Redmi Note 5 Pro Cover - Buy Redmi Note 5 Pro Back Cover online in India with cool & trendy designs at Beyoung.in. Shop the best Redmi Note 5 Pro cover and cases from Beyoung's exclusive range of protective Mi Note 5 Pro back cover at just Rs 199.
Redmi Note 5 pro Back Cover : In the diverse range of Redmi note 5 pro covers, Beyoung is the best online store that deals in protective and designer Redmi Note 5 Pro back covers at an affordable price. Moreover, an urge of having a wide collection of covers that slays the occasion is very common these days. Simultaneously, Beyoung brings the most iconic look in the Redmi Note 5 Pro mobile cover that showcases the fashion sense in a creative way. The best part is Beyoung fulfills all the desires that you see in an ideal Mi note 5 pro back cover that is great rigidity and durability. Therefore, there is no as such constraint in the Redmi Note 5 Pro back cover online shopping as Beyoung.in has got you covered in all aspects.
The Redmi Note 5 Pro cover and cases online at Beyoung are equipped with all the features which are required to be an ideal mobile cover. There is no doubt in saying that the Redmi note 5 pro back cover is best to buy at Beyoung as here you will not only experience the funky designs but also some more additional features like zero shipping cost, no hidden charges, customization process and diverse genres in Redmi note 5 pro cover and cases designs. Therefore, retrospect all the designs in the note 5 pro cover and shop the one that actually relates your persona.
Why Choose Beyoung to Buy Best Cover for Redmi Note 5 Pro?
Buying best cover for Redmi Note 5 Pro hard case online at beyoung has become quite blissful and worthy. Basically, the factors that are sole responsible for productive shopping in Note 5 Pro back cover is premium quality and low price. In addition to this you can also indulge yourself in a creative process as well that is Beyoung provides a customization tool where you can design the Redmi Note 5 Pro cases and covers online in just a few clicks. Lastly, the print quality makes it the best cover for Redmi note 5 Pro.
Along with the stylish & funky Redmi note 5 pro mobile cover you will come up with the phenomenon factor that is the quality which is the heart of the cover. Speaking about the quality of the case then it comprises of the polycarbonate material that eventually shields the device in the 360 degrees. Basically, it has a sturdy feature to absorb the abrupt impacts that can comprise the functionality and appearance. Speaking about the price of the Redmi note 5 pro mobile cover online shopping at Beyoung then it is highly pocket-friendly and the desire of having a cool collection of best cover for Redmi Note 5 Pro is fulfilled at ease. Therefore, you can deduce that the options are immense; it is just that you need to take a perfect call in order to buy a worthy Redmi Note 5 Pro mobile cover.
The Redmi Note 5 Pro back cover online shopping has become worthy and significant. Basically, Beyoung has introduced an amazing collection of Redmi Note 5 Pro back case that denotes both the aspects of cool appearance and protective feature. Beyoung is the best online shopping sites in India where you can explore a vivid collection of Redmi Note 5 Pro mobile cover, for instance, Apna Time Aaeyega and How's the Josh. To be precise you can come up with various themes in Redmi Note 5 Pro back cover designs like one-liners, cartoon, motivational quotes, images, superhero, Bollywood, PUBG and likewise. So without much ado shop, Redmi Note 5 Pro back cover online at Beyoung before it gets out of reach.
The best quality of the Redmi Note 5 Pro phone covers gives a sturdy aspect that keeps the device prone to all abrupt damages.
The matte finish Redmi Note 5 pro hard case is the secret touch to covers that gives a lively texture that enhances the appearance of the design.
Price of the Redmi Note 5 Pro back covers online at Beyoung is such pocket-friendly that you may tend to have a numerous collection of designer Redmi note 5 pro back cases.
Awesome collection of trendy designs in Redmi Note 5 Pro back cover for girls & boys will avail you much liberty to choose the phone that flaunts about your fashion sense.
There are also beatific designer mobile covers of other Xiaomi models and also most loved design options is available in more than 120 models like Nokia 6.1 Plus back covers, Oneplus 6 Covers, Redmi Note 6 pro Back Cover, Vivo V9 back Cover, Honor 9 Lite back cover, Redmi 6 Pro back cover, Redmi Note 5 back cover, Redmi Note 7 back cover and many more so that you always have contented shopping experience. The variety of designs available in the Redmi note 5 pro back case will lure you to buy one. The designs in the Redmi Note 5 Pro phone cover is very adorable and funky. You can also have mobile cover printing at Beyoung's customization tool So what are you waiting for, visit Be young and from the exclusive range of mobile cases choose the best Redmi Note 5 Pro back cover. Moreover, You can Also Visit Some of The Most Popular Categories of Beyoung like T Shirts for girls, T-shirts for Men, Full Sleeve T-shirts, Couple T Shirts & Custom T Shirts.
By Dikshit Maheshwari - I purchased Redmi Note 5 Pro back cover from Beyoung. The print quality and cover is really awesome. When I case it on my phone it is looking very nice.
By Shivani Vyas - I like the shipping process as it is easy and the order was placed in just few clicks. The overall experience of buying Redmi Note 5 Pro cover is very nice and I am satisfied from this mobile cover.
By Mohit Jain - The Redmi Note 5 Pro case is looking awesome and the quality is just commendable. I am happy to shop this product from Beyoung.


text:
using System.IO;

namespace NetworkGameCopier
{
    public class Utilities
    {

        public static long DirSize(DirectoryInfo d)
        {
            long size = 0;
            // Add file sizes.
            FileInfo[] fis = d.GetFiles();
            foreach (FileInfo fi in fis)
            {
                size += fi.Length;
            }
            // Add subdirectory sizes.
            DirectoryInfo[] dis = d.GetDirectories();
            foreach (DirectoryInfo di in dis)
            {
                size += DirSize(di);
            }
            return size;
        }

    }
}

redpajama_set_name: RedPajamaGithub


"Extreme cut jeans" have the internet in a frenzy.
As if crotchless and ripped jeans weren't enough, one denim company decided to take the shredded look to a whole new level.
Carmar Denim is selling a pair of "extreme cut out" jeans that leave very little to the imagination.
The jeans are selling for $168, despite the slim amount of fabric that actually went into making them.
Remember when jeans were just denim trousers? Then came the advent of skinny jeans, followed by the high-waisted and the "mom" cut... all fashion revolutions in their own right, at the time.
Oh, how things have changed. Now, there are all sort of kooky reincarnations of the humble wardrobe staple, from crotchless styles to pairs that have quite literally been ripped to shreds.
However, the latest sartorial scandal to take the jean world by storm comes courtesy of LA-based brand, Carmar Denim, whose "extreme cut out" jeans leave so little to the imagination, they're practically an anatomy lesson.
A bit pricey for the lack of fabric.
The unique garment is quite literally "cut out" in almost every area, hiding one's modesty only very slightly via a few denim threads - the model wearing the jeans on the website also happens to pair them with a cut out swimsuit, just in case you didn't get the memo.
The irony is of course that, despite how little fabric they consist of — just 20 grams — the jeans are rather pricey, setting adventurous consumers back $168 (£122).
Interestingly, the website describes the jeans as a "high rise pant with large statement cutouts on front and back."
It adds that they come in a "relax fit." However, it is unclear as to just how relaxed you have to be in order to wear them in public.
Naturally, people have responded to the jeans with dismay on social media.
"Please somebody tell me this is a joke shop and there aren't people dumb enough to pay $168 for this," wrote one person on Twitter.
"It's almost like they ran out of ideas," added another, "'Hey guys lets come up with some ideas for some new fashion!' *guy in the back raises hand* 'I know! Lets make pants that look like a harness.' Literally, the stupidest thing I've ever seen."
One thing's for sure, if you do decide to don these jeans, underwear is most certainly advisable, particularly if adhering to the standards of common decency is something you stand by.


text:
IT EN ES
Home > Documentation
DIAS Card. Ivan
Cardinal Ivan Dias, Prefect emeritus of the Congregation for the Evangelization of Peoples, Archbishop emeritus of Bombay, India, was born on 14 April 1936 in Mumbai, India. He was ordained for the Archdiocese of Bombay on 8 December 1958 and holds a doctorate in canon law from the Pontifical Lateran University, Rome.
He entered the Holy See's diplomatic service in 1964 and was posted to the Nordic countries, Indonesia, Madagascar, La Réunion, the Comorros, Mauritius and the Secretariat of State. On 8 May 1982 he was appointed titular Archbishop of Rusubisir and Apostolic Pro-Nuncio in Ghana, Togo and Benin, and received episcopal ordination on 19 June. He later served as Apostolic Nuncio in Korea (1987-91) and Albania (1991-97).
On 8 November 1996 he was appointed Archbishop of Bombay.
President Delegate of the 10th Ordinary General Assembly of the Synod of Bishops (October 2001).
He was appointed Prefect of the Congregation for the Evangelization of Peoples and Grand Chancellor of the Pontifical Urbanian University from On 20 May 2006 until 10 May 2011.
He participated in the conclave of April 2005, which elected Pope Benedict XVI and in the conclave of March 2013, which elected Pope Francis.
Created and proclaimed Cardinal by St. John Paul II in the consistory of 21 February 2001, of the Title of Spirito Santo alla Ferratella (Holy Spirit at Ferratella).
Cardinal Ivan Dias died in Rome on June 19th 2017.

redpajama_set_name: RedPajamaCommonCrawl


text:
NFL Scouts' Takes: Biggest Risers in the 2019 Draft Class
Matt Miller@nfldraftscoutTwitter LogoNFL Draft Lead WriterJanuary 22, 2019
Which 2019 prospects are soaring up draft boards?Gregory Shamus/Getty Images
Every college football season, we see players enter the scouting process who weren't on draft lists over the summer. Even the scouting services most NFL teams use to help prepare for the upcoming draft class before the college football season will miss a few.
The players listed below weren't all missed by scouts—some are underclassmen who popped onto the radar after winning starting jobs over the summer—but they've all seen their stock rise considerably through great play and exceptional NFL-caliber traits.
The job of scouting this class is far from done, though, and each player could still sink his stock with an off-field incident, injury or poor testing time. One bad interview with a general manager at the combine or poor effort at the Senior Bowl could change everything. The bet is that the character, wiring and work ethic of the following players are all high-level enough to prevent letdowns and allow them to ride their rising stock into the first round come draft night.
Kyler Murray, Quarterback, Oklahoma
Lynne Sladky/Associated Press
Scout's Quote: "I don't know how you evaluate someone that height and that build, but he's electric. I don't think he's as accurate or as smart as Baker [Mayfield], but he could absolutely go first overall."
Kyler Murray went from competing with Austin Kendall for the starting quarterback job in August to winning the Heisman Trophy in December. Now he's considered one of the best quarterback prospects in the 2019 draft class…if he sticks with football.
Murray, who was a top-10 pick by the Oakland A's in the 2018 MLB draft, has options. That could mean he reports to spring training in mid-February, or he could participate in the NFL Scouting Combine later that month and signal his intention to make professional football his job.
There will be wide ranges of evaluations on Murray given his stature, but there is a general excitement for him to get under the NFL's microscope should he choose a pigskin over a bat.
Dwayne Haskins, Quarterback, Ohio State
Mark J. Terrill/Associated Press
Scout's Quote: "Kyler Murray entering the draft kind of stole his thunder, but he'll get back into the top-five talk. I bet someone trades into the first four picks to get him. He's NFL-ready."
Another first-year starter, this time at Ohio State, Dwayne Haskins checks all the boxes for a future NFL starter. He's big, strong-armed, poised in the pocket and has the accuracy to all levels to attack the field in any offensive scheme.
Haskins doesn't have the mobility of Kyler Murray or the experience of Daniel Jones (Duke) or Drew Lock (Missouri), but he finished his college career with exceptional showings against Michigan, Northwestern and Washington while taking on some of the best defenses in the country.
Haskins' ceiling is high, and while he might not be ready to start in the NFL from Day 1, he's gone from first-time starter to sure-fire first-rounder in the course of a season.
Josh Jacobs, Running Back, Alabama
Wilfredo Lee/Associated Press
Scout's Quote: "The biggest question is height. He looks pretty short, but he's incredibly powerful and quick. Catches the ball well. If you're going to take a running back in the first [round], it would be him."
When the 2018 college football season started, there were no running backs considered first-round prospects. The closest thing to a first-rounder was Alabama's Damien Harris. Then Josh Jacobs got on the field for Alabama and it all changed.
The 5'10", 215-pound Jacobs dominated SEC defenses in his junior season en route to becoming a legitimate first-rounder. He's No. 9 overall on my most recent board and was the No. 5 pick in Daniel Jeremiah's first mock draft.
Jacobs' ascent from third on the Alabama depth chart behind Damien Harris and Najee Harris to a potential top-10 pick makes him this season's biggest riser.
Mecole Hardman, Wide Receiver, Georgia
Mike Stewart/Associated Press
Scout's Quote: "He can move! I don't know that his 40 time will be amazing, but he's super agile and can cut on a dime. Plus his acceleration is amazing. Bet you he goes Round 2."
The Georgia junior wide receiver entered the 2019 NFL draft class and has scouts excited. The feedback received during NFLPA Collegiate Bowl practices was that Hardman received a Day 2 grade from the NFL draft advisory board. One scout I spoke to believes that is low in a draft "full of big, slow receivers."
Hardman is an electric playmaker on offense with the ability to make an impact running the ball, catching it or working as a return man. As every team looks for its version of Tyreek Hill, Hardman will be in demand.
Quinnen Williams, Defensive Lineman, Alabama
Brynn Anderson/Associated Press
Scout's Quote: "For a redshirt sophomore, he's a powerful player with a high football IQ. Everyone there loves the kid. He's a no-brainer as a 3-technique. Top-five pick."
The NFL is shifting into a league that values interior pressure to get after the quarterback. That lines up perfectly with the rise of Quinnen Williams this season.
A first-year starter at Alabama in 2018, Williams quickly overshadowed his more well-known defensive line counterparts Raekwon Davis and Isaiah Buggs to earn a reputation as a legitimate top-five player in the upcoming class. At 6'4" and 295 pounds, Williams has the perfect build to play as an interior rusher but is athletic enough to move around to find the best matchup on the line.
Williams might not be Aaron Donald, but those watching his film might be reminded of Gerald McCoy or Ndamukong Suh as prospects.
T.J. Hockenson, Tight End, Iowa
Holly Hart/Associated Press
Scout's Quote: "He's the best tight end in the class. Better than his teammate [Noah Fant] as a three-down player. He's a real tight end. Powerful, fast, good hands, great blocker. I'd draft him top-15 in this class."
The Mackey Award for the best tight end in college football went to a player few outside Iowa City knew when the season began. T.J. Hockenson became a household name soon after the opening kickoff.
Sharing targets with another first-round talent in Noah Fant, Hockenson still became recognized as a unique talent and dominant player in the Hawkeyes offense. He's versatile enough to move around the formation while showing excellent ability as a blocker.
His 49 catches in 2018 might not sound like a lot, but Hockenson showed excellent feel and hands to earn those receptions and his six touchdowns. It's a statement about his overall game that with only 49 catches he was still named the best tight end in football.
Nasir Adderley, Safety, Delaware
Scout's Quote: "He's a little undersized and might get knocked for that, but he's a top-tier athlete who can really run. He has ball skills. Flexible positionally since he's played some nickel. I think he goes top-50."
A commonality among the previous players on this list is that all play at major schools. Delaware senior Nasir Adderley has been hidden behind the Taylor Rapps and Deionte Thompsons of the college world until recently, but those who break down the film are seeing an explosive athlete with serious coverage skills and range.
Senior Bowl week is big for Adderley as he competes against the best quarterbacks and pass-catchers in the country. He can show that coming from a small school won't slow him down and that he's more than "just" an athlete.
Don't be surprised if Adderley comes out of the Senior Bowl with a Round 1 grade.
Deionte Thompson, Safety, Alabama
Butch Dill/Associated Press
Scout's Quote: "He's just so skinny, and Clemson beat him up in the title game. Still a first-rounder because of this safety class and he has really good range. But probably late first."
There is a running theme of first-year starters in this class who rose through the ranks to become stud prospects. Alabama's Deionte Thompson is another who took his first major snaps in 2018 and now looks like a first-rounder.
As mentioned, Thompson's build may get him knocked behind Taylor Rapp (Washington) or Johnathan Abram (Mississippi State), but he's the best coverage safety in this class. Many teams favor a two-deep safety look and crave deep coverage help, and Thompson is the best fit for them.
Thompson needs a strong combine to lock into Round 1 on my board after the struggles shown in the Clemson game. The fact that he played with a loaded defense in front of him might have shielded Thompson from plays the other top safeties had to handle alone.


Robert L. Millet, "What We Worship," in My Redeemer Lives! ed. Richard Neitzel Holzapfel and Kent P. Jackson (Provo, UT: Religious Studies Center, Brigham Young University, 2011).
Robert L. Millet was a professor of ancient scripture at Brigham Young University when this book was published.
In his great Intercessory Prayer, Jesus pleaded with the Father to make his disciples one as they (the Father and the Son) were one. It was earlier in that same prayer that Jesus uttered the following timeless words: "And this is life eternal, that they might know thee the only true God, and Jesus Christ, whom thou hast sent" (John 17:3). This is a tremendously important message. In the end, it will matter precious little what we know about a myriad of things if we have not come to know, through the power of the Holy Spirit, the God we worship. This is one of the great purposes of mortality and thus the quest of a lifetime. It is in this life that we prepare to meet God (see Alma 34:32), that we come to know him and thereby to grasp what we worship, to comprehend whom we worship (see D&C 93:19).
St. Thomas Aquinas, the great Christian philosopher and theologian, taught that in the long run we cannot really know what God is, only what he is not.  While I tend to disagree with Aquinas on this—especially in light of the above passage in John 17—it may be just as important to know what the Lord is not as to know what he is. In that light, let us consider some things that the Lord Jesus Christ is not.
1. Jesus Christ is not a cosmic errand boy. I mean no disrespect or irreverence in so saying, but I do intend to convey the idea that while he loves us deeply and dearly, Christ the Lord is not perched on the edge of heaven, anxiously anticipating our next wish. When we speak of God being good to us, we generally mean that he is kind to us. In the words of the inimitable C. S. Lewis, "What would really satisfy us would be a god who said of anything we happened to like doing, 'What does it matter so long as they are contented?' We want, in fact, not so much a father in heaven as a grandfather in heaven—a senile benevolence who as they say, 'liked to see young people enjoying themselves,' and whose plan for the universe was simply that it might be truly said at the end of each day, 'a good time was had by all.'"  You know and I know that our Lord is much, much more than that.
2. Jesus was not just a Galilean guru, nor was he a Samaritan Socrates. It is indeed fascinating to read the New Testament Gospels, looking specifically for such things as what Jesus said, how he said it, how he responded to questions, and how he dealt with criticism and ridicule. To be sure, Jesus Christ was one bright man. He seemed to always have the right answer for the situation. But he was more than a teacher, more than an inspiring teacher, more than a great moral teacher. He was the Son of God, God the Son. That means he was more than a composite of intelligent answers, more than walking wisdom. In him was understanding and insight, but, more important, inside him were the powers of godliness, the powers of immortality.
seems rather like holding up Tiger Woods as an example of how to hit a golf ball. Even if I started now and practiced for eight hours a day, it is highly unlikely that I would ever be able to do what Woods can do; and there are many people out there, younger and fitter than I, who are trying their hardest to do it and still find they can't. Similarly, watching Jesus . . . makes most of us, all but the most proud or ambitious, feel like we do when watching Tiger Woods hit a golf ball. Only more so. . . .
What's more, that we treat Jesus as a moral example can be, and in some people's thinking has been, a way of holding at arm's length the message of God's kingdom on the one hand and the meaning of his death and resurrection on the other. Making Jesus the supreme example of someone who lived a good life may be quite bracing to contemplate, but it is basically safe: it removes the far more dangerous challenge of supposing that God might actually be coming to transform this earth, and us within it, with the power and justice of heaven. . . . Jesus as "moral example" is a domesticated Jesus, a kind of religious mascot.
3. Jesus is not "religious." Let me explain what I mean. The Latin word religio originally referred to a binding obligation, a rather special obligation. In that sense, religion represents man's effort to apply true principles and doctrines in order to enable people to keep their obligations, their covenants with the Almighty. You may recognize in the word religion the root for another word that we know quite well—the word ligament. A ligament is a fibrous tissue that binds or links bone to cartilage or that holds organs in place. Thus the true purpose of God-ordained religion is to link or tie mortal men and women to an immortal, glorified Deity. As the Apostle James taught, "Pure religion and undefiled before God and the Father is this, To visit the fatherless and widows in their affliction, and to keep [oneself] unspotted from the vices of the world" (Joseph Smith Translation, James 1:27). In other words, pure religion deals with two main aspects of our lives: how we treat other people, and to what extent we strive to remain free from sin in a sinful world. In this sense, God clearly is religious.
Too often, however, religion has been separated off from everyday life and simply made into another aspect or dimension of our lives. Thus we speak of our intellectual life, our athletic life, our social life, and our religious life. Religion is thus one of the pieces of the larger pie. Authentic Christianity, Brennan Manning observed, is not "a code of do's and don'ts, not a tedious moralizing, not a list of forbidding commandments, and certainly not the necessary minimum requirement for avoiding the pains of hell. Life in the Spirit is the thrill and the excitement of being loved by and falling in love with Jesus Christ."  True religion represents our link with God, our tie to the Infinite, and thus should and must inform and impact every other phase of our existence. In other words, religion is not something we do on Sunday, while we go about our business the other six days of the week. As Latter-day Saints, our religion is life, a 24/7 life.
One of the great challenges we Latter-day Saints face as a Church is a happy challenge—the challenge of Church growth. Not only does this mean we have to prepare more and more young people to serve full-time missions and more new converts to serve as leaders, but we must also see to it that we do not allow the Church or our personal lives to be directed solely by rules and written regulations. This is the Lord's Church, and he is at the head. Because it is his Church, it is to be conducted according to his plan and under his direction, through the Holy Spirit. If we become so sterile and fossilized in the way we do things in the Church, we may lose that marvelous spiritual spontaneity that should and must accompany the body of Christ (see Moroni 6:9; D&C 46:2). Pure religion is far less concerned with what we do than with who and what we are and what we are becoming.  Pure religion is a thing of the heart.
4. Jesus is not just my elder brother. Latter-day Saints have insight into our eternal existence that persons of other faiths do not possess. For example, many of the things that happen to us in this life, including traumas and tragedies, are viewed with a more elevated perspective, given what we know about the fact that we lived as spirits before we were born in mortality. In our first estate, our premortal existence, we were taught, trained, and prepared to come to earth and take on a wondrous mortal body, all as a very significant part of the overall plan of salvation. John the Beloved opened his Gospel with this statement: "In the beginning was the Word, and the Word was with God, and the Word was God. The same was in the beginning with God. All things were made by him, and without him was not any thing made that was made" (John 1:1–3). That is, in the premortal life Christ, here designated as the Word, was with our Heavenly Father. In fact, Christ was God in that first estate. As God and leader of the "noble and great ones" (Abraham 3:22; 4:1), he created "worlds without number" (Moses 1:33; 7:30; compare Hebrews 1:1–2). It is theologically appropriate, therefore, to refer to Jesus Christ as our elder brother as pertaining to the premortal life.
It is of great interest to me, however, that of the almost one hundred names given to our Lord and Savior by the Nephite prophets, the phrase "elder brother" is never used once. He is called the Almighty, the Almighty God, Alpha and Omega, Creator, Eternal Father, Eternal God, Eternal Head, Eternal Judge, Everlasting God, Father of heaven and of earth, God, God of Abraham, God of nature, Holy Messiah, Holy One of Israel, Immanuel, Keeper of the Gate, Lamb of God, Lord God Almighty, Lord God Omnipotent, Lord God of Hosts, Mighty One of Israel, Most High God, Redeemer of Israel, Supreme Being, True and Living God, and True Messiah but never "elder brother." In other words, I am convinced that because the Nephites looked upon Christ with such awe and viewed him with such majesty, it may not have seemed appropriate to them to refer to him as "elder brother"; he was God.
We occasionally hear some members refer to Jesus as our Elder Brother, which is a true concept based on our understanding of the premortal life with our Father in Heaven. But like many points of gospel doctrine, that simple truth doesn't go far enough in terms of describing the Savior's role in our present lives and His great position as a member of the Godhead. Thus, some non-LDS Christians are uncomfortable with what they perceive as a secondary role for Christ in our theology. They feel that we view Jesus as a spiritual peer. They believe that we view Christ as an implementer for God, if you will, but that we don't view Him as God to us and to all mankind, which, of course, is counter to biblical testimony about Christ's divinity. Let me help us understand, with clarity and testimony, our belief about Jesus Christ. We declare He is the King of Kings, Lord of Lords, the Creator, the Savior, the Captain of our Salvation, the Bright and Morning Star. He has taught us that He is in all things, above all things, through all things and round about all things, that He is Alpha and Omega, the Lord of the Universe, the first and the last relative to our salvation, and that His name is above every name and is in fact the only name under heaven by which we can be saved. . . .
5. Christ is not "my buddy." There is a natural tendency, and it is a dangerous one, to seek to bring Jesus down to our level in an effort to draw closer to him. This is a problem among people both in and outside the LDS faith. Of course we should seek with all our hearts to draw near to him. Of course we should strive to set aside all barriers that would prevent us from closer fellowship with him. And of course we should pray and labor and serve in an effort to close the gap between what we are and what we should be. But drawing close to the Lord is serious business; we nudge our way into intimacy at the peril of our souls.
There are a number of gospel ironies in the scriptures. One of them is the irony that only those who lose their lives in service to the Lord find eternal life (see Matthew 16:25–26). Jesus said on one occasion that he came not to bring peace but to bring a sword "to set a man at variance against his father and the daughter against her mother. . . . And a man's foes shall be they of his own household" (Matthew 10:34–36). How odd! Jesus is the Prince of Peace, and everyone knows that he above all would want family members to be close, to be united. He is teaching, however, that sometimes there is a cost associated with receiving and living the gospel, a price that may indeed separate you from those you love most.
Another gospel irony is that the way to get close to the Lord is not by attempting in any way to shrink the distance between us, to emphasize more of his humanity than his divinity, or to speak to him or of him in casual, colloquial language. In fact, as King Benjamin explained, you and I are enabled to retain a remission of sins from day to day by recognizing Christ's magnificence, majesty, and power and the fact that we are unprofitable servants who are less than the dust of the earth (see Mosiah 2:20–25; 4:11–12). Perhaps the greatest lesson that the mighty lawgiver Moses learned was a lesson that followed a transcendent encounter with Deity and a panoramic vision of God's creations. "And it came to pass that it was for the space of many hours before Moses did again receive his natural strength like unto man; and he said unto himself: Now, for this cause I know that man is nothing, which thing I never had supposed" (Moses 1:10).
This is a terribly difficult balance to strike. We want ever so much to be close to the Lord. We seek, ever so diligently, spiritual fellowship with the Father and the Son (see 1 John 1:3). And yet we dare not presume upon the dignity and divinity of Deity.
6. Jesus Christ is not the "god of the gaps." Several years ago my colleague Stephen Robinson and I were invited to spend the day in conversation with representatives of the Southern Baptist Convention. We talked together for about seven hours that day, and some of the time was pleasant. At a certain point in the conversation, however, one of our Baptist friends made the comment, "But of course you folks do not believe in the grace of Jesus Christ." Steve and I both leaned forward in our chairs and proceeded to try to convince our friend that in fact we did believe in and teach the importance of salvation by the grace of Christ. At that point one of his associates responded, "Yes, we understand, you believe in and worship the god of the gaps." I replied, "I have never heard that before in my life. Who or what is the god of the gaps?" He went on to explain that it was his understanding that Latter-day Saints believed in a kind of works righteousness, that men and women are to do everything they can and expend all of their efforts and then Jesus would fill in the remaining deficit. An hour later, and after seeking again and again to dissuade them from their caricature of Mormonism, we realized that we had failed.
I have met members throughout the Church who suppose that this means that Christ can only help us on the Day of Judgment when we have expended our best efforts and done everything we know how to do. First of all, who will have done everything they could have done? Who will have spent every waking hour of every day of every year serving God tirelessly and tenaciously? Only one person fits this bill, and that was the Lord Jesus Christ himself; he was the only one to live a perfectly sinless life. I sincerely believe that what Nephi is trying to teach is that we are saved by the grace of Jesus Christ—his unmerited divine favor, his unearned divine assistance, his enabling power—above and beyond all we can do, notwithstanding all we can do, in spite of all we can do. Too often we are prone to think of grace as the Lord's final boost into celestial glory hereafter. To be sure, we will need all the divine help we can get in order to qualify to go where God and angels are. But the grace of God is extended to you and me every hour of every day and is not limited to the bar of judgment.
Now having emphasized that one of the burdens of holy scripture, ancient and modern, is that we are saved by merit, but not our own merit; that salvation is free (see 2 Nephi 2:4); that eternal life is a gift, indeed, the greatest of all the gifts of God (see D&C 6:13; 14:7); and that there is consummate peace in trusting in and relying upon the goodness of God our Savior, I hasten to add that one of the scandals of the Christian world, a scandal into which we cannot afford to slide, is a seeming disregard for the simple statement of the Master: "If ye love me, keep my commandments" (John 14:15). In an effort not to supplement in any way the finished work of Jesus Christ, many of our Christian friends have created their own hyperorthodoxy of language and practically outlawed such words as work or labor or obedience or commandment keeping. The result is that the Christian message in the world has become less and less appealing, for the men and women who claim Christian status do not seem to live, on the whole, any differently than worldly people.  Their speech is impressive, but their personal lives leave much to be desired. Easy believism and cheap grace have replaced the depth of discipleship demanded by Deity.
Nephi taught that the words of the prophets are "sufficient to teach any man the right way; for the right way is to believe in Christ and deny him not; for by denying him ye also deny the prophets and the law. And now behold, I say unto you that the right way is to believe in Christ, and deny him not; and Christ is the Holy One of Israel; wherefore ye must bow down before him, and worship him with all your might, mind, and strength, and your whole soul; and if ye do this ye shall in nowise be cast out" (2 Nephi 25:28–29).
We learn from both ancient and modern scripture of the Person and powers of Jesus the Christ: that he was God before he was born (see John 1:1–2); that through a grand condescension (see 1 Nephi 11:16–33) he left his "throne divine"  so as to "be like man almost" ; that he came into the world to be crucified and to bear the sins of the world (see D&C 76:41); that he received grace for grace and continued from grace to grace until, in the resurrection, he received a fulness of the glory and power of the Father (see D&C 93:16; compare Matthew 28:18);  and that as we follow our Lord's pathway, we too will receive grace for grace, we too will progress from one level of divine acceptance to a higher. We thereby tread on holy ground and emulate the sacred walk of this Prototype of all saved beings.  "I give unto you these sayings," the Savior declared in a modern revelation, "that you may understand and know how to worship, and know what you worship, that you may come unto the Father in my name, and in due time receive of his fulness" (D&C 93:19).
Jesus is not only central to the plan of salvation; he is vital. We cannot save ourselves. We cannot earn our exaltation. We cannot exercise the sufficient grit and willpower to do the works of righteousness and battle against Satan on our own. Christ is our Lord, our Savior, our Redeemer, and our King. Through the medium of his Spirit, he is the Agent of the mighty change that comes to those who come unto him. He is the Lord of Hosts, the Lord of Armies, the Captain of our Salvation. He is God, and if it were not so, he could not save us. Without him, we have nothing. With him, we have everything.
 Thomas Aquinas, Summa Theologiae, Article 10.
 C. S. Lewis, The Problem of Pain (New York: Touchstone, 1996), 35–36.
 D. Stephen Long, "God Is Not Nice," in God Is Not . . . , ed. D. Brent Laytham (Grand Rapids, MI: Brazos Press, 2004), 49–50.
 Brennan Manning, Ruthless Trust (San Francisco: HarperSanFrancisco, 2000), 107.
 Robert Stein, The Method and Message of Jesus' Teachings (Philadelphia: Westminster Press, 1978), 118–19.
 John Meier, A Marginal Jew: Rethinking the Historical Jesus (Garden City, New York: Doubleday, 1991), 1:177.
 C. S. Lewis, Mere Christianity (New York: Touchstone, 1996), 56.
 Joseph Smith, comp., Lectures on Faith (Salt Lake City: Deseret Book, 1985), 7:9.
 N. T. Wright, After You Believe: Why Christian Character Matters (New York: HarperOne, 2010); 126–27; emphasis in original.
 Brennan Manning, The Importance of Being Foolish: How to Think Like Jesus (San Francisco: HarperSanFrancisco, 2005), 174.
 See Dallin H. Oaks, "The Challenge to Become," Ensign, November 2000, 32–34.
 Wright, After You Believe, 26–27.
 Barbara Brown Taylor, "Working People," Christian Century 127, no. 6 (March 23, 2010): 35.
 M. Russell Ballard, "Building Bridges of Understanding," Ensign, June 1998, 66.
 Manning, Importance of Being Foolish, 51.
 Bruce R. McConkie, "Our Relationship with the Lord," Brigham Young University devotional address, March 2, 1982; in Doctrines of the Restoration: Sermons and Writings of Bruce R. McConkie, ed. Mark L. McConkie (Salt Lake City: Bookcraft, 1989), 67–68.
 Ballard, "Building Bridges of Understanding," 65.
 Dallin H. Oaks, With Full Purpose of Heart (Salt Lake City: Deseret Book, 2002), 75; see also Jeffrey R. Holland, "The Atonement of Jesus Christ," Ensign, March 2008, 35–36.
 See Ronald J. Sider, The Scandal of the Evangelical Conscience: Why Are Christians Living Just Like the Rest of the World? (Grand Rapids, MI: Baker Books, 2005); Dallas Willard, The Great Omission: Reclaiming Jesus's Essential Teachings on Discipleship (San Francisco: HarperSanFrancisco, 2006); Robert Jeffress, Grace Gone Wild: Getting a Grip on God's Amazing Gift (Colorado Springs: WaterBrook Press, 2005).
 Wright, After You Believe, 62–63.
 "I Stand All Amazed," Hymns (Salt Lake City: The Church of Jesus Christ of Latter-day Saints, 1985), no. 193.
 "O God, the Eternal Father," Hymns, no. 175.
 See Joseph Fielding Smith, Doctrines of Salvation, comp. Bruce R. McConkie (Salt Lake City: Bookcraft, 1954–56), 2:269; Bruce R. McConkie, Mormon Doctrine, 2nd ed. (Salt Lake City: Bookcraft, 1966), 339.
 Smith, Lectures on Faith, 7:9.


text:
\section{Introduction}
\label{intro}
Polyominoes and polyiamonds and their tiling properties have been the subject of
computational geometry research that investigated
which polyominoes can tile the plane isohedrally and
which can tile by translations alone. \cite{rhoads,Kev}.
In earlier papers\cite{fukuda2006,fukuda2008},
we gave algorithms to create polyominoes and
polyiamonds that were fundamental domains for isohedral tilings having {\bf p3}, {\bf p4}, or {\bf p6} symmetry
groups.
In this paper, we consider the expanded task of producing polyomino and polyiamond
tiles that generate isohedral tilings of types {\bf p3}, {\bf p3m1}, {\bf p31m}, {\bf p4}, {\bf p4m}, {\bf p4g}, {\bf p6}
or {\bf p6m} and for
which the tiles are fundamental domains of the tiling.
Recently an extension of this study was published as a separate paper\cite{fukuda2011}
in which we carried out these investigations for symmetry groups of
types {\bf pmm}, {\bf pmg}, {\bf pgg}, and {\bf cmm}.

A polyomino (or $n$-omino) is a tile homeomorphic to a disk, made up of $n$ unit squares that
are connected at their edges; that is, the intersection of two unit squares in the polyomino is
either empty or an edge of both squares.
Similarly a polyiamond (or $n$-iamond) is a tile
homeomorphic to a disk, made up of $n$ unit equilateral triangles that are connected at their edges;
the intersection of two unit triangles in the polyiamond is either empty or an edge of both
triangles.

An isohedral tiling of the plane is a tiling by congruent tiles in which the symmetry group of
the tiling acts transitively on the tiles. A fundamental domain for an isohedral tiling is a region
of least area that generates the whole tiling when acted on by the symmetry group of the tiling.
A fundamental domain for an isohedral tiling cannot contain two points that are identical under
the action of the symmetry group of the tiling.
This leads immediately to the following observations:

\begin{romanlist}
\item
{\it In an isohedral tiling in which each tile is a fundamental domain, no tile can contain a
rotation center or axis of reflection for the whole tiling except on its boundary.}
\item
{\it In an isohedral tiling having a polyomino as fundamental domain, a 4-fold center of the tiling
can only occur at a vertex of a square that is a ``corner" of the polyomino;
that is, only one unit square of the polyomino contains that vertex.}
\item
{\it In an isohedral tiling having a polyiamond as fundamental domain, 3-fold or 6-fold centers of
the tiling can only occur at vertices of unit triangles on the boundary of the polyiamond.
Moreover, a 3-fold center can only occur at a vertex at which exactly two unit triangles meet. A
6-fold vertex can only occur at a vertex that is a ``corner" of the tile, that is, exactly one unit
triangle of the polyiamond contains that vertex.}
\item
{\it There are no 3-or 6-fold centers in a tiling by polyominoes.
Hence, there are no \mbox{\bf p3}, \mbox{\bf p3m1},
\mbox{\bf p31m}, \mbox{\bf p6} or \mbox{\bf p6m} isohedral tilings by polyominoes.
There are no 4-fold centers in a tiling by
polyiamonds. Hence, there are no \mbox{\bf p4}, \mbox{\bf p4m}, or \mbox{\bf p4g} isohedral tilings by polyiamonds. }
\end{romanlist}

In each of the sections that follow, we begin with a fixed lattice of symmetry elements for a
symmetry group $G$ (that is, a fixed array of rotation centers, reflection axes and glide-reflection
axes for elements of $G$) and give a backtracking procedure to produce a complete set of
polyominoes (or polyiamonds) that are fundamental domains for $G$.
A tile $T$ is {\it a fundamental domain for a symmetry group $G$}
if the action of $G$ on $T$ produces an isohedral tiling and $T$ is a
region of minimal area for which $G$ can generate that tiling. $G$ will be contained in (or equal to)
the full symmetry group of the tiling.

As we consider symmetries of our isohedral tilings, the following theorem will be useful. \cite{fukuda2011}

\begin{theorem}
\label{th1}
Let $G$ be one of the 17 two-dimensional symmetry groups and ${\cal T}$ an isohedral tiling
generated by $G$ acting on a tile $T$ that is a fundamental domain for $G$.
Let $G'$ be the full
symmetry group of ${\cal T}$.
If $G$ is a proper subgroup of $G'$, there is element of $G'$ (other than the
identity) that leaves $T$ invariant. In this case, a fundamental domain for $G'$ has area smaller
than $T$.
\end{theorem}

\section{{\bf p4}}
\label{p4}
\subsection{Creating polyominoes as fundamental domains for {\bf p4} symmetry groups}
\label{p4:create}
We begin with a lattice of unit squares as shown in Fig.~\ref{fig:p4lattice}.
Our $n$-omino tiles that are fundamental domains for a {\bf p4} group will be built from these unit squares.
By observation (ii) in section \ref{intro},
the 4-fold rotation centers for a {\bf p4} symmetry group must be located at lattice points.
So, first we place a 4-fold rotation center, (a black circle) at a lattice point and call this the origin.
We then place orthogonal unit vectors $\mathbf u$ and $\mathbf v$ at the origin (see Fig.~\ref{fig:p4lattice}).
Next we place a second
4-fold rotation center, the white circle in Fig.~\ref{fig:p4lattice}, at $x{\mathbf u} + y{\mathbf v}$,
where $x$ and $y$ are nonnegative
integers, not both $0$. These two choices of 4-fold rotation centers determine a whole {\bf p4} lattice of
rotation centers; rotations about the two placed black and white centers generate a {\bf p4} group $G$.
\begin{figure}[h]
\centerline{\psfig{file=Fig1.eps,width=2.7cm}}
\caption{
A lattice of unit squares with black and white 4-fold rotation centers that generate a {\bf p4} group.
The shaded isosceles right triangle is a standard fundamental domain;
its unmarked third vertex is a 4-fold center equivalent to
the white center. Here $x = 1$, $y = 3$; the area of the fundamental domain is $5$ sq. units.
\label{fig:p4lattice}
}
\end{figure}

The area $S$ of a fundamental domain for this {\bf p4} group $G$ is given by
\begin{equation}
S=\frac{x^2+y^2}{2}
\end{equation}
taking the area of a unit square as $1$.
Since we want our $n$-omino to be a fundamental domain, $n = S$ and $S$ must be an integer.
Therefore, $x$ and $y$ must have the same parity, and
\begin{equation}
n=1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25,	25,	26,	29,	32,	34,	36,	37,	40,	41,	45,	49,
\ldots
\end{equation}
where two pairs $(x,y)$, namely, $(5,5)$ and $(7,1)$ correspond to $n = 25$.
All unit squares in the lattice
are classified into $n$ equivalence classes by the action of the {\bf p4} group $G$.
We denote the equivalence class of a unit square $e$ as $C(e)$.

We construct a set $\mathscr{T}_n$ of $n$-ominoes
that are fundamental domains for the {\bf p4} group $G$ and a
given $n$ by following Procedure 1 below, using these definitions:

\begin{enumerate}
\item
$T$ is a set of unit squares; $B(T)$ is a set of unit squares that are edge-adjacent to the squares
in $T$; $\mathscr{T}_n$ is a set of $n$-ominoes.
\item
When $T$ is the empty set, we define $B(\emptyset)$ as the set of four squares around the origin (the
four unit squares at the lower left in Fig.~\ref{fig:p4lattice}.)
\item
We define $E(T)$, the boolean function of $T$, which is true if $\# T = n$ and the white circle is on
the boundary of $T$. Otherwise $E(T)$ is false.
\item
$ B'(T)=\{e| e \in B(T), C(e)\ne C(f), \mbox{for all} f \in T \} $.
This is the set of all unit squares that are
edge-adjacent to those in $T$, but not equivalent to any unit squares in $T$.
\end{enumerate}

Procedure 1 creates a sequence of pairs of sets $(T, U_T)$,
in which $U_T$ is the set of unit squares
that can be added to $T$ to create the next set $T$ in the sequence.
The sets $T$ are built up, one
element at a time, until a set $T$ is achieved for which $E(T)$ is true,
at which time $T$ is added to the
collection $\mathscr{T}_n$.
When this takes place, the procedure backtracks to the most recent previous pair
$(T, U_T)$ for which $U_T \ne \emptyset$, and repeats the process.

\vspace{1em}
\noindent
{\bf Procedure 1.}
\begin{enumerate}
\item
Fix $n$ (from the list in (2) above). Begin with $\mathscr{T}_n$ empty.
\item
Make $T$ empty. Make $U_T = B'(T) = B(\emptyset)$. Make $k = 0$. Make $S_k = \{(T, U_T)\}$.
\item
For $(T, U_T)$ in $S_k$, if $U_T \ne \emptyset$,
choose an element $e$ (a unit square) of $U_T$. Remove $e$ from $U_T$
and save the pair $(T, U_T)$ in $S_k$.
\item
Increase $k$ by $1$. Add $e$ (from step 3) to $T$ to create a new $k+1$-omino $T$,
and let $U_T = B'(T)$ for the new $T$. Add the new pair $(T, U_T)$ to $S_k$.
\item
If $T = \emptyset$ and $U_T = \emptyset$, quit the procedure.
\item
If $E(T)$ is true, an $n$-omino tile is completed and add $T$ to $\mathscr{T}_n$ provided there is no equivalent
tile in $\mathscr{T}_n$. We regard two $n$-ominoes equivalent if conditions (a) and (b) below are satisfied.
\begin{enumerate}
\item
The tiles are congruent (including mirror reflection).
\label{proc1a}
\item
When the tiles are superimposed, the positions of $m$-fold rotational centers on the
boundaries of the tiles are the same.
If there are several $m$-fold rotation centers for the same $m$,
they can be permuted appropriately before comparison.
\label{proc1b}
\end{enumerate}
\item
If $U_T = \emptyset$, decrease $k$ by $1$.
\item
Go back to step 3.
\end{enumerate}

Fig.~\ref{p4fighi} shows the set of inequivalent $n$-ominoes in $\mathscr{T}_n$
for $n \le 8$.
From each of these $n$-ominoes
we can obtain the associated {\bf p4} tiling by using the black and white circles as 4-fold rotation
centers.
Fig.~\ref{p4tiling} shows the corresponding isohedral tilings produced by n-ominoes in Fig.~\ref{p4fighi} for $n \le 5$.
\begin{figure}[h]
\centerline{\psfig{file=p4fighi.eps,width=\linewidth}}
\caption{
List of $n$-ominoes produced by the procedure in section 2.1 for $n \le 8$.
The tiles are fundamental domains for
the {\bf p4} group used to construct them.
The labels indicate $n$ followed by the tile number for that $n$. Parentheses
indicate that the tiles produce tilings having more symmetries than the {\bf p4} group that generates the tilings.
\label{p4fighi}
}
\end{figure}
\begin{figure}[h]
\centerline{\psfig{file=p4tiling.eps,width=0.9\linewidth}}
\caption{
List of {\bf p4} isohedral tilings by $n$-ominoes in Fig.~\ref{p4fighi}, for $n \le 5$,
generated by a given {\bf p4} group. Labels
correspond to those in Fig.~\ref{p4fighi}.
The symmetry group of each tiling is the given {\bf p4} group,
except for tilings whose labels are in parentheses.
\label{p4tiling}
}
\end{figure}

\subsection{Symmetries of tiles}
The list of isohedral tilings corresponding to tiles in Fig.~\ref{p4fighi} includes some tilings having
symmetry group larger than the {\bf p4} group $G$ generated by 4-fold rotations about the black
and white rotation centers. By Theorem~\ref{th1}, in every case where this occurs, the $n$-omino
that generates the tiling must have reflection and/or rotation symmetry.

We outline below how to identify polyominoes whose corresponding isohedral tilings
have symmetry groups larger than the group $G$ that generated the tiling.
Such tiles in Fig.~\ref{p4fighi} and their tilings in Fig.~\ref{p4tiling} are identified by
parentheses around their labels.
\begin{itemlist}
\item
Select a polyomino that has rotation and/or reflection symmetry, and examine its
tiling $\cal T$ generated by $G$.
\item
Look at all vertices and centers of unit squares in a polyomino in $\cal T$ except for the
original 4-fold centers we have chosen (black and white centers), and determine whether
or not $\cal T$ is invariant under a 4-fold rotation about such a point.
If so, then it is a new 4-fold center for $\cal T$, and this symmetry is not in $G$.
For example, in Fig.~\ref{p4tiling}, tilings (5-2-2)
and (5-2-3), while generated differently, are the same tiling and have 4-fold centers at
every vertex where 4 tiles meet.
Tiling (5-3) has 4-fold rotation centers at the centers of
the ``cross" tiles and at every vertex where 4 tiles meet.
These tilings have full symmetry groups of type {\bf p4}, with $G$ as a proper subgroup.
\item
Other new symmetry elements of $\cal T$ can be sought by using chart 2 in\cite{schatt}.
If the line joining the black and white 4-fold centers we placed is
an axis of mirror reflection for $\cal T$, then $\cal T$ is type {\bf p4m}.
For example, in Fig.~\ref{p4tiling}, tilings (1-1)
and (4-3) are type {\bf p4m}.
If this line is not a mirror reflection axis, but the line connecting
two adjacent (nearest) 2-fold centers for $\cal T$ is a mirror reflection axis for $\cal T$,
then $\cal T$ is type {\bf p4g}.
For example, in Fig.~\ref{p4tiling}, tiling (2-1) is type {\bf p4g}.
\end{itemlist}

In sections~\ref{p4g} and \ref{p4m} that follow,
we indicate how to modify our Procedure 1 so as to
produce directly polyominoes that are fundamental domains for {\bf p4g} or {\bf p4m} symmetry
groups.

\section{{\bf p4g}}
\label{p4g}
\subsection{Creating polyominoes as fundamental domains for {\bf p4g} symmetry groups}
\label{p4gcre}
A {\bf p4g} symmetry group contains 4-fold rotations, reflections, and glide-reflections; the
subgroup generated by its 4-fold rotations is type {\bf p4}.
Thus we begin as in section~\ref{p4},
with a lattice of unit squares.
Following observation (ii) in section~\ref{intro},
we place a 4-fold rotation center (a black circle) at a lattice point and call this the origin.
Orthogonal unit vectors $\mathbf u$ and $\mathbf v$ are then placed at the origin (see Fig.~\ref{fig:p4glattice}).
\begin{figure}[h]
\centerline{\psfig{file=Fig4.eps,width=2.75cm}}
\caption{
A lattice of unit squares with symmetry elements for a {\bf p4g} group.
Black and white circles are 4-fold rotation centers, thick solid lines are reflection axes,
and dashed lines are glide-reflection axes. 2-fold
rotation centers are at the intersections of reflection axes.
4-fold rotations about the center black circle (the
origin) and reflections about one axis nearest the origin generate the {\bf p4g} group.
A standard fundamental
domain is shaded. Here $x = 2$; the area of a fundamental domain for the {\bf p4g} group is 4.
\label{fig:p4glattice}
}
\end{figure}

Next, we place the reflection axes that are nearest to the origin;
these must lie along the edges of unit squares according to observation (i) in section~\ref{intro}.
These axes are vertical and horizontal lines at $x{\mathbf u}$ and at $x{\mathbf v}$, respectively,
where $x$ is a positive integer.
The placement of the origin and choice of $x$ determine the whole {\bf p4g} lattice of rotation
centers, reflection axes and glide-reflection axes since the {\bf p4g} group $G$ is generated
by 4-fold rotations about the origin and reflections in one of the reflection axes nearest the
origin.

The area $S$ of a fundamental domain for the group $G$ is given by
\begin{equation}
S=x^2
\end{equation}
where the area of a unit square is 1. Since we want our $n$-omino to be a fundamental
domain, $n = S$. Therefore,
\begin{equation}
n=1, 4, 9, 16, 25, 36, 49, 64, \ldots.
\label{p4g:n}
\end{equation}

Action by the group $G$ partitions the unit squares into $n$ equivalence classes; as before, we
denote the equivalence class of a unit square $e$ as $C(e)$.
We construct the set $\mathscr{T}_n$ of $n$-ominoes
that are fundamental domains for $G$ using Procedure 1 in section~\ref{p4},
with $n$ chosen from the list in (\ref{p4g:n}) and the additional constraint that unit squares in the set $T$ must
be in the region bounded by the mirror reflection axes nearest to the origin.

Fig.~\ref{p4gfighi} shows the set of inequivalent tiles in $\mathscr{T}_n$ for $n \le 9$.
We can obtain the associated {\bf p4g} tilings by using the origin as a 4-fold rotation center to fill out the square
bounded by the reflection axes nearest the origin, then reflecting this square in its edges.
Fig.~\ref{p4gtiling} shows the corresponding isohedral tilings produced by polyominoes in Fig.~\ref{p4gfighi}.
\begin{figure}[h]
\centerline{\psfig{file=p4gfighi_bk.eps,width=\linewidth}}
\vspace*{8pt}
\caption{
List of $n$-ominoes produced as described in section 3.1 for $n \le 9$;
these are fundamental domains for
the {\bf p4g} group used to construct them.
The labels indicate $n$ followed by the tile number for that $n$.
Parentheses indicate that the tiles produce tilings having more symmetries than the {\bf p4g} group that
generates the tilings.
\label{p4gfighi}
}
\end{figure}

\begin{figure
\centerline{\psfig{file=p4gtiling1.eps,width=0.9\linewidth}}
\vspace*{8pt}
\caption{
List of {\bf p4g} isohedral tilings by $n$-ominoes in Fig.~\ref{p4gfighi},
generated by a given {\bf p4g} group.
Labels correspond to those in Fig.~\ref{p4gfighi}.
The symmetry group of each tiling is the given {\bf p4g} group,
except for tilings whose labels are in parentheses.
\label{p4gtiling}
}
\end{figure}
\begin{figure
\centerline{\psfig{file=p4gtiling2.eps,width=0.9\linewidth}}

    \vspace*{8pt}
    \fontsize{8pt}{0pt}\selectfont
    {\it Fig.~\ref{p4gtiling}.} $(${\it Continued}$)$
\end{figure}

\subsection{Symmetries of tiles}
The only tilings in Fig.~\ref{p4gtiling} having symmetry group larger than the {\bf p4g} group $G$ that generated
them are tilings with square polyomino tiles.
In fact, this is always true.

\begin{theorem}
\label{th2}
Every tiling by a polyomino $T$ produced by our algorithm in section~\ref{p4gcre} has as its
full symmetry group the {\bf p4g} group that generated it and $T$ is a fundamental domain for the tiling,
except in the case when the shape of $T$ is square.
In that case, the tiling has symmetry group {\bf p4m} and $T$ is not a fundamental domain.
\end{theorem}

\begin{proof}
Let $G'$ be the full symmetry group of a tiling $\cal T$ produced by a {\bf p4g} group $G$ acting on a
polyomino $T$ that is a fundamental domain for $G$, as described in section~\ref{p4gcre}.
If $G$ is a proper
subgroup of $G'$, then by Theorem~\ref{th1},
a rotation or reflection symmetry in $G'$ is also a symmetry of $\cal T$.
By its construction, one corner of $T$ is a 4-fold center for $G$ at the origin;
without loss of generality, we may assume that $T$ contains the unit square shown in black in Fig.~\ref{X}.
\begin{figure}[h]
\centerline{
\psfig{file=Fig7a.eps,width=1.7cm}\psfig{file=Fig7b.eps,width=1.7cm}\psfig{file=Fig7c.eps,width=1.7cm}\psfig{file=Fig7d.eps,width=1.7cm}
}
\centerline{(a) \hspace{1.0cm} (b) \hspace{1.0cm} (c) \hspace{1.0cm} (d)}
\vspace*{8pt}
\caption{
The derivation of polyomino fundamental domains $T$ for a {\bf p4g} group by removing unit squares from and
adding equivalent unit squares to the original square polyomino fundamental domain in (a). Here $x = 4$.
\label{X}
}
\end{figure}
$T$ can also be obtained beginning with an $x \times x$ square polyomino tile as in Fig.~\ref{X}(a),
then removing unit squares and adding other unit squares equivalent (by a rotation of $90^\circ$ or $270^\circ$ about
the origin) to
these, always keeping the new tile homeomorphic to a disc.
Figures~\ref{X}(b), (c), and (d) illustrate this.
From this, we see that one edge of $T$ that abuts a reflection axis in $G$ has at least $x$ unit
squares in a straight row. Since the two edges of $T$ that join the origin to each of the reflection
axes at $x {\mathbf u}$ and $x {\mathbf v}$ are related by a $90^\circ$ rotation and
are not straight unless $T$ is a square,
the only possible $T$ that can have rotation symmetry is the square polyomino in Fig.~\ref{X}(a).

Suppose that $T$ is not square but has reflection symmetry.
If the two edges of $T$ that abut the
reflection axes at $x{\mathbf u}$ and $x {\mathbf v}$ are the same length,
that reflection must map those edges to each other.
But that diagonal reflection will not leave $T$ invariant (see Fig.~\ref{X}(b)).
If the two edges of $T$ that abut the reflection axes at $x {\mathbf u}$ and $x {\mathbf v}$ have
different lengths, a reflection symmetry of $T$
must leave fixed the longer edge, so must have its axis a horizontal or vertical line.
But this reflection cannot be a symmetry in $G'$.
For example, if $T'$ is the image of $T$ rotated by $90^\circ$ about
the origin, this reflection will not map $T'$ onto another tile in $\cal T$ (see Fig.~\ref{X}(d)).
\end{proof}

\section{{\bf p4m}}
\label{p4m}
The symmetry elements of a {\bf p4m} group are shown in Fig.~\ref{fig:p4m}.
\begin{figure}[h]
\centerline{
\psfig{file=p4m.eps,width=3.5cm}}
\vspace*{8pt}
\caption{
The symmetry elements of a {\bf p4m} group.
Black and white circles are 4-fold centers, solid lines are reflection
axes and dashed lines are glide-reflection axes.
Points where glide-reflection axes intersect are 2-fold rotation
centers. The shaded area is a fundamental domain for the {\bf p4m} group.
\label{fig:p4m}
}
\end{figure}
Black and white circles are 4-fold centers, solid lines are mirror reflection axes,
and dashed lines are glide-reflection axes. Points
where glide-reflection axes intersect are 2-fold centers. The shaded region is a fundamental
domain for the {\bf p4m} group. A polyomino that is a fundamental domain for the tiling must have
its edges on the reflection axes, by observation (i) in section~\ref{intro}.
But this is clearly impossible.

\begin{theorem}
There are no {\bf p4m} isohedral tilings having polyominoes as fundamental domains.
\end{theorem}

\section{{\bf p3}}
\label{p3}
\subsection{Creating polyiamonds as fundamental domains for {\bf p3} symmetry groups}
\label{p3:create}
To build our $n$-iamond tiles that will be fundamental domains for a {\bf p3} isohedral tiling, we begin
with a lattice of unit equilateral triangles.
By observation (iii) in section~\ref{intro},
the 3-fold rotation centers for a {\bf p3} symmetry group are located at the lattice points.
So first we place a 3-fold
rotation center, a black circle, at a lattice point and call this the origin, then place vectors
$\mathbf u$ and $\mathbf v$
at the origin along edges of a unit triangle.
Next we place a second 3-fold rotation center, a
white circle, at $x{\mathbf u} + y{\mathbf v}$, where $x$ and $y$ are nonnegative integers,
not both zero. See Fig.~\ref{fig:p3}(a).
\begin{figure}[h]
\centerline{
\psfig{file=Fig9a.eps,width=2.7cm}\psfig{file=Fig9b.eps,width=2.52cm}}
\centerline{(a) \hspace{2.1cm} (b)}
\vspace*{8pt}
\caption{
(a) A lattice of unit triangles with two 3-fold rotation centers that generate a {\bf p3} group.
Here $x = 2$, $y = 1$.
(b) The shaded region is a standard fundamental domain for the {\bf p3} group generated by the black and white
3-fold centers; in this example, the area of a fundamental domain is 14 triangular units.
\label{fig:p3}
}
\end{figure}
These two choices of 3-fold rotation centers determine the whole {\bf p3} lattice of rotation centers;
3-fold rotations about the black and white centers generate the whole {\bf p3} symmetry group.

A standard fundamental domain for the {\bf p3} group $G$ generated by the black and white 3-fold
centers is a $120^\circ$ rhombus having as short diagonal the segment that joins the black and white
circles, as shown in Fig.~\ref{fig:p3}(b). The other two vertices of this rhombus (gray in Fig.~\ref{fig:p3}(b)) are also
3-fold centers for $G$. The area $S$ of a fundamental domain for $G$ is given by
\begin{equation}
S=2 (x^2+y^2+xy)
\end{equation}
taking the area of a unit triangle as 1.
Since we want our $n$-iamond to be a fundamental
domain, $n = S$. Therefore,
\begin{equation}
n=2, 6, 8, 14, 18, 24, 26, 32, 38, 42, 50, 54, 56, 62, 72, 74, 78, 86, 96, 98, 98, \ldots,
\label{p3:n}
\end{equation}
where two pairs $(x,y)$, namely, $(5,3)$ and $(7,0)$ correspond to $n = 98$.

All unit triangles are classified into n equivalence classes by the action of $G$, and we denote
the equivalence class of a unit triangle $e$ as $C(e)$.

We construct a set $\mathscr{T}_n$ of $n$-iamonds
that are fundamental domains for $G$ by modifying
Procedure 1 in section~\ref{p4},
choosing $n$ from the list in (\ref{p3:n}), and replacing definitions in section~\ref{p4} with these:

\begin{enumerate}
\item
$T$ is a set of unit triangles; $B(T)$ is a set of unit triangles that are edge-adjacent to the
triangles in $T$; $\mathscr{T}_n$ is a set of $n$-iamonds.
\item
When $T$ is the empty set, we define $B(\emptyset)$ as the set of six triangles around the origin in
Fig.~\ref{fig:p3}(a).
\item
We define $E(T)$, the boolean function of $T$, which is true if $\# T = n$ and the white circle is on
the boundary of $T$ . Otherwise $E(T)$ is false.
\item
$B'(T) = \{e | e \in B(T),C(e) \ne C(f), \mbox{for all} f \in T \}$.
This is the set of all unit triangles that are edge-adjacent to those in $T$,
but not equivalent to any unit triangles in $T$.
\end{enumerate}
The procedure creates a sequence of pairs of sets $(T, U_T)$,
in which $U_T$ is the set of unit triangles that can be added to $T$ to
create the next set $T$ in the sequence.

Fig.~\ref{p3fighi} shows the set of inequivalent $n$-iamonds in $\mathscr{T}_n$
for $n \le 8$. From each of these we can
obtain the associated {\bf p3} tiling by using the black and white circles as 3-fold rotation centers.
Fig.~\ref{p3tiling} shows the corresponding isohedral tilings produced by $n$-iamonds in Fig.~\ref{p3fighi}.
\begin{figure}[h]
\centerline{
\psfig{file=p3fighi_ed.eps,width=0.9\linewidth}
}
\vspace*{8pt}
\caption{
List of $n$-iamonds produced as described in section~\ref{p3:create} for $n \le 8$.
The tiles are fundamental domains for the
{\bf p3} group used to construct them.
The labels indicate $n$ followed by the tile number for that $n$.
Parentheses indicate that the tiles produce tilings having more symmetries
than the {\bf p3} group that generates the tilings.
\label{p3fighi}
}
\end{figure}
\begin{figure}[h]
\centerline{
\psfig{file=p3tiling.eps,width=0.8\linewidth}}
\vspace*{8pt}
\caption{
List of {\bf p3} isohedral tilings by $n$-iamonds in Fig.~\ref{p3fighi},
generated by a given {\bf p3} group.
Labels correspond to those in Fig.~\ref{p3fighi}.
The symmetry group of each tiling is the {\bf p3} group that generated it,
except for tilings whose labels are in parentheses.
\label{p3tiling}
}
\end{figure}

\subsection{Symmetries of tiles}
\label{p3:sym}
The list of isohedral tilings in Fig.~\ref{p3tiling} includes some tilings having symmetry group larger than
the {\bf p3} group $G$ generated by 3-fold rotations about the black and white rotation centers.
By Theorem~\ref{th1}, when this occurs, the $n$-iamond that generates the tiling must have reflection and/or
3-fold rotation symmetry.
The full symmetry group of the tiling could be of any of these types:
{\bf p3}, {\bf p3m1}, {\bf p31m}, {\bf p6}, and {\bf p6m}.
The tilings in Fig.~\ref{p3tiling} that have additional symmetries, and the
$n$-iamonds in Fig.~\ref{p3fighi} that generate them are indicated by parentheses in their labels.
We outline below how to identify these polyiamonds.

\begin{itemlist}
\item
Select a polyiamond $T$ that has rotation and/or reflection symmetry, and examine its tiling
$\cal T$ generated by $G$.
\item
Look at all vertices and centers of unit triangles in $T$ except for those centers of rotation in
$G$ (black, white and gray centers), and determine whether or not any can be new 3-fold centers
for $\cal T$.
If a new 3-fold center is found, then the full symmetry group $G'$ of $\cal T$ contains $G$ as a
proper subgroup and $T$ is not a fundamental domain for $\cal T$.
If a new 3-fold center is a 6-fold
center, $G'$ is type {\bf p6m} if there is a reflection axis joining it to any nearest 3-fold center;
otherwise, $G'$ is type {\bf p6}.
Tiling (6-4) in Fig.~\ref{p3tiling} is type {\bf p6m}, and has a 6-fold center at the center
of the hexagonal 6-iamond.
If there are new 3-fold centers but none are 6-fold centers, then $G'$ is type {\bf p3}.
\item
If the tiling contains only 3-fold centers for $G$,
we seek new symmetry elements by using chart 2 in\cite{schatt}.
\begin{enumerate}
\item
If some 3-fold center for $G$ is a 6-fold center,
the tiling $\cal T$ will have {\bf p6m} symmetry if the
line joining a 6-fold center to a nearest 3-fold center is a reflection axis for $\cal T$,
otherwise it will have {\bf p6} symmetry.
In Fig.~\ref{p3tiling}, tilings (2-1) and (8-4) have {\bf p6m} symmetry,
and tiling (6-1) has {\bf p6} symmetry.
\item
Otherwise, if the line connecting two adjacent different 3-fold centers
(say, black to white) is a reflection axis, $\cal T$ has {\bf p3m1} symmetry.
\item
Otherwise, if the line connecting adjacent 3-fold centers of the same kind
(black-black, white-white, or gray-gray) is a reflection axis,
$\cal T$ has {\bf p31m} symmetry.
In Fig.~\ref{p3tiling}, tiling (6-2) has {\bf p31m} symmetry.
\end{enumerate}
\end{itemlist}

\section{{\bf p31m}}
\label{p31m}
\subsection{Creating polyiamonds as fundamental domains for {\bf p31m} symmetry groups}
\label{p31m:create}
We begin with a lattice of unit triangles as in section~\ref{p3:create},
since a {\bf p31m} symmetry group contains 3-fold rotations, reflections, and glide-reflections;
the subgroup generated by its 3-fold rotations is type {\bf p3}.
\begin{figure}[h]
\centerline{
\psfig{file=Fig12rev.eps,width=0.3\linewidth}}
\vspace*{8pt}
\caption{
A lattice of unit triangles with a 3-fold rotation center at the origin (black circle), reflection axes
(thick solid lines) and glide-reflection axes (dashed lines) that are symmetries in a {\bf p31m} group.
The 3-fold rotation about the
black circle and reflections about one axis generate the {\bf p31m} group,
which also has 3-fold centers where reflection
axes intersect (one of these is shown as a white circle).
A typical fundamental domain for the group is shaded.
Here $x = 2$; the area of the fundamental domain is 12 triangular units.
\label{fig:p31m}
}
\end{figure}
To build an $n$-iamond tile that is a fundamental domain for a {\bf p31m} isohedral tiling,
we first place a 3-fold rotation center, a black circle, at a lattice point and call this the origin.
Then we place vectors $\mathbf u$ and $\mathbf v$ at the origin along edges of a unit triangle (see Fig.~\ref{fig:p31m}).

Next, we place reflection axes that are nearest the origin;
these must lie along the edges of
unit triangles according to observation (i) in section \ref{intro}.
These three axes intersect at the points
$x ({\mathbf u}+{\mathbf v})$, $x(-2{\mathbf u} +{\mathbf v})$, and $x({\mathbf u} -2{\mathbf v})$,
where $x$ is a positive integer;
their intersections are equivalent 3-fold rotation centers
(one of these is shown as a white circle in Fig.~\ref{fig:p31m}).
The placement of the origin and choice of $x$ determine the whole {\bf p31m} lattice of
rotation centers, reflection axes and glide-reflection axes.
The {\bf p31m} group $G$ is generated by 3-fold rotations about the origin and
reflections in one of the three reflection axes.

The area $S$ of a fundamental domain for the tiling is one-third the area enclosed by the
three reflection axes closest to the origin, which is
\begin{equation}
S=3 x^2
\end{equation}
where the area of a unit triangle is 1.
Since we want our $n$-iamond to be a fundamental domain, $n = S$.
Therefore,
\begin{equation}
n=3, 12, 27, 48, 75, 108, 147, 192, 243, \ldots.
\label{p31m:n}
\end{equation}

The action of $G$ classifies all unit triangles into $n$ equivalence classes
and we denote the equivalence class of a unit triangle $e$ as $C(e)$.
We construct a set $\mathscr{T}_n$ of $n$-iamonds
that are
fundamental domains for $G$ as described in section~\ref{p3:create},
choosing $n$ from the list in (8) and
having the additional constraint that the unit triangles in the set $T$
must be in the region bounded by the mirror reflection axes in Fig.~\ref{fig:p31m}.

Fig.~\ref{p31mfighi} shows the set of inequivalent $n$-iamonds in $\mathscr{T}_n$  for $n \le 12$.
Fig.~\ref{p31mtiling} shows the corresponding {\bf p31m} tilings obtained
by performing a 3-fold rotation about the origin (black circle) to fill out
the triangle bounded by reflection axes (Fig.~\ref{fig:p31m}),
then reflecting this triangle in its edges.
\begin{figure}[h]
\centerline{
\psfig{file=p31mfighi_bk.eps,width=1.0\linewidth}}
\vspace*{8pt}
\caption{
List of $n$-iamonds produced as described in section \ref{p31m:create} for $n \le 12$.
The tiles are fundamental domains for the {\bf p31m} group used to construct them.
The labels indicate $n$ followed by the tile number for that $n$.
\label{p31mfighi}
}
\end{figure}
\begin{figure
\centerline{
\psfig{file=p31mtiling_ed.eps,width=0.9\linewidth}}
\vspace*{8pt}
\caption{
List of {\bf p31m} isohedral tilings by $n$-iamonds in Fig.~\ref{p31mfighi},
generated by a given {\bf p31m} group, for $n \le 12$.
Labels correspond to those in Fig.~\ref{p31mfighi}.
Every $n$-iamond in Fig.~\ref{p31mfighi} is a fundamental domain for its tiling in Fig.~\ref{p31mtiling}.
\label{p31mtiling}
    }
\end{figure}

\subsection{Symmetries of tiles}

\begin{theorem}
Every isohedral tiling by a polyiamond $T$ produced by our algorithm in section \ref{p31m:create}
has as its symmetry group the {\bf p31m} group that generated it and $T$ is a fundamental domain for
the tiling.
\end{theorem}

\begin{proof}
Our proof is analogous to that for Theorem~\ref{th2}.
Let $G'$ be the full symmetry group of a
tiling $\cal T$ produced by a {\bf p31m} group $G$ acting on a polyiamond $T$ that is a fundamental domain for $G$,
as described in section \ref{p31m:create}.
If $G$ is a proper subgroup of $G'$, then by Theorem~\ref{th1},
a reflection symmetry or 3-fold rotation symmetry in $G'$ is also a symmetry of $T$.
$T$ can be obtained from the shaded fundamental domain shown in Fig.~\ref{fig:p31m}
by removing unit triangles and adding unit triangles equivalent to these
(by a $120^\circ$ rotation about the origin),
always keeping the new tile homeomorphic to a disc.
This process shows that one edge of $T$ that lies on a reflection axis will
have length at least $x+1$, and at most one other edge of $T$ (lying on an adjacent reflection axis of $G$)
will have this same length (see Fig.~\ref{p31mfighi}).
From this, it follows that $T$ cannot have 3-fold rotation symmetry.

Suppose that $T$ has reflection symmetry.
If the two straight edges of $T$ that lie on reflection
axes for $G$ have the same length, then the reflection that leaves $T$ fixed must map these edges to each other.
But this reflection cannot leave $T$ invariant, since the other edges of $T$ are related by
a 3-fold rotation about the origin (see, for example, tiles 12-1, 12-2, 12-3 in Fig.~\ref{p31mfighi}).
If the two straight edges of $T$ that lie on reflection axes for $G$ have different lengths,
then a reflection that fixes $T$ must leave the longer edge fixed,
and so is perpendicular to that edge.
This can only happen if $T$ is the rhombus fundamental domain in Fig.~\ref{fig:p31m},
and in this case, the reflection symmetry for $T$ is not a symmetry for the tiling $\cal T$
(see tiling 12-4 in Fig.~\ref{p31mtiling}).
\end{proof}

\section{{\bf p3m1}}

The lattice of reflections and 3-fold rotation centers of a {\bf p3m1} symmetry group is shown
in Fig.~\ref{fig:p3m1};
black, white, and grey circles denote three inequivalent 3-fold rotation centers.
Here, unlike the {\bf p31m} case,
all 3-fold centers lie on reflection axes.
The shaded region bounded by reflection axes is a fundamental domain for the {\bf p3m1} group
that generates the tiling by reflections those axes.
\begin{figure}[h]
\centerline{
\psfig{file=p3m1.eps,width=0.23\linewidth}}
\vspace*{8pt}
\caption{
The lattice of reflection axes and 3-fold rotation centers of a {\bf p3m1} group
(glide-reflection axes are not shown).
Three inequivalent 3-fold centers are black, grey, and white circles.
Lines are reflection axes.
The shaded region is a fundamental domain for the {\bf p3m1} group generated by reflections
in the axes surrounding the region.
\label{fig:p3m1}
    }
\end{figure}
By observation (i) in section \ref{intro},
this is the only polyiamond tile possible having the area of a fundamental domain.
But the full symmetry group of this tiling is type {\bf p6m},
which has a fundamental domain with area 1/6 that of the shaded tile. Thus,

\begin{theorem}
There are no {\bf p3m1} isohedral tilings having polyiamonds as fundamental domains.
\end{theorem}

We note that if the shaded region in Fig.~\ref{fig:p3m1} is decorated with an asymmetric motif
then there are $n$-iamonds (where $n = k^2$, $k$ a positive integer) having the shape of an equilateral triangle,
for which the decorated triangle is a fundamental domain for a {\bf p3m1} isohedral tiling.

\section{{\bf p6}}
\label{p6}
\subsection{Creating polyiamonds as fundamental domains for {\bf p6} symmetry groups}
\label{p6:create}
A {\bf p6} symmetry group contains 6-fold, 3-fold, and 2-fold rotations;
the subgroup generated by its 3-fold rotations is type {\bf p3}.
Thus we begin as in section \ref{p3:create}, with a lattice of equilateral triangles,
to build $n$-iamond tiles that are fundamental domains for a {\bf p6} group.
By observation (iii) in section \ref{intro},
the 6-fold and 3-fold rotation centers for a {\bf p6} symmetry group are located at lattice points.
So first we place a 6-fold rotation center, a black circle, at a lattice point and call this the origin, then
place vectors $\mathbf u$ and $\mathbf v$ at the origin, along edges of a unit triangle.
Next we place a 3-fold rotation center,
a white circle, at $x{\mathbf u} + y{\mathbf v}$,
where $x$ and $y$ are nonnegative integers,
not both $0$.
\begin{figure}[h]
\centerline{
\psfig{file=Fig16a.eps,width=2.88cm}
\hspace{1em}
\psfig{file=Fig16b.eps,width=2.7cm}
}
\centerline{(a) \hspace{3cm} (b)}
\vspace*{8pt}
\caption{
(a) A lattice of unit triangles with a black 6-fold rotation center and white 3-fold rotation center that
generate a {\bf p6} group.
Here $x = 2$, $y = 1$. (b). A standard fundamental domain for the {\bf p6} group generated by the
black and white rotation centers;
in this example, the area of the fundamental domain is $7$ triangular units.
\label{p6lattice}
}
\end{figure}
See Fig.~\ref{p6lattice}.
These choices of 6-fold and 3-fold rotation centers determine the whole {\bf p6} lattice of rotation
centers;
rotations about the two chosen black and white centers generate the {\bf p6} group.

The shaded triangular region shown in Fig.~\ref{p6lattice}(b) is a standard fundamental domain for the {\bf p6}
group generated by 6-fold rotations about the black circle and 3-fold rotations about the white circle.
Hence the area of a fundamental domain for this {\bf p6} group is given by
\begin{equation}
S=x^2+y^2+xy
\end{equation}
taking the area of a unit triangle as 1.
Since we want our $n$-iamond to be a fundamental
domain, $n = S$. Therefore,
\begin{equation}
n=1, 3, 4, 7, 9, 12, 13, 16, 19, 21, 25, 27, 28, 31, 36, 37, 39, 43, 48, 49, 49, \ldots,
\label{p6:n}
\end{equation}
where two pairs of $(x,y)$, namely, $(5,3)$ and $(7,0)$ correspond to $n = 49$.

The action of the {\bf p6} group classifies all unit triangles into $n$ equivalence classes and we denote
the equivalence class of a unit triangle $e$ as $C(e)$.
We construct a set $\mathscr{T}_n$ of $n$-iamonds that are
fundamental domains for the given {\bf p6} group as in section~\ref{p3:create},
with $n$ chosen from the list in (\ref{p6:n}).
Fig.~\ref{p6fighi} shows the set of inequivalent $n$-iamonds in $\mathscr{T}_n$
for $n \le 9$.
From each of these $n$-iamonds we obtain the associated {\bf p6} tiling
by using the black circles as 6-fold rotation centers
and white circles as 3-fold rotation centers.
Fig.~\ref{p6tiling} shows the corresponding isohedral tilings
produced by $n$-iamonds in Fig.~\ref{p6fighi}.
\begin{figure
\centerline{
\psfig{file=p6fighi.eps,width=1.0\linewidth}}
\vspace*{8pt}
\caption{
List of $n$-iamonds produced as described in section \ref{p6:create} for $n \le 9$.
The tiles are fundamental domains for the {\bf p6} group used to construct them.
The labels indicate $n$ followed by the tile number for that $n$.
Parentheses indicate that the tiles produce tilings having more symmetries than the {\bf p6} group that generates the tilings.
\label{p6fighi}
}
\end{figure}
\begin{figure
\centerline{
\psfig{file=p6tiling.eps,width=0.9\linewidth}}
\vspace*{8pt}
\caption{
List of {\bf p6} isohedral tilings by $n$-iamonds in Fig.~\ref{p6fighi} for $n \le 7$,
generated by a given {\bf p6} group.
Labels correspond to those in Fig.~\ref{p6fighi}.
The symmetry group of each tiling is the given {\bf p6} group,
except for tilings whose labels are in parentheses.
\label{p6tiling}
}
\end{figure}
\begin{figure
\centerline{
\psfig{file=p6tiling2.eps,width=0.9\linewidth}}
\vspace*{8pt}
\fontsize{8pt}{0pt}\selectfont
    {\it Fig.~\ref{p6tiling}.} $(${\it Continued}$)$
\end{figure}

\subsection{Symmetries of tiles}
Some tilings in Fig.~\ref{p6tiling} have symmetries in addition to those in the {\bf p6} group $G$
that generated them.
These tilings and their polyiamond tiles in Fig.~\ref{p6fighi} are indicated by parentheses
in their labels. We can identify these polyiamonds as follows.
\begin{itemlist}
\item
Select a polyiamond that has rotation and/or reflection symmetry, and examine its tiling $\cal T$
generated by $G$.
\item
If the line connecting adjacent 6-fold and 3-fold centers (black to white) is a reflection axis
for $\cal T$, the tiling has {\bf p6m} symmetry.
Tilings (1-1) and (4-1) in Fig.~\ref{p6tiling} have {\bf p6m} symmetry.
These tilings also have 3-fold centers for the tiling at the centers of the polyiamond tiles.
\item
Otherwise, look at all vertices and centers of unit triangles in a polyiamond in $\cal T$ except for
those black and white rotation centers in the {\bf p6} group we have generated and determine whether
or not they can be new 3-fold centers for $\cal T$.
If new 3-fold centers are found, the full symmetry
group of the tiling contains $G$ as a proper subgroup.
Tiling (7-1) in Fig.~\ref{p6tiling} has a larger {\bf p6} symmetry group than $G$
since there are 3-fold rotation centers for $\cal T$ at the centers of the rotorlike polyiamond tiles.
Note that the same tiles appear in tiling 7-1-2, but in this tiling, the centers
of the tiles are not 3-fold centers of rotation for $\cal T$.
\end{itemlist}

\section{{\bf p6m}}
\label{p6m}
Fig.~\ref{fig:p6m} shows the symmetry elements of a {\bf p6m} group.
\begin{figure}[h]
\centerline{
\psfig{file=p6m.eps,width=0.27\linewidth}}
\vspace*{8pt}
\caption{
The lattice of symmetry elements for {\bf p6m}.
Solid lines are reflection axes and dashed lines are glide reflections axes.
Black and white circles are 6-fold and 3-fold centers, respectively, and smaller grey circles where
two reflection axes intersect are 2-fold centers. The shaded region is a fundamental domain.
\label{fig:p6m}
}
\end{figure}
Black, white, and gray circles are the 6-, 3-, and 2-fold centers, respectively.
Solid lines are reflection axes and dashed lines are glidereflection axes.
The shaded $30^\circ$- $60^\circ$- $90^\circ$-triangle is a fundamental domain.
A polyiamond that is a fundamental domain for the tiling must have its edges on the reflection axes,
by observation (i) in section \ref{intro}.
But this is clearly impossible. Thus,
\begin{theorem}
There are no {\bf p6m} isohedral tilings having polyiamonds as fundamental domains.
\end{theorem}

\section{Enumeration Tables}
\label{tables}
Tables 1--3 give
the number of tiles and isohedral tilings that we have generated in sections \ref{p4}
through \ref{p6m}.
$N_n$ is the number of inequivalent tiles $T$ in $\mathscr{T}_n$
for a given symmetry group $G$ of type
{\bf p3}, {\bf p31m}, {\bf p3m1}, {\bf p4}, {\bf p4m}, {\bf p4g}, {\bf p6},
and {\bf p6m} that generate an isohedral tiling having
the $n$-omino or $n$-iamond tiles as fundamental domain.
Tiles are equivalent only if they generate the same tiling by the action of the group $G$
when each tile is marked with an asymmetric motif.
For example, the tiles 5-2 and 5-2-2 in Fig.~\ref{p4tiling} are congruent and their
corresponding tilings are the same, but the placement of their 4-fold centers is different,
and so if the tiles are marked with an asymmetric motif, they generate different isohedral tilings.

Marking each tile with an asymmetric motif also guarantees that the group that generates the
tiling is the full symmetry group of the tiling.
So $N_n$ is also the number of isohedral tilings
having $G$ as full symmetry group and having $n$-omino or $n$-iamond tiles as fundamental domain,
when each tile is marked with asymmetric motif.

$S_n$ is is the corresponding number of the $N_n$ tilings when the asymmetric motif of each tile is
removed. That is, $S_n$ is the number of isohedral tilings having full symmetry group $G$ and having
(unmarked) $n$-ominoes or $n$-iamonds as fundamental domains.
In our figures that depict the
isohedral tilings for small values of $n$, these tilings do not have parentheses around their labels.
These are the most important ``counting" results in this article.
$N'_n$ is the number of non-congruent
tiles in the $N_n$ tilings, counted by ignoring rotation centers attached to tiles; similarly,
$S'_n$ is the number of non-congruent tiles in the $S_n$ tilings.
\begin{table}[h]
\tbl{{\bf p3}, {\bf p31m} and {\bf p3m1}.}{
\begin{tabular}{lrrrr}
\multicolumn{5}{l}{
{\bf p3} $n$-iamonds }\\
\toprule
$n$    & 2 & 6 & 8 & 14 \\
\colrule
$N_n$  & 1 & 4 & 7 & 306 \\
$S_n$  & 0 & 1 & 6 & 294 \\
$N'_n$ & 1 & 4 & 7 & 288 \\
$S'_n$ & 0 & 1 & 6 & 277 \\
\botrule
\end{tabular}
\hspace{1em}
\begin{tabular}{lrrr}
\multicolumn{4}{l}{
{\bf p31m} $n$-iamonds }\\
\toprule
$n$    & 3 & 12 \\
\colrule
$N_n$  & 1 & 20 \\
$S_n$  & 1 & 20 \\
$N'_n$ & 1 & 20 \\
$S'_n$ & 1 & 20 \\
\botrule
\end{tabular}
\hspace{1em}
\begin{tabular}{lrrrr}
\multicolumn{5}{l}{
{\bf p3m1} $n$-iamonds }\\
\toprule
$n$    & 1 & 4 & 9 & $\ldots$ \\
\colrule
$N_n$  & 1 & 1 & 1 & $\ldots$ \\
$S_n$  & 0 & 0 & 0 & $\ldots$ \\
$N'_n$ & 1 & 1 & 1 & $\ldots$ \\
$S'_n$ & 0 & 0 & 0 & $\ldots$ \\
\botrule
\end{tabular}
}
\label{tbl:p3}
\end{table}

\begin{table}[h]
\label{tbl:p4}
\tbl{{\bf p4}, {\bf p4g} and {\bf p4m}.}{
\begin{tabular}{lrrrrrrr}
\multicolumn{8}{l}{
{\bf p4} $n$-ominoes}\\
\toprule
$n$    & 1 & 2 & 4 & 5  & 8  & 9  & 10 \\
\colrule
$N_n$  & 1 & 1 & 3 & 12 & 45 & 82 & 300 \\
$S_n$  & 0 & 0 & 2 & 9  & 38 & 77 & 296 \\
$N'_n$ & 1 & 1 & 3 & 8  & 45 & 80 & 277 \\
$S'_n$ & 0 & 0 & 2 & 7  & 38 & 76 & 275 \\
\botrule
\end{tabular}
\hspace{1em}
\begin{tabular}{lrrrr}
\multicolumn{5}{l}{
{\bf p4g} $n$-ominoes}\\
\toprule
$n$    & 1 & 4 & 9 & 16    \\
\colrule
$N_n$  & 1 & 3 & 26 & 596  \\
$S_n$  & 0 & 2 & 25 & 595  \\
$N'_n$ & 1 & 3 & 26 & 596  \\
$S'_n$ & 0 & 2 & 25 & 595  \\
\botrule
\end{tabular}
\hspace{1em}
\begin{tabular}{lrrr}
\multicolumn{4}{l}{
{\bf p4m} $n$-ominoes}\\
\toprule
$n$    & 1 & 2 & $\ldots$ \\
\colrule
$N_n$  & 0 & 0 & $\ldots$ \\
$S_n$  & 0 & 0 & $\ldots$ \\
$N'_n$ & 0 & 0 & $\ldots$ \\
$S'_n$ & 0 & 0 & $\ldots$ \\
\botrule
\end{tabular}
}
\end{table}

\begin{table}[h]
\label{tbl6}
\tbl{{\bf p6} and {\bf p6m}.}{
\begin{tabular}{lrrrrrr}
\multicolumn{7}{l}{
{\bf p6} $n$-iamonds }\\
\toprule
$n$    & 1 & 3 & 4 & 7 & 9 & 12 \\
\colrule
$N_n$  & 1 & 1 & 3 & 20 & 29 & 195 \\
$S_n$  & 0 & 1 & 2 & 19 & 28 & 194 \\
$N'_n$ & 1 & 1 & 3 & 16 & 27 & 191 \\
$S'_n$ & 0 & 1 & 2 & 16 & 26 & 190 \\
\botrule
\end{tabular}
\hspace{1em}
\begin{tabular}{lrrrr}
\multicolumn{5}{l}{
{\bf p6m} $n$-iamonds }\\
\toprule
$n$    & 1 & 2 & 3 & $\ldots$ \\
\colrule
$N_n$  & 0 & 0 & 0 & $\ldots$ \\
$S_n$  & 0 & 0 & 0 & $\ldots$ \\
$N'_n$ & 0 & 0 & 0 & $\ldots$ \\
$S'_n$ & 0 & 0 & 0 & $\ldots$ \\
\botrule
\end{tabular}

}
\end{table}

For example, for symmetry group {\bf p6} and $n = 7$,
there are 20 tiles in Fig.~\ref{p6fighi}, with
corresponding isohedral tilings in Fig.~\ref{p6tiling}, so $N_7 = 20$.
Since tiling 7-1 of Fig.~\ref{p6tiling} has parentheses
around its label, $S_7 = 19$.
From Fig.~\ref{p6fighi},
we can see that four pairs of tiles are congruent: 7-1 and
7-1-2; 7-3 and 7-3-2; 7-5 and 7-5-2; 7-12 and 7-12-2.
Thus $N'_7 = 20 - 4 = 16$.
Among the 19 tiles counted for $S_7$, there are also 16 non-congruent tiles, so $S'_7 = 16$.

\section{Summary}
We have described computer algorithms that can enumerate and display isohedral tilings by
$n$-omino or $n$-iamond tiles for given $n$ in which the tiles are fundamental domains and the tilings
have 3-, 4-, or 6-fold rotational symmetry.
Their symmetry groups are of types {\bf p3}, {\bf p31m}, {\bf p4}, {\bf p4g}, and {\bf p6}.
We have shown that there are no isohedral tilings with symmetry groups of types
{\bf p3m1}, {\bf p4m}, or {\bf p6m} that have polyominoes or polyiamonds as fundamental domains.
For symmetry groups of types {\bf p3}, {\bf p31m}, {\bf p4}, {\bf p4g},
and {\bf p6} we used the backtracking Procedure 1 to obtain a set $\mathscr{T}_n$ of $n$-omino
or $n$-iamond tiles where each tile produced one isohedral tiling,
generated by a given symmetry group $G$ of one of these five types.
We can denote $\mathscr{T}_{n}(G)$ as the set $\mathscr{T}_n$ for that symmetry group $G$
and $\mathscr{T}^{*}_{n}(G)$ the corresponding set of isohedral tilings.

We investigated the symmetries of tilings in the set $\mathscr{T}^{*}_{n}(G)$ and noted those tilings that
satisfy the following two conditions:
(1) the full symmetry group of the tiling is $G$, and
(2) the tiles are fundamental domains for $G$.
We denote the subset of $\mathscr{T}^{*}_{n}(G)$ that satisfies (1) and (2)
as $\mathscr{S}^{*}_{n}(G)$ and the corresponding set of tiles as $\mathscr{S}_{n}(G)$.
(For small values of $n$, these tiles and their tilings were displayed with labels without parentheses.)
The enumeration of $\mathscr{S}^{*}_{n}(G)$ is the main counting result of this article.
Although the $n$-omino or $n$-iamond tiles produced by our algorithm are not
always fundamental domains for the isohedral tilings they generate, if we mark
these tiles with an asymmetric motif, then the set $\mathscr{T}^{*}_{n}(G)$ is the set of all isohedral tilings with
symmetry group $G$ in which the corresponding tiles in $\mathscr{T}_{n}(G)$ are fundamental domains.
The set $\mathscr{T}^{*}_{n}(G)$ can then also include a marked fundamental domain for a {\bf p3m1} symmetry group.
In Tables 1--3 of section \ref{tables}, we used the notation $N_n=\#\mathscr{T}^{*}_{n}(G)$ and
$S_n=\#\mathscr{S}^{*}_{n}(G)$.

redpajama_set_name: RedPajamaArXiv


Strong performance and low noise make this gas blower a great choice. The 25.4cc professional grade engine produces 375 CFM and 158 MPH at the nozzle at only 64 dB (A). Product offers a 25.4cc professional grade, 2-stroke engine.
Product has 375 CFM air volume at the nozzle. Product offers 158 MPH maximum air speed. Product offers a 64 dB(A) noise level.
Item offers the i-30 starting system for easy starts. Product has a hip-mounted throttle with cruise control. : 16.5 Product Width in.
: 16 Blower Type: Backpack Blower CA (CARB) Compliant: CARB Compliant Commercial: Yes Engine Displacement (cc): 25.4 Engine Type: 2 Cycle Features: Adjustable Speed, Antivibration, Translucent Fuel Tank Fuel Tank Capacity oz. If you are not happy with your item please return it in the original packaging with all tags attached within 14 days. Our policy is to provide complete satisfaction with the product. We provide open, honest, prompt and clear communication with our buyers.
Please Note: Products are designed and rated for use in the US using 120V current. With or without a transformer, this product is not recommended for use outside the US. The item "Echo Commercial Professional Powerful Gas Leaf Blower Lightweight Backpack Easy" is in sale since Thursday, September 28, 2017. This item is in the category "Home & Garden\Yard, Garden & Outdoor Living\Outdoor Power Equipment\Leaf Blowers & Vacuums". The seller is "gianthomesupplies" and is located in Whitestone, New York.