text:
Interpret "Pink Floyd"
Label: Warner Music Group
The Endless River (Deluxe Edition - Remastered)
The Early Years 1967-72 Cre/ation (Remastered)
Label: Pink Floyd Records
The Division Bell (2011 Remastered)
Biographie Pink Floyd
Inductees: Syd Barrett (guitar, vocals; born January 6, 1946; died July 7, 2006), David Gilmour (guitar, vocals; born March 6, 1944), Nick Mason (drums; born January 27, 1945), Roger Waters (bass, synthesizer, vocals; born September 9, 1944), Rick Wright (keyboards, synthesizers; born July 28, 1945; died September 15, 2008).
Pink Floyd's hallucinatory presentation of lights and music at London's Roundhouse in 1966 brought psychedelia to the U.K. scene. The group carried rock and roll into a dimension that was more cerebral and conceptual than what preceded it. What George Orwell and Ray Bradbury were to literature, Pink Floyd is to popular music, forging an unsettling but provocative combination of science fiction and social commentary. In their early years, with vocalist, guitarist and songwriter Syd Barrett at the helm, Pink Floyd were the psychedelic Pied Pipers of the "London underground" scene. In the Seventies, with bassist Roger Waters providing more of the songwriting and direction, Pink Floyd became one of the most influential rock bands of all time.
Before they settled on Pink Floyd, the group went by the names Sigma 6 and the Architectural Abdabs, and they mainly performed rhythm and blues covers. Singer-guitarist Syd Barrett provided Pink Floyd with most of its original early material, including the British hits "See Emily Play" and "Arnold Layne." Barrett's elfin, tuneful psychedelia made him the Lewis Carroll of the pop scene. Pink Floyd's debut album, Piper at the Gates of Dawn, is a classic of psychedelic whimsy that epitomized the remarkable year of 1967 at its most playful and creative. As the British music magazine Q opined in 1995, "Piper at the Gates of Dawn is, even counting Sgt. Pepper, possibly the defining moment of English psychedelia and Syd Barrett's magnum opus." Among its highlights was a nine-minute instrumental, "Interstellar Overdrive," that represented one of rock's first forays into deep space. It was a preoccupation of Pink Floyd's that would later surface in songs like "Set the Controls for the Heart of the Sun" (from A Saucerful of Secrets) and the group's masterwork, Dark Side of the Moon.
Intense experimentation with LSD unfortunately transported Barrett from enlightenment to mental instability, and increasingly unpredictable behavior necessitated his departure from Pink Floyd in 1968. Among the prime "acid casualties" of the Sixties, Barrett subsequently released two magnificent, if eccentric, solo albums – The Madcap Laughs and Barrett, both from 1970 – with considerable input from his erstwhile bandmates in Pink Floyd. Thereafter, however, Barrett became one of rock's most legendary hermits and the subject of Roger Waters' tributary opus "Shine On You Crazy Diamond." It was the side-long centerpiece of Pink Floyd's Wish You Were Here (1975) and a sterling example of what the group has referred to as its recurring "theme of absence."
With guitarist David Gilmour on-board as Barrett's replacement, Pink Floyd's lineup remained constant for the next 15 years. In the wake of Piper, they recorded psychedelic soundscapes such as A Saucerful of Secrets and the double album Ummagumma, which comprised one disc of live performances and one of individual works by each band member. Laid-back but experimental, Pink Floyd kicked off the Seventies with the pastoral, atmospheric albums Atom Heart Mother (1970) and Meddle (1971). Each featured a side-long epic, "Atom Heart Mother Suite" and "Echoes," respectively. Fittingly for a band with who took a cinematic approach to music, Pink Floyd provided music for three films. Their work as film scorers can be heard on the soundtrack albums More (1969), Zabriskie Point (1970) and Obscured by Clouds: Music from La Vallee (1972).
Their 1973 release Dark Side of the Moon hit Number One on the Billboard charts and ultimately broke all records by remaining on the Top 200 album charts for 741 weeks. Dark Side of the Moon did not drop off Billboard's Top 200 album chart until 1988. The album signaled rock's willingness to move from adolescence into adulthood, conceptually addressing such subjects as aging, madness, money and time. From its prismatic cover artwork to the music therein, Dark Side of the Moon is a classic-rock milestone. The subject of alienation was further explored in Wish You Were Here (1975), an album whose central preoccupation was the band members' distance from each other ("Wish You Were Here") and erstwhile leader Syd Barrett's distance from reality ("Shine On You Crazy Diamond"). They turned their gaze outward yet again on the Orwellian Animals (1977), whose songs bore the titles "Pigs," "Sheep" and "Dogs."
Success continued into the Eighties with The Wall, a four-sided epic about a rock star named Pink who suffers a nervous breakdown while on tour. Much of it reflected chief architect Roger Waters' dim view of the concert experience as rock expanded into arenas and stadiums. "I wanted to make comparisons between rock and roll concerts and war," Roger Waters toldRolling Stone in 1982. He elaborated on this central tenet in the liner notes forThe Wall Live: 1980-81: "The idea that we, as individuals, generally find it necessary to avoid or deny the painful aspects of our experience, and in fact often use them as bricks in a wall behind which we may sometimes find shelter, but behind which we may just as easily become emotionally immured, relatively simply stated and easy to grasp." That, in a nutshell, is the theme pursued by Pink Floyd from Dark Side of the Moon forward.
Possibly the most pessimistic album ever to reach #1, The Wall also addressed childhood, education and marriage, finding all of these passages to be dehumanizing. The Wall, the most theatrical and complex stage show that rock had ever seen, was performed 24 times in multi-night stands at four places - London, Los Angeles, Long Island and Dortmund, Germany. During the performance, an actual "wall" was constructed in front of the band, and its collapse at the end provided a fitting denouement. The Wall was subsequently revived by Roger Waters for a star-studded staging in Berlin in 1990, to commemorate the unification of East and West Germany. Performances from the Pink Floyd's original staging of the epic saw release in 2000 as The Wall Live: 1980-81.
In the wake of The Wall, Pink Floyd itself gradually seemed to collapse, at least temporarily. The Wall turned out to be the last album the foursome of Waters, Gilmour, Wright and Mason recorded together. The Final Cut, which was recorded under extreme duress, found Wright absent from the group. Almost wholly Waters' vision, it was an antiwar album triggered by Britain's 1982 conflict in the Falkland Islands. The group unofficially disbanded after its release, and that seemed to mark the end of Pink Floyd, as the members involved themselves in endeavors, including solo projects, outside the band.
Throughout their history, the members of Pink Floyd have projected a rather static personal image, allowing music, lyrics, lighting and theatrical settings to communicate for them. Consequently, they've largely avoided the sort of public scrutiny that typifies the lives of rock stars. Little was known or reported about their personal lives. Only when a bitter war of words and a court battle erupted between Roger Waters and the others after Gilmour, Mason and Wright reconvened Pink Floyd was the silence broken.
Pink Floyd released Momentary Lapse of Reason in 1987 and followed it up a year later with Delicate Sound of Thunder, a live album drawn from an extensive tour. The group reconvened in the Nineties with Gilmour again at the helm, releasing The Division Bell in 1994 and another tour souvenir,Pulse, a year later. Both albums went to the top of the charts, proving that the public's fascination with this most unconventional supergroup had not dimmed in the least. (Source: www.rockhall.com)


Exhaust threads repaired, new top end gaskets, freshened up the head, carbs cleaned, a few parts here an there and its a runner.
I believe this is a '69 Commando S that has been modified a bit. It's been sitting outside since 1990. Engine is stuck solid. I have the go ahead to explore the situation and see how bad the damage is.
1. At the rise of the hand by Policeman, stop rapidly.
Do not pass him by or otherwise disrespect him.
tootel him with vigor and express by word of mouth, warning Hi, Hi.
fright as you pass him. Do not explode the exhaust box at him.
4. Give big space to the festive dog that makes sport in roadway.
Avoid entanglement of dog with wheel spokes.
Great on-board footage of the same course.


Santa Rosa Junior College is set to host special guest Belva Davis, Emmy Award-Winning Broadcast Journalist, for a College Forum, Thursday, April 26, 2012 at 12:30 p.m. at the Carole L. Ellis Auditorium, Petaluma Campus.
Davis will be speaking about her career in journalism and her book entitled "Never in My Wildest Dreams: A Black Woman's Life in Journalism." This event is free and open to the public. The book, co-written by Petaluma-based journalist Vicki Haddock, is available at the SRJC Bookstore on both the Santa Rosa and Petaluma Campuses.
Belva Davis, who has a home in the Petaluma area, is recognized as the first black woman TV journalist in the West and one of the most respected media figures in the nation. She has indisputably changed the face of American journalism.
In her book Never in My Wildest Dreams, Davis recounts a harrowing personal history interwoven with the volatile cultural upheaval of the '60s, '70s, and '80s. In her five decades of reporting, she covered the most explosive local and national stories of the last half-century, including the Berkeley student protests, the rise of feminism, the birth of the Black Panthers, the Moscone/Milk murders, the onset of the AIDS epidemic and the terrorist attacks of 9/11.
This extraordinary woman will share her story of courage, grace, and determination as she witnessed and reported on many of the most transformative events of her generation. Having won eight local Emmys and a number of lifetime achievement awards, Davis has established herself as a cultural icon intimately tied to American history.


- In general, I support discussing how systemic failures and economic injustice can partially explain some individual violent crimes, including murders and suicides. But it seems like the only widely accepted context for such discussions is with the murder of disabled children. Similar discussions involving other kinds of people and situations … such as gang violence and school shootings … are generally despised as excusing and coddling criminals. Somehow, though, killing disabled people is in some way "understandable". How can that be anything but galling and scary for people who have disabilities?
- We absolutely should talk more about the unacceptable delays and gaps in support services to parents of kids with disabilities ... but never in connection with the murder of disabled children. We cannot afford to undermine the idea that disabled children … even the most difficult and baffling … are sentient human beings.
- We shouldn't assume that parents who go off the deep end are always ill-served. Some fumble or even reject opportunities for support, out of confusion, exhaustion, or failure to recognize promising pathways when they appear. Some parents also find it hard to engage support because they feel it's a weakness to ask for help, or because they are terrified of being judged or losing parental authority. All of these can be legitimate concerns, but are never enough to justify, or even properly explain, murder.
- When we do discuss systemic failures, we also need to discuss the influence of well-funded, popular, but unhelpful, wrong, and destructive ideas about disability and child rearing that are regularly fed to parents who often don't even know there are other points of view. With autism, especially, parents urgently need to be told that no matter how mystifying their behavior and communication might be, autistic children are first and foremost people, not wild animals or tortured, miserable monsters.
- One reason why a note of sympathy keeps coming up in cases where parents murder their disabled children is that by and large, people find it easier to identify with frustrated parents at the end of their rope, than with children who have what seem like mysterious, frightening disabilities.
- To repeat … discussing the lack of support for parents is fine and necessary, but NOT IN CONNECTION WITH ACTUAL MURDER OR ATTEMPTED MURDER OF DISABLED CHILDREN OR ADULTS. It should be a taboo, indecent, simply not done.


text:
Summary win32k.sys in the kernel-mode drivers in Microsoft Windows XP SP2 and SP3, Windows Server 2003 SP2, Windows Vista SP2, Windows Server 2008 SP2, R2, and R2 SP1, Windows 7 Gold and SP1, and Windows 8 Consumer Preview does not properly manage Keyboard Layout files, which allows local users to gain privileges via a crafted application, aka "Keyboard Layout File Vulnerability."
cpe:2.3:o:microsoft:windows_7:*:*:*:*:*:*:*:*
cpe:2.3:o:microsoft:windows_server_2008:-:sp2:*:*:*:*:*:*
cpe:2.3:o:microsoft:windows_server_2008:r2:*:*:*:*:*:*:*
cpe:2.3:o:microsoft:windows_server_2008:r2:sp1:*:*:*:*:*:*
Manipulating Web Input to File System Calls
An attacker manipulates inputs to the target software which the target software passes to file system calls in the OS. The goal is to gain access to, and perhaps modify, areas of the file system that the target software did not intend to be accessible.
Leverage Executable Code in Non-Executable Files
An attack of this type exploits a system's trust in configuration and resource files. When the executable loads the resource (such as an image file or configuration file) the attacker has modified the file to either execute malicious code directly or manipulate the target process (e.g. application server) to execute based on the malicious configuration parameters. Since systems are increasingly interrelated mashing up resources from local and remote sources the possibility of this attack occurring is high.
Using Malicious Files
An attack of this type exploits a system's configuration that allows an attacker to either directly access an executable file, for example through shell access; or in a possible worst case allows an attacker to upload a file and then execute it. Web servers, ftp servers, and message oriented middleware systems which have many integration points are particularly vulnerable, because both the programmers and the administrators must be in synch regarding the interfaces and the correct privileges for each interface.
Target Programs with Elevated Privileges
This attack targets programs running with elevated privileges. The attacker would try to leverage a bug in the running program and get arbitrary code to execute with elevated privileges. For instance an attacker would look for programs that write to the system directories or registry keys (such as HKLM, which stores a number of critical Windows environment variables). These programs are typically running with elevated privileges and have usually not been designed with security in mind. Such programs are excellent exploit targets because they yield lots of power when they break. The malicious user try to execute its code at the same level as a privileged system call.
Restful Privilege Elevation
Rest uses standard HTTP (Get, Put, Delete) style permissions methods, but these are not necessarily correlated generally with back end programs. Strict interpretation of HTTP get methods means that these HTTP Get services should not be used to delete information on the server, but there is no access control mechanism to back up this logic. This means that unless the services are properly ACL'd and the application's service implementation are following these guidelines then an HTTP request can easily execute a delete or update on the server side. The attacker identifies a HTTP Get URL such as http://victimsite/updateOrder, which calls out to a program to update orders on a database or other resource. The URL is not idempotent so the request can be submitted multiple times by the attacker, additionally, the attacker may be able to exploit the URL published as a Get method that actually performs updates (instead of merely retrieving data). This may result in malicious or inadvertent altering of data on the server.
description win32k.sys in the kernel-mode drivers in Microsoft Windows XP SP2 and SP3, Windows Server 2003 SP2, Windows Vista SP2, Windows Server 2008 SP2, R2, and R2 SP1, Windows 7 Gold and SP1, and Windows 8 Consumer Preview does not properly manage Keyboard Layout files, which allows local users to gain privileges via a crafted application, aka "Keyboard Layout File Vulnerability."
title Keyboard Layout File Vulnerability

redpajama_set_name: RedPajamaCommonCrawl


text:
Shearwater, the Mullumbimby Steiner School
349 Left Bank Road, Mullumbimby, NSW 2482
Origin and Foundation >
Eurythmy
Learning Program
Steiner's Spiritual Science
The Arts and Steiner Education
Shearwater Steiner School >
A bit of history and geography
A special community
Structure of the School Community
Why Shearwater? >
Why Is Our School Special
Working at Shearwater
The School Year
Safe on Social Media
Early Childhood >
Early Childhood Enrolments
Student Exits
Shearwater Wearable Arts
School contact details
Class 5 Greek Olympics 2019
If you have practiced hard for Olympia
And if you have not been lazy or done
Anything dishonourable, then
Go forward with confidence.
We will judge before the gods
with truth and honour
All those who compete before us.
Judges' verse
Last Friday, athletes from the mighty city-states of ancient Greece (Shearwater, Cape Byron, Aetaomah, Casuarina, Chrysalis, Tallowood and Rainbow Ridge Steiner schools) came together, in the shadow of Olympus, to prove their strength, grace and agility before the gods. Irascible Zeus opened the proceedings, imploring the athletes to lift their efforts into the realm of truth, beauty and goodness. With gifts of song, strings, verse and gymnastics, let the games begin!
Mixed-school groups went around the pentathlon stations where rigorous standards of sportsmanship were apparent in each and every encounter. New friends encouraged each other to greater and more graceful efforts.
Discus events aimed at 'floating' the discus aerodynamically; Running was approached as 'graceful technique.;
the Standing Long Jump employed 'coordination and weight shift', Wrestling worked on 'balance, strength and poise.; Javelin, demonstrated by Zeus with lightning bolts, required parabolic arcing and point-perfect landings.
Throughout the day, the athlete's inner qualities of determination, effort and sportsmanship were as highly valued as strength and skill.
After suitable midday repasts, Zeus called all athletes to relive the exploits of Pheidippides, the first Marathon runner. Mighty praise was showered on all athletes and medals presented. Athletes returned in glory to their city-states fulfilled and honoured.
Accurate Historic Account rendered by
Rheumatoid Pseudo Zeus the Imposter (aka Michael Lester)
Class 5 Teacher
Class 7 Trystan and Essylt
Class 7J has recently completed their class play, 'Trystan and Essylt'. This tale of love and sorrow, woven into Arthurian history, was taken up by the students, and over a two-and-a-half week period, they brought the story together and performed for their peers and parents.
This was an immense challenge for the students. Key actors fell ill. The script had to be extended and edited along the way, something many of the students worked upon, ensuring a connection with their characters. Costumes were created (and often re-created) as characters were developed. The drama room had to be set up to cater for the various scenes in the play, and the final touches of lights, music and backstage activity all combined to form the final piece in a very short amount of time.
As well as writing a number of scenes, two key students, Olive and Evie, took on the tasks of costuming and directing. Their interest and dedication to those roles supported the final production, beautifully.
Julie Marx
Spring Fair 2019
Class 6 student Matilda Vial (centre) picks up three gold medals at the CIS State Athletics last week.
Two of our Primary School athletes, Keahi Avery-Numa (Class 5) and Matilda Vial (Class 6) travelled to Sydney to compete in the CIS State Athletics last week, where Matilda continued her winning streak, bringing home gold in the 100m, 200m and Long Jump, with personal best times in all three events. Congratulations to both of them and good luck to Matilda who is now competing in the national championships in Darwin.
With a recent injection of funding to deliver extra physical activity to students in the Primary School and junior High School this term, years 3 to 8 have been enjoying gymnastics with Spaghetti Circus and classes 6 and 8 have had tennis lessons with Infinity Tennis Coaching.
Gymnastics and tennis provide children with opportunities to develop their fundamental movement skills by participating in a range of safe and engaging activities. The lesson plans were developed by coaches and teachers and are aligned with the newly endorsed Australian Curriculum. With the Term 4 funding, which has recently been received, we plan to establish an orienteering map of the School campus.
Also in Term 4, we are getting prepared for the regional tennis competition in Grafton and the Primary School basketball regionals in Byron Bay.
Magnum Brotto
PDHPE Teacher
Year 10 Vanuatu Camp
PICS by Jala and Endre
Pele Moments by Ellen Bartlett
The boat sits low in the water. Low enough to make us feel like we are sitting in the gentle lap of the sea. The deep, quiet thrum of the motor, constant and reassuring, reminding us that we are dry, safe, and privileged. Pushing us forward to new places yet to be experienced. A random handful of people, happy, satisfied, and on the start of their journey to contentment.
Today we danced. We danced at dusk on the sand, feeling the weight of the night slowly sink over us. Heat emanating from the center of the circle, warming over our bodies and our spirits. Taking us to a place of familiar unknown, bringing back nostalgia and filling in the empty slots with new dance. Movement compact with decades of culture and centuries of warm-hearted generosity. Different people from every way the wind blows brought together by the warm light of the fire that the soil is always craving. Sparks of joy flying, crackles of curiosity flickering and our energy slowly, slowly, burning.
Children by Finn Morris and Yanik Scheffers
The kids are full of energy and quickly grow attached to you, they are often bustling with excitement from the moment you wake to even late at night. Each family has at least two children who are introduced to us as our siblings.
The family we stayed with was large and we had many younger siblings between us. Ronny our eight-year-old brother was constantly demanding attention and always wanted to play a game, Jaylene our youngest sister, somewhere between the age of five and six, was extremely shy and would attach herself to Yanik and I, then cling for as long as she could. Rex, our smallest brother was hands-down the cutest kid on the island. He spent the first few days with us but soon left for the mainland with his father.
Taz and Ariel's house was a hub for the younger kids and was nearly always surrounded, which at times did get irritating. Due to the language barrier, it was sometimes difficult to communicate boundaries and ask them questions but the majority of the time it wasn't a huge problem.
Hard at Work by Finn Morris and Yanik Scheffers
The first days on the worksite, consisted of trips down the island shorelines collecting coral for the wall; mixing concrete; constructing the framework for roofs and putting together the window frames. The first half of the day, most of the class would come and go between the worksite and their bungalows, although some worked constantly, only resting when lunch came around. After lunch, there was usually a solid hour before everyone would have to return for the afternoon meeting. At these meetings there was often a firm pep talk from Nick, followed by a solid 15 minutes of hard work from the class before everyone would scatter across the island.
When the building was nearing completion more kids jumped on board to help finish off the building - carvings were engraved into the coral walls, patterns were chiseled into the wooden support beams and the final nails were hit into place.
Banners, flowers, and ribbons decorated the building when it reached completion, the islanders laid out an enormous feast of traditional meals along a huge centre table. After a two-hour ceremony of appreciation for the community's new meeting place, the feast commenced. The class lined up first and tried to fill their plates with the most recognisable foods. After us, the entire island began to gather around and form long spiraling lines.
During our time on Pele Island, it was clear to see just how much the building meant to the community. From the workers' dedicated enthusiasm to the project to the sheer joy expressed when the building reached completion - they saw the entire experience as a gift. In 12 days our class helped to create a building that would serve the community and bring joy for years to come.
We've had a wonderful year to date, full of laughter and learning together. Over the mid-year holidays the maintenance team manifested a fabulous sand pit in the little courtyard outside our classroom. It's under shade and private and the children adore playing in there - often all together. They are so creative; it's a joy to watch.
We began Term 3 with Animal Fables, Aesop's and others. At this stage of development, children relate easily to tales in which animals bring pictures of human foibles and attributes to strive towards, and they loved acting out each story and working out the inherent morals.
In our most recent Main Lesson we are discovering how indigenous peoples also related to animals, to share history through stories. During this block the children have attempted their first writing project, learning to put their developing skills together in texts, rewriting and illustrating, to create their own books.
During Book Week the children brought in their favourite books to share in class, and animals featured strongly. On Book Day many children dressed up as animals from their books. It was a delightful sight and tricky to remember everyone's name, all day.
We have just completed a significant Main Lesson introducing place value. Watching the children grasp the conservation of number has been wonderful, as they worked collaboratively using hands-on materials, gemstones and coloured sticks, bundled to grasp hundreds, tens and ones.
Tying our literacy skills together, we have just spent six months deeply immersed in our reading program, supported by volunteering parents each morning. Everyone in class has made significant leaps in reading and comprehension. Many blessings for the school and home help from families and a warm thank you to our class assistant Sophie, who is always diligent and patient, caring and insightful.
Linda Mayer
HSC Major Work Showcase
The Shearwater HSC Major Work Showcase is an exciting opportunity for our community to see what the inspired young adults in our class of 2019 have achieved this year. The event will be a feast for your senses and a great way to celebrate the hard work of our senior students. The event will run over two nights, with the displays open 5-6.30pm in the art rooms and IT lab and the performances in the hall 6.30-8.30pm - Film and Drama on Wednesday and Music on Thursday. We look forward to seeing you there!
Hummingbee Kindergarten
After our beautiful Winter Festival and happy holidays, our Hummingbee Kindies returned refreshed to a garden filled with Dancing Daffodils, and veggie gardens bursting with fresh ingredients for our baking days... One of our favourites was potatoes baked in our fire pit. We collected wood on an exciting walk to the Big School, prepared our toppings of herbs and grated cheese and wrapped the potatoes for the fire dancers to bake. Another was Afghani filled flat bread Bolani Katchalu... rolling dough, chopping potatoes and silver beet and herbs, filling and cooking in our outdoor restaurant kitchen in the frying pan.
For Book Week we dressed up from our baskets of silks and muslins, and even used our finger-knitting to dress up as characters from our Kindy stories... Bobber Blue Wren, Jingle Jangle Jellyfish, Flutterby Butterfly and many more. The children loved being a part of it. We watched the Primary School parade with interest and delight.
The children have been enjoying putting on shows for each other, re-telling our stories and creating their own. We are all learning quickly how to be good audience members!
Our busy fingers have been weaving our finger-knitting to make the beautiful rainbow shell of Grandmother Shellyback Turtle.
We've been enjoying the sunshine with extra outside play, making cubbies, obstacle courses, and learning new group games. Some favourites are Puddledock Farm, Ultimate Tips and Hide and Sneak. Our mud pit has seen many alterations... including speed bumps, trenches, pipe laying, form work and concreting. It's always a hive of activity with our big garden trolleys out collecting gravel, timber, pipes, mulch and furniture for cubby building!
We continue to extend our skills on the monkey bars - backwards, twisting and turning, skipping bars, and combinations of all three, as well as practicing balancing and hopping for hopscotch.
Our guinea pigs, Daisy and Jasmine are enjoying their new home, and caring for them is a focus every day, with the children lovingly picking fresh greens, filling their water bowl, cleaning out their home and putting in fresh straw for their bedroom and, of course, lots of cuddles!
This week, we are looking forward to our first class outing, to see Wink Blink Lighthouse and watch out for Wash Wallow Whale and his family. Next week we have our Spring Festival to end the term with song and dance, flower crowns and a feast for morning tea before our next happy holidays.
Joanne Owens
A tale of 20,000 trees
During the first three weeks of Term 3, Julie's Class 7 students planted 530 trees. 208 of these were donated by Mullumbimby's Rainforest 4, to coincide with a visit to Shearwater by an Indonesian delegation dedicated to saving orangutans through education, conservation and plantings in their country. These trees were planted in a sharp bend of the creek beyond the music rooms and Grandfather Fig.
The other 322 trees were planted along the watercourse of our ephemeral wetland that was altered to facilitate the construction of the Kindergarten and Preschool.
Cathy's Class 7 plantings from just six months ago behind the Kindy have absolutely thrived as a result of the wonderful wet weather that followed the dry hot summer.
These latest plantings behind the Preschool will have a harder time of it as we appear to be heading into another long, hot and probably dry spring and early summer.
As I am retiring after 19 years of environmental restoration at Shearwater, this will be my last Class 7 planting and I congratulate this class on the magnificent effort they put in. There was no shade for this planting, just full sun and some very windy days. Thank you Julie for what you have given these bright young students.
The School received a for $6,600 grant from the Biodiversity Conservation Trust (you've seen their Land for Wildlife signs all around the shire) to cover this planting and the next two years of wetland restoration.
With the help of farmer Beau (who probably planted some of the more established trees when he was a student of the School), we have also managed to plant 150 trees along the road verge, beyond the Farm dam, with Classes 8, 9 and 10.
All told, we have planted over 20,000 trees at the School since 2001 - an epic project and one I will miss being a part of. But I am confident Shearwater will continue its environmental restoration well into the future.
Thank you Shearwater, for letting me be the guardian of our forest.
Ken Ohlsson
Coordinator Gardening and Bush Regeneration
Fresh Faced Friday
Students from classes 6 to 9 enjoyed Fresh Faced Friday last week, with a series of workshops run by the Year 10 students, aimed at building body and self positivity. Fresh Faced Friday is an annual event held in schools and organisations around Australia, developed by a group of young people involved in the ABC Heywire program, whose concept was to create a day full of celebrating being who we want to be, judgement free.
Pet Day
P&F
© Shearwater The Mullumbimby Steiner School


text:
\section{Introduction}
\sloppy We study the alternating quantifier depths of a class of recursively defined first order properties on rooted, locally finite trees. A tree $T$ is a connected graph on either a finite or an infinite vertex set that does not contain any cycle. We in particular shall consider trees as directed graphs, with clear distinction of parent and child between adjacent vertices. A rooted tree has a special vertex called the root, which we denote by $R$. A tree is called locally finite when each vertex of the tree has finite degree. We denote by $V = V(T)$ the vertex set of the tree $T$. For a given rooted tree $T$ and a vertex $v \in V(T)$, we let $T(v)$ denote the subtree of $T$ that consists of $v$ and all its descendants.

We now define the first order (FO) language on rooted trees. This consists of the root $R$ of the tree as a constant symbol, the equality of vertices (denoted $x = y$ which means that the vertices $x$ and $y$ coincide), and the parent-child relation (denoted $\pi(y) = x$ which means that the vertex $x$ is the parent of the vertex $y$). We denote vertices in the tree other than the root by letters such as $x, y, z, \ldots$ usually in lower case. Sentences are finite and comprise Boolean connectives such as $\neg, \wedge, \vee, \implies, \Leftrightarrow$ etc., and existential (denoted $\exists$) and universal (denoted $\forall$) quantification over vertices. We refer the reader to any of \cite{02}, \cite{03} and \cite{04} for detailed references on first order logic. The \emph{quantifier depth}, abbreviated as q.d., of an FO sentence $A$ is the minimum number of nested quantifiers required to express $A$, whereas the \emph{alternating quantifier depth} of $A$, which we abbreviate henceforth as the a.q.d.\ of $A$, is the minimum number, the minimum being taken over all formulas that express $A$, of the maximum number of alternating nested quantifiers required to write each such formula. That is, a.q.d.\ gives us the minimum number of times we have to switch from existential to universal or universal to existential quantification in a nested sequence of quantifiers in order to express the sentence. Trivially, the a.q.d.\ of any FO sentence is always bounded above by its q.d. Purely existential or purely universal FO sentences are defined to have a.q.d.\ $0$.

\begin{example}
Consider the FO sentence: there exists a vertex with precisely one child. This can be written as
\begin{equation}
\exists x \left[\exists y \left[\left[\pi(y) = x\right] \wedge \forall z \left[\pi(z) = x \implies z = y\right]\right]\right].
\end{equation}
The q.d.\ of this sentence is $3$ whereas the a.q.d.\ is $1$, because we have two existential followed by one universal quantifier in the nested sequence.

\end{example}

One of the classically studied family of questions by mathematical logicians comprise the model-theoretic results about the expressive power of logical languages. In general, given two languages $A$ and $B$ such that $A$ is a subset of $B$, the task is to show that the two are not tautologically equal, i.e.\ there exists a sentence which is expressible in $B$ but not in $A$. Quantifier alternation hierarchy is a natural tool for proving such results. We now discuss some literature on the a.q.d.\ of FO, and also touch briefly on a.q.d.\ of monadic second order (MSO) logic. \cite{05} is a survey paper that discusses the usefulness of quantifier alternation hierarchy in membership algorithms -- algorithms that are used to decide whether a given regular language of finite words is definable by a sentence from FO logic or not. They beautifully express the necessity to understand quantifier alternation hierarchy as a measure of the difficulty of defining a language -- a language is considered complicated if many switches need to be made between blocks of existential quantifiers and blocks of universal quantifiers. \cite{06} considers quantifier alternation hierarchy within FO language comprising two variables over finite words with linear order and binary successor predicate. They show that for a given regular language and any non-negative integer $m$, it is decidable whether the language is definable by an FO sentence of a.q.d.\ at most $m$. \cite{07} also considers quantifier alternation hierarchy in FO logic on finite words -- they prove that one can decide membership of a regular language to the levels of FO sentences that have a.q.d.\ $1$ or FO sentences with a.q.d.\ $2$ and beginning with an existential quantifier. \cite{08} provides an effective characterization of tree languages that are definable by FO sentences with a.q.d.\ $1$. \cite{15} proves the strictness of FO quantifier alternation hierarchy over the class of finite labeled graphs  -- for each positive integer $k$, they exhibit a property of finite labeled directed graphs that is expressible as an FO sentence of a.q.d.\ $k+1$ but not as any FO sentence of a.q.d.\ $k$. \cite{16} shows that FO quantifier alternation hierarchy is equivalent to dot-depth alternation hierarchy for FO formulas over word models with a total ordering on the alphabet rather than the successor relation on word positions. \cite{13}, \cite{14} and \cite{17} discuss the strictness of MSO quantifier alternation, i.e.\ sentences in prenex normal form having a prefix of $k+1$ many alternations of set quantifiers can describe strictly more graph properties than those having a prefix of only $k$ many alternations of set quantifiers.

All the results we have been able to find in the literature pertaining to quantifier alternation hierarchy as an important tool to understand the expressive power of languages, are in the premise of graphs, trees, grids or pictures where each vertex or position is labeled by some element from a finite alphabet. In particular, in case of trees, some form of ordering is considered on the vertices, and each vertex is assigned a label from the pre-fixed, finite alphabet, and the unary predicate that specifies the label of a vertex is considered. We hope that our attempt to investigate a.q.d.'s of FO properties of rooted trees without the involvement of any alphabet will serve to begin a new direction of study with new definitions of FO and MSO languages on rooted, locally finite trees. Studies in this direction may reveal strictness, or lack thereof, of quantifier alternation hierarchy of the FO language, as we define it, on rooted trees. Continuing with this hope, we cite here \cite{12}, where the authors define the FO language on graphs using the binary relations of adjacency and equality of vertices (without the involvement of any alphabet). They show that the minimum number of quantifier alternations that an FO sentence $A$ must have in order to fail to have a $0-1$ law on $G(n, n^{-\alpha})$ for infinitely many values of $\alpha$ is $3$. We also hope that future work in this area would bring to light many more classes of recursively defined properties on rooted trees whose analysis may reveal more information on quantifier alternation hierarchy.

\subsection{Organization of the paper:} Our paper is organized in the following manner. In Section~\ref{sec:description} we describe the class of recursively defined properties on rooted, locally finite trees that we examine. In Section~\ref{sec:Ehrenfeucht} we describe the main tool we use to understand the a.q.d.\ of these properties -- the Ehrenfeucht games, also known as the Ehrenfeucht-Fra\"{i}ss\'{e} games. In Section~\ref{sec:construction} we describe the detailed inductive construction of the trees where the Ehrenfeucht games are played, and finally, in Section~\ref{sec:winning}, we describe, also along an inductive argument, the winning strategy for Duplicator, that gives us the final conclusion.

\section{Description of the problem}\label{sec:description}
We define, for any rooted, locally finite tree $T$ with root $R$, and any $x \in V(T)$, the property $P_{0}(x)$ which states that $x$ has no child, which can be expressed as
\begin{equation}
P_{0}(x) = \forall y \neg\left[\pi(y) = x\right].
\end{equation}
We now define the class of properties $P_{i}(x)$, for $i \in \mathbb{N}$ and any $x \in V(T)$, recursively as follows:
\begin{equation}\label{recur_prop}
P_{i}(x) := \forall \ y \left[\pi(y) = x \implies \neg P_{i-1}(y)\right].
\end{equation}
For example, $P_{1}(x)$ denotes the property that $x$ has no child with no child. In particular, we define the property $\kein_{i} = P_{i}(R)$ for every $i \in \mathbb{N}$. The aim of this paper is to show that the a.q.d.\ of $\kein_{i}$ is $i$ for every $i \in \mathbb{N} \cup \{0\}$. It is immediate to see that $\kein_{0}$ has a.q.d.\ $0$, since it is a purely universal sentence. From our recursive definition \eqref{recur_prop}, we can also see that the a.q.d.\ of $\kein_{i}$ is at least $i$. It remains to be shown that $i$ is indeed the minimum number of alternations of nested quantifiers required to express the sentence $\kein_{i}$ for every $i \in \mathbb{N}$.

We mention here an analogous family of properties defined for graphs. On graphs, the corresponding FO language will consist of vertices denoted by $x, y, z, \ldots$ etc.\, the equality of vertices (again, denoted $x = y$) and the adjacency of vertices (denoted $x \sim y$), the Boolean connectives and the existential and universal quantifiers over vertices. One can define, for any vertex $x$, the property $Q_{0}(x)$ that $x$ has no neighbour, or, in other words, $x$ being an isolated vertex, and set $N_{0} = \exists x [Q_{0}(x)]$. Then one can define
\begin{equation}
Q_{1}(x) = \forall y [y \sim x \implies [\exists z [\neg[z = x] \wedge [z \sim y]]]],
\end{equation}
or, in other words, for every neighbour $y$ of $x$, there exists some neighbour $z$ of $y$ which is distinct from $x$, or, in other words, no neighbour of $x$ is a degree $1$ vertex. We then define $N_{1} = \exists x [Q_{1}(x)]$. Finally, we recursively define
\begin{equation}
Q_{i}(x) = \forall y[y \sim x \implies \neg[Q_{i-1}(y)]],
\end{equation}
and set $N_{i} = \exists x [Q_{i}(x)]$. Notice that although defined in a rather similar fashion to the tree properties $\kein_{s}$, the properties $N_{s}$ are harder to analyze. This is because of the following reason. As we shall see in the subsequent sections, our argument hinges upon the use of Ehrenfeucht games on two trees that we construct recursively on $s$, such that one of them satisfies $\kein_{s}$ and the other satisfies $\neg \kein_{s}$. The attempt to do a similar recursive construction in the graph case fails, because there, the property $\neg N_{s}$ claims something much stronger than $\neg \kein_{s}$ -- it claims that a certain property fails to hold for \emph{every} vertex in the graph. The study of the properties $N_{s}$ for $s \in \mathbb{N}$ and their a.q.d.\ remains of keen interest to us in future research in this area. The comparison of such classes of properties on graphs and rooted trees may reveal significant difference between FO on graphs and FO on rooted trees as we have defined them.

\section{The Ehrenfeucht game for alternating quantifiers}\label{sec:Ehrenfeucht}
As mentioned above, the main objective of this paper is to show that $\kein_{s}$ has a.q.d.\ precisely equal to $s$ for every $s \in \mathbb{N}$. The rigorous proof of this statement relies on a special version of the well-known Ehrenfeucht-Fra\"{i}ss\'{e} games, which we henceforth refer to as simply the Ehrenfeucht games. For standard definition of these combinatorial games and their connection to mathematical logic, we refer the reader to any one of the references \cite{01} and \cite{03}. For the special version of the game we are about to make use of in this paper, we refer the reader to Definition~2.7, Definition~2.8 and Theorem~2.9 of \cite{10}, and a formal proof of the connection of this special version with a.q.d.\ of FO sentences in \cite{11}.

We first state here, in the premise of rooted trees, the version of the Ehrenfeucht games described in \cite{10} and \cite{11}, that determines the maximum among the a.q.d.'s of all FO sentences that hold true in both the structures on which the game is being played. We first state here a few general rules and terminology which apply to all three of Definitions~\ref{EHR_first}, \ref{EHR_alt} and \ref{weaker_EHR}. The game is played, for a given number of rounds, on two given rooted trees $T_{1}$ and $T_{2}$, by two players known as Spoiler and Duplicator. Each round of the game consists of two parts: a \emph{move} by Spoiler followed by a move by Duplicator. By a move, we mean the action by any player of choosing a vertex from one of the two trees. In each round, once Spoiler has made his selection of a vertex from one of the trees, Duplicator must make her selection of a vertex from the \emph{other} tree. Thus, in every round, there is precisely one vertex chosen from $T_{1}$ and one from $T_{2}$.

If in the $i$-th round, for some $1 \leq i \leq r-1$ where $r$ is the total number of rounds, Spoiler makes his move on $T_{1}$ and in the $(i+1)$-st round makes his move on $T_{2}$, or vice versa, then we say that a \emph{switch} has happened.
\begin{defn}\label{EHR_first}[Ehrenfeucht game with given maximum number of alternations]
Given two rooted trees $T_{1}$ with root $R_{1}$ and $T_{2}$ with root $R_{2}$, and two positive integers $r$ and $s$ with $r \geq s$, this game, denoted $\EHR\left[T_{1}, T_{2}, s, r\right]$, consists of $r$ many rounds. Spoiler is allowed to make his move in the first round on \emph{any} of the two trees $T_{1}$ and $T_{2}$, but throughout the game, he is allowed to make at most $s$ many switches.

Let $x_{i}$ be the vertex selected from $T_{1}$ and $y_{i}$ that from $T_{2}$ in round $i$, for $1 \leq i \leq r$. We set $x_{0} = R_{1}$ and $y_{0} = R_{2}$. Duplicator wins the game if \emph{all} of the following conditions hold: for all $i, j \in [r]$,
\begin{enumerate}[label={(Main \arabic*)},leftmargin=*]
\item \label{main_1} $\pi(x_{j}) = x_{i} \Leftrightarrow \pi(y_{j}) = y_{i}$;
\item \label{main_2} $x_{i} = x_{j} \Leftrightarrow y_{i} = y_{j}$,
\end{enumerate}
where for any positive integer $n$, we denote by $[n]$ the set $\left\{0,1, \ldots, n\right\}$.
\end{defn}

The importance of this version of the Ehrenfeucht games is illustrated in the following theorem, which is Theorem~2.9 of \cite{10} stated for rooted, locally finite trees.
\begin{theorem}\label{EHR_logic}
For any two rooted trees $T_{1}$ and $T_{2}$, for positive integers $r$ and $s$, Duplicator wins $\EHR\left[T_{1}, T_{2}, s, r\right]$ if and only if for every FO sentence $A$ of q.d.\ at most $r$ and a.q.d.\ at most $s$, we have $T_{1} \models A \Leftrightarrow T_{2} \models A$.
\end{theorem}
Here, for any rooted tree $T$ and any FO sentence $A$, the notation $T \models A$ implies that $A$ holds in $T$. The conclusion of Theorem~\ref{EHR_logic} is that, if Duplicator wins $\EHR\left[T_{1}, T_{2}, s, r\right]$, then for every FO sentence $A$ of q.d.\ at most $r$ and a.q.d.\ at most $s$, either $A$ holds for both $T_{1}$ and $T_{2}$, or it holds for neither.

The version of the Ehrenfeucht games that we use is slightly different from that given in Definition~\ref{EHR_first}. We show in Lemma~\ref{EHR_implies_EHR_alt} that if, on two given trees $T_{1}$ and $T_{2}$, Duplicator wins the Ehrenfeucht game described in Definition~\ref{EHR_alt} with sufficiently large values of the parameters concerned, then she also wins the Ehrenfeucht game described in Definition~\ref{EHR_first}.
\begin{defn}\label{EHR_alt}[Ehrenfeucht game with alternation after every $k$ rounds for fixed $k$]
Given two rooted trees $T_{1}$ with root $R_{1}$ and $T_{2}$ with root $R_{2}$, and two positive integers $k$ and $s$, this game, denoted $\ehr\left[T_{1}, T_{2}, s; k\right]$, consists of $s k$ rounds. The $s k$ many rounds are divided into $s$ many batches of $k$ rounds each. Spoiler, before the very first round, selects any one of $T_{1}$ and $T_{2}$ and in the first $k$ rounds makes his moves on that tree, in the next $k$ rounds he makes his moves on the other tree, and so on, i.e.\ he makes a switch after every $j k$-th round for every $1 \leq j \leq s-1$ (in particular, there can be no switch possible if $s = 1$).

If $x_{i}$ is the vertex selected from $T_{1}$ and $y_{i}$ that from $T_{2}$ in round $i$, for $i \in [s k]$, setting $x_{0} = R_{1}$ and $y_{0} = R_{2}$, the winning conditions for Duplicator are \ref{main_1} and \ref{main_2}, as described in Definition~\ref{EHR_first}.
\end{defn}

\begin{lemma}\label{EHR_implies_EHR_alt}
For any two positive integers $r$ and $s$, and two given rooted trees $T_{1}$ and $T_{2}$, if Duplicator wins $\ehr\left[T_{1}, T_{2}, s+1; r\right]$, then she also wins $\EHR\left[T_{1}, T_{2}, s, r\right]$.

\end{lemma}

\begin{proof}
Suppose Spoiler makes $t$ many switches during the game $\EHR\left[T_{1}, T_{2}, s, r\right]$, where $t \leq s$. Without loss of generality, let us assume that Spoiler starts playing on $T_{1}$. Let the first switch happen after the $i_{1}$-th round, the second switch after the $(i_{1}+i_{2})$-th round, and so on, where $t, i_{1}, \ldots, i_{t}$ are entirely dependent on Spoiler's decision and are unknown to Duplicator a priori. Clearly, for each $1 \leq j \leq t$, we have $i_{j} \leq r$.

Having a winning strategy $W$ for Duplicator for the game $\ehr\left[T_{1}, T_{2}, s+1; r\right]$ means that, whatever sequence of $(s+1) r$ moves Spoiler plays according to the rules of the game, Duplicator has a sequence of $(s+1) r$ responses such that she can maintain all the winning conditions. Without loss of generality, suppose Spoiler starts playing $\ehr\left[T_{1}, T_{2}, s+1; r\right]$ on $T_{1}$, and let $x_{1}, \ldots, x_{r}$ denote his moves in $T_{1}$ in the first $r$ rounds, and let $y_{1}, \ldots, y_{r}$ denote the corresponding responses of Duplicator in $T_{2}$ according to $W$; let $y_{r+1}, \ldots, y_{2r}$ denote the moves made by Spoiler in $T_{2}$ in rounds $r+1, \ldots, 2r$, and let $x_{r+1}, \ldots, x_{2r}$ denote the corresponding responses of Duplicator in $T_{1}$ according to $W$, and so on.

Now, we construct the winning strategy for Duplicator for the game $\EHR\left[T_{1}, T_{2}, s, r\right]$. As mentioned above, let Spoiler start playing on tree $T_{1}$. Let $a_{1}, \ldots, a_{i_{1}}$ denote the moves made by Spoiler in rounds $1, \ldots, i_{1}$ in tree $T_{1}$. Setting $x_{j} = a_{j}$ for all $1 \leq j \leq i_{1}$, Duplicator replies with moves $b_{j} = y_{j}$ for all $1 \leq j \leq i_{1}$ where the $y_{j}$'s, as mentioned above, are chosen according to strategy $W$. Note that Duplicator can do this because $i_{1} \leq r$. Next, for $i_{1}+1 \leq j \leq i_{1} + i_{2}$, Spoiler plays on $T_{2}$ and selects vertices $b_{j}, i_{1}+1 \leq j \leq i_{1}+i_{2}$. Duplicator, setting $y_{r+j} = b_{i_{1}+j}$ for all $1 \leq j \leq i_{2}$, selects vertices $a_{i_{1}+j} = x_{r+j}$, for $1 \leq j \leq i_{2}$, in $T_{1}$ according to winning strategy $W$. Once again, this is possible since $i_{2} \leq r$. She continues to play like this until all $r$ rounds are done.

Clearly, the winning conditions listed in Definition~\ref{EHR_first} now only need to hold for a subset $\left\{(x_{j}, y_{j}), 1 \leq j \leq i_{1}\right\} \cup \left\{(x_{j}, y_{j}), r+1 \leq j \leq r+i_{2}\right\} \cup \cdots \left\{(x_{j}, y_{j}), (t-1)r+1 \leq j \leq (t-1)r + i_{t}\right\} \cup \left\{(x_{j}, y_{j}), tr+1 \leq j \leq tr + \left(r - \sum_{\ell=1}^{t} i_{\ell}\right)\right\}$ of the set $\left\{(x_{j}, y_{j}), 1 \leq j \leq r s\right\}$, and we know that the winning conditions already hold for the bigger set because Duplicator chose her responses according to the winning strategy $W$. This shows that she wins $\EHR\left[T_{1}, T_{2}, s, r\right]$.

\end{proof}

We give here yet another version of the Ehrenfeucht games, which comes in handy in the description of winning strategies for Duplicator in Subsection~\ref{start on T_2}.
\begin{defn}\label{weaker_EHR}
Given rooted trees $T_{1}$ with root $R_{1}$ and $T_{2}$ with root $R_{2}$, and positive integers $s$, $i_{1}, i_{2}, \ldots i_{s}$, this version of the Ehrenfeucht game, denoted $\ehrw\left[T_{1}, T_{2}; i_{1}, \ldots i_{s}\right]$, consists of $i_{1} + i_{2} \ldots + i_{s}$ rounds. First, Spoiler chooses any of $T_{1}$ and $T_{2}$, and makes his first $i_{1}$ moves on that tree, while Duplicator makes the corresponding $i_{1}$ moves on the other tree. Spoiler makes the first switch after the $i_{1}$-th round, the second switch after the $(i_{1}+i_{2})$-th round, $\ldots$, and finally, the $(s-1)$-st switch after the $(i_{1}+ \cdots + i_{s-1})$-th round.

As before, if $x_{i}$ is the vertex selected from $T_{1}$ and $y_{i}$ that from from $T_{2}$ in round $i$, for $1 \leq i \leq \sum_{j=1}^{s}i_{j}$, then, setting $x_{0} = R_{1}$ and $y_{0} = R_{2}$, Duplicator wins the game if \ref{main_1} and \ref{main_2} hold.
\end{defn}
Notice that this is weaker than the game described in Definition~\ref{EHR_first} because here the values $i_{1}, \ldots, i_{s}$ are known to both Spoiler and Duplicator before the start of the game.

\begin{lemma}\label{EHR_alt_implies_EHR_weak}
For any two positive integers $s$ and $k$ and rooted trees $T_{1}$ and $T_{2}$, if Duplicator wins $\ehr[T_{1}, T_{2}, s; k]$, then she also wins $\ehrw[T_{1}, T_{2}; i_{1}, \ldots i_{s}]$ for any $1 \leq i_{1}, \ldots i_{s} \leq k$.
\end{lemma}

\begin{proof}
The proof of this lemma is very similar to that of Lemma~\ref{EHR_implies_EHR_alt}, and is in fact simpler, and the details are therefore omitted.
\end{proof}

We now describe the way we put the Ehrenfeucht game described in Definition~\ref{EHR_alt} to use in proving that $\kein_{s}$ has a.q.d.\ $s$ for each $s \in \mathbb{N}$. For any positive integer $k$, we construct two rooted trees $T_{1}$ and $T_{2}$ (these trees will obviously depend on $k$ and $s$), such that $T_{1} \models \kein_{s}$ and $T_{2} \models \neg \kein_{s}$, and Duplicator wins $\ehr\left[T_{1}, T_{2}, s; k\right]$. From this, we can draw the following conclusion: using Lemma~\ref{EHR_implies_EHR_alt}, we know that Duplicator wins $\EHR\left[T_{1}, T_{2}, s-1, k\right]$, and hence, from Theorem~\ref{EHR_logic}, we can tell that for every FO sentence $A$ that has q.d.\ at most $k$ and a.q.d.\ at most $s-1$, either $A$ holds for both $T_{1}$ and $T_{2}$, or $\neg A$ holds for both $T_{1}$ and $T_{2}$. But notice that $\kein_{s}$ holds for $T_{1}$ whereas $\neg \kein_{s}$ holds for $T_{2}$. This clearly means that $\kein_{s}$ either fails to have q.d.\ at most $k$, or it fails to have a.q.d.\ at most $s-1$. Since we are able to construct $T_{1}$ and $T_{2}$ and provide a winning strategy for Duplicator for arbitrary $k$, clearly it is the upper bound on the a.q.d.\ of $\kein_{s}$ that fails. This shows that $\kein_{s}$ must have a.q.d.\ at least $s$. Since we have noted before that its a.q.d.\ is at most $s$, it must be precisely equal to $s$. This completes the proof of our main result.

\section{Construction of the trees}\label{sec:construction}
In this section, we describe the construction of the trees $T_{1}$ and $T_{2}$ depending on the given parameters $s$ and $k$. We take into account an additional parameter $m$, whose role becomes clear from the construction. The constructions are described inductively on $s$, starting with the base case of $s = 1$, and arbitrary $k$, which we now describe.

For any positive integer $m \geq k$, we construct the rooted trees $T_{1}^{(1,k,m)}$ and $T_{2}^{(1,k,m)}$ as follows:
\begin{enumerate}
\item In $T_{1}^{(1,k,m)}$, the root $R_{1}$ has $m+1$ children $u_{1}, \ldots u_{m+1}$, and each of them has $m$ childless children of its own.
\item In $T_{2}^{(1,k,m)}$, the root $R_{2}$ has $m+1$ children $v_{1}, \ldots v_{m+1}$. Each of $v_{1}, \ldots v_{m}$ has $m$ childless children of its own; $v_{m+1}$ has no child.
\end{enumerate}
An illustration is given in Figure~\ref{base_2}.
\begin{figure}
\centering
\includegraphics[width = 0.8\textwidth]{./KEIN_2_image}
\caption{$T_{1}^{(1,k,m)}$ and $T_{2}^{(1,k,m)}$}
\label{base_2}
\end{figure}

Suppose we know how to construct the trees $T_{1}^{(s',k,m)}$ and $T_{2}^{(s',k,m)}$ for all $s' \leq s$ where $s$ is some positive integer, for arbitrary $k$, and for all $m \geq s' k$, such that $T_{1}^{(s',k,m)} \models \kein_{s'}$ and $T_{2}^{(s',k,m)} \models \neg \kein_{s'}$, and Duplicator wins $\ehr\left[T_{1}^{(s',k,m)}, T_{2}^{(s',k,m)}, s'; k\right]$. Now, fixing an arbitrary $k$, we construct $T_{1}^{(s+1,k,m)}$ and $T_{2}^{(s+1,k,m)}$ for all $m \geq (s+1)k$, such that $T_{1}^{(s+1,k,m)} \models \kein_{s+1}$ whereas $T_{2}^{(s+1,k,m)} \models \neg \kein_{s+1}$, and Duplicator wins $\ehr\left[T_{1}^{(s+1,k,m)}, T_{2}^{(s+1,k,m)}, s+1; k\right]$.

Note that we make use of $T_{1}^{(s,k,m)}$ and $T_{2}^{(s,k,m)}$ in constructing $T_{1}^{(s+1,k,m)}$ and $T_{2}^{(s+1,k,m)}$ for $m \geq (s+1)k$, and since this last condition ensures that $m > s k$, hence it makes sense to talk about $T_{1}^{(s,k,m)}$ and $T_{2}^{(s,k,m)}$. The inductive construction is as follows:
\begin{enumerate}
\item In $T_{1}^{(s+1,k,m)}$, the root $R_{1}$ has $m+1$ children $u_{1}, \ldots u_{m+1}$, and from each of them hangs a copy of $T_{2}^{(s,k,m)}$.
\item In $T_{2}^{(s+1,k,m)}$, the root $R_{2}$ has $m+1$ children, $v_{1}, \ldots v_{m+1}$. From each of $v_{1}, \ldots v_{m}$, hangs a copy of $T_{2}^{(s,k,m)}$, and from $v_{m+1}$ hangs a copy of $T_{1}^{(s,k,m)}$.
\end{enumerate}
An illustration is given in Figure \ref{fig:s+1}.
\begin{figure}[h!]
    \includegraphics[width=0.8\textwidth]{KEIN_image_s+1}
  \caption{Trees $T_{1}^{(s+1,k,m)}$ and $T_{2}^{(s+1,k,m)}$}
\label{fig:s+1}
\end{figure}

We make sure here that indeed we have $T_{1}^{(s+1,k,m)} \models \kein_{s+1}$ and $T_{2}^{(s+1,k,m)} \models \neg \kein_{s+1}$. Recall from \eqref{recur_prop} that a rooted tree $T$ satisfies $\kein_{s+1}$ if for every child $x$ of the root $R$, the property $\neg P_{s}(x)$ holds. By induction hypothesis, $T_{2}^{(s,k,m)} \models \neg \kein_{s}$. Since for each child $u_{i}$ of the root in $T_{1}^{(s+1,k,m)}$, we have
$$T_{1}^{(s+1,k,m)}(u_{i}) \cong T_{2}^{(s,k,m)}$$
(where the notation $\cong$ indicates a tree ismorphism map which maps the root of one tree to the root of the other), hence $\neg P_{s}(u_{i})$ holds for every $1 \leq i \leq m+1$. On the other hand, also by induction hypothesis, $T_{1}^{(s,k,m)} \models \kein_{s}$. In $T_{2}^{(s+1,k,m)}$, the root has one child $v_{m+1}$ such that
$$T_{2}^{(s+1,k,m)}(v_{m+1}) \cong T_{1}^{(s,k,m)},$$
and hence $P_{s}(v_{m+1})$ holds. This tells us that $\neg \kein_{s+1}$ must hold for $T_{2}^{(s+1,k,m)}$.

\section{Winning strategy for Duplicator for arbitrary $s$}\label{sec:winning}

We first describe Duplicator's winning strategy for $s = 1$. For $s = 1$, no alternation is allowed in $\ehr\left[T_{1}, T_{2}, s; k\right]$, i.e.\ either Spoiler plays the entire game on $T_{1}$ while Duplicator answers on $T_{2}$, or Spoiler plays the entire game on $T_{2}$ and Duplicator answers on $T_{1}$. Also, there are a total of $k$ many rounds now. We now describe the winning strategy for Duplicator in either scenario.

We introduce here a terminology that we shall use in our exposition of the winning strategy for Duplicator henceforth.
\begin{defn}\label{free so far}
Suppose $i$ rounds of the game $\ehr\left[T_{1}, T_{2}, s; k\right]$ have been played. For any vertex $u \in T_{1}$, we call $u$ \emph{free up to round $i$} if no $x_{j}$, $1 \leq j \leq i$, has been selected from $T_{1}(u)$. Similarly, for any vertex $v \in T_{2}$, we call $v$ free up to round $i$ if no $y_{j}$, $1 \leq j \leq i$, has been selected from $T_{2}(v)$.
\end{defn}

First, consider the case where Spoiler plays on tree $T_{2}^{(1,k,m)}$. Throughout the game, Duplicator maintains the following conditions, and we show that she is able to maintain them using an inductive argument. Suppose $i$ rounds of the game have been played. Then, for $1 \leq j \leq i$,
\begin{enumerate}
\item if Spoiler selects $y_{j} = v_{\ell}$ for some $1 \leq \ell \leq m+1$, then Duplicator selects $x_{j} = u_{\ell}$;
\item if Spoiler selects $y_{j}$ to be a child of the vertex $v_{\ell}$, for some $1 \leq \ell \leq m$, then Duplicator selects $x_{j}$ to be a child of $u_{\ell}$, making sure that \ref{main_2} is maintained (note that these two conditions together imply that $x_{j} = R_{1}$ if and only if $y_{j} = R_{2}$).
\end{enumerate}
Suppose Duplicator has been able to maintain these conditions up to and including round $i$. Now suppose Spoiler selects $y_{i+1}$ to be equal to some $v_{\ell}$ for $1 \leq \ell \leq m+1$. Then Duplicator sets $x_{i+1} = u_{\ell}$. If Spoiler selects $y_{i+1}$ to be a child of $v_{\ell}$ for some $1 \leq \ell \leq m$, and this child was not chosen before, then Duplicator selects $x_{i+1}$ to be a child of $u_{\ell}$ that has not been chosen before, and notice that such a child she can always find because each $u_{\ell}$ for $1 \leq \ell \leq m$ has $m$ children and $m \geq k$, and there are only $k$ many rounds. If Spoiler selects $y_{i+1}$ to be a child of $v_{\ell}$ that was already chosen before, say in the $j$-th round for some $j \leq i$, then Duplicator, to maintain \ref{main_2}, simply sets $x_{i+1} = x_{j}$. It is straightforward to see that these choices do guarantee the satisfaction of both \ref{main_1} and \ref{main_2} at the end of the game.

Now suppose Spoiler plays on tree $T_{1}^{(1,k,m)}$. Once again, Duplicator maintains the following conditions throughout the game, which we prove via inductive arguments that she is able to. Suppose $i$ rounds of the game have been played. Then for all $j, j' \in [i]$,
\begin{enumerate}
\item for $j \neq j'$, $x_{j} = x_{j'}$ if and only if $y_{j} = y_{j'}$ (this implies that $x_{j} = R_{1}$ if and only if $y_{j} = R_{2}$);
\item $x_{j} \in \left\{u_{1}, \ldots, u_{m+1}\right\}$ if and only if $y_{j} \in \left\{v_{1}, \ldots, v_{m}\right\}$;
\item $x_{j}$ is a child of $u_{\ell}$ for some $1 \leq \ell \leq m+1$ if and only if $y_{j}$ is a child of $v_{\ell'}$ for some $1 \leq \ell' \leq m$;
\item for $j \neq j'$, $\pi(x_{j'}) = x_{j}$ if and only if $\pi(y_{j'}) = y_{j}$;
\item $x_{j} = u_{\ell}$ for some $1 \leq \ell \leq m+1$ such that $u_{\ell}$ has been free up to round $j-1$ if and only if $y_{j} = v_{\ell'}$ for some $1 \leq \ell' \leq m$ such that $v_{\ell'}$ has been free up to round $j-1$;
\item for $j \neq j'$, $x_{j}$ and $x_{j'}$ are siblings, i.e.\ they share a common parent, if and only if $y_{j}$ and $y_{j'}$ are siblings.
\end{enumerate}
Suppose Duplicator has been able to maintain all of these conditions up to and including round $i$, for some $i \leq k-1$. Of course, if Spoiler selects $x_{i+1} = x_{j}$ for some $j \leq i$, then Duplicator sets $y_{i+1} = y_{j}$. So, for the rest of the inductive argument, assume that $x_{i+1}$ is distinct from all previously chosen vertices.

Suppose Spoiler selects $x_{i+1}$ to be some $u_{\ell}$ for $1 \leq \ell \leq m+1$ such that $u_{\ell}$ has been free up to round $i$. Note that $m \geq k$ and there are $k$ rounds in total, hence there has to be at least one vertex $v_{\ell'}$ for some $1 \leq \ell' \leq m$ free up to round $i$. Hence Duplicator finds such a $v_{\ell'}$ and sets $y_{i+1} = v_{\ell'}$. Suppose Spoiler selects $x_{i+1}$ to be $u_{\ell}$ such that there exists $j \leq i$ with $x_{j}$ a child of $u_{\ell}$; by induction hypothesis, we know that $y_{j}$ is a child of $v_{\ell'}$ for some $1 \leq \ell' \leq m$, and we set $y_{i+1} = v_{\ell'}$.

If Spoiler selects $x_{i+1}$ to be a child of some vertex $u_{\ell}$ for $1 \leq \ell \leq m+1$ such that $u_{\ell}$ has been free up to round $i$, then Duplicator once again finds $v_{\ell'}$ for some $1 \leq \ell' \leq m$ that has been free up to round $i$ (possible for the same reason as argued above), and then selects $y_{i+1}$ to be any child of $v_{\ell'}$. Suppose Spoiler selects $x_{i+1}$ to be a child of $u_{\ell}$, for some $1 \leq \ell \leq m+1$, such that for some $j \leq i$, we have $x_{j} = u_{\ell}$. Then by induction hypothesis, we know that $y_{j}$ must equal $v_{\ell'}$ for some $1 \leq \ell' \leq m$. Duplicator selects a child of $v_{\ell'}$, distinct from all previously chosen vertices, as $y_{i+1}$. As there are $m$ children of $v_{\ell'}$ and a total of $k$ rounds with $m \geq k$, hence she is able to make such a selection. Suppose Spoiler selects $x_{i+1}$ to be a child of $u_{\ell}$ for some $1 \leq \ell \leq m+1$ such that for some $j \leq i$, the vertex $x_{j}$ is also a child of $u_{\ell}$. Again by induction hypothesis we know that $y_{j}$ is a child of $v_{\ell'}$ for some $1 \leq \ell' \leq m$, and Duplicator sets $y_{i+1}$ to be a child of $v_{\ell'}$ that is different from $y_{j}$ as well as any other previously chosen vertex. Note that, since each $u_{\ell}$ and each $v_{\ell'}$ has $m$ children and $m \geq k$, where $k$ is the total number of rounds, there is always a choice of $y_{i+1}$ as a child of $v_{\ell'}$ which does not coincide with any previously chosen vertex.

This exhausts all possible moves by Spoiler and we have shown that Duplicator is able to respond in each case such that all the conditions mentioned above are maintained.

We now come to the winning strategy for Duplicator on $\ehr\left[T_{1}^{(s,k,m)}, T_{2}^{(s,k,m)}, s; k\right]$ for arbitrary $k$ and $m \geq s k$, and this is described as an inductive strategy where the induction happens on $s$. Thus, we assume that we already have a winning strategy for Duplicator for the game $\ehr\left[T_{1}^{(s,k,m)}, T_{2}^{(s,k,m)}, s; k\right]$, and then devise a strategy for her for the game $\ehr\left[T_{1}^{(s+1,k,m)}, T_{2}^{(s+1,k,m)}, s+1; k\right]$ for the same $k$ and $m \geq (s+1)k$. The analysis is split into two parts, according to which tree Spoiler starts playing on, and these are described separately, in detail, in Subsections \ref{start on T_2} and \ref{start on T_1}.

\subsection{Spoiler starts playing on $T_{2}^{(s+1,k,m)}$}\label{start on T_2}
Suppose Spoiler starts playing the game $\ehr\left[T_{1}^{(s+1,k,m)}, T_{2}^{(s+1,k,m)}, s+1; k\right]$ on $T_{2}^{(s+1,k,m)}$. Recall, from the inductive construction of the trees in Section~\ref{sec:construction}, that $T_{1}^{(s,k,m)}$ and $T_{2}^{(s,k,m)}$ are as illustrated in Figure~\ref{fig:s}.
\begin{figure}[h!]
  \centering
    \includegraphics[width=0.8\textwidth]{KEIN_image_s}
  \caption{Trees $T_{1}^{(s,k,m)}$ and $T_{2}^{(s,k,m)}$}
  \label{fig:s}
\end{figure}

\begin{ind}\label{induction hyp 1}
For every $1 \leq t \leq m+1$ and $1 \leq t' \leq m$, fix a tree isomorphism $\varphi^{(s)}_{t, t'}: T_{1}^{(s,k,m)}(u_{t}) \rightarrow T_{2}^{(s,k,m)}(v_{t'})$ (this is possible as both are copies of $T_{2}^{(s-1,k,m)}$), such that $\varphi^{(s)}_{t,t'}(u_{t}) = v_{t'}$. Fix $\ell$ pairs of vertices $(x_{1}, y_{1}), \ldots (x_{\ell}, y_{\ell})$ in $T_{1}^{(s,k,m)} \times T_{2}^{(s,k,m)}$ such that $\ell \leq k$ and they satisfy the following conditions:
\begin{enumerate}[label=(IH\arabic{*})]
\item \label{ind hyp 1} for each $1 \leq i \leq \ell$, we have $x_{i} \in \bigcup_{t=1}^{m+1} T_{1}^{(s,k,m)}(u_{t})$ and $y_{i} \in \bigcup_{t'=1}^{m} T_{2}^{(s,k,m)}(v_{t'})$.
\item \label{ind hyp 2} For $1 \leq i \neq j \leq \ell$, if $x_{i}, x_{j} \in T_{1}^{(s,k,m)}(u_{t})$ for some $1 \leq t \leq m+1$, then we can find some $1 \leq t' \leq m$ such that $y_{i}, y_{j} \in T_{2}^{(s,k,m)}(v_{t'})$. The converse also holds, i.e.\ if for $1 \leq i \neq j \leq \ell$, if $y_{i}, y_{j} \in T_{2}^{(s,k,m)}(v_{t'})$ for some $1 \leq t' \leq m$, then we can find some $1 \leq t \leq m+1$ such that $x_{i}, x_{j} \in T_{1}^{(s,k,m)}(u_{t})$.
\item \label{ind hyp 3} For every $1 \leq i \leq \ell$, if $x_{i} \in T_{1}^{(s,k,m)}(u_{t})$ and $y_{i} \in T_{2}^{(s,k,m)}(v_{t'})$ for some $1 \leq t \leq m+1$ and $1 \leq t' \leq m$, then $y_{i} = \varphi^{(s)}_{t,t'}(x_{i})$.
\end{enumerate}
As in Definitions~\ref{EHR_first} and \ref{EHR_alt}, we set $x_{0} = R_{1}$, the root of $T_{1}^{(s,k,m)}$, and $y_{0} = R_{2}$, the root of $T_{2}^{(s,k,m)}$. Then Duplicator will be able to win $\ehr\left[T_{1}^{(s,k,m)}, T_{2}^{(s,k,m)}, s; k\right]$, where Spoiler starts playing on $T_{2}^{(s,k,m)}$, with the pairs $(x_{0}, y_{0}), \ldots (x_{\ell}, y_{\ell})$ considered as designated pairs.
\end{ind}

\begin{rem}
It is helpful to imagine designated pairs of vertices as pairs chosen \emph{before} the game has even begun. For example, in Induction hypothesis~\ref{induction hyp 1}, one could visualize the designated pairs $(x_{0}, y_{0}), \ldots (x_{\ell}, y_{\ell})$ as pairs already selected / provided to the players from $\ell+1$ many rounds that occurred \emph{before the actual game}, and then the game $\ehr\left[T_{1}^{(s,k,m)}, T_{2}^{(s,k,m)}, s; k\right]$ of $s k$ many rounds begins, which Duplicator is able to win, taking into account these $(\ell+1)$ many pairs as well, if the conditions given in Induction hypothesis~\ref{induction hyp 1} hold. In other words, suppose in $\ehr\left[T_{1}^{(s,k,m)}, T_{2}^{(s,k,m)}, s; k\right]$, the pair selected in round $i$ from $T_{1}^{(s,k,m)} \times T_{2}^{(s,k,m)}$ is denoted by $(x_{i+\ell}, y_{i+\ell})$, for all $1 \leq i \leq s k$. Then Duplicator has to maintain \emph{all} of the following conditions to win $\ehr\left[T_{1}^{(s,k,m)}, T_{2}^{(s,k,m)}, s; k\right]$ with $(x_{0}, y_{0}), \ldots (x_{\ell}, y_{\ell})$ as designated pairs:
\begin{enumerate}
\item $x_{i} = \pi(x_{j}) \Leftrightarrow y_{i} = \pi(y_{j})$ for all $0 \leq i \neq j \leq s k + \ell$;
\item $x_{i} = x_{j} \Leftrightarrow y_{i} = y_{j}$ for all $0 \leq i \neq j \leq s k + \ell$.
\end{enumerate}
\end{rem}

\par Armed with Induction hypothesis~\ref{induction hyp 1}, we now prove that the corresponding claim holds on $T_{1}^{(s+1,k,m)}$ and $T_{2}^{(s+1,k,m)}$ where $m \geq (s+1)k$. First, we need to fix $\ell \leq k$ designated pairs that satisfy conditions analogous to \eqref{ind hyp 1} through \eqref{ind hyp 3}. So, referring to Figure \ref{fig:s+1}, we fix, for every $1 \leq t \leq m+1$ and every $1 \leq t' \leq m$, an isomorphism $\varphi_{t,t'}^{(s+1)}: T_{1}^{(s+1,k,m)}(u_{t}) \rightarrow T_{2}^{(s+1,k,m)}(v_{t'})$ (again possible because both $T_{1}^{(s+1,k,m)}$ and $T_{2}^{(s+1,k,m)}(v_{t'})$ are both copies of $T_{2}^{(s,k,m)}$); next, we fix an arbitrary collection of pairs $(x_{1}, y_{1}), \ldots (x_{\ell}, y_{\ell})$ for any non-negative integer $\ell \leq k$ (if $\ell = 0$, then we choose only the roots $R_{1}$ and $R_{2}$ as $x_{0}$ and $y_{0}$ respectively), such that the following hold:
\begin{enumerate}[label=(C\arabic{*})]
\item \label{cond 1 designated} for each $1 \leq i \leq \ell$, we have $x_{i} \in \bigcup_{t=1}^{m+1} T_{1}^{(s+1,k,m)}(u_{t})$ and $y_{i} \in \bigcup_{t'=1}^{m} T_{2}^{(s+1,k,m)}(v_{t'})$.
\item \label{cond 2 designated} For $1 \leq i \neq j \leq \ell$, if $x_{i}, x_{j} \in T_{1}^{(s+1,k,m)}(u_{t})$ for some $1 \leq t \leq m+1$, then we can find some $1 \leq t' \leq m$ such that $y_{i}, y_{j} \in T_{2}^{(s+1,k,m)}(v_{t'})$. The converse also holds, i.e.\ if for $1 \leq i \neq j \leq \ell$, if $y_{i}, y_{j} \in T_{2}^{(s+1,k,m)}(v_{t'})$ for some $1 \leq t' \leq m$, then we can find some $1 \leq t \leq m+1$ such that $x_{i}, x_{j} \in T_{1}^{(s+1,k,m)}(u_{t})$.
\item \label{cond 3 designated} For every $1 \leq i \leq \ell$, if $x_{i} \in T_{1}^{(s+1,k,m)}(u_{t})$ and $y_{i} \in T_{2}^{(s+1,k,m)}(v_{t'})$, then $y_{i} = \varphi^{(s+1)}_{t,t'}(x_{i})$.
\end{enumerate}

We now provide a strategy for Duplicator to win $\ehr\left[T_{1}^{(s+1,k,m)}, T_{2}^{(s+1,k,m)}, s+1; k\right]$ with $(x_{0}, y_{0}), (x_{1}, y_{1}) \ldots (x_{\ell}, y_{\ell})$ as designated vertices. Recall again that $x_{0} = R_{1}$, the root of $T_{1}^{(s+1,k,m)}$, and $y_{0} = R_{2}$, the root of $T_{2}^{(s+1,k,m)}$.

\begin{assume}\label{assume 1}
Since $\ell \leq k$ and $m \geq (s+1)k$ with $s \geq 2$, there exists at least one $1 \leq i_{0} \leq m+1$ such that no $x_{j}, 1 \leq j \leq \ell$, has been chosen from $T_{1}^{(s+1,k,m)}(u_{i_{0}})$. Because for each $1 \leq i \leq m+1$, the subtree $T_{1}^{(s+1,k,m)}(u_{i})$ is a copy of $T_{2}^{(s,k,m)}$, we can therefore assume without loss of generality that $i_{0} = m+1$.
\end{assume}
Assumption \ref{assume 1} is made throughout Subsection~\ref{start on T_2}.

\par Let us now take a closer look at the trees $T_{1}^{(s+1,k,m)}$ and $T_{2}^{(s+1,k,m)}$, especially in more detail the subtrees $T_{1}^{(s+1,k,m)}(u_{m+1})$ and $T_{2}^{(s+1,k,m)}(v_{m+1})$. These are illustrated in Figures \eqref{fig:s+1:1:detailed} and \eqref{fig:s+1:2:detailed}.

\begin{figure}[h!]
\includegraphics[width = 0.7\textwidth]{./KEIN_s+2_detailed_image_1}
\caption{$T_{1}^{(s+1,k,m)}$ with detailed view of $T_{1}^{(s+1,k,m)}(u_{m+1})$}
\label{fig:s+1:1:detailed}
\end{figure}

\begin{figure}[h!]
\includegraphics[width = 0.7\textwidth]{./KEIN_s+2_detailed_image_2}
\caption{$T_{2}^{(s+1,k,m)}$ with detailed view of $T_{2}^{(s+1,k,m)}(v_{m+1})$}
\label{fig:s+1:2:detailed}
\end{figure}

\par Let $(x_{i+\ell}, y_{i+\ell})$ be the pair selected from $T_{1}^{(s+1,k,m)} \times T_{2}^{(s+1,k,m)}$ in round $i$ for $1 \leq i \leq (s+1) k$. As is our convention for Subsection~\ref{start on T_2}, Spoiler plays the first $k$ rounds on $T_{2}^{(s+1,k,m)}$. In the following paragraph, we state some conditions Duplicator maintains on the configuration $\left\{(x_{0}, y_{0}), \ldots, (x_{\ell}, y_{\ell})\right\} \cup \{(x_{i+\ell}, y_{i+\ell}): 1 \leq i \leq k\}$ (i.e.\ on the pairs of vertices resulting from the first $k$ rounds, along with the designated pairs). And we prove that she can indeed maintain these conditions by using an inductive argument (within the first $k$ rounds).

First, Duplicator fixes any tree isomorphism $\varphi_{t,t'}^{(s)}: T_{1}^{(s+1,k,m)}(u_{m+1,t}) \rightarrow T_{2}^{(s+1,k,m)}(v_{m+1,t'})$, with $\varphi_{t,t'}^{(s)}(u_{m+1,t}) = v_{m+1,t'}$, for all $1 \leq t \leq m$ and $1 \leq t' \leq m+1$. This is possible as each $T_{1}^{(s+1,k,m)}(u_{m+1,t})$, $1 \leq t \leq m$, as well as each $T_{2}^{(s+1,k,m)}(v_{m+1,t'})$, $1 \leq t' \leq m+1$, is a copy of $T_{2}^{(s-1,k,m)}$. Suppose $p \leq k-1$ rounds of the game have been played. The conditions on the configuration $\left\{(x_{0}, y_{0}), \ldots, (x_{\ell}, y_{\ell})\right\} \cup \left\{(x_{\ell+1}, y_{\ell+1}), \ldots (x_{\ell+p}, y_{\ell+p})\right\}$ are as follows:

\begin{enumerate}[label=(A\arabic{*})]

\item\label{first k rounds 1} $x_{i+\ell} = R_{1} \Leftrightarrow y_{i+\ell} = R_{2}$. If $x_{i+\ell} \in T_{1}^{(s+1,k,m)}(u_{t})$ for some $1 \leq t \leq m$, then $y_{i+\ell} \in T_{2}^{(s+1,k,m)}(v_{t'})$ for some $1 \leq t' \leq m$, and vice versa; moreover, in this case, $y_{i+\ell} = \varphi^{s+1}_{t, t'}(x_{i+\ell})$.

Note that if $x_{i+\ell} \in T_{1}^{(s+1,k,m)}(u_{m+1})$, then actually $x_{i+\ell} \in \bigcup_{t=1}^{m} \in T_{1}^{(s+1,k,m)}(u_{m+1,t}) \bigcup \{u_{m+1}\}$. If $x_{i+\ell} = u_{m+1}$ then $y_{i+\ell} = v_{m+1}$, and vice versa. We have $x_{i+\ell} \in T_{1}^{(s+1,k,m)}(u_{m+1,t})$ for some $1 \leq t \leq m$ if and only if $y_{i+\ell} \in T_{2}^{(s+1,k,m)}(u_{m+1,t'})$ for some $1 \leq t' \leq m+1$, and moreover, in this case, $y_{i+\ell} = \varphi^{(s)}_{t,t'}(x_{i+\ell})$.

\item\label{first k rounds 2} Suppose we have $x_{i+\ell}, x_{j} \in T_{1}^{(s+1,k,m)}(u_{t})$ for some $1 \leq i \leq p$, some $1 \leq j \leq p+\ell$, and some $1 \leq t \leq m$. Then there exists some $1 \leq t' \leq m$ such that $y_{i+\ell}, y_{j} \in T_{2}^{(s+1,k,m)}(v_{t'})$.

The converse of this statement is true as well. That is, if for some $1 \leq i \leq p$, some $1 \leq j \leq p+\ell$, and some $1 \leq t' \leq m$, we have $y_{i+\ell}, y_{j} \in T_{2}^{(s+1,k,m)}(v_{t'})$, then there exists some $1 \leq t' \leq m$ such that $x_{i+\ell}, x_{j} \in T_{1}^{(s+1,k,m)}(u_{t})$.

\item\label{first k rounds 3} Suppose for $1 \leq i \neq j \leq p$ we have $x_{i+\ell}, x_{j+\ell} \in T_{1}^{(s+1,k,m)}(u_{m+1, t})$ for some $1 \leq t \leq m$. Then there exists $1 \leq t' \leq m+1$ such that $y_{i+\ell}, y_{j+\ell} \in  T_{2}^{(s+1,k,m)}(v_{m+1,t'})$. The converse of this statement is also true, i.e.\ if for $1 \leq i \neq j \leq p$, we have $y_{i+\ell}, y_{j+\ell} \in T_{2}^{(s+1,k,m)}(v_{m+1, t'})$ for some $1 \leq t' \leq m+1$, then there exists $1 \leq t \leq m$ such that $x_{i+\ell}, x_{j+\ell} \in T_{1}^{(s+1,k,m)}(u_{m+1,t})$.
\end{enumerate}

Suppose Duplicator has been able to maintain all these conditions up to and including round $p$. Now, Spoiler chooses $y_{\ell+p+1}$ from $T_{2}^{(s+1,k,m)}$. Here is how Duplicator replies:
\begin{enumerate}
\item If $y_{p+\ell+1} = R_{2}$, the root of $T_{2}^{(s+1,k,m)}$, then Duplicator chooses $x_{p+\ell+1} = R_{1}$, the root of $T_{1}^{(s+1,k,m)}$.
\item Suppose $y_{p+\ell+1} \in \bigcup_{t'=1}^{m} T_{2}^{(s+1,k,m)}(v_{t'})$. Then there are a few possible cases, as follows:
\begin{enumerate}
\item \label{needed later 1} Suppose there exists $1 \leq j \leq \ell+p$ such that $y_{j}$ and $y_{p+\ell+1}$ both belong to the same $T_{2}^{(s+1,k,m)}(v_{t'})$ for some $1 \leq t' \leq m$. Then Duplicator finds $t$ such that $x_{j} \in T_{1}^{(s+1,k,m)}(u_{t})$. Note that if $1 \leq j \leq \ell$, then $1 \leq t \leq m$ because of Assumption~\ref{assume 1}), and if $\ell+1 \leq j \leq p+\ell$, then $1 \leq t \leq m$ because of induction hypothesis \ref{first k rounds 1}. Then Duplicator sets $x_{p+\ell+1} = \left\{\varphi^{(s+1)}_{t,t'}\right\}^{-1}(y_{p+\ell+1})$.

\item \label{needed later 2} Suppose $y_{p+\ell+1} \in T_{2}^{(s+1,k,m)}(v_{t'})$ for some $1 \leq t' \leq m$, such that $v_{t'}$ is free up to round $p$, i.e.\ $y_{j} \notin T_{2}^{(s+1,k,m)}(v_{t'})$ for all $1 \leq j \leq \ell+p$ (recall Definition~\ref{free so far}). Then Duplicator finds a $1 \leq t \leq m$ such that $u_{t}$ is free up to round $p$. She can always find such a $t$ as $m \geq (s+1)k > 2k > p+\ell$. She then sets $x_{p+\ell+1} = \left\{\varphi^{(s+1)}_{t,t'}\right\}^{-1}(y_{p+\ell+1})$.

\end{enumerate}

\item Now suppose $y_{p+\ell+1} \in T_{2}^{(s+1,k,m)}(v_{m+1})$. Again, there are a few possible cases:
\begin{enumerate}
\item If $y_{p+\ell+1} = v_{m+1}$, then Duplicator sets $x_{p+\ell+1} = u_{m+1}$.
\item If $y_{p+\ell+1} \in T_{2}^{(s+1,k,m)}(v_{m+1,t'})$ for some $1 \leq t' \leq m+1$ such that there exists some $1 \leq i \leq p$ with $y_{i+\ell} \in T_{2}^{(s+1,k,m)}(v_{m+1,t'})$, then Duplicator finds the $t$ (where $1 \leq t \leq m$ by induction hypothesis \ref{first k rounds 1}) such that $x_{i+\ell} \in T_{1}^{(s+1,k,m)}(u_{m+1,t})$. Then she sets $x_{p+\ell+1} = \left\{\varphi^{(s)}_{t,t'}\right\}^{-1}(y_{p+\ell+1})$.
\item If $y_{p+\ell+1} \in T_{2}^{(s+1,k,m)}(v_{m+1,t'})$ for some $1 \leq t' \leq m+1$ such that $y_{i+\ell} \notin T_{2}^{(s+1,k,m)}(v_{m+1,t'})$ for all $1 \leq i \leq p$, then Duplicator finds a $1 \leq t \leq m$ such that $u_{m+1,t}$ is free up to round $p$ (again, possible since $m \geq (s+1)k > 2k > p+\ell$) and sets $x_{p+\ell+1} = \left\{\varphi^{(s)}_{t,t'}\right\}^{-1}(y_{p+\ell+1})$.

\end{enumerate}
\end{enumerate}

Let us focus now on $T_{2}^{(s+1,k,m)}(v_{m+1})$ and $T_{1}^{(s+1,k,m)}(u_{m+1})$ only. Suppose the only pairs selected up to round $k$ that are in $T_{1}^{(s+1,k,m)}(u_{m+1}) \times T_{2}^{(s+1,k,m)}(v_{m+1})$ are $\left(x_{i_{1}+\ell}, y_{i_{1}+\ell}\right), \ldots \left(x_{i_{r}+\ell}, y_{i_{r}+\ell}\right)$. Now, $T_{1}^{(s+1,k,m)}(u_{m+1})$ is a copy of $T_{2}^{(s,k,m)}$ and $T_{2}^{(s+1,k,m)}(v_{m+1})$ is a copy of $T_{1}^{(s,k,m)}$. From \ref{first k rounds 1} and \ref{first k rounds 3}, we can see that the pairs $(y_{i_{1}+\ell}, x_{i_{1}+\ell}), \ldots (y_{i_{r}+\ell}, x_{i_{r}+\ell})$ satisfy Conditions \ref{ind hyp 1} through \ref{ind hyp 3}.
\begin{conc}\label{conc 1}
So Duplicator, by Induction hypothesis \ref{induction hyp 1}, will win $\ehr[T_{2}^{(s+1,k,m)}(v_{m+1}), T_{1}^{(s+1,k,m)}(u_{m+1}), s; k]$ with designated pairs $(y_{i_{1}+\ell}, x_{i_{1}+\ell}), \ldots (y_{i_{r}+\ell}, x_{i_{r}+\ell})$ (notice the deliberate writing of $T_{2}^{(s+1,k,m)}(v_{m+1})$ before $T_{1}^{(s+1,k,m)}(u_{m+1})$).
\end{conc}

\par Now consider the subsequent game of remaining $s k$ rounds. We call this the \emph{second part of the game}. For this part of the game, we really can split the tree $T_{1}^{(s+1,k,m)}$ into
\begin{equation}\label{split 1}
\{R_{1}\} \cup S_{1} \cup T_{1}^{(s+1,k,m)}(u_{m+1}), \text{ where } S_{1} = \left\{\bigcup_{t=1}^{m} T_{1}^{(s+1,k,m)}(u_{t})\right\},
\end{equation}
and the tree $T_{2}^{(s+1,k,m)}$ into
\begin{equation}\label{split 2}
\{R_{2}\} \cup S_{2} \cup T_{2}^{(s+1,k,m)}(v_{m+1}), \text{ where } S_{2} = \left\{\bigcup_{t'=1}^{m} T_{1}^{(s+1,k,m)}(v_{t'})\right\}.
\end{equation}
Suppose $p$ rounds of the game have been played, with $k < p \leq (s+1)k$. Duplicator maintains the following conditions on the configuration $\left\{(x_{i}, y_{i}): 0 \leq i \leq p+\ell\right\}$:
\begin{enumerate}[label=(B\arabic{*})]
\item \label{b_1} For $0 \leq i \leq p$, we have $x_{i} = R_{1} \Leftrightarrow y_{i} = R_{2}$.
\item \label{b_2} For $0 \leq i \leq p$, we have $x_{i} \in S_{1} \Leftrightarrow y_{i} = S_{2}$. In this case, if $x_{i} \in T_{1}^{(s+1,k,m)}(u_{t})$ for some $1 \leq t \leq m$ and $y_{i} \in T_{2}^{(s+1,k,m)}(v_{t'})$ for some $1 \leq t' \leq m$, then $y_{i} = \varphi^{(s+1)}_{t,t'}(x_{i})$.
\item \label{b_3} For $0 \leq i, i' \leq p$, if $x_{i}, x_{i'} \in T_{1}^{(s+1,k,m)}(u_{t})$ for some $1 \leq t \leq m$, then there exists $1 \leq t' \leq m$ with $y_{i}, y_{i'} \in T_{2}^{(s+1,k,m)}(v_{t'})$.
\item \label{b_4} For $0 \leq j \leq p$, we have $x_{j} \in T_{1}^{(s+1,k,m)}(u_{m+1}) \Leftrightarrow y_{j} \in T_{2}^{(s+1,k,m)}(v_{m+1})$. Suppose $(i-1) k + 1 \leq p \leq ik$, where $2 \leq i \leq s+1$. Suppose the number of pairs of vertices selected from $T_{2}^{(s+1,k,m)}(v_{m+1}) \times T_{1}^{(s+1,k,m)}(u_{m+1})$ between rounds $k+1$ and $2k$ is $j_{1}$, between rounds $2k+1$ and $3k$ is $j_{2}$, $\ldots$, between rounds $(i-2)k +1$ and $(i-1)k$ is $j_{i-2}$, between rounds $(i-1)k+1$ and $p$ is $j'$. Then Duplicator wins $\ehrw\left[T_{2}^{(s+1,k,m)}(v_{m+1}), T_{1}^{(s+1,k,m)}(u_{m+1}); j_{1}, j_{2}, \ldots j_{i-2}, j'+ik-p, \underbrace{k, \ldots, k}_\text{$(s+1-i)$ many}\right]$ with designated pairs $(y_{i_{1}+\ell}, x_{i_{1}+\ell}), \ldots (y_{i_{r}+\ell}, x_{i_{r}+\ell})$, where the first $j_{1} + j_{2} + \ldots j_{i-2} + j'$ rounds have been played out, with Spoiler starting on $T_{1}^{(s+1,k,m)}(u_{m+1})$.

\end{enumerate}

It is straightforward to see that Duplicator is able to maintain all the conditions stated above, via a similar inductive argument as has been detailed in this paper before. She is able to maintain Condition \ref{b_4} because of Conclusion \ref{conc 1} and Lemma \ref{EHR_alt_implies_EHR_weak}. This brings us to the end of the inductive proof. \qed

\subsection{When Spoiler starts playing on $T_{1}^{(s,k,m)}$}\label{start on T_1}
The construction remains the same as before. The induction hypothesis in this case is simpler to state.
\begin{ind}\label{induction hyp 2}
Duplicator wins $\ehr\left[T_{1}^{(s,k,m)}, T_{2}^{(s,k,m)}, s; k\right]$ when Spoiler starts on $T_{1}^{(s,k,m)}$.
\end{ind}

We show here the proof by induction. Refer to Figures \ref{fig:s+1:1:detailed} and \ref{fig:s+1:2:detailed}. Again, we chalk out the detailed strategy of Duplicator for the first $k$ rounds, where we do not use Induction hypothesis \ref{induction hyp 2}. For every $1 \leq t \leq m+1$ and $1 \leq t' \leq m$, she fixes any tree isomorphism $\varphi^{(s+1)}_{t,t'}: T_{1}^{(s+1,k,m)}(u_{t}) \rightarrow T_{2}^{(s+1,k,m)}(v_{t'})$ with $\varphi^{(s+1)}_{t,t'}(u_{t}) = v_{t'}$. This is possible since each of $T_{1}^{(s+1,k,m)}(u_{t})$ and $T_{2}^{(s+1,k,m)}(v_{t'})$ is a copy of $T_{2}^{(s,k,m)}$ for all $1 \leq t \leq m+1$ and $1 \leq t' \leq m$. For the first $p$ rounds, for all $1 \leq p \leq k$, she maintains the following conditions on the configuration $\left\{(x_{i}, y_{i}): 1 \leq i \leq p\right\}$:
\begin{enumerate}[label=(A'\arabic{*})]
\item For all $1 \leq i \leq p$, we have $x_{i} = R_{1} \Leftrightarrow y_{i} = R_{2}$.
\item For all $1 \leq i \leq p$, we have $x_{i} \in T_{1}^{(s+1,k,m)}(u_{t})$ for some $1 \leq t \leq m+1$ if and only if $y_{i} \in T_{2}^{(s+1,k,m)}(v_{t'})$ for some $1 \leq t' \leq m$. Furthermore $y_{i} = \varphi^{(s+1)}_{t,t'}(x_{i})$.
\item If $x_{i}, x_{j} \in T_{1}^{(s+1,k,m)}(u_{t})$ for some $1 \leq i \neq j \leq p$ and some $1 \leq t \leq m+1$, then there exists some $1 \leq t' \leq m$ such that $y_{i}, y_{j} \in T_{2}^{(s+1,k,m)}(v_{t'})$. The converse is also true, i.e.\ if $y_{i}, y_{j} \in T_{2}^{(s+1,k,m)}(v_{t'})$ for some $1 \leq i \neq j \leq p$ and some $1 \leq t' \leq m$, then there exists some $1 \leq t \leq m+1$ such that $x_{i}, x_{j} \in T_{1}^{(s+1,k,m)}(u_{t})$.
\end{enumerate}

That Duplicator is able to maintain these conditions can be shown by induction on $p$ where $1 \leq p \leq k-1$. So, suppose the first $p$ rounds have been played. In the $(p+1)$-st round Spoiler chooses $x_{p+1}$ from $T_{1}^{(s+1,k,m)}$. Duplicator replies as follows:
\begin{enumerate}
\item If $x_{p+1} = R_{1}$, then Duplicator sets $y_{p+1} = R_{2}$.
\item If $x_{p+1} \in T_{1}^{(s+1,k,m)}(u_{t})$ for some $1 \leq t \leq m+1$ such that there exists some $1 \leq j \leq p$ with $x_{j} \in T_{1}^{(s+1,k,m)}(u_{t})$ as well, then Duplicator finds the $1 \leq t' \leq m$ such that $y_{j} \in T_{2}^{(s+1,k,m)}(v_{t'})$ and sets $y_{p+1} = \varphi^{(s+1)}_{t,t'}(x_{p+1})$.
\item If $x_{p+1} \in T_{1}^{(s+1,k,m)}(u_{t})$ for some $1 \leq t \leq m+1$ such that $x_{j} \notin T_{1}^{(s+1,k,m)}(u_{t})$ for all $1 \leq j \leq p$, then Duplicator finds $1 \leq t' \leq m$ such that $v_{t'}$ has been free up to round $p$. She can always find such a $v_{t'}$ since $ m \geq (s+1)k > k > p$. Then she sets $y_{p+1} = \varphi^{(s+1)}_{t,t'}(x_{p+1})$.
\end{enumerate}

\begin{assume}\label{assume 2}
Since $m \geq (s+1)k > k$, hence there will be at least one $1 \leq i_{0} \leq m+1$ such that $u_{i_{0}}$ is free up to round $k$. Since for all $1 \leq t \leq m+1$, $T_{1}^{(s+1,k,m)}(u_{t})$ is a copy of $T_{2}^{(s,k,m)}$, hence we can assume without loss of generality that $i_{0} = m+1$.
\end{assume}
We make Assumption~\ref{assume 2} throughout the rest of Subsection~\ref{start on T_1}.

Now we come to the second part of the game, where there are $s k$ many rounds, and Spoiler plays the first $k$ rounds of this part on $T_{2}^{(s+1,k,m)}$. Note that $T_{1}^{(s+1,k,m)}(u_{m+1})$ is a copy of $T_{2}^{(s,k,m)}$ whereas $T_{2}^{(s+1,k,m)}(v_{m+1})$ is a copy of $T_{1}^{(s,k,m)}$.

\begin{conc}\label{conc 2}
By induction hypothesis, Duplicator wins $\ehr\left[T_{2}^{(s+1,k,m)}(v_{m+1}), T_{1}^{(s+1,k,m)}(u_{m+1}), s; k\right]$ when Spoiler starts playing on $T_{2}^{(s+1,k,m)}(v_{m+1})$.
\end{conc}

We can again split up the two trees as follows:
\begin{equation}
T_{1}^{(s+1,k,m)} = \{R_{1}\} \cup S_{1} \cup T_{1}^{(s+1,k,m)}(u_{m+1}), \text{ where } S_{1} = \bigcup_{t=1}^{m} T_{1}^{(s+1,k,m)}(u_{t}),
\end{equation}
and
\begin{equation}
T_{2}^{(s+1,k,m)} = \{R_{2}\} \cup S_{2} \cup T_{2}^{(s+1,k,m)}(v_{m+1}), \text{ where } S_{2} = \bigcup_{t'=1}^{m} T_{2}^{(s+1,k,m)}(v_{t'}).
\end{equation}
In the second part of the game, for every $k \leq p \leq (s+1)k$, Duplicator maintains the following conditions on the configuration $\left\{(x_{i}, y_{i}): 0 \leq i \leq p\right\}$:
\begin{enumerate}[label=(B'\arabic{*})]
\item \label{b'_1} For $0 \leq i \leq p$, we have $x_{i} = R_{1} \Leftrightarrow y_{i} = R_{2}$.
\item \label{b'_2} For $0 \leq i \leq p$, we have $x_{i} \in S_{1} \Leftrightarrow y_{i} \in S_{2}$. If $x_{i} \in T_{1}^{(s+1,k,m)}(u_{t})$ for some $1 \leq t \leq m$ and $y_{i} \in T_{2}^{(s+1,k,m)}(v_{t'})$ for some $1 \leq t' \leq m$, then $y_{i} = \varphi^{(s+1)}_{t,t'}(x_{i})$.
\item \label{b'_3} For $1 \leq i \neq j \leq p$, we have $x_{i}, x_{j} \in T_{1}^{(s+1,k,m)}(u_{t})$ for some $1 \leq t \leq m$ if and only if there exists some $1 \leq t' \leq m$ with $y_{i}, y_{j} \in T_{2}^{(s+1,k,m)}(v_{t'})$.
\item \label{b'_4} For $0 \leq j \leq p$, we have $x_{j} \in T_{1}^{(s+1,k,m)}(u_{m+1}) \Leftrightarrow y_{j} \in T_{2}^{(s+1,k,m)}(v_{m+1})$. Suppose $(i-1) k + 1 \leq p \leq ik$, where $2 \leq i \leq s+1$. Suppose the number of pairs of nodes selected from $T_{2}^{(s+1,k,m)}(v_{m+1}) \times T_{1}^{(s+1,k,m)}(u_{m+1})$ between rounds $k+1$ and $2k$ is $j_{1}$; between rounds $2k+1$ and $3k$ is $j_{2}$; $\ldots$ between rounds $(i-2)k +1$ and $(i-1)k$ is $j_{i-2}$; between rounds $(i-1)k+1$ and $p$ is $j'$. Then Duplicator wins $\ehrw\left[T_{2}^{(s+1,k,m)}(v_{m+1}), T_{1}^{(s+1,k,m)}(u_{m+1}); j_{1}, j_{2}, \ldots j_{i-2}, j'+ik-p, \underbrace{k, \ldots, k}_\text{$(s+1-i)$ many}\right]$, where the first $j_{1} + j_{2} + \ldots j_{i-2} + j'$ rounds have been played out, with Spoiler starting on $T_{2}^{(s+1,k,m)}(u_{m+1})$.
\end{enumerate}

Again, it is straightforward to see that Duplicator is able to maintain all the conditions stated above. She is able to maintain the Condition~\ref{b'_4} because of Conclusion \ref{conc 2} and Lemma~\ref{EHR_alt_implies_EHR_weak}. This brings us to the end of the inductive proof. \qed

\section{Acknowledgements}
The author expresses her sincere gratitude to her doctoral advisor Joel Spencer for suggesting this beautiful combinatorial problem to her. She also extends her thanks to Maksim Zhukovskii for carefully perusing through the proofs in this paper for correctness of content, and for suggesting further scopes of research in this area. This research has been supported partially by the grant NSF DMS-1444084.

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text:
Home – UncategorizedJames Collins has stated that Wales need to win their opening game at the Euros
James Collins has stated that Wales need to win their opening game at the Euros
Posted on	June 24, 2016 by content
West Ham and Wales defender has said that Wales need to win their opening game at the Euros if they want to have a successful tournament.
The first game will be against Slovakia, and the defender believes that they will need to be ready for the opening game if they want to make it out of the group stage. Indeed, the other two matches will not be easier as they will be facing England and Russia.
James Collins said that it will be a new experience for the Wales team and that they are excited to be part of the tournament. This will be the first international competition that Wales will be participating in after the 1958 World Cup, and they will be hoping that they have a successful one.
He believes that a win and a draw could be enough for them to go through the next round and this is what they are hoping to achieve. Collins would want to think that they can kick off their tournament with a win against Slovakia and get a result either against England or Russia to put them through to the next round.
He said that playing England in a friendly is in itself a big game, and now they will be playing in a major tournament. He believes that the game opposing England will be a great one and that both teams will be ready to win the game. However, he wants to stay focus on the first game and ensure that they could get a positive result to put them in the best position.
Wales will be hoping that Gareth Bale will inspire them to a good tournament and that they could reach the second round where anything can happen.
← JAMES COLLINS HAPPY WITH THE PERFORMANCE OF CALLUM SAUNDERS DESPITE MISS AGAINST OLDHAM James Collins Stays With Crawley →


text:
Brian McAllister, Ph.D., CPA
Wilcox Endowed Professor of Accounting and Department Chair
bmcallis@uccs.edu
Ph.D., Business University of Nebraska, Lincoln
B.S., Business, University of Colorado at Colorado Springs, CO
Professor, Accounting, 2020 to present
Associate Professor, 2011-2020
Assistant Professor, 2007-2011
Wichita State University, KS
Assistant Professor, Accounting
Certified Public Accountant, State of Colorado, 1994 - present
Audit Supervisor, Senior, and Staff Accountant, BKD, LLP, Colorado Springs, CO
AABY (Altruistic & Behavioral Yardstick) Award, College of Business: 2014-2015
Outstanding Faculty Award, College of Business: 2008-2009 and 2012-2013
Outstanding Intellectual Contributions Award, College of Business: 2012-2013
Outstanding Teacher Award, College of Business, 2011-2012 and 2015-2016
Outstanding Service Award, College of Business, 2014-2015
Best Paper Award, American Accounting Association Government and Nonprofit Section Annual Meeting, 2014
Best Paper Award, American Accounting Association Government and Nonprofit Section Midyear Meeting, 2009
Lawler Award for the Best Article in the Journal of Accountancy in 2008, American Institute of Certified Public Accountants, 2009
American Accounting Association, Member, 2001 - present
American Institute of Certified Public Accountants, Member, 1994 - present
Colorado Society of Certified Public Accountants, Member, 1994 -1998, 2007 - present
Not-for-Profit Accounting
View Full VITA PDF
UCCS Articles about Brian McAllister, Ph.D., CPA
Adapting for the Not-For-Profit - June 1, 2018
Monitor your KPIs with a performance dashboard - March 24, 2017
Brian McAllister, Journal of Accountancy - August 1, 2016


Summer was originally designed and the pattern written in 1999, during my first year of pattern writing. Looking back on some of my earlier patterns that have not made the rank of retirement, I cringe. So at the beginning of 2009 I decided to take some time here and there and rewrite these older patterns in the style that I have evolved to through time and experience. With Summer I have re-designed the doll herself, and just her concept and look remains the same. It is now a much better pattern and I hope you enjoy making this little doll!
Summer is 15", with a 3-piece head, articulated hands and sculpted toes. Water color pencils are applied for her features which are also needle sculpted. Summer's hair is made from unraveled sweater wool. her costume is quick and easy!


text:
Q: How to include and use new fonts in wxWidgets projet? How to include and use new fonts in wxWidgets projet?
I am using VS2005.
I just want to print text using new ttf font.
Thanks in advance!!

A: Unless you're willing to link against something like FreeType:
http://en.wikipedia.org/wiki/FreeType
...most any program is going to require the font to be installed to the operating system, by the user or by some OS-specific script.  You can't just load it by filename off the cuff in your app.
Because of the platform dependence of naming and accessing custom fonts, the path of least resistance is not to try and hardcode a font...but to let the user pick one out of a dialog.  You would use a wxFontDialog for this:
http://docs.wxwidgets.org/stable/wx_wxfontdialog.html
It will let you retrieve the wxFontData, from which you can get the chosen wxFont:
http://docs.wxwidgets.org/stable/wx_wxfontdata.html#wxfontdatagetchosenfont
Once you have that, you can save and reload an identity of the font via the native string interface:
http://docs.wxwidgets.org/stable/wx_wxfont.html#wxfontgetnativefontinfodesc
Trying to formulate these strings on your own or work with the "face name" is a little dodgier:
http://docs.wxwidgets.org/stable/wx_wxfont.html#wxfontsetfacename
Generally speaking a lot of the same problems arise here as dealing with fonts in HTML.  If you have a very specific idea about the cross-platform appearance of some text, your best bet is often to make an image out of that text and use that instead of going through the hoops to get the font you want in the app.  If you're more flexible and have a lot of text the user is interested in, then they may be interested in changing the font too.  So just use a default but offer the user a choice to pick anything they want which is installed on their system.
(Note: I personally consider the handling of fonts in pretty much every OS or document system to be a disgrace.  Imagine a world where in order to get a graphic to display in your program you had to register it with the operating system through a complex process and it would not copy from machine to machine when you copied a document in which it was embedded.  We're dealing now with graphics that are orders of magnitude larger than font files, and yet they are handled seamlessly while people seem to accept the lack of seamless font transfer as "normal".  Archaic DRM mindsets of font vendors is one side of the problem, but lame technology is another big component.)

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text:
Auditor & Elections
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View the dates listed below for upcoming elections in Marion County.
PCM School Special Election: February 15, 2022
Marion County voters who live in the PCM Community School District may vote to fill a vacancy for school board member director district four. The Otley Reformed Church, located at 700 Johnson Street in Otley, is the only applicable voting location for Marion County voters who live in the PCM School District. Polling place hours are from 7:00 a.m. - 8:00 p.m.
Primary Election: June 7, 2022
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Candidate Filing Period: March 7 - March 25.
The Marion County Auditor's Office is the filing officer for the positions listed below.
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General Election: November 8, 2022
The General Election is the most prominent election conducted by the Auditor's Office as it has the largest turn out of all elections. This election is held in even numbered years, to elect federal, state, county and township officials and other officials as provided by law. It is also held to vote on statewide and county public measures and constitutional amendments. All countywide polling locations are utilized for this election.
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