Australians love outdoor entertaining, and more of us are embracing the idea of the 'outdoor room' as an extension of the home. Much like indoor rooms, a ceiling fan in exterior areas creates a relaxed ambience while keeping the area cool on those scorching days and humid nights.
An added bonus is that the breeze created by the ceiling fan deters flying insects. You can install a ceiling fan just about anywhere; such as covered decks, porches, gazebos and verandas all make great spots for a ceiling fan.
Here's a few tips for hassle free installation of exterior ceiling fans.
All our fans display symbols indicating their use; Indoor Only, Alfresco or Indoor, or Coastal use. Fans for Alfresco areas are suitable provided the fan is fully undercover with a minimum of 2 walls. The fan must be positioned in a location protected from the water, wind, dust and salt.
The fan should never be installed under an open pergola where rain or moisture will find its way into the motor. Ceiling fans not installed undercover will be subject to weather and moisture, and most likely will not have been installed by an electrician.
An exterior fan needs enough solid surfaces to allow the air to bounce off the walls and actually create a breeze. When the area has at least three walls and a ceiling, the fan can provide enough breeze to assist with personal cooling, much like a mild breeze on a summers day.
Stainless steel metal blade fans are a great choice for exterior applications as they are less likely to rust. The pitch of the blade also moves air more effectively in a large space.
Whilst considered a good choice for exterior use, it is highly recommended that ceiling fans installed in exterior situations should be thoroughly treated before installation with neutral oil such as 'Steel Kleen' and this should be repeated at least 4 times a year. If serious tea staining or corrosion occurs, use 'Steel Power' designed for use on stainless steel and follow with a 'Steel Kleen' oil treatment.
The Altitude Eco & Revolution Ceiling Fans are both IP rated to IP44 and are specifically designed for use outdoors.
It is widely considered that timber bladed ceiling fans are not suitable for exterior use. This is not necessarily so, as a timber bladed ceiling fan with an Aluminium motor, such as the Futura and Concept, are both suitable for exterior use.
However, you must ensure the fan is fully undercover, and ynote that the wooden blades will need to be cleaned more regularly and may need to be replaced.
What are suitable locations for ceiling fans?
Recently we introduced 3 new categories for our fans, which grade their suitability in different locations.
The 3 categories are Indoor, Alfresco, and Coastal. All of our fans fall into one of these categories.
These can easily be identified by the symbols below.
Indoor Use Only: Suitable for indoor use only.
Indoor & Outdoor: Suitable for indoor and outdoor areas provided the fan is undercover with a minimum of 2 walls.
Fan must be positioned in a location protected from water, wind, dust, and salt.
Coastal: Suitable for indoor, alfresco and coastal areas provided the fan is undercover with a minimum of 1 wall.
Fan must be positioned in a location protected from water, wind, and dust.


text:
Federal Liberals to send Dave Birdi to the campaign trail
Dave Birdi has been selected as the
Liberal Party of Canada candidate for
Skeena-Bulkley Valley
There is one more name to add to the list of candidates for the upcoming Federal Election, with the Federal Liberals announcing that Dave Birdi will be contesting the Skeena-Bulkley Valley seat on behalf of the "Trudeau Team".
Hailing from Fort St. James, Mr. Birdi is President of the Fort St. James TV and Radio Society, as well as having served on the town council for over ten years.
His resume includes six years as Director of Community Futures Stuart Nechako branch, with his biography noting that he is currently the Economic Development Officer of Binche Whut'enne.
Mr. Birdi introduced himself and announced some of his campaign themes yesterday with a fairly lengthy press release by way of his Facebook page, which you can review here.
He notes that he is anxious to get on the campaign trail across Skeena-Bulkley Bally to speak to the many important issues facing residents of the region.
The Liberal Party's official announcement of Mr. Birdi's selection was released on July 23rd as well.
Federal Liberal candidate announcement
While the candidate is doing an admirable job of getting the word out ...
The Liberal head office however does not seem to know too much about their own candidate just yet, with his place marker on the official Team Trudeau site, somewhat limited in Biography, social media links or even a photo of their chosen candidate.
You can however donate to the Liberal campaign, with the donation button in good working order.
Mr. Birdi makes for the fifth entry so far in the upcoming electoral contest, already nominated and out on the campaign trail are:
Taylor Bacharach -- NDP
Jody Craven -- People's Party of Canada
Claire Rattée -- Conservative Party
Rod Taylor -- Christian Heritage Party
The Green Party has yet to announce who will be their candidate in the fall election.
You can keep up to date on developments along the campaign trail from our Skeena-Bulkley Valley Archive page here.
Labels: Federal Liberals name candidate for Skeena-Bulkley Valley campaign

redpajama_set_name: RedPajamaCommonCrawl


It has been more than a year since I last read a book from this series. I totally forgot about this one, and then I was looking at my manga list on another site and found out that I still had one volume to read.
Yay! A new volume is out and I could finally continue reading this spectacular and gorgeous series.
☆, Boarding School Wonderful or Not, Did Not Finish, For Girls One Shelf, OMG! It's an ARC!
I recently finished L.A. Candy and the Fame Game and I was looking for a new book or series about reality shows and people participating in said shows.
Wow, what a disappointment this series turned out to be. It had all elements of a great series. Boarding School + Shoujo Ai = (normally) Love!
I just found out that today is my blog's first anniversary! I totally forgot (I thought it was in 3 days and I also blame having the flu for memory lapse), but luckily I was just in time!
I want to thank everyone that visits this site and those who read my reviews. Really thank you all, you are all amazing! I am sorry that I can't do much more than just post this message. I know some people do really grand things with their anniversaries, and I would love to do it too, but due to reasons I can't. Maybe next year I will do something grand. For now I will just thank you all. And I hope I will be able to write a lot of reviews, meme posts and other fun posts that people will like reading this coming new year.
Again, thanks to everyone. I hope everyone will continue to visit my blog. And here's to another year!
Not really the best one. It is clearly suffering from the Second Book In a Series Syndrome, which is truly a shame.


The easiest way to subscribe is by pressing either "on iTunes" or "on Android" button above. If you have an Android device and already have one or more podcast apps, you'll be prompted to choose which one you'd like to use.
If you have an Android device and don't already have a podcast app installed, there are several (free or premium) ones you can download. The Listening Brief has been tested and works on all of the following. Clicking on any of these will take you to each app's site – once you have it on your device, come back to this page, and select "on Android" above – you should be prompted to select which app you want the subscription to work on. (Some are also available in versions for Apple and Windows devices).


Two sheets held together with a gold elastic pawl. Cover includes red foiled dove that matches napkin (item 29-5672). Certificate reads: Certificate of Baptism. This certifies that ... was baptized in the name of the Father and of the Son and of the Holy Spirit on this ... day of ... etc. Includes space for parent's names, place and date of birth. Includes envelope.


Coach Valorie Kondos Field said she does not fear going after the Bruin gymnasts' weaknesses.
No. 2 UCLA gymnastics (10-2, 3-1 Pac-12) will take on No. 16 Nebraska (6-1, 5-0 Big Ten), No. 23 Iowa State (15-2, 1-1 Big 12) and Kent State (5-4, 2-3 MAC) in the Masters Classic on Sunday in Lincoln, Nebraska.
The Bruins will be traveling out of state for the first time in over a month. They have been undefeated on the road so far this season, going 8-0 with two conference victories.
Kondos Field said that the team will be focused, but it has to manage the gymnasts' fatigue.
Redshirt freshman Grace Glenn agreed with her coach, but said her team will be ready to go Sunday.
UCLA will be trying to build off a season-high score of 198.075 from its last meet against No. 8 Oregon State (4-3, 3-3 Pac-12). Even with a great overall score, the Bruins only earned a 49.125 on the vault, their second-worst score on that event this season. Kondos Field said her team still needs to have a meet in which it has consistent performances throughout.
UCLA got a boost last week with sophomore Madison Kocian competing in the floor exercise for the first time this season. After having only competed on the balance beam so far this season, she said she was happy to be back in the event.
This meet is the first of three in 10 days for the Bruins. Kondos Field said the team will use its large roster to try to combat any fatigue, but she feels her team is ready for the challenge.


The Murano Millefiori Collection displays colorful unique masterpieces. The 1000 years old Murano glass blowing tradition meets with the modern fresh design in these truly inspiring glass-works. Distinct variety of artworks shimmering in a surprisingly delicate colors. For the modern and luxurious living. Attention to detail, exquisite craftsmanship along with innovative design characterizes the enchanting pieces of the collection. Browse through an exquisite compiliation of form and function. Shop with confidence! We guarantee that all our displayed artworks are 100% authentic. Deliver right to your doorstep. Provide 14 days money back guarantee. Your satisfaction is 100% guaranteed. Shop now!


I've been a fangirl before that was a word. I've been a fangirl for almost twenty years, long before the internet made fans accessible to each other, before there was a community to find online the way there is now. Before Tumblr and Instagram and Twitter. I say that and it makes me feel old.
I stood in long lines outside Filenes to get concert tickets when they went on sale. I wore a blue wig just to feel like I stood out in a crowd of thousands. I wrote letters that I never sent. I pasted posters from Bop magazine on my bedroom door. I scoffed at anyone who called me obsessed, dismissed people who refused to understand. At night I dreamed up scenarios where real people turned into characters in my mind.
I was a fangirl during the days of dial-up internet and AOL chats. I made e-mail newsletters in crazy fonts and colors before there were e-mail spam issues.
I wrote fanfiction for friends and strangers, emailed chapters of stories as I completed them. I met some of my best friends that way. We didn't have website where we could upload images we created and make own our merchandise, and there weren't fan sites that sold pretty things. No Etsy or Redbubble. We used Geocities to make fan pages with terrible flashy graphics instead.
The first form of "social media" and online expression that I immersed myself in was OpenDiary. After that there was Xanga, Myspace and Livejournal. All were online worlds where I could connect with people I didn't know who loved the same things I loved. In these worlds I could open my heart up in an anonymous way, like writing in a journal that others would see. I could know that someone out there read what I read and felt that, too.
Being a fangirl in the early 2000's meant you didn't have a cell phone to text people, and you had to rely on long-distance phone calls to hear someones' voice. We took pictures with regular cameras, selfies with digital cameras (before "selfie" was a word). Before digital cameras we had to wait for them to be developed and hope they came out okay. We recorded television appearances on VHS because we didn't know if we could ever see those recordings, hear the silly stupid words said again.
Now there is a lot of community online. You can find community for any sort of thing you can fangirling over). You can immerse yourself and meet new people and relate. And I'm glad about this. I know lots of people scoff at everyone being too connected and addicted to their phones, but they discount the experiences of these people. They discount the connections made long distance with friends you'd never have met another way. I have a few very dear friends that I still talk to, and I met them because we connected over the Internet. We dreamed about Hanson. We wanted to write like Francesca Lia Block. When I was in high school I still felt disconnected. No one else had liked Francesca Lia Block, no one knew who she was. Everyone made fun of me for loving three blonde boys who, at one point, could have been mistaken for girls (who cares about their gender, anyway?). I'm glad we can get together and post in a group on Facebook, use hashtags and share fanart in ways that used to be more difficult.
A couple years ago I read the book Fangirl by Rainbow Rowell and it was the first time I read a book with a main character so much like me. I couldn't believe it. All those years fangirling over books and music and falling in love with songs and boys – both fictional and real – and writing endless stories and finally I had found a character that I could relate to so much.
So being a fangirl has changed a lot over the years. In a way, I think it's a lot cooler now, but maybe it just feels more cool to me. There's more access to community, shops that make candles based on character and story scents, artists that sell fanart in the form of prints and clothing among other things, subscription boxes to fill your fandom needs and so much more.
Previous Post Do That Thing.
Oh my goodness, you just threw me through a sea of memories! Bop magazine?! Stop it! I was never a "fangirl" really, I liked Backstreet Boys but my posters were of the Beatles, Looney Tunes, and anything blacklight. My friends though always practically had wallpapered the rooms with clips from the magazine!!
Yes! Blacklights and those posters from Spencer gifts, lava lamps and any sort of posters!
Glow in the dark stars were awesome, too! And of course the Disco ball! So 90's (even though they were throwbacks to the 70's. It's funny how trends recycle).
You were a fan girl before fan girls were cool. It has changed drastically, I mean now fan girls make that into a stable income with all the social medium at our fingertips.
hehe the hipster fangirl, basically 😉 It's amazing that people can now make that into a stable income! I feel like I missed the bus on that one, and I should have figured out how I could do something like that years ago. I watch BookTube videos all the time, and I am amazed by the content that is produced!
Growing up in the 90's I am well aware of the Hansons. My sister is a big fan of them, me not so much. I was more into Nsync and in my fourth grade I discovered Ricky Martin. I had jelly sandals. Multiply pairs of overalls.
There definitely was something special about those simpler times! It was so different and we were left dreaming a lot more than we are these days, it seems. For books there are short stories that authors release, for bands – the members are active members of the social media communities. Authors Tweet all the time. It's incredible how things have changed!


Via the Critical Path Initiative, the FDA identified the need for evidentiary standards to qualify biomarkers, determining this to be "essential to improving the efficiency and effectiveness of drug development." According to the FDA, the evidentiary standards associated with different types of biomarkers vary from one to another. As such, FDA has stated that there are a number of specific challenges involved in qualifying drug safety biomarkers, which is one of the primary reasons there is such an immediate need for a system to do so.
In holding this workshop, FDA's goal is to create alignment among scientific stakeholders to support a proposed framework to determine the evidence needed to qualify biomarkers for use in drug development. In order to achieve this goal, attendees will discuss the development of a general framework for biomarker qualification, as well as specific application of the framework to different context(s) of use related to drug safety.
The stakeholders involved in this workshop include FDA, NIH, the biopharmaceutical industry, academic researchers, and patient groups.
The workshop will be held in Bethesda, Maryland, on April 14th and 15th from 8am to 5pm. Although there is no fee associated with attendance, the Agency is requiring participants to register in advance. The deadline for registration is April 1st. Due to the limited space that is available, registration is on a first-come, first-served basis. As such, if you are interested in attending the workshop it is advised that you register online as soon as possible.


text:
\section{Introduction}

\baselineskip=18pt

A Fox $m$-coloring \cite{Fox} is an assignment of elements from $\{0, 1, \ldots, m-1\}$ to
the arcs of a link diagram such that at each crossing twice the integer assigned
to the over-arc equals to the sum of the integers assigned to the two under-arcs
mod $m$. For each link diagram and each modulus $m>1$, there are always $m$
trivial colorings, namely by assigning the same integer mod $m$ to every arc of
the diagram. A coloring with at least two distinct colors (i.e., two distinct integers mod $m$
assigned to two arcs) is called a non-trivial coloring. It is easy to check
that if one diagram of a link has a non-trivial $m$-coloring, then each diagram
of that link has a non-trivial $m$-coloring. A link is called $m$-colorable if
it admits a diagram with non-trivial $m$-colorings. The following well-known theorem presents a
criterion for checking if a given link is $m$-colorable.

\begin{theorem}[\cite{LM}]\label{determinant}
A link $L$ is $m$-colorable if and only if the determinant of $L$ ($\det L$)
and $m$ are not relatively prime.
\end{theorem}

The following definition was introduced by Harary and Kauffman in \cite{HK}.

\medskip

DEFINITION 1. ~Given an integer $m$ greater than $1$. Let $L$ be a link admitting non-trivial
$m$-colorings. Let $D$ a diagram of $L$, let $n_{m, D}$ be the minimum
number of colors mod $m$ it takes to construct a non-trivial $m$-coloring on $D$.
Set
$$\text{mincol}_m L \doteq \min \{n_{m, D} \mid D\ \text{is a diagram of}\ L \}.$$
We call $\text{mincol}_m L$ the minimum number of colors of $L$, mod $m$.

\medskip

We call any non-trivial $m$-coloring of $L$ using $\text{mincol}_m L$ colors a
minimal $m$-coloring of $L$.

\bigbreak

In this article we give a shorter proof of the following Theorem.

\begin{theorem}[\cite{LM}]
Let $p$ be a prime greater than $7$, let $L$ be a link with $p\mid \det L \neq 0$. Then $mincol_p\, L \geq 5$.
\end{theorem}

We also prove the following fact.

\begin{theorem}
Let $p$ be a prime greater than $13$, let $L$ be a link with $p\mid \det L \neq 0$. Then $mincol_p\, L \geq 6$.
\end{theorem}

We give these proofs because we think they are both insightful and instructive. In fact some of our methods lead to new research problems which we will detail later in this paper.

Nakamura, Nakanishi, and Satoh proved a more general theorem in \cite{NNS} for knots, which we were unaware of at the time of the writing of our article using a completely different approach.

\begin{theorem}\label{NNS}
Let $p$ be an odd prime. Any $p$-colorable knot $K$ satisfies
\begin{equation}
\text{mincol}_p (K)\geq \lfloor \log_2 p \rfloor+2
\label{inequality}
\end{equation}
where $\lfloor x \rfloor$ is the largest integer less than or equal to $x$.
\end{theorem}

It is worth pointing out that Nakamura et al.'s proof for Theorem \ref{NNS} can not be
naturally extended to $p$-colorable links with non-zero determinant. We would like to
understand whether links with non-zero determinant also have such a good lower bound or not.

Let $c(K)$ denote the crossing number of $K$. Since $c(K)\geq \text{mincol}_p (K)$, any
$p$-colorable knot $K$ satisfies $c(K)\geq \lfloor \log_2 p \rfloor +2$. In \cite{NNS},
Nakamura, Nakanishi, and Satoh ask if the equality only holds for the trefoil knot ($p=3$) and
the figure-eight knot ($p=5$). We will give a positive answer to this question for classical knots
in Section 5, Theorem \ref{equality}.

We also define an equivalence relation among sets of colors over a given modulus, see Definition 5, and count the number of such equivalence classes for the least $mincol_p$ for $p=11$ or $13$. This equivalence relation among sets of colors unveils somewhat more the fascinating topic of Fox colorings.

\section{Definitions and Preliminaries}

REMARK 1. All split links (which have zero determinant) can be non-trivially colored with
two colors in any modulus. That is to say, for any split link and any modulus,
the minimun number of colors is $2$. Moreover, links with zero determinant are colorable over any modulus.
On the other hand, the fact that a link is colorable
only over a finite number of moduli introduces a certain regularity that allows
to prove Theorems like Theorem \ref{mincol} below.
For this reason all links referred to in this
paper are links with non-zero determinant.

\medskip

Clearly, $\text{mincol}_m L$ is an invariant of link $L$. For small primes,
we have certain results for this invariant, listed as Theorem \ref{mincol}
\cite{KL1, Satoh, Oshiro, Saito, LM, CJZ}. However, the minimum number of
colors is very difficult to compute in general, even for torus knots
$T(2, n)$ \cite{KL1, KL2}.

\begin{theorem}\label{mincol}
Let $L$ be a link with non-zero determinant.

(1) If $2\mid \det L$, then $\text{mincol}_2 L=2$.

(2) If $3\mid \det L$, then $\text{mincol}_3 L=3$.

(3) If $5\mid \det L$, then $\text{mincol}_5 L=4$.

(4) If $7\mid \det L$, then $\text{mincol}_7 L=4$.

(5) If $p$ is a prime greater than $7$ and $p\mid \det L$, then $\text{mincol}_p L\geq 5$.

(6) If $11\mid \det L$, then $5\leq \text{mincol}_{11} L\leq 6$.
\end{theorem}

More precisely, Satoh \cite{Satoh} proved that any 5-colorable link
with non-zero determinant can be colored by $\{1, 2, 3, 4\}$. Oshiro
\cite{Oshiro} proved that any 7-colorable link with non-zero determinant
can be colored by \{0, 1, 2, 4\}. And Cheng et al. \cite{CJZ} proved that
any 11-colorable link with non-zero determinant can be colored by 5 or 6
colors in \{0, 1, 4, 6, 7, 8\}. These three papers used similar techniques
developed by Satoh.

For $p=11$, the fourth named author of this article found an interesting behavior recently (\cite{Lopes}). In order
to describe this discovery, we list some terminology introduced in \cite{Lopes} first.

\medskip

DEFINITION 2. ~Let $m$ be a positive integer greater than $1$ and let $L$ be a link
admitting non-trivial $m$-colorings.

$\bullet$ An $m$-sufficient set of colors (for $L$) is a set of integers
mod $m$ such that a non-trivial $m$-coloring can be realized on a diagram
of $L$ with colors from this set.

$\bullet$ An $m$-minimal sufficient set of colors (for $L$) is an $m$-sufficient
set of colors (for $L$) whose cardinality is $\text{mincol}_m L$.

\medskip

\medskip

DEFINITION 3. ~Suppose $m$ is a positive integer greater than $1$ such that for any two links,
$L$ and $L'$, admitting non-trivial $m$-colorings, $\text{mincol}_m L = \text{mincol}_m L'$;
in this case, let $\text{mincol} (m)$ denote this common minimum number of colors,
$\text{mincol} (m) = \text{mincol}_m L = \text{mincol}_m L'$.

If there is a set $S$ of $\text{mincol} (m)$ elements from $\mathbb{Z}_m$ such
that for any $m$-colorable link $L''$, there exists a non-trivial $m$-coloring
with colors from this set, then we call $S$ a common $m$-minimal sufficient set
of colors.

If there is a set $S'$ of $\text{mincol} (m)$ elements from $\mathbb{Z}_m$ such
that any diagram supporting minimal $m$-colorings can be colored with colors from
this set, then we call $S'$ a universal $m$-minimal sufficient set of colors.

\medskip

Combining results proved in \cite{KL1, Satoh, Oshiro, Saito, LM, Lopes}, it is proved in \cite{Lopes}
that for each prime $p< 11$, there is a universal $p$-minimal sufficient set of colors.
Moreover, for any of these primes, any common $p$-minimal sufficient set of colors is a
universal $p$-minimal sufficient set of colors.

Changes happen at $p=11$. It is also proved in \cite{Lopes} that

\begin{theorem}\label{Lopes}
There is no universal 11-minimal sufficient set of colors.
\end{theorem}

We remark it remains unknown whether there is a common 11-minimal sufficient set of colors
or not.

To make our results easier to understand, we now introduce more terminology and
results already in the literature.

\medskip

DEFINITION 4. ~Given a positive integer $m$, we define an $m$-coloring automorphism to be a
permutation $f$, of the set $\mathbb{Z}_m$, such that
$$f(a\ast b)=f(a)\ast f(b)$$
for all $a, b\in \mathbb{Z}_m$, with $x\ast y=2y-x\ (\text{mod}\ m)$,
for every $x, y\in \mathbb{Z}_m$.

\medskip

\begin{proposition}
$m$-coloring automorphisms preserve $m$-colorings.
\end{proposition}

\begin{proposition} \cite{EMR}
Given a positive integer $m$, each $m$-coloring automorphism is given by:
$$f_{\lambda, \mu}(x)=\lambda x+\mu$$
with $\mu \in \mathbb{Z}_m$ and $\lambda \in \mathbb{Z}_m^\ast$, the set of
units of $\mathbb{Z}_m$. Furthermore, the set of $m$-coloring automorphisms
equipped with composition of functions constitutes a group.
\end{proposition}

For a permutation $f$ and a set $\{a_i\}_{i\in T}$, we denote $f(\{a_i\}_{i\in T})=\{f(a_i)\}_{i\in T}$.
Let $D$ be an $m$-colorable diagram with arcs $x_1, \ldots, x_n$. Suppose
$C$ is an $m$-coloring of $D$ with arc $x_i$ colored by $a_i$. Let $f$ be
an $m$-coloring automorphism. Similarly, we denote $f(C)$ the $m$-coloring
of $D$ such that arc $x_i$ is colored by $f(a_i)$.

Now it is natural for us to define an equivalence relation among color sets.

\medskip

DEFINITION 5. ~Given a positive integer $m$. Two color sets $S_1$ and $S_2$ are called
equivalent mod $m$ if and only if there exists an $m$-coloring automorphism
$f_{\lambda, \mu}(x)=\lambda x+\mu$ (mod $m$), such that $f(S_1)=S_2$.

\medskip

Since coloring automorphisms preserve colorings, if a link $L$ can be colored
with colors in a color set $S$, then it can be colored with colors in any color
set $S'$ equivalent to $S$. Furthermore, if a color set $S$ is an $m$-sufficient
set ($m$-minimal sufficient set) for $L$, then any color set $S'$ equivalent to
$S$ is an $m$-sufficient set ($m$-minimal sufficient set) for $L$.

\begin{lemma}\label{012}
Let $L$ be a non-split link. Let $p\geq 3$ be a prime number. If a diagram
$D$ of $L$ admits a non-trivial $p$-coloring $C$, then there exist a $p$-coloring
automorphism $f$ such that $f(C)$ is a non-trivial $p$-coloring of $D$ containing colors
0, 1 and 2.
\end{lemma}

\medskip

Proof. ~Since $L$ is a non-split link and $p$ is an odd prime, $D$ must have a non-trivially
colored crossing. Suppose at this crossing, the over-arc is colored by $b$ and the
two under-arcs are colored by $a$ and $c$. Then $2b=a+c$ (mod $p$) and $a, b, c$ are
distinct.

Let $f(x)=(b-a)^{-1}(x-a)$, where $x^{-1}$ is the inverse element of $x$ in $\mathbb{Z}_p$.
Then $f(a)=0$, $f(b)=1$ and $f(c)=\frac{c-a}{b-a}=\frac{2b-a-a}{b-a}=2$. Hence $f(C)$ is
a non-trivial $p$-coloring of $D$ containing colors 0, 1 and 2. \hfill{} $\Box$

\medskip

The fourth author of this article studied what kind of color set would never be a sufficient
set or minimal sufficient set. The following two theorems are useful.

\begin{theorem}\label{test1}\cite{Lopes}
Let $k$ be a positive integer and $L$ a link with non-zero determinant, admitting
non-trivial $(2k+1)$-colorings. If $S\subseteq \{0, 1, 2, \ldots, k\}$, then $S$
is not a $(2k+1)$-sufficient set of colors for $L$. Moreover, let $f$ be any
$(2k+1)$-coloring automorphism. Then any set $S'\subseteq \{f(0), f(1), . . . f(k)\}$
cannot be a $(2k+1)$-sufficient set of colors for L.
\end{theorem}

\begin{theorem}\label{test2}\cite{Lopes}
Let $k$ be a positive integer and $L$ a link with non-zero determinant, admitting
non-trivial $(2k+1)$-colorings. Suppose $S=\{c_1, \ldots, c_n\}$ is a $(2k+1)$-sufficient
set of colors for $L$. For each $c_i$ $(i=1, 2, \ldots, n)$, let $S_i$ be the set
of unordered pairs $\{c_i^1, c_i^2\}$ (with $c_i^1\neq c_i^2$ from $S$) such that
$2c_i=c_i^1+c_i^2 ~\mod~ 2k+1$.

$(1)$. There is at least one $i$ such that the expression $2c_i=c_i^1+c_i^2$ does not
make sense over the integers.

$(2)$. If there is $i\in \{1, 2, \ldots, n\}$ such that for all $j\in \{1, 2, \ldots, n\}\backslash \{i\}$,
$c_i$ is not in any of the unordered pairs in $S_j$, then $n> \text{mincol}_{2k+1} L$
and $S\backslash \{c_i\}$ is also a $(2k+1)$-sufficient set of colors for $L$.
\end{theorem}

\medskip

REMARK 2. Part $(1)$ of Theorem \ref{test2} is a generalization of Theorem \ref{test1}.

\medskip

Motivated by Theorem \ref{test2}, we define an associated edge-colored
simple graph for a color set.

\medskip

DEFINITION 6. ~Let $m$ be a positive integer. For a mod $m$ color set
$S=\{c_1, c_2, \ldots, c_n\}\subseteq \mathbb{Z}_{m}$, the associated edge-colored
(or briefly, colored) graph $G_c(S)$ is a simple graph constructed as follows.
Let $\{c_1, c_2, \ldots, c_n\}$ be the vertex set of $G_c(S)$. There is a red edge
between $c_i$ and $c_j$ if and only if there exists a positive integer $k\leq n$,
such that $2c_k=c_i+c_j$ over the integers. There is a blue edge between $c_i$ and
$c_j$ if and only if there exists an positive integer $k\leq n$, such that
$2c_k=c_i+c_j$ mod $m$ but $2c_k\neq c_i+c_j$ over the integers. We denote $G(S)$ the
underlying graph of $G_c(S)$, i.e., the graph obtained by replacing each colored
edge by a normal edge.

\medskip

\medskip

REMARK 3. Nakamura et al. defined a ``pallet graph'' for a $p$-coloring in \cite{NNS}. Their
graph and ours are similar in some respects. But our definition was made independently
and we have different motivations.

\medskip

Now we rewrite and extend Theorem \ref{test2} by using the associated graph.

\begin{theorem}\label{test3}
Let $k$ be a positive integer and $L$ a link with non-zero determinant, admitting
non-trivial $(2k+1)$-colorings. Suppose $S=\{c_1, \ldots, c_n\}$ is a $(2k+1)$-sufficient
set of colors for $L$. Let $G_c(S)$ ($G(S)$) be the associated colored (underlying)
graph.

$(1)$. $G_c(S)$ contains at least one blue edge.

$(2)$. If $c_i$ is an isolated vertex in $G(S)$, then $n> \text{mincol}_{2k+1} L$
and $S\backslash \{c_i\}$ is also a $(2k+1)$-sufficient set of colors for $L$.

$(3)$. Let $c(G(S))$ be the number of connected components of $G(S)$. Let $c(L)$
be the component number of $L$. If $c(G(S))> c(L)$, then $n> \text{mincol}_{2k+1} L$
and there exist $c(L)$ components of $G(S)$ such that the vertex set of these
$c(L)$ components is also a $(2k+1)$-sufficient set of colors for $L$.
\end{theorem}

\begin{corollary} (Corollary to both Theorem \ref{test2} and \ref{test3})\label{11}
$\{0, 1, 2, 3, 7\}$ is not a 11-minimal sufficient set of colors for any link
with non-zero determinant.
\end{corollary}

\medskip

Proof. The graph associated with $\{0, 1, 2, 3, 7\}$ is shown in Fig. \ref{fig1}. According
to part (1) of Theorem \ref{test2} or Theorem \ref{test3}, $\{0, 1, 2, 3, 7\}$
is not a 11-minimal sufficient set of colors for any link with non-zero determinant. \hfill{} $\Box$

\medskip

\begin{figure}[htbp]
\centering
\includegraphics[width=10cm]{G_S.eps}
\renewcommand{\figurename}{Fig.}
\caption{{\footnotesize The left is the colored graph associated with $\{0, 1, 2, 3, 7\}$,
the right is the colored graph associated with $\{0, 1, 2, 4, 10\}$.
}}\label{fig1}
\end{figure}

\begin{corollary} (Corollary to both Theorem \ref{test2} and \ref{test3})\label{13}
$\{0, 1, 2, 4, 10\}$ is not a 13-minimal sufficient set of colors for any link
with non-zero determinant.
\end{corollary}

\medskip

Proof. The graph associated with $\{0, 1, 2, 4, 10\}$ is shown in Fig. \ref{fig1}. According
to either part (1) or part (2) of Theorem \ref{test2} or Theorem \ref{test3},
$\{0, 1, 2, 4, 10\}$ is not a 13-minimal sufficient set of colors for any link
with non-zero determinant. \hfill{} $\Box$

\medskip

The following proposition is obvious and we omit its proof here.

\begin{proposition}
Let $m$ be a positive integer. Let $S_1$ and $S_2$ be two mod $m$ color sets with the same
cardinality. If $S_1$ and $S_2$ are equivalent (mod $m$), then the associated underlying
graphs $G(S_1)$ and $G(S_2)$ are isomorphic.
\end{proposition}

\section{Minimal Sufficient Sets of Colors}
The fourth author of this article proved that unlike $p\leq 7$, there is no universal
$11$-minimal sufficient set of colors \cite{Lopes}. But it still remains unknown whether
there is a common $11$-minimal sufficient set of colors or not. We now give further results
on $11$-, $13$-, and $17$-sufficient or minimal sufficient sets of colors.

\subsection{Minimal Sufficient Sets of Colors with Cardinality $5$}

\begin{theorem}\label{p=11}
Let $L$ be an 11-colorable link with non-zero determinant. If a diagram of $L$ can be colored
by a color set of 5 colors, then the color set must be either $\{0, 1, 2, 3, 6\}$
or $\{0, 1, 2, 4, 7\}$ in the sense of equivalence of color sets induced by
coloring automorphism.
\end{theorem}

\medskip

Proof. It was shown in \cite{Lopes} that, in the sense of equivalence class of
color sets, knot $6_2$ has unique 11-minimal sufficient set $\{0, 1, 2, 3, 6\}$
and knot $7_2$ has unique 11-minimal sufficient set $\{0, 1, 2, 4,7\}$.
It was also shown in \cite{Lopes} that $\{0, 1, 2, 3, 6\}$ is not equivalent
to $\{0, 1, 2, 4, 7\}$. Recalling Lemma \ref{012}, we only need to consider
color sets of type $\{0, 1, 2, x, y\}$. Table 1 shows all instances of
color sets of type $\{0, 1, 2, x, y\}$. ``type 1'' (``type 2'') means it
is equivalent to $\{0, 1, 2, 3, 6\}$ ($\{0, 1, 2, 4, 7\}$). ``N, Th \ref{test1}''
(``N, Co \ref{11}'') means there is no link which can be 11-colored by it due
to Theorem \ref{test1} (Corollary \ref{11}). We pick three color sets as
examples of how to read Table 1.

(1) $\{0, 1, 2, 3, 8\}$. The coloring automorphism $f_{10, 3}(x)=10x+3$
transforms the color set $\{0, 1, 2, 3, 8\}$ into $\{0, 1, 2, 3, 6\}$,
so $\{0, 1, 2, 3, 8\}$ is equivalent to $\{0, 1, 2, 3, 6\}$.

(2) $\{0, 1, 2, 3, 9\}$. The coloring automorphism $f_{1, 2}(x)=x+2$
transforms the color set $\{0, 1, 2, 3, 9\}$ into $\{0, 2, 3, 4, 5\}$, a
subset of $\{0, 1, \ldots, 5\}$. According to Theorem \ref{test1},
$\{0, 1, 2, 3, 9\}$ is not $11$-sufficient set of colors for any link
with non-zero determinant.

(3) $\{0, 1, 2, 4, 9\}$. The coloring automorphism $f_{6, 1}(x)=6x+1$
transforms the color set $\{0, 1, 2, 4, 9\}$ into $\{0, 1, 2, 3, 7\}$.
According to Corollary \ref{11}, $\{0, 1, 2, 3, 7\}$ is not a $11$-minimal
sufficient set of colors for any link with non-zero determinant. Hence
$\{0, 1, 2, 4, 9\}$ is not a $11$-minimal sufficient set of colors for
any link with non-zero determinant either. \hfill{} $\Box$

\medskip

\begin{table}[ht]
\center{Table 1. 5 color sets mod 11.}
\vskip 5 true pt
{\begin{tabular}{|l|r|c||l|r|c|}
 \hline
0,1,2,3,4  &                          & N, Th \ref{test1} & 0,1,2,5,7  & ($\times$2+1)=0,1,3,4,5   & N, Th \ref{test1} \\ \hline
0,1,2,3,5  &                          & N, Th \ref{test1} & 0,1,2,5,8  & ($\times$7)=0,1,2,3,7     & N, Co \ref{11}    \\ \hline
0,1,2,3,6  &                          & type 1            & 0,1,2,5,9  & ($\times$8+6)=0,1,2,3,6   & type 1            \\ \hline
0,1,2,3,7  &                          & N, Co \ref{11}    & 0,1,2,5,10 & (+1)=0,1,2,3,6            & type 1            \\ \hline
0,1,2,3,8  & ($\times$10+3)=0,1,2,3,6 & type 1            & 0,1,2,6,7  & ($\times$2)=0,1,2,3,4     & N, Th \ref{test1} \\ \hline
0,1,2,3,9  & (+2)=0,2,3,4,5           & N, Th \ref{test1} & 0,1,2,6,8  & ($\times$2)=0,1,2,4,5     & N, Th \ref{test1} \\ \hline
0,1,2,3,10 & (+1)=0,1,2,3,4           & N, Th \ref{test1} & 0,1,2,6,9  & ($\times$2)=0,1,2,4,7     & type 2            \\ \hline
0,1,2,4,5  &                          & N, Th \ref{test1} & 0,1,2,6,10 & (+1)=0,1,2,3,7            & N, Co \ref{11}    \\ \hline
0,1,2,4,6  & ($\times$6)=0,1,2,3,6    & type 1            & 0,1,2,7,8  & ($\times$2)=0,2,3,4,5     & N, Th \ref{test1} \\ \hline
0,1,2,4,7  &                          & type 2            & 0,1,2,7,9  & ($\times$5+1)=0,1,2,3,6   & type 1            \\ \hline
0,1,2,4,8  & ($\times$3)=0,1,2,3,6    & type 1            & 0,1,2,7,10 & ($\times$10+2)=0,1,2,3,6  & type 1            \\ \hline
0,1,2,4,9  & ($\times$6+1)=0,1,2,3,7  & N, Co \ref{11}    & 0,1,2,8,9  & (+3)=0,1,3,4,5            & N, Th \ref{test1} \\ \hline
0,1,2,4,10 & (+1)=0,1,2,3,5           & N, Th \ref{test1} & 0,1,2,8,10 & (+3)=0,2,3,4,5            & N, Th \ref{test1} \\ \hline
0,1,2,5,6  & ($\times$2+1)=0,1,2,3,5  & N, Th \ref{test1} & 0,1,2,9,10 & (+2)=0,1,2,3,4            & N, Th \ref{test1} \\ \hline
\end{tabular}}
\end{table}

\begin{theorem}\label{p=13}
Let $L$ be a $13$-colorable link with non-zero determinant. If a diagram of $L$ can be
colored by 5 colors, then it can be only colored by $\{0, 1, 2, 4, 7\}$ in
the sense of equivalence. Specifically, if $\text{mincol}_{13} L=5$ holds
for all 13-colorable links with non-zero determinant, then $\{0, 1, 2, 4, 7\}$
is the only common 13-minimal sufficient set of colors in the sense of
equivalence of color sets induced by coloring automorphism.
\end{theorem}

\medskip

Proof. It was shown in \cite{Lopes} that, in the sense of equivalence class
of color sets, knots $6_3$, $7_3$ and $10_{154}$ has unique 13-minimal
sufficient set $\{0, 1, 2, 4, 7\}$. Recalling Lemma \ref{012}, we only
need to consider color sets of type $\{0, 1, 2, x, y\}$. Table 2 shows
circumstances of all color sets of type $\{0, 1, 2, x, y\}$. ``Y'' means
it is equivalent to $\{0, 1, 2, 4, 7\}$. ``N, Th \ref{test1}'' (``N, Co \ref{13}'')
means there is no link can be 11-colored by it due to Theorem \ref{test1}
(Corollary \ref{13}). \hfill{} $\Box$

\medskip

\begin{table}[ht]
\center{Table 2. 5 color sets mod 13.}
\vskip 5 true pt
{\begin{tabular}{|l|r|c||l|r|c|}
\hline
0,1,2,3,4  &                            & N, Th \ref{test1} & 0,1,2,5,12  & (+1)=0,1,2,3,6            & N, Th \ref{test1} \\ \hline
0,1,2,3,5  &                            & N, Th \ref{test1} & 0,1,2,6,7   & ($\times$2+1)=0,1,2,3,5   & N, Th \ref{test1} \\ \hline
0,1,2,3,6  &                            & N, Th \ref{test1} & 0,1,2,6,8   & ($\times$2+1)=0,1,3,4,5   & N, Th \ref{test1} \\ \hline
0,1,2,3,7  & ($\times$2)=0,1,2,4,6      & N, Th \ref{test1} & 0,1,2,6,9   & ($\times$3)=0,1,3,5,6     & N, Th \ref{test1} \\ \hline
0,1,2,3,8  & ($\times$2)=0,2,3,4,6      & N, Th \ref{test1} & 0,1,2,6,10  & ($\times$3)=0,3,4,5,6     & N, Th \ref{test1} \\ \hline
0,1,2,3,9  & ($\times$2)=0,2,4,5,6      & N, Th \ref{test1} & 0,1,2,6,11  & ($\times$10+7)=0,1,2,4,7  & Y                 \\ \hline
0,1,2,3,10 & (+3)=0,3,4,5,6             & N, Th \ref{test1} & 0,1,2,6,12  & ($\times$2+2)=0,1,2,4,6   & N, Th \ref{test1} \\ \hline
0,1,2,3,11 & (+2)=0,2,3,4,5             & N, Th \ref{test1} & 0,1,2,7,8   & ($\times$2)=0,1,2,3,4     & N, Th \ref{test1} \\ \hline
0,1,2,3,12 & (+1)=0,1,2,3,4             & N, Th \ref{test1} & 0,1,2,7,9   & ($\times$2)=0,1,2,4,5     & N, Th \ref{test1} \\ \hline
0,1,2,4,5  &                            & N, Th \ref{test1} & 0,1,2,7,10  & ($\times$2)=0,1,2,4,7     & Y                 \\ \hline
0,1,2,4,6  &                            & N, Th \ref{test1} & 0,1,2,7,11  & ($\times$6+1)=0,1,2,4,7   & Y                 \\ \hline
0,1,2,4,7  &                            & Y                 & 0,1,2,7,12  & ($\times$2+2)=0,2,3,4,6   & N, Th \ref{test1} \\ \hline
0,1,2,4,8  & ($\times$7)=0,1,2,4,7      & Y                 & 0,1,2,8,9   & ($\times$2)=0,2,3,4,5     & N, Th \ref{test1} \\ \hline
0,1,2,4,9  & ($\times$3+1)=0,1,2,4,7    & Y                 & 0,1,2,8,10  & ($\times$11+4)=0,1,2,4,10 & N, Co \ref{13}    \\ \hline
0,1,2,4,10 &                            & N, Co \ref{13}    & 0,1,2,8,11  & ($\times$12+2)=0,1,2,4,7  & Y                 \\ \hline
0,1,2,4,11 & (+2)=0,2,3,4,6             & N, Th \ref{test1} & 0,1,2,8,12  & ($\times$2+2)=0,2,4,5,6   & N, Th \ref{test1} \\ \hline
0,1,2,4,12 & (+1)=0,1,2,3,5             & N, Th \ref{test1} & 0,1,2,9,10  & (+4)=0,1,4,5,5            & N, Th \ref{test1} \\ \hline
0,1,2,5,6  &                            & N, Th \ref{test1} & 0,1,2,9,11  & (+4)=0,2,4,5,6            & N, Th \ref{test1} \\ \hline
0,1,2,5,7  & ($\times$2)=0,1,2,4,10     & N, Co \ref{13}    & 0,1,2,9,12  & (+4)=0,3,4,5,6            & N, Th \ref{test1} \\ \hline
0,1,2,5,8  & ($\times$11+4)=0,1,2,4,7   & Y                 & 0,1,2,10,11 & (+3)=0,1,3,4,5            & N, Th \ref{test1} \\ \hline
0,1,2,5,9  & ($\times$3)=0,1,2,3,6      & N, Th \ref{test1} & 0,1,2,10,12 & (+3)=0,2,3,4,5            & N, Th \ref{test1} \\ \hline
0,1,2,5,10 & ($\times$3)=0,2,3,4,6      & N, Th \ref{test1} & 0,1,2,11,12 & (+2)=0,1,2,3,4            & N, Th \ref{test1} \\ \hline
0,1,2,5,11 & ($\times$12+2)=0,1,2,4,10  & N, Co \ref{13}    &             &                           &                   \\ \hline
\end{tabular}}
\end{table}

\subsection{Possible Minimal Sufficient Sets of Colors with Cardinality $6$}

Now we determine which color sets with cardinality $6$ may be minimal sufficient sets of colors
for primes $p=11$, $13$, and $17$. Our strategy is as follows.

{\bf Step 1.} List all subsets of $\{0, 1, \ldots, p\}$ with cardinality $6$ and
containing $0, 1$, and $2$.

{\bf Step 2.} Classify these color sets into equivalence classes (recall
Definition 5).

{\bf Step 3.} Use Theorem \ref{test1} and Theorem \ref{test2} to check the color
sets in order. If $S$ can not be a $p$-minimal sufficient set of colors
for any link with non-zero determinant, then delete all those color sets
equivalent to $S$. We call the remaining ones possible $p$-sufficient
sets of colors with cardinality $6$.

Our results follow. We use a C program to achieve Step 1 and 2.

For $p=11$, there are $\binom{8}{3}=56$ color sets with cardinality $6$ and
containing $0, 1$, and $2$. They can be classified into $6$ equivalence classes.
Among them, there are $4$ possible $11$-minimal sufficient sets of colors up
to the equivalence relation: $\{0, 1, 2, 3, 4, 6\}$, $\{0, 1, 2, 3, 4, 7\}$,
$\{0, 1, 2, 3, 5, 6\}$, and $\{0, 1, 2, 3, 5, 9\}$.

\begin{proposition}
To the extent encompassed by Theorems \ref{test1} and \ref{test2},
there are no obstructions for an $11$-colorable link $L$ with non-zero
determinant to satisfy $\text{mincol}_{11} L = 5$.
\end{proposition}

\medskip

Proof. By \cite{CJZ}, $\text{mincol}_{11} L$ is either $5$ (in which case
the proof is concluded) or $6$. In the latter instance, there is a
diagram of $L$ equipped with a non-trivial $11$-coloring using $6$
colors either from $\{0, 1, 2, 3, 4, 6\}$, or from $\{0, 1, 2, 3, 4, 7\}$,
or from $\{0, 1, 2, 3, 5, 6\}$, or from $\{0, 1, 2, 3, 5, 9\}$, up
to the equivalence relation. We now prove that it is always possible
to remove one color from any of these sets in order to obtain
$\{0, 1, 2, 3, 6\}$ or $\{0, 1, 2, 4, 7\}$. Using the tests described in
Theorems \ref{test1} and/or \ref{test2} along with the help of Table 1,
we note that $3$ or $4$ can be removed from the set $\{0, 1, 2, 3, 4, 6\}$
giving rise to a set equivalent to $\{0, 1, 2, 3, 6\}$;
$0$ or $1$ ($3$) can be removed from the set $\{0, 1, 2, 3, 4, 7\}$ giving
rise to a set equivalent to $\{0, 1, 2, 3, 6\}$ ($\{0, 1, 2, 4, 7\}$);
$5$ can be removed from the set $\{0, 1, 2, 3, 5, 6\}$ giving rise to
$\{0, 1, 2, 3, 6\}$; $0$ or $3$ can be removed from the set $\{0, 1, 2, 3, 5, 9\}$
giving rise to a set equivalent to $\{0, 1, 2, 3, 6\}$. End of proof. \hfill{} $\Box$

\medskip

For $p=13$, there are $\binom{10}{3}=120$ color sets with cardinality 6 and
containing 0, 1, and 2. They can be classified into 14 equivalence classes.
Among them, there are 8 possible 13-minimal sufficient sets of colors up to
the equivalence relation: \{0, 1, 2, 3, 4, 7\}, \{0, 1, 2, 3, 4, 8\},
\{0, 1, 2, 3, 5, 7\}, \{0, 1, 2, 3, 5, 8\}, \{0, 1, 2, 3, 5, 9\},
\{0, 1, 2, 3, 5, 11\}, \{0, 1, 2, 3, 6, 10\}, and \{0, 1, 2, 4, 5, 8\}.

For $p=17$, there are $\binom{14}{3}=364$ color sets with cardinality 6 and
containing 0, 1, and 2. They can be classified into 49 equivalence classes.
Among them, there are 9 possible 17-minimal sufficient sets of colors up to
the equivalence relation: \{0, 1, 2, 3, 5, 9\}, \{0, 1, 2, 3, 5, 10\},
\{0, 1, 2, 3, 5, 12\}, \{0, 1, 2, 3, 6, 9\}, \{0, 1, 2, 3, 6, 10\},
\{0, 1, 2, 3, 6, 11\}, \{0, 1, 2, 3, 6, 13\}, \{0, 1, 2, 3, 6, 14\},
and \{0, 1, 2, 3, 7, 10\}.

\section{Minimum Number of Colors}
In this section, we study the lower bound of the minimum number of colors.
First we give a short proof of $\text{mincol}_p L\geq 5$ for any
$p$-colorable link $L$ with $\det L\neq 0$ and prime $p\geq 11$. Then
we show that we can go further by using a similar approach.

\begin{theorem}\label{lowerbound5}\cite{LM}
Let $L$ be a link with non-zero determinant. If there is a prime $p\geq 11$
such that $L$ admits non-trivial $p$-colorings, then $\text{mincol}_p L\geq 5$.
\end{theorem}

\medskip

Proof. It is easy to prove the mod $p$ minimum number of colors is at least $4$ for
any link with non-zero determinant and any prime $p\geq 5$. See Lemma 2.1 in
\cite{Satoh} for example. So we only need to prove $L$ cannot be colored by
4 colors. Recalling Lemma \ref{012}, it is enough to consider color set $S$ of
type $\{0, 1, 2, a\}$, where $3\leq a \leq p-1$. If $L$ can be colored by $S$,
then $\frac{p+1}{2} \leq a \leq p-1$ due to Theorem \ref{test1}. The coloring
automorphism $f(x)=x+(p-a)$ transforms $S$ into $\{0, p-a, p-a+1, p-a+2\}$.
Hence $p-a+2\geq \frac{p-1}{2}+1$ due to Theorem \ref{test1}, thus $a\leq \frac{p+3}{2}$.
So $a\in \{\frac{p+1}{2}, \frac{p+3}{2}\}$. We use the same notations as in
Theorem \ref{test2}. Let $c_1=0$, $c_2=1$, $c_3=2$, $c_4=a$. Since
$a\in \{\frac{p+1}{2}, \frac{p+3}{2}\}$, we have
$6\leq \frac{p+1}{2} \leq c_4 \leq \frac{p+3}{2} = p-\frac{p-3}{2} \leq p-4$.
It is easy to check $S_1=\emptyset$, $S_2=\{\{0, 2\}\}$ and $S_3=\emptyset$.
So $c_4$ is not in any of the unordered pairs in $S_j$, $j\in \{1, 2, 3\}$.
According to part $(2)$ of Theorem \ref{test2}, $4> \text{mincol}_p L$, which
never happens. \hfill{} $\Box$

\medskip

\begin{theorem}\label{lowerbound6}
Let $L$ be a link with non-zero determinant. If there is a prime $p\geq 17$
such that $L$ admits non-trivial $p$-colorings, then $\text{mincol}_p L\geq 6$.
\end{theorem}

\medskip

Proof. Recalling part $(5)$ of Theorem \ref{mincol}, we only need to prove $L$ cannot
be colored by 5 colors. Suppose $L$ can be colored by a color set $S$ with
$|S|=5$. Then $S$ is a $p$-minimal sufficient set of colors for $L$.

\medskip

{\bf Claim.} Suppose $S'=\{x_1, x_2, x_3, x_4\}\subseteq S$ with $x_1< x_2< x_3< x_4$.
Then $x_4-x_1\geq 5$.

\medskip

Proof of the Claim. We prove the Claim by contradiction. If there is an $S'$ such that $x_4-x_1\leq 4$,
then the coloring automorphism $f_{1, p-x_1}(x)=x+p-x_1$ transforms $S$ into a
set $S''=\{0, y_1, y_2, y_3, y_4\}$, where $y_1< y_2< y_3\leq 4$ and
$y_3< y_4$. According to Theorem \ref{test1}, $y_4\geq \frac{p+1}{2}$ and
$p-y_4+4\geq \frac{p+1}{2}$, which yields $9\leq y_4 \leq p-5$.
So $y_4$ is not in any of the unordered pairs in any $S_j$ (which means $y_4$
cannot be the color of any under-arc at a polychromatic crossing). Recalling
Theorem \ref{test2}, $S''$ (so that $S$) is not a $p$-minimal sufficient set
of colors for $L$ and $\text{mincol}_p L\leq 4$, which is impossible. The
proof is complete.

\medskip

Recalling Lemma \ref{012}, it is enough to consider color set $S$ of type
$\{0, 1, 2, a, b\}$, where $3\leq a \leq b \leq p-1$. Then
$\frac{p+1}{2} \leq b \leq p-1$ due to Theorem \ref{test1}.

The Claim indicates that $a\geq 5$ immediately. It also indicates that
$b \leq p-3$, otherwise the coloring automorphism $f_{1, p-b}(x)=x+p-b$
will transform $\{0, 1, 2, b\}$ into $\{0, p-b, p-b+1, p-b+2\}$ where
$p-b+2\leq 4$. Hence $a\geq 5$ and $\frac{p+1}{2}\leq b \leq p-3$. So
there is no crossing with the over-arc colored by one color from $\{0, 1, 2\}$,
whose under-arcs are colored with one color from $\{0, 1, 2\}$, and the other from $\{a, b\}$.

We keep all the notations used in Theorem \ref{test2}. Let $c_1=0$, $c_2=1$,
$c_3=2$, $c_4=a$, $c_5=b$. Now we divide all the possibilities into the
following two cases.

\noindent{\bf Case 1.} $a+b\notin \{0, 2, 4\}\mod p$.

In this case, both $a$ and $b$ are not in any of the unordered pairs in
$S_1$, $S_2$ and $S_3$. So $a$ must be in one of the unordered pairs in
$S_5$ and $b$ must be in one of the unordered pairs in $S_4$ according
to Theorem \ref{test2}, i.e., $\exists \alpha, \beta \in \{0, 1, 2\}$($\alpha$ and $\beta$ are not
necessarily different), such that
\begin{equation}\label{1}
\left\{ \begin{array}{l}
2b=a+\alpha+p \\
2a=b+\beta
\end{array}. \right.
\end{equation}
Hence
\begin{equation}\label{2}
\left\{ \begin{array}{l}
a=\frac{p+\alpha+2\beta}{3} \\
b=\frac{2p+2\alpha+\beta}{3}
\end{array}. \right.
\end{equation}

The system of equations (\ref{1}) is over integers, so are
(\ref{3}) and (\ref{5}) below. Since
$\alpha, \beta \in \{0, 1, 2\}$, the coloring automorphism $f_{3, 0}(x)=3x$
transforms $S$ into $\{0, 3, 6, \alpha+2\beta, 2\alpha+\beta\}\subseteq \{0, 1, \ldots, \frac{p-1}{2}\}$.
So $S$ is not a $p$-sufficient set of colors for $L$.

\noindent{\bf Case 2.} $a+b=\gamma \in \{0, 2, 4\}\mod p$.

It is easy to see $c_2=1$ is not in any of the unordered pairs in $S_1$
and $S_3$. So $1$ must be in one of the unordered pairs in $S_4$ or $S_5$,
i.e., either $2a=1+b$ or $2b=1+a+p$ (over integers). So there are two subcases.

\noindent{\bf Case 2.1.} $2a=1+b$.

In this subcase,
\begin{equation}\label{3}
\left\{ \begin{array}{l}
2a=1+b \\
a+b=\gamma+p
\end{array}. \right.
\end{equation}
Hence
\begin{equation}\label{4}
\left\{ \begin{array}{l}
a=\frac{p+\gamma+1}{3} \\
b=\frac{2p+2\gamma-1}{3}
\end{array}. \right.
\end{equation}
Since $\gamma \in \{0, 2, 4\}$, the coloring automorphism $f_{3, 1}(x)=3x+1$
transforms $S$ into $\{1, 4, 7, \gamma+2, 2\gamma\}\subseteq \{0, 1, \ldots, \frac{p-1}{2}\}$.
So $S$ is not a $p$-sufficient set of colors for $L$.

\noindent{\bf Case 2.2.} $2b=1+a+p$.

In this subcase,
\begin{equation}\label{5}
\left\{ \begin{array}{l}
2b=1+a+p \\
a+b=\gamma+p
\end{array}. \right.
\end{equation}
Hence
\begin{equation}\label{6}
\left\{ \begin{array}{l}
a=\frac{p+2\gamma-1}{3} \\
b=\frac{2p+\gamma+1}{3}
\end{array}. \right.
\end{equation}
Since $\gamma \in \{0, 2, 4\}$, the coloring automorphism $f_{3, 1}(x)=3x+1$
transforms $S$ into $\{1, 4, 7, 2\gamma, \gamma+2\}\subseteq \{0, 1, \ldots, \frac{p-1}{2}\}$.
So $S$ is not a $p$-sufficient set of colors for $L$.

Therefore, $S$ is not a $p$-minimal sufficient set of colors for $L$, a contradiction. \hfill{} $\Box$

\medskip

As we pointed out before, for any split link and any modulus, the
minimum number of colors is $2$. So Theorem \ref{lowerbound5},
Theorem \ref{lowerbound6} and Theorem \ref{NNS} cannot be extended
to all links. But can we use non-split links instead of links with
non-zero determinant in these theorems? The following lemma gives
a negative answer.

\begin{lemma}
For any modulus $m\geq 5$, there are infinitely many non-split links
with zero determinant having minimum number of colors at most $4$.
\end{lemma}

\medskip

Proof. Figure \ref{L8n8} shows that for any modulus $m\geq 5$, the non-split link
$L8n8$ in the Thistlethwaite link table has a non-trivial coloring with
color set $\{0, 1, 2, 3\}$. Note that $L8n8$ is $(2, -2, 2, -2)$-pretzel
link. It is easy to see that for any $n$,
$(\underbrace{2, -2, \ldots, 2, -2}_{2n~\text{strands}})$-pretzel link
has the same property.

For pretzel link $P(p_1, p_2, \ldots, p_n)$, the determinant is
$\Big|\sum_{j=1}^n \frac{\prod_{i=1}^n p_i}{p_j}\Big|$ \cite{DFKLS, GZ}.
Hence $\det (P(2, -2, \ldots, 2, -2))=0$. This completes the proof. \hfill{} $\Box$

\medskip

\begin{figure}[htbp]
\centering
\includegraphics[width=5cm]{nonsplit.eps}
\renewcommand{\figurename}{Fig.}
\caption{{\footnotesize A non-trivial coloring of $L8n8$ with $4$ colors.
}}\label{L8n8}
\end{figure}

\section{A Question Raised By Nakamura, Nakanishi, and Satoh}
We recall that Theorem \ref{NNS} which states that
$\text{mincol}_p (K) \geq \lfloor \log_2 p \rfloor +2$ is proved in
Nakamura et al.' \cite{NNS}. Since the crossing number of knot $K$,
$c(K)$, satisfies $c(K)\geq \text{mincol}_p (K)$, for any $p$-colorable
knot $K$, these authors wonder if the equality $c(K)= \lfloor \log_2 p \rfloor +2$
only holds for the trefoil and the figure-eight knots, see (iii) in Remark
3.3 on page 96 of \cite{NNS}. Here we settle this matter with Theorem \ref{equality}.

\begin{theorem}\label{equality}
Let $p$ be an odd prime. Let $K$ be a $p$-colorable classical knot. Then the equality in
$c(K)\geq \lfloor \log_2 p \rfloor +2$ only holds for the trefoil knot ($p=3$)
and the figure-eight knot ($p=5$).
\end{theorem}

Let $D$ be a link diagram. Let
$$d_n^\infty := \max \{\det(D)\ |\ D\ \text{is a link diagram of $n$ crossings}\}.$$
In \cite{Sto}, Stoimenow showed
\begin{eqnarray}
d_n^\infty\leq d_{n-1}^\infty + d_{n-2}^\infty + d_{n-3}^\infty\ \ (n>2), \label{f1}
\end{eqnarray}
and then proved the following theorem.

\begin{theorem}\cite{Sto}\label{Sto}
Let $\delta\approx 1.83929$ be the real positive root of $x^3-x^2-x-1=0$.
There exists a constant $C>0$ such that for any link diagram $D$ of $c(D)$
crossings
$$\det(D)\leq C\cdot \delta^{c(D)}.$$
\end{theorem}

Stoimenow also pointed out that $C=1$ is always valid.

\medskip

\noindent{Proof of Theorem \ref{equality}.}

Let $\tilde{D}$ be a minimal diagram of $K$. Since $K$ is a $p$-colorable
knot, we have $p \mid \det(K)$ and $\det K>0$. By Theorem \ref{Sto},
$$\log_2 p \leq \log_2 \det(K)=\log_2 \det(\tilde{D})\leq c(\tilde{D})\log_2 \delta <0.87915\cdot c(K).$$
It is easy to see, for $c(K)\geq 17$,
$$c(K)>0.87915\cdot c(K)+2>\log_2 p+2>\lfloor \log_2 p \rfloor +2.$$

Table 3 shows the numerical results of $d_n^\infty$ and
$\lfloor \log_2 d_n^\infty \rfloor +2$ for $3\leq n\leq 16$. The first
four values of $d_n^\infty$ ($n\geq 3$) appeared in \cite{Sto} and other
values are estimated by formula (\ref{f1}).

\begin{table}[ht]
\center{Table 3. $d_n^\infty$ and $\lfloor \log_2 d_n^\infty \rfloor +2$ for $3\leq n\leq 16$.}
\vskip 5 true pt
{\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$n$                                     &    $3$     &    $4$     &    $5$     &    $6$      &    $7$      &    $8$      &     $9$     \\  \hline
$d_n^\infty$                            &    $3$     &    $5$     &    $8$     &    $16$     & $\leq 29$   & $\leq 53$   & $\leq 98$   \\  \hline
$\lfloor \log_2 d_n^\infty \rfloor +2$  &    $3$     &    $4$     &    $5$     &    $6$      & $\leq 6$    & $\leq 7$    & $\leq 8$    \\  \hline
$n$                                     &    $10$    &    $11$    &    $12$    &    $13$     &    $14$     &    $15$     &     $16$    \\  \hline
$d_n^\infty$                            & $\leq 180$ & $\leq 331$ & $\leq 609$ & $\leq 1120$ & $\leq 2060$ & $\leq 3789$ & $\leq 6969$ \\  \hline
$\lfloor \log_2 d_n^\infty \rfloor +2$  & $\leq 9$   & $\leq 10$  & $\leq 11$  & $\leq 12$   & $\leq 13$   & $\leq 13$   & $\leq 14$   \\  \hline
\end{tabular}}
\end{table}

Hence, for any knot $K$ with crossing number between $7$ and $16$, we obtain
$$c(K)>\lfloor \log_2 d_n^\infty \rfloor +2\geq \lfloor \log_2 \det(K) \rfloor +2\geq \lfloor \log_2 p \rfloor +2.$$

For any knot $K$ with crossing number $5$ or $6$, it is easy to check that
$c(K)>\lfloor \log_2 p \rfloor +2$. The proof is complete. {\hfill$\Box$}

\vskip 20 true pt

\noindent ACKNOWLEDGEMENTS. ~X. Jin and J. Ge acknowledge support from the National Natural Science
Foundation of China (No. 11271307). P. Lopes acknowledges partial
funding from FCT (Portugal) through projects PEst-OE/EEI/LA0009/2013,
and EXCL/MAT-GEO/0222/2012 (``Geometry and Mathematical Physics'').
L. Zhang acknowledges support from the National Natural Science Foundation
of China (Nos. 11171279 and 11471273). The authors thank Prof. Satoh for
kindly sending us \cite {NNS}. J. Ge thanks Jingkang Zhou for helping
writing the program.

redpajama_set_name: RedPajamaArXiv


text:
After Pantaleo firing: 'The job has been dying and today, it's dead'
By Rafael A. Mangual
NYPD Commissioner James O'Neill AP
73-year-old man blinded after vicious sucker punch in Crown Heights
Man's body found bound and burned in Queens house fire
Man stabbed in brawl outside Brooklyn steakhouse
New York AG Letitia James' obscene attack on the NYPD
"Policing is dead," a member of the NYPD told me Monday, after NYPD Commissioner James O'Neill decided to fire Officer Daniel Pantaleo over the death of Eric Garner. "Morale was already low. This was just the nail in the coffin."
Another member of the department told me on the condition of anonymity, "It's just disheartening."
"It's hard to say whether cops will back off," he continued, "but I think the writing is definitely on the wall. Based on how things went in other cities [Chicago and Baltimore] I wouldn't be surprised if we went in that same direction."
The reactions echoed that of Police Benevolent Association President Pat Lynch, who told reporters, "The job has been dying; and today, the job is dead." Lynch went on to assert Pantaleo's dismissal was based "on the politics."
As is often the case with these things, context and timing are all-important. The decision to fire Pantaleo will not be perceived in a vacuum, but rather through the lens of these politically charged times.
Mayor Bill de Blasio says 'justice was done' with Pantaleo firing
Mayor Bill de Blasio said "justice was done" with the...
The decision came in the middle of a presidential-election cycle in which nearly all the candidates pursuing the Democratic Party's nomination have openly demonized police as cogs in a machine of racial oppression and injustice.
Indeed, just a couple of weeks ago, protesters interrupted a prime-time debate to demand that candidate (and, when he has time, New York City Mayor) Bill de Blasio fire Pantaleo. The political implications of the decision were not lost on the officers I spoke to, one of whom referred to Pantaleo as "the victim of a presidential run."
Unfortunately, Pantaleo's firing also came in the wake of a series of escalating instances of NYPD officers being taunted, threatened and assaulted while in the field.
Not long after two videos of cops being doused with water went viral, another video showed cops being swarmed by a crowd in The Bronx. That melee resulted in one cop being hit with an open container of Chinese takeout; and in yet another incident reported by The Post over the weekend, officers in Brooklyn were hit with " 'air mail' objects thrown down at cops from rooftops."
After Pantaleo's dismissal, a chance to move past cop-hating
As expected, Police Commissioner James O'Neill fired Officer Daniel Pantaleo...
Police morale shouldn't dictate the outcome of investigations properly within the purview of a police department's oversight mechanisms. But neither should such decisions be colored by the politics of our moment.
The pursuit of justice does not require sending police the message that their presence is not appreciated or wanted in Gotham's high-crime neighborhoods. Will politicians stop doing precisely that?
Rafael A. Mangual is a fellow and deputy director of legal policy at the Manhattan Institute for Policy Research, and a contributing editor at City Journal.
Filed under daniel pantaleo , eric garner , james o'neill , nypd , pat lynch , policing
The antifa wars are all about machismo
Profanity at Women's March proves protesters aren't learning: Devine
Cuomo outplays de Blasio again, this time in the budget wars
Karol Markowicz
How to not lose friends over politics during 2020 election
Browse covers archive


Are you in Picton and need a Lawyer?
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Missouri Gov. Jay Nixon has signed legislation putting an end to a controversial policy in which driver's license offices were making electronic copies of applicants' personal documents. Nixon announced yesterday that he had signed the legislation, and the policy change took effect immediately.


The popularity of television crime dramas has made CSI, or crime scene investigation, a household term. Crime scene investigation involves collecting, preserving, and analyzing evidence and other data found at crime scenes. Forensics experts who participate in these processes play vital roles in the legal system. Schools offering Forensic Science degrees can also be found in these popular choices.
The term 'crime scene investigation' has become largely recognizable, thanks to pop culture and television media. Crime scene investigation, often referred to as CSI, involves collecting, preserving, documenting, and analyzing evidence found at crime scenes. Forensics, the application of science in the legal realm, plays a big role in crime scene investigation and a number of forensic professionals are involved in the process.
While the TV drama version of crime scene investigation involves expensive, high-end equipment, real-life investigators generally work with limited budgets and must rely on their creativity to get the job done. On-screen depictions of crime scene investigation also occasionally skew the reality of working in the field by misrepresenting the career profiles of real-life professionals.
Forensics experts are responsible for two main tasks: retrieving and processing evidence found at crime scenes and acting as expert witnesses. The first of these tasks involves studying physical evidence, analyzing clues recovered from crime scenes, and drawing conclusions based on evidence and one's expertise. The second part of a forensic expert's job requires communicating his or her complex findings to judges and juries in understandable terms.
According to the Bureau of Labor Statistics (BLS, www.bls.gov), the employment of detectives and criminal investigators is expected to decline by 1% between 2014 and 2024, while forensic science technicians could see employment opportunities increase by about 27% during the same decade.


Whether he's DJ'ing festivals under his own name, recording as Plastikman, or traveling with the ENTER events, Richie Hawtin remains connected and committed to stretching, morphing and breaking the limits put on making music, especially with his normal working genres of techno, electro-house and deep dance music.
Growing up in Windsor, Ontario, right across the Canadian border from Detroit, Hawtin felt the pull of the Motor City from an early age, though the initial attraction wasn't musical. "I was one of the Windsor kids who wanted to go to Detroit," Hawtin says. "I wanted to search for more and that search didn't start with music. It just was going resale shopping and things like that, but it led me to music and eventually going beyond the music that I was into at that point, like Electro and breakdancing stuff, and finding music that was actually coming from Detroit, the early Detroit—Juan Atkins, Derrick May, Kevin Saunderson, Eddie Fowlkes, all these guys. So that time of being young, and kind of innocent and just sucking everything in, still somehow is the blueprint of what inspires me and how I approach making music, how I approach DJ-ing, and how I approach being creative with technology. Those early days watching Derrick May DJ, Jeff Mills, even actually before I watched Jeff Mills I used to listen to him on the radio. Because in Windsor, we'd get all the Detroit stations and you could even close your eyes and imagine what Jeff was doing. His name was "Wizard" and that was like industrial music and techno and hip-hop all smashed together and you're like, 'how the hell is that possible?' And one of the things that still resonates with me today was [with] all those guys, it was about mutating and ripping something apart, [then] putting it back together in a new way as a DJ. But even in their music, it was something based upon [what's happening] now, and I guess that would have been their own influences, but it was always looking forward. So if I'm in the studio or if I'm creating something from scratch or if I'm in front of a crowd and I'm DJing, I want to create something that sounds new and from the future. I don't [want] people to have a reference point. I'm not a painter trying to do a landscape of what it looks like out of my window. I want people to close their eyes and feel that they're getting taken away and that they're getting pulled inside my imagination.
"Techno to me—then, now and in the future—will always be about throwing the rules away and looking forward. And remember, like 20 years ago, 25 years ago, you didn't have to throw the rules away because there were no rules. Most of us were buying used synthesizers from the pawn shops. They very rarely came with a manual. You get a DX-100 or DX-7 without a manual and try to figure that [stuff] out. Probably you'll never figure it out, but somehow you figure out enough to make your own sound. That was part of the fascination and discovery with technology going into the studio to see what kind of crazy sound you could make.
"At that time, the machines I used and all of us used were what we found at the pawn shop. Of course, we were like, 'Wow, it'd be great to get one of these Moogs," but they never came up and if they did, they were way expensive. We were looking for things that were like 50 bucks. Like a TR, a TB-303, 75 bucks — an 808.
"Some guy called me up, 'Hey, I think I found some stuff that you may like.' You know, some guy in Windsor came over, opened it and was like a brand new 808. "I don't know what the hell you can make with this but how much you want to give me?' 90 bucks, Canadian. Okay. It was nothing. You know?
"So the machines I love now—even the textures of sound—weren't something that I had a choice about. [For instance,] I love Dave Smith instruments because in the beginning I had a Pro-1, a Sequential six-track and I just use them all the time. Same with the 101, the Roland [SH-]101, and 202, 303, 808, 909, 707, 727. This is stuff that I found and that just became my sound. I couldn't afford a Moog and it wasn't until years [later] that I got [one], and I love the sound of that instrument. But when I'm very honest, other people use a Moog much better than me because they've spent time and that's their sound.
"It goes back down to being an artist. You know, for me, Dave Smith is an artist. Moog was an artist. I'm an artist. We all hear things differently. It's not that's better or that's worse. It's just like I resonate with those frequencies over there and as much as I resonate and get turned on by certain records, a lot of these records that I like have a certain sound that I like which actually quite often has some of the synthesizers I like by some of the designers, who are actually some of my friends now, who I like. So it all really makes sense. It's all about frequencies and resonating and finding a vibration that's real and honest and you.
There's always been a strong connection between DJing and production.
"I've never been one to sit too long. Let me just say that again—I've never been one to really enjoy sitting in front of the computer during the creative moment for too long. Later on, editing and stuff—I love editing, back edits and this and that. But drawing notes on a computer screen—I've always enjoyed grabbing something and manipulating and moving it and seeing what happens. To me, there's always been a strong connection between DJing and production. I think that's why I still love hand hold, grabbing things and manipulating. My studio work is, of course, programming drum machines and setting up a number of rhythms and then leaving on the machines and the computers all running and jamming on the mixing console and those machines nearly in a dub kind of style. You know, pressing 'record' and jamming. Sometimes you're 20, 30, sometimes an hour and then finding a great chunk within that which is a song.
"I used to prepare for my exams and tests the night before, so I was never that guy. I love the energy that is nearly like I'm back on the dance floor. When you're a participant and you don't know… I think I know what record's coming in or is that something, that frequency is being made by something else I never heard before and it's like this kind of uneasiness and this excitement. It's just something magic and that's the moment I'm trying to create on the dance floor or even with the records.
"So that's the situation I put myself in. I have Ableton [Live] going. I have TRAKTOR. I have thousands of records. I have certain drum sounds in my Push that I've decided I like to use. So I guess I have all the sounds and the frequencies in place, but Push and Ableton start empty. My record box basically starts empty. I may have one or two records that are a good intro and then I go for it.
"I don't go back and look at what I play or how I play it. I record. We tweet out what I'm playing live as a playlist for everybody, but I try to remember and if I feel that I need to try and go back to a point on the next gig, maybe I will, but you don't really remember exactly what happened anyway, so it's kind of up to chance. You know, when I'm playing I'm listening through my playlist and 90% of the time I don't think, 'Ah, I know I'm going to play that record next.' I'm listening and I'm, 'Okay, those frequencies sound good to what is out here right now — boom.' And by the time I realize what I'm playing, you're already hearing it. I often describe it as a stream of consciousness, an audio stream. And yeah, that's the best way I can explain that — whatever's happening.


- Gems of California scenery, no. 646.
Lawrence & Houseworth, Publisher. Fallen Leaf Lake and Tahlac Mountain, Valley of Lake Tahoe . , 1866. [Published] Photograph. https://0-www.loc.gov.oasys.lib.oxy.edu/item/2002723569/.
Lawrence & Houseworth, P. (1866) Fallen Leaf Lake and Tahlac Mountain, Valley of Lake Tahoe . , 1866. [Published] [Photograph] Retrieved from the Library of Congress, https://0-www.loc.gov.oasys.lib.oxy.edu/item/2002723569/.
Lawrence & Houseworth, Publisher. Fallen Leaf Lake and Tahlac Mountain, Valley of Lake Tahoe . [Published] Photograph. Retrieved from the Library of Congress, <www.loc.gov/item/2002723569/>.


Please join the MVCC Homeless Solutions Committee "Welcome Home Event".
We can make a difference! Please see the full flyer attached - help us spread the word.
This event will be hosted at the Mar Vista home of Susan Klos.


The Beaufort Scale, used throughout the marine world, has developed over many years since it was first devised by Admiral Francis Beaufort in 1806.
Today, the Beaufort scale is defined for seamen in terms of sea state. It's emphasis is more on the observed effect of the wind, rather than the actual wind speed.
Francis Beaufort was born in 1774 in County Meath, Ireland and began his nautical career at 13 as a cabin boy in the Navy. He realised that it was important whilst at sea to be aware of, and to know about weather conditions, so started to keep a meteorological journal, which he continued to do until he died.
By the age of 22 Beaufort had risen to the position of lieutenant. Whilst on assignment in 1812, which was a combined hydro graphic study and patrol mission against the pirates operating out of the Levant, the mission came under attack and Beaufort got hit in the groin by sniper fire, leaving him with a fractured hip. His ship was called back and Beaufort's Navy career changed direction at this point. He was given the position of Hydrographer to the Admiralty, and never returned to the sea. Beaufort rose to the position of Rear Admiral, and worked in the Navy until two years before his death in 1857, serving for 68 years.
Beaufort invented the Beaufort Scale in 1806, and used it from then onwards. The Royal Navy then adopted his method in 1838. Although over the next hundred years the Beaufort scale was slightly adapted, it is still based around Francis Beaufort's original concept.


text:
var Attachment = require('../../models/Attachment');

module.exports = {
    'Attachment': function(it, setup) {
        it('should get fullName from filename + ext', function(expect) {
            var att = new Attachment({ filename: 'example.mode.html' });
            expect(att.get('fullName')).toBe('example.mode.html.js');

            att.set('filename', 'example/mode.html');
            att.set('ext', 'txt');
            expect(att.get('fullName')).toBe('example/mode.html.txt');
        });

        it('should be able to get shortName', function(expect) {
            var att = new Attachment({ filename: 'really/long/path/file' });

            expect(att.get('shortName')).toBe('file.js');
        });

        it('should be editable if a text type', function(expect) {
            var att = new Attachment();
            expect(att.isEditable()).toBe(true);
            att.set('ext', 'css');
            expect(att.isEditable()).toBe(true);
        });

        it('should not be editable if img', function(expect) {
            var att = new Attachment({ ext: 'png' });
            expect(att.isEditable()).toBe(false);
        });
    }
};

redpajama_set_name: RedPajamaGithub


i am a guy, fun loving , like to socialise meet new people and make friends. I am looking for a companion, soulmate, someone to share my loneliness and boredom. honesty , simple, loving, caring and understanding are some of my charasteristics.
Just a normal guy, not trying to be special.
I am exciting,love challenges,reliable,great sense of humour,good listener.
fun loving,active,funny,honest and loves a good party.
Im very decent , academicaly active & outgoing man. Im Handsome I guess, that's what all people say. Im Currently a student at the University of Free State.
i am a man of good, courteous conduct.
Love the outdoors, animals, friends and family.


text:
HomeCongressional DelegationHouse of Representatives
Congressional Delegation (House of Representatives)
Representative Diana DeGette
First Congressional District
Rep. DeGette serves on two committees: the House Committee on Energy and Commerce and the House Committee on Natural Resources. She also serves on several subcommittees and is chair of the Oversight and Investigations panel of the House Committee on Energy and Commerce.
2111 Rayburn House Office Building
Denver District Office
600 Grant Street, Suite 202
Representative Joe Neguse
Second Congressional District
Congressman Joe Neguse serves as a member of the House Judiciary Committee, House Natural Resources Committee and the House Select Committee on the Climate Crisis. His priorities in Congress include lowering health care costs and prescription drug prices, raising workers' wages, ensuring greater accountability in government, protecting our treasured public lands, and fighting the existential threat of climate change.
2503 Walnut Street, Suite 300
1220 S. College Avenue, Unit 100A
Scott Tipton
Third Congressional District
Scott was first elected to the U.S. House of Representatives in 2010. He currently serves on the House Committee on Financial Services. He is Co-chairman of the Congressional Small Business Caucus and Vice Chair of the Congressional Western Caucus. Throughout his time as a U.S. Representative, Scott's priority has been to advance policies that grow jobs, support the economy, and keep Colorado's families safe and secure.
218 Cannon HOB
835 E. Second Avenue
Representative Ken Buck
Fourth Congressional District
Congressman Ken Buck (CO-04) serves on the House Judiciary Committee and the House Foreign Affairs Committee. He serves as the Ranking Member on the House Judiciary Subcommittee on Immigration and Citizenship, and he also serves on the Subcommittee on Antitrust, Commercial and Administrative Law. He also serves on the House Foreign Affairs Subcommittees on Oversight and Investigations; and Western Hemisphere, Civilian Security, and Trade.
900 Castleton Rd.
1023 39th Ave, Suite B
Representative Doug Lamborn
Fifth Congressional District
Doug Lamborn serves on the the Strategic Forces and Emerging Threats and Capabilities Subcommittees of the House Armed Services Committee. Doug also serves as the Chairman of the House Natural Resources Committee's Subcommittee on Water, Power, and Oceans. Throughout his time in public service, Doug has been a leader on pro-family and small business issues, a strong advocate for our veterans and our military, and a supporter of lower taxes, immigration reform, less government spending, and protection of the Second Amendment.
2371 Rayburn HOB
1125 Kelly Johnson Blvd. Suite 330
Representative Jason Crow
Sixth Congressional District
Congressman Jason Crow serves on the Armed Services Committee, Subcommittee on Intelligence, Emerging Threats and Capabilities, Subcommittee on Readiness, Small Business Committee. He is also Chairman of the Subcommittee on Innovation and Workforce Development and serves on the Subcommittee on Economic Growth, Tax and Capital Access, Subcommittee on Rural Development, Agriculture, Trade and Entrepreneurship.
3300 S Parker Rd #100
Representative Ed Perlmutter
Seventh Congressional District
Representative Ed Perlmutter is proud to serve on three committees in the House: Financial Services; Science, Space and Technology; and Rules. Ed is also a member of the New Democrat Coalition which is focused on harnessing American ingenuity and innovation to create new jobs, greater economic prosperity and a safer and more secure future for our county.
12600 West Colfax Avenue,


text:
(function() {

  'use strict';

  var app = angular
    .module('devPortal')
    .controller('SectionsCtrl', SectionsCtrl);

  /* @ngInject */
  function SectionsCtrl($log) {
    var vm = this;

    vm.getSectionItems = getSectionItems;
    vm.getTitle = getTitle;
    vm.getIntro = getIntro;
    vm.getBody = getBody;

    function getSectionItems() {
      return homeSectionItems;
    }

    function getTitle(index) {
      return homeSectionItems[index].title;
    };

    function getIntro(index) {
      return homeSectionItems[index].intro;
    };

    function getBody(index) {
      return homeSectionItems[index].body;
    };
  };

  app.filter("sanitize", ['$sce', function($sce) {
    return function(htmlCode){
      return $sce.trustAsHtml(htmlCode);
    }
  }]);

  var homeSectionItems = [
    {
      title: '#ProtectNotHide',
      intro: 'Introducing Web Security as a Service ...',
      body: '
Develop, build and deploy web/cloud connected applications with enterprise grade security that scales as your application and customer base grows.
'+
      '
Integratingfactor.com IDP service provides following Web Security as a Service offerings
'+
      '
AuthN/AuthZ framework using OAuth2</il>'+
      '
RBAC customized to your application needs
'+
      '
Usage throttling customized to your application needs
'+
      '
Secured user on-boarding and multi-factor authentication
'+
      '
Protection against liability from security breach
'+
      '

Develop secured products with RBAC from blogs to banks.
'
    },
    {
      title: '#BYOI',
      intro: 'Integratingfactor.com framework is built with "Bring Your Own Identity" model...',
      body: '
End users should not have to create multiple different accounts to access multiple different applications. Also, applications should be able to work together for a single user, with appropriate authorizations from the end user.'+
      '
This translates to following end user benefits:'+
      '

easier to adopt strong credentials because user does not have to remember multiple different credentials</il>'+
      '
centralized multi-factor authentication for all services, reducing/mitigating account hijacking and spoofing
'+
      '
simplified user experience with single identity to access different applications

' +
      '

Application developer benefits:
' +
      '
BYOI model reduces cost of security implementation significantly with amortization acrosss multiple applications
' +
      '
no additional federation work required to implement inter-operability between 3rd party applications' +
      '
applications have no additional cost to leverage continuous imporovements in security features for user identity management
'+
      '

Reduce your liability, increase your profitability and build secured products faster with no upfront cost.
'
    },
    {
      title: 'Free Projects <button disabled class="btn btn-success pull-right">Coming soon</button>',
      intro: 'Developers can create free projects with following features...',
      body:'

2 free Endpoint app registrations	Endpoint apps are the clients in your project distributed to end users to access your application

1 free Backend app registration	Backend apps are the services in your project that endpoint clients interact with

1k registered users	Upto 1000 users can be registered for free with each project

Custom user roles	Projects can define custom roles for users, in addition to standard roles

'
    },
    {
      title: 'Premium Projects<button disabled class="btn btn-success pull-right">Coming soon</button>',
      intro: 'Upgrade projects as they grow for following premium features...',
      body: '

additional Endpoint app registrations	$1 / mo / app

additional Backend app registrations	$10 / mo / app

additional active users	$10 / mo / 1k users

Service app registration for user management	$100 / mo

'
    }
  ]
})();


Abstract: Meat production has come under increasing scrutiny from consumers and citizens who feel that certain practices are unethical and negatively affect farm-animal welfare. Animal welfare can be viewed as both a scientific and social concept, and purchasing products with animal welfare claims can be considered an act of 'ethical consumption'. The present paper reviews research that examines consumer attitudes to animal welfare and highlights tensions between consumer and citizen attitudes and behaviours, and assumptions that are made within these studies. We present our own research into motivations to purchase free-range eggs as an example of research that attempts to unpack these assumptions, in particular, that such purchases are made out of concern for animal welfare. We present a further example of our own research that attempts to identify how attitudes to meat production are socially constructed. We conclude with recommended strategies to engage the broader community in discussions about animal production, so as to improve industry–community communication about farm-animal welfare in meat-production industries.


So many of us pride ourselves on our ability to lift heavy. Maybe you're even one of the best workout programmers - but have you ever thought about joint therapy?
If you're unfamiliar with how muscles work on a very specific level than your joints may be taking a big hit. Every time you lift something, contract a muscle, you put stress on the joints - and more specifically the bones.
You see, muscles are attached to tendons which then attach to the bones. When you lift something heavy you put stress through this system of connections and can put consistent stress on bones and joints, which, if gone untreated can lead to injuries such as avulsion fractures, stress fractures and even muscle tears.
This is worst case scenario - and it's not something you want to get into. Taking care of your joints can be one of the best ways to ensure both strong training sessions and faster recovery.
Here are Three things you can do to take care of your joints.
Perhaps one of the most overlooked components of any training program is hydration. Far too many of us just drink coffee and a couple cups of water daily.
Hydration means healthy muscles and healthy joints. Do yourself a favour and start to weigh yourself before and after each workout. For every pound you lose you should drink 20 ounces of water in order to compensate for water loss and keep performance up.
The entire benefits of eating fish and soy are to get as much omega 3 and protein as possible. Yes, you may have heard that calcium forms strong bones, but chances are you already obtain more than you need through milk, nuts/seeds and other grains (sesame seeds, for example, contain more calcium per gram than most foods).
Eating fish is a great way to obtain the clean protein that is relatively low in fat, and high in omega fatty acids. If you're not a fan of fish, then grab yourself some soy products such as edamame beans or tofu.
Soy products will also provide you with a decent source of fibre to help limit inflammation to joints and other soft tissue.
These are the three kings of joint therapy that should be included in your training program - especially if you lift heavy. Supplementing with Glucosamine, chondroitin and MSM will ensure that your joints are getting the specific inflammation reduction and healing properties that are needed in order to continue to grow stronger without creeping into injury territory.
If you're serious about your training and want to ensure growth without becoming injured then check out our post-workout formula. FALLOUT contains all the necessary amino acids to help sustain muscle growth, hydration tools such as MCT's and a matrix of glucosamine, chondroitin and MSM to ensure strong joints and fast recovery.
This is not a supplement that you want to miss. Take care of your joints and sustain muscle growth with FALLOUT.


If you are struggling, you are not alone. God hears your prayers and cares about your needs. We work to assist people facing hard times and to equip men, women and families to overcome a cycle of lack and defeat. When you visit with us, you are given the utmost respect and care, knowing that perhaps this is a very sensitive time of transition or crisis in your life. All of the resources we provide are for residents of New Jersey, free of charge, and no strings attached. If you plan to come to our Food & Clothing Distribution, simply bring two proofs of your current residence (Driver's Licence, current mail, etc.), and one proof of income (pay stub, state award letter or ID). We also encourage you to bring carts and bags to transport your items.


Portsmouth Cathedral has committed to giving entrepreneurs a helping hand as one way of responding to the difficult economic climate in Britain.
The cathedral unveiled plans in the House of Lords on Tuesday to help entrepreneurs launch small businesses.
The Cathedral Innovation Centre will open up spare office space to 14 fledgling businesses and offer them a package that includes desk space, a start-up loan from the Parity Trust, administrative support and the chance to meet business leaders.
Local business people, many from within the churches, will also be invited to act as mentors for the new entrepreneurs.
The project is a partnership between Portsmouth Cathedral, the University of Portsmouth's Business School, and the Joint Venture, a new social enterprise that works with local government and community organisations to make the most of their assets, resources and people.
If successful, the project could be started in other cathedrals and large churches across the country to create a network of innovation centres.
There are already plans afoot to launch the project in Southampton, where a building has been offered to run the venture.
Unveiling the plans in the House of Lords yesterday, Baroness Berridge said: "Perhaps the major contribution needed from our faith communities today is job creation.
Francis Davis, the chairman of the Cathedral Innovation Centre, said: "When St Paul's and other cathedrals saw the Occupy protests recently, there were very few practical alternatives being suggested.
"But the Church is in a position to help businesses to launch, develop and expand. This is a concrete response at a time of economic crisis, when talk is obviously not enough.
"We can only have growth in our economy if real firms are creating real jobs, and we can help to attain social justice if we are creating businesses that are run along responsible lines."
Mr Davis is looking for 50 mentors from across Hampshire and the Isle of Wight to give advice on all aspects of business, from management, to recruitment, marketing and human resources.
"I'd like to tap into the expertise that's available in our churches," he said. "It will be a new way of volunteering for members of our churches.
"And in a new approach to philanthropy, we're looking for 500 people to invest £300 to get these businesses off the ground. They would be helping people back into work and stimulating the economy."
Among the first businesses looking to move in are technology start-ups founded by local students, a healthcare firm spun out from the local hospital, and a community leadership and training foundation.
Another firm that has already signed up is the Turnaround Foundation, which offers management consultancy to small firms in trouble, and ethical business advice to firms facing insolvency.


Problems on app crashes are usually tied to corrupted files and data from within the errant app. But in some cases, an app can become a bit shaky due to other factors including faulty updates, invalid settings, system bugs and malware. Generally, the main issue is from within the software structure and is therefore rectifiable.


Now onto point 13 from my earlier post.
So...... I have a Moby Wrap and have used it since day one. At the beginning you simply do the Newborn hug hold, where the legs are in lotus, where his feet are tucked up inside like a little frog. He used to go to sleep within seconds of me putting him into this. Or you can do the cradle hold (which I only did once to be honest). Both of those holds are shown on this You Tube instructions video.
THEN, you move onto the legs out stage... which you do not do at a certain weight, rather... when they start to kick out of the earlier tie. I was out shopping one day and when he woke up, he was kicking to get out. I said Mum, when we get home, I need to read the instructions and work out how to tie it, so that his legs are out. I was so nervous about it BUT it was easy. The legs out stage is very well demostrated on this You Tube instructions video. I agree with this lady on two counts (1) I also put the arms and head in when he falls asleep (2) I also can have him in there for 1-2 hours without my back feeling any strain. Went on a one hour walk and felt like I could do it again. The Moby Wrap really does distribute the weight all over the body so that no one part takes all the strain - in fact, what strain?!?!?!
You can do so many things around the house while your little one is in there, if they're in a mood where they are demanding to be carried.
If you are not yet convinced, read 10 reasons to Wear Your Baby.


text:
Q: Error: invalid operand for instruction using indexed addressing and Clang I'm catching compile problems using Clang. GCC compiles the program fine. The program uses indexed addressing.
The errors are:
$ clang++ -g2 -O1 test.cxx -c
test.cxx:19:10: error: invalid operand for instruction
        "movq     (%[idx],%[in]), %[x]   ;\n"
         ^
<inline asm>:5:23: note: instantiated into assembly here
movq     (%rbx,%rsi), -8(%rsp)   ;
                      ^~~~~~~~
test.cxx:20:10: error: invalid operand for instruction
        "movq     (%[idx],%[out]), %[y]  ;\n"
         ^
<inline asm>:6:23: note: instantiated into assembly here
movq     (%rbx,%rdi), -16(%rsp)  ;
                      ^~~~~~~~~
test.cxx:21:10: error: invalid operand for instruction
        "cmovnzq  %[x], %[y]             ;\n"  // copy in to out if NZ
         ^
<inline asm>:7:20: note: instantiated into assembly here
cmovnzq  -8(%rsp), -16(%rsp)             ;
                   ^~~~~~~~~
test.cxx:22:10: error: invalid operand for instruction
        "movq     %[y], (%[idx],%[out])  ;\n"
         ^
<inline asm>:8:21: note: instantiated into assembly here
movq     -16(%rsp), (%rbx,%rdi)  ;
                    ^~~~~~~~~~~
4 errors generated.

How do I fix the problem? (Or how do I tell Clang to stop defining __GNUC__ to keep it out of GCC code paths).

$ cat test.cxx
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdint>

void test_cmov(uint8_t in[96], uint8_t out[96], uint64_t flag)
{
#if defined(__GNUC__)
    const uint32_t iter = 96/sizeof(uint64_t);
    uint64_t* optr = reinterpret_cast<uint64_t*>(out);
    uint64_t* iptr = reinterpret_cast<uint64_t*>(in);
    uint64_t idx=0, x, y;

    __asm__ __volatile__ (
        ".att_syntax                     ;\n"
        "cmpq     $0, %[flag]            ;\n"  // compare, set ZERO flag
        "movq     %[iter], %%rcx         ;\n"  // load iteration count
        "1:                              ;\n"
        "movq     (%[idx],%[in]), %[x]   ;\n"
        "movq     (%[idx],%[out]), %[y]  ;\n"
        "cmovnzq  %[x], %[y]             ;\n"  // copy in to out if NZ
        "movq     %[y], (%[idx],%[out])  ;\n"
        "leaq     8(%[idx]), %[idx]      ;\n"  // increment index
        "loopnz   1b                     ;\n"  // does not affect flags
        : [out] "+D" (optr), [in] "+S" (iptr), [idx] "+b" (idx),
          [x] "=g" (x), [y] "=g" (y)
        : [flag] "g" (flag), [iter] "I" (iter)
        : "rcx", "cc"
    );
#else
    if (flag)
        std::memcpy(out, in, 96);
#endif
}

int main(int argc, char*argv[])
{
    uint8_t in[96], out[96];
    uint64_t flag = (argc >=2 && argv[1][0] == 'y');

    std::memset(in, 0x00, 96);
    std::memset(out, 0x00, 96);

    std::memcpy(in, argv[0], std::min(96ul, std::strlen(argv[0])));

    test_cmov(in, out, flag);
    std::cout << (const char*)out << std::endl;

    return 0;
}

$ gcc --version
gcc (GCC) 8.3.1 20190223 (Red Hat 8.3.1-2)
...

$ clang --version
clang version 7.0.1 (Fedora 7.0.1-6.fc29)
Target: x86_64-unknown-linux-gnu
...

$ lsb_release -a
LSB Version:    :core-4.1-amd64:core-4.1-noarch
Distributor ID: Fedora
Description:    Fedora release 29 (Twenty Nine)
Release:        29
Codename:       TwentyNine

A: You used "=g" constraints that let the compiler pick a mem operand for %[x] and %[y].
Use "=r" instead.
Your template uses movq  (%[idx],%[in]), %[x], which obviously fails to assemble if %[x] is memory, because x86 doesn't support 2 explicit memory operands for any instruction.
The clang difference from gcc here is that it likes to pick memory operands if given the choice.  (This is an optimizer bug, IMO, but not a correctness problem.  Your inline asm is buggy and only happens to work with GCC because it picks a register for [x] "=g" (x))
This is obvious if you read the error messages:
<inline asm>:5:23: note: instantiated into assembly here
movq     (%rbx,%rsi), -8(%rsp)

...

cmovnzq  -8(%rsp), -16(%rsp)

Obviously these are not valid instructions.

If you care about clang, generally avoid giving it the choice of a memory operand for inline asm unless it will definitely help in the normal case.
When you write a whole loop in inline asm, you definitely want to make the compiler spill something else to free up registers if necessary for some loop temporaries.  Or really any time you use and operand multiple times in the same inline asm block.  GCC doesn't look at this, and won't know the cost of choosing memory.  (And clang is just dumb and chooses memory even when there are lots of free regs.)

redpajama_set_name: RedPajamaStackExchange


A week before my labour started I was admitted to hospital for high blood pressure and stress to my baby. I was monitored carefully and given tablets to bring down my blood pressure and a few other tablets every 3-4 hours, 24/7. I was told I'd be induced and my water would be broken early the next morning; however my water broke naturally at about 1-2 am. I had been given an injection of pain meds as I could not handle the pain of being induced, then after that wore off I was taken to the birthing room for an epidural as my blood pressure was too high and my little baby was very stressed. If my blood pressure was not brought down and my baby was not "de-stressed" then I would have had an emergency caesarean. But lucky for me, the epidural worked and my blood pressure dropped. I calmed down and so did my baby.
I was then able to give birth to my baby boy with help of some forceps as I was completely exhausted and could not push by myself.
Next time I hope I can have a natural birth from the start to finish, all of that pain exhaustion needles etc was worth it. I love my son more than anything in the world!


Hamburg is, without a doubt, one of the most popular travel destinations in Europe. No wonder for Hamburg is a city that not only has countless impressive attractions; it also has breathtaking musical productions.
The STAGE THEATER IM HAFEN & STAGE THEATER AN DER ELBE are even within sight of the MADISON. Disney's musical THE LION KING as well as MARY POPPINS are performed here. And you don't need a lengthy journey to arrive punctually for the beginning of the show.
- The last ferry departs 1.5 hours after the end of the musical.
Right in the heart of Hamburg's most sinful mile, the Reeperbahn, lies the STAGE THEATER OPPERETTENHAUS. It is the most famous street in the world next to Broadway and 5th Avenue in New York. It's no surprise that it is a genuine tourist attraction and has a true gem to offer musical fans: The musical TINA - a Tina Turner musical is performed here.
- A taxi takes about 10 minutes and our recepiton team would be happy to order one for you.
In the center of the Hamburg Altona is where you will find the THEATER NEUE FLORA. Here you can immerse yourself in a world of magic costumes, magnificent stage productions and captivating music. The musical PARAMOUR is sure to enchant you. Of course we have personally tested all the musicals and can whole heartedly recommend this Cirque du Soleil musical.
- By car you need 10-15 minutes for the 3.5 km. For a fee you can park in the car park next to the theater in the Alsenstrasse.
- To travel with public transportation you take the U3 from Baumwall in the direction of Barmbek, then at station Feldstrasse transfer to bus 3 in the direction of Trabrennbahn Bahrenfeld and get off at the bus stop Sternbrücke. From here it is just 200m jaunt.