be? Toronto is considered an unfashionable market by the U.S. media establishment – with or without Kawhi Leonard.
But no one was taking them seriously last year, either. How did that turn out?
Disclaimer: The opinions expressed in this article do not necessarily reflect the views of CCN.com.
Kawhi LeonardNBAToronto Raptors
Aubrey Hansen
I'm a full-time royal & entertainment commentator at CCN. I've had my opinions and links to my work on the royal family featured by outlets such as The Daily Mail, International Business Times, and The Express. Reach me at aubrey.hansen@ccn.com, visit my LinkedIn profile here or my Twitter profile here.
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Alrighty, we shall begin unpacking that wildly enigmatic paragraph of teasers. The first six are: Final. Pack. Late for taxi to airport. Angry lines. Late plane. Trapped in paradise.
Final – Jacelyn took her final and finished her first semester of medical school! Over the break she received the email informing her that she not only TOOK the final, but she PASSED it, as well! So, it's on to #2 (not to be confused with my prior writings concerning Valsalva).
Pack – That's where you put your clothes and various belongings in a travel container of sorts in order to proceed to a new destination. For further reading regarding "pack", please consult THIS PAGE, specifically number 20.
Late for taxi to airport – Our flight was scheduled to leave at 7, so, because of the mass exodus of students, we had to be at the airport at 5am at the latest. As a result of miscommunication between my ears and everything else in the world, we woke up with enough time to scream, change, and drag our puffy eyes down to the taxi. We made it to the airport at right around 5am, at which point we and a hundred other students sat down and waited until 6, which (we found out) was when the airport officially opens and the staff get there.
Angry Lines – After much screaming by particularly frazzled first semester med students, we all did our best Waldorf impression and plodded our way through the ticket line, the "pay us so you can leave" line, the "take your clothes off" security line, and the "you can't sit in those four chairs; they're for other people" line. We made it to our gate (one of two) by 7, at which point we were told……….nothing.
Late Plane – We all sat and waited, and waited, and waited. We watched the 7:40 flight leave. We watched the 8:20 flight leave. Still no word. Finally, our 7am flight arrived to get us at 8:35, which, ironically, was the exact time our connecting flight was scheduled to take of an hour away on Barbados. When we arrived on Barbados, no one at our airline's desks knew that we were late or had any idea what to do with us. In fact, the customs man wouldn't let us IN to Barbados, since we had no visa or reason to be there; yet we couldn't get OUT because, well, the plane had left the station. It was much like Hotel California, minus the pink champagne and freaky looking bass player.
Trapped in Paradise – So, missing our connection to Barbados made us miss our connection to St. Vincent, which made us miss our connection to San Juan, which made us miss our connection to Atlanta, which made us miss our connection to Minneapolis. The delay caused us to have to purchase $828 more in plane tickets (when I say "us", I most definitely mean Ralph Frederick, who is quoted as saying, "Whenever you guys travel, I prepare myself to bail you out") (Ralph is not only generous, but possesses deep wisdom of the ages). Needless to say, our benevolent airline, though in NO WAY prepared to help us in ANY WAY with the predicament they had put us in by having pilots with an unfortunate combination of procrastination, constipation, and narcolepsy, DID put us up in a small hotel 30 minutes from the airport. I will write much, much more about Liat Airlines in the future.
I'm off to get lining for our curtains so that I can stop freaking out the 4th semester student whose apartment window faces the window that faces the room I face when I'm naked.
Holy smokes!! LOL, that may be the funniest thing I've read all year. Not what happened–i'm really sorry about all of that. But, funny stuff Nic, funny stuff.
So sorry all that happened to you both but hey it makes for great story telling, right? What about Burt? Is he doing okay now?
Hey nic, saw that you're coming home, or working on it anyways (sorry 2 hear bout the extra wait and cost, man that would get me). Tonite is on the 20. I'll be praying for you and the kids. Praise God for you guys.


text:
\section{Introduction}

Let $ G $ be an undirected simple graph $ G $, where $ V(G) $ and $ E(G) $ denote the set of vertices and the set of edges of $ G $, respectively. For two vertices $ u,v \in V(G) $, $ u $ and $ v $ are \textit{neighbors} if they are adjacent, that is, if there is an edge $ e=uv \in E(G) $. The \textit{open neighborhood} of a vertex $ v $ is the set $ N_G(v) = \{u \in V(G): uv \in E(G)\} $ and the \textit{degree} of $ v $ is given by the cardinality of $ N_G(v) $. Two vertices $ u,v \in V(G) $ are \textit{false twins} if $ N_G(u) = N_G(v) $. Let $ \delta(G) $ and $ \Delta(G) $ denote the minimum and the maximum degree of $ G $, respectively. For a subset $ S\subseteq V(G) $, let $ G[S] $ denote the subgraph induced by $S$. An edge $ e=uv \in E(G) $ is \textit{subdivided} when it is deleted from $ G$ and a new vertex $ x $ is added with two new edges $ xu $ and $ xv $. Here, the vertex $ x $ is a \textit{subdivision vertex}. Let $ [r] $ denote  the set of integers $ \{1,\dots,r\} $ throughout the paper.

A subset $ D\subseteq V(G) $ is \textit{dominating} in $ G $ if every vertex of $ V(G) \setminus D $ has at least one neighbor in $ D $. Similarly, a subset $ D \subseteq V(G) $  is \textit{$ k $-dominating} in $ G $ if every vertex of $ V(G)\setminus D $ has at least $ k $ neighbors in $ D $. The \textit{domination number} $ \gamma(G) $ and the \textit{$ k $-domination number} $\gamma_k(G) $ of $ G $ are the minimum cardinalities of a dominating and a $ k $-dominating set of $ G$, respectively.

We say that a connected graph $G$ is a \emph{$(\gamma, \gamma_k)$-graph} if $\gamma_k(G)=\gamma(G) +k-2$ and $\Delta(G)\ge k$.
A connected graph $G$ is \emph{$(\gamma, \gamma_k)$-perfect} if  $\delta(G) \ge k$ and  every connected induced subgraph $H$ of $G$ with $\delta(H) \ge k$ satisfies the equality $\gamma_k(H)=\gamma(H) + k-2$.

Hypergraphs are set systems  that are conceived as a natural generalization of graphs.  A \emph{hypergraph} $H = (V,E)$ contains a finite set $V$ of vertices  together with a collection $E$ of nonempty subsets of $V$, called \emph{hyperedges} or simply \emph{edges}. The number of vertices, that is $|V|$, is called the \emph{order} of $H$. Throughout this paper, we suppose that  $|e| \ge 2 $ holds  for every $e\in E$. The \emph{degree} of a vertex $v$ in $H$, denoted by $d_H(v)$, is the number of edges containing the vertex $v$. The hypergraph $H$ is $k$-\emph{uniform} if every edge contains exactly $k$ vertices. Thus, every (simple) graph is a $2$-uniform hypergraph. A hypergraph $H'=(V', E')$ is an \emph{induced subhypergraph} of $H=(V,E)$ if $V' \subseteq V$ and $E'$ contains all edges $e \in E$ satisfying $e \subseteq V'$. We also use the notation $H[V']$ for the subhypergraph induced by $V'$. Given a collection ${\cal F}$ of $k$-uniform hypergraphs, we say that a $k$-uniform hypergraph $H$ is \emph{${\cal F}$-free} if no induced subhypergraph of $H$ is isomorphic to any hypergraph contained in ${\cal F}$. The complete $k$-uniform hypergraph of order $n$ is the hypergraph ${\cal K}_n^k=(V,E)$, where $|V|=n\ge k$ and $E$ contains all $k$-element subsets of $V$.

A set $T \subseteq V$  is a \emph{transversal}
(or \emph{vertex cover}) in the hypergraph $H=(V,E)$ if $|e\cap T|\ge 1$ holds for every edge $e\in E$.  The complement $V\setminus T$ of a transversal $T$ is called a \emph{(weakly) independent} vertex set. The minimum cardinality of a transversal and the maximum cardinality of a weakly independent vertex set in $H$ are denoted by $\tau(H)$ and $\alpha_w(H)$, respectively. The minimum number of edges that together cover every vertex of $H$ is the \emph{edge cover number} of $H$ and denoted by $\rho(H)$. For instance, in a complete $k$-uniform hypergraph ${\cal K}_n^k$, any $k-1$ vertices form a maximum weakly independent vertex set and any $n-k+1$ vertices form a minimum transversal. In particular, we have $\tau({\cal K}_n^k)=n-k+1$, $\alpha_w({\cal K}_n^k)=k-1$, and $\rho({\cal K}_n^k) = \lceil\frac{n}{k}\rceil$.

Two vertices $u$ and $v$ of $H$ are \emph{adjacent} if there is an edge $e$ of $H$ such that $\{x,y\}\subseteq e$. A hypergraph is \emph{connected} if for every two vertices $x$ and $y$ there exists a sequence $x=x_0,x_1, \dots, x_j=y$ such that $x_{i-1}$ and $x_i$ are adjacent vertices for every $i \in [j]$. The \emph{$2$-section graph} of $H=(V,E)$ is the graph $G$ on the same vertex set $V$ such that two vertices  form an edge in $G$ if and only if they are adjacent in $H$. The \emph{incidence graph} of the hypergraph $H=(V,E)$ is the bipartite graph $G'=(A \cup B, E')$ where the vertices in $A$ and $B$ represent the vertices and edges of $H$, respectively. Moreover, two vertices, $a\in A$ and $b\in B$ are adjacent in $G'$ if the vertex of $H$ that is represented by $a$ is contained in the hyperedge represented by $b$. Note that $H$ is a connected hypergraph, if and only if, its $2$-section graph is connected and, if and only if, its incidence graph is connected.

\vspace{0.3cm}\noindent\textbf{Structure of the paper.} In Section~\ref{sec:2}, we first cite some previous results and prove some general lemmas. Then, for each $k \ge 2$, three graph classes  ${\cal B}_k^*$, ${\cal B}_k$ and  ${\cal G}_k$ are defined such that every connected $(\gamma, \gamma_k)$-graph belongs to ${\cal G}_k$. Moreover, in Section~\ref{sec:3}, we concentrate on the case of $k \ge 3$ and show that every connected bipartite  $(\gamma, \gamma_k)$-graph is contained in ${\cal B}_k$. We also prove that a connected bipartite graph $ G \in {\cal B}_k $ is a $(\gamma, \gamma_k)$-graph if and only if its $ \gamma_k $-simplified graph $ G^* \in {\cal B}_k^* $ is a $(\gamma, \gamma_k)$-graph. Then, we present a characterization of  bipartite $(\gamma, \gamma_k)$-graphs in terms of the properties of the underlying hypergraph. In Section~\ref{sec:4}, we work on the hereditary version of the problem and give a characterization for all bipartite $(\gamma, \gamma_3)$-perfect graphs. Later, in Section~\ref{sec:5}, we prove that it can be decided in polynomial time whether a given bipartite graph is a $(\gamma, \gamma_k)$-graph, while the corresponding decision problem is NP-hard on $ {\cal G}_k $.

\section{Preliminary Results on $ (\gamma, \gamma_k)$-graphs}
\label{sec:2}

Fink and Jacobson \cite{Fink85, Fink85-2} introduced  $ k $-domination in graphs as a generalization of the concept of domination. Motivated by this definition, related problems have been studied extensively by many researchers (see for example \cite{Bujtas2017, Caro1990-2, Caro1990, Favaron1988, Favaron2008, Hansberg2015, Hansberg2013, Shaheen2009}). For more details, we refer the reader to the books on domination by Haynes, Hedetniemi and Slater \cite{ DominationBook2, DominationBook1} and to the survey on $ k $-domination and $ k $-independence by Chellali \textit{et al.} \cite{ChellaliSurvey}.

Fink and Jacobson \cite{Fink85} proved the following result on the relation between the domination number and the $ k $-domination number of $ G $.

\begin{thm} \cite{Fink85}
	\label{thm:Fink}
	For any graph $ G $ with $ \Delta(G)\geq k\geq 2 $, $ \gamma_k(G) \geq \gamma(G)+k-2$.
\end{thm}

Although it is proved that the above inequality  is sharp for every $k\ge 2$, the characterization of graphs attaining the equality is still open, even for the small values of $ k $.  The corresponding characterization problem was studied in~\cite{Hansberg2015, Hansberg2016, Hansberg2008}. Recently, we considered a large class of graphs and gave a characterization for the members satisfying the equality $ \gamma_2(G) = \gamma(G)$. We also proved that it is NP-hard to decide whether this equality holds for a graph. Moreover, we gave a necessary and sufficient condition for a graph to satisfy $ \gamma_2(G) = \gamma(G)$ hereditarily \cite{manu}. Some
similar problems involving different domination-type graph and hypergraph invariants were considered for example in~\cite{Arumugam2013, Blidia2006, Dettlaff2016, Hartnell1995, Brause2016, Randerath1998}.

\begin{lemma}
	\label{lem:1}
	Let $D$ be a minimum $ k $-dominating set of a graph $G$. If  $ \gamma_k(G)=\gamma(G) +k-2$, then $\gamma(G[D])\geq \gamma_k(G)-(k-2)$.
\end{lemma}
\begin{proof}
	Suppose, to the contrary, that $\gamma(G[D])\leq \gamma_k(G)-k+1$. Let $ S $ be a dominating set of cardinality $|D|-k+1$ in $ G[D] $. We claim that $ S $ is also a dominating set in $ G $. Indeed, as we removed $k-1$ vertices from the $k$-dominating set $D$, every vertex in $ V(G)\setminus D $ is still dominated by at least one vertex of $ S $. Note that $S$ dominates all the vertices in $ D $ by the choice of $S$. Since $ \gamma(G) \leq |S| = \gamma_k(G) - k +1 $, we get a contradiction. Thus, $\gamma(G[D])\geq \gamma_k(G)-k+2$.
\end{proof}
Since $\gamma(G) \leq |V(G)|-\Delta(G)$ holds for every graph $G$, the previous lemma directly implies the following result obtained by Hansberg (see Theorem 5 in \cite{Hansberg2015}).
\begin{lemma} \cite{Hansberg2015}
	\label{lem:AH}
	Let $D$ be a minimum $ k $-dominating set of a graph $G$. If  $ \gamma_k(G)=\gamma(G) +k-2$, then $\Delta(G[D])\leq k-2 $.
\end{lemma}
\medskip

For each $k\ge 2$, we define three graph classes which will have crucial role in our study.
\begin{definition}
\label{Def:B_k*}
	Given an arbitrary connected $ k $-uniform hypergraph $ F $ with vertex set $ V(F) = D = \{v_1,\dots, v_s\} $ and edge set $E(F)=\{e_1, \dots ,e_p\}$, we define the following graph classes.

	\begin{itemize}
	\item[$ (i) $] Define a pair of vertices $ X_{i} = \{x_i^1, x_i^2\} $ for every hyperedge $ e_i \in E(F)$. Let $X=\bigcup_{i \in [p]}X_{i}$ and  let
	$$V(G_1)=X \cup V(F),  \quad E(G_1)=\{x_i^jv_\ell: \enskip v_\ell \in e_i, \enskip i \in [p], \enskip j \in [2], \enskip \ell \in [s]\}.$$
	The graph class $\mathcal{B}^*(F)$ contains only this graph $G_1$ which is also called the double incidence graph of $F$.

	\item[$(ii)$] The class ${\cal B}(F)$ contains a graph $G_2$ if it can be obtained from the double incidence graph $G_1$  in the following way. We keep all vertices and edges of $G_1$. For each edge $e_i \in E(F)$, we create some (maybe zero) false twins of the vertex $x_i^1$. Further, if $S \subseteq V(F)$ induces a complete subhypergraph in $F$, then we may supplement $G_1$ with some (maybe zero) new vertices which are adjacent to all vertices in $S$. We denote by $Y$ the set of the new vertices that is $Y=V(G_2)\setminus V(G_1)$. Putting it the other way around, we say that $G_1$ is the \emph{$\gamma_k$-simplified graph} of $G_2$

	\item[$(iii)$] The class ${\cal G}(F)$ contains $G_3$ if it can be obtained from a graph $G_2\in {\cal B}(F)$ by supplementing it with some (maybe zero) new edges inside $ D $ and $X\cup Y$. These edges can be chosen arbitrarily, but each $X_{i}$ must remain independent.
	\end{itemize}
	For every $ k\geq 2 $, the graph classes ${\cal B}^*_k$, ${\cal B}_k$, and  ${\cal G}_k$ contains those graphs $G$ for which there exists a $ k $-uniform hypergraph $F$ such that $G$ belongs to ${\cal B}^*_k(F)$, ${\cal B}_k(F)$, and  ${\cal G}_k(F)$, respectively. We say that $ F $  is the underlying hypergraph of $ G $ if $ G \in {\cal G}(F) $.
\end{definition}
 It is clear that each member of ${\cal B}_k$ is bipartite and that ${\cal B}_k^* \subseteq {\cal B}_k \subseteq {\cal G}_k$ holds for every $k \ge 2$. In particular, for each $k$-uniform hypergraph $F$, we have exactly one graph in ${\cal B}_k^*$  having $F$ as its underlying hypergraph, but there are infinitely many such graphs in ${\cal B}_k$.   Note that in \cite{manu}, where $ (\gamma,\gamma_2) $-graphs were studied, a graph class $ {\cal G} $ was introduced that exactly corresponds to the class $ {\cal G}_2 $ defined here. Moreover, by the results in \cite{Arumugam2013, Hartnell1995, Lingas2018}, the class ${\cal B}_2$ is the collection of those connected graphs which satisfy $\tau(G)=\gamma(G)$ and $\delta(G) \ge 2$.

Hansberg proved the following lemma which directly implies that the class ${\cal G}_k$ contains all $(\gamma, \gamma_k)$-graphs.
\begin{lemma} \cite{Hansberg2015}
	\label{lem:AH2}
	Let $ G $ be a $(\gamma, \gamma_k)$-graph for an integer $k \ge 2$ and  suppose that $ D $ is a minimum $ k $-dominating set of $ G $. Then, for every $ k $ vertices $ v_1,v_2, \dots, v_k$ from $D$, if $ \bigcap_{i=1}^k N(v_i) \neq \emptyset $, then there is a nonadjacent pair $ x,y \in V\setminus D $ such that $ N_G(x) \cap D =  N_G(y)\cap D = \{v_1,v_2, \dots, v_k\}$.
\end{lemma}
\begin{corollary} \label{lem:2} If $ G $ is a connected $ (\gamma,\gamma_k) $-graph with $\Delta(G) \ge k \ge 2$ and $D$ is a minimum $k$-dominating set of $G$, then there exists a $k$-uniform underlying hypergraph on the vertex set $D$ such that $ G\in {\cal G}_k(F)$.
	\end{corollary}

\section{Bipartite  ($ \gamma,\gamma_k $)-graphs} \label{sec:3}

In this section we characterize connected bipartite graphs satisfying $\Delta(G) \ge k$ and $\gamma_k(G)=\gamma(G)+k-2$, for every $k \ge 3$. Observe that  $\gamma_k (G) \ge \gamma(G)$ holds  for every graph without imposing conditions on the vertex degrees. Moreover, assuming $k \ge 3$, the equality $\gamma_k(G)=\gamma(G)$ is satisfied if and only if $G$ consists of isolated vertices. Consequently, if $\Delta(G) \ge k \ge 3$ and if $G$ is disconnected, then  $\gamma_k(G)=\gamma(G)+k-2$ holds if and only if $G$ contains exactly one component which is a $(\gamma, \gamma_k)$-graph and all the further components are isolated vertices. Therefore, it is enough to concentrate on the connected $(\gamma, \gamma_k)$-graphs.

In the first subsection, we prove that all connected bipartite $(\gamma, \gamma_k)$-graphs belong to ${\cal B}_k$ and it is enough to consider the graphs from the subclass ${\cal B}_k^*$ where the underlying hypergraph uniquely represents the graph $G$. In the second subsection, we introduce a hypergraph invariant called `vertex-edge cover number'\ or shortly `TC-number'\ and derive the characterization of connected bipartite $(\gamma, \gamma_k)$-graphs via the properties of the underlying hypergraph.
\medskip

\subsection{Reducing the problem to $\gamma_k$-simplified graphs}
\begin{thm}
	\label{thmPartite}
	Let $k \ge 3$ and let $G$ be a connected bipartite graph that satisfies $\gamma_k(G)=\gamma(G)+k-2$ and $\Delta(G) \ge k$. Then, $G \in {\cal B}_k$ and every minimum $ k $-dominating set corresponds to a partite class of $G$.
\end{thm}
\begin{proof}
	Let $D$ be a minimum $ k $-dominating set of $G$. The degree condition implies that $V(G)\setminus D$ is not empty and, therefore, the underlying hypergraph $F$ contains at least one edge.  Let $e_i=\{v_1, \dots, v_k\}$ be an edge of $F$. Since all the vertices in $ e_i $ have common neighbors in $X_i$, they belong to the same partite class of $G$. This implies that the vertices of a connected component of $F$ are in the same partite class. If $F$ is connected, the same is true for the entire $D$. Moreover, since every vertex $x\in V(G)\setminus D$ has some neighbors in $D$, they all must be contained in the other partite class. It finishes the proof for the case when $F$ is connected.

	If $F$ is not connected, consider a nontrivial component $F_1$ and the remaining part $F_2=F-F_1$ of the underlying hypergraph. By Corollary~\ref{lem:2}, no vertex of $V(G)\setminus D$ can be adjacent to a vertex from $V(F_1)$ and to a vertex from $V(F_2)$ simultaneously. By our assumption, $G$ is connected. It is enough to consider the following two cases.

	\vspace{0.3cm}\noindent\textbf{Case 1.} There exists an edge $uv$ in $G[D]$ such that $u\in V(F_1)$ and $v\in V(F_2)$.

	\noindent Since $F_1$ is connected and nontrivial, there is an edge $e_i \in E(F_1)$ containing $ u $. In this case, the $ k $ vertices from $ e_i \setminus \{u\} \cup \{v\} $ cannot have a common neighbor in $V(G)\setminus D$ and consequently, \[D'=D\setminus (e_i \setminus \{u\} \cup \{v\}) \cup \{x_i^1\},\]

	\noindent where $ x_i^1$  is from the pair $ X_i $ associated with $ e_i $. Observe that $ D' $ is a dominating set of $G$. As $|D'| <|D|-k+2$, this is a contradiction.

	\vspace{0.3cm}\noindent\textbf{Case 2.} There exists an edge $xy$ in $G[V(G)\setminus D]$ such that $x$ has neighbors from $V(F_1)$ and $y$ has neighbors from $V(F_2)$.

	\noindent Consider $ k $ vertices, namely $v_1,\dots, v_k$ from $N(x)\cap V(F_1)$ and $ k $ vertices, namely $u_1,\dots,u_k$ from $N(y)\cap V(F_2)$ and then define $D'= D\setminus \{v_1,\dots, v_{k-1},u_1,\dots, \allowbreak u_{k-1}\}\cup \{x,y\}$. Observe that $D'$ is a dominating set of $G$, since $\{v_1,\dots, v_{k-1},u_1,\allowbreak \dots, u_{k-1}\}\cup \{x,y\}$ does not contain any edges of $F$. By our condition $ k\geq 3 $, it follows that  $|D'|= |D| - (2k-2) + 2 <|D|-k+2$. We have a contradiction again which completes the proof of the theorem.

\end{proof}

Note that Theorem \ref{thmPartite} does not hold for $ k=2 $. As an example, consider the graph $ H $ constructed by two vertex-disjoint copies of $ K_{2,3} $ by adding exactly one edge between them such that the maximum degree remains three. Observe that this graph is bipartite and satisfies the equality $ \gamma_2 (H) = \gamma(H) = 4$, but neither of the partite classes corresponds to a minimum dominating set or minimum $ 2 $-dominating set.

\begin{thm}
	\label{thm:simpl}
	Let $ G $ be a connected bipartite graph from ${\cal B}_k $ and let $G^*$ be its $\gamma_k$-simplified graph from ${\cal B}_k^*$, where $ k\geq 3 $. Then $\gamma_k(G)=\gamma(G)+k-2$ if and only if  $\gamma_k(G^*)=\gamma(G^*)+k-2$.
\end{thm}
\begin{proof}
	First, we prove that $\gamma_k(G)=\gamma_k(G^*)$. Since $D$ is supposed to be a minimum  $ k $-dominating set in $G$ and we did not delete any vertex from it when $G^*$ was obtained, $D$ is a $ k $-dominating set in $G^*$. Hence, $\gamma_k(G^*) \le |D|$. On the other hand, each vertex of $D$ can be $ k $-dominated in $ G^* $ by itself or by at least $ k $ neighbors from $X$. Since every vertex from $X$ is of degree $ k $, the $ k $-domination of all vertices in $D$ needs at least $|D|$ vertices in $G^*$. So, $\gamma_k(G^*)\ge |D|$ and $D$ is a minimum  $ k $-dominating set in $G^*$ as well.

	Now, we prove that $\gamma(G)=\gamma(G^*)$. Let $Q'$ be a minimum dominating set in $G^*$ such that $|Q'\cap D|$ is maximum. Then, $Q'$ contains at most one vertex from each pair $\{x_i^1,x_i^2\}$, since otherwise $x_i^2$ can be replaced by a vertex of $e_i$. Since $x_i^1$ and $x_i^2$ are false twins, we may suppose that the vertex $x_i^2$ does not belong to $Q'$ for any $ i $. Note that $Q'\cap D$ is a transversal in the underlying hypergraph $F$. We claim that $Q'$ is a dominating set in $G$ as well. Indeed, all vertices in $D\cup X$ are dominated. Further, for any vertex $y\in Y$ we may consider $ k $ arbitrarily chosen neighbors from $D$, they must form a hyperedge $e_j$ in $F$. The vertex $x_j^2$ is not in $Q'$ but it is dominated by a vertex from $Q'\cap e_j$. This vertex dominates $y$ as well. This proves $\gamma(G) \leq |Q'| = \gamma(G^*)$.

	On the other hand, consider a minimum dominating set $Q$ in $G$ such that $|Q\cap D|$ is maximum.
	Since every $X_i$ contains false twin vertices, we have $|Q\cap X_i|\leq 1$ for every $i$. If  $Q\cap X_i$  contains a vertex different from $x_i^1$, we may replace it with $x_i^1$ in the dominating set. Again, $Q\cap D$ must be a transversal in $F$. Now assume that $Q$ contains a vertex $y$ from $Y$. Let $z_1, \dots, z_\ell$ be those vertices from $D$ which are privately dominated by $y$ (they have no further neighbors in the dominating set). Note that all of $z_1, \dots, z_\ell$ are outside of $Q$. By Corollary~\ref{lem:2} and the definition of $F$, if $\ell \ge k$, then the set $\{z_1, \dots, z_\ell\}$ must contain at least one edge from $F$. This contradicts the fact that $Q$ is a transversal in $F$.
	If $\ell \le  k-1$, then there is an edge $e_i \in E(F)$ containing  $z_1, \dots, z_\ell$. As $x_i^1$ can dominate all these vertices  $z_1, \dots, z_\ell$ and the vertex which dominates $x_i^2$ in $Q$ also dominates $y$, the set $Q\setminus \{y\}\cup \{x_i^1\}$ is a dominating set of $G$.

	Perform these changes while there is a vertex  from $Y$  in the dominating set. At the end, we have a minimum dominating set $Q^*\subseteq (D\cup X)$ in $G$. Clearly, $Q^*$ is a dominating set of $G^*$, and we have $\gamma(G^*)\leq  |Q^*| =\gamma(G)$.  This implies $\gamma(G)=\gamma(G^*)$ which completes the proof.
\end{proof}

\subsection{Characterization via Underlying Hypergraphs}

\begin{definition}
For a hypergraph $H$, a set $S \subseteq V(H) \cup E(H)$  is a \emph{vertex-edge cover} or shortly a \emph{TC-set} of $H$ if $S\cap V(H)$ is a transversal (vertex cover) in $H$ and the edges in $S\cap E(H)$ together cover all vertices outside $S\cap V(H)$. The smallest cardinality of a TC-set in $H$ is called the \emph{TC-number} of $H$ and denoted by $tc(H)$.
\end{definition}

\begin{prop}
	\label{prop:tc}
	For every hypergraph $H$ of order $n$, the following statements hold and the upper bounds given in $(i)$ and $(ii)$ are sharp.
\begin{itemize}
	\item[$(i)$] If $r$ is the smallest size of an edge in $H$, then $tc(H) \le n-r+2$.
	\item[$(ii)$] If $H$ is $k$-uniform, then $tc(H) \le n-k+2$.
	\item[$(iii)$] If $H$ is $2$-uniform, that is a simple graph, then $tc(H)=n$.
\end{itemize}
\end{prop}
\begin{proof}
Let $H$ be a hypergraph and $e\in E(H)$ an edge of minimum cardinality $r$. If $v$ is an arbitrary vertex from $e$, the $r-1$ vertices in $e\setminus \{v\}$ does not contain any edges of $H$. Hence, $T=(V(H)\setminus e) \cup \{v\}$ is a transversal and $T \cup \{e\}$ is a TC-set in $H$. Then, we have $tc(H) \le |T|+1 = n-(r-1)+1=n-r+2$ that proves $(i)$. From $(i)$, the statement $(ii)$ can be obtained as a direct consequence. Observe further that every complete $k$-uniform hypergraph gives a sharp example as every transversal of ${\cal K}_n^k$ contains at least $n-k+1$ vertices and, if it is not the entire vertex set, we also have to put an edge into the TC-set. Thus, $tc({\cal K}_n^k)= n-k+2$ holds for every $n \ge k \ge 2$.

Now, let $H$ be an arbitrary graph and let $T$ be its transversal set. Since every edge of $H$ intersects $V(H)\setminus T$ in at most one vertex, we cannot cover $V(H)\setminus T$ with less than $|V(H)\setminus T|=n-|T|$ edges. This gives $tc(H)\ge n$ and it is clear, or concluded from part $(ii)$, that $tc(H) \le n$. This proves $(iii)$.
\end{proof}

Concerning the extremal cases in part $(ii)$ of Proposition~\ref{prop:tc}, we prove the following.

\begin{thm}
	\label{thm:TC-extr}
 A $k$-uniform hypergraph $H$ of order $n$ satisfies $tc(H)=n-k+2$, if and only if, for every $\ell \le k-1$ and for every $\ell$  edges $e_{i_1}, \dots, e_{i_\ell}$ of $H$, the union $L=\bigcup_{j=1}^\ell e_{i_j}$ contains at most $\ell+k-2$ weakly independent vertices.
\end{thm}
\begin{proof}
	We prove the equivalence for the  negations.
	First suppose that there exist $\ell$ edges, say $e_{1}, \dots, e_{\ell}$, in $H$ such that $L=\bigcup_{j=1}^\ell e_{j}$ contains $s=\ell+k-1$ (weakly) independent vertices. Let $X=\{v_{1}, \dots, v_{s}\}$ be such an independent subset of $L$. Then, the set $T=V(H) \setminus X$ is a transversal of cardinality $n-s$ and the remaining vertices in $X$ can be covered by using the $\ell$ edges $e_{1}, \dots, e_{\ell}$. This TC-set contains $n-s+\ell=n-(\ell+k-1)+\ell=n-k+1$ elements and, as follows, we have $tc(H) \le n-k+1 <n-k+2$. This proves the first direction of the equivalence.

	Now we assume that $tc(H) \le n-k+1$ and  show the existence of $\ell \le k-1$ edges such that the union $L$ contains more than $\ell+k -2$ independent vertices. Consider a TC-set $S$ with $|S|=n-k+1$ and, renaming the edges and vertices if necessary, let $S\cap E(H)=\{ e_{1}, \dots, e_{\ell}\}$ and $X=V(H) \setminus (S \cap V(H))=\{v_{1}, \dots, v_{s}\}$ where $s=n-|S\cap V(H)|=n-(n-k+1-\ell)=k+\ell-1$. By the definition of a TC-set, $ X$ is an independent set in $H$ and all vertices in $X$ can be covered by the $\ell$ edges $e_{1}, \dots, e_{\ell}$. Therefore, the union $L$ of these $\ell$  edges contains a set $X$ of $s> \ell+k-2$ independent vertices.

	Now it suffices to prove that we can ensure $\ell \le k-1$. Suppose that $\ell \ge k$ and the union $L$ of $e_{1}, \dots, e_{\ell}$ contains a set $X$ of $s=k+\ell -1$ independent vertices. Then, $\ell = s-k+1 \ge s-\ell +1$ holds and we may derive that $2\ell -1 \ge s$. By the pigeonhole principle, there exists an edge $e_q$, where $q \in [\ell]$, which contains at most one `private'\ vertex from $X$ that is, $$|(e_q \setminus \bigcup_{j \in [\ell], \: j\neq q} e_j ) \cap X| \le 1.$$
	Removing this edge $e_q$ from the list, we have $\ell -1$ edges such that their union contains at least $(\ell-1)+k-1$ independent vertices. If $\ell-1$ is still greater than $k-1$, we may repeat the procedure. At the end, we have $\ell' \le k-1$ edges such that their union contains $k + \ell'-1$ independent vertices. Consequently, it is enough to check the property for at most $k-1$ edges.
		\end{proof}

	Motivated by Theorem~\ref{thm:TC-extr}, we define the following classes of hypergraphs.
	\begin{definition}
	For every $k\ge 3$, a $k$-uniform hypergraph $H$ belongs to the class ${\cal H}_k$, if and only if, $\rho(H) \le k-1$ and $\alpha_w(H) \ge \rho(H)+k-1$.
	\end{definition}
	The condition  $\rho(H) \le k-1$ implies that for each $k$ and every $H \in {\cal H}_k$, the order of $H$ is at most $k(k-1)$ and, consequently,  ${\cal H}_k$ is a finite set of hypergraphs. Having the definition of ${\cal H}_k$ in hand, we can formulate the following corollary.

	\begin{corollary}
		\label{cor:forbid-F}
	For each $k \ge 3$,  a $k$-uniform hypergraph $F$ satisfies $tc(F)=|V(F)|-k+2$ if and only if $F$ is ${\cal H}_k$-free.
	\end{corollary}
\medskip

    The next theorem and its consequences give a characterization for connected bipartite $(\gamma, \gamma_k)$-graphs. In fact, these characterizations refer to the properties of the underlying hypergraph but, since we know that every bipartite $(\gamma, \gamma_k)$-graph $G$ belongs to a class ${\cal B}_k(F)$, these results also characterize the structure of $G$.

    \begin{thm}
    	\label{thm:char-gamma-k}
   For each $k \ge 3$, a connected bipartite graph $G$ satisfies $\gamma_k(G)=\gamma(G)+k-2$, if and only if, $G \in {\cal B}_k$ and the underlying hypergraph $F$ of $G$ satisfies $tc(F)=|V(F)|-k+2$.
    \end{thm}
\begin{proof}
If $G$ is a connected bipartite $(\gamma,\gamma_k)$-graph then, by Theorem~\ref{thmPartite}, we have $G \in {\cal B}_k$. Further, by Theorem~\ref{thm:simpl}, $G$ is a $(\gamma, \gamma_k)$-graph if and only if  its $\gamma_k$-simplified graph $G^*$ is a $(\gamma,\gamma_k)$-graph as well. Note that $G$ and  $G^*$ admit the same underlying hypergraph $F$ where $|V(F)|= \gamma_k(G)= \gamma_k(G^*)$. Hence, it suffices to prove that $\gamma(G^*)=tc(F)$.

Consider now a minimum dominating set $Q$ of $G^*$ such that $|Q\cap V(F)|$ is maximum under this condition. First, observe that $Q$ does not contain two false twins from $V(G)\setminus V(F)$. Indeed,  $x_i^1 \in Q$ dominates itself and all vertices from the associated edge $e_i$ of $F$ and then, $x_i^2$ can be replaced by any vertex of $e_i$ in the dominating set. We may suppose, without loss of generality, that $x_i^2 \notin Q$ holds for each $e_i \in E(F)$. Since $N_{G^*}(x_i^2)=e_i$ and $|N_{G^*}(x_i^2) \cap Q| \ge 1$, we infer that $Q$ contains at least one vertex from each $e_i \in E(F)$. Thus, $Q\cap V(F)$ is a transversal of $F$. Further, if a vertex $v_j \in V(F)$ does not belong to $Q$, it is dominated by a vertex $x_s^1\in Q$ where the edge $e_s$ of the underlying hypergraph is incident with $v_j$. It means that the edge set
$$R=\{e_s: e_s \in E(F)    \mbox{ and } x_s^1 \in Q\}$$
covers all vertices of $V(F)$ which are outside $Q$. Then, $(Q \cap V(F))\cup R$ is a TC-set in $F$ and since $|R|$ equals the number of vertices in $Q \setminus V(F)$, the cardinality of this TC-set is $|Q|= \gamma(G^*)$. Therefore, $tc(F) \le |Q| =\gamma(G^*)$.
\medskip

To prove the other direction, suppose that $S'= T'\cup R'$ is a minimum TC-set of $F$ where $T'= S' \cap V(F)$ and $R'=S' \cap E(F)$. Now, determine the set $Q^* \subseteq V(G^*)$ as $Q^*=T' \cup R^*$ where $R^*$ consists of those vertices $x_i^1$ which represent the edges $e_i \in  R'$ that is,
$$R^*=\{x_i^1: N_{G^*}(x_i^1) \in R'\}.$$

Hence, we have $|Q^*|= |S'|= tc(F)$. Observe that, by the definition of the underlying hypergraph, the transversal $T'$ of $F$ dominates all vertices in $V(G) \setminus V(F)$ in the graph $G^*$. Moreover, each vertex $u$ of $V(F)$ which is outside $T'$ is covered by an edge $e_i \in R'$ in the underlying hypergraph and, therefore, $u$ is dominated by the vertex $x_i^1 \in R^*$ in $G$. We conclude that $Q^*$ is a dominating set of $G^*$ and we have $\gamma(G^*) \le |Q^*| =tc(F)$. This finishes the proof of the theorem.
\end{proof}

Theorem~\ref{thm:char-gamma-k}, Theorem~\ref{thm:TC-extr}, and Corollary~\ref{cor:forbid-F} immediately imply  other formulations of the characterization.

\begin{corollary}
	\label{cor:TC-extr}
For each $k \ge 3$, a connected bipartite graph $G$ is a $(\gamma, \gamma_k)$-graph if and only if $G \in {\cal B}_k$ and the underlying hypergraph $F$ of $G$ satisfies the following property: for every $\ell \le k-1$ and for every $\ell$  edges $e_{i_1}, \dots, e_{i_\ell}$ of $F$, the union $L=\bigcup_{j=1}^\ell e_{i_j}$ contains at most $\ell+k-2$ weakly independent vertices.
\end{corollary}

\begin{corollary}
	\label{cor:forbid-gamma-k}
	For each $k \ge 3$, a connected bipartite graph $G$ is a $(\gamma, \gamma_k)$-graph if and only if $G \in {\cal B}_k$ and the underlying hypergraph of $G$ is ${\cal H}_k$-free.
\end{corollary}

For the case of $k=3$, Corollary~\ref{cor:TC-extr} can be reformulated as follows.
\begin{corollary}
	\label{cor:2-2Prop}
A connected bipartite graph $G$ is a $(\gamma, \gamma_3)$-graph if and only if $G \in {\cal B}_3$ and the underlying hypergraph $F$ satisfies the following property:
\begin{itemize}
\item[$(\star)$] Every four different vertices that can be split into two pairs of adjacent vertices induce at least one hyperedge in $F$.
\end{itemize}
	\end{corollary}
 \begin{prop}
 If $G$ is a connected bipartite $(\gamma, \gamma_3)$-graph and $D$ is a minimum $3$-dominating set of $G$, then  $G^2[D]$ is a threshold graph.
\end{prop}
 \begin{proof}
 Suppose that $G$ and $D$ satisfies the conditions of the proposition and consider the underlying hypergraph $F$ with $V(F)=D$. Since $G \in {\cal B}_3$, any two vertices $u,v\in D$ having distance $2$ in $G$ belong to a common hyperedge of $F$. In other words, if $uv \in E(G^2[D])$, then $uv$ is an edge in the $2$-section graph $H$ of $F$. Similar argumentation shows that the other direction is valid, too. Thus, $H=G^2[D]$.
By Corollary~\ref{cor:2-2Prop}, the underlying hypergraph $F$ satisfies $(\star)$. This implies that  for every two vertex disjoint edges $vv'$ and $uu'$ of $H$, the vertex set $\{v,v',u,u'\}$ contains at least one hyperedge of $F$. We infer that there exist three vertices in $\{v,v',u,u'\}$ which induces a triangle in $H$. Checking all possible extensions of a $2K_2$ on four vertices, we obtain that $2K_2$, $P_4$ and $C_4$ are the forbidden induced subgraphs. Since  $(2K_2, P_4, C_4)$-free graphs are exactly the threshold graphs, the statement follows.
 \end{proof}

\section{Characterization of Bipartite $( \gamma, \gamma_3 )$-perfect Graphs} \label{sec:4}

In this section, we prove a characterization for bipartite $( \gamma, \gamma_3 )$-perfect graphs. First, we define the graph classes $\mathcal{S}_k$ and prove that all members of $\mathcal{S}_k$ are $(\gamma, \gamma_k)$-perfect  for every $k\ge 2$. Then, the main theorem of this section states that $\mathcal{S}_3$ is the set of all bipartite $(\gamma, \gamma_3)$-perfect graphs.

\begin{definition}
	\label{def:Sk}
  Let $k, r, i_1, \dots, i_r$ be integers which satisfy  $k \ge 2$, $r \ge 1$, and $i_j \ge k$ for every $j \in [r]$. We define $ S_k(i_1, \dots, i_r) $ as the graph that is obtained from the star $K_{1,r}$ in the following way. First replace the edges $e_1, \dots, e_r$ of the star with $i_1, \dots, i_r$ parallel edges, respectively, and subdivide each edge exactly once. Finally, supplement the graph by $k-2$ new vertices which are false twins with the center. For a fixed integer $k \ge 2$, let $ \mathcal{S}_k $ be the graph class that contains  all  $S_k(i_1, \dots, i_r) $ with $r \ge 1$ and $i_j \ge k$ for every $j \in [r]$.
\end{definition}
With this notation, $S_2(\ell)$ corresponds to $K_{2, \ell}$ whenever $\ell \ge 2$. In general, $S_k (\ell)$ is just $K_{k, \ell}$ if $\ell \ge k \ge 2$.  For another example that is not a complete bipartite graph see Fig.~\ref{fig:S3}.

	\begin{figure}[t]
\centerline{\includegraphics[width=0.5\textwidth]{figureC_v2.eps}}
	\captionsetup{width=0.8\textwidth}
	\centering
	\caption{\protect An illustration for the graph $ S_3(3,3,4) $. The thick red box represents two false twin vertices.}
	\label{fig:S3}
\end{figure}

\begin{prop}
	\label{prop:Sk}
	If $ G \in \mathcal{S}_k $, then $ G $ is ($ \gamma,\gamma_k $)-perfect.
\end{prop}
\begin{proof}
	Consider a graph $G=S_k(i_1,\dots, i_r)$ from the graph class $\mathcal{S}_k$.  It is straightforward to check that the minimum vertex degree is $k$ and that $\gamma_k(G)= r+k-1$ and $\gamma(G)=r+1$ hold. Alternatively, the equality $\gamma_k(G)=\gamma(G)+k-2$ can be verified directly by Theorem~\ref{thm:char-gamma-k}. Thus, $G$ is a $(\gamma, \gamma_k)$-graph. Note that every subdivision vertex has degree $k$. Now, consider an induced subgraph $H$ of $G$ that satisfies  $\delta(H) \ge k$. Suppose that a vertex $u \in V(G)$ does not belong to $V(H)$. If $u$ is the center or its false twin  then, since $\delta(H) \ge k$,   no subdivision vertices can be present in $H$. This contradicts the degree condition.  If $u$ is a leaf in the original star then,  by the degree condition again, no neighboring subdivision vertices belong to $H$. If $u$ is a subdivision vertex, then consider the neighbor of $u$ which was a leaf in the star, say $u'$, and denote the degree $d_G(u')$ by $i_j$. When $u$ does not belong to $H$, we have two cases: either $u'$ and at least $k$ of its neighbors are still present in $H$; or  the entire $N_G[u']$ is omitted from $H$. We infer that every induced subgraph $H$ of $G$ with $\delta(H) \ge k$ is isomorphic to a graph $S_k(i_1',\dots , i_{r'}')$ where  $i_j' \ge k$ for all $j' \in [r']$. Therefore, $H$ is a $(\gamma, \gamma_k)$-graph again. This proves the $(\gamma, \gamma_k)$-perfectness of $G$.
\end{proof}

As proved in \cite{manu}, ${\cal S}_2$ is the set of all  $(\gamma, \gamma_2)$-perfect graphs. Since each ${\cal S}_k$ contains only bipartite graphs, it is equivalent to saying that ${\cal S}_2$ is the set of all bipartite $(\gamma, \gamma_2)$-perfect graphs. Here we  prove an analogous statement for ${\cal S}_3$.

\begin{thm} \label{thm: perfect-3}
	$ G $ is a bipartite $( \gamma,\gamma_3 )$-perfect graph if and only if $ G \in \mathcal{S}_3 $.
\end{thm}
\begin{proof}
	By Definition~\ref{def:Sk} and Proposition~\ref{prop:Sk}, every member of  $\mathcal{S}_3$ is  $(\gamma, \gamma_3)$-perfect and bipartite. To prove the other direction, suppose that $G$ is a bipartite $( \gamma,\gamma_3 )$-perfect graph, $D$ is a minimum $3$-dominating set and $F$ is the $3$-uniform underlying hypergraph of $G$ (with respect to $D$). Under these conditions, we first prove a couple of claims on the structure of $F$ and $G$.

 \begin{unnumbered}{Claim~A.}
 		Every edge of $F$ contains a vertex of degree $1$.
 \end{unnumbered}
\noindent
\textit{Proof of Claim A.} \enskip Suppose, to the contrary, that $e_1=\{u,u',u''\}$ is an edge in $F$ such that $d_F(u)\ge d_F(u')\ge d_F(u'')\ge 2$.	Let $Y_4$ be the set of vertices of degree at least $4$ in $V(G)\setminus D$ and let $Z= (N_G(u) \cap N_G(u') \cap N_G(u'')) \setminus \{x_1^1\}$.

Consider the induced subgraph $H=G[V(G)\setminus (Y_4 \cup Z) ]$ of $G$. Since $G$ is bipartite and no vertex from $D$ was deleted, every vertex $y \in  V(H)\setminus D$ remains of degree $3$ in $H$. If $v$ is a vertex from $D$ such that  $d_F(v)=1$, then $v$ cannot be contained in a clique of order at least $4$ in $F$ and, by Corollary~\ref{lem:2}, $v$ cannot be adjacent to any vertex from $Y_4$.
By our assumption, $e_1$ does not contain the degree-1 vertex $v$ and therefore, $v$ is adjacent to none of the vertices in $Z$. Thus, $d_F(v)=1$ implies $d_{H}(v)=d_G(v) \ge 3$.  Now, consider a vertex $v \in D$ that satisfies $d_F(v) \ge 2$. This vertex $v$ is incident with at least two different edges in $F$, say $v$ is incident with $e_i$ and $e_j$. Observe that $d_G(v) \ge 4$ as  $v$ is adjacent to all vertices from $X_i\cup X_j$. Since $Y_4 \cup Z$ contains at most one vertex, namely $x_1^2$, from $X_i\cup X_j$, the degree of $v$ is still at least $3$ in $H$. We conclude $\delta(H)=3$.

It is straightforward to show that, under our assumptions, $(D\cup \{x_1^1\}) \setminus \{u, u', u''\}  $ is a dominating set of $H$ and hence we have $\gamma(H) \le |D|-2$. We now prove that $\gamma_3(H)\ge |D|$ that  gives the desired contradiction. Consider an arbitrary $3$-dominating set $Q$ in $H$. Denote $|D\setminus Q|$ by $s$ and $|Q\cap (V(H) \setminus D)|$ by $\ell$. To $3$-dominate all vertices in $D\setminus Q$, we need at least $3s$ edges between the vertex sets $D\setminus Q$ and $Q\cap (V(H) \setminus D)$. On the other hand, we may have at most $3\ell$ edges between the two sets because every vertex in the second set is of degree $3$. Consequently, $\ell \ge s$ holds and this implies
$$|Q|=|Q\cap D|+ |Q\cap (V(H) \setminus D)|= |D|-s+\ell \ge |D|$$
for every $3$-dominating set $Q$ of $H$. Therefore, we have $\gamma_3(H)\ge |D| >\gamma(H) +1$ for an induced subgraph of minimum degree $3$ that contradicts the $(\gamma, \gamma_3)$-perfectness of $G$. This contradiction finishes the proof of Claim A.

\medskip
\begin{unnumbered}{Claim~B.}
	Every vertex from $V(G) \setminus D$ is of degree $3$ in $G$.
\end{unnumbered}
\noindent
\textit{Proof of Claim B.} \enskip By Claim A, there is no complete subhypergraph in $F$ on four (or more) vertices. Together with Corollary~\ref{lem:2} these imply that there exist no vertices of degree greater than $3$ in $V(G)\setminus D$. Since $\delta(G) \ge 3$, this results in $d_G(y) = 3$ for every $y \in V(G)\setminus D$ as stated.

\medskip
\begin{unnumbered}{Claim~C.}
 If three vertices form an edge in the underlying hypergraph $F$, then they have at least three common neighbors in $G$.
\end{unnumbered}
\noindent
\textit{Proof of Claim~C.} \enskip By Claim~A, every edge $e=\{u,u',u''\}$ of $F$ contains a vertex of degree 1. We may assume that $u$ is such a vertex in $e$. By the definition of the underlying hypergraph and by Claim B, all neighbors of the degree-1 vertex $u$ are associated with the edge $e$ that is,
$$N_G(u)= \{y \in V(G)\setminus D: N_G(y)=\{u,u',u''\} \}.$$
Since $|N_G(u)|=d_G(u) \ge 3$, there exist at least three common neighbors of $u$, $u'$, and $u''$.

\medskip
\begin{unnumbered}{Claim~D.}
	$F$ does not have two edges that share exactly one vertex.
\end{unnumbered}
\noindent
\textit{Proof of Claim~D.} \enskip
First suppose that two edges, namely $e_i$ and $e_j$, of $F$ share exactly one vertex $w$. Let  $e_i=\{u,u',w\}$ and $e_j = \{v,v', w\}$.  By Claim~C  $u$, $u'$ and $w$ have at least three common neighbors, denote them by $x_i^1$, $x_i^2$ and $x_i^3$. Similarly, let the common neighbors of $v$, $v'$ and $v''$ be $x_j^1$, $x_j^2$ and $x_j^3$. Consider the subgraph $H$ of $G$ induced by these $11$ vertices. Observe that $H$ has minimum degree $3$,  $\gamma_3({H})=5$ and, as $w$, $x_i^1$, $x_j^1$ form a dominating set,  $\gamma(H) \le 3$. Thus, we have $\gamma_3(H) > \gamma(H)+1$ for a connected induced subgraph of $G$ with minimum degree 3. This contradicts our condition on the $(\gamma, \gamma_3)$-perfectness of $G$. Consequently, $F$ cannot contain two edges intersecting in exactly one vertex.

\medskip

Concerning the statement and proof of Claim~D, remark that even if $e_i \cup e_j$ contains further edges in $F$, the considered subgraph $H$ is an induced subgraph of $G$.

\medskip

To complete the proof of Theorem~\ref{thm: perfect-3} we make the following observations. If $F$ contains only one edge, then $G$ is a $K_{3,\ell}$ that is isomorphic to $S_3(\ell)$ for an integer $\ell \ge 3$. So, we may assume that $|D|\ge 4$ and $F$ contains at least two edges. Since $G$ is a connected bipartite graph and $D$ is one of the partite classes, the underlying hypergraph $F$ must be connected as well. Thus, $F$ has two intersecting edges $e_i$ and $e_j$. By Claim~D, $e_i$ and $e_j$ share two vertices, say $v_0^1$ and $v_0^2$.  Claim~A implies that the third vertex of $e_i$ is of degree $1$ and the same is true for $e_j$. The connectivity of $F$ then requires that each edge of $F$ is incident with  $v_0^1$ and $v_0^2$ and contains a further private vertex. Therefore, the underlying hypergraph $F$ of a   $(\gamma,\gamma_3)$-perfect $G$ can always be obtained (up to isomorphism) in the following way:
$$V(F)=\{v_0^1, v_0^2, v_1, \dots , v_r\};$$
$$E(F)=\{e_1, \dots , e_r\} \quad \mbox{where} \quad e_i=\{v_0^1, v_0^2, v_i\} \enskip \mbox{for every $i \in [r]$.}$$
Then, $G$ can be constructed by assigning at least three, say $i_j$, new vertices to each edge $e_j$ of $F$  and making adjacent the vertices in $e_j$ to the new associated vertices. This clearly results in the graph $S_3(i_1,\dots, i_r) $ with $i_j\ge 3$ for all $j \in [r] $. Note that the condition $\delta(G)\ge 3$ implies that $F$ is not edgeless and consequently, $r \ge 1$ must hold. We conclude that every $(\gamma, \gamma_3)$-perfect graph belongs to $\mathcal{S}_3$.
\end{proof}
\medskip

We show that the statement analogous to Theorem~\ref{thm: perfect-3} holds neither for $k=4$ nor for any even integer $k>4$. First, consider the lexicographic product $C_6 \circ \overline{K}_{2}$ that is, every vertex of the cycle $C_6$ is replaced with $2$ independent vertices. Then, $\gamma_4(C_6 \circ \overline{K}_{2})=6=\gamma(C_6 \circ \overline{K}_{2})+2$ and, since the graph is $4$-regular, every proper subgraph has a vertex of degree smaller than $4$. Thus, $C_6 \circ \overline{K}_{2}$ is $(\gamma, \gamma_4)$-perfect and bipartite but does not belong to $\mathcal{S}_4$. As an infinite class of similar examples we propose the following construction.

\vspace{0.3cm}\noindent\textbf{Example.} For each even integer $k=2\ell \ge 4$ consider the bipartite graph $G_k \in G_k^*$ obtained as the double incidence graph of the following underlying hypergraph $F_k$. The vertex set of $F_k$ is $V=W \cup U$ where $W=\{w_1, \dots, w_\ell\}$ and $U=\{u_1, \dots, u_{\ell+1}\}$; the edge set is $E=\{e_1, \dots , e_{\ell+1}\}$ where $e_i = V \setminus \{u_i\}$ for all $i\in [\ell+1]$. One can check that $F_k$ satisfies the condition given in Corollary~\ref{cor:TC-extr} and, therefore, $G_k$ is a $(\gamma, \gamma_k)$-graph. On the other hand, one can check that every proper subgraph of $G_k$ has minimum degree less than $k$. We may conclude that $G_k$ is a bipartite $(\gamma, \gamma_k)$-perfect graph but does not belong to $\mathcal{S}_k$.

\section{Complexity Results} \label{sec:5}

\begin{prop}
	Let $k$ be a fixed integer with $k \ge 3$ and let $G $ be a bipartite graph with $\Delta(G) \ge k$. It can be decided in polynomial time whether the graph $ G $ satisfies the equality $ \gamma_k(G) = \gamma(G) +k-2$.
\end{prop}

\begin{proof} If $G$ is a connected bipartite $(\gamma, \gamma_k)$-graph with $\Delta(G) \ge k$, then, by Theorem~\ref{thmPartite}, every minimum $k$-dominating set is a partite class of $G$ and $G \in {\cal B}_k$. This can be checked efficiently and, assuming that the conditions are satisfied, the $k$-uniform underlying hypergraph $F$ can be determined in polynomial time as well. Then, by Corollary~\ref{cor:forbid-gamma-k}, it is enough to decide whether $F$ is ${\cal H}_k$-free. Recall that ${\cal H}_k$ is a finite set of $k$-uniform hypergraphs each of which is of order at most $k(k-1)$. Thus, this step can also be performed in polynomial time.

Finally, note that if $G$ is disconnected, it satisfies $ \gamma_k(G) = \gamma(G) +k-2$, if and only if, $G$ contains one component which is a $(\gamma,\gamma_k)$-graph and the further components are isolated vertices. Hence, the statement remains true over the class of disconnected bipartite graphs.
\end{proof}

\begin{thm}
	For each $ k \geq 2 $, it is NP-hard to decide whether the equality $ \gamma_k(G) = \gamma(G)+k-2 $ holds for $G$ over the class  $ {\cal G}_k$.
\end{thm}

\begin{proof}
In order to prove that the corresponding decision problem is NP-hard, we construct a polynomial time reduction from 3-SAT problem, a classical NP-complete problem \cite{Garey1979}. Note that we adopt the approach we used in \cite{manu} to prove NP-hardness. Since the construction we need here is more general, we only present the construction explicitly and give a sketch of the rest of the proof, for the brevity.

\begin{figure}[t]
	\centerline{\includegraphics[width=1.2\textwidth]{construction_v1.eps}}
	\captionsetup{width=0.8\textwidth}
	\centering
	\caption{\protect An illustration of the construction for $ 3 $-SAT reduction: The clauses $ C_1 $ and $ C_\ell $ corresponding to the vertices $ c_1 $ and $ c_\ell $, resp., are $ C_1=(x_1 \vee \neg x_3 \vee \neg x_s) $ and  $ C_\ell=(x_1 \vee \neg x_2 \vee x_s) $. The thick red box represents $k-1$ independent vertices with the same open neighborhood.}
	\label{fig:construction}
\end{figure}
Let $ \mathcal{C} $ be a $ 3 $-SAT instance with clauses $ C_1,C_2,\dots, C_\ell $ over the Boolean variables $ X=\{x_1,x_2, \dots, x_s\} $. We may assume that for every three variables $ x_{i_1}, x_{i_2}, x_{i_3} $ there exists a clause $ C_j $, where $ j\in [\ell] $, such that $ C_j $ does not contain any of the variables $ x_{i_1}, x_{i_2}, x_{i_3} $ (neither in positive form, nor in negative form). Otherwise, the problem could be reduced to at most eight (separated) $ 2$-SAT problems, which are solvable in polynomial time.

We construct a graph $ G \in  {\cal G}_k$, such that the given instance $ \mathcal{C} $ of $ 3 $-SAT problem is satisfiable if and only if $ \gamma(G) < \gamma_k(G)-k+2$.

For every variable $ x_i $, we create three vertices $ \{x_i^t, x_i^f, v_i\} $ and then we add the edges $ x_i^tv_i $ and $ x_i^fv_i $. For every clause $ C_j \in \mathcal{C}$, we create a vertex $ c_j $, and if $ x_i $ is a literal in $ C_j $, then $ x_i^tc_j \in E(G) $; if $ \neg x_i $ is a literal in $ C_j $, then    $ x_i^fc_j \in E(G) $. Moreover, we add a vertex $ c^* $ and the edges $ c^*x_i^t  $ and $  c^*x_i^f $ for every $ i\in [s] $. We also add a vertex $ v_{s+1} $ and the edge set $  \{c_iv_{s+1}:1\le i\le\ell\} \cup \{c^*v_{s+1}\}$.  Finally, we add $ k-1 $ new vertices $ v^1_0, \dots, v^{k-1}_0$, such that each vertex $ v^r_0 $ is adjacent to every vertex in $ V(G)\setminus \{v_1,v_2\dots,v_{s+1}\} $ for $ r \in [k-1] $ (for an illustration of the construction, see Fig.~\ref{fig:construction}). The order of $ G $ is obviously $ 3s+\ell +k+1 $ and this construction can be done in polynomial time. Note that $ G \in {\cal G}(F)$ and any two hyperedges in $ F $ share $ k-1 $ vertices, namely $ v_0^1, \dots, v_0^{k-1} $. It is straightforward to prove that $\{v_1, \dots, v_{s+1}, v_0^1, \dots, v_0^{k-1}\}$ is a minimum $k$-dominating set in $G$ and, therefore, $\gamma_k(G)=k+s$.

In order to finish the proof, it suffices to show that $ \mathcal{C}  $ is satisfiable if and only if $ \gamma(G) < \gamma_k(G)-k+2 $. First, assume that $ \mathcal{C} $ is satisfiable and let $ \varphi: X \rightarrow \{t,f\} $ be  the corresponding truth assignment. The set $ D' = D_1 \cup D_2 \cup \{c^*\} $ is a dominating set, where  $D_1=\bigcup_{i\in [s] }\{x_i^t :\varphi(x_i)=t\} $ and  let $D_2 = \bigcup_{i\in [s] }\{x_i^f : \varphi(x_i)=f\}$. Since $ |D'|=s+1 $, we have $ \gamma(G) < \gamma_k(G)-k+2=(k+s)-k+2 $ for $ k\geq 2 $.

For the other direction, assume that $ \gamma(G) < \gamma_k(G)-k+2 $ and consider a dominating set $D'$ such that $ |D'|\leq s+1 $. In order to dominate $ v_i $, the set $ D' $ contains at least one vertex from the set $ \{x_i^t, x_i^f, v_i\} $, for each $ i\in [s] $. Similarly, to dominate $ v_{s+1} $, the set $ D' $ contains at least one vertex from the set $ \{c_1,c_2,\dots,c_\ell,c^*,v_{s+1}\}$. Since $ |D'|\le s+1 $, we have $|D' \cap \{x_i^t, x_i^f, v_i\}| = 1 $ for every $ i\in [s] $. Moreover, $ |D'\cap \{c_1,c_2,\dots,c_\ell,c^*,v_{s+1}\}| = 1$ and $ \{v_0^1, \dots, v_0^{k-1}\} \cap D' = \emptyset $. There are three cases to consider and for a detailed discussion, we refer the reader to \cite{manu}.
\begin{itemize}
	\item[$ (i) $] If $ v_{s+1} \in D' $ then, to dominate $x_i^t$ and $x_i^f$ for each $i \in [s]$, every $v_i$ must be contained in $D'$. Hence,  the vertices $ \{v_0^1, \dots, v_0^{k-1}\} $ are not dominated by a vertex from $ D' $, a contradiction.
	\item[$ (ii) $] If $ c_j \in D' $ for some $ j \in [\ell] $, then for all but at most three $i \in [s]$ we have  $v_i \in D'$. Then, there is a vertex $ c_q $, such that $ q \in [\ell] $ and $ c_q $ is not dominated by a vertex from $ D' $, a contradiction.
	\item[$(iii) $] If $ c^* \in D' $, then $c_j$ must be dominated by a vertex $x_i^t$ or $x_i^f$ for every $j \in [\ell]$. Now, consider the truth assignment $ \varphi: X \rightarrow \{t,f\} $  where $\varphi(x_i)=t$ if and only if $x_i^t \in D'$ and observe that $\varphi$ satisfies the 3-SAT instance $ \mathcal{C} $.
\end{itemize}
This finishes the proof.
\end{proof}

\section*{Acknowledgements}
  This research was started when the second author visited the Ege University in Izmir, Turkey; the authors thank the financial support of T\" UBİTAK under the grant  BİDEB 2221 (1059B211800686). The second author also acknowledges the financial support from the Slovenian Research Agency under the project N1-0108.

\bibliographystyle{siamnodash}

redpajama_set_name: RedPajamaArXiv


text:
Home CARIBBEAN COMMUNITY Former Jamaican Prime Minister Hails Kofi Annan as "a Visionary"
Former Jamaican Prime Minister Hails Kofi Annan as "a Visionary"
Says the Caribbean Will Remember him for his "Constant Attention"
KINGSTON, Jamaica – Former Prime Minister of Jamaica, P. J. Patterson has hailed the late former UN Secretary General, Kofi Annan Kofi Annan, as a visionary who conceived the Millennium Development Goals as the engine for moving forward the international society in its entirety.
Patterson, in tribute to Annan who died last month, said he was a man of "unique and yet many distinctions".
"He was the first and to date the only person who passed through the labyrinth of the United Nations and its many agencies to ascend to the post of Secretary General by virtue of his brilliance at every working stage; and the first son of subSaharan Africa to become chief of that global institution."
PJ Patterson (Stabroek News photo)
In a statment on Friday, Patterson noted that Kofi Annan perceived the U.N. as a catalyst for change and an instrument for human progress.
"His style was one of quiet but resolute diplomacy which would forestall conflict whenever possible and be ready to respond promptly and effectively whenever tension could not be defused or conflict averted."
Patterson said the Caribbean will remember Annan for the "constant attention he paid to our severe resource constraints and vulnerability to a spate of natural disasters."
"His passionate advocacy was highlighted in the Barbados Programme of Action that the UN Conference on the Sustainable Development of Small Island Developing States adopted at the start of his incumbency.
"For this urbane and astute Secretary General, the concept of retirement meant primarily a transfer of his focus and attention from his New York Headquarters to roam the continent of his descent and broken agreements which would foster harmony and economic growth. As Chairman of the Group of Elders, he sought to rid Africa of the dogmas of the past in order to embrace pragmatic progress in working democracies under the rule of law."
The former Prime Minister said the rich legacy of the late former UN Secretary general must inspire "a new generation of men and women to carry on the relentless struggle into a new era when the hopes and aspirations of their martyrs and heroes will be realized once and for all."
"Mine was the privilege of having known Kofi throughout the years and the honour of a productive working relationship which grew into a close brotherhood.Kofi Anan has now joined our noble ancestors and entered that realm of eternal rest and enjoy perpeptual peace at last."
Kofi Annan, the first black African to serve as Secretary General of the UN was originally from Ghana.
He died in hospital in Switzerland following a brief illness. – CMC
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text:
The Academy of Sacred Drama's 2019–2020 season features the work of Antonio Gianettini (1648–1721): the once-celebrated composer of Baroque oratorio and opera whose name is now almost entirely forgotten in the world of music.
Join us for the modern premiere of Gianettini's 1690 oratorio
Encounter rare gems of the Baroque violin repertoire curated by the Academy artistic director, Jeremy Rhizor
Experience the American premiere of Gianettini's 1687 oratorio.
Oratorio performances and salon concerts feature the music and libretti of works rarely or never heard in modern times accompanied by lectures and refreshments. Stories from Scripture and the lives of the saints are imagined through seventeenth and eighteenth century perspectives with historically-informed singing and instruments styled after those of the Baroque era.
Oratorio performances and salon concerts are ticketed public events in New York City and White Plains, NY, and other performances are produced in conjunction with partner organizations and presenters in other cities.