Retrieved January 23, Marketers focused their efforts on connecting with consumers via universal themes, rather than outdoing Halo 3 ' s push. Glad to be helpful: Retrieved October 2, Reach sold well in most matchmakiing, moving more than three million units its first month in North America. Retrieved May 14, How does halo reach matchmaking work can follow the question or vote as helpful, but you cannot reply to this thread. How does halo reach matchmaking work "headhunter", players drop skulls upon death, which other players can pick up and deposit at special zones for points.
Halsey informs Noble Team that the Covenant forces matchmsking the relay were searching for important information. Archived from the original on December 3, Game modes like generator defense are also playable in firefight. Eventually you will earn enough credits macthmaking get the next rank.
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Answered How do you get the firefight voice to work? Matchmsking a player that quits or reaach kick from hal game will they loose all the exp points they earn for that game? Retrieved July 24, Thanks for marking this hwo the answer. Reach' multiplayer beta lands in May".
Archived from the hlao on January 22, Decades later, Six's helmet remains on the grassy plains of a now-restored Reach with a ship landing in the background to bring new colonists to the planet. Will I be able to hide my rank? Reach' Hits Stores Sept.
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It would take place rezch the planet Reach, during a pivotal time in the war. Tried and tried but nothing how does it work? How does it work? Will I be able to see my and other people over all stats by selecting there name?
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Reach ending posted on YouTube after review version hacked, shared". Matfhmaking Locks In September Launch". Answered Why doesn't forge work? Retrieved October 6, Reach ; Bungie's Astonishing Prequel Revealed".
Why doesn't Firefight Matchmaking ever work? Sorry this didn't help. They retreat to an underground bunker when the Covenant begin matxhmaking bombard the city with plasma, but Kat is killed by a Covenant sniper before they reach it. The how does halo reach matchmaking work redesigned the game enginethe software that handles rendering and much mqtchmaking gameplay. Retrieved October 5, Well the thing is, whenever I start the matchmaking for firefight, it starts searching for players, then half an hour later it's still searching.
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Thanks for your feedback, it dating site for free without payment us improve the site. This would depend if they set up the play modes as separate comparisons within the TrueSkill system.
The matchmaking data from Halo 2 would be used to help develop TrueSkill, the skill-based ranking system used in future Halo titles and other video games. En route to the Autumn ' s dry dock, Carter is critically wounded.
You may want to check router firewall, if it causing the issue. Most anything you do in game gives you credits killing enemies, getting medals or commendations, etc. Here is Microsoft's article for the match making system of How does halo reach matchmaking work Live So contrary dating with depression and anxiety other answers, yes, there is an answer to this, but you wont get the math specifics.
In Firefight mode, your character will vocalize when you perform certain actions earn a multikill, for example. Combat Evolved PC Version. Retrieved April 25, As Alex stated, I doubt you'll get an exact answer. Archived from the original on September 13, how does halo reach matchmaking work If a player needs to get a ranking on five magchmaking to get a ranking for a season, then the player will have to play four games per night for five individual days.
When Emile is slain by Elites, Matchma,ing remains behind to control the gun, ensuring the Autumn ' s escape. What We Know So Far". Retrieved January 22, As the article reads, it does not try to pick dead match team games, but will try to pair dead matched players and put them into opposite teams. Archived from the original on September 18, Is this similar to the ranking number in Halo 2?
Archived from the original on July 9, Answered How does cortanas firefight voice work? Will there be a way to restart your rank once you get the highest rank in whatever playlist that its in or will you have to get a 50 in every playlist to restart at Rank 1.
Matchmaking is a multiplayer system how does halo reach matchmaking work provides players the ability to enter into a game with less effort. Is there a difference between the cooperative modes ie firefight and the competitive modes ie slayer? Computer and Video Games. Retrieved December 10, hwo The agencies were involved with Halo 3 ' s marketing.
Reach received critical acclaim upon its release. The update also contained playlists for Halo: Log In Sign Up. Will I how does halo reach matchmaking work able to keep the same armor from Halo 4 across every playlist?
Вопрос 1/3 Jun 24, · If these % of Halo Reach online players have attained a level of technique so advanced that my friends and I can only ascribe this to cheating, then I tip my hats to these players and will endeavor to work harder toward achieving this ability myself. hi, when i click onto halo reach matchmaking it comes up with the following.' The halo: Reach server is unavailable at this time, please try again later.' It has been doing this for the past week, i took my xbox around to my friends house and it works over there. Last time I seriously played Halo multiplayer was Halo 2. The system gave you a number that ranked your skill and it tried to match you with people of that number. One common jerky thing skilled players did was get themselves downranked back to a 1 and beat up on new players for a while.


text:
RACES, RESULTS & TRAINING TIMES
TRY LUGE
SLIDER SEARCH
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ABOUT LUGE
HISTORY AND FAST FACTS
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Norton USA Luge Youth and Junior National Championships set for Feb. 28-Mar. 1 and Mar. 7-8
Youth Olympic Games qualification Stage 1: 3 days of racing in Lake Placid through Nov. 9
By Sandy Caligiore | Nov. 07, 2019, 1:25 p.m. (ET)
LAKE PLACID, N.Y. – Matt Greiner and McKenna Mazlo swept men's and women's singles qualifying, respectively, Saturday as USA Luge began its Youth Olympic Games (YOG) selection process this week in Lake Placid.
The two lead qualifiers maxed out with 300 seeding points after winning all three events that were held Thursday, Friday and Saturday at the Olympic Sports Complex.
The top two men's and women's singles finishers advanced to the final stage of selection in December.
Greiner, of Park City, Utah, and Hunter Harris, of East Fairfield, Vt., moved on with Mazlo, of Souderton, Pa., and Sophia Gordon, of Sussex, Wis. Gordon rallied Saturday to overtake Emma Erickson, of Park City, for the final slot by just five points. Harris, third after Thursday's qualifier, moved into second place over Aiden Mueller, of W. Islip, N.Y., on Friday and held on the final day to finish runner-up.
The next and final series of YOG qualifying occurs at the four December Youth A World Cups, with two taking place in Igls, Austria and the final two in Koenigssee, Germany. The entire process will conclude Dec. 8.
The women's doubles team of Maya Chan, of Chicago, and Reannyn Weiler, of Whitesboro, N.Y., and the men's doubles unit of Sam Eckert and Sam Day, have also advanced to the December events. Chan and Weiler, who medaled in St. Moritz last season, captured the lone race on Thursday against Erickson and Elana Morrison, of Muskegon, Mich.
Eckert and Day are the lone age eligible doubles team on the U.S. roster and were exempt from qualifying in Lake Placid.
Also on the ice at Mount Van Hoevenberg were Caitlyn Pettys, of York, Pa., Jessa Muller, of New London, Wis., Mary Grace Richardson, of Mount Vernon, Ohio, Marcus Mueller, of Brookfield, Wis., Ansel Haugsjaa, of Framingham, Mass., and Frank Ike, of Lititz, Pa. They raced for experience and were not part of the YOG procedure.
The 2020 YOG sliding events will be held on the natural ice track in St. Moritz, about a five-hour drive from the event's central location in Lausanne, Switzerland. The track is built yearly from natural ice pulled from nearby Lake St. Moritz and hand-placed along its entire length, starting in the ritzy tourist destination and concluding two miles later in nearby Celerina. The YOG luge events will not use the full course.
The YOG, for athletes ages 15-18, is scheduled for Jan. 9-22.
1) Matt Greiner, 1:29.643, 300 points
2) Hunter Harris, 1:30.593, 240 points
3) Aiden Mueller, 1:31.180, 210 points
4) Sean Rogers, 1:31.669, 190 points
5) Jeff Sichler, 1:32.320, 170 points
1) McKenna Mazlo, 1:29.514, 300 points
2) Sophia Gordon, 1:30.011, 230 points
3) Emma Erickson, 1:30.131, 225 points
4) Elana Morrison, 1:30.929, 190 points
1) Maya Chan and Reannyn Weiler, DNS Friday and Saturday, 100 points
Emma Erickson and Elana Morrison, DNS Friday and Saturday, 0 points
1) Maya Chan and Reannyn Weiler, DNS, 100 points
2) Emma Erickson and Elana Morrison, DNS, 0 points
2) Aiden Mueller, 1:30.817, 85 points
3) Hunter Harris, 1:31.031, 70 points
4) Sean Rogers, 1:31.275, 60 points
5) Jeff Sichler, 1:33.477, 55 points
2) Sophia Gordon, 1:29.818, 85 points
3) Emma Erickson, 1:30.245, 70 points
4) Elana Morrison, 1:30.534, 60 points
1) Maya Chan and Reannyn Weiler, 1:30.601, 100 points
Emma Erickson and Elana Morrison, DNF
With just one age eligible team, the twosome of Sam Eckert and Sam Day are exempt from qualifying in Lake Placid.
Currently, the USA Luge National Team is in Sochi for training on the 2020 World Championship track, followed by final preparations in Igls for the World Cup opener.
The junior national team, meanwhile, will return to Sigulda, Latvia for the continuation of pre-season training. The Junior World Cup tour opener, originally scheduled for Park City, Utah Nov. 21-22, has been canceled due to refrigeration problems at the Utah Olympic Park. Their Youth A World Cup has been added to the program in Koenigsee, Germany next month.

redpajama_set_name: RedPajamaCommonCrawl


51' Matt Phillips with two corners into the Wolves box, the first cleared behind, the second fired into the arms of Ikeme by Seb Polter.
My lucky Sparkly Scarf's done good!!
Corner seemed to be met by Fer who fired over. But ref says corner. Must have been a Wolves touch.
Cross deflected and Ikeme gathers.
im doing my goal dance BOOWADAHDIDDYBOOMOOBOO !
Wolves need to be stronger in possession.
48' Henry looks for goal from 20-yards, his effort takes a big deflection, wrong-foots Alex Smithies and finds the back of the net.
The sides are back out at Loftus Road and we're back underway. Wolves now defending the goal in front of the away support.
paul is mystic meg if it happens !
Crowd today is 15266 with 1356 from Wolves.
There are 1,356 travelling Wolves fans here at Loftus Road today. Thank you for your support. #COYW .
Well done Wendy - keep the candle burning, and show us the way!


text:
Sexual Chronicles of a French Family (2012)
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text:
\subsection{K3 Surfaces}

A \emph{K3 surface} is a compact complex surface $X$ with trivial canonical bundle and $\dim H^1(X,\mathcal O_X)=0$. All K3 surfaces considered here will be projective and minimal.

It is well-known that all K3 surfaces are diffeomorphic and K\"ahler.
Given a K3 surface $X$, $H^2(X,\mathbb{Z})$ is free of rank 22,
the Hodge numbers of $X$ are $h^{2,0}(X)=h^{0,2}(X)=1$, $h^{1,1}(X)=20$ and $h^{1,0}(X)=h^{0,1}(X)=0$,
and the Euler characteristic is $24$.
The Picard group of $X$ coincides with the N\'eron--Severi group,
and both are torsion free.

From the facts above, we see that $H^{2,0}(X)$ is one--dimensional.
In fact, it is generated by a nowhere--vanishing two--form $\omega_X$,
which satisfies $\langle \omega_X,\omega_X \rangle=0$ and $\langle \omega_X,\overline{\omega}_X \rangle>0$.

Given an automorphism $\sigma$ of the K3 surfaces $X$,
we get an induced Hodge isometry $\sigma^*$, which preserves $H^{2,0}(X)$,
i.e. $\sigma^*\omega_X=\lambda_\sigma \omega_X$ for some $\lambda_\sigma\in \mathbb{C}^*$.
We call $\sigma$ \emph{symplectic} if $\lambda_\sigma=1$
and \emph{non-symplectic} otherwise.
If $\sigma$ is an automorphism with nonprime order $m$,
we say $\sigma$ is \emph{purely non-symplectic} if $\lambda_\sigma=\xi_m$
with $\xi_m$ a primitive $m$-th root of unity.

\subsection{Lattice theory}\label{s:lattice}

A \emph{lattice} is a free abelian group $L$ of finite rank
together with a non-degenerate symmetric bilinear form $B\colon L \times L \to \mathbb{Z}$.
A lattice $L$ is \emph{even} if $B(x,x)\in 2\mathbb{Z}$ for each $x\in L$.
The \emph{signature} of $L$ is the signature $(t_+,t_-)$ of $B$.
A lattice $L$ is \emph{hyperbolic} if its signature is $(1,\rk(L)-1)$.
A sublattice $L\subset L'$ is called \emph{primitive} if $L'/L$ is free.
On the other hand, a lattice $L'$ is an \emph{overlattice} of finite index of $L$
if $L\subset L'$ and $L'/L$ is a finite abelian group.
We will refer to it simply as an \emph{overlattice}.

Given a finite abelian group $A$,
a \emph{finite quadratic form} is a map $q:A\to \mathbb{Q}/2\mathbb{Z}$ such that
for all $n\in \mathbb{Z}$ and $a,a'\in A$
$q(na)=n^2q(a)$
and
$q(a+a')-q(a)-q(a')\equiv 2b(a,a') \imod{2\mathbb{Z}}$
where $b:A\times A\to \mathbb{Q}/\mathbb{Z}$ is a finite symmetric bilinear form.
We define orthogonality on subgroups of $A$ via $b$.

Given a lattice $L$, the corresponding bilinear form $B$ induces an embedding
$L \hookrightarrow L^\ast$, where $L^\ast:= \Hom(L,\mathbb{Z})$.
The \emph{discriminant group} $A_{L}:= L^\ast/L$ is a finite abelian group.
In fact, if we write $B$ as a symmetric matrix in terms of a minimal set of generators of $L$,
then the order of $A_{L}$ is equal to $|\det(B)|$.
The bilinear form $B$ can be extended to $L^*\times L^*$ taking values in $\mathbb{Q}$.
If $L$ is even, this induces a finite quadratic form $q_L: A_L\to \mathbb{Q}/2\mathbb{Z}$.

The minimal number of generators of $A_L$ is called the \emph{length} of $L$.
If $A_L$ is trivial, $L$ is called \emph{unimodular}.
For a prime number $p$, $L$ is called \emph{$p$-elementary}
if $A_L \simeq (\mathbb{Z}/p\mathbb{Z})^a$ for some $a\in\mathbb N_0$;
in this case, $a$ is the length of $A_L$.

Two lattices $L$ and $K$ are said to be \emph{orthogonal},
if there exists an even unimodular lattice $S$ such that $L\subset S$ and $L^\perp_S\cong K$.
Orthogonality will be a key ingredient in the definition of mirror symmetry for K3 surfaces. The following fact will also be useful.

\begin{prop}[cf. {\cite[Corollary 1.6.2]{nikulin}}]\label{p:orth}
Two lattices $L$ and $K$ are orthogonal if and only if $q_L\cong -q_K$.
\end{prop}

We recall the definition of several lattices that we will encounter later.
The lattice $U$ is the hyperbolic lattice of rank 2 whose bilinear form is given by the matrix $\left( \begin{array}{cc}
0 & 1 \\
1 & 0  \\
\end{array} \right)$.

The lattices $A_n, D_m, E_6,E_7, E_8, n\geq 1, m\geq 4$ are the even negative definite lattices associated to the respective Dynkin diagrams.
For $n\geq 1$, the lattice $A_n$ has rank $n$ and its discriminant group is $\mathbb{Z}/({n+1})\mathbb{Z}$. If $p$ is prime, $A_{p-1}$ is $p$-elementary (with $a=1$).
For $m\geq 4$, the lattice $D_m$ has rank $m$ and its discriminant group is $\mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$ for $m$ even, and $\mathbb{Z}/4\mathbb{Z}$ for $m$ odd.
Finally, $E_6,E_7,E_8$ have ranks 6, 7, and 8 and discriminant groups of order 3, 2, and 1, respectively.

For $p\equiv 1 \pmod 4$ the lattice $H_p$ is the hyperbolic even lattice of rank 2, whose bilinear form is given by the matrix
\[
H_p=\left( \begin{array}{cc}
\frac{(p+1)}2& 1\\
1&2\end{array} \right).
\]
The discriminant group of $H_p$ is $\mathbb{Z}/p\mathbb{Z}$.

There are two non--isomorphic hyperbolic lattices of rank 2 with discriminant group $\mathbb{Z}/9\mathbb{Z}$ defined by the matrices
\[ L_9=\left( \begin{array}{cc}
-2& 1\\
1&4\end{array} \right),\qquad
M_9=\left( \begin{array}{cc}
-4&5\\
5&-4\end{array} \right).\]

Following \cite{belcastro} we recall that $T_{p,q,r}$ with $p,q,r\in \mathbb{Z}$
is the lattice determined by a graph which has the form of a T,
and $p,q,r$ are the respective lengths of the three legs.
The rank of $T_{p,q,r}$ is $p+q+r-2$ and the discriminant group has order $pqr-pq-qr-pr$.

Given a lattice $L$, we denote by $L(n)$, the lattice with the same rank as $L$, but whose values under the bilinear form $B$ are multiplied by $n$.

Many even lattices are uniquely determined by their rank and the discriminant quadratic form.
To make this statement precise, we introduce the following finite quadratic forms.
The notation follows \cite{belcastro} and the results are proven in \cite{nikulin}.

We define three classes of finite quadratic forms forms, $w_{p,k}^\epsilon$, $u_k$, $v_k$ as follows:

\begin{enumerate}
\item For $p\neq 2$ prime, $k\geq 1$ an integer, and $\epsilon\in \set{\pm 1}$, let $a$ be the smallest even integer that has $\epsilon$ as quadratic residue modulo $p$. Then we define $w_{p,k}^\epsilon:\mathbb{Z}/{p^k}\mathbb{Z}\to \mathbb{Q}/2\mathbb{Z}$ via $w_{p,k}^\epsilon(1)= ap^{-k}$.

\item For $p=2$, $k\geq 1$ and $\epsilon\in \set{\pm 1, \pm 5}$, we define $w_{2,k}^\epsilon: \mathbb{Z}_{2^k}\to \mathbb{Q}/2\mathbb{Z}$ on the generator via $w_{2,k}^\epsilon (1)=\epsilon\cdot 2^{-k}$.

\item For $k\geq 1$ an integer, we define the forms $u_k$ and $v_k$ on $\mathbb{Z}/{2^k}\mathbb{Z}\times \mathbb{Z}/{2^k}\mathbb{Z}$ via the matrices:
\[
u_k=\left(\begin{matrix}
0 & 2^{-k} \\
2^{-k} & 0
\end{matrix}\right) \quad
v_k=2^{-k}\left(\begin{matrix}
2 & 1 \\
1 & 2
\end{matrix}\right)
\]

\end{enumerate}

For example, if we consider the lattice $L=A_2$, then $A_L\cong \mathbb{Z}/3\mathbb{Z}$ and $q_L$ has value $\tfrac{4}{3}$ on the generator. Thus $q_L\cong \omega_{3,1}^1$.

\begin{thm}[cf. {\cite[Thm. 1.8.1]{nikulin}}]\label{t:relations}
The forms $w_{p,k}^\epsilon$, $u_k$, $v_k$ generate the semi--group of finite quadratic forms.
\end{thm}

In other words every finite quadratic form can be written (not uniquely)
as a direct sum of the generators $w_{p,k}^\epsilon$, $u_k$, $v_k$.
Relations can be found in \cite[Thm. 1.8.2]{nikulin}.

For a finite quadratic form $q$, and a prime number $p$,
we denote $q$ restricted to the $p$--component $(A_q)_p$ of $A$ by $q_p$.
The following results describe the close link between discriminant quadratic forms and even lattices.

\begin{thm}[cf. {\cite[Thm. 1.13.2]{nikulin}}]\label{t:lattice_unique}
An even lattice $S$ with invariants $(t_+,t_-,q)$ is unique if, simultaneously,
\begin{enumerate}
\item $t_+\geq 1, t_-\geq 1, t_++t_-\geq 3$;
\item for each $p\neq 2$, either $\rk S \geq 2+l((A_q)_p)$ or $
q_p\cong w_{p,k}^\epsilon\oplus w_{p.k}^{\epsilon'}\oplus q_p'$;

\item for $p= 2$, either $\rk S \geq 2+l((A_q)_2)$ or one of the following holds
\[
q_2\cong u_k\oplus q_2',\quad
q_2\cong v_k\oplus q_p',\quad
q_2\cong w_{2,k}^\epsilon\oplus w_{2.k}^{\epsilon'}\oplus q_2'.
\]

\end{enumerate}
\end{thm}

\begin{cor}[cf. {\cite[Corollary 1.13.3]{nikulin}}]\label{c:latunique}
An even lattice $S$ with invariants $(t_+,t_-,q)$ exists and is unique
if $t_+ -t_-\equiv \sign q \imod 8$, $t_+ +t_-\geq 2+ l(A_q)$, and $t_+, t_- \geq 1$.
\end{cor}

\begin{cor}[cf. {\cite[Corollary 1.13.4]{nikulin}}]\label{c:U+T}
Let $S$ be an even lattice of signature $(t_+,t_-)$. If $t_+\geq 1$, $t_-\geq 1$ and $t_+ +t_-\geq 3+l(A_S)$, then $S\cong U
\oplus T$ for some lattice $T$.
\end{cor}

In Table \ref{tab-forms}, we list the discriminant form associated to each of the lattices a
ppearing in our calculations (see Sections \ref{sec-ex},\ref{sec:tables}).
A complete description can be found in \cite[Appendix A]{belcastro}.

\begin{table}[h!]\centering
\begin{tabular}{ l c c || l c c|| l c c}
 $L$&$\sign L$ &$q_L$& $L$&$\sign L$ &$q_L$& $L$&$\sign L$ &$q_L$\\
  \hline
  $U$ 		&(1,1)	& trivial   			&$D_6$		&(0,6)	& $(w_{2,1}^1)^2$	& $T_{4,4,4}$	&(1,9)	& $v_2$\\
  $U(2)$		&(1,1)	& $u$ 			&$D_9$		&(0,9)	& $w_{2,2}^{-1}$ 	& $T_{3,4,4}$	&(1,8)	& $w_{2,3}^5$\\
  $A_1$ 		&(0,1) 	& $w^{-1}_{2,1}$ 	&$E_6$		&(0,6)	& $w_{3,1}^{-1}$	&  $T_{2,5,6}$	&(1,10)	& $w_{2,3}^{-5}$\\
  $A_2$		&(0,2)	& $w_{3,1}^{1}$ 	&$E_7$		&(0,7)	& $w_{2,1}^{1}$   	&   $<2>$		&(1,0)	& $w_{2,1}^1$\\
  $A_3$		&(0,3)	& $w_{2,2}^5$		&$E_8$		&(0,8)	& trivial			&  $<4>$		&(1,0)	& $w_{2,2}^1$\\
  $A_1(2)$		&(0,1)	& $w_{2,2}^{-1}$ 	&$H_5$		&(1,1)	& $w_{5,1}^{-1}$	&  $<8>$		&(1,0)	& $w_{2,3}^1$\\
  $D_4$		&(0,4)	& $v$ 			&$L_9$		&(1,1)	& $w_{3,2}^1$		&  $<-8>$		&(0,1)	& $w_{2,3}^{-1}$\\
  $D_5$		&(0,5)	& $w_{2,2}^{-5}$	&$M_9$		&(1,1)	& $w_{3,2}^{-1}$	&	&	&\\
 \end{tabular}
\caption{Lattices and forms} \label{tab-forms}
\end{table}

Let $L'$ be an overlattice $L'$ of the lattice $L$.
We call $H_{L'}:=L'/L$.
By the chain of embeddings $L\subset L'\subset (L')^*\subset L^*$
one has $H_{L'}\subset A_L$ and $A_{L'}=((L')^*/L)/H_{L'}$.

\begin{prop}[cf. {\cite[Prop. 1.4.1]{nikulin}}]\label{p:overl}
The correspondence $L'\leftrightarrow H_{L'}$ is a 1:1 correspondence
between overlattices of finite index of $L$ and $q_L$-isotopic subgroups of $A_L$,
i.e. subgroups on which the form $q_L$ is 0.
Moreover, $H_{L'}^\perp=(L')^*/L$ and $q_{L'}=({q_L}_{|H_{L'}^\perp})/H_{L'}$.
\end{prop}

\subsection{K3 lattices} \label{sec-K3lattice}
Let $X$ be a K3 surface.
It is well--known that $H^2(X,\mathbb{Z})$ is an even unimodular lattice of signature $(3,19)$.
As such, it is isometric to the \emph{K3-lattice} $\klat = U^3\oplus (E_8)^2$.

We let
\[
S_X = H^2(X,\mathbb{Z}) \cap H^{1,1}(X,\mathbb{C})
\]
denote the {\em Picard} lattice of $X$ in $H^2(X,\mathbb{Z})$ and $T_X = S_X^\perp$ denote the {\em transcendental lattice}.

Let $\sigma$ be a non-symplectic automorphism of $X$.
We let $S_X(\sigma)\subseteq H^2(X,\mathbb{Z})$ denote the $\sigma^\ast$-invariant sublattice of $H^2(X,\mathbb{Z})$:
\[S_X(\sigma)=\{x\in H^2(X,\mathbb{Z}): \sigma^*x=x\}.\]
One can check that it is a primitive sublattice of $H^2(X,\mathbb{Z})$. In fact, $S_X(\sigma)$ is a primitive sublattice of $S_X$ and in general $S_X(\sigma)\subsetneq S_X$.
We let $T_X(\sigma) = S_X(\sigma)^\perp$ denote its orthogonal complement. The signature of $S_X(\sigma)$ is $(1,t)$ for some $t\leq 19$, i.e. $S_X(\sigma)$ is hyperbolic.

\section*{Introduction}
\input{intronew}

\section{Background}
\label{sec-background}
\input{background}

\section{Mirror symmetry}
\label{sec-mirror}
\input{mirror}

\section{Methods}
\label{sec-method}
\input{methods}

\section{Examples}
\label{sec-ex}
In this section we will first give examples to illustrate each of the methods that was used to determine $S_X(\sigma_m)$. Then we will describe the subgroups in the two cases where $\SL_W/J_W$ is not cyclic.

\input{ex9}
\input{ex16}

\input{ex-noncyclic}

\section{Tables}
\label{sec:tables}
\footnotesize{\input{tables}}
\normalsize

\section{Computer code for computing lattices}

In order to compute the lattices using the configuration of curves, we used the following Magma code, developed by Antonio Laface and added here with his permission.

This first function takes an even bilinear form $B$, and outputs generators of the discriminant group and the values of $q_B$ on these generators.

\begin{alltt}
disc:=function(M)
 S,A,B:=SmithForm(M);
 l:=[[S[i,i],i]: i in [1..NumberOfColumns(S)]| S[i,i] notin {0,1}];
 sA:=Matrix(Rationals(),ColumnSubmatrixRange(B,l[1][2],l[#l][2]));
 for i in [1..#l] do
  MultiplyColumn(\(\sim\)sA,1/l[i][1],i);
 end for;
 Q:=Transpose(sA)*Matrix(Rationals(),M)*sA;
 for i,j in [1..NumberOfColumns(Q)] do
  if i ne j then
   Q[i,j]:=Q[i,j]-Floor(Q[i,j]);
  else
   Q[i,j]:=Q[i,j]-Floor(Q[i,j])+ (Floor(Q[i,j]) mod 2);
  end if;
 end for;
 return [l[i][1]: i in [1..#l]], Q;
end function;
\end{alltt}

The next function determines whether a given even bilinear form has overlattices. Input is a matrix $M$ and a number $n$. The output is the subgroup of $A_L$ that takes values equal to $n$ modulo $2\mathbb{Z}$. For isotropic subgroups of the discriminant group, use $n=0$.
\begin{verbatim}
isot:=function(M,n)
 v,U:=disc(M);
 Q:=Rationals();
 A:=AbelianGroup(v);
 return [Eltseq(a) : a in A |
 mod2(Matrix(Q,1,#v,Eltseq(a))*U*Matrix(Q,#v,1,Eltseq(a)))[1,1] eq n];
end function;
\end{verbatim}

The function {\verb mod2 } is as follows:
\begin{verbatim}
mod2:=function(Q);
 for i,j in [1..Nrows(Q)] do
  if i ne j then Q[i,j]:=Q[i,j]-Floor(Q[i,j]);
  else Q[i,j]:=Q[i,j]-2*Floor(Q[i,j]/2);
  end if;
 end for;
 return Q;
end function;

\end{verbatim}

Finally, the following function compares two discriminant quadratic forms, and lets us know if they are the same finite quadratic form or not. This is not always easy to check due to the relations in Proposition~\ref{t:relations}.
\begin{verbatim}
dicompare:=function(M,Q)
 v,U:=disc(M);
 w,D:=disc(Q);
 if v ne w then return false; end if;
 A:=AbelianGroup(v);
 Aut:=AutomorphismGroup(A);
 f,G:=PermutationRepresentation(Aut);
 h:=Inverse(f);
 ll:=[Matrix(Rationals(),[Eltseq(Image(h(g),A.i)) : i in [1..Ngens(A)]]) : g in G];
 dd:=[mod2(a*U*Transpose(a)) : a in ll];
 return D in dd;
end function;
\end{verbatim}

\subsection{Method for $m$ not prime}\label{sec-notprime}

If $m$ is not prime, $S_X(\sigma_m)$ is no longer $p$--elementary for any prime $p$.
The best we can say about $A_{S_X(\sigma)}$ is that it is a subgroup of $(\mathbb{Z}_m)^a$, as in the following lemma.

\begin{lem}
Let $\sigma$ be a purely non-symplectic automorphism of order $m$. The discriminant group $A_{S_X(\sigma)}$ is a subgroup of $(\mathbb{Z}_m)^a$ for some $a\geq 0$.
\end{lem}

\begin{proof}
We know $A_{S_X(\sigma)}$ is a finite abelian group. Thus it suffices to show that for each $x\in A_{S_X(\sigma)}$, we have $mx=0$. By Maschke's Theorem, the vector space $H^2(X,\mathbb{Z})\otimes \mathbb{Q}$ is completely reducible. Therefore it decomposes as a sum of irreducible $\mathbb{Q}[\mathbb{Z}/m\mathbb{Z}]$-modules. Each direct summand is of the form $Q(\xi_d)$ where $d|m$. So we have $\sum_{i=0}^{m-1} (\sigma^*)^i=0$ on $T_X(\sigma)$. Now we have $A_{S_X(\sigma)}\cong H^2(X,\mathbb{Z})/(S_X(\sigma)\oplus T_X(\sigma))\cong A_{T_X(\sigma)}$, and these automorphisms commute with the action of $\sigma^*$. Thus $\sigma^*$ acts by identity on $A_{T_X(\sigma)}$, but since $\sum_{i=0}^{m-1} (\sigma^*)^i=0$ on $T_X(\sigma)$, $mx=0 \imod {T_X(\sigma)}$ for $x\in T_X(\sigma)^*$.
\end{proof}

We see that $A_{S_X(\sigma_m)}$ can have any possible order dividing a power of $m$. Even in the best of situations (e.g. $A_{S_X(\sigma_m)}\cong \mathbb{Z}_m^a$) the lattice is no longer determined by $(r,a)$.
Consider for example the two lattices $L_1=U\oplus D_5^2$ and $L_2=U\oplus A_3\oplus D_7$: they both have discriminant group $\mathbb{Z}_4^2$ and $(r,a)=(12,2)$, but $q_{L_1}=(w_{2,2}^3)^2$ whereas $q_{L_2}=w_{2,2}^1\oplus w_{2,2}^5$. In other words, we have two different lattices with the same rank, and isomorphic discriminant groups.

It is true, however, that the lattice is determined in all cases considered here by
the rank and the discriminant quadratic form
(cf. Corollary \ref{c:latunique}).
Thus, rather than computing $(r,a)$ it suffices to compute $(r,q_{S_X(\sigma_m)})$.
We will be able to compute these invariants for all K3 surfaces $X_{W,G}$ with $W$ as in \eqref{eq-W} and $G\subset \SL_W$ except in case $m=4,8,12$.

The first step to compute  $S_X(\sigma_m)$ is to compute its rank $r$.
In order to do so, we describe an important set of curves on $X_{W,G}$, the orbits of the exceptional curves.
Consider the hypersurface $Y_{W,G}$ as in Section \ref{BHK_sec}, and the resolution of singularities $X_{W,G}\to Y_{W,G}$. Because $W$ is nondegenerate, all of the singularities of $Y_{W,G}$ lie on the coordinate curves.

Let $\mathcal{E}$ denote the set of the exceptional curves. Because $\sigma_m$ leaves the coordinate curves invariant, the action of $\sigma_m$ on $X_{W,G}$ induces an action of $\sigma_m$ on $\mathcal{E}$. Given an exceptional curve $E$, let $G_E$ denote the isotropy group for $E$ and let $\mathcal{E}/\sigma_m$ denote the set of orbits of this action. Now we set
\[
b_E=\tfrac 1{|G_E|}\sum_{i=0}^{m-1}\sigma_m^iE.
\]
This is simply the sum of all curves in the orbit of $E$. In particular, if $E$ is fixed by $\sigma_m$, then $b_E=E$. We can use this topology to determine $r$:

\begin{lem}\label{l:rank}
The rank $r$ of $S_X(\sigma)$ is equal to 1 plus the number of orbits of exceptional curves in the blow-up $X_{W,G}\to Y_{W,G}$, i.e.
\[
\rk S_X(\sigma_m)=1+|\mathcal{E}/\sigma_m|
\]
\end{lem}

\begin{proof}

Consider the quotient $X_{W,G}/\sigma_m$.
We have the following diagram

\[
\begin{tikzcd}
X_{W,G}   \arrow{r}{} \arrow{d}{\pi}  &  Y_{W,G} \arrow{d}{\pi}\\
X_{W,G}/\sigma_m\arrow{r}{}  &  \mathbb{P}(w_2,w_3,w_4) \\
\end{tikzcd}
\]
where the bottom horizontal arrow is obtained by blowing up isolated singular points of $\mathbb{P}(w_2,w_3,w_4)$ and singular points on the branch locus $\set{f(x_2,x_3,x_4)=0}\subset \mathbb{P}(w_2,w_3,w_4)$. Notice in the quotient $\mathbb{P}(w_2,w_3,w_4)\cong Y_{W,G}/\sigma_m$, the branch locus corresponds to the curve $y=0$ in $Y_{W,G}$. Furthermore, each orbit of exceptional curves will correspond to exactly one exceptional curve of the bottom horizontal map.

Since $h^2(\mathbb{P}(w_2,w_3,w_4))=1$ (see e.g. \cite{twistedproj} or \cite{dolgachev2}), and $H^2(X_{W,G}/\sigma_m,\mathbb{Q})$ is generated by the exceptional curves and the pullback from $\mathbb{P}(w_2,w_3,w_4)$, we have
\[
\dim H^2(X_{W,G}/\sigma_m,\mathbb{Q})=1+|\mathcal{E}/\sigma_m|.
\]

Finally, we know by \cite{macdonald} that $H^2(X_{W,G}/\sigma_m,\mathbb{Q})=H^2(X_{W,G}, \mathbb{Q})^{\sigma_m}$. This implies the result.
\end{proof}

\begin{rem}
Although $H^2(X_{W,G}, \mathbb{Q})^{\sigma_m}$ is generated by the orbits of the exceptional curves and one coordinate curve, this is not always the case for the lattice $S_X(\sigma_m)$.
\end{rem}

The second step in determining $(r,q)$ requires another set of curves.
Let $\mathcal{K}$ be the set of smooth irreducible components of the (strict transforms of the) coordinate curves ${x_i=0}$
(we omit curves that have worse singularities than a separating nodes between two or more smooth components).
As before $\sigma_m$ acts on this set.
For $C\in \mathcal{K}$, we define $G_C$ as the isotropy group, $\mathcal{K}/\sigma_m$ as the set of orbits, and
\[
b_C=\tfrac 1{|G_C|}\sum_{i=0}^{m-1}\sigma_m^iC.
\]
Notice that $\sigma_m$ leaves the coordinate curves invariant, so $b_C$ is the sum of some of the smooth irreducible components of one of the coordinate curves. In cases where the coordinate curve is smooth, $b_C$ is the coordinate curve. There are cases, however, where the coordinate curve is not irreducible.

Finally we define the set of orbit curves $\mathcal{B}=\set{b_E\mid E\in \mathcal{E}/\sigma_m}\cup \set{b_C\mid C\in \mathcal{K}/\sigma_m}$. Let $L_{\mathcal{B}}$ denote the sublattice of $S_X$ generated by $\mathcal{B}$. This lattice is clearly a sublattice of $S(\sigma_m)$ since $\sigma_m$ fixes each of the generators. Therefore, the rank is at most $r$.
By Lemma~\ref{l:rank}, we see that $\rk L_\mathcal{B}=r$. Hence $S(\sigma_m)$ is an overlattice of $L_\mathcal{B}$. Because of this explicit description, we can actually determine the lattice $L_{\mathcal{B}}$ in each case. (See the appendix for Magma code, which allows us to compute $L_\mathcal{B}$.)

Determining $(r,q)$ now breaks into four different cases, each of which requires a different method.
We now describe those four methods. In most cases, we show $L_\mathcal{B}=S_X(\sigma_m)$

\subsubsection{Method I}
The first method is the most common.
Using Proposition~\ref{p:overl}, we can determine all of the overlattices of $L_\mathcal{B}$. It often happens that the discriminant group has no isotropic subgroups, and so there are no nontrivial overlattices. Since $S_X(\sigma_m)$ is an overlattice of $L_\mathcal{B}$, we can conclude $S_X(\sigma_m)=L_\mathcal{B}$. An example of this situation is shown in Example \ref{ex-method1}.

This covers all possibilities when $m=18,20,24,30,42$, and parts of the other orders, namely whenever $m$ and $r$ are one of the following possibilities: $m=10$,  $r=3,8,12,17$; $m=9$, $r=4,16$; and $m=6$, $r=1,4,6,9,11,14,16,19$. One may notice in Table~\ref{tab-6} that there are two possibilities for  $m=6, r=10$. One of them has quadratic form $4\omega_{3,1}^1$ and therefore has no overlattices, so this method works. But the other $v+4\omega_{2,1}^{-1}$ does have overlattices, so we use another method.

\subsubsection{Method II}
The second method does not use $L_\mathcal{B}$. Instead, we notice that some power $\sigma_m^k$ has order $p$ for some prime $p$ and $S_X(\sigma_m)\subset S_X(\sigma_m^k)$ is a primitive embedding of lattices. If it happens that $\rk S_X(\sigma_m)= \rk S_X(\sigma_m^k)$, then indeed $S_X(\sigma_m)= S_X(\sigma_m^k)$. In this case $S_X(\sigma_m)$ is $p$--elementary and we can use the methods in Section~\ref{s:methodsprime} to compute it.
See Example~\ref{ex-method2} for an example of this situation.

This covers the remaining cases with $m=6,10, 15, 22$, the cases with $m=9$, $r=16$, as well as the two cases with $m=14$, $r=13$.

\subsubsection{Method III}
For the third method, we use the results of Belcastro in \cite{belcastro}. Belcastro has computed the Picard lattice $S_{X_t}$ of the general member of a family of K3 surfaces $X_t$ for each of the 95 weight systems. Each invertible polynomial $W$ yields a special member of this family, but often with bigger Picard lattice. It often happens that $S_{X_t}\cong L_\mathcal{B}$. When this happens, because $S_{X_t}\subset H^2(X_t,\mathbb{Z})$ is a primitive embedding of lattices, the composition of embeddings
\[
L_\mathcal{B}\subset S_X\subset H^2(X_W,\mathbb{Z})
\]
remains a primitive embedding for the special member, i.e. $L_\mathcal{B}$ is primitively embedded in $H^2(X_W,\mathbb{Z})$. Here we require the group $G=J_W$.

By the isomorphism theorems for groups, $S_{X_W}/L_\mathcal{B}\subset H^2(X_W.\mathbb{Z})/L_\mathcal{B}$, and the latter is a free abelian group. Therefore $L_\mathcal{B}$ is a primitive sublattice of $S_X$. By similar argument, $L_\mathcal{B}$ is a primitive sublattice of $S_X(\sigma_m)$. Since they have the same rank, we see $L_\mathcal{B}=S_X(\sigma_m)$. We illustrate this method in Example~\ref{ex-meth3}.

This method is used whenever $m$ and $r$ are one of the following possibilities: $m=9$, $r=8$; $m=14$, $r=7$; and $m=16$, $r=9$.

\subsubsection{Method IV} This leaves two cases:  $m=16$, $r=11$ and $m=9$, $r=12$. As each of these cases is different we will work out each case in Examples ~\ref{ex-meth41} and \ref{ex-meth42}. This method involves finding $S_{X_{W,G}}$ and showing explicitly that $L_\mathcal{B}$ embeds primitively into $S_{X_{W,G}}$. Then, as in Method II, we can conclude that $L_\mathcal{B}=S_X(\sigma_m)$.

\begin{rem}
What we see when employing these four methods is that the invariant lattice seems to be the lattice $L_\mathcal{B}$. This is certainly true for every invariant lattice computed by methods I,III and IV, though it also seems to be true for each of the others (including the invariant lattice for the automorphisms of prime order and of order 4, 8, 12), though we have not checked every case explicitly.

\end{rem}

In what follows, we will make use of the following lemma.

\begin{lem}[cf. \cite{order_six, order_four}] \label{l:rationaltree}
Let $R_1,R_2,\dots, R_s$ be a tree of smooth rational curves on a K3 surface $X$ and $\sigma$ a non-symplectic automorphism of finite order on $X$ leaving each of the $R_i$ invariant. Then the intersection points of the $R_i$'s are fixed by the automorphism and it suffices to know the action at one intersection point to know the action on the entire tree.
\end{lem}

We will also make use of the following formula, giving the genus of a curve of degree $d$ in weighted projective space $\mathbb{P}(w_1,w_2,w_3)$ in order to compute the genus of the coordinate curves. This formula can be found, e.g., in \cite[Theorem 12.2]{fletcher}.
\begin{equation*}
g(C)=\frac{1}{2}\left(\frac{d^2}{w_1w_2w_3}-d\sum_{1\leq j<i}\frac{\gcd(w_i,w_j)}{w_iw_j}+\sum_{i=1}^3\frac{\gcd(w_i,d)}{w_i}-1\right).
\end{equation*}
With the genus, one can also compute self--intersection numbers for curves on a K3 surface via Riemann--Roch:
\[
2-2g(C)= C\cdot C
\]

\subsection{Method for $m=p$ prime}\label{s:methodsprime}

As mentioned, the argument given in \cite{ABS} and \cite{CLPS} essentially boils down to determining the invariant lattice $S_X(\sigma_p)$ for $X_{W,G}$.
The method for determining this lattice relies on the following powerful theorems.

\begin{thm}[\cite{other_primes}] Given a $K3$ surface with a non-symplectic automorphism $\sigma$ of order $p$, a prime, the invariant lattice $S_X(\sigma)$ is $p$--elementary, i.e. $A_{S_X(\sigma)}\cong (\mathbb{Z}/p\mathbb{Z})^a$.
\end{thm}

\begin{thm}[\cite{RS,nikulin}] For a prime $p\neq 2$, a hyperbolic, $p$--elementary lattice $L$ with rank $r\geq 2$ is completely determined by the invariants $(r,a)$, where $a$ is the length. An indefinite 2--elementary lattice is determined by the invariants $(r,a,\delta)$, where $\delta\in{0,1}$ and $\delta=0$ if the discriminant quadratic form takes values 0 or 1 only and $\delta=1$ otherwise.
\end{thm}
By Proposition \ref{p:orth}, the orthogonal complement in $L_{K3}$ of a $p$--elementary lattice with invariants $(r,a)$ is a $p$--elementary lattice with invariants $(22-r,a)$. For both a given 2--elementary lattice and its orthogonal complement, the third invariant $\delta$ agrees.
In the setting of Theorem \ref{t:main_thm}, Corollary \ref{c:U+T} shows that we also have $S_X(\sigma_p)^\perp\cong U\oplus M$ where $M$ is a hyperbolic $p$--elementary with invariants $(20-r,a)$. Thus for $p\neq 2$ it is enough to verify that $(r,a)$ for $X_{W,G}$ and $(r^T,a_T)$ for $X_{W^T,G^T}$ satisfy $r=20-r^T$ and $a=a_T$. For $p=2$ the third invariant $\delta$ must also be compared, which the authors checked in \cite{ABS}.

In order to compare $(r,a)$, we first look at the topology of the fixed point locus.

\begin{thm}[cf. \cite{nikulin2, other_primes}]\label{summary_invariants}
Let $X$ be a K3 surface with a non-symplectic automorphism $\sigma$ of prime order $p\neq 2$. Then the fixed locus $X^\sigma$ is nonempty, and consists of either isolated points or a disjoint union of smooth curves and isolated points of the following form:
\begin{equation}
X^\sigma=C\cup R_1\cup\ldots\cup R_k\cup \{p_1,\ldots,p_n\}\label{e:fixedlocus}.
\end{equation}
Here $C$ is a curve of genus $g\geq 0$, $R_i$ are rational curves and $p_i$ are isolated points.

If $p=2$, then the fixed locus is either empty, the disjoint union of two elliptic curves, or is of the form \eqref{e:fixedlocus} with $n=0$.
\end{thm}

In \cite{ABS} and \cite{CLPS}, $\sigma_p$ always fixes a curve. Furthermore, the case of two elliptic curves does not appear in this setting described there. Therefore, the fixed locus is determined by the invariants $(g,k,n)$.
In \cite{other_primes}, the authors give formulas to calculate $(r,a)$ given $(g,k,n)$ for each prime $p$ (see \cite[Theorem 0.1]{other_primes}).
Thus, in order to prove Theorem \ref{t:main_thm} for $p$ prime, one first computes the invariants $(g,k,n)$,
and from them computes the invariants $(r,a)$ (if $p=2$, additional computation are required to obtain $\delta$).
Then one compares the invariants for BHK mirrors as described above.

\input{methodnotprime}

\subsection{Proof of main Theorem}
We now provide the details of the proof of Theorem \ref{t:main_thm}.
For each K3 surface $X_{W,G}$ we have the invariants $(r,q_{S_X(\sigma_m)})$ for the invariant lattice $S_X(\sigma_m)$, as discussed in the previous section. We know the invariant lattice has signature $(1,r-1)$, and so the orthogonal complement has signature $(2,20-r)$. We check that the conditions of Corollary \ref{c:U+T} are fulfilled so that $S_X(\sigma_m)^\perp_{L_{K3}}\cong U\oplus M$ for some lattice $M$\footnote{In one case, the conditions are not fulfilled, namely $m=6$, $r=1$. However, in this case, we know the lattice $U\oplus\langle4\rangle$ has the given invariants, and by Corollary~\ref{c:latunique} this is the only such lattice}. This lattice $M$ is hyperbolic with signature $(1,19-r)$ and has discriminant quadratic form $-q_{S_X(\sigma_m)}$. One can see from the tables that the conditions of Theorem~\ref{t:lattice_unique} are satisfied. Hence, there is exactly one lattice with these invariants. To prove the theorem, therefore, we need simply to check that $(20-r, -q_{S_X(\sigma_m)})$ are the invariants for the invariant lattice $S_{X^T}(\sigma_m^T)$ for $X_{W^T,G^T}$. This can be checked by consulting the tables in Section \ref{sec:tables}.
This concludes the proof.

Tables contain all possible invertible polynomial of the from \eqref{eq-W} with non--symplectic automorphism of order $m$, and for each polynomial, we list the orders of the possible groups $G/J_W$ satisfying $J_W\subset G\subset \SL_W$.
In most cases $\SL_W/J_W$ is cyclic, so the properties of $G^T$ make it clear what the dual group is. However for two examples, one can see by the multiplicity of subgroups with the same order that the group $\SL_W/J_W$ is not cyclic. These two examples are $x^3+y^3+z^6+w^6$ (number 3d) and $x^2+y^4+z^6+w^{12}$ (number 8d) in Table \ref{tab-6}. In these cases, we will clear up any ambiguities in the following sections.

From now on we will make a change of notation from $(x_1,x_2,x_3,x_4)$ to $(x,y,z,w)$ for the variables of $W$, so that the variables are arranged with the weights in nonincreasing order. In other words, it is possible that $x_1$ corresponds to any of $x,y,z$, or $w$. This convention is also used in Tables of Section \ref{sec:tables}.

\begin{rem}
It is possible that a given K3 surface admits a purely non--symplectic automorphism of different orders. It turns out that it doesn't matter which automorphism one uses to exhibit LPK3 mirror symmetry, the notion still agrees with BHK mirror symmetry, in the sense of Theorem~\ref{t:main_thm}, as long as the defining polynomial is of the proper form \eqref{eq-W}. We expect that the theorem still holds for K3 surfaces that don't take the form of \eqref{eq-W}, but that is a topic for further investigation.
\end{rem}

\subsection{Mirror symmetry for K3 surfaces}
\label{sec-LPK3}
Mirror symmetry for a Calabi--Yau manifold $X$ and its mirror $X^\vee$ can be thought of as an exchanging of the K\"ahler structure on $X$ for the complex structure of $X^\vee$. Thus a first prediction of mirror symmetry is the rotation of the Hodge diamond:
\begin{equation*
H^{p,q}(X,\mathbb{C})\cong H^{q,N-p}(X,\mathbb{C})
\end{equation*}
where $N$ is the dimension of $X$.

For K3 surfaces, however, the Hodge diamond
is symmetric under the rotation mentioned above. So we need to consider a refinement of this idea. This is accomplished by the notion of lattice polarization. Roughly, we choose a primitive lattice $M\hookrightarrow S_X$, which plays the role of the K\"ahler deformations, and the mirror lattice $M^\vee$, which we now define, plays the role of the complex deformations.
We will refer to this formulation of mirror symmetry simply as LPK3 mirror symmetry.

Following \cite{dolgachev}, let $X$ be a K3 surface and suppose that $M$ is a lattice of signature $(1,t)$. If $j\colon M\hookrightarrow S_X$ is a primitive embedding into the Picard lattice of $X$, the pair $(X,j)$ is called an \emph{$M$-polarized K3 surface}.  There is a moduli space of $M$--polarized K3 surfaces with dimension $19-t$.

We will not be concerned about the embedding. As in \cite{CLPS}, we will call the pair $(X,M)$ an \emph{$M$-polarizable} K3 surface if such an embedding $j$ exists.  Note that for an $M$-polarizable K3 surface $(X,M)$, the lattice $M$ naturally embeds primitively into $\klat$.

\begin{defn}\label{mirror_defn}
Let $M$ be a primitive sublattice of $\klat$ of signature $(1,t)$ with $t\leq 18$ such that $M^\perp_{L_{K3}}\cong U\oplus \mirror{M}$.
We define $\mirror{M}$ to be (up to isometry) the \emph{mirror lattice} of $M$.\footnote{As in \cite{CLPS}, our definition in this restricted setting is slightly coarser than the one used by Dolgachev in \cite{dolgachev}, since we do not keep track of the embedding $U\hookrightarrow M^\perp$ and instead only consider $M^\vee$ up to isometry.}
\end{defn}

By Theorem \ref{t:lattice_unique}
this definition is independent of the embedding $M$ into $\klat$. Furthermore under some conditions, (see e.g. Corollary~\ref{c:U+T} and Theorem~\ref{t:lattice_unique}) this definition is also independent of the embedding $U$ into $M^\perp$. One can check that these conditions are satisfied for the lattices we consider here.

Note that $\mirror{M}$ also embeds primitively into $\klat$ and has signature $(1,18-t)$.  Furthermore, $q_M\cong -q_{\mirror{M}}$. One easily checks that $(\mirror{M})^\perp_{\klat}\cong U\oplus M$.

Given $(X,M)$ an $M$-polarizable K3 surface and $(X',M')$ an $M'$-polarizable K3 surface, with $M$ and $M'$ primitive sublattices of $S_X$ and $S_{X'}$, resp.,
we say that $(X,M)$ and $(X',M')$ are \emph{LPK3 mirrors} if $M' = \mirror{M}$ (or equivalently $M=\mirror{(M')}$).

Notice that if $M$ has rank $t+1$, then the dimension of the moduli space of $M^\vee$ polarized K3 surfaces is $19-(18-t)$ which agrees with the rank of $M$. Returning to the question of K\"ahler deformations and complex deformations, we see that this definition of mirror symmetry matches the idea behind rotation of the Hodge diamond, as mentioned earlier.

  \subsection{Quasihomogeneous polynomials and diagonal symmetries}\label{quasi_sec}
We recall a few facts and definitions (cf. \cite{CLPS} for details).
A \emph{quasihomogeneous} map of degree $d$ with integer weights $w_1, w_2, \dots, w_n$
is $W:\mathbb{C}^n\to \mathbb{C}$ such that for every $\lambda \in \mathbb{C}$,
\[
W(\lambda^{w_1}x_1, \lambda^{w_2}x_2, \dots, \lambda^{w_n}x_n) = \lambda^dW(x_1,x_2, \dots, x_n).
\]
One can assume
$\gcd(w_1, w_2, \dots, w_n)=1$ and say $W$ has the \emph{weight system} $(w_1, w_2, \ldots, w_n; d)$.
Given a quasihomogeneous polynomial $W:\mathbb{C}^n  \rightarrow \mathbb{C}$ with a critical point at the origin,
we say it is \emph{non-degenerate} if the origin is the only critical point of $W$ and
the fractional weights $\frac{w_1}{d}, \ldots, \frac{w_n}{d}$ of $W$ are uniquely determined by $W$.

A non-degenerate quasihomogeneous polynomial $W$ (also called \emph{potential} in the literature) is \emph{invertible} if it has the same number of monomials as variables.

If $W$ is invertible we can rescale variables so that $W = \sum_{i=1}^n \prod_{j=1}^n x_j^{a_{ij}}$.
This polynomial can be represented by the square matrix $\A{W} = (a_{ij})$,
which we will call the \emph{exponent matrix} of the polynomial.
Since $W$ is invertible, the matrix $\A{W}$ is an invertible matrix.

The \emph{group $G_W$ of diagonal symmetries} of an invertible polynomial $W$ is
\begin{equation*}
\Gmax{W} = \{(c_1, c_2, \ldots, c_n) \in (\mathbb{C}^*)^n:  W(c_1x_1, c_2x_2, \ldots, c_nx_n) = W(x_1, x_2, \ldots, x_n)\}.
\end{equation*}

Observe that, given $\gamma=(c_1, c_2, \ldots, c_n)\in \Gmax{W}$,
the $c_i$'s are roots of unity.
Thus one can consider $\Gmax{W}$ as a subgroup of $(\mathbb{Q} / \mathbb{Z})^n$, using addivite notation and identifying
$(c_1, c_2, \ldots, c_n) = (e^{2 \pi i g_1}, e^{2 \pi i g_2}, \ldots, e^{2 \pi i g_n})$ with $(g_1, g_2, \ldots, g_n)\in(\mathbb{Q} / \mathbb{Z})^n$.
Observe that the order of $\Gmax{W}$ is $|\Gmax{W}| = \det(\A{W})$.

Since $W$ is quasihomogeneous, the \emph{exponential grading operator} $\J{W} = \left(\frac{w_1}{d}, \frac{w_2}{d}, \ldots, \frac{w_n}{d}\right)$ is contained in $\Gmax{W}$.
We denote by $J_W$ the cyclic group of order $d$ generated by $\J{W}$: $J_W=\inn{\J{W}}$.
Moreover, each $\gamma=(g_1,\dots,g_n)$ define a diagonal matrix and thus
$G_W$ is embedded in  $\GL_n(\mathbb{C})$.
We define $$\SLgp{W}:=\Gmax{W}\cap \SLn{n},$$
i.e. $\gamma=(g_1,\dots,g_n)\in \SLgp{W}$ if and only if $\sum_i g_i\in \mathbb{Z}$.
The group $\SLgp{W}$ is called the {\em symplectic group} since, by \cite[Proposition 1]{ABS},
an automorphism $\sigma\in \Gmax{W}$ is symplectic if and only if $\det \sigma =1$, that is, if and only if $\sigma\in \SLgp{W}$.

\subsection{K3 surfaces from $(W,G)$}\label{K3_sec}

Reid (in an unpublished work) and Yonemura \cite{yonemura}
have indipendently compiled a list of the 95 normalized weight systems $(w_1,w_2,w_3,w_4;d)$
(``the 95 families'') such that $\mathbb{P}(w_1,w_2,w_3,w_4)$
admits a quasismooth hypersurface of degree $d$ whose minimal resolution is a K3 surface.
We consider one of these weight systems $(w_1,w_2,w_3,w_4;d)$
and an invertible quasihomogeneous polynomial of the form
\begin{equation}\label{eq-W}
  W=x_1^m+f(x_2,x_3,x_4).
\end{equation}
Moreover, let $G$ be a group of symmetries such that $J_W\subseteq G \subseteq \SLgp{W}$
and let $\widetilde{G}=G/J_W$.
The polynomial $W$ defines a hypersurface $Y_{W,G}\subset \mathbb{P}(w_1,w_2,w_3,w_4)/\widetilde{G}$
and one shows that the minimal resolution $X_{W,G}$ of $Y_{W,G}$ is a K3 surface (see \cite{ABS,CLPS}).

The group $\Gmax{W}$ acts on $Y_{W,G}$ via automorphisms, which extend to automorphisms on the K3 surface $X_{W,G}$.
The given form of $W$ ensures that the K3 surface $X_{W,G}$
admits a purely non-symplectic automorphism of order $m$:
\[
\sigma_m:\pp{x_1}{x_2}{x_3}{x_4}\mapsto \pp{\zeta_mx_1}{x_2}{x_3}{x_4}
\]
where $\zeta_m$ is a primitive $m$-th root of unity.
With additive notation, it is $\sigma_m=\left(\tfrac 1m,0,0,0\right)$.

  \subsection{BHK mirror symmetry}\label{BHK_sec}
Now we can describe the second relevant formulation of mirror symmetry coming from mirror symmetry for Landau--Ginzburg models and which we call BHK (from Berglund-H\"ubsch-Krawitz) mirror symmetry.
This particular formulation of mirror symmetry was developed initially by Berglund--H\"ubsch in \cite{berghub}, and later refined by Berglund--Henningson in \cite{berghenn} and Krawitz in \cite{krawitz}. Because of the LG/CY correspondence and a theorem from Chiodo--Ruan \cite{BHCR}, this mirror symmetry of LG models can be translated into mirror symmetry for Calabi--Yau varieties (or orbifolds).

We consider $(W,G)$ with $W$ invertible and $W=\sum_{i=1}^n \prod_{j=1}^n x_j^{a_{ij}}$
and define another pair $(W^T,G^T)$, called the BHK mirror.
We first define the polynomial $W^T$ as $$W^T = \sum_{i=1}^n \prod_{j=1}^n x_j^{a_{ji}},$$
i.e. the matrix of exponents of $W^T$ is $A_W^T$.
By the classification of invertible polynomials (cf. \cite[Theorem 1]{KrSk}),
$W^T$ is invertible.

Next, using additive notation, one defines the dual group $G^T$ of $G$ as
\begin{equation}\label{dualG_def}
G^T = \set{\; g \in \Gmax{W^T} \; | \;\;g\A{W} h^T \in \mathbb{Z} \text{ for all } h \in G \; }.
\end{equation}

The following useful properties of the dual group can be found in \cite[Proposition 3]{ABS}:
\begin{prop}[cf. {\cite[Proposition 3]{ABS}}]
Given $G$ and $G^T$ as before, one has:
\begin{enumerate}
\item $(G^T)^T = G$.
\item If $G_1\subset G_2$, then $G_2^T\subset G_1^T$ and $G_2/G_1\cong G_1^T/G_2^T$.
\item $(\Gmax{W})^T = \triv$, $(\triv)^T = \Gmax{W^T}$.
\item $(J_W)^T=\SLgp{W^T}$. In particular, if $J_W\subset G$, then $G^T\subset \SLgp{W}$.
\end{enumerate}
\end{prop}
Given the pair $(W,G)$ with $W$ invertible with respect to one of the 95 weight systems,
we associated to it the K3 surface $X_{W,G}$.
One can  check that in this case the weight system of $W^T$ also belongs to the 95.
By the previous result, $J_{W^T}\subseteq G^T\subseteq \SL_{W^T}$, so that $X_{W^T,G^T}$ is again a K3 surface.
We call $X_{W^T,G^T}$ the \emph{BHK mirror of $X_{W,G}$}.

\subsection{Main theorem}\label{main_sec}
We have  described two kinds of mirror symmetry for K3 surfaces: LPK3 mirror symetry and BHK one.
Since mirror symmetry describes a single physical phenomenon,
we expect the two constructions to be compatible in situations where both apply.
We will now state our main theorem, which shows that BHK and LPK3 mirror symmetry agree for the K3 surfaces $X_{W,G}$, when $W$ is of the form \eqref{eq-W}.
When no confusion arises, we will denote the mirror K3 surfaces $X_{W,G}$ and $X_{W^T,G^T}$ simply by $X$ and $X^T$.

Consider the data $(W,G,\sigma_m)$,
where \begin{itemize}
\item $W$ is an invertible polynomial of the form \eqref{eq-W}
whose weight system belong to the 95 families of Reid and Yonemura,
\item $\sigma_m=(\frac 1m,0,0,0)$ is the non-symplectic automorphism of order $m$,
\item $G$ is a group of diagonal symmetries of $W$ such that $J_W\subseteq G\subseteq \SL_W$.\end{itemize}
By section \ref{sec-K3lattice}, the invariant lattice $S_X(\sigma_m)$ is a primitive sublattice of $S_X$
and $(X_{W,G}, S_X(\sigma_m))$ is a $S_X(\sigma_m)$--polarizable K3 surface.
Let $r$ be the rank of $S_X(\sigma_m)$.
The BHK mirror is given by $(W^T,G^T,\sigma_m^T)$, where $\sigma_m^T$ is the non-symplectic automorphism of order $m$ on $X_{W^T,G^T}$.
Notice that $\sigma_m$ and $\sigma_m^T$ have the same form, namely $(\tfrac{1}{m},0,0,0)$, but they act on different surfaces.

\begin{thm}\label{t:main_thm}
Suppose $m\neq 4,8,12$. If $W$ is a polynomial of the form \eqref{eq-W}, quasihomogeneous with respect to one of the 95 weight systems for K3 surfaces as in Section~\ref{K3_sec} and $G$ is a group of diagonal symmetries satisfying $J_W\subseteq G\subset\SL_W$, then $\left(X_{W^T,G^T}, S_{X^T}(\sigma_p^T\right))$ is an LPK3 mirror of $\left(X_{W,G}, S_X(\sigma_p)\right)$.
\end{thm}

The theorem is proved by showing that $$S_X(\sigma_m)^\vee\cong S_{X^T}(\sigma_m^T).$$
 As we have seen in section \ref{s:lattice}, this amounts to checking that the invariants $(r,q_{S_X(\sigma_m)})$ for $X_{W,G}$ and $(r^T,q_{S_{X^T}(\sigma_m^T)})$ for $X_{W^T,G^T}$ satisfy $r=20-r^T$ and $q_{S_X(\sigma_m)}\cong -q_{S_{X^T}(\sigma_m^T)}$. Thus the heart of the proof is determining $q_{S_X(\sigma_m)}$ (or equivalently in our case $S_X(\sigma_m)$).
In the following section, we will describe how this is done.
It involves computing the invariant lattice and its overlattices.
We list the results in tables in Section \ref{sec:tables}.
Unfortunately, our method does not work for $m=4,8,12$ due to the presence of many overlattices, so that we cannot exactly pinpoint the invariant lattice.

\section{Order 42}
\begin{longtable}{p{12pt}|c|c|c|c|c|c|c}
\multicolumn{7}{c}{}\\
No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\
 \hline
  \midrule
 \endfirsthead

 No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\\hline    \midrule
 \endhead
14& (21,14,6,1;42)&	$x^2+y^3+z^7+w^{42}$&	1&1&$(10,<0>)$&14&I\\

\caption{Table for $m=42$} \label{tab-42} \end{longtable}

\begin{longtable}{p{12pt}|c|c|c|c|c|c|c}
\multicolumn{7}{c}{}\\
No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\
 \hline
  \midrule
 \endfirsthead

 No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\\hline    \midrule
 \endhead
 38&	(15,8,6,1;30)		&$x^2+y^3z+z^5+w^{30}$&		1&1&$(11,w_{2,1}^{-1})$	&50&I	\\
 50&	(15,10,4,1;30)		&$x^2+y^3+yz^5+w^{30}$&		1&1&$(9,w_{2,1}^{1})$	&38&I	\\
 \caption{Table for $m=30$} \label{tab-30} \end{longtable}

\begin{longtable}{p{12pt}|c|c|c|c|c|c|c}
\multicolumn{7}{c}{}\\
No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\
 \hline
  \midrule
 \endfirsthead
 No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\\hline    \midrule
 \endhead
13a	&(12,8,3,1;24)	&$x^2+y^3+xz^4+w^{24}$	&1	&1	&$(8,w_{3,1}^{-1})$	&20&I\\
13b	&(12,8,3,1;24)	&$x^2+y^3+z^8+w^{24}$	&2	&2	&$(12, w_{3,1}^{1})$	&13b&I\\
	&			&					&	&1	&$(8,w_{3,1}^{-1})$	&13b&I\\
20	&(9,8,6,1;24)	&$x^2z+y^3+z^4+w^{24}$	&1	&1	&$(12, w_{3,1}^{1})$	&13a&I\\
\caption{Table for $m=24$} \label{tab-24}
\end{longtable}

\begin{longtable}{p{12pt}|c|c|c|c|c|c|c}
\multicolumn{7}{c}{}\\
No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\
 \hline
  \midrule
 \endfirsthead
 No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\\hline    \midrule
 \endhead
 78	&(11,6,4,1;22)	&$x^2+y^3z+yz^4+w^{22}$	&1	&1	&$(10,w_{2,1}^{-1}\oplus w_{2,1}^{1})$	&78&II\\
\caption{Table for $m=22$} \label{tab-22}  \end{longtable}

\begin{longtable}{p{12pt}|c|c|c|c|c|c|c}
\multicolumn{7}{c}{}\\
No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\
 \hline
  \midrule
 \endfirsthead
 No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\\hline    \midrule
 \endhead
 9a	&(10,5,4,1;20)	&$x^2+xy^2+z^5+w^{20}$	&1	&1	&$(10,w_{5,1}^{-1})$		&9a&I\\
9b	&(10,5,4,1;20)	&$x^2+y^4+z^5+w^{20}$	&2	&2	&$(10,w_{5,1}^{-1})$		&9b&I\\
	&			&					&	&1	&$(10,w_{5,1}^{-1})$		&9b&I\\
 \caption{Table for $m=20$} \label{tab-20} \end{longtable}

\begin{longtable}{p{12pt}|c|c|c|c|c|c|c}
\multicolumn{7}{c}{}\\
No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\
 \hline
  \midrule
 \endfirsthead
 No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\\hline    \midrule
 \endhead
12a	&(9,6,2,1;18)	&$x^2+y^3+yz^6+w^{18}$		&2	&2	&$(11,w_{2,1}^1\oplus w_{3,1}^1)$		&39a&I\\
	&			&						&	&1	&$(6,v)$							&39a&I\\
12b	&(9,6,2,1;18)	&$x^2+y^3+z^9+w^{18}$		&3	&3	&$(14,v)$							&12b&I\\
	&			&						&	&1	&$(6,v)$							&12b&I\\
39a	&(9,5,3,1;18)	&$x^2+y^3z+z^6+w^{18}$		&2	&2	&$(14,v)$							&12a&I\\
	&			&						&	&1	&$(9,w_{2,1}^{-1}\oplus w_{3,1}^{-1})$	&12a&I\\
39b	&(9,5,3,1;18)	&$x^2+y^3z+xz^3+w^{18}$	&1	&1	&$(9,w_{2,1}^{-1}\oplus w_{3,1}^{-1})$	&60&I\\
60	&(7,6,4,1;18)	&$x^2z+y^3+yz^3+w^{18}$	&1	&1	&$(11,w_{2,1}^1\oplus w_{3,1}^1)$		&39b&I\\

\caption{Table for $m=18$} \label{tab-18}  \end{longtable}

\begin{longtable}{p{12pt}|c|c|c|c|c|c|c}
\multicolumn{7}{c}{}\\
No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\
 \hline
  \midrule
 \endfirsthead
 No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\\hline    \midrule
 \endhead
37a	&(8,4,3,1;16)	&$x^2+xy^2+yz^4+w^{16}$	&1	&1	&$(9,w_{2,3}^{5})$					&58&III\\
37b	&(8,4,3,1;16)	&$x^2+y^4+yz^4+w^{16}$		&2	&2	&$(11,w_{2,3}^{-5})$					&37b&IV\\
	&			&						&	&1	&$(9,w_{2,3}^{5})$					&37b&III\\
58	&(6,5,4,1;16)	&$x^2z+xy^2+z^4+w^{16}$	&1	&1	&$(11,w_{2,3}^{-5})$					&37a&IV\\
\caption{Table for $m=16$} \label{tab-16}  \end{longtable}

\begin{longtable}{p{12pt}|c|c|c|c|c|c|c}
\multicolumn{7}{c}{}\\
No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\
 \hline
  \midrule
 \endfirsthead
 No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\\hline    \midrule
 \endhead
11a	&(15,10,3,2;30)		&$x^2+y^3+xz^5+w^{15}$ 	&1	&1	&$(10,w_{3,1}^{-1}\oplus w_{3,1}^{1})$	&22a&II\\
11b	&(15,10,3,2;30)		&$x^2+y^3+z^{10}+w^{15}$		&1	&1	&$(10,w_{3,1}^{-1}\oplus w_{3,1}^{1})$	&11b&II\\
22a	&(6,5,3,1;15)		&$x^2z+y^3+z^5+w^{15}$		&1	&1	&$(10,w_{3,1}^{-1}\oplus w_{3,1}^{1})$	&11a&II\\
22b	&(6,5,3,1;15)		&$x^2z+y^3+xz^3+w^{15}$	&1	&1	&$(10,w_{3,1}^{-1}\oplus w_{3,1}^{1})$	&22b&II\\
 \caption{Table for $m=15$} \label{tab-15}
 \end{longtable}

\begin{longtable}{p{12pt}|c|c|c|c|c|c|c}

 \multicolumn{7}{c}{}\\
No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\
 \hline
  \midrule
 \endfirsthead
 No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\\hline    \midrule
 \endhead
40a	&(7,4,2,1;14)	&$x^2+y^3z+z^{7}+w^{14}$	&1	&1	&$(7,v\oplus w_{2,1}^{-1})$	&47	&III\\
40b	&(7,4,2,1;14)	&$x^2+y^3z+yz^{5}+w^{14}$	&2	&2	&$(13,v\oplus w_{2,1}^{1})$	&40b&II	\\
	&			&						&	&1	&$(7,v\oplus w_{2,1}^{-1})$	&40b	&III\\
47	&(21,14,4,3;42)	&$x^2+y^3+yz^{7}+w^{14}$	 &1	&1	&$(13,v\oplus w_{2,1}^{1})$	&40a&II	\\
 \caption{Table for $m=14$} \label{tab-14} \end{longtable}

\begin{longtable}{p{12pt}|c|c|c|c|c|c|c
\multicolumn{7}{c}{}\\
No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\
 \hline
  \midrule
 \endfirsthead
 No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\\hline    \midrule
 \endhead
6a	&(5,2,2,1;10)	&$x^2+y^4z+z^5+w^{10}$		&2	&2	&$(8,w_{5,1}^{-1}\oplus 2w_{2,1}^1)$	&36a&I	\\
	&			&						&	&1	&$(6,u\oplus v)$					&36a&II	\\
6b	&(5,2,2,1;10)	&$x^2+y^5+z^5+w^{10}$		&5	&5	&$(14,u\oplus v)$					&6b	&II\\
	&			&						&	&1	&$(6,u\oplus v)$					&6b	&II\\
6c	&(5,2,2,1;10)	&$x^2+y^4z+yz^4+w^{10}$	&3	&3	&$(14,u\oplus v)$					&6c	&II\\
	&			&						&	&1	&$(6,u\oplus v)$					&6c	&II\\
11a	&(15,10,3,2;30)	&$x^2+y^3+z^{10}+yw^{10}$	&2	&2	&$(17,w_{2,1}^{1})$				&42a&I	\\
	&			&						&	&1	&$(10,v\oplus v)$					&42a&II	\\
11b	&(15,10,3,2;30)&$x^2+y^3+z^{10}+w^{15}$		&1	&1	&$(10,v\oplus v)$					&11b&II	\\

36a	&(10,5,3,2;20)	&$x^2+y^4+yz^5+w^{10}$		&2	&2	&$(14,u\oplus v)$				&6a	&II\\
	&			&						&	&1	&$(12,w_{5,1}^{-1}\oplus 2w_{2,1}^{-1})$					&6a&I	\\
36b	&(10,5,3,2;20)	&$x^2+xy^2+yz^5+w^{10}$	&1	&1	&$(12,w_{5,1}^{-1}\oplus 2w_{2,1}^{-1})$					&63&I	\\
42a	&(5,3,1,1;10)	&$x^2+y^3w+z^{10}+w^{10}$	&2	&2	&$(10,v\oplus v)$					&11a&II	\\
	&			&						&	&1	&$(3,w_{2,1}^{-1})$				&11a&I	\\
42b	&(5,3,1,1;10)	&$x^2+y^3z+xz^5+w^{10}$	&1	&1	&$(3,w_{2,1}^{-1})$				&68&I	\\
42c	&(5,3,1,1;10)	&$x^2+y^3z+yz^7+w^{10}$	&4	&4	&$(17,w_{2,1}^{1})$				&42c&I	\\
	&			&						&	&2	&$(10,v\oplus v)$					&42c&II\\
	&			&						&	&1	&$(3,w_{2,1}^{-1})$				&42c&I	\\
63	&(4,3,2,1;10)	&$x^2z+y^2x+z^5+w^{10}$	&1	&1	&$(8,w_{5,1}^{-1}\oplus 2w_{2,1}^1)$	&36b&I	\\
68	&(13,10,4,3;30)	&$x^2z+y^3+yz^5+w^{10}$	&1	&1	&$(17,w_{2,1}^{1})$				&42b&I	\\
\caption{Table for $m=10$} \label{tab-10}  \end{longtable}

\begin{longtable}{p{12pt}|c|c|c|c|c|c|c}
\multicolumn{7}{c}{}\\
No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\
 \hline
  \midrule
 \endfirsthead
 No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\\hline    \midrule
 \endhead
 12a	&(9,6,2,1;18)	&$x^2+y^3+z^9+xw^9$		&3	&3	&$(16,w_{3,1}^{-1})$				&25a&I	\\
	&			&						&	&1	&$(4,w_{3,1}^1)$				&25a&I	\\
12b	&(9,6,2,1;18)	&$x^2+y^3+z^9+yw^{12}$		&1	&1	&$(4,w_{3,1}^1)$				&43a&I	\\
12c	&(9,6,2,1;18)	&$x^2+y^3+z^9+w^{18}$		&3	&3	&$(16,w_{3,1}^{-1})$				&12c&I	\\
	&			&						&	&1	&$(4,w_{3,1}^1)$				&12c&I	\\
18a	&(3,3,2,1;9)	&$x^3+y^3+xz^3+w^9$		&3	&3	&$(12,w_{3,1}^1\oplus w_{3,2}^1)$		&18a&IV	\\
	&			&						&	&1	&$(8,w_{3,1}^{-1}\oplus w_{3,2}^{-1})$	&18a&III	\\
18b	&(3,3,2,1;9)	&$x^3+xy^2+yz^3+w^9$		&2	&2	&$(12,w_{3,1}^1\oplus w_{3,2}^1)$		&18b&IV	\\
	&			&						&	&1	&$(8,w_{3,1}^{-1}\oplus w_{3,2}^{-1})$	&18b&III	\\
25a	&(4,3,1,1;9)	&$x^2w+y^3+z^9+w^9$		&3	&3	&$(16,w_{3,1}^{-1})$				&12a&I	\\
	&			&						&	&1	&$(4,w_{3,1}^1)$				&12a&I	\\
25b	&(4,3,1,1;9)	&$x^2w+y^3+z^9+yw^6$		&1	&1	&$(4,w_{3,1}^1)$				&43b&I	\\
25c	&(4,3,1,1;9)	&$x^2w+y^3+z^9+xw^5$		&3	&3	&$(16,w_{3,1}^{-1})$				&25a&I	\\
	&			&						&	&1	&$(4,w_{3,1}^1)$				&25a&I	\\
43a	&(18,11,4,3;36)	&$x^2+y^3w+z^9+w^{12}$	&1	&1	&$(16,w_{3,1}^{-1})$				&12b&I	\\
43b	&(18,11,4,3;36)	&$x^2+y^3w+z^9+xw^6$		&1	&1	&$(16,w_{3,1}^{-1})$				&25b&I	\\
\caption{Table for $m=9$} \label{tab-9}  \end{longtable}

\begin{longtable}{p{12pt}|c|c|c|c|c|c|c}
\multicolumn{7}{c}{}\\
No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\
 \hline
  \midrule
 \endfirsthead
 No. &Weights&Polynomial&SL/J&G/J&$(r,q)$&BHK dual&Method\\\hline    \midrule
 \endhead
2a	&(4,3,3,2;12)	&$x^3+y^3z+z^4+w^6$	&3	&3	&$(16,v\oplus w_{3,1}^1)$				&3a&I	\\
	&			&					&	&1	&$(10,4w_{3,1}^1)$				&3a&II	\\
2b	&(4,3,3,2;12)	&$x^3+y^3z+yz^3+w^6$	&1	&1	&$(10,4w_{3,1}^1)$				&2b	&II\\
2c	&(4,3,3,2;12)	&$x^3+y^4+z^4+w^6$	&2	&2	&$(10,4w_{3,1}^1)$				&2c&II	\\
	&			&					&	&1	&$(10,4w_{3,1}^1)$				&2c&II	\\

3a	&(2,2,1,1;6)	&$x^3+y^3+yz^4+w^6$	&3	&3	&$(10,4w_{3,1}^1)$				&2a	&II\\
	&			&					&	&1	&$(4,v\oplus w_{3,1}^{-1})$			&2a	&I\\
3b	&(2,2,1,1;6)	&$x^2y+y^3+z^6+w^6$	&6	&6	&$(19,w_{2,1}^{-1})$				&5a	&I\\
	&			&					&	&3	&$(16,v\oplus w_{3,1}^1)$				&5a&I	\\
	&			&					&	&2	&$(9,3w_{2,1}^{-1}\oplus 2w_{3,1}^{-1})$	&5a&I	\\
	&			&					&	&1	&$(4,v\oplus w_{3,1}^{-1})$			&5a	&I\\
3c	&(2,2,1,1;6)	&$x^3+xy^2+yz^4+w^6$	&1	&1	&$(4,v\oplus w_{3,1}^{-1})$			&57	&I\\
3d	&(2,2,1,1;6)	&$x^3+y^3+z^6+w^6$	&9	&9	&$(16,v\oplus w_{3,1}^1)$				&3d&I	\\
	&			&					&	&3	&$(16,v\oplus w_{3,1}^1)$				&3d	&I\\
	&			&					&	&3	&$(10,4w_{3,1}^1)$				&3d&II	\\
	&			&					&	&3	&$(10,4w_{3,1}^1)$				&3d&II	\\
	&			&					&	&3	&$(4,v\oplus w_{3,1}^{-1})$			&3d	&I\\
	&			&					&	&1	&$(4,v\oplus w_{3,1}^{-1})$			&3d	&I\\
3e	&(2,2,1,1;6)	&$x^2y+xy^2+z^6+w^6$	&3	&3	&$(16,v\oplus w_{3,1}^1)$				&3e	&I\\
	&			&					&	&1	&$(4,v\oplus w_{3,1}^{-1})$			&3e	&I\\

5a	&(3,1,1,1;6)	&$x^2+xy^3+z^6+w^6$	&6	&6	&$(16,v\oplus w_{3,1}^1)$				&3b&I	\\
	&			&					&	&3	&$(11,3w_{2,1}^1\oplus 2w_{3,1}^{1})$	&3b&I	\\
	&			&					&	&2	&$(4,v\oplus w_{3,1}^{-1})$			&3b	&I\\
	&			&					&	&1	&$(1,w_{2,1}^1)$				&3b&I	\\
5b	&(3,1,1,1;6)	&$x^2+y^5w+z^6+w^6$	&2	&2	&$(8,6w_{2,1}^{-1})$				&29	&II\\
	&			&					&	&1	&$(1,w_{2,1}^1)$				&29&I	\\
5c	&(3,1,1,1;6)	&$x^2+xy^3+yz^5+w^6$	&1	&1	&$(1,w_{2,1}^1)$				&56	&I\\
5d	&(3,1,1,1;6)	&$x^2+y^6+z^5w+zw^5$	&8	&8	&$(19,w_{2,1}^{-1})$				&5d	&I\\
	&			&					&	&4	&$(12,6w_{2,1}^1)$				&5d&II	\\
	&			&					&	&2	&$(8,6w_{2,1}^{-1})$				&5d	&II\\
	&			&					&	&1	&$(1,w_{2,1}^1)$				&5d&I	\\
5e	&(3,1,1,1;6)	&$x^2+y^6+z^6+w^6$	&12	&12	&$(19,w_{2,1}^{-1})$				&5e	&I\\
	&			&					&	&6	&$(12,6w_{2,1}^1)$				&5e	&II\\
	&			&					&	&4	&$(9,3w_{2,1}^{-1}\oplus 2w_{3,1}^{-1})$	&5e&I	\\
	&			&					&	&3	&$(11,3w_{2,1}^1\oplus 2w_{3,1}^{1})$	&5e	&I\\
	&			&					&	&2	&$(8,6w_{2,1}^{-1})$				&5e&II	\\
	&			&					&	&1	&$(1,w_{2,1}^1)$				&5e	&I\\

8a	&(6,3,2,1;12)	&$x^2+y^4+z^6+xw^6$	&2	&2	&$(14,2w_{2,1}^{-1}\oplus w_{3,1}^1)$	&23	&I\\
	&			&					&	&1	&$(6,2w_{2,1}^1\oplus w_{3,1}^{-1})$	&23	&I\\
8b	&(6,3,2,1;12)	&$x^2+y^4+z^6+yw^{9}$	&2	&2	&$(10,v\oplus 4w_{2,1}^{-1})$			&33a&I	\\
	&			&					&	&1	&$(6,2w_{2,1}^1\oplus w_{3,1}^{-1})$	&33a	&I\\
8c	&(6,3,2,1;12)	&$x^2+xy^2+z^6+yw^{9}$	&1	&1	&$(6,2w_{2,1}^1\oplus w_{3,1}^{-1})$	&70	&I\\
8d	&(6,3,2,1;12)	&$x^2+y^4+z^6+w^{12}$	&4	&4	&$(14,2w_{2,1}^{-1}\oplus w_{3,1}^1)$	&8d&I	\\
	&			&					&	&2	&$(14,2w_{2,1}^{-1}\oplus w_{3,1}^1)$	&8d	&I\\
	&			&					&	&2	&$(6,2w_{2,1}^1\oplus w_{3,1}^{-1})$	&8d&I	\\
	&			&					&	&2	&$(10,v\oplus 4w_{2,1}^{-1})$			&8d&II	\\
	&			&					&	&1	&$(6,2w_{2,1}^1\oplus w_{3,1}^{-1})$	&8d	&I\\
8e	&(6,3,2,1;12)	&$x^2+xy^2+z^6+w^{12}$	&2	&2	&$(14,2w_{2,1}^{-1}\oplus w_{3,1}^1)$	&8e&I	\\
	&			&					&	&1	&$(6,2w_{2,1}^1\oplus w_{3,1}^{-1})$	&8e	&I\\

23	&(5,3,2,2;12)	&$x^2w+y^4+z^6+w^6$	&2	&2	&$(14,2w_{2,1}^{-1}\oplus w_{3,1}^1)$	&8a	&I\\
	&			&					&	&1	&$(6,2w_{2,1}^1\oplus w_{3,1}^{-1})$	&8a	&I\\
29	&(15,6,5,4;30)	&$x^2+y^5+z^6+yw^6$	&2	&2	&$(19,w_{2,1}^{-1})$				&5b&I	\\
	&			&					&	&1	&$(12,6w_{2,1}^1)$				&5b&II	\\

33a	&(9,4,3,2;18)	&$x^2+y^4w+z^6+w^{9}$	&2	&2	&$(14,2w_{2,1}^{-1}\oplus w_{3,1}^1)$	&8b	&I\\
	&			&					&	&1	&$(10,v\oplus 4w_{2,1}^{-1})$			&8b	&II\\
33b	&(9,4,3,2;18)	&$x^2+y^4w+z^6+yw^{7}$&1	&1	&$(10,v\oplus 4w_{2,1}^{-1})$			&33b&II	\\
56	&(11,8,6,5;30)	&$x^2y+y^3z+z^5+w^6$	&1	&1	&$(19,w_{2,1}^{-1})$				&5c	&I\\
57	&(9,6,5,4;24)	&$x^2y+y^4+xz^3+w^6$	&1	&1	&$(16,v\oplus w_{3,1}^1)$				&3c	&I\\
70	&(8,5,3,2;18)	&$x^2w+xy^2+z^6+w^{9}$&1	&1	&$(14,2w_{2,1}^{-1}\oplus w_{3,1}^1)$	&8c	&I\\
\caption{Table for $m=6$} \label{tab-6}    \end{longtable}

redpajama_set_name: RedPajamaArXiv


Good communication skill is an essential aspect of business. Being able to communicate well to clients and employees can make a big difference on how you are perceived as a company and as a business owner. Using the proper format in creating a business letter, whether it be a formal or an informal one, will allow you to broadcast your ideas clearly and accurately. This ensures that the recipient of your letter understands your plans and thoughts.
When you are trying to bid for a customer's purchase of your service or product, a quotation is typically given to give the client a description of the specific goods and services that are provided as well as its corresponding costs. A quotation, in business, can also be referred to as a bid, an estimate, or a tender and proposal.
Provide salutations and a formal greeting for your quotation letter. Because you are constructing a formal business letter, you have to use proper titles that include a Mr., Ms., Atty., etc. Be sure to use proper titling as some people are very sensitive to it. For example, do not use Mrs. on a Ms. or a Mr. on a Dr.
Provide a refresher or a background information to let your client know of the purpose of the letter. Give a brief summary of the quotation or the proposal in the first paragraph to help your client have a quick idea about it and to help him understand the proposal better.
Provide clear details that highlight the goal of your letter. Do not add unnecessary details or an overload of information. Try to focus on the basic terms of the proposal for your client to understand.
End your letter by providing words of appreciation. Thank the person for taking the time to consider you and your company and encourage him to contact you if he has any questions. End your letter with a valediction such as "Sincerely" or "Respectfully" with your name appearing beneath the valediction.
Include any type of supporting documents to your letter by denoting them into your letter on the left side of you signature. Supporting documents may include proposals or quotation forms that outline the cost and expenses of the materials and the services that the client is needing.
You can also view our Sample Letters, Business Letter Format Samples, or Project Quote Samples to give you an idea on how to properly and effectively create your own quotation letters and business letters.


A drift is an urban game developed by game designer/poet Thom Kiraly and artist/curator Sebastian Quack . It's based on A Drift, a format originally developed by Thom with Mikael Vesavuori and Niklas Persson and featured at several urban play festivals around 2014. It is a technique with a wide range of historical influences and traditions. Drift Club is a group of people who organize drifts and share experiences picked up while drifting. So far, we've drifted through cities all over Europe, in all types of weather and all times of day. There's a lot more to try out. Anyone can join!


text:
Q: Group members cannot upload files I have an interesting problem involving group permissions and possibly the setgid bit.
A set of directories is set as the following:
drwxrwsr-x  2 smith web-ww2 18 Oct  4 12:05 grids
drwxrwsr-x 16 smith web-ww2 16 Oct  4 12:05 jpgs

Here is an example of the files in the grids directory:
-rwxrwxr-x 1 smith web-ww2 393K Feb 24  2016 grid2003.zip
-rwxrwxr-x 1 smith web-ww2 366K Feb 24  2016 grid2004.zip

The owner, smith, is able to upload files in both directories. Members of group web-ww2 however cannot. All users are using WinSCP to upload files. No extended attributes are set.
Any idea what would be keeping the members of web-ww2 from uploading files within those directories?

A: Turns out this was a PEBKAC problem.
The users were trying to upload to those sub-directories in a different location than where they existed.
Once they were pointed to the proper place they were able to upload with no problem.

redpajama_set_name: RedPajamaStackExchange


text:
Music » Music Lead
Sondre Lerche embraces pop music without boundaries
Sondre Lerche with Fancy Colors
8 :30 p.m. Wednesday, Sept. 12 | Beachland Ballroom, 15711 Waterloo Rd.
Tickets: $17, $15 in advance | 216-383-1124 | beachlandballroom.com
For many people, turning 30 represents the inevitable transition into middle-age. It's often a difficult adjustment, especially for those who aren't entirely satisfied with what they've accomplished so far. Sondre Lerche, who turned 30 this week, says he's looking forward to the next decade. In fact, he celebrated his birthday on September 5 by kicking off his fall tour.
"I guess the tour is a desperate attempt to prolong the celebration, since it's going to last for an entire month," laughs the Norwegian singer-songwriter. "I feel great about turning 30. I've been counting the years, and there's no way outside of it. I feel it's a privilege to do what I do and to realize I have been doing it for 10 to 12 years. It feels good to be able to look back a little and discover songs that I've forgotten about and bring them back into the repertoire."
The tour coincides with reissues of Lerche's first four albums and the release of his first official live record, Bootlegs. Not bad for a guy who struggled with music lessons when he was younger. "I signed up for guitar lessons, and I thought I could learn what I wanted," he recalls. "It was classical guitar, and it was not for me. I kept trying to do that for a couple of years, and I was terrible at it. I wish now that I could impress people by playing some classical pieces, but I remember nothing.
"I'm thankful that I had a good teacher who eventually realized it was a dead-end road, so he taught me Brazilian music and bossa nova, and that became a way into the music that I now appreciate."
Influenced by exotic Latin American music, Lerche started writing songs in his early teens; growing up in Norway, his influences were much more varied than they might have been if he had been born in the U.S., where radio stations adhere more strictly to formats. Living on what he calls the "outskirts of popular culture," he was introduced to a wild mix of music that has subsequently informed his pop sensibilities.
"I think I was excited and influenced by the stuff my mom would play in the car, which was the soundtrack to Grease and bands like Wham!," he says. "My sister was going through the goth thing, so she was playing the Cure and Depeche Mode. My brother was playing A-ha and the Pet Shop Boys. And by the time the '90s started, I had access to MTV, and they were playing Nirvana and Elton John side by side. None of the boundaries that I see clearly now existed. I was just into pop music, and I realized it could also be jazz and grunge and electronic."
That mix comes across in his own music, which recalls everything from Fleetwood Mac to Erasure to Elliott Smith. It didn't take long for someone to notice Lerche; he scored a record deal when he was 19. "I feel lucky, because when I released my first record, I was signed to a major label, and it was right before everything went to hell for the major labels," he says. "I benefited from the last peak of that world, when they had a lot of resources and money to build my career. I don't know that I could have lived up to their expectations, but we were never to find out, because everything fell apart."
Everything fell apart for the record companies, but not for Lerche. After parting ways with Astralwerks, he rebounded with the soundtrack to the indie film Dan in Real Life and a tour with Elvis Costello. Last year's self-titled album, the first record on Lerche's own Mona imprint, is a departure of sorts for the singer-songwriter, who usually takes his time making records. Everything on Sondre Lerche was written and recorded rather quickly. So songs like the brisk "Private Caller" and the folksy "Living Dangerously" sound a little rougher than his past work.
"It was a record that I didn't prepare a lot for it," notes Lerche. "When I had the songs, there was no preparation — we just went in the studio. I gave us two weeks to record it. If it didn't go well, I wouldn't have released it, but it went really well. That limitation sparked a certain energy and the desire to strip things down and make do with less in a sense. It was energizing."
But Lerche says he probably won't take the same approach the next time around. "I've been looking back at the different periods of the last 10 years, and what I was really trying to do, and what I succeeded at, and what I still haven't achieved artistically," he says. "I have some songs I want to pursue, so we'll see. I probably won't put the same limitation on this one. It might take more than two weeks to record."


Lewis Hamilton is Formula One World Champion. Barack Obama is President of the USA. Blur are getting back together. All is well. But born misanthropes that we are, Crave is done surfing the last week's wave of positivity and is hitting the sugarcrash. Hard luck then for the Fujifilm FinePix J150w, a perfectly serviceable 10-megapixel point-and-shoot camera that should have clocked what kind of mood we were in and sidled back out before we noticed it.
It has a boxy, simple design with a subtle brushed effect on the face. We like the angled blades of the lens cover, making the J150 look a lot like the Ricoh R7 -- remember that? Now that was a camera.
The screen is a substantial 76mm (3-inch) with a resolution of roughly 230,000 dots. For some reason Fujifilm informs us that the screen gives 97 per cent coverage: eh? What's the point of a live view screen that leaves any of the frame out? That 3 per cent may contain something really interesting.
Scene modes are the usual: portrait, landscape, beach, sport, fireworks, blah blah blah. Crave hates fireworks. Oooh. Aaah. Repeat. Where are the bonkers scene modes? Baby mode: yeah, whatever. It's no firefighting mode, is it?
Video is the bog-standard 640x480-pixel VGA resolution at 30 frames per second, with the option to dial down to 320x240. It's only monaural sound, and you can't zoom during movie recording. It's not high definition, is it?
The J150w does pack a creditable 5x optical zoom lens, with a 28mm wide angle, equivalent to a 35mm camera. Oh all right, that is impressive.
The Fujifilm FinePix J150w is available now for just under £100. We're off to scowl at anyone who mentions Christmas. We really hate Christmas.


Douek, M; Winehouse, J; Deng, J; Holz, K; Lakhani, SR; Lees, W; Davidson, T; ... Baum, M; + view all Douek, M; Winehouse, J; Deng, J; Holz, K; Lakhani, SR; Lees, W; Davidson, T; Taylor, I; Baum, M; - view fewer (1998) Detection of breast cancer recurrence by contrast-enhanced Doppler ultrasound: implications for breast biopsy. In: (pp. p. 628).


text:
RRS Sir David Attenborough
Posted on 26 August 2022 in Company News
The RRS Sir David Attenborough was docked in Belfast for the first time this week and representatives from Todd Navigation, William Todd (MD) and Paul Esler (GM), went along to pay it a visit.
The vessel, the UK's most advanced polar research ship, owned by the Natural Environmental Research Council and operated by the British Antarctic Survey, has recently been to the Antarctic carrying out trials, testing sea ice and looking at the effects of climate change, all while using nautical charts and publications, as well as digital navigation supplied by Todd Navigation.
International senior scientific advisers are using the ship as a platform to issue a statement about the urgent need to address the climate crisis. They want to see a concerted drive to develop, and use, the technologies that will keep global temperature rise to a limited 1.5C and underpin the net zero economies of the future.
These technologies include better ways of creating, storing and using low-emissions, including improving semiconductors, batteries and low-emitting fuel production, as well as work on heating and cooling, and carbon capture and storage.
The vessel was in Belfast to pick up new crew members ahead of carrying out further scientific studies and sea trials around the UK coastline before heading on to Frederikshavn in Denmark.
We here at Todd Navigation would like to wish the vessel and all its crew safe passage on all its future voyages and the very best in carrying out its vital work to help us all make the World a better place now and for generations to come.
Follow @toddnavigation
Leisure Boating
About Todd Navigation.
We are a worldwide supplier of nautical charts from a variety of publishers including Admiralty, Imray, Navionics and NV Charts.
What type of customer are you?
Leisure Customer
Ship Chandler
Ship Owner
Shipping Agent


Teaching approach: I always believe to follow the basic. How much the foundation will be strong the future is going to be brighter and brighter for a leaner. I always try to make my students to think in that language which they are learning. Rather than giving only few phrases to memorize I always try to make them think in that language so that they can prepare their own sentences. Therefore please learn the language not the phrases!


text:
(5036) Tuttle ist ein Asteroid des Hauptgürtels, der am 31. Oktober 1991 von den japanischen Astronomen Seiji Ueda und Hiroshi Kaneda am Observatorium von Kushiro (IAU-Code 399) entdeckt wurde.

Der Asteroid ist nach dem amerikanischen Astronomen und Kometenentdecker Horace Parnell Tuttle benannt.

Siehe auch
 Liste der Asteroiden

Weblinks
 Asteroid Tuttle: Discovery Circumstances gemäß dem Minor Planet Center der Internationalen Astronomischen Union bei dem Harvard-Smithsonian Center for Astrophysics, Cambridge, USA
 Asteroid Tuttle in der Small-Body Database des Jet Propulsion Laboratory der NASA bei dem California Institute of Technology, Pasadena, USA

Hauptgürtelasteroid unter 50 km Durchmesser

redpajama_set_name: RedPajamaWikipedia


text:
Q: Jooq generates classes for tables without fields responsible for columns in tables For example all generated classes for tables look like that:
SuppressWarnings({"all", "unchecked", "rawtypes"})
public class TableName extends TableImpl<TableNameRecord> {

    private static final long serialVersionUID = 1L;

    public static final TableName TABLE_NAME = new TableName();

    @Override
    public Class<TableNameRecord> getRecordType() {
        return TableNameRecord.class;
    }
   ... (more contructors)
}

Maven:
Code-generation:
<plugin>
    <groupId>org.jooq</groupId>
    <artifactId>jooq-codegen-maven</artifactId>
    <version>3.16.1</version>
    <executions>
        <execution>
            <phase>generate-sources</phase>
            <goals>
                <goal>generate</goal>
            </goals>
        </execution>
    </executions>
    <dependencies>
        <dependency>
            <groupId>org.postgresql</groupId>
            <artifactId>postgresql</artifactId>
            <version>42.3.1</version>
        </dependency>
    </dependencies>
    <configuration>
        <jdbc>
            <driver>org.postgresql.Driver</driver>
            <url>'db-url'</url>
            <user>'user-name'</user>
            <password>'pswd'</password>
        </jdbc>
        <generator>
            <database>
                <name>org.jooq.meta.postgres.PostgresDatabase</name>
                <includes>.*</includes>
                <inputSchema>'my-schema'</inputSchema>
                <excludes/>
            </database>
            <generate>
                <records>true</records>
            </generate>
            <target>
                <packageName>'package-name'</packageName>
                <directory>target/generated-sources/jooq</directory>
            </target>
        </generator>
    </configuration>
</plugin>

Dependencies:
jooq, jooq-meta, jooq-codegen
version: 3.16.1
What's wrong?

A: I added this
<database>
      <includeTables>true</includeTables>
</database>

to configuration and it solved my problem


The numbers on my profile don't look right?
You would've seen a change in some of your profile stats, namely - 'Total Profile views', 'Total Artwork Views' and 'Number of followers' in the recent past.
Your own profile and artwork views are not being counted now.
The data display is live (i.e. if someone favs your artwork, you can see the change in your dashboard immediately).
Comparison of # of artworks uploaded and sold each month to show the co-relation between uploads and sales.
This change in numbers happened early Jan-19. Given the dynamic nature of these numbers (and the magnitude), we had to hold back some of the historic data (i.e. profile and artwork views from before Nov-18 are currently not visible). We have been running and monitoring the implemented system for stability, precision and accuracy. This means we're trialling out the best ways of keeping and displaying these numbers and making sure in the long run nothing breaks and the numbers are helpful and accurate.
We would've preferred to test this in a way where you, the artist, only gets the finished product so there's no confusion around this. Although, given the complexity of the project, it's near impossible to test out all the factors involved without testing it with live data (i.e. real time interaction on real profiles). This is very helpful but at the same time means if something doesn't work - you get to see it. This isn't a bad this - this only means we've found something that doesn't work and will make the system better in the long run.
2019 is the year of change at Bluethumb and a lot of projects are churning in the pipeline. You'll see these coming one after the other. A few big ones are directly related to the artist experience (i.e. dashboard, artwork upload, artwork management, profile and artwork listings). We have some more goodies which will make your experience on Bluethumb a bit smoother.


text:
This is what happens when you download everything Facebook knows about you
Every message, every event and just about everyone you have ever spoken to — on Facebook or elsewhere
The Cambridge Analytica scandal, and subsequent apology by Facebook chief executive Mark Zuckerberg, have drawn attention to a little-known feature of the social network — downloading everything the social media network knows about you.
Available via a link on the front page of Facebook's settings menu, downloading your data was once known as an easy way to grab all of the photos and videos you have ever posted to Facebook.
But in the wake of reports that Facebook data on 50 million users fell into the wrong hands — and fresh concerns over how much access the social network has to your data and that of your friends and family — understanding what the company knows is more relevant than ever.
Head to the settings page of your Facebook account, or click here, then click 'Download a copy'.
Click the green Download Archive button
Enter your password when asked
It took around 25 minutes for our archive to be ready for downloadGearBrain
After these steps, you receive an email from Facebook confirming the request for your data. Then, around 25 minutes later or less (or more), a second email arrives to say your data is ready to download. This morning I did exactly that, and downloaded a 920MB folder of everything Facebook had about me. Here's what I found out:
Clicking on the index.html file opens up a simple way to navigate through the data you have just downloaded, instead of wading through folders of images, messages, video and other data.
A list of everyone you have ever known
First, I headed to the Contact Info tab, which lists — in no particular order — just about everyone I have ever digitally communicated with since I was about 17 years old. In this list there are friends, family members and work colleagues, plus people from school, college and university, and some names I simply don't remember.
To be clear, this is not a list of my Facebook friends. This is a list of everyone in my life whom Facebook knows the name of, along with their phone number and email address if Facebook knows it — which is almost always does — because you wouldn't have a contact in your phone book without these details.
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Many of the phone numbers and email addresses have come from the work email account I used at my former employer, and I would say everyone who ever contacted me on that address is listed here. Most of these email addresses are for work and so can probably be found online if you look hard enough. It is still discomforting to see this huge list and know it came from Facebook.
Other details have come from Facebook itself, such as email addresses of people I once knew and added into my smartphone. One, the mother of an ex-girlfriend, I added to my iPhone a couple of years ago and have since deleted. Facebook, however, has kept it.
Facebook's justification for snooping around like this is the same as its defense this week for recording Android call and text message metadata — to help you connect with people.
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It shouldn't feel weird to see these phone numbers and email addresses. They are not complete strangers and everyone here has communicated with me in some form or another over the last decade or so. But there are people on this list who are no longer in my life — and in some cases no longer alive. Giving Facebook users the ability to rediscover phones numbers of people they had deleted — or people they had been asked to delete, like after a breakup — is clearly a problem.
The social network says "you are always in control of the information you share with Facebook," but how many of the people on this list know I still have their phone number and/or email address?
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'Advertisers with your contact details'
Because advertising is how Facebook earns its money, I went to the 'Ads' tab next and scrolled down to the ominously-titled 'Advertisers with your contact details.'
Here I found a list of 19 companies and websites, most of which I know I've given my details to over the years. The BBC is here, along with Airbnb, Uber, Deliveroo and The Economist. EBay.co.uk is there too, which is fine. But more confusing is why eBay Germany and eBay Canada are also there.
Clicking on the 'Profile' tab shows all the email addresses I have used to log into Facebook, the date (to the exact minute) when I joined, information from my public profile like my gender, location and hometown, my education and workplaces. Facebook also keeps a list of 'previous relationships', based on fellow users you've noted as such on the site. However, I notice that if that person has since deleted their profile, they do not appear here.
Facebook's data collection is prompting some users to consider leavingiStock
The 'Timeline' tab is a complete history of anything ever written on your wall, by you or your friends. For me, this was mostly uninteresting status updates and an annual smattering of birthday wishes. This section is fairly innocent, but serves as a reminder of how, between about 2007 and 2010, we all used to post statuses in the third person, and that subjects which would now be sent in a private message were plastered over each others' walls instead.
The 'Photos' and 'Videos' section are self-explanatory. They don't include anything you are tagged in, but every image you have posted yourself, plus any comments they received.
'Friends' is a list of every friend you have on Facebook, ordered by the date you connected. Below this is a section for friend requests you have received but not acted on, requests you have declined, friends you have removed, and people who follow you but are not your friends.
Every Facebook conversation you have ever had
Next up is the Messages tab, which is a record of every private conversation you have ever had on Facebook. As with the rest of this data haul, it is difficult to navigate and is in no real order. I found it easier to view this data as a list of local files rather than through the html browser; that way, I can order by file size and expect to see my closest friends (those who I communicate with the most) at the top.
Conversations with people you are no longer friends with — and with those who no longer have a Facebook account — are here but their names are replaced with 'Facebook User', so finding specific chat logs is tricky. People you have spoken to, but have never been friends with, are here too but they also appear as 'Facebook User.' It turns out that unfriending someone is not enough to remove your chat history with them from Facebook's server.
The 'Events' tag is where Facebook lists every event you have ever attended, and every event you have been invited to. Even if you didn't interact with the event in any way, it's in this list. I don't really have a problem with this, but it seems strange that this data still exists years after the events — events I didn't attend — took place.
A couple of years ago, this data probably wouldn't make many Facebook users feel uncomfortable. Friends lists and chat logs are made by us, after all. But in the wake of the Cambridge Analytica scandal, and amid the #DeleteFacebook movement, seeing this data could be enough to make at least some of Facebook's two billion members think twice about continuing to use it.
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