text:
Q: Can you use a vaadin-button in upload? All the Upload examples show using NativeButton when programmatically creating an Upload. I want to use a Button (vaadin-button) that is part of a Designer layout. When I try this I see the following error:
Caused by: java.lang.IllegalArgumentException: Trying to detach an element from parent that does not have it.
    at com.vaadin.flow.dom.impl.AbstractNodeStateProvider.removeChild(AbstractNodeStateProvider.java:123)
    at com.vaadin.flow.dom.Node.removeChild(Node.java:376)
    at com.vaadin.flow.dom.Element.removeFromParent(Element.java:577)
    at com.vaadin.flow.dom.Node.insertChild(Node.java:293)
    at com.vaadin.flow.dom.Node.appendChild(Node.java:141)
    at com.vaadin.flow.component.upload.GeneratedVaadinUpload.addToAddButton(GeneratedVaadinUpload.java:1344)
    at com.vaadin.flow.component.upload.Upload.setUploadButton(Upload.java:276)

We are using Vaadin 14 in CDI mode.

A: To change the upload button with Designer, you need to drag the button into the upload in the hierarchy, then add slot="add-button" attribute.
Your templates should end up looking like this:
<vaadin-upload>
 <vaadin-button slot="add-button">
   Button
 </vaadin-button>
</vaadin-upload>

Note that you must do this in the hierarchy panel.

If this does not help (maybe you did this already), could you perhaps include some code that shows what you are trying to do?

A: I've just created an issue for this improvement in our Designer repository https://github.com/vaadin/designer/issues/2399.
Feel free to comment.


A.B. Paterson College has a wonderful history of high academic achievement and an outstanding commitment to the education and development of young people since its Foundation in 1991.The College is focused on meeting the contemporary needs of our current and future students, preparing them for the world they live in, and providing our society with the leaders of tomorrow, whilst preserving the Australian heritage of the College.
We aim to provide opportunities throughout the College, and in all areas of endeavour, for developing the character and leadership of our young men and women through their active participation in a wide range of educational experiences. These opportunities are provided by a committed and professionally focused staff, who model the College's Motto of Excellence, Care and Commitment in all that they do within the College. A culture of leadership, sense of community, integrity and service are the integral elements of our Key Focus Areas.


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[Discussion] Guitar Smashing and Instrument Destruction
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By ritalin98, March 12 in General Chat
ritalin98 0
This is sort of a poll, I guess.
I always thought everyone just kinda went crazy for instrument destruction during a show, until I saw all these comments on YouTube videos for bands like The Dillinger Escape Plan, even Deep Purple, Jimi Hendrix, The Who, and Nirvana, all legends that all of a sudden people are trashing and calling "overrated and terrible" just because they destroyed an inanimate object. "It's so disrespectful, give it to me" all that whiny BS.
I vote badass. I understand why some don't like it, but that's a part of the point. And overall, people love it precisely because it's disrespectful and brutal. It's fun to riot and come on, who doesn't like to destroy shit, or watch shit get destroyed? That's why most people like it. It satisfies an aggressive desire for sadism in a typically high adrenaline/high testosterone environment. A random and epic outburst of violence.
I think it's a bit different today - I can't imagine artists like Jimi Hendrix or Pete Townsend smashing instruments if they came about today, it's floated up the genre ladder as rock gradually produced more aggressive genres. I would expect to see a guitar smashed in a fit of adrenaline at a hardcore show, maybe not a blues or a folk rock show, for example...though it's all over the place.
Off the top of my head, The Who, Jimi Hendrix, Nirvana, Kiss, Metallica, Saxon, Yngwie Malmsteen, The Yardbirds, Skid Row, Nine Inch Nails, Deep Purple, Krokus, Warbeast, Dillinger Escape Plan, The Clash, The Misfits, ad infinitum, made a habit of doing this. They're still awesome musicians anyways, and anyone trashing them because they destroy their own property for entertainment is being a pussy killjoy, imo.
But I can't stand people that whine and say it's "cliché". Um, yeah, that's because it's almost as old as the guitar solo...showmanship, specifically instrument destruction, began in '63 with The Who (though with a few smaller, isolated acts before then) - that's going on 60 years, more than half a century, of instrument destruction being an oft practiced staple across the rock music scene and its many, many sub-genres. Millions of bands across the word, small local ones and big, arena-selling ones have done it. It's badass. It's fun. It's not supposed to be some artsy expression that "was only acceptable when Jimi did it" - it's fun to destroy shit. Its awesome and fun and it always will be.
Anyone else wanna offer their take? Is it badass or just bad?
ChainsawAkimbo 176
I've done it once, was much younger and drunk.. and wasn't my instrument.. 😅
I do think it's really a worn out cliche. Not once have i seen someone smash an instrument and thought it was cool or badass. To me it's disrespecting the instrument, and every person who would love to own a guitar as good but just can't afford one. That said everyone is entitled to do whatever the hell they want with their stuff, just doesn't make you any cooler. In my books makes someone a bit of an idiot. I know i was back when i did it. I've never been huge on showmanship either, quite possibly because i hate all kinds of pretending. I prefer that the music does the talking. Everything else around it to me is pointless nonsense.
Requiem 1,095
The True Requiem
LocationSouthern Darkness, Australia
I've been to hundreds of gigs and I can't recall anyone doing this. I think it sort of reached its apex in the 70s and slowly dwindled from there.
RelentlessOblivion 1,600
Lovecraft's Horror
LocationThe Great Southern Land
The smashing of instruments had its time. You don't see it often these days and I'm thankful for that. Not once in the decade of playing guitar have I thought smashing one would be fun or cool. It's just a played out trope, Spinal Tap shit, and there are more lasting, impactful ways to leave an impression on your audience.
MaxFaust 37
When Jimi Hendrix did it I think it was between him and the guitar. They had a "special relationship" that almost certainly included some hate. Pete Townsend was the certified bad boy of his time. The Who were always the worst. They had a reputation to keep up. The first time I saw someone smash up their guitar on stage was in 1980. It was Paul Stanley of Kiss. I even secured a part as he was tossing them into the audience. But it was of course not his good guitar. Everybody could see that.
My thoughts on this is that if they are your instruments, they are yours to do whatever you want with, including smashing them up. However, I fail to see what's "badass" about destroying your own property just for the hell of it. If it's a part of some greater show of message, sure, why not, but the general rule of thumb is that musicians should be careful with theatrical antics. It doesn't always play out as they had imagined it would.
The Who were shockers, they just destroyed any and everything they came into contact with. I get destroying instruments for them and Hendrix surely had a love/hate relationship with the guitar. I hate seeing it these days.
On 3/16/2019 at 1:00 AM, Requiem said:
It's a specific type of theatric. Does everyone do it? No. Even in the 70s, only a handful of bands were noted to have done it, because they were majorly succesful
But it's a fairly common practice, and has remained so since the 60s
FatherAlabaster 3,444
Give Me All Your Coffee
Location: In The Canal
On 3/16/2019 at 5:32 AM, MaxFaust said:
the general rule of thumb is that musicians should be careful with theatrical antics. It doesn't always play out as they had imagined it would.
My favorite example of this is a local show I went to about fifteen years ago where a guitarist tried and utterly failed to break his guitar in any satisfying way. It was a non-sequitur at that show to begin with, but it all came together for me as he impotently rammed the headstock down into the floor over and over with a "please, please break" expression on his face.
Tortuga 83
LocationSomewhere in Dalarna, Sweden
I have been to more than my fair share of gigs, too and no, I cannot recall any imstruments being smashed up on stage either, or that I simply found it so unimpressiv that I have forgotten all about it.
As to how I feel about it: punk rock, hard rock and metal is energy to me and especially live, it makes me feel high. My way of absorbing this energy feels like entering another realm of exsistance...and I do this in a quiet way. For others this energy might make them feeling very energetic and some what aggressive...so rather take it out on some guitar or drumset than anything else. So yes, if this is the case... granted, I am not a musician myself... then fine by me.
And then there are instruments...and instruments: I assume that most people would be pretty pissed off if someone smashed up a Stradivarius violin, but more relax about the same happening to a mass produced instrument.
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Vltimas 'Something Wicked Marches In'
MacabreEternal posted a blog entry in Album Reviews, April 3
https://www.metalforum.com/blogs/entry/52-vltimas-something-wicked-marches-in/
MacabreEternal, April 3
Queensryche "The Verdict"
MacabreEternal posted a blog entry in Album Reviews, March 10
https://www.metalforum.com/blogs/entry/50-queensryche-the-verdict/
MacabreEternal, March 11
Candlemass "The Door to Doom"
MacabreEternal posted a blog entry in Album Reviews, February 24
https://www.metalforum.com/blogs/entry/48-candlemass-the-door-to-doom/
MacabreEternal, February 24
Overkill "The Wings of War"
Full length number 19 from overkill certainly makes a splash in the energy stakes, I mean there's some modern thrash bands that are a good two decades younger than Overkill who can only hope to achieve the levels of spunk that New Jersey's finest produce here. That in itself is an achievement, for a band of Overkill's stature and reputation to be able to still sound relevant four decades into their career is no mean feat. Even in the albums weaker moments it never gets redundant and the energy levels remain high. There's a real sense of a band in a state of some renewed vigour, helped in no small part by the addition of Jason Bittner on drums. The former Flotsam & Jetsam skinsman is nothing short of superb throughout "The Wings of War" and seems to have squeezed a little extra out of the rest of his peers.
The album kicks of with a great build to opening track "Last Man Standing" and for the first 4 tracks of the album the Overkill crew stomp, bash and groove their way to a solid level of consistency. The lead work is of particular note and Blitz sounds as sneery and scathing as ever. The album is well produced and mixed too with all parts of the thrash machine audible as the five piece hammer away at your skull with the usual blend of chugging riffs and infectious anthems.
There are weak moments as mentioned but they are more a victim of how good the strong tracks are. In it's own right "Distortion" is a solid enough - if not slightly varied a journey from the last offering - but it just doesn't stand up well against a "Bat Shit Crazy" or a "Head of a Pin". As the album draws to a close you get the increasing impression that the last few tracks are rescued really by some great solos and stomping skin work which is a shame because trimming of a couple of tracks may have made this less obvious.
Rotting Christ "The Heretics"
It is fair to say that the Greek stalwarts of the Black Metal scene now have a sound that rarely ventures into the territory so well tread on "Thy Mighty Contract" and "Non Serviam" but it is also fair to say that "The Heretics" sounds from start to finish like a band who firmly stamp themselves on the metal map for 2019. What album number thirteen from Rotting Christ does is take a measured approach to variation and repetition to build a memorable and lasting experience.
The sound itself goes from dark and brooding metal and hard rock to almost Gothic metal proportions. Yet at the same time that familiar melodic warmth remains in situ throughout. More often than not there's that big soaring melodic riff riding the vocals like a surfer with his board. "The Voice of the Universe" does this superbly as does "The Sons of Hell". The overall feel the record as a whole leaves me with is one of ritualistic allure, tracks like "Hallowed Be Thy Name" and "Fire God and Fear" build well like powerful incantations being cited from voluminous tomes, flooding forwards intermittently with flowing dark melodies.
It does remind me a lot of Behemoth ("The Satanist" Behemoth, not last years pompous bollocks Behemoth), with the theme of religion so heavily referenced but at the same time it feels like a much more cohesive effort than recent Behemoth keeping a level of consistency and restraint to the songwriting which holds the attention well. Only once does the album visit anything like Black Metal on "I Believe", here they base a poem by Nikolaos Kazantzakis at the centre of the chaotic and frantic pace and sadly it fails miserably on my ears, coming across as just filler. My only other criticism is that it does get a bit samey in places (namely the intros to each track). Otherwise it is a perfectly respectable piece of dark metal. C'mon, we can't this black metal anymore folks.
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text:
Union says arrows and gasoline tossed at striking B.C. assisted living workers
BCGEU president Stephanie Smith says arrows and gasoline were thrown at the picketers early last week
Assisted living workers will be "beefing up" their Courtenay picket line after picketers were allegedly attacked early in the week.
Casa Loma Seniors Village and Comox Valley Seniors Village workers have been taking targeted job action at the two assisted living facilities since Sunday, Dec. 2, including manning a picket line along Headquarters Road.
READ MORE: Assisted living workers go on strike for fair wages
According to BC Government and Service Employees' Union (BCGEU) president Stephanie Smith, workers were out on the picket lines during an evening shift last week when arrows were shot at the line and gasoline was thrown in their direction. She says the RCMP were called.
"Picket lines can be places of heightened emotion for sure," said Smith. "We've had incidents of cars trying to run through our lines… there's often verbal assaults, that sort of thing, but certainly never like this."
The picketers have resumed their rotating strikes and the BCGEU has put out the call for more members to come out and support the picketing workers.
"Immediately, we sent an email out to all our members in the area asking them to come and support the line, particularly on the evening shifts. We've asked that we are closely monitoring to ensure the safety of our picketers," she said, adding that there is safety in numbers.
Smith could not give an update on the status of the negotiations between the BCGEU and the employer, Pacific Reach Seniors Housing Management. However, she hopes that Comox Valley residents will remember these workers are taking job action in order to offer better support to the people in their care.
"These workers are striking to improve the working conditions there, but also to improve the quality of care that these residents receive by ensuring that they have proper staffing levels and that they are able to recruit and retain workers at these centres," said Smith. "I can't even begin to try and understand the mindset of people who would do this – risk the lives and safety of workers who are exercising the constitutional right to run a picket line in support of their bargaining."
A request for comment has been sent to the Comox Valley RCMP. The story will be updated when more information is available.
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Early screening for Breast,Cervical,Ovarian and Womb issues has saved many womens lives and this is a vital part of primary care medicine.
Sexually transmitted diseases are an increasing feature of everyday general practice. We offer comprehensive sexual health screening service including full STI testing, and treatment where required.
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Weight loss and achieving fitness levels for some is easy but for others is a lifelong struggle. Fad, extreme diets and brief sudden exercise surges are customary for those trying to achieve immediate and dramatic weight loss and immediate fitness.
We are happy to provide medical services to companies for medical screening, pre-employment medicals or for an independent medical opinion regarding a work issue.


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W. Ernestine Teen Hoffman
Birth Date: Saturday, May 15, 1926$Date of Passing: Tuesday, August 19, 2008$
WOOSTER – W. Ernestine Teen HOFFMAN, 82 of Sunrise Assisted Living, formerly of Barbara Dr., Wooster died Tuesday, August 19, 2008 at Sunrise.
Friends will be received at the McIntire, Davis & Greene Funeral Home, 216 E. Larwill Street, Wooster, OH, on Friday evening from 6-8:00 P.M.
Services will be held at the Funeral Home, Saturday, August 23, 2008, at 1:00 P.M. with Rev. Larry Black officiating. Interment will follow in Moreland Cemetery.
Those who wish may make memorial contributions to People to People Ministries, 454 E. Bowman St., Wooster, OH 44691 or to Wooster Youth Baseball, P.O. Box 1134, Wooster, OH 44691.
Born Wilma Ernestine Johnson on May 15, 1926 in Kentucky, Teen later made her home with the James family of Moreland. She married her childhood sweetheart, Paul A. Hoffman, on September 14, 1946. He died on August 7, 2007.
Following graduation from Wooster High School in 1943 and marriage to Paul after his dischargge from the Navy, Teen accompanied Paul to Ohio State University where he earned a Pharmacy Degree. The couple then returned to Wooster where they built the family home on Barbara Drive and raised sons, Scott and Steve.
During Paul's 40 year career as a pharmacist and store manager for Gray Drug/Rite Aid, Teen held a variety of jobs including stints as a medical records librarian at Wooster Community Hospital and as a departmental secretary at the College of Wooster. For many years, she ran a brisk home business featuring stained glass lamps, windows and other items which she designed and created. Together she and Paul volunteered for many years with Wooster Youth Baseball. Teen also regularly donated dozens of home baked cookies to People to People Ministries.
After retiring, Teen and Paul enjoyed walking, hiking, golfing and traveling, spending several months each year in Alabama and Georgia where both loved visiting their grandchildren. They spent their final few years together at Sunrise Assisted Living.
Teen always found time to reach out to others with a cheery card or visit, a word of encouragement, a touch of humor, or a batch of freshly made cowboy cookies. Above all, she valued her family, and she especially prized her relationship with her grandsons. Her generous spirit, unfailing enthusiasm and contagious sense of fun will be greatly missed.
Teen is survived by sons Scott of Conyers, GA and Steve (Sharon) of Millersburg; grandsons Brennan and Brian of Georgia; sisters-in-law Ruth Butler of Wooster and Helen Culbertson of Akron; and nieces and a nephew.
In addition to her husband, Teen was preceded in death by brothers-in-law Wayne Hoffman, Glenn Butler and John Culbertson, sister-in-law Treva Hoffman, niece Mabel Hoffman and daughter-in-law Rebecca Hoffman.
The family wishes to thank the staff of Sunrise Assisted Living for their loving care.
People to People Ministries, 454 E. Bowman St., Wooster, OH 44691 or to Wooster Youth Baseball, PO Box 1134, Wooster, OH 44691
Saturday at 1:00 P.M. at the funeral home
Friday evening, August 22, 2008 from 6-8:00 P.M.


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Oh, no! In this cartoon story, a character has come to life! A brother and sister have to think fast to corral Chewy the bubble gum monster!
Do you love the covers of Boys' Quest? Meet the talented cartoonist who makes you laugh with every issue.
Caleb goes to a fair and comes back home with a new interest. It's a good things his family likes Caleb's funny drawings of them!
Learn 10 tips for making your cartoons really great.
It's fun and easy to make a cartoon flip book. Fan the pages, and your character seems to move!
Wait till you see this cartoon featuring a wacky baseball game! See how the cartoonist uses cartooning devices to add more life to the game.
Did you know you can create characters with molding clay and make your own claymation movie? Here are easy step-by-step instructions. Cool!
You'll never run out of ideas for drawing funny faces after reading this article!
Who doesn't like Charlie Brown, Lucy, Snoopy, and the other Peanuts characters? Find out how Charles Schultz, the strip's creator, came up with the idea for Snoopy.
ALL THIS PLUS MORE IDEAS FOR MAKING GREAT CARTOONS, FUNNY CARTOONS TO LAUGH OVER, AN AMAZING SCIENCE PROJECT, MORE STORIES AND ARTICLES, PUZZLES, AND MUCH MORE!!


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\section{Introduction}

Perfect codes are important objects of study in coding theory ever since the beginning of information theory. Roughly speaking, a code is perfect if it achieves maximum possible error correction without ambiguity. In the classical setting, much work has been focused on perfect codes under the Hamming or Lee metric. Solving a long-standing conjecture, it was proved in the 1970s \cite{Va73, Tie73, ZL73} that the well-known Hamming and Golay codes are the only nontrivial linear perfect codes under the Hamming metric. (A \emph{linear code} is a subspace of some linear space $\mathbb{F}_q^n$, where $\mathbb{F}_q$ is the field with $q$ elements, $q$ being a prime power and $n$ a positive integer.) In contrast to the linear case, there are many nonlinear perfect codes under the Hamming metric, and the study of them has long been an active research area in coding theory. The reader is referred to the survey papers \cite{Heden1, Va75} for a large number of results on perfect codes under the Hamming metric. With regard to the Lee metric, the famous Golomb-Welch conjecture asserts that for any $n > 2$, $e > 1$ and $q \ge 2t+1$  there is no $q$-ary perfect $t$-codes of length $n$ under the Lee metric. A central problem for Lee codes, this 50-year-old conjecture is still wide open \cite{HK18} despite extensive research on the topic.

From a mathematical point of view, perfect codes can be defined for any finite metric space: Given an integer $t \ge 1$, a subset of a finite metric space is called a \emph{perfect $t$-code} \cite{Va75} if the balls of radius $t$ with centres in the subset form a partition of the space. In particular, since any graph is a metric space under the usual graph distance, we can talk about perfect $t$-codes in graphs. Let $\Gamma$ be a graph with vertex set $V(\Gamma)$ and edge set $E(\Gamma)$. The \emph{distance} in $\Gamma$ between two vertices $u, v \in V(\Gamma)$, denoted by $d(u, v)$, is the length of a shortest path between $u$ and $v$ in $\Gamma$, and is defined to be $\infty$ if no path between $u$ and $v$ exists. In view of the definition above, a subset $C$ of $V(\Gamma)$ is a \emph{perfect $t$-code} \cite{Kr86} in $\Gamma$ if every vertex of $\Gamma$ is at distance no more than $t$ to exactly one vertex of $C$. In what follows a perfect $1$-code is simply called a {\em perfect code}. It is readily seen that a subset $C$ of $V(\Gamma)$ is a perfect code in $\Gamma$ if and only if $C$ is an independent set of $\Gamma$ and every vertex in $V(\Gamma) \setminus C$ is adjacent to exactly one vertex in $C$. In graph theory, a perfect code in a graph is also called an efficient dominating set \cite{DeS} or independent perfect dominating set \cite{Le} of the graph.

The {\em Cartesian product} of $n$ graphs $\Gamma_1, \Gamma_{2}, \ldots, \Gamma_{n}$ is the graph with vertex set $V(\Gamma_1) \times V(\Gamma_{2}) \times \cdots \times V(\Gamma_{n})$ such that two vertices $(u_1, u_2, \ldots, u_n), (v_1, v_2, \ldots, v_n)$ are adjacent if and only if $u_{i} \neq v_{i}$ for exactly one subscript $i$, and for this $i$, $u_{i}$ and $v_{i}$ are adjacent in $\Gamma_i$. The \emph{Hamming graph} $H(n, q)$ is the Cartesian product of $n$ copies of the complete graph $K_q$ with $q$ vertices. Denote by $C_q^{\Box n}$ the Cartesian product of $n$ copies of the cycle $C_q$ of length $q$. In particular, $H(n, 2)$ is the $n$-dimensional cube $Q_n$ and $C_q^{\Box 2}$ is the grid graph on a torus. Alternatively, we may define $H(n, q)$ and $C_q^{\Box n}$ to be the graphs with vertex set $\mathbb{Z}_q^n$ such that two elements $(u_1, u_2, \ldots, u_n), (v_1, v_2, \ldots, v_n)$ of $\mathbb{Z}_q^n$ are adjacent in $H(n, q)$ if and only if they differ at exactly one coordinate and adjacent in $C_q^{\Box n}$ if and only if $u_{i} \neq v_{i}$ for exactly one $i$ and moreover $u_{i} \equiv v_{i} \pm 1 \mod q$ for this $i$.

It is well known that the Hamming and Lee metrics over $\mathbb{Z}_q^n$ are exactly the graph distances in $H(n, q)$ and $C_q^{\Box n}$, respectively. Therefore, perfect $t$-codes under the Hamming or Lee metric are exactly those in $H(n, q)$ or $C_q^{\Box n}$, respectively. It is also well known that all Hamming graphs are distance-transitive. (A graph $\Gamma$ is called \emph{distance-transitive} if for any $u, v, u', v' \in V(\Gamma)$ with $d(u, v) = d(u', v')$ there exists an automorphism of $\Gamma$ which maps $(u, v)$ to $(u', v')$.) This motivated Biggs \cite{Big} to study perfect codes in distance-transitive graphs as a generalization of perfect codes under the Hamming metric. Among other things he generalized the celebrated Lloyd's Theorem \cite{Lens} to perfect codes in any distance-transitive graph. The seminal paper of Biggs \cite{Big} and the fundamental work of Delsarte \cite{Del} inspired much work on perfect codes in distance-transitive graphs and, in general, in distance-regular graphs and association schemes. It is known \cite{Chi, MZ95} that many infinite families of classical distance-regular graphs have no nontrivial perfect codes, including the Grassmann graphs and the bilinear forms graphs. Doob graphs are an important family of distance-regular Cayley graphs. A necessary and sufficient condition for a Doob graph to admit perfect codes was recently given in \cite{Kro20}, and all possible parameters of subgroup perfect codes in Doob graphs were described in \cite{SHK19}. It is known \cite{Best} that if $q$ is not a prime power then a perfect $t$-code in $H(n,q)$ exists only when $t \in \{1, 2, 6, 8\}$, but the existence of such codes is a long-standing open problem. More results on perfect codes in distance-regular graphs can be found in, for example, \cite{HS, SE, Thas80}.

\medskip
\textbf{Perfect codes in Cayley graphs.}~
As observed in \cite{HXZ18}, perfect codes in Cayley graphs are another generalization of perfect codes in the classical setting. This is so because $H(n, q)$ and $C_q^{\Box n}$ are both Cayley graphs of $\mathbb{Z}_q^n$. In general, given a group $G$ with identity element $e$ and an inverse-closed subset $S$ of $G$ with $e \notin S$, the {\em Cayley graph} ${\rm Cay}(G, S)$ of $G$ with \emph{connection set} $S$ is defined to be the graph with vertex set $G$ such that two distinct elements $x,y$ are adjacent if and only if $yx^{-1}\in S$, where a subset $A$ of $G$ is called {\em inverse-closed} if $A^{-1}:=\{a^{-1}: a \in A\} = A$. Obviously, ${\rm Cay}(G, G \setminus \{e\})$ is the complete graph with vertex set $G$, and ${\rm Cay}(G, \emptyset)$ is the graph on $G$ with no edges.

In recent years, perfect codes in Cayley graphs have received considerable attention \cite{D14, DSLW16, FHZ, HXZ18, KM13, Le, MBG07, OPR07, Ta13, Z15}. The reader is referred to \cite[Section 1]{HXZ18} for a brief account of results on perfect codes in Cayley graphs and connections between such codes and factorizations and tilings of the underlying groups. In general, a \emph{tiling} \cite{Dinitz06} of a group $G$ is a pair of subsets $(A, B)$ of $G$ such that $e \in A \cap B$ and every element of $G$ can be expressed uniquely as $ab$ with $a \in A$ and $b \in B$. It is readily seen that $(A, B)$ is a tiling of $G$ such that $A$ is inverse-closed if and only if $B$ is a perfect code of ${\rm Cay}(G, A \setminus \{e\})$ such that $e \in B$.

In \cite{HXZ18}, Huang, Xia and Zhou introduced the following concept: A subset $C$ of a group $G$ is called a {\em perfect code} of $G$ if there exists a Cayley graph ${\rm Cay}(G, S)$ of $G$ which admits $C$ as a perfect code. In particular, a perfect code of $G$ which is also a subgroup of $G$ is called a {\em subgroup perfect code} of $G$. In the same paper, Huang, Xia and Zhou obtained a necessary and sufficient condition for a normal subgroup of a group $G$ to be a perfect code of $G$, and determined all subgroup perfect codes of all dihedral groups and some abelian groups. As explained in \cite{HXZ18}, in some sense subgroup perfect codes are an analogue of linear perfect codes.

\medskip
\textbf{Code-perfect groups.}~
It may happen that every subgroup of a given group is a perfect code. We call a group with this property a code-perfect group. More explicitly, a group $G$ is said to be \emph{code-perfect} if for every subgroup $H$ of $G$ there exists a Cayley graph ${\rm Cay}(G, S)$ of $G$ which admits $H$ as a perfect code (that is, $(S \cup \{e\}, H)$ is a tiling of $G$ with $S \cup \{e\}$ inverse-closed). Note that the trivial subgroup $\{e\}$ is a perfect code in the complete Cayley graph ${\rm Cay}(G, G \setminus \{e\})$ and the whole group $G$ is a perfect code in the empty Cayley graph ${\rm Cay}(G, \emptyset)$. So a code-perfect group can also be defined as a group in which every proper subgroup is a perfect code in the group.

It is natural to ask which groups are code-perfect. In this paper, we answer this question by giving a complete characterization of code-perfect groups. As we will see shortly, by this characterization not all abelian groups are code-perfect. So one may ask when a given subgroup of an abelian group is a perfect code. We give an answer to this question by reducing the problem of determining when a subgroup of an abelian group is a perfect code to the case of abelian $2$-groups. It turns out that no generalized quaternion group can be code-perfect. We determine all subgroup perfect codes together with corresponding Cayley graphs in any generalized quaternion group.

\medskip
\textbf{Notation and terminology.}~Before stating our results let us introduce some notation and terminology first. All groups considered in the paper are finite, and all graphs considered are finite and undirected with no loops or multiple edges. So we will omit the adjective ``finite'' before the words ``group'' and ``graph''. We always use $e$ to denote the identity element of the group under consideration. We use $G_2$ and $G_{2'}$ to denote the Sylow $2$-subgroup and Hall $2'$-subgroup of a group $G$, respectively. Note that, for any abelian group $G$, $G_2$ consists of the elements of $G$ with order a power of $2$, and $G_{2'}$ consists of the elements of $G$ with odd order. Denote
$$
G^2 = \{g^2: g \in G\}.
$$
An abelian group $G$ is said to be \emph{$2$-divisible} if $G^2 = G$. For an abelian group $G$, a subgroup $H$ of $G$ is called a {\em $2$-pure subgroup} of $G$ if $G^2 \cap H=H^2$. As usual, we use $A \times B$ to denote the direct product of two groups $A$ and $B$. We use $Q_{4n}$ to denote the generalized quaternion group of order $4n$, where $n \ge 2$. It is well known (see, for example, \cite[pp. 44--45]{Jon}) that
\begin{equation}
\label{eq:q4n}
Q_{4n}=\langle x,y: x^n=y^2, x^{2n}=e, y^{-1}xy=x^{-1}\rangle
\end{equation}
and the order of $x^i y$ in $Q_{4n}$ is $4$ for $0 \le i \le n$.

\medskip
\textbf{Main results.}~The first main result in this paper is as follows.

\begin{thm}
\label{newthm1}
A group is code-perfect if and only if it has no elements of order $4$.
\end{thm}

The sufficiency of this result will be proved by construction: Given any group $G$ with no elements of order $4$ and any proper subgroup $H$ of $G$, we will construct an inverse-closed subset $S$ of $H$ with $e \notin S$ such that ${\rm Cay}(G, S)$ admits $H$ as a perfect code.

In \cite[Corollary 2.4(a)]{HXZ18}, it was proved that every normal subgroup of any group of odd order is a perfect code of the group. The following corollary of Theorem \ref{newthm1} generalizes this result from all normal subgroups to all subgroups.

\begin{cor}
\label{newthm1-cor1}
Any group of odd order is code-perfect.
\end{cor}

Theorem \ref{newthm1} also implies the following result, in which $\mathbb{Z}_2^d \times Q$ is interpreted as $Q$ when $d=0$.

\begin{cor}
\label{newthm1-cor2}
An abelian group is code-perfect if and only if it is isomorphic to $\mathbb{Z}_2^d \times Q$ for some integer $d \ge 0$ and abelian group $Q$ of odd order.
\end{cor}

All simple groups with no elements of order $4$ have been classified in \cite{Wa69}. Combining this and Theorem \ref{newthm1}, we obtain the following result.

\begin{cor}
\label{newthm1-cor3}
A simple group is code-perfect if and only if it is isomorphic to one of the following groups:
\begin{itemize}
\item[{\rm (a)}] a cyclic group of prime order;

\item[{\rm (b)}] ${\rm PSL}(2,2^e)$, $e\ge 2$;

\item[{\rm (c)}] ${\rm PSL}(2,q)$, $q\equiv \pm 3~({\rm mod}~8)$, $q\ge 5$;

\item[{\rm (d)}] a Ree group $^2G_2(3^{2n+1})$, $n\ge 1$;

\item[{\rm (e)}] the Janko group $J_1$.
\end{itemize}
\end{cor}

Theorem \ref{newthm1} implies that not every abelian group is code-perfect. So one may ask when a given subgroup of an abelian group is a perfect code of the group. The following result shows that this problem can be reduced to the case of abelian $2$-groups.

\begin{thm}
\label{abethm1}
Let $G$ be an abelian group and $H$ a proper subgroup of $G$. Then $H$ is a perfect code of $G$ if and only if $H_2$ is a $2$-pure subgroup of $G_2$, which in turn is true if and only if $H_2$ is a perfect code of $G_2$.
\end{thm}

A property which is diagonally opposite to the one of being a code-perfect group is that no nontrivial proper subgroup is a perfect code. Our next result gives all abelian non-simple groups with this property. It would be interesting if one can obtain a characterization of non-abelian groups with this property.

\begin{thm}
\label{newrem}
Let $G$ be an abelian group which is not a simple group. Then every nontrivial proper subgroup of $G$ is not a perfect code of $G$ if and only if $G$ is isomorphic to the cyclic group $\mathbb{Z}_{2^m}$ for some $m\ge 2$.
\end{thm}

In the special case when $G$ is a cyclic group, Theorems~\ref{abethm1} and \ref{newrem} together yield \cite[Corollary 2.8(a)]{HXZ18}, which asserts that a proper subgroup $H$ of a cyclic group $G$ is a perfect code of $G$ if and only if either $|H|$ or $[G:H]$ is odd.

Theorem~\ref{newthm1} also implies that $Q_{4n}$ is not code-perfect. We determine all subgroup perfect codes of $Q_{4n}$ together with corresponding Cayley graphs in the following result.

\begin{thm}
\label{newmainth2}
Let $Q_{4n}$ be the generalized quaternion group as presented in \eqref{eq:q4n}, and let $H$ be a proper subgroup of $Q_{4n}$. Then $H$ is a perfect code of $Q_{4n}$ if and only if one of the following holds:
\begin{itemize}
\item[{\rm (a)}] $H=\langle x^t\rangle$, where $t$ is a positive integer dividing $2n$ such that $\frac{2n}{t}$ is odd;

\item[{\rm (b)}] $H=\langle x^t, x^sy\rangle$, where $t \ge 3$ is an odd integer dividing $2n$ and $s$ is an integer between $0$ and $t-1$.
\end{itemize}
Moreover, if (a) occurs, then $\langle x^t\rangle$ is a perfect code of ${\rm Cay}(Q_{4n}, S)$, where
\begin{equation}
\label{eq:s1}
S = \left\{x^n, x^i, x^{-i}: 1 \le i \le (t/2)-1\right\} \cup \left\{x^i y, x^{n+i} y: 0 \le i \le (t/2)-1\right\};
\end{equation}
and if (b) occurs, then $\langle x^t, x^sy\rangle$ is a perfect code of ${\rm Cay}(Q_{4n}, S)$, where
\begin{equation}
\label{eq:s2}
S = \left\{x^i, x^{-i}: 1 \le i \le (t-1)/2\right\}.
\end{equation}
\end{thm}

\medskip
\textbf{Structure of the paper.}~We will present some preliminary results in the next section. The proofs of Theorems \ref{newthm1} and \ref{newmainth2} will be given in Sections \ref{Sec1} and \ref{Se}, respectively, and the proofs of Theorems \ref{abethm1} and \ref{newrem} will be given in Section \ref{abeg}. An example to illustrate Theorem \ref{newmainth2} will be given in Section \ref{Se}.

\section{Preliminaries}
\label{sec:pre}

We will use the following result in our proofs of Theorems \ref{abethm1} and \ref{newmainth2}.

\begin{thm}{\rm (\cite[Theorem 2.2(a)]{HXZ18})}
\label{basic}
Let $G$ be a group and $H$ a normal subgroup of $G$. Then $H$ is a perfect code of $G$ if and only if the following holds: for any $g\in G$, $g^2\in H$ implies $(gh)^2=e$ for some $h\in H$.
\end{thm}

The following lemma is an extension of \cite[Lemma 2.1]{HXZ18}, where the equivalence between the first two statements was established.

\begin{lem}
\label{mainlem0}
Let $G$ be a group and $H$ a subgroup of $G$. Let $S$ be an inverse-closed subset of $G$ such that $e \not\in S$.
The following statements are equivalent:
\begin{itemize}
\item[{\rm (a)}] $H$ is a perfect code of ${\rm Cay}(G, S)$;

\item[{\rm (b)}] $S\cup\{e\}$ is a left transversal of $H$ in $G$;

\item[{\rm (c)}] $S\cup\{e\}$ is a right  transversal of $H$ in $G$.
\end{itemize}
\end{lem}

\proof
The equivalence between (a) and (b) was proved in \cite[Lemma 2.1]{HXZ18}. It remains to prove that (b) and (c) are equivalent.

Suppose that $S\cup\{e\}$ is a left transversal of $H$ in $G$. We claim that $Ha\ne Hb$ for distinct $a, b \in S$. Suppose to the contrary that $Ha = Hb$. Then $a^{-1}H=b^{-1}H$. Since $a^{-1}, b^{-1}\in S$ (as $S$ is inverse-closed) and $S\cup\{e\}$ is a left transversal of $H$ in $G$, we deduce that $a^{-1}=b^{-1}$, but this contradicts our assumption that $a\ne b$. Therefore, $\{Hs:s\in S\}$ consists of $|S|$ right cosets of $H$ in $G$. Since $S\cup\{e\}$ is a left transversal of $H$ in $G$, we have $[G:H]=|S|+1$ and $s \notin H$ for each $s \in S$. This together with $e\notin S$ implies that $S\cup\{e\}$ is a right  transversal of $H$ in $G$. So (b) implies (c). Similarly, we can prove that (c) implies (b).
\hfill$\Box$\vspace{11pt}

Lemma \ref{mainlem0} implies the following result.

\begin{cor}
\label{gtlem0}
Let $G$ be a group and $H$ a proper subgroup of $G$. Then $H$ is a perfect code of $G$ if and only if there exists a left or right transversal of $H$ in $G$ which contains $e$ and is inverse-closed.
In particular, if there exists an element $x \in G\setminus H$ such that $xH$ or $Hx$ is inverse-closed and contains no involutions, then $H$ is not a perfect code of $G$.
\end{cor}

Of course a left or right transversal $T$ of $H$ in $G$ contains $e$ if and only if $T \cap H = \{e\}$.

\section{Proof of Theorem \ref{newthm1}}
\label{Sec1}

We will prove two lemmas before giving the proof of Theorem \ref{newthm1}. An element $x$ of a group is called a \emph{square} if it can be expressed as $x = y^2$ for some element $y$ of the group.

\begin{lem}
\label{mainlem1}
Let $G$ be a group and $x$ an involution of $G$. Then $\langle x\rangle$ is a perfect code of $G$ if and only if $x$ is not a square of $G$.
\end{lem}

\proof
Denote $H=\langle x\rangle = \{e, x\}$. If $x$ is a square, say, $x=y^2$ for some $y\in G$, then $Hy=\{y, xy\}=\{y,y^{-1}\}$ is inverse-closed and contains no involutions. Hence, by the second statement in Corollary \ref{gtlem0}, $H$ is not a perfect code of $G$.

Now assume that $x$ is not a square. We will apply induction to construct a right transversal $T$ of $H$ in $G$ which contains $e$ and is inverse-closed. Once this is achieved, we then obtain from Corollary \ref{gtlem0} that $H$ is a perfect code of $G$.

To begin with, we process initially the coset $H$ and put $e$ into $T$ to represent $H$. Inductively, suppose that we have processed some but not all right cosets of $H$ in $G$ and selected a representative for each of them, in such a way that the set of representatives selected so far is inverse-closed. Take an element $y \in G$ which is not in any right coset already processed. (For example, when only the coset $H$ has been processed, we simply take any $y \in G \setminus H$.) According to the orders of $y$ and $xy$, we now process one, two or four right cosets of $H$ in the following way.

\smallskip
{\bf Case 1.} $y$ is an involution.
\smallskip

In this case we only process $Hy$ and put $y$ into $T$ as the representative of $Hy$. (Alternatively, if $xy$ is also an involution, we can put $xy$ but not $y$ into $T$ to represent $Hy$.)

\smallskip
{\bf Case 2.} $y$ has order greater than $2$ but $xy$ is an involution.
\smallskip

In this case we only process $Hy = Hxy$ and put $xy$ into $T$ as the representative of $Hy$. We can do so because $xy$ has not been selected, for otherwise $y$ would be in a previously processed right coset of $H$ in $G$, which is a contradiction.

Note that $Hy^{-1}=\{y^{-1}, xy^{-1}\}$ and $xy^{-1}=(yx)^{-1}$ is an involution. Hence our rules in Cases 1 and 2 applied to $Hy^{-1}$ implies that $xy^{-1}$ but not $y^{-1}$ is selected to represent $Hy^{-1}$ when processing $Hy^{-1}$ (which can take place before or after $Hy$ is processed). Therefore, the undesired situation where $y^{-1}$ is a representative but $y$ is not (or $y$ is a representative but $y^{-1}$ is not) cannot happen.

\smallskip
{\bf Case 3.} Both $y$ and $xy$ have order greater than $2$.
\smallskip

Assume that $xy=yx$ first. Then $xy^{-1}=y^{-1}x$ and so $H(xy)^{-1}=\{y^{-1}x,xy^{-1}x\}=\{y^{-1}x,y^{-1}\}=Hy^{-1}$. Since $x$ is not a square, we have $y \neq y^{-1} x$. Hence $Hy=\{y,xy\}$ and $Hy^{-1}$ are distinct cosets. We process both $Hy$ and $Hy^{-1}$, and put $y$ and $y^{-1}$ into $T$ to represent $Hy$ and $Hy^{-1}$, respectively.

Now we assume that $xy\ne yx$. We have $Hy=\{y,xy\}$, $H(xy)^{-1}=\{y^{-1}x,xy^{-1}x\}$, $H(xy^{-1}x)^{-1}=\{xyx,yx\}$ and $H(yx)^{-1}=\{xy^{-1},y^{-1}\}$, and one can easily verify that these cosets are distinct and their union is inverse-closed. We process these four cosets and put $y, xy^{-1}x, xyx$ and $y^{-1}$ into $T$ as their representives, respectively.

After the treatment above, we have processed at least one more right coset of $H$ in $G$ and obtained a larger set of representatives. By our selection of representatives and based on the hypothesis, this larger set of representatives remains to be inverse-closed. If all right cosets of $H$ have been processed, we stop and output $T$. Otherwise we repeat the procedure above. By induction we can eventually obtain a transversal $T$ of $H$ in $G$ which contains $e$ and is inverse-closed, as required.
\hfill$\Box$\vspace{11pt}

\begin{lem}\label{mainlem2}
Suppose that $G$ is a group with no elements of order $4$. Then for every subgroup $H$ of $G$ there exists a right transversal of $H$ in $G$ which contains $e$ and is inverse-closed.
\end{lem}

\proof
Let
$$
\Lambda_1 = \{Hx: x \in G \text{ and $Hx$ contains an element of order $2$}\} \setminus \{H\}
$$
and
$$
\Lambda_2=\{Hx: x\in G \text{ and $Hx$ contains no elements of order $2$}\}\setminus \{H\}.
$$

\smallskip
{\bf Claim 1.} Let $Hx\in \Lambda_2$. Then $(hx)^2\notin H$ for any $h\in H$.
\smallskip

Suppose for a contradiction that $(hx)^2\in H$ for some $h\in H$. We claim that there exist elements $y_1$ and $y_2$ of $G$ such that $y_1^2=e$, $y_2$ is of odd order and $hx=y_1y_2 = y_2y_1$. In fact, if $hx$ is of odd order, then we can take $y_1 = e$ and $y_2 = hx$. Assume that $hx$ is of even order. Then the order of $hx$ is of the form $2t$ for some odd integer $t = 2m - 1 \ge 1$ as $G$ has no elements of order $4$. Setting $y_1 = (hx)^{-t}$ and $y_2 = (hx)^{2m}$, we have $hx = y_1y_2 = y_2y_1$, $y_1^2 = e$, and $y_2$ has odd order $t$, as required.

By the assumption $(hx)^2\in H$ and the claim above, we obtain $(hx)^2=(y_1y_2)^2=y_2^2\in H$. Since the order of $y_2$ is odd, it follows that $y_2\in \langle y_2^2\rangle\subseteq H$. Now $Hx=(Hh^{-1})y_1y_2=Hy_1y_2=(Hy_2)y_1=Hy_1$. Hence $y_1 \in Hx$. Since $x\notin H$ (as $Hx \neq H$) and $y_1^2=e$, it follows that $y_1$ has order $2$, but this contradicts the assumption that $Hx\in \Lambda_2$.

\smallskip
{\bf Claim 2.} Let $Hx\in \Lambda_2$. Then $Hx\ne Hx^{-1}$ and $Hx^{-1}\in \Lambda_2$.
\smallskip

In fact, if $Hx=Hx^{-1}$, then $x^2\in H$ and so $(x^2 x)^2 = x^6\in H$, which contradicts Claim 1. Hence $Hx\ne Hx^{-1}$. Since $x\notin H$, we have $Hx^{-1}\ne H$. We claim that $Hx^{-1}\in \Lambda_2$. Suppose otherwise. Then $Hx^{-1}\in \Lambda_1$. So there exists $h\in H$ such that $(hx^{-1})^2=e$, that is, $xh^{-1} = h x^{-1}$. We then have $(h^{-1}x)^2=h^{-1}(xh^{-1})x = h^{-1}(hx^{-1})x = e\in H$, which contradicts Claim 1. Hence $Hx^{-1}\in \Lambda_2$.

\smallskip
{\bf Claim 3.} The operation $Hx \cdot h=H(xh)$ for $Hx \in \Lambda_2$ and $h\in H$ defines an action of $H$ on $\Lambda_2$.
\smallskip

In fact, for $Hx\in \Lambda_2$ and $h \in H$, since $Hxh=h^{-1}(Hx)h$, the set of orders of the elements in $Hx$ is the same as the set of orders of the elements in $Hxh$. In particular, like $Hx$, $Hxh$ contains no elements of order $2$. Moreover, since $x\notin H$ (as $Hx \ne H$), we have $Hxh \ne H$. Thus $Hxh\in \Lambda_2$ and the operation above defines an action of $H$ on $\Lambda_2$.

By Claim 2, whenever $Hx\in \Lambda_2$, we have $Hx^{-1} \in \Lambda_2$. Denote by ${\rm orb}_H(Hx)$ and ${\rm orb}_H(Hx^{-1})$ the orbits of $Hx$ and $Hx^{-1}$ under the action of $H$ defined in Claim 3, respectively.

\smallskip
{\bf Claim 4.} Let $Hx\in \Lambda_2$. Then ${\rm orb}_H(Hx) \cap {\rm orb}_H(Hx^{-1})=\emptyset$.
\smallskip

Suppose otherwise. Then $Hx^{-1}\in {\rm orb}_H(Hx)$ and so $Hx^{-1}=Hxh$ for some
$h\in H$. Hence $xhx\in H$. We then have $(hx)^2\in H$, but this contradicts Claim 1. Therefore, ${\rm orb}_H(Hx) \cap {\rm orb}_H(Hx^{-1})=\emptyset$.

\smallskip
{\bf Claim 5.} Let $Hx\in \Lambda_2$. Then $|{\rm orb}_H(Hx)|=[H:H\cap H^x]=|{\rm orb}_H(Hx^{-1})|$.
\smallskip

In fact, the stabilizer of $Hx$ under the action of $H$ is equal to $\{h\in H: Hxh=Hx\} =\{h\in H: xhx^{-1}\in H\} = \{h\in H: h\in H^x\} = H\cap H^x$. Hence $|{\rm orb}_H(Hx)|=[H:H\cap H^x]$. Similarly, $|{\rm orb}_H(Hx^{-1})|=[H:H\cap H^{x^{-1}}]$. However, as
$H\cap H^{x^{-1}}=(H\cap H^x)^{x^{-1}}$, we have $[H:H\cap H^x]=[H:(H\cap H^x)^{x^{-1}}]=[H:H\cap H^{x^{-1}}]$. Hence $|{\rm orb}_H(Hx)|=|{\rm orb}_H(Hx^{-1})|$.

\smallskip
{\bf Claim 6.} If $Hx \in \Lambda_2$ and $Hy\in \Lambda_2\setminus ({\rm orb}_H(Hx)\cup {\rm orb}_H(Hx^{-1}))$, then $Hy^{-1} \in \Lambda_2\setminus ({\rm orb}_H(Hx)\cup {\rm orb}_H(Hx^{-1}))$.

In fact, if $Hy^{-1} \in {\rm orb}_H(Hx^{\epsilon})$ where $\epsilon=\pm 1$, then $Hy^{-1}=Hx^{\epsilon}h$ for some $h \in H$. It follows that $x^{\epsilon}hy=h_1 \in H$. Hence $hy=x^{-\epsilon}h_1$. Therefore, $Hy=Hx^{-\epsilon}h_1 \in {\rm orb}_H(Hx^{-\epsilon})$, a contradiction.

We are now ready to construct a right transversal of $H$ in $G$ which contains $e$ and is inverse-closed. First, we put the identity element $e$ into the transversal to represent coset $H$. Then, for each $Hx \in \Lambda_1$, we choose an element of $Hx$ with order $2$ and put it into the transversal. It remains to select an appropriate representative for each coset in $\Lambda_2$.

By Claims 2, 4 and 6, there exist elements $x_1, x_2, \ldots, x_m$ of $G$ such that $\Lambda_2$ is partitioned into
$$
{\rm orb}_H(Hx_1),\ {\rm orb}_H(Hx_1^{-1}),\ {\rm orb}_H(Hx_2),\ {\rm orb}_H(Hx_2^{-1}),\ \ldots,\ {\rm orb}_H(Hx_m),\ {\rm orb}_H(Hx_m^{-1}).
$$
By Claim 5, for $1 \le i \le m$, we may assume that
$$
{\rm orb}_H(Hx_i)=\{Hx_i, Hx_i h_{i1}, Hx_i h_{i2},\ldots,Hx_i h_{ik_i}\}
$$
and
$$
{\rm orb}_H(Hx_i^{-1})=\{Hx_i^{-1},Hx_i^{-1}g_{i1},
Hx_i^{-1}g_{i2},\ldots,Hx_i^{-1}g_{ik_i}\}
$$
for some $h_{i1},\ldots,h_{ik_i} \in H$ and $g_{i1},\ldots,g_{ik_i}\in H$, where $k_i = [H:H\cap H^{x_i}] - 1$. Note that
$$
{\rm orb}_H(Hx_i)=\{Hx_i, Hg_{i1}^{-1} x_i h_{i1}, Hg_{i2}^{-1}x_i h_{i2},\ldots,Hg_{ik_i}^{-1}x_i h_{ik_i}\}
$$
and
$$
{\rm orb}_H(Hx_i^{-1})=\{Hx_i^{-1},Hh_{i1}^{-1}x_i^{-1}g_{i1},
Hh_{i2}^{-1}x_i^{-1}g_{i2},\ldots,Hh_{ik_i}^{-1}x_i^{-1}g_{ik_i}\}.
$$
So we can add
$$
x_i, g_{i1}^{-1} x_i h_{i1}, g_{i2}^{-1}x_i h_{i2},\ldots g_{ik_i}^{-1}x_i h_{ik_i}
$$
and
$$
x_i^{-1}, h_{i1}^{-1}x_i^{-1}g_{i1}, h_{i2}^{-1}x_i^{-1}g_{i2},\ldots, h_{ik_i}^{-1}x_i^{-1}g_{ik_i}
$$
to the transversal to represent the cosets in ${\rm orb}_H(Hx_i)$ and ${\rm orb}_H(Hx_i^{-1})$, respectively. Note that $(g_{ij}^{-1}x_i h_{ij})^{-1} = h_{ij}^{-1}x_i^{-1}g_{ij}$ for $1 \le i \le m$ and $1 \le j \le k_i$.

So far we have chosen a representative for each right coset of $H$ in $G$ and thus constructed a right transversal of $H$ in $G$. The construction itself ensures that this transversal contains $e$ and is inverse-closed.
\hfill$\Box$\vspace{11pt}

\noindent {\em Proof of Theorem~{\rm\ref{newthm1}}.} Let $G$ be a group. If $G$ contains an element of order $4$, say, $y$, then $x = y^2$ is an involution and by Lemma~\ref{mainlem1}, $\langle x\rangle$ is not a perfect code of $G$. Hence $G$ is not code-perfect.

Now assume that $G$ has no elements of order $4$. By Lemma~\ref{mainlem2}, for every proper subgroup $H$ of $G$, there is a right transversal $T$ of $H$ in $G$ such that $e\in T$ and $T^{-1} = T$. Hence, by Corollary~\ref{gtlem0}, $H$ is a perfect code of $G$. Since this holds for any proper subgroup of $G$, we conclude that $G$ is a code-perfect group. This proves the sufficiency.
\hfill$\Box$\vspace{11pt}

The proof of Lemma \ref{mainlem2} gives an algorithm for constructing a Cayley graph which admits a given proper subgroup of a group with no elements of order $4$ as a perfect code. In fact, if $S \cup \{e\}$ is the right transversal of $H$ in $G$ constructed in the proof of Lemma \ref{mainlem2}, then this Cayley graph is ${\rm Cay}(G, S)$.

\section{Proofs of Theorems \ref{abethm1} and \ref{newrem}}
\label{abeg}

The next lemma is obtained by applying Theorem \ref{basic} to abelian groups.

\begin{lem}
\label{new-2-group}
Let $G$ be an abelian group and $H$ a subgroup of $G$. Then $H$ is a perfect code of $G$ if and only if $H$ is a $2$-pure subgroup of $G$.
\end{lem}

\proof
We first prove the sufficiency. Suppose that $H$ is a $2$-pure subgroup of $G$. Then $G^2\cap H=H^2$. Thus, for any $g \in G$ with $g^2\in H$, we have $g^2\in H^2$ and hence there exists $h\in H$ such that $g^2=h^2$, which implies that $(gh^{-1})^2=e$. Now by Theorem~\ref{basic}, $H$ is a perfect code of $G$.

We next prove the necessity. Suppose that $H$ is a perfect code of $G$. Let $g\in G$ be such that $g^2\in G^2\cap H$. Since $H$ is normal in $G$ and $g^2 \in H$, by Theorem~\ref{basic} there exists $h\in H$ such that $(gh)^2=e$. So $g^2=(h^{-1})^2\in H^2$. It follows that $G^2\cap H\subseteq H^2$. Also, it  is clear that
$H^2 \subseteq G^2\cap H$. Hence $G^2\cap H=H^2$ and $H$ is a $2$-pure subgroup of $G$.
\hfill$\Box$\vspace{11pt}

The following lemma is well known. We include its proof as we are unable to find a specific reference for it.

\begin{lem}
\label{2-pure}
Let $G$ be an abelian group and $H$ a subgroup of $G$. Then $H$ is a $2$-pure subgroup of $G$ if and only if $H_2$ is a $2$-pure subgroup of $G_2$.
\end{lem}

\proof
Suppose that $H_2$ is a $2$-pure subgroup of $G_2$. Since $G_2$ is a $2$-pure subgroup of $G$ and the property of being a $2$-pure subgroup is transitive, we obtain that $H_2$ is a $2$-pure subgroup of $G$. Since $H_{2'}$ has odd order, it is $2$-divisible and hence a $2$-pure subgroup of $G$. It follows that $H$ is a $2$-pure subgroup of $G$.

Now suppose that $H$ is a $2$-pure subgroup of $G$. We aim to prove that $H_2$ is a $2$-pure subgroup of $G_2$. Let $g \in G_2$ be such that $g^2 \in H_2$. Since $H$ is a 2-pure subgroup of $G$, there exists $h \in H$ such that $g^2 = h^2$. We can write $h= xy$ for some $x \in H_2$ and $y \in H_{2'}$. So $g^2 = (xy)^2 = x^2 y^2$ as $G$ is abelian. Hence $y^2 \in H_2 \cap H_{2'} = \{e\}$. Now that $y^2 = e$, we have $g^2 = x^2 \in H_2^2$. It follows that $G_2^2 \cap H_2 \subseteq H_2^2$ and therefore $H_2$ is a $2$-pure subgroup of $G_2$.
\hfill$\Box$\vspace{11pt}

We are now ready to prove Theorem~\ref{abethm1}.
\medskip

\noindent {\em Proof of Theorem~{\rm\ref{abethm1}}.}
The first statement in Theorem~\ref{abethm1} follows from Lemmas \ref{new-2-group} and \ref{2-pure} immediately. Applying Lemma \ref{new-2-group} to the subgroup $H_2$ of $G_2$, we then obtain the second statement in Theorem~\ref{abethm1}.
\hfill$\Box$\vspace{11pt}

A {\em complement} of a subgroup $H$ in a group $G$ is a subgroup $K$ of $G$ such that $G=HK$ and $H\cap K=\{e\}$. We will use the following lemma in our proof of Theorem \ref{newrem}.

\begin{lem}\label{comp}
Let $G$ be a group and $H$ a subgroup of $G$. If $H$ has a complement in $G$, then $H$ is a perfect code of $G$.
\end{lem}
\proof
Let $K$ be a complement of $H$ in $G$. It is easy to see that $K$ is a right transversal of $H$ in $G$. Of course $K$ contains $e$ and is inverse-closed. So, by Corollary~\ref{gtlem0}, $H$ is a perfect code of $G$.
\hfill$\Box$\vspace{11pt}

\medskip

\noindent {\em Proof of Theorem~{\rm\ref{newrem}}.}
Suppose that $G\cong \mathbb{Z}_{2^m}$, where $m\ge 2$. Let $G=\langle g\rangle$ and let $H=\langle g^t\rangle$ be any nontrivial proper subgroup of $G$. It is clear that $t$ is even. It follows that $g^t\in G^2\cap H$. Note that $2|H^2|=|H|$ and $H^2$ is a subgroup of $G$. We have $g^t\notin H^2$ and hence $G^2\cap H\ne H^2$. This means that $H$ is not a $2$-pure subgroup of $G$, and so $H$ is not a perfect code of $G$ by Theorem~\ref{abethm1}.

Conversely, suppose that for any nontrivial proper subgroup $H$ of $G$, $H$ is not a perfect code of $G$. Then by Corollary~\ref{newthm1-cor1}, $G$ has even order.
If $G=H\times K$ with $|H|\ge 2$ and $|K|\ge 2$, then from Lemma~\ref{comp} it follows that $H$ is a perfect code of $G$, a contradiction. Therefore, $G$ is a cyclic group with order a power of $2$, as desired.
\hfill$\Box$\vspace{11pt}

\section{Proof of Theorem \ref{newmainth2}}
\label{Se}

It is known that the subgroups of $Q_{4n}$ (where $n \ge 2$) are $\langle x^t\rangle$ and $\langle x^t, x^sy\rangle$, where $t$ is a positive integer dividing $2n$ and $s$ is an integer with $0\le s \le t-1$. Clearly, $\langle x^t, x^sy\rangle$ is either the cyclic group $\langle y\rangle$ or a generalized quaternion group.

We observe that, for any odd integer $q \ge 3$ and any odd integer $i$ between $1$ and $q-2$, the number $j = \frac{q-i}{2}$ is an integer between $1$ and $\frac{q-1}{2}$ satisfying $i+2j=q$.

\begin{lem}\label{g1}
Let $n \ge 2$ be an integer and $t$ a positive integer dividing $2n$. Then the proper subgroup $\langle x^t\rangle$ of $Q_{4n}$ is a perfect code of $Q_{4n}$ if and only if $\frac{2n}{t}$ is odd. Moreover, if $\frac{2n}{t}$ is odd, then ${\rm Cay}(Q_{4n}, S)$ with $S$ as given in \eqref{eq:s1} admits  $\langle x^t\rangle$ as a perfect code.
\end{lem}

\proof
Denote $G=Q_{4n}$ and $H=\langle x^t\rangle$.

Suppose that $H$ is a perfect code of $G$. Suppose to the contrary that $\frac{2n}{t}$ is even. That is, $|H|$ is even and so $x^n\in H$.
Take $g\in G\setminus\langle x\rangle$. Then $g^2=x^n\in H$.
Since $H$ is normal in $G$, by Theorem~\ref{basic} we have $(gh)^2=e$ for some $h\in H$.
Since $g\notin H$, we deduce that $gh$ is an involution.
It follows that $gh=x^n$, which implies that $g=x^nh^{-1}\in H$, a contradiction. Hence $\frac{2n}{t}$ must be odd.

Suppose that $\frac{2n}{t}=q$ is odd. Then $t$ is even and $|H|=q$. Using Theorem~\ref{basic}, we are going to prove that $H$ is a perfect code of $G$. Clearly, if $q=1$, then $H=\langle e\rangle$ is a perfect code of $G$. Assume that $q\ge 3$ in the sequel. Consider any $g\in G$ such that $g^2\in H$. Say, $g^2=x^{it}$ for some $1\le i \le q$. The order of $g$ is not equal to $4$. So we have $g\in \langle x\rangle$ and $g=x^k$ for some $1\le k \le 2n$. Hence $x^{2k}=x^{it}$. If $g\in H$, then $(gh)^2=e$ for $h = g^{-1}\in H$. Assume that $g\in \langle x\rangle\setminus H$ in the remaining proof. Set $k' = k$ if $1 \le k \le n$ and $k' = k-n$ if $n < k \le 2n$. Since $x^{2n}=e$ and $x^{2k}=x^{it}$, we have $x^{2k'}=x^{it}$. Since $1 \le k' \le n$, it follows that $2k' = it$. So $i$ is odd as $g\notin H$. If $i=q$, then $g=x^n$ and taking $h=e$ we obtain that $(gh)^2=e$. Now assume $1 \le i \le q-2$. Setting $j = \frac{q-i}{2}$, by the observation before Lemma \ref{g1} we see that $j$ is an integer satisfying $i+2j=q$ and $1 \le j \le \frac{q-1}{2}$. Since $2k' = it$ and $tq=2n$, we have $k' + j t = n$. Thus $(gh)^2=e$ for $h=x^{jt}\in H$. In summary, we have proved that for any $g\in G$ with $g^2\in H$ there exists $h \in H$ such that $(gh)^2=e$. Therefore, by Theorem~\ref{basic}, $H$ is a perfect code of $G$.

We now construct an inverse-closed subset $S$ of $G \setminus \{e\}$ such that $H=\langle x^t\rangle$ is a perfect code in ${\rm Cay}(G, S)$ under the condition that $\frac{2n}{t}=q$ is odd. If $q = 1$, then we can take $S = G \setminus \{e\}$ (which agrees with \eqref{eq:s1} as $t=2n$). Assume that $q \ge 3$ in the sequel. It is clear that $\{x^i: 0 \le i \le t-1\}$ is a right transversal of $H$ in $\langle x \rangle$.
Since $x^{2n-i} = x^{(q-1)t}x^{t-i} \in Hx^{t-i}$ for each $i$, it follows that $\left\{e, x^i, x^{2n-i}: 1 \le i \le \frac{t}{2}-1\right\}$ is a right transversal of $H$ in $\langle x\rangle$. Since $x^{\frac{t(q-1)}{2}}\in H$, we have $x^n = x^{\frac{t(q-1)}{2}}x^{\frac{t}{2}} \in Hx^{\frac{t}{2}}$ and hence $A := \left\{e, x^n, x^i, x^{2n-i}: 1 \le i \le \frac{t}{2}-1\right\}$ is a right transversal of $H$ in $\langle x\rangle$. Moreover, $A$ is inverse-closed as $x^{2n} = e$. Let $B := \{x^i y: 0 \le i \le t-1\}$. Then $\cup_{b \in B} Hb = G\setminus\langle x\rangle$ and $Hb\ne Hb'$ for distinct $b, b' \in B$. Since $q \ge 3$ is odd, we have $x^{\frac{(q-1)t}{2}}\in H$. Hence $x^{n+j}y = x^{\frac{(q-1)t}{2}} (x^{\frac{t}{2}+j}y) \in H(x^{\frac{t}{2}+j}y)$ for $0\le j \le \frac{t}{2}-1$. It follows that $C := \left\{x^i y, x^{n+i} y: 0 \le i \le \frac{t}{2}-1\right\}$ satisfies $\cup_{c \in C} Hc = G\setminus\langle x\rangle$ and $Hc \ne Hc'$ for distinct $c, c' \in C$. Moreover, $C$ is inverse-closed as $(x^ry)^{-1}=x^{n+r}y$ for any integer $r$. Therefore, $A \cup C$ is an inverse-closed right transversal of $H$ in $G$. Setting $S := (A \cup C) \setminus\{e\}$, we obtain from Lemma~\ref{mainlem0} that $H$ is a perfect code in ${\rm Cay}(G, S)$. Note that $S$ is equal to the subset defined in \eqref{eq:s1}.
\hfill$\Box$\vspace{11pt}

We can also prove the sufficiency of Lemma \ref{g1} using \cite[Lemma 2.10(a)]{HXZ18}. As usual, for a group $G$, let $\mathbb{Z}[G]$ be the group ring of $G$ over $\mathbb{Z}$. For a subset $A$ of $G$, write
$$
\overline{A}=\sum_{g\in G}\mu_A(g)g\in \mathbb{Z}[G],
$$
where
\begin{equation*}
\mu_A(g)=
\left\{
       \begin{array}{ll}
         1, & \hbox{$g\in A$;} \\
         0, & \hbox{$g\notin A$.}
       \end{array}
      \right.
\end{equation*}

\begin{lem}{\rm (\cite[Lemma 2.10(a)]{HXZ18})}
\label{basichuan}
Let $G$ be a group and ${\rm Cay}(G,S)$ a Cayley graph of $G$.
Let $C$ be a subset of $G$. Then $C$ is a perfect code in ${\rm Cay}(G,S)$ if and only if $\overline{S\cup\{e\}}\cdot\overline{C}=\overline{G}$.
\end{lem}

It is straightforward to verify that, for any positive divisor $t$ of $2n$ such that $\frac{2n}{t}$ is odd, we have $\overline{S\cup\{e\}}\cdot \overline{\langle x^t\rangle}=\overline{Q_{4n}}$, where $S$ is as defined in \eqref{eq:s1}. Hence, by Lemma \ref{basichuan}, $\langle x^t\rangle$ is a perfect code in ${\rm Cay}(Q_{4n}, S)$, proving the sufficiency of Lemma \ref{g1}. Using the same method, we can also prove the following lemma.

\begin{lem}\label{g2}
Let $n \ge 2$ be an integer, $t \ge 3$ an integer dividing $2n$, and $s$ an integer with $0\le s \le t-1$. Then the proper subgroup $\langle x^t, x^sy\rangle$ of $Q_{4n}$ is a perfect code
of $Q_{4n}$ if and only if $t$ is odd. Moreover, for any odd divisor $t \ge 3$ of $2n$ and any integer $s$ with $0\le s \le t-1$, ${\rm Cay}(Q_{4n}, S)$ with $S$ as given in \eqref{eq:s2} admits $\langle x^t, x^sy\rangle$ as a perfect code.
\end{lem}

\proof
Denote $G=Q_{4n}$ and $H=\langle x^t, x^sy\rangle$. Then $|H|$ is even and $|G:H|=t$. First, suppose that $H$ is a perfect code of $G$. That is, $H$ is a perfect code of some Cayley graph ${\rm Cay}(G,S)$ of $G$. Then, by Lemma~\ref{mainlem0}, $S\cup\{e\}$ is a left transversal of $H$ in $G$. Hence $|S|=t-1$. Suppose to the contrary that $t$ is even. Then $|S|$ is odd. Since $S^{-1}=S$, it follows that $x^n\in S$. On the other hand, as $|H|$ is even, we have $x^nH=H$, which contradicts the fact that $S\cup\{e\}$ is a left transversal of $H$ in $G$. Hence $t$ must be odd.

Conversely, suppose that $t \ge 3$ is odd. Note that
$$
\overline{H}=(e+x^t+x^{2t}+\cdots+x^{2n-t})(e+x^sy).
$$
Set $m = \frac{t-1}{2}$. Let $S=\{x,x^{-1},x^2,x^{-2},\ldots,x^m,x^{-m}\}$ be the subset of $G$ as given in \eqref{eq:s2}. Observe that $|S|=2m=t-1$.
It follows that
\begin{eqnarray*}
\overline{S\cup\{e\}}\cdot\overline{H}&=&(e+x+x^2+\cdots+x^{m}+
x^{-1}+x^{-2}+\cdots+x^{-m})
\\
& &(e+x^t+x^{2t}+\cdots+x^{2n-t})(e+x^sy)\\
&=&(e+x+x^2+\cdots+x^{2n-1})(e+x^sy)\\
&=&(e+x+x^2+\cdots+x^{2n-1})+(e+x+x^2+\cdots+x^{2n-1})x^sy\\
&=&\overline{G}.
\end{eqnarray*}
Therefore, by Lemma~\ref{basichuan}, $H$ is a perfect code of ${\rm Cay}(G,S)$.
\hfill$\Box$\vspace{11pt}

\noindent {\em Proof of Theorem~{\rm\ref{newmainth2}}.}
Theorem~\ref{newmainth2} follows from Lemmas~\ref{g1} and \ref{g2} immediately.
\hfill$\Box$\vspace{11pt}

We conclude the paper by the following example to illustrate Theorem \ref{newmainth2}.

\begin{ex}
\label{Q24}
Let $G=Q_{24}$. By Theorem~\ref{newmainth2}, we know that $G$, $\{e\}$, $\langle x^{4}\rangle$ and $\langle x^3,x^s y\rangle$ ($0\le s \le 2$) are all subgroup perfect codes of $G$. More explicitly, ${\rm Cay}(G, S_1)$ with $S_1=\{x,x^6,x^{11},y,xy,x^6y,x^7y\}$ admits $\langle x^{4}\rangle$ as a perfect code, and ${\rm Cay}(G, S_2)$ with $S_2=\{x,x^{11}\}$ admits $\langle x^3,x^s y\rangle$ as a perfect code for each $0\le s \le 2$. We draw ${\rm Cay}(G, S_1)$ in Figure~\ref{Q24-1}, where for each $i$ with $0 \le i \le 5$, $x^i$ is joined to $x^{i+6}$ by an edge. From this drawing one can easily see that $\langle x^{4}\rangle = \{e, x^4, x^8\}$ is a perfect code of ${\rm Cay}(G, S_1)$; that is, the vertices $e$, $x^4$ and $x^8$ are pairwise non-adjacent and every other vertex is adjacent to exactly one of these three vertices. One can also see that ${\rm Cay}(G, S_2)$ is disconnected with two connected components, namely the $12$-cycles $(e, x, x^2, \ldots, x^{11}, e)$ and $(y, xy, x^2 y, \ldots, x^{11}y, y)$.

\smallskip
\begin{figure}[hptb]
  \centering
  \includegraphics[width=10cm]{Q1}\\
  \caption{The graph ${\rm Cay}(G, S_1)$ in Example \ref{Q24}.}
  \label{Q24-1}
\end{figure}
\end{ex}

\bigskip
\noindent \textbf{Acknowledgements}~~We are grateful to the anonymous referees for their careful reading and helpful comments, and to Binzhou Xia for informing us an error in an early version of this paper. Ma was supported by the National Natural Science Foundation of China (Grant No. 11801441)
and the Natural Science Basic Research Program of Shaanxi (Program No. 2020JQ-761).
Wang was supported by the National Natural Science Foundation of China (Grant No. 11671043). Zhou was supported by the National Natural Science Foundation of China (Grant No. 61771019) and the Research Grant Support Scheme of The University of Melbourne.

{\small

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text:
Carriera

Club

Akabe si trasferì nel Trabzon İdmanocagı il 4 dicembre 2015. Ha segnato il suo primo gol due giorni dopo contro il Kireçburnu Spor in una vittoria per 2 a 0.

Nazionale
Henriette Akaba  gioca nell'edizione 2012 della Coppa delle Nazioni Africane femminile, in cui la nazionale arriva al terzo posto, grazie alla vittoria per 1-0 sulla Nigeria nella finale per il terzo posto, e ottenendo la finale nel 2014 avvenuta grazie alla sconfitta per 2-0 contro la Nigeria garantirà a lei e alle sue compagne la storica qualificazione alla fase finale di un campionato mondiale di calcio femminile all'edizione di Canada 2015.

Viene inserita nella rosa della Nazionale in partenza per il Campionato mondiale di calcio femminile 2015 che si gioca in Canada. In questa edizione dopo aver superato il girone eliminatorio con sei punti; frutto di due vittorie e una sconfitta, viene eliminata agli ottavi di finale dalla Nazionale di calcio femminile della Cina con il risultato di 1 a 0 a favore delle asiatiche.

Nel 2016 partecipa alla Coppa delle Nazioni Africane femminile in cui la sua nazionale era qualificata di diritto essendone l'organizzatrice. Il Camerun arriverà fino in finale dove però viene sconfitta dalla nazionale Nigeriana per 1 a 0.

Note

Altri progetti

Collegamenti esterni

Calciatrici della Nazionale camerunese

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Offer ends in 9hrs 15min 52sec!
The minute anyone sees you in this Super Bowl LIII Bound Tom Brady Eligible Receiver Name & Number V-Neck T-Shirt from NFL Pro Line by Fanatics Branded, your allegiance to one of the best teams on the gridiron will be noticed. The vibrant New England Patriots graphics on this top are sure to supply you with ample team spirit during the big game. You're sure to love this New England Patriots gear the minute you put it on.


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Dreamfall Chapters
Dreamfall Chapters is a third-person 3D adventure and sequel to award-winning adventure games The Longest Journey and Dreamfall. Set in parallel worlds — a dark cyberpunk vision of the future, and a magical fantasy realm — the story follows university dropout Zoë and disgraced apostle Kian working together to unravel a conspiracy spanning multiple realities.
Dreamfall Chapters is available in a complete edition for PlayStation 4 and Xbox One. The Final Cut edition is available for PC and Mac via Steam, GOG and the Humble Store. Both versions feature improved art, sound and music, and new special features.
Dreamfall Chapters was developed with support from the Norwegian Film Institute.
Draugen is a first-person, single-player psychological Fjord Noir mystery, set on the northwestern coast of Norway in 1923.
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Over the course of seven days in the remote village of Graavik, Edward — together with Lissie — will investigate his sister's disappearance and learn the truth of the eponymous Draugen.
Draugen is in production for Windows PC, PlayStation 4 and Xbox One.
Draugen is created with the support of the Creative Europe programme of the European Union, the Norwegian Film Institute and Viken Filmsenter.
Svalbard is a post-apocalyptic Polar Punk survival RPG that takes place on the island of Svalbard at the end of a nuclear winter, a thousand years from now.
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Q: Unit testing Abstract classes in Groovy I am new to unit testing and mocking. I am trying to unit test an abstract domain class in Grails. How should I mock an implementation so I can unit test the constraints of the domain class? Is there a way to use the mock libraries that come with groovy or grails? Should I just implement a class that simply extends the abstract class?

A: One cool thing about groovy (among many) is that you can use a map of method names with closures as values to mock out a class.  This includes abstract classes.
abstract class Foo {
    def foo() {
       return bar() + 1
    }

    abstract int bar()
}

def fooInst = [bar: {-> return 1 }] as Foo
assert 2 == fooInst.foo()

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text:
package net.canadensys.harvester.occurrence.processor;

import java.util.Map;

import net.canadensys.dataportal.occurrence.model.ResourceContactModel;
import net.canadensys.harvester.ItemProcessorIF;
import net.canadensys.harvester.exception.ProcessException;
import net.canadensys.harvester.exception.TaskExecutionException;
import net.canadensys.harvester.occurrence.SharedParameterEnum;

import org.apache.log4j.Logger;
import org.gbif.metadata.eml.Address;
import org.gbif.metadata.eml.Agent;
import org.gbif.metadata.eml.Eml;

/**
 * Process org.gbif.metadata.eml.Eml to extract info into a ResourceContactModel
 * @author canadensys
 *
 */
public class ResourceContactProcessor implements ItemProcessorIF<Eml, ResourceContactModel>{

	//get log4j handler
	private static final Logger LOGGER = Logger.getLogger(ResourceContactProcessor.class);

	@Override
	public void init() {

	}

	@Override
	public ResourceContactModel process(Eml eml, Map<SharedParameterEnum, Object> sharedParameters)
			throws ProcessException {
		String sourceFileId = (String)sharedParameters.get(SharedParameterEnum.SOURCE_FILE_ID);

        if(sourceFileId == null){
			LOGGER.fatal("Misconfigured processor : needs  sourceFileId");
			throw new TaskExecutionException("Misconfigured processor");
		}
        Agent agent = eml.getContact();
        ResourceContactModel resourceContactModel = new ResourceContactModel();
        resourceContactModel.setSourcefileid(sourceFileId);
        resourceContactModel.setResource_name(eml.getTitle());
        resourceContactModel.setName(agent.getFullName());
        resourceContactModel.setPosition_name(agent.getPosition());
        resourceContactModel.setOrganization_name(agent.getOrganisation());

        Address agentAddress = agent.getAddress();
        resourceContactModel.setAddress(agentAddress.getAddress());
        resourceContactModel.setCity(agentAddress.getCity());
        resourceContactModel.setAdministrative_area(agentAddress.getProvince());
        resourceContactModel.setPostal_code(agentAddress.getPostalCode());
        resourceContactModel.setCountry(agentAddress.getCountry());

        resourceContactModel.setEmail(agent.getEmail());
        resourceContactModel.setPhone(agent.getPhone());

		return resourceContactModel;
	}

	@Override
	public void destroy() {

	}
}

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D.A. Pennebaker, Director of Dylan, Bowie Films and More, Dead at 94
August 4, 2019 by admin 0 Comments
Filmmaker D.A. Pennebaker, whose work includes the 1967 Bob Dylan tour film Don't Look Back and the 1973 David Bowie documentary Ziggy Stardust and the Spiders From Mars—has died. He was 94 years old, according to Variety and The Hollywood Reporter. Variety reports that Pennebaker died Thursday night (August 1) of natural causes.
Donn Alan Pennebaker began his career in 1953, when he filmed his first short Daybreak Express. The brief clip was shot in New York City, and offered a colorful glimpse into the bustling daily routine of its commuters. For the next 65 years, Pennebaker continued to make films, and though his work covered a wide range of topics, he is best known for capturing some of the most legendary rock musicians during pivotal moments in their careers.
In 1965, Pennebaker joined Bob Dylan on his final acoustic tour in England, filming the now-iconic concert documentary Don't Look Back. The 1967 film is now considered one of the great rock documentaries of the 20th century, as well as an early example of cinema verité filmmaking. In addition to filming Dylan, Pennebaker went on to shoot footage of countless other artists, including Jimi Hendrix playing the 1967 Monterey International Pop Festival (for Pennebaker's film Monterey Pop), Jerry Lee Lewis, Little Richard, Chuck Berry, and Bo Diddley (for 1969's Keep On Rockin'), and David Bowie (for 1973's Ziggy Stardust and the Spiders From Mars).
In 1993, Pennebaker released The War Room, an award-winning documentary that chronicled the minds behind Bill Clinton's 1992 presidential campaign. His body of work continued to evolve throughout his long career, taking on subjects as disparate as Broadway actors, Egyptian revolutionaries, and more. In 2013, Pennebaker received a Lifetime Achievement Oscar for his filmography.
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The Geometrical Serigraphy Of The Backlited Mirror Matches With The Front Texture, Making The Vanit Harmony Complete.
Waves Made By The Wind, Like A Velvet Beach, Memory Of Foreign Lands.
Handle For Oasi Front: Small Graphic Symbol On A Geometric Texture.


text:
For yderligere oplysninger, se Politik i Region Midtjylland. For generelle oplysninger, se Region Midtjylland.
Efter kommunal- og regionsrådsvalget, der blev afviklet den 17. november 2009, blev det afgjort, at Bent Hansen (S) stod til at genindtage posten som regionsrådsformand. Socialdemokraterne (S) blev det største parti med 15 mandater, Venstre (V) 11 og Socialistisk Folkeparti (SF) 6 mandater som de tre største partier.

Resultat

Kilde: Fintællingsresultat fra KMD Valg

Regionsrådsmedlemmer
Fordelingen af de 41 mandater mellem de nedenstående partier.

Kilde: Den nye Regionsråd  på KMD Valg.

Se også
 Regionsrådet i Region Midtjylland 2005-2009
 Regionsrådet i Region Midtjylland 2013-2017
 Region Midtjylland

Referencer

Region Midtjylland
Kommunal- og regionsrådsvalg 2009
Regionsrådsmedlemmer fra Danmark


text:
Q: Grails hibernate many to many issue My Domain has a relationship as per this:
class Offices {

String officeLocation
String officeName

static belongsTo = Department
static hasMany = [departments:Department, employees:Employee]

}

When the user is on the page that allows them to add and remove location this is done by loading templates multiple times depending on the users choice. When the user hits "Add Location" I need to make sure the instance remains. My function in the controller:
def addLocation() {

def officesInstance = new Offices(officeLocation: params.officeLocation, officeName: params.officeName)

render template: "form" , model: [officesInstance: officesInstance]

}

view code:
<g:each var="entry" in="${departmentList}">

                    <g:message code="department.locked.label" default="${entry.encodeAsHTML()}" />

                    <g:checkBox name="department_${entry}" value="${officesInstance.value}" />

        </g:each>

The problem I'm having is I'm getting the following error when the addLocation function is pressed.

| Error 2015-02-27 12:51:52,412 [http-bio-8080-exec-7]
ERROR hibernate.AssertionFailure  - an assertion failure occured (this may indicate a bug in Hibernate, but is more likely due to unsafe use of the session)
  Offices: null id in com.myapp.Offices entry (don't flush the Session after an exception occurs)
Line | Method
     200 | doFilter  in grails.plugin.cache.web.filter.PageFragmentCachingFilter

63 | doFilter  in grails.plugin.cache.web.filter.AbstractFilter
53 | doFilter  in . grails.plugin.springsecurity.web.filter.GrailsAnonymousAuthenticationFilter
49 | doFilter  in grails.plugin.springsecurity.web.authentication.RequestHolderAuthenticationFilter
82 | doFilter  in grails.plugin.springsecurity.web.authentication.logout.MutableLogoutFilter
1145 | runWorker in java.util.concurrent.ThreadPoolExecutor
615 | run . . . in java.util.concurrent.ThreadPoolExecutor$Worker
745 | run       in java.lang.Thread

| Error 2015-02-27 12:51:52,416 [http-bio-8080-exec-7]
  ERROR errors.GrailsExceptionResolver  - AssertionFailure occurred when processing request: [POST] /myapp/offices/addLocation - parameters:
officeLocation: London
formId: createForm
elementToReplace: elementToReplace
officeName: Office Name
null id in com.smstool.Offices entry (don't flush the Session after an exception occurs). Stacktrace follows:
  Offices: null id in com.myapp.Offices entry (don't flush the Session after an exception occurs)
     Line | Method
->>  200 | doFilter  in grails.plugin.cache.web.filter.PageFragmentCachingFilter

63 | doFilter  in grails.plugin.cache.web.filter.AbstractFilter
53 | doFilter  in grails.plugin.springsecurity.web.filter.GrailsAnonymousAuthenticationFilter
49 | doFilter  in grails.plugin.springsecurity.web.authentication.RequestHolderAuthenticationFilter
82 | doFilter  in grails.plugin.springsecurity.web.authentication.logout.MutableLogoutFilter
1145 | runWorker in java.util.concurrent.ThreadPoolExecutor
615 | run . . . in java.util.concurrent.ThreadPoolExecutor$Worker
745 | run       in java.lang.Thread


text:
https://alphauniverse.media.zestyio.com/ilce-6500_sel24f18z_right-2.be110857376e1c1dc5afaa178864837f.jpg
Endgadget's 24 Hours With The Sony α6500
Author Alpha Universe
Date 2016.11.18
Endgadget.com spent some time with the new Sony α6500 mirrorless camera shooting fast action. Edgar Alvarez took some sample shots while at a press event in Austin, Texas and noted that the camera "was a pleasure to use." Sony's new flagship mirrorless touts speed as a main feature and according to Alvarez, "Sony wasn't kidding when it said its α6500 was all about speed."
"I used the camera mostly with the Vario-Tessar T* E 16-70mm f/4 and FE 70-200mm f/2.8 lenses, which made it easy to take shots in the dark and at sports games. It's one of the few times I've been able to capture decent shots at a sporting event. Indeed, Sony says its goal is to appeal to appeal to sports photographers, and even some people who aren't professionals." - Edgar Alvarez, Endgadget.com
See what else Alvarez had to say about the Sony α6500 along with his sample shots at Endgadget.com
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text:
Walther Hellmuth Karl Otto Graßmann (* 1. Mai 1885 Neu-Luckow; † 27. September 1918 gefallen in den Vorgefechten zur Schlacht von Cambrai) war ein deutscher Landwirt und Ornithologe.

Leben und Wirken
Walther Graßmann wurde als drittes Kind des Landwirtes Adolf Graßmann und Anna geb. von Domarus geboren. Seine älteste Schwester war Elisabeth, die nach ihrer Heirat Thierbach hieß. Nach seiner Schulzeit begann Graßmann im Herbst 1902 eine Ausbildung als Landwirt in Ringenwalde. Es folgten die Stationen Parmen, Grauwinkel und Kloster Hadmersleben. Am Kloster Hadmersleben schien er eine gehobenere Stelle innegehabt zu haben, denn 1914 wurde er zu großen Wanderausstellung der Deutschen Landwirtschafts-Gesellschaft nach Hannover entsandt.

Mit Beginn des Ersten Weltkriegs diente er zunächst als Leutnant der Reserve an die Westfront. Es erfolgt die Versetzung nach Ost-Galizien und schließlich vom Sommer 1915 bis in den Sommer 1917 in das Gebiet der Prypjatsümpfe. Im Winterhalbjahr 1917/18 erfolgte die Versetzung ins Elsass und andere Gebiete der Westfront. Als er am 27. September 1918 als Batterieführer bei den Englisch-amerikanischen Massenangriffe Richtung Cambrai auf Erkundungsritt unterwegs war, kehrte er von diesem nicht mehr zurück und galt seither als vermisst.

Den noch neun erhaltenen Briefen von Graßmann an seine Eltern, kann man nur wenige Schilderungen über naturhistorische Beobachtungen entnehmen. Somit ist anzunehmen, dass er seine Begeisterung für die Ornithologie nicht durch seine Eltern vermittelt bekommen hat. Trotz allem erwähnt er in einem Brief aus dem Jahr 1914 stolz, dass er auf Kosten von Ferdinand Heine junior zur Versammlung des Deutschen Vereins zum Schutz der Vogelwelt nach Quedlinburg eingeladen worden war. So berichtete er begeistert von seiner ersten Begegnung mit Alwin Voigt (1852–1922), dessen Buch Excursionsbuch zum Studium der Vogelstimmen er erst kürzlich zu seinem Geburtstag geschenkt bekommen habe. Außerdem lernte er dort andere bekannte Ornithologen wie Hans Freiherr von Berlepsch, Johannes Thienemann, Oskar und Magdalena Heinroth, Carl Richard Hennicke und andere bei der Versammlung kennen. Während der Bericht zur Versammlung in der Ornithologische Monatsschrift relativ nüchtern geschrieben war, sprühte Graßmanns Brief geradezu voller Begeisterung über das Treffen.

In seinen Ausführungen zur Vogelwelt in den Prypjatsümpfen bezog sich Otto Eduard Graf von Zedlitz und Trützschler auf viele Beobachtungen aus Manuskripten von Graßmann und der dessen Publikationen zu den Rokitnosümpfen (ein Synonym für Prypjatsümpfe). Am Ende seiner ausführlichen Publikationen über das Gebiet schrieb von Zedlitz und Trützschle, wohl auch im Gedenken an Graßmann.

Graßmanns ornithologischen Schriften erschienen in den Fachzeitschriften Journal für Ornithologie, Ornithologische Monatsberichte und Ornithologische Monatsschrift.

Noch heute befindet sich eine Eiersammlung von Graßmann im Phyletischen Museum in Jena. Die oologische Sammlung aus 24 Gelegen und Einzeleiern von 15 Arten übergab Ursula Thierbach dem Museum gemeinsam mit einem Sonderdruck über die Vogelwelt des 420 ha umfassenden Guts- und Gemeindebezirk Grauwinkel b. Schönwalde.

Graßmanns Tod wurde erst am 23. April 1920 in Hadmersleben offiziell registriert.

Dedikationsnamen
Graf von Zedlitz und Trützschler widmete ihm 1920 die Haselhuhn-Unterart Tetrastes bonasia grassmanni, die heute aber als Synonym zur Nominatform gesehen wird.

Publikation (Auswahl)

Literatur

Einzelnachweise

Ornithologe
Leutnant (Preußen)
Oologe
Deutscher
Geboren 1885
Gestorben 1918
Mann


text:
Video: Tim Rosaforte hands out his "Rosies" on Golf Channel
By Sam Weinman
What's bigger than winning a Wanamaker Trophy and a Ryder Cup in the same year? How about throwing in a year-end "Rosie" from our own Tim Rosaforte as well?
Perhaps that's overstating it. Nonetheless, the Golf World senior writer saw fit to include Rory McIlroy among his honorees in his 2012 edition of The Rosies on Golf Channel's "Morning Drive" -- and not just for the world No. 1's exploits on the golf course.
See for yourself in the video below:


text:
Often asked: How Many Professional Soccer Players Are There In The World?
1 How many professional soccer players are there?
2 How many professional football players are in the world?
3 How many professional soccer teams are there in the world?
4 How many soccer players are there globally?
5 Which country has best footballers?
6 Which country is soccer most popular in?
7 Who is the god of football?
8 Who is the king of football in the world?
9 Is it hard to become a professional soccer player?
10 Who invented soccer?
11 Who won MLS 2020?
12 What is the highest governing body in soccer?
How many professional soccer players are there?
It is played by approximately 250 million players in over 200 countries and dependencies, making it the world's most popular sport. The game is played on a rectangular field called a pitch with a goal at each end.
How many professional football players are in the world?
It has been estimated that there are 250 million association football players in the world, and many play the other forms of football.
How many professional soccer teams are there in the world?
Major League Soccer's 27 teams are divided between the Eastern and Western Conferences.
How many soccer players are there globally?
Soccer is the world's most popular sport, played by over 250 million people in more than 200 countries. It is particularly popular in Europe, Central and South America, and Africa, though has a growing influence in North America and Asia.
Which country has best footballers?
Top 10 exporting countries in world football
Uruguay (288 footballers)
Nigeria (292 footballers)
Croatia (323 footballers)
Germany (335 footballers)
Spain (362 footballers)
England (451 footballers)
Serbia (460 footballers)
Argentina (753 footballers)
You might be interested: Question: How To Teach Your Dog To Play Soccer?
Which country is soccer most popular in?
By the number of fans, soccer is most popular in China, with 187 million people reported to have an interest in the sport. The country that's the greatest percentage number interested in soccer is The United Arab Emirates!
Who is the god of football?
He was none other than Diego Maradona, one of the world's greatest football players, also called 'The God of Football'. He saw heaven and hell on Earth and died Wednesday at the age of 60. Maradona was a player who, apart from scoring goals, also made mistakes.
Who is the king of football in the world?
430 Leo Messi: King of football ideas in 2021 | leo messi, messi, lionel messi.
Is it hard to become a professional soccer player?
Becoming a professional footballer is an incredibly difficult thing to accomplish and there are many things that a player must master to be successful. Everyone has a different football journey, it is unique to each player and the most important thing is to enjoy it.
Who won MLS 2020?
Philadelphia Union won the Supporters' Shield on the final day of the regular season, the first major trophy in the club's history. Columbus Crew SC defeated defending champions Seattle Sounders FC in MLS Cup 2020 3–0 to win a second MLS Cup title, and their first since 2008.
You might be interested: FAQ: How Many Teams Are In Major League Soccer?
What is the highest governing body in soccer?
Founded in 1904 to provide unity among national soccer associations, the Federation Internationale de Football Association (FIFA) boasts 209 members, rivaling that of the United Nations, and is arguably the most prestigious sports organization in the world.
Quick Answer: How Many Players Are On A Middle School Soccer Team?
Quick Answer: How Much Do Average Soccer Players Make?


text:
ALA Unveils 2017 Andrew Carnegie Medals Shortlist
Six books listed for excellence in fiction and nonfiction
The American Library Association (ALA) today announced six books shortlisted for the esteemed Andrew Carnegie Medals for Excellence in Fiction and Nonfiction, awarded for the previous year's best fiction and nonfiction books written for adult readers and published in the United States. The two medal winners will be announced at the Reference and User Services Association's (RUSA) Book and Media Awards event at ALA Midwinter Meeting & Exhibits in Atlanta, Georgia, on Sunday, January 22, 2017.
2017 Andrew Carnegie Medals for Excellence in Fiction and Nonfiction shortlist titles include:
Andrew Carnegie Medal for Excellence in Nonfiction 2017 Shortlist
The Firebrand and the First Lady: Portrait of a Friendship: Pauli Murray, Eleanor Roosevelt, and the Struggle for Social Justice, by Patricia Bell-Scott, published by Alfred A. Knopf, Penguin Random House LLC. Bell-Scott meticulously chronicles the boundary-breaking friendship of Pauli Murray and Eleanor Roosevelt, telling each remarkable woman's story within the context of the crises of the times, from ongoing racial violence to WWII and the vicious battle over school integration.
Evicted: Poverty and Profit in the American City, by Matthew Desmond, published by Crown, Penguin Random House LLC. Desmond shares harrowing stories of eight families who find themselves facing home evictions in Milwaukee, Wisconsin, shining a light on how eviction sets people up to fail.
Blood at the Root: A Racial Cleansing in America, by Patrick Phillips, published by W. W. Norton. Phillips presents a precise and disquieting account of long underreported tyranny and violence against African Americans in a farming community in Forsyth Country, Georgia, in 1912, which resulted in nothing less than racial cleansing.
Andrew Carnegie Medal for Excellence in Fiction 2017 Shortlist
Moonglow, by Michael Chabon, published by Harper, an imprint of HarperCollins. A young writer listens in breath-held astonishment as his ailing grandfather, whose lifelong reticence has been vanquished by strong painkillers, tells the hidden stories of his hardscrabble boyhood, WWII military service, obsession with moon missions, and love for a French Holocaust survivor.
Swing Time, by Zadie Smith, published by Penguin Press, Penguin Random House LLC. Two "brown girls" growing up in London public housing share a passion for dance but follow divergent paths, which lead to adventures in America and Africa, and raise complex questions about family, friendship, race, creativity, and celebrity.
The Underground Railroad, by Colson Whitehead, published by Doubleday, Penguin Random House LLC. Whitehead reimagines the Underground Railroad in this powerful tale about smart and resilient Cora, a young third-generation slave who escapes the brutality of a Georgia cotton plantation and seeks sanctuary throughout the terrorized South.
The Carnegie Medal winner announcements will be made at the Midwinter Meeting & Exhibits, which now offers 24 hours of ALA book and media award excitement with the Book and Media Awards preceding the ALA Youth Media Award (YMA) announcements. Carnegie Medal winners will each receive $5,000, and the four finalists each receive $1,500. All the finalists will be honored during a celebratory event at ALA's 2017 Annual Conference in Chicago.
The awards, established in 2012, recognize the best fiction and nonfiction books for adult readers published in the US in the previous year and serve as a guide to help adults select quality reading material. They are the first single-book awards for adult books given by the ALA and reflect the expert judgment and insight of library professionals who work closely with adult readers.
The medals are made possible, in part, by a grant from Carnegie Corporation of New York in recognition of Andrew Carnegie's deep belief in the power of books and learning to change the world, and are cosponsored by ALA's Booklist and RUSA.
Annotations and more information on the finalists and the awards can be found at http://www.ala.org/carnegieadult. Book cover artwork is available for download at http://tinyurl.com/Carnegieshortlist .
#alamw17
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