Once he had settled down definitively in Paris in 1904, Picasso got back in touch with several of his old friends from Barcelona. Without doubt, it was the ties to Ricard Canals which were strengthened the more in these new circumstances, and the portrait of Benedetta Bianco, the sentimental partner of Canals, testifies to that. At the Bateau Lavoir the two couples – Picasso and Fernande, and Canals and Benedetta – established a very close friendship: according to Fernande, Picasso would spend the days in the studio of Canals and Benedetta would make use of her culinary skills to feed everyone when the economic resources were scarce.
Carried out in the autumn of 1905, the portrait of Benedetta Bianco is particularly interesting, as it perfectly illustrates one of the two main aspects that can be distinguished throughout the artistic trajectory of Picasso. On the one hand there are the works belonging to the intellectual-stylistic development of the artist, those which are the result of his growth and progress. On the other hand, there are the works which result from the biographical storyline of Picasso, which can be understood as the specific visual notes in his personal diary. This portrait clearly belongs to the second aspect.
In 1904 Picasso abandoned the Blue Period to go into the so-called Rose Period, in which the circus world would become the main subject of his works. Despite maintaining the stylistic characteristics of warm pastel colors, the portrait of Madame Canals -with a serious face and adorned with a black mantilla- is at odds with the topics of that time. In this painting, the artist uses elements which were typically Spanish, which were a recurring topic in the work of his friend Canals and very popular in the French art market of the time, fascinated by the exoticism of everything Spanish. Such was the fascination for these scenes, that the art dealer Durand-Ruel funded a trip for Canals around the Spanish geography to provide him with inspiration in 1904.
Similarly revealing are the clearly visible influences of the Grand Masters of painting in this work in particular. The harmonious spirituality that emerges from the portrait of Benedetta, which brings to mind The Lady with a Fan by Velázquez (Wallace Collection, London), constitutes for Picasso the exploration of a new limit: the psychological portrait. In a similar way, placing the focal point of the work on the face of the personage on a less worked background, evokes the canvasses of the French artist Ingres, whose work Picasso saw for the first time one year before painting Benedetta's portrait. During what would be the definitive trip to Paris, which he undertook with the artist Junyer i Vidal, they stopped in Montauban, the city of birth of the painter that hosts the Ingres Museum. This event is included in the series of drawings carried out by Picasso in the form of an auca, a sort of comic strip very much in vogue in the Barcelona of the turn of the century, and which is now part of the collection of our museum. The following year he would see the work of French genius again, as one of his paintings was exhibited in the Salon d'Automne of that year.
Both Velázquez and Ingres would be a source of inspiration for Picasso throughout his artistic career, as shown the series Las Meninas, also part of our collection, or the famous Les Demoiselles d'Avignon (MoMA, New York).
The trip that Picasso made to Holland in the summer of 1905 led him towards a more sober and classical style. The portrait of Benedetta constitutes an extraordinary work, as it not only exemplifies this new direction and the influences of other artists on Picasso, but also, along with the portrait of Fernande with a black mantilla (Guggenheim Museum, New York), it constitutes a fantastic visual testimony of the close relationship established between Picasso and Canals, and their respective partners, during those hard but emotional first years of Picasso in the French capital. This magnificent portrait will be shown in the exhibition «Picasso Portraits », organized in collaboration with the National Portrait Gallery, London, which will open at the Museu Picasso in March 2017, when we will have the chance to see both works here.


text:
Черношина () — річка в Словаччині, права притока Топлі, протікає в окрузі Бардіїв.

Довжина — 12 км. Витік знаходиться в масиві Ондавська височина — на висоті 440 метрів. Протікає територією сіл Вишня Воля; Нижня Воля; Ольшавце; Кохановце і Гаргай.

Впадає у Топлю на висоті 192 метри.

Примітки

Посилання

Бардіїв (округ)
Річки Словаччини
Річки басейну Ондави

redpajama_set_name: RedPajamaWikipedia


COACHELLA VALLEY – More than 140 volunteers rolled up their sleeves and got their hands dirty logging more than 475 hours of sweat equity in an unprecedented Day of Action for the United Way of the Desert.
Throughout the Coachella Valley, the tireless volunteers used elbow grease at 12 schools in three school districts to create the largest United Way of the Desert Day of Action to date.
The day started promptly at 6:45 a.m. with team leaders showing up on site to meet with school coordinators and get all the assignments in order. Team volunteers showed up at 7 a.m. and got right to work. While volunteers in Coachella planted flower beds with Supervisor V. Manuel Perez at Peter Pendleton Elementary School, others such as Gay for Good Palm Springs were cleaning up playgrounds at John F. Kennedy Elementary School.
Other organizations/companies involved were: Democratic Assemblymember Eduardo Garcia's office, Target (Cathedral City, Palm Desert, and La Quinta), Wells Fargo, Diamond Resorts, Desert Arc, US Bank, and The Rilington Group. The Day of Action wrapped up by 11 a.m.
United Way of the Desert CEO, Kristal Granados, also expressed her gratitude to the volunteers and schools.


text:
What Is a Bevel Cut?
By Chris Deziel Updated October 5, 2018
You make a bevel when you cut the edge of a board at any angle other than 90 degrees.
Image Credit: Wonkee Donkee Tools
Dimensioned lumber comes with square edges, which means the angle between the face of a board and its edge is 90 degrees. If you want to change this angle to anything other than 90 degrees, you have to make what is known as a bevel cut. You can do this with several tools, including a table saw, a circular saw, a router and even a jigsaw, although the latter won't give very accurate results.
Bevel cuts are often decorative. Cabinet trim, interior moldings and picture frames look more elegant with beveled edges than they do with straight ones. Beveling isn't just for looks, though. It's often a necessary procedure to fit moldings, and it's a common technique in joinery. One purely functional woodworking standby, the French cleat, relies on a bevel cut to do its job, which is to support heavy objects from a wall.
A Bevel Cut Isn't a Miter Cut
A compound miter saw makes a miter cut and a bevel cut at the same time.
Image Credit: Miter Saw Hub
Like a bevel cut, a miter cut is also an angle cut, and it's one that carpenters frequently have to make. The most common miter cut is a 45-degree angle at the end of a trim board for a door or window. A miter cut doesn't alter the angle of the edge of the wood with the respect to the face. Two pieces of wood that have been mitered to form a 90-degree corner have edges that are parallel to each other. That's because the angle of the blade that cuts a miter is parallel to the edge of the wood and perpendicular to the face.
If you want to make a bevel cut, you have to change the angle of the blade with respect to the edge and face of the piece of wood you're cutting. You can do this at the same time as you make a miter cut by using a compound miter saw, which has blade angle adjustments in two different planes. First, rotate the saw on its vertical pivot to set the miter angle, then rotate it on its horizontal pivot to set the bevel angle. Finish carpenters often need to do this when installing baseboards and crown molding.
Best Ways to Cut a Bevelled Edge
Using a chamfer bit, you can make a beveled edge with your router.
Image Credit: Carbide Processors Inc.
The most accurate way to bevel the ends of a piece of wood is to use a compound miter saw, but that won't work for beveling along the length. For that, you need either a table saw, a circular saw or a router. Of course, you first need to calculate the angle you need, and that might take a little math.
To make a bevel cut with a table saw, you first change the blade angle from 90 degrees to the desired bevel angle, using the angle adjustment control on the front of the saw. The control is graduated to tell you the blade angle, but the scale isn't always accurate, so it's best to check the blade angle with a protractor or a bevel, which is a tool that carpenters use to measure angles. Once the blade angle is set, just push the wood through to rip it to the proper bevel angle.
You can also rip a bevel with a circular saw, which has an adjustment that allows you to change blade angle. If you want accurate results, be sure to clamp the wood firmly to a workbench, check the blade angle with a protractor or bevel and cut slowly. ,
You can use a router to convert the straight edge of a wood piece to a beveled edge by fitting the tool with an angled cutting bit called a chamfer bit. Clamp the wood firmly and use a bit with the desired angle to shape the edge and produce the desired bevel angle.
Wood Magazine: Bevel Basics and Beyond
Rockler: Woodworking Tools Angle Scales and Terminology
Chris Deziel
Chris Deziel is a contractor, builder and general fix-it pro who has been active in the construction trades for 40 years. He has degrees in science and humanities and years of teaching experience. An avid craftsman and musician, Deziel began writing on home improvement topics in 2010. He worked as an expert consultant with eHow Now and Pro Referral -- a Home Depot site. A DIYer by nature, Deziel regularly shares tips and tricks for a better home and garden at Hunker.com.
How to Set Fence Posts in a Straight Line
How to Figure Angles for Cutting Wood
How to Use a Miter Saw
How to Cut Molding With a Circular Saw
How to Make an Angle Greater Than 45 Degrees on a Table Saw
How to Make Coped Joints in Molding

redpajama_set_name: RedPajamaCommonCrawl


Recreational kayaks are designed with both the beginning paddler and flat water in mind. They are shorter, wider, more stable, and easier to maneuver than touring kayaks but they take more effort to paddle than a touring kayak. They are great for flat water in the rivers and harbor, exploring smaller inlets, or on local ponds. Best choice for the beginner.
Touring kayaks are sportier and a step up from a recreational kayak. They are longer and sleeker than recreational kayaks and have smaller cockpits suitable for attachment of a skirt. They are more efficient in moving through the water, but they are somewhat less maneuverable and less stable than the recreational kayaks. Most touring kayaks have storage for gear. They are perfect for people who are comfortable on the water and want to paddle farther or longer.
Perfect for two adults with room for child or pet as well as picnic lunch and plenty of gear.
Easy-in, easy-out kayak great for the beginner (or claustrophobic). No worries about tipping over because you can just climb back in and the boat is self-draining.
Looking for the latest in paddling – this is it. A full body workout which is becoming a favorite cross-training activity, and with the advantage of height, you'll get great views of the wildlife and what's on the horizon. It's almost like walking on water!


Belmont University will hold its winter 2011 commencement ceremony for graduate and undergraduate students at 7:30 p.m. Dec. 16 in the Curb Event Center.
The University will celebrate the graduation of a total of 364 students. During the graduation ceremony, 275 undergraduate and 89 master's degrees will be conferred.
Dr. Todd Lake, vice president for spiritual development, will present the commencement address.
The ceremony can be viewed live by webcast from the Belmont homepage at 7:30 p.m. Dec. 16.


Discussion in 'Alienware 18 and M18x' started by IXvirusI, Aug 7, 2018.
I got alienware 18 with bios A12 860 sli in it and an i7 4700m,i bought i7 4940mx, and found a seller has 2 980m graphic card or 980m gsync with cheap prices maximum 800 US for both, i need help in how to install them and which one to choose and why, i would really appreciate the help, will it brick any thing or burn my laptop and i hope some one guide me i have 330 power adapter also. and can i use the same heatsink i use for my 860 sli. thankssssss so much for ur time and effort i know answering to this is not easy but i realy love alieanware 18 dnt like the modern laptops design.
G-sync won't work with AW18, need to be non G-sync 980m's.
On average, 4930mx actually overclock better than 4940mx (at least from my experience overall). Either way, you're going to want to use conductonaut liquid metal to properly cool that hotwell 4940mx properly. You can always use something like Kryonaut if you want to be safe but your overclocks will suffer.
The Alienware 18's weren't of the high standard of the M18x r1/r2 and many have blown their motherboards trying to do too much and/or the system ends up blowing out their slave GPUs. Some, not all. Many are running them happily decked out to this day. If you were running dual 860m's, I'd, personally, start with a single 980m along with that 4940mx w/ a passable stable overclock (>=4Ghz on all cores), run it for awhile and re-asses the need for a second 980m.
I have 2 x 980M for sale, also have special alienware 18 980M heatsinks that were made special by Dell for Alienware 18, let me know if interested.
Combination of BIOS for base settings and XTU or Throttlestop to really make it shine (as much as Hotwell can shine ).
You don't want coolabratory liquid pro. You want to use Thermal Grizzly Conductonaut liquid metal. It has supplanted CLU as the go to liquid metal to use.
If the seller can, make sure they have the cards flashed and ready to go with Dell's 980m master and slave vbios (you can always do it yourself if needed). But if the seller has them properly flashed as master and slave, then you should be good to go.
Better yet, if you can, deal with @Raidriar He's a long standing Alienware user and knows his stuff and can provide you with competent heatsinks to boot. You would get much better customized help. I'd still lean towards one 980m to be safe overall with the draw of a 4940mx OC'd on top of 2x 980m's, but that's your call.
Which 4930mx work better between qs and official edition with overclock?
i already have a 4940mx with liquid-cooling.But i can't achieve 4.5ghz or more. Is it necessary take place of 4940mx by 4930mx?qs or official?
Alienware 17 (early 2014) GPU Upgrade?


text:
Q: Why do we need a SynchronizationContext in the EventAggregator for WPF I have read quiet a few implementations of the EventAggregator (Messenger) to communicate between different ViewModels in WPF but I am having difficulties why we necessarily need to use a SynchronizationContext.
I understand that the SynchronizationContext is used to ensure that a Post/Send action runs on the same context's thread. What advantage does it provide in the case of an EventAggregator. Rather, could someone provide and example where the EA would fail if a SynchronizationContext was not used.

redpajama_set_name: RedPajamaStackExchange


Scientists at the Ottawa Hospital Research Institute (OHRI) and the University of Ottawa (uOttawa) have discovered that mice that lack a gene called Snf2l have brains that are 35 per cent larger than normal. The research, led by Dr. David Picketts and published in the prestigious journal Developmental Cell, could lead to new approaches to stimulate brain regeneration and may provide important insight into developmental disorders such as autism and Rett syndrome.
Dr. Picketts and his team created the mice to learn more about Snf2l, which is known to play a role in packaging DNA and determining which genes are active versus inactive. They found that the mutant mice were completely normal, except that they had larger brains, more cells in all areas of the brain, and more actively dividing brain stem cells.
The Snf2l mutant mice are also providing insight into developmental disorders that are associated with changes in brain size. For example, by studying these mice, Dr. Picketts and his team found that Snf2l controls the expression of a gene called Foxg1, which causes the intellectual disability disorder Rett syndrome in some people. While the mutant mice have high levels of Foxg1 and large brains, people with Rett syndrome lack Foxg1 and have small brains. This research shows that Snf2l and Foxg1 work against each other to balance brain size. Autism is also commonly associated with changes in brain size (one third of autistic individuals have a larger brain), however no studies have yet provided a direct link between Snf2l and autism.
The full publication is titled "Snf2l regulates foxg1-dependent progenitor cell expansion in the developing brain", and the authors include: Darren J. Yip, Chelsea P. Corcoran, Matías Alvarez-Saavedra, Adriana DeMaria, Stephen Rennick, Alan J. Mears, Michael A. Rudnicki, Claude Messier and David J. Picketts.
The Ottawa Hospital Research Institute (OHRI) is the research arm of The Ottawa Hospital and is an affiliated institute of the University of Ottawa, closely associated with the University's Faculties of Medicine and Health Sciences. The OHRI includes more than 1,500 scientists, clinical investigators, graduate students, postdoctoral fellows and staff conducting research to improve the understanding, prevention, diagnosis and treatment of human disease. Research at OHRI is supported by The Ottawa Hospital Foundation. www.ohri.ca.


text:
Assembly and Annotation of Expressed Sequence Tags (ESTs) for Functional Analysis of G. arboreum Transcriptome
MEGHA SIHAG, SURESH KUMAR GAHLAWAT AND PRIYANKA SIWACH
Department of Biotechnology, Chaudhary Devi Lal University, Sirsa-125 055 (Haryana), India
*(e-mail : Psiwach20@gmail.com; Mobile : 01666247143)
(Received : March 1, 2019; Accepted : May 8, 2019)
Expressed Sequence Tags (ESTs) analysis from different tissues and organs provides the basis for functional genomics research. In the present work, a total of 33,387 unigenes (7,530 in consensus contigs and 25,857 as singletons) were assembled from all 64,903 cotton ESTs currently available. After systematic assembly of unigenes, these were subjected to annotation, comparative genomics, Blast, Gene Ontology (GO) analysis, and pathway enrichment by KEGG to investigate global functions of the cotton transcriptome. Over 29,658 (88.83%) unigenes were found to show similarities with publicly available proteins. KEEG annotations from GO proteins, results showed enrichment for the category 'metabolism' (79.76%), followed by– for the categories organismal systems (11.23%) and cellular processes (9%). Further, 73.47% unigenes showed 100% similarity to coding sequences (CDS) of G. arboreum. In a comparison with six model plant species, cotton ESTs showed the highest overall similarity to Vitis vinifera. 739 and 1346 simple sequence repeats (SSRs) were identified using two different softwares, SSRIT and WebSat, respectively, of which 628 SSRs were common. The latter were subjected to designing of viable primer pairs for future applications.
Key words :Expressed sequence tags (ESTs), coding sequences (CDS), genome sequence, simple sequence repeats (SSRs), gene ontology (GO)
Functional Analysis of MIR159b Promoter in Leaf and Callus of Transgenic Brassica juncea var. Varuna Using Promoter : GUS Transcriptional Reporter
Estimation of Length-weight and Length-length Relationship of Catfish, Rita rita Inhabiting in the Middle Stretch of River Ganga
Constraints in Dairy Entrepreneurship among Youth in Uttar Pradesh
Endophytic Bacteria of Artemisia as Biological Control Agent for Cercospora Leaf Spot in Paddy
Diversity and Status of Avifauna in Bhindawas Bird Sanctuary, Haryana, India
Overexpression of Calpain and its Effect on Autophagy in Dictyostelium discoideum
Role of Azotobacter, PSB and Sulphur in Yield Improvement for Wheat Crop
Suppression Effectivity of Rice Sheath Blight Disease by Endophytic Bacteria Isolate Artemisia
Studies on Intercropping of Turmeric and Ginger in Mango Orchard
A Review on Anaerobic Digestion of Wastewater


text:
\section{Introduction}

This note is a sequel to \cite{FX}. We assume that the reader is familiar with the basic theory of association schemes as can be found in \cite{BI, BCN}. For background in strongly regular graphs, we refer the reader to \cite{bh, godsilr}. All association schemes considered in this paper are commutative and symmetric. Let $(X, \{R_i\}_{0\leq i\leq d})$ be an association scheme with $d$ classes. For $i\in \{0,1,\ldots ,d\}$, let $A_i$ be the adjacency matrix of the relation $R_i$, and let $E_0=\frac {1} {|X|}J, E_1, \ldots , E_d$ be the primitive idempotents of the Bose-Mesner algebra of the scheme $(X, \{R_i\}_{0\leq i\leq d})$, where $J$ is the all-one matrix of size $|X|\times |X|$. The basis transition matrix from $\{E_0,E_1,\ldots ,E_d\}$ to $\{A_0,A_1,\ldots ,A_d\}$ is denoted by $P=\left(p_j(i)\right)_{0\le i,j\le d}$, and usually called the {\it  first eigenmatrix} (or {\it character table}) of the scheme. Explicitly $P$ is  the $(d+1)\times (d+1)$ matrix with rows and columns indexed by $0,1,2,\ldots
,d$ such that
$$(A_0,A_1, \ldots ,A_d)=(E_0,E_1, \ldots ,E_d)P.$$
Let $k_i=p_i(0)$ and $m_i={\rm rank}(E_i)$. The $k_i$'s and $m_i$'s are called {\it valencies} and {\it multiplicities} of the scheme, respectively. We say that the scheme $(X, \{R_i\}_{0\leq i\leq d})$ is {\it pseudocyclic} if there exists an integer $t$ such that $m_i=t$ for all $i\in \{1,\ldots , d\}$. A classical example of pseudocyclic association schemes is the cyclotomic association scheme over a finite field, which we define below.

Let $q=p^f$, where $p$ is a prime and $f$ a positive integer. Let $\gamma$ be a fixed primitive element of $\mathbb{F}_q$ and $N|(q-1)$ with $N>1$. Let $C_0=\langle \gamma^N\rangle$, and $C_i=\gamma^i C_0$ for $1\leq i\leq N-1$. Assume that $-1\in C_0$. Define $R_0=\{(x,x) \mid  x\in \mathbb{F}_q\}$, and for $i\in \{1,2,\ldots ,N\}$, define $R_i=\{(x,y)\mid x,y\in \mathbb{F}_q, x-y\in C_{i-1}\}$. Then $(\mathbb{F}_q, \{R_i\}_{0\leq i\leq N})$ is an association scheme. We will call this scheme {\it the cyclotomic association scheme of class $N$ over $\mathbb{F}_q$}. The first eigenmatrix $P$ of the cyclotomic scheme of class $N$ is the following $(N+1)$ by $(N+1)$ matrix (with the rows of $P$ arranged in a certain way)
\begin{equation}\label{eigenmatrix}
P=\left(\begin{array}{cccccc}
1& \frac{N-1}{q}&\frac{N-1}{q}&\frac{N-1}{q}&\cdots &\frac{N-1}{q}\\
1&\eta_{_{N-1}}  &\eta_0   &\eta_1 & \cdots     &\eta_{_{N-2}}   \\
1&\eta_{_{N-2}}  &\eta_{_{N-1}} & \eta_0& \cdots  &\eta_{_{N-3}} \\
\vdots & & & & \\
1&\eta_0 &\eta_1 &\eta_2& \cdots  &\eta_{_{N-1}}\\
\end{array}\right)
\end{equation}
where the $\eta_i$'s are the cyclotomic periods  (or Gauss periods) of order $N$ defined by
$$\eta_i=\sum_{x\in C_i}\psi(x).$$
In the above defintion, $\psi$  is the additive character of $\mathbb{F}_q$ defined by
\begin{equation}\label{defaddchar}
\psi: \mathbb{F}_{q} \rightarrow {\mathbb C}^{*}, \quad \psi(x)=\xi_{p}^{{\rm Tr}(x)},
\end{equation}
where $\xi_p=e^{2\pi i/p}$ and ${\rm Tr}$ is the absolute trace from $\mathbb{F}_q$ to $\mathbb{F}_p$.

The following theorem gives combinatorial characterizations of pseudocyclic association schemes.

\begin{thm}\label{pseudocyc}
Let $(X, \{R_i\}_{0\leq i\leq d})$ be an association scheme, and for $x\in X$ and $1\leq i\leq d$, let $R_i(x)=\{y\mid (x,y)\in R_i\}$. Then the following are equivalent.
\begin{enumerate}
\item $(X, \{R_i\}_{0\leq i\leq d})$ is pseudocyclic.
\item For some constant $k$, we have $k_j=k$ and $\sum_{i=1}^{d}p_{ii}^j=k-1$, for $1\leq j\leq d$.
\item $(X, {\mathcal B})$ is a $2-(v,k,k-1)$ design, where ${\mathcal B}=\{R_i(x)\mid x\in X, 1\leq i\leq d\}$.
\end{enumerate}
\end{thm}
For a proof of this theorem, we refer the reader to \cite[p.~48]{BCN} and \cite[p.~84]{henkthesis}. Part (2) of the above theorem will be useful in Section 3.

Let $(X, \{R_i\}_{0\leq i\leq d})$ be an association scheme.
For a partition $\Lambda_0:=\{0\}, \Lambda_1,\ldots ,\Lambda_{d'}$ of $\{0,1,\ldots ,d\}$, let $R_{\Lambda_i}=\cup_{k\in \Lambda_i}R_k$,
for $0\leq i\leq d'$. If $(X, \{R_{\Lambda_i}\}_{0\leq i\leq d'})$ forms an association scheme,
then we say that $(X, \{R_{\Lambda_i}\}_{0\leq i\leq d'})$ is a {\it fusion scheme} of the original scheme.
If $(X, \{R_{\Lambda_i}\}_{0\leq i\leq d'})$ is an association scheme for every partition $\{\Lambda_i\}_{0\leq i\leq d'}$ of $\{0,1,2,\ldots ,d\}$ with $\Lambda_0=\{0\}$,
then we call the original scheme $(X, \{R_i\}_{0\leq i\leq d})$ {\it amorphic}. For a recent survey on amorphic association schemes, we refer the reader to \cite{vanDamM}. Given a partition $\{\Lambda_i\}_{0\leq i\leq d'}$ of $\{0,1,2,\ldots ,d\}$ with $\Lambda_0=\{0\}$,  there is a simple criterion in terms of the first eigenmatrix $P$ of $(X, \{R_i\}_{0\leq i\leq d})$ for deciding whether $(X, \{R_{\Lambda_i}\}_{0\leq i\leq d'})$ forms an association scheme or not. We state this criterion below.\\

\noindent{\bf The Bannai-Muzychuk Criterion.} Let $P$ be the first eigenmatrix of an association scheme $(X, \{R_i\}_{0\leq i\leq d})$. Let $\Lambda_0:=\{0\}, \Lambda_1,\ldots ,\Lambda_{d'}$ be a partition of $\{0,1,\ldots ,d\}$. Then $(X, \{R_{\Lambda_i}\}_{0\leq i\leq d'})$ forms an association scheme if and only if there exists a partition $\{\Delta_i\}_{0\leq i\leq d'}$ of $\{0,1,2,\ldots ,d\}$ with $\Delta_0=\{0\}$ such that each $(\Delta_i, \Lambda_j)$-block of $P$ has a constant row sum. Moreover, the constant row sum of the $(\Delta_i, \Lambda_j)$-block is the $(i,j)$ entry of the first eigenmatrix of the fusion scheme. (For a proof of this criterion we refer the reader to \cite{BannaiSub, Muzthesis}.)

A. V. Ivanov conjectured in \cite{IP} that if each nontrivial relation in an association scheme is strongly regular, then the association scheme must be amorphic.
This conjecture turned out to be false. A counterexample was given by Van Dam \cite{vanDamC1} in the case where the association scheme is imprimitive.
Later on, Van Dam \cite{vanDamC2} also gave a counterexample  in the case where the association scheme is primitive.
More counterexamples were given by Ikuta and Munemasa \cite{ikutam} in the primitive case. However it should be noted that there are only a few known counterexamples to Ivanov's conjecture in the primitive case (cf. \cite{ikutam}).

The purpose of this note is to generalize the counterexamples to Ivanov's conjecture by Ikuta and Munemasa \cite{ikutam} into infinite families. Along the way, we obtain many more infinite families of counterexamples to Ivanov's conjecture in the primitive case. The counterexamples we came up with are all pseudocyclic fusion schemes of the cyclotomic schemes. One of the main tools that we use is the theory of Gauss sums, which we review in the next section.

\section{Gauss sums}
Let $p$ be a prime, $f$ a positive integer, and $q=p^f$. Let $\xi_{p}=e^{2\pi i/p}$ and let $\psi$ be the additive character of $\mathbb{F}_{q}$ defined in (\ref{defaddchar}). Let
$$\chi:\mathbb{F}_{q}^* \rightarrow {\mathbb C}^{*} $$
be a character of $\mathbb{F}_{q}^{*}$. We define the {\it Gauss sum} by
$$ g(\chi)=\sum_{a \in \mathbb{F}_{q}^*} \chi(a)\psi(a).$$
Note that if $\chi_0$ is the trivial multiplicative character of
$\mathbb{F}_q$, then $g(\chi_0)=-1$. We are usually concerned with nontrivial Gauss sums $g(\chi)$, i.e., those with $\chi\neq \chi_0$.

While it is easy to show that the absoulte value of a nontrivial Gauss sum $g(\chi)$ is equal to $\sqrt{q}$, the explicit determination of Gauss sums
is a difficult problem. However, there are a few cases where the Gauss sums $g(\chi)$ can be explicitly evaluated.
The simplest case is the so-called {\it semi-primitive case}, where there exists an integer $j$ such that $p^j\equiv -1$ (mod $N$) ($N$ is the order of $\chi$ in ${\widehat \mathbb{F}_q^*}$, the character group of $\mathbb{F}_q^*$). Some authors \cite{BMW, bew} also refer to this case as uniform cyclotomy, or pure Gauss sums. We refer the reader to \cite[p.~364]{bew} for the precise evaluation of Gauss sums in this case.

The next interesting case is the index 2 case, where $-1$ is not in the subgroup $\langle p\rangle$, the cyclic group generated by $p$,
and $\langle p\rangle$ has index 2 in $(\mathbb{Z}/N\mathbb{Z})^*$ (again here $N$ is the order of $\chi$ in ${\widehat \mathbb{F}_q^*}$). Many authors have investigated this case, including Baumert and Mykkeltveit \cite{bmy}, McEliece \cite{McE}, Langevin \cite{Lang}, Mbodj \cite{Mbo}, Meijer and Van de Vlugt \cite{mv}, and Yang and Xia  \cite{yx}. In the index 2 case, it can be shown that $N$ has at most two odd prime divisors. Assume that $N$ is odd, we have the following three possibilities in the index 2 case (see \cite{yx}): Below both $p_1$ and $p_2$ are primes.
\begin{enumerate}
\item $N=p_1^m$, $p_1\equiv 3$ (mod 4);
\item $N=p_1^mp_2^{n}$, $\{p_1\; ({\rm mod}\; 4), p_2\; ({\rm mod}\; 4)\}=\{1,3\}$, ${\rm ord}_{p_1^m}(p)=\phi(p_1^m)$, ${\rm ord}_{p_2^{n}}(p)=\phi(p_2^{n})$;
\item $N=p_1^mp_2^{n}$, $p_1\equiv 1,3$ (mod 4), ${\rm ord}_{p_1^m}(p)=\phi(p_1^m)$ and $p_2\equiv 3$ (mod 4), ${\rm ord}_{p_2^{n}}(p)=\phi(p_2^{n})/2$.
\end{enumerate}

We state below the results on evaluation of Gauss sums in Case (1) and (2)  from the above list.

\begin{thm}\label{lang}{\em (Langevin \cite{Lang})}
Let $N=p_1^m$, where $m$ is a positive integer, $p_1$ is a prime such that $p_1>3$ and $p_1\equiv 3$ (mod 4). Let $p$ be a prime such that $[(\mathbb{Z}/N\mathbb{Z})^*:\langle p\rangle]=2$ (that is, $f:={\rm ord}_N(p)=\phi(N)/2$) and let $q=p^f$. Let $\chi$ be a  multiplicative character of order $N$ of $\mathbb{F}_q$, and $h$ be the class number of $\mathbb{Q}(\sqrt{-p_1})$. Then the Gauss sum $g(\chi)$ over $\mathbb{F}_q$ is determined up to complex conjugation by
$$g(\chi)=\frac{b+c\sqrt{-p_1}}{2}p^{h_0},$$
where
\begin{enumerate}
\item $h_0=\frac {f-h}{2}$,
\item $b,c\not\equiv 0$ (mod $p$),
\item $b^2+p_1c^2=4p^{h}$,
\item $bp^{h_0}\equiv -2$ (mod $p_1$).
\end{enumerate}
\end{thm}

\begin{thm}\label{mbodj}{\em (Mbodj \cite{Mbo})}
Let $N=p_1^mp_2^{n}$, where $m,n$ are positive integers, $p_1$ and $p_2$ are prime such that $\{p_1\; ({\rm mod}\; 4), p_2\; ({\rm mod}\; 4)\}=\{1,3\}$, ${\rm ord}_{p_1^m}(p)=\phi(p_1^m)$, ${\rm ord}_{p_2^{n}}(p)=\phi(p_2^{n})$. Let $p$ be a prime such that $[(\mathbb{Z}/N\mathbb{Z})^*:\langle p\rangle]=2$ (that is, $f:={\rm ord}_N(p)=\phi(N)/2$) and let $q=p^f$. Let $\chi$ be a multiplicative character of order $N$ of $\mathbb{F}_q$, and $h$ be the class number of $\mathbb{Q}(\sqrt{-p_1p_2})$. Then the Gauss sum $g(\chi)$ over $\mathbb{F}_q$ is determined up to complex conjugation by
$$g(\chi)=\frac{b+c\sqrt{-p_1p_2}}{2}p^{h_0},$$
where
\begin{enumerate}
\item $h_0=\frac {f-h}{2}$,
\item $b,c\not\equiv 0$ (mod $p$),
\item $b^2+p_1p_2c^2=4p^{h}$,
\item $b\equiv 2p^{h/2}$ (mod $\ell$), here $\ell\in\{p_1,p_2\}$ is the prime congruent to 3 modulo 4.
\end{enumerate}
\end{thm}

\section{Pseudocyclic fusion schemes of the cyclotomic schemes}

Let $p$ be a prime, $f$ be a positive integer and $q=p^f$. Let $\gamma$ be a fixed primitive element of $\mathbb{F}_q$, and $N>1$ be an integer such that $N|(q-1)$.
As we did in Section 1, let $C_0=\langle \gamma^N\rangle$ and $C_i=\gamma^iC_0$ for $1\leq i\leq N-1$. Assume that $-1\in C_0$.
Define $R_0=\{(x,x) \mid  x\in \mathbb{F}_q\}$, and for $i\in \{1,2,\ldots ,N\}$, define $R_i=\{(x,y)\mid x,y\in \mathbb{F}_q, x-y\in C_{i-1}\}$.
Then $(\mathbb{F}_q, \{R_i\}_{0\leq i\leq N})$ is the cyclotomic association scheme of class $N$ on $\mathbb{F}_q$.
It was proven by Baumert, Mills and Ward \cite{BMW} that $(\mathbb{F}_q, \{R_i\}_{0\leq i\leq N})$ is amorphic if and only if $-1$ is congruent to a power of $p$ modulo $N$
(i.e., the so-called semi-primitive condition holds). See also \cite{BanMun} for a proof of this fact. Below we will show that even though in the index 2 case the cyclotomic association
scheme $(\mathbb{F}_q, \{R_i\}_{0\leq i\leq N})$ is not amorphic, we can still have interesting fusion schemes of $(\mathbb{F}_q, \{R_i\}_{0\leq i\leq N})$.

\subsection{The index 2 case with $N=p_1^mp_2$}

In this subsection, we assume that $N=p_1^mp_2$ ($m\geq 1$),  $p_1$, $p_2$ are primes such that $\{p_1\; ({\rm mod}\; 4), p_2\; ({\rm mod}\; 4)\}=\{1,3\}$,
$p$ is a prime such that $\gcd(p, N)=1$, ${\rm ord}_{p_1^m}(p)=\phi(p_1^m)$ and ${\rm ord}_{p_2}(p)=\phi(p_2)$, and $f:={\rm ord}_N(p)=\phi(N)/2$. Let $q=p^f$, and as before let $C_0,C_1,\ldots ,C_{N-1}$ be the $N$-th cyclotomic classes of $\mathbb{F}_q$. Note that  here we have $-C_i=C_i$ for all $0\leq i\leq N-1$ since either $2N|(q-1)$ or $q$ is even. For convenience, we define $d:=p_1p_2$. For $0\leq k\leq d-1$, define
\begin{equation}
D_k=\bigcup_{i=0}^{p_1^{m-1}-1}C_{ip_2+kp_{1}^{m-1}}
\end{equation}
Note that $D_k=\gamma^{kp_1^{m-1}}D_0$ and $\{0\}, D_0, D_1, \ldots ,D_{d-1}$ form a partition of $\mathbb{F}_q$. Now define $R'_0=R_0$ and
\begin{equation}\label{defREL}
R'_k=\{(x,y)\mid x,y\in \mathbb{F}_q, x-y\in D_{k-1}\}.
\end{equation}
We will show that $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq d})$ is a fusion scheme of $(\mathbb{F}_q,\{R_i\}_{0\leq i \leq N})$. The proof depends on the following evaluation of Gauss sums in the index 2 case, and results from \cite{FX}.

Let $\chi_1$ be the multiplicative character of order $p_1^m$ of $\mathbb{F}_q$ defined by $\chi_1(\gamma)=\textup{exp}(\frac{2\pi i}{p_1^m})$, and let $\chi_2$ be the multiplicative character of order $p_2$ of $\mathbb{F}_q$ defined by $\chi_2(\gamma)=\textup{exp}(\frac{2\pi i}{p_2})$. By Theorem~\ref{mbodj}, we have
\begin{equation}\label{evalGauss}
g({\bar \chi_1}{\bar \chi_2})=\frac{b+c\sqrt{-p_1p_2}}{2}p^{h_0},
\end{equation}
where $h_0=\frac {f-h}{2}$ ($h$ is the class number of $\mathbb{Q}(\sqrt{-p_1p_2})$), $b,c\not\equiv 0$ (mod $p$), $b^2+p_1p_2c^2=4p^{h}$, and $b\equiv 2p^{h/2}$ (mod $\ell$), here $\ell\in \{p_1,p_2\}$ is the prime congruent to 3 modulo 4.

\begin{thm}\label{pseudoFusion1}
With the definition of $R'_k$ given in (\ref{defREL}), $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq d})$ is a pseudocyclic association scheme.
\end{thm}

\noindent{\bf Proof:}\  We will first prove that $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq d})$ is an association scheme by using the Bannai-Muzychuk criterion discussed in Section 1.

For each $a$, $0\le a\le N-1$, there exists a unique $i_{a}\in \{0, 1, ..., p_{1}^{m-1}-1\}$ such that $p_{1}^{m-1}\mid (a+p_{2}i_{a})$. It follows that there is a unique $j_{a}$, $0\leq j_a\leq p_1p_2-1$, such that $a\equiv -p_{2}i_{a}+p_{1}^{m-1}j_{a}$ (mod $N$). It is now easy to check that $-ip_2+jp_1^{m-1}$, $0\leq i\leq p_1^{m-1}-1$ and $0\leq j\leq p_1p_2-1$, form a complete set of residues modulo $N$.

The group of additive characters of $\mathbb{F}_q$ consists of $\psi_0$ and $\psi_{\gamma^a}$, $0\leq a\leq q-2$,  where $\psi_0$ is the trivial character and $\psi_{\gamma^a}$ is defined by
\begin{equation}
\psi_{\gamma^a}: \mathbb{F}_{q} \rightarrow {\mathbb C}^{*}, \quad \psi_{\gamma^a}(x)=\xi_{p}^{{\rm Tr}(\gamma^ax)}.
\end{equation}
We usually write $\psi_1$ simply as $\psi$. The character values of $D_0$ were computed in the proof of Theorem 5.1 \cite{FX}. Since $D_k$ is a (multiplicative) translate of $D_0$, we know the character values of $D_k$ as well. Explicitly, for each $a$, $0\leq a\leq N-1$, write
$$a\equiv -p_{2}i_{a}+p_{1}^{m-1}j_{a}\; ({\rm mod}\; N),$$
with $0\leq i_a\leq p_1^{m-1}-1$ and $0\leq j_a\leq p_1p_2-1$. For convenience we introduce the Kronecker delta $\delta_{a,p_1}$, which equals 1 if $p_1|a$, 0 otherwise. Also we define $\delta_{a,p_2}$ by setting it equal to 1 if $p_2|a$, 0 otherwise. By the results in \cite{FX}, we have $$\psi_{\gamma^a}(D_k)=\psi(\gamma^{a+p_1^{m-1}k}D_0)=\frac{1}{N}T_{a+p_1^{m-1}k},$$
where
\begin{align*}
T_{a+p_1^{m-1}k}&=-p_1^{m-1}-(-1)^{\frac{p_1-1}{2}}p_1^{m-1}p_2\sqrt{q}\delta_{a+p_1^{m-1}k,p_2}-(-1)^{\frac{p_2-1}{2}}p_1^m\sqrt{q}\delta_{j_a+k,p_1}\\
&\quad+\frac{b}{2}p^{h_0}p_1^{m-1}(p_1\delta_{j_a+k,p_1}-1)(p_2\delta_{a+p_1^{m-1}k,p_2}-1)\\
&\quad -\bigg(\frac{a+p_1^{m-1}k}{p_2}\bigg)\bigg(\frac{j_a+k}{p_1}\bigg)\frac{c}{2}p^{h_0}p_1^mp_2
\end{align*}
In the above formula, $b,c$ are given by (\ref{evalGauss}), $(\frac{.}{p_2})$ and $(\frac{.}{p_1})$ are Legendre symbols.
Observe that $a+p_1^{m-1}k\equiv -p_{2}i_{a}+p_{1}^{m-1}(j_{a}+k)$ (mod $N$). So $\delta_{a+p_1^{m-1}k, p_{2}}=\delta_{j_{a}+k,p_2}$,
and $\bigg(\frac{a+p_1^{m-1}k}{p_{2}}\bigg)=\bigg(\frac{p_1}{p_2}\bigg)^{m-1}\bigg(\frac{j_a+k}{p_2}\bigg)$. Therefore, $\psi_{\gamma^a}(D_k)$ is independent of $i_a$.\\

In order to apply the Bannai-Muzychuk criterion, we define the following partition of $\{\psi_{\gamma^a}\mid a\in \mathbb{Z}/N\mathbb{Z}\}$. For each $j$, $0\leq j\leq d-1$, define
$$\Delta_{j+1}=\{\psi_{\gamma^{-p_2i + p_1^{m-1} j}}\mid  0\leq i\leq p_1^{m-1}-1\},$$
and $\Delta_{0}=\{\psi_0\}$. Clearly $\Delta_{0}, \Delta_{1}, \ldots ,\Delta_{d}$ form a partition of $\{\psi_{\gamma^a}\mid a\in \mathbb{Z}/N\mathbb{Z}\}$.   For each $0\leq k\leq d-1$, since $\psi_{\gamma^a}(D_k)$ is independent of $i_a$ (here $a\equiv -p_{2}i_{a}+p_{1}^{m-1}j_{a}\; ({\rm mod}\; N)$),  we see that $\psi_{\gamma^a}(D_k)$ is a constant for those $a$ in the same subset of the above partition. By the Bannai-Muzychuk criterion (with $\Lambda_0=\{0\}$, $\Lambda_{j+1}=\{1+ip_2+p_1^{m-1}j\mid 0\leq i\leq p_1^{m-1}-1\}$, $0\leq j\leq d-1$), we see that $(\mathbb{F}_q, \{R'_{0}$, $R'_1,\ldots , R'_{d}\})$ is an association scheme.\\

Next we show that the association scheme $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq d})$ is pseudocyclic. To this end, we show that the following group ring equation holds in $\mathbb{Z}[(\mathbb{F}_q,+)]$.\\

\noindent{\it Claim:} $\sum_{k=0}^{d-1}D_k^2=(q-1)\cdot 0_{\mathbb{F}_q}+\frac{q-1}{p_1p_2}(\mathbb{F}_q-0_{\mathbb{F}_q})$, where $0_{\mathbb{F}_q}$ is the zero element in $\mathbb{F}_q$. \\

For any $a$, $0\leq a\leq N-1$, we write $a\equiv -i_ap_2+j_ap_1^{m-1} \pmod{N}$ with $i_a\in\{0,1,\ldots,p_1^{m-1}-1\}$ and $j_a\in\{0,1,2,\ldots ,d-1\}$. Since $\psi_{\gamma^a}(D_k)$ is independent of $i_a$, we may assume that $i_a=0$. We now compute
\begin{align*}
 \sum_{k=0}^{d-1}\psi_{\gamma^a}(D_k)^2=\frac{1}{N^2}\sum_{k=0}^{d-1}T_{p_1^{m-1}(j_a+k)}^2=\frac{1}{N^2}\sum_{k=0}^{d-1}T_{kp_1^{m-1}}^2
\end{align*}
Since the last expression above is independent of $a$, we see that $\sum_{k=0}^{d-1}\psi_{\gamma^a}(D_k)^2$ are equal to the same constant for all $0\leq a\leq N-1$. Since each $D_k$ is a union of some $N$-th cyclotomic classes, it follows that $\sum_{k=0}^{d-1}\psi_{\gamma^a}(D_k)^2$ are equal to the same constant for all $0\leq a\leq q-2$. Therefore, by the inversion formula, we have
\[\sum_{k=0}^{d-1}D_k^2=(n-\lambda)\cdot 0_{\mathbb{F}_q}+\lambda\mathbb{F}_q,\]
for some integers $n,\lambda$. Now applying the principal character to both sides, and computing the coefficients of $0_{\mathbb{F}_q}$ on both sides, we have
\begin{align*}
 n&=p_1p_2\cdot\frac{q-1}{p_1p_2},\\
 n+(q-1)\lambda&=d\cdot\bigg(\frac{q-1}{p_1p_2}\bigg)^2.
\end{align*}
It follows that $n=q-1$, and $\lambda=\frac{q-1}{p_1p_2}-1$. The claim is now established. A direct consequence is that $\sum_{i=0}^{d-1}p_{i,i}^j=\frac{q-1}{N}-1$, for all $j$, where $p_{i,i}^j$ are the intersection parameters given by  $D_i^2=\sum_{j=0}^{d-1}p_{i,i}^jD_j+p_{i,i}^{0}\cdot 0_{\mathbb{F}_q}$. By Part (2) of Theorem~\ref{pseudocyc}, the association scheme $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq d})$ is pseudocyclic. The proof is complete.
\hfill$\Box$\vspace{4mm}

In order to obtain counterexamples to Ivanov's conjecture, we need to have each $R'_k$ ($1\leq k\leq d$) in Theorem~\ref{pseudoFusion1} to be strongly regular.
Note that $R'_k$ is just the Cayley graph ${\rm Cay}(\mathbb{F}_q, D_{k-1})$, and ${\rm Cay}(\mathbb{F}_q, D_{k-1})\cong {\rm Cay}(\mathbb{F}_q, D_0)$ for all $1\leq k\leq d$ since $D_{k-1}=\gamma^{(k-1)p_1^{m-1}}D_0$.
It follows that if ${\rm Cay}(\mathbb{F}_q, D_0)$ is strongly regular, then all $R'_k$, $1\leq k\leq d$, are strongly regular. In \cite{FX}, we obtained necessary and sufficient conditions for ${\rm Cay}(\mathbb{F}_q, D_0)$ to be strongly regular, which we quote below.

\begin{thm}{\rm (Corollary 5.2 in \cite{FX})}\label{srg2}
With $b,c,h$ given in (\ref{evalGauss}), ${\rm Cay}(\mathbb{F}_q,D_0)$ is a strongly regular graph if and only if $b,c\in \{1,-1\}$, $h$ is even and $p_1=2p^{h/2}+(-1)^{\frac{p_1-1}{2}}b$, $p_2=2p^{h/2}-(-1)^{\frac{p_1-1}{2}}b$.
\end{thm}

In \cite{FX}, we used a computer to search for $p, p_1, p_2$ satisfying the conditions in Theorem~\ref{srg2}. We found six infinite families of strongly regular graphs in this way. By the discussion preceding Theorem~\ref{srg2}, and since the parameters of each of the six examples of srg are neither Latin square type nor negative Latin square type, each of the six families of srg gives rise to an infinite class of counterexamples to Ivanov's conjecture. Below we list the parameters of these examples. For the detailed reasons why we have strongly regular graphs, we refer the reader to \cite{FX}.

\begin{example}\label{21st}
Let $p=2$, $q=2^{4\cdot 3^{m-1}}$, $p_1=3$, $p_2=5$, $N=3^m\cdot 5$, with $m\geq 1$. Then we have a 15-class pseudocyclic fusion scheme $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq 15})$ in which each relation $R'_k$, $1\leq k\leq 15$, is strongly regular.
\end{example}

We remark that when $m=2$, Example~\ref{21st} is the same as Example 1 in \cite{ikutam}.

\begin{example}\label{22st}
Let $p=2$, $q=2^{4\cdot 5^{m-1}}$, $p_1=5$, $p_2=3$, $N=5^m\cdot 3$, with $m\geq 1$. Then we have a 15-class pseudocyclic fusion scheme $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq 15})$ in which each relation $R'_k$, $1\leq k\leq 15$, is strongly regular.
\end{example}

We remark that when $m=2$, Example~\ref{22st} is the same as Example 2 in \cite{ikutam}.

\begin{example}\label{23st}
Let $p=3$, $q=3^{12\cdot 5^{m-1}}$, $p_1=5$, $p_2=7$, $N=5^m\cdot 7$, with $m\geq 1$. Then we have a $35$-class pseudocyclic fusion scheme $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq 35})$ in which each relation $R'_k$, $1\leq k\leq 35$, is strongly regular.
\end{example}

\begin{example}\label{24st}
Let $p=3$, $q=3^{12\cdot 5^{m-1}}$, $p_1=7$, $p_2=5$, $N=7^m\cdot 5$, with $m\geq 1$. Then we have a $35$-class pseudocyclic fusion scheme $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq 35})$ in which each relation $R'_k$, $1\leq k\leq 35$, is strongly regular.
\end{example}

\begin{example}\label{25st}
Let $p=3$, $q=3^{144\cdot 17^{m-1}}$, $p_1=17$, $p_2=19$, $N=17^m\cdot 19$, with $m\geq 1$. Then we have a $323$-class pseudocyclic fusion scheme $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq 323})$ in which each relation $R'_k$, $1\leq k\leq 323$, is strongly regular.
\end{example}

\begin{example}\label{26st}
Let $p=3$, $q=3^{144\cdot 19^{m-1}}$, $p_1=19$, $p_2=17$, $N=19^m\cdot 17$, with $m\geq 1$. Then we have a $323$-class pseudocyclic fusion scheme $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq 323})$ in which each relation $R'_k$, $1\leq k\leq 323$, is strongly regular.
\end{example}

We remark that by using Corollary 3.2 in \cite{ikutam}, one can further obtain 3-class fusion schemes of the above pseudocyclic association schemes, in which two relations are strongly regular graphs, while the third relation is not (see the character table of these 3-class fusion schemes in the statement of Corollary 3.2 of \cite{ikutam}).

\subsection{The index 2 case with $N=p_1^m$}

In this subsection, we assume that $N=p_1^m$  (here $m\geq 1$,  $p_1>3$ is a prime such that $p_1\equiv 3$ (mod 4)), $p$ is a prime such that $\gcd(N,p)=1$, and  $f:={\rm ord}_{N}(p)=\phi(N)/2$. Let $q=p^f$, and as before let $C_0,C_1,\ldots ,C_{N-1}$ be the $N$-th cyclotomic classes of $\mathbb{F}_q$. Note that $-C_i=C_i$ for all $0\leq i\leq N-1$ since either $2N|(q-1)$ or $q$ is even. For $0\leq k\leq p_1-1$, define
\begin{equation}
D_k=\bigcup_{i=0}^{p_1^{m-1}-1}C_{i+kp_1^{m-1}}
\end{equation}
Note that $D_k=\gamma^{kp_1^{m-1}}D_0$ and $\{0\}, D_0, D_1, \ldots ,D_{p_1-1}$ form a partition of $\mathbb{F}_q$. Now define $R'_0=R_0$ and
\begin{equation}\label{defREL2}
R'_k=\{(x,y)\mid x,y\in \mathbb{F}_q, x-y\in D_{k-1}\}.
\end{equation}
We will show that $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq p_1})$ is a fusion scheme of $(\mathbb{F}_q,\{R_i\}_{0\leq i \leq N})$. The proof depends on the following evaluation of Gauss sums in the index 2 case, and results from \cite{FX}.

Let $\chi$ be the multiplicative character of $\mathbb{F}_q$ defined by $\chi(\gamma)=\text{exp}(\frac{2\pi i}{N})$. By Theorem~\ref{lang}, we have
\begin{equation}\label{evalGauss2}
g({\bar \chi})=\frac{b+c\sqrt{-p_1}}{2}p^{h_0},\;\; b,c\not\equiv 0\pmod{p},
\end{equation}
where $h_0=\frac {f-h}{2}$ and $h$ is the class number of $\mathbb{Q}(\sqrt{-p_1})$, $b^2+p_1c^2=4p^{h}$, and $bp^{h_0}\equiv -2\pmod{p_1}$.

\begin{thm}\label{pseudoFusion2}
With the definition of $R'_k$ given in (\ref{defREL2}), $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq p_1})$ is a pseudocyclic association scheme.
\end{thm}

\noindent{\bf Proof:}\  The proof is similar to that of Theorem~\ref{pseudoFusion1}. For each $a$, $0\leq a\leq N-1$, there is a unique $i_a\in\{0,1,\ldots,p_1^{m-1}-1\}$, such that $p_1^{m-1}|(a+i_a)$. It follows that there is a unique $j_a, 0\leq j_a\leq p_1-1$, such that $a\equiv -i_a+p_1^{m-1}j_a\; ({\rm mod}\; N).$ It is now easy to check that $-i+jp_1^{m-1}$, $0\leq i\leq p_1^{m-1}-1$ and $0\leq j\leq p_1-1$, form a complete set of residues modulo $N$.

The group of additive characters of $\mathbb{F}_q$ consists of $\psi_0$ and $\psi_{\gamma^a}$, $0\leq a\leq q-2$. The character values of $D_0$ were computed in the proof of Theorem 4.1 \cite{FX}.  Since $D_k$ is a (multiplicative) translate of $D_0$, we know the character values of $D_k$ as well. Explicitly, for each $a$, $0\leq a\leq N-1$, write
$$a\equiv -i_a+p_1^{m-1}j_a\; ({\rm mod}\; N),$$
with $0\leq i_a\leq p_1^{m-1}-1$ and $0\leq j_a\leq p_1-1$. For convenience, we also introduce the Kronecker delta $\delta_{j_a}$, which equals $1$ if $p_1|j_a$, and $0$ otherwise. By the results in \cite{FX}, we have
$$\psi_{\gamma^a}(D_k)=\psi(\gamma^{a+kp_1^{m-1}}D_0)=\frac{1}{N}T_{a+kp_1^{m-1}},$$
where
$$T_{a+kp_1^{m-1}}=-p_1^{m-1}+\frac{p^{h_0}p_1^{m-1}b}{2}(p_1\delta_{j_a+k}-1)-\frac{p^{h_0}p_1^mc}{2}\bigg(\frac{j_a+k}{p_1}\bigg).$$
In the above formula, $b,c$ are given in (\ref{evalGauss2}), and $(\frac{.}{p_1})$ is the Legendre symbol. It is important to note that $\psi_{\gamma^a}(D_k)$ is independent of $i_a$. \\

We define the following partition of $\{\psi_{\gamma^a}\mid a\in \mathbb{Z}/N\mathbb{Z}\}$. For each $j$, $0\leq j\leq p_1-1$, we define
$$\Delta_{j+1}=\{\psi_{\gamma^{-i + p_1^{m-1} j}}\mid  0\leq i\leq p_1^{m-1}-1\},$$
and $\Delta_{0}=\{\psi_0\}$. Then clearly $\Delta_{0}, \Delta_{1}, \ldots ,\Delta_{p_1}$ form a partition of $\{\psi_{\gamma^a}\mid a\in \mathbb{Z}/N\mathbb{Z}\}$.   For each $0\leq k\leq p_1-1$, since $\psi_{\gamma^a}(D_k)$ is independent of $i_a$ (here $a\equiv -i_{a}+p_{1}^{m-1}j_{a}\; ({\rm mod}\; N)$),  we see that $\psi_{\gamma^a}(D_k)$ is a constant for those $a$ in the same subset of the above partition. By the Bannai-Muzychuk criterion (with $\Lambda_0=\{0\}$, $\Lambda_{j+1}=\{1+i+p_1^{m-1}j\mid 0\leq i\leq p_1^{m-1}-1\}$, $0\leq j\leq p_1-1$), we see that $(\mathbb{F}_q, \{R'_{0}$, $R'_1\ldots , R'_{p_1}\})$ is an association scheme.

Similarly we can show that the following group ring equation holds in $\mathbb{Z}[(\mathbb{F}_q,+)]$:$$\sum_{k=0}^{p_1-1}D_k^2=(q-1)\cdot 0_{\mathbb{F}_q}+\frac{q-1}{p_1}(\mathbb{F}_q-0_{\mathbb{F}_q}),$$
from which the pseudocyclicity of the scheme $(\mathbb{F}_q, \{R'_{0}$, $R'_1,\ldots , R'_{p_1}\})$ follows. We omit the details of the proof of the above group ring equation. The proof is now complete.
\hfill$\Box$\vspace{4mm}

In order to obtain counterexamples to Ivanov's conjecture, we need to have each $R'_k$ ($1\leq k\leq p_1$) in Theorem~\ref{pseudoFusion2} to be strongly regular.
Note that $R'_k$ is just the Cayley graph ${\rm Cay}(\mathbb{F}_q, D_{k-1})$, and ${\rm Cay}(\mathbb{F}_q, D_{k-1})\cong {\rm Cay}(\mathbb{F}_q, D_0)$ for all $1\leq k\leq p_1$ since $D_{k-1}=\gamma^{(k-1)p_1^{m-1}}D_0$.
Again it follows that if ${\rm Cay}(\mathbb{F}_q, D_0)$ is strongly regular, then all $R'_k$, $1\leq k\leq p_1$, are strongly regular.
In \cite{FX}, we obtained necessary and sufficient conditions for ${\rm Cay}(\mathbb{F}_q, D_0)$ to be strongly regular, which we quote below.

\begin{thm}{\rm (Corollary 4.2 in \cite{FX})}\label{srg1}
With $b,c$ given in (\ref{evalGauss2}), ${\rm Cay}(\mathbb{F}_q,D)$ is a strongly regular graph if and only if $b,c\in \{1,-1\}$.
\end{thm}

In \cite{FX}, we used a computer to search for $p, p_1$ satisfying the conditions in Theorem~\ref{srg1}. We found six infinite families of strongly regular graphs in this way.
By the discussion preceding Theorem~\ref{srg1}, each of the six examples of srg gives rise to a class of infinitely many counterexamples to Ivanov's conjecture. Below we list the parameters of these examples. For the detailed reasons why we have strongly regular graphs, we refere the reader to \cite{FX}.

\begin{example}\label{1st}
Let $p=2$, $q=2^{3\cdot 7^{m-1}}$, $p_1=7$, $N=p_1^m$, $m\geq 1$ is an integer. Then we have a $7$-class pseudocyclic fusion scheme $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq 7})$ in which each relation $R'_k$, $1\leq k\leq 7$, is strongly regular.
\end{example}

We remark that when $m=2$, Example~\ref{1st} is the same as Example 3 of \cite{ikutam}.

\begin{example}\label{2nd}
Let $p=3$, $q=3^{53\cdot 107^{m-1}}$, $p_1=107$, $N=p_1^m$, $m\geq 1$ is an integer. Then we have a $107$-class pseudocyclic fusion scheme $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq 107})$ in which each relation $R'_k$, $1\leq k\leq 107$, is strongly regular.
\end{example}

\begin{example}\label{3rd}
Let $p=5$, $q=5^{9\cdot 19^{m-1}}$, $p_1=19$, $N=p_1^m$,  $m\geq 1$ is an integer. Then we have a $19$-class pseudocyclic fusion scheme $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq 19})$ in which each relation $R'_k$, $1\leq k\leq 19$, is strongly regular.
\end{example}

\begin{example}\label{4th}
Let $p=5$, $q=5^{249\cdot 499^{m-1}}$, $p_1=499$, $N=p_1^m$, $m\geq 1$ is an integer. Then we have a $499$-class pseudocyclic fusion scheme $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq 499})$ in which each relation $R'_k$, $1\leq k\leq 499$, is strongly regular.
\end{example}

\begin{example}\label{5th}
Let $p=17$, $q=17^{33\cdot 67^{m-1}}$, $p_1=67$,  $N=p_1^m$, $m\geq 1$ is an integer. Then we have a $67$-class pseudocyclic fusion scheme $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq 67})$ in which each relation $R'_k$, $1\leq k\leq 67$, is strongly regular.
\end{example}

\begin{example}\label{6th}
Let $p=41$, $q=41^{81\cdot 163^{m-1}}$, $p_1=163$, $N=p_1^m$, $m\geq 1$ is an integer. Then we have a $163$-class pseudocyclic fusion scheme $(\mathbb{F}_q, \{R'_k\}_{0\leq k\leq 163})$ in which each relation $R'_k$, $1\leq k\leq 163$, is strongly regular.
\end{example}

Again we remark that by using Corollary 3.2 in \cite{ikutam}, one can further obtain 3-class fusion schemes of the above pseudocyclic association schemes, in which two relations are strongly regular graphs, while the third relation is not.

redpajama_set_name: RedPajamaArXiv


The event ticket is not included in the meet & greet upgrade . Children age 12 and under must be accompanied by a Parent/Guardian. Each Child, Parent or Guardian must have an event ticket as well as a Meet & Greet Upgrade to attend. CLICK HERE to purchase an event ticket.
Please make sure to bring your camera, as you will want to capture every superhero moment! Parents, please note that our Meet & Greet characters come in costume directly from the stage. While most children are excited to meet the characters up-close, others can feel frightened. If your child is scared or hesitant to meet the characters, we recommend moving to the back of the line; This will allow your child the extra time he/she may need to warm up to the characters and see how they interact with other children present. Please remember, children are not allowed access without a Meet & Greet paid adult.


(Admission Office – Van Lang, Oct/24th/2018) - In October, 2018, the lecturers of Business and Commerce of Van Lang University have organized many university - businesses activities: working and visiting the industrial zones, organizing seminars to improve professional skills.
The Board of Faculty of Commerce and Business Administration have been constantly seeking and building more and more contacts with businesses. From the collaboration with the businesses, the members of the faculty can enrich their knowledge and have extensive and multi-dimensional view in many fields so that they can continuously update the actual business situations into the lessons for the students. This is also an opportunity to strengthen the links between the School and the businesses, between training and practice, and expand the career options for students.
The delegation is led by Dr. Phan Quan Viet - Vice Dean of Faculty of Commerce & Business Administration, with the participation of teachers in Business and Commerce of Van Lang University.
The delegation of Van Lang is warmly and thoughtfully welcomed by leaders of Vinh Loc A Industrial Zone, as well as leaders of Cholimex Company and Sai Gon Beer Company. The entire scale of Vinh Loc Industrial Park is several hundred hectares in area. Firms, factories and warehouses are hired by businesses that are engaged in a wide range of production and business lines, including construction, infrastructure and food processing… Workers' accommodation is also being put into operation and use, along with attractive incentives and benefits for middle managers, senior staff, and workers of the zone.
Frozen food, with 90% of its output exported to international markets, is an important partner of the well-known brand "Mark and Spencer"
The teachers tour Sai Gon Beer Factory, "SAGOTA" (Sai Gon Ta), with closed brewery production process and strict inspection from input, material preparation, cook, mix, and split. The factory implements technological system to strictly control each stage of beer production. Each product line has different blending rates and uses different technologies. The process of packaging and warehouse system of the brewery also helps the teachers of Logistics have more practical knowledge in the field of beverage production.
Through this visit, leaders of Vinh Loc Industrial Park are ready to create favorable conditions for students of Faculty of Commerce and Business Administration of Van Lang University to practice in the field of Business, Commerce, Import and Export, Warehouse, and Logistics. The Faculty and Department will continue to promote cooperation with businesses to ensure that the theory comes along with practice in teaching and expand the job opportunities for students.
21/ 10/ 2018: Workshop on "Enhancing e-customs declaration on ECUSSS-VNACCS software"
On Sunday, October 21, 2018, the Faculty of Commerce & Business Administration coordinates with the Group of Import-Export Customs to hold a seminar on "Enhancing e-customs declaration on software ECUSSS-VNACCS software ". Seminars takes place with 2 sessions: from 9 am to 12 pm and from 13:00 to 16:00. Two main speakers in the program are Ngo Hoang Huy and Tran Dang Khoa - specialists of Thai Son Technology Development Company.
Many teacher, professionals, and students of Faculty of Commerce & Business Administration are interested in participating in the seminar. The seminar focuses on the key issues such as Customs declaration of export and import forms on ECUSS-VNACCS software and ECUS AUDIT software manual for checking data before the final report of auditing and clearance.
With many years of experience in their jobs and proficiency in development and software knowledge, experts have guided and answered many questions of participants in the workshop.
• Afternoon session: All participants are instructed to declare customs for processing goods (All-in, starting from the registration of processing contract ...); Customs declaration for export production (All-in); the report on final settlement of processed goods, export production, and export processing on ECUSS; Guidelines on checking the settlement report on Audit.
With a scale of 100 participants, the seminar brings many benefits to lecturers and professionals and also prepares the professional background for students in their upcoming work. It is expected that the Faculty of Commerce and Business Administration will continue to collaborate with the business units to organize more seminars to improve professional knowledge and skills for teaching staff and learners.


text:
Алексей Сергеевич Агрба (1897, с.Лдзаа, Сухумский округ, Кутаисская губерния, Российская империя — 21 апреля 1938, Москва, «Коммунарка») — советский государственный деятель Абхазии, Грузии, Азербайджана, Закавказской Федерации. Председатель ЦИК Абхазской АССР (1937). Расстрелян в 1938 году, реабилитирован посмертно.

Биография
Родился в бедной многодетной крестьянской семье, получил педагогическое образование (2-классная учительская школа в Гудаутах и 2-годичные учительские курсы в Сухуми, по окончании которых ему было присвоено звание народного учителя). С 1917 г. работал учителем в родном селе. В 1917 г. был председателем Лидзавского комитета общественной безопасности, до 1920 г. — участник подпольного революционного движения в Абхазии. Член РКП(б) с 1920 г.

В период правления меньшевиков в Грузии по заданию Гудаутского подпольного большевистского комитета служил заместителем начальника уездной милиции. В июне 1920 был арестован по обвинению в связи с большевиками, заключен в сухумскую тюрьму и предан военно-полевому суду. В связи с наступлением Красной Армии был переведён в тюрьму Поти, затем — в тюрьмы Кутаиси и Батума, где в марте 1921 г. был освобождён после установления Советской власти в Абхазии и назначен председателем Кодорского уездного ревкома (с 1922 г. — уездного исполнительного комитета. В мае 1922 г. назначен наркомом по военным и военно-морским делам ССР Абхазии. С 1922—1924 гг. — председатель Чрезвычайной комиссии при СНК ССР Абхазия, одновременно с марта 1923 г. — народный комиссар внутренних дел ССР Абхазия, с конца 1923 г. — заместитель председателя СНК ССР Абхазии.

В 1923 г. переведен на работу в ГПУ при СНК ССР Грузия, ЗСФСР. В 1929 г. становится членом Коллегии ГПУ Грузии и Закавказья. С февраля 1929 по ноябрь 1931 г. — заместитель председателя ГПУ при СНК Азербайджанской ССР (сменил на этом посту Л. П. Берия).
В ноябре 1931 — марте 1933 г. — полномочный представитель ОГПУ по ЗСФСР, одновременно председатель ГПУ при СНК ЗСФСР,
 1933—1934 гг. — председатель ГПУ при СНК Азербайджанской ССР,
 1934—1936 гг. — второй секретарь ЦК КП(б) Азербайджана (первый — Мир Джафар Багиров),
 1936—1937 гг. — первый секретарь Абхазского областного комитета КП(б) Грузии, также введен в состав Бюро ЦК КП(б) Грузии.
С февраля по сентябрь 1937 г. — председатель ЦИК Абхазской АССР. При нем в Абхазии начались массовые аресты.

Избирался членом ЦИК СССР 7-го созыва (1935−1937).

В сентябре 1937 г. арестован по обвинению в «создании в республике контрреволюционной буржуазно-националистической организации». Внесен в список «Москва-центр» от 19 апреля 1938 г. по 1-й категории («за» Сталин, Молотов, Каганович, Жданов). Осужден формальным приговором ВКВС СССР 21 апреля 1938 г. и расстрелян в тот же день. Место захоронения- Московская обл., «Коммунарка». Посмертно реабилитирован.

А. С. Агрба установлен памятник в Пицунде и названа улица в его честь.

Награды и звания
Награждён орденом Ленина (22.03.1936) и двумя орденами Трудового Красного Знамени (ЗСФСР и Грузинской ССР), двумя знаками «Почётный сотрудник госбезопасности».

Примечания

Литература

Ссылки

 http://stalin.memo.ru/spiski/pg08006.htm

Члены Бюро ЦК КП Грузии
Первые секретари Абхазского обкома КП Грузии
Руководители партийных органов, репрессированные в СССР
Расстрелянные и похороненные на полигоне Коммунарка
Казнённые политики
Чекисты
Министры Абхазской АССР
Сотрудники Министерства национальной безопасности Азербайджана
Члены ЦИК СССР
Вторые секретари ЦК КП Азербайджана
Казнённые в 1938 году
Председатели ЦИК Абхазской АССР
Члены ВКП(б)


The Penny Black designers are playing with tags again and have added them to cards this time. I chose the 'brush pines' stamp and some cold-pressed watercolour paper for my layered pinecones card. I stamped the pine cones several times so I would have extra images for the tags. I also stamped a few times without re-inking to create paler impressions. The ink was applied with Memento markers and blended with water to soften and spread the colours. I die cut the tags with the smaller die from the 'tagged' set popped them up over the stamped panel.
Beautiful, just beautiful! I love it! What a good idea Heather… just you can do it!!! J'adore!!!
What a lovely way to add dimension to this card! I also like the perfect balance between the images and the big sentiment.
Loved this card on the PB Blog–thank you, Heather!


Robot vs Boxer print BOILERPLATE steampunk poster Jack Johnson 12"x16" signed FREE shipping in U.S.
I love the mixture of steampunk and action adventure.
Service was great, no complaints!
Built in a small Chicago laboratory, Boilerplate was a prototype soldier built for "preventing the deaths of men in the conflicts of nations". Although it was the only such prototype, Boilerplate was eventually able to exercise its proposed function in several combat actions.
Boilerplate fought alongside such notables as Teddy Roosevelt and Lawrence of Arabia. Campion and his robot also circled the planet with the U.S. Navy, trekked to the South Pole, made silent movies, and hobnobbed with the likes of Mark Twain and Nikola Tesla.
Inkpot Award-winning multimedia artist Paul Guinan's best known creation is Boilerplate. The book "Boilerplate: History's Mechanical Marvel," is in development as a feature film from producer J.J. Abrams.


text:
Tradeskills » es blocking out any discussion about negot
Guildwars 2 » Forum » Tradeskills » es blocking out any discussion about negot
#1 | es blocking out any discussion about negot 18.05.2019 02:42
CLEVELAND -- The Browns arent letting lockdown cornerback Joe Haden get away. Haden signed a five-year, $68 million extension with Cleveland on Monday, the total value of the contract surpassing cornerback Richard Shermans four-year, $57 million agreement last week with the Seattle Seahawks. Hadens contract runs through the 2019 season and includes $45 million guaranteed, agent Drew Rosenhaus told The Associated Press. The package also contains a $14 million signing bonus. One of the AFCs top defensive backs and one of Clevelands most popular athletes, Haden made his first Pro Bowl last season, his fourth with the team. The Browns selected him with the No. 7 overall pick in 2010. "Joes a good, young player whos made a commitment to our organization, and hes somebody with whom we want to move forward in order to advance our football team," general manager Ray Farmer said in a statement. "Hes a great person. Hes the right guy to both build with and build around as we attempt to become an elite football team. "When we talk about identifying talent, building through the draft, developing solid players and retaining them, Joe Hadens at the top of the list." Haden posted a photo on Twitter of him signing his new contract. Signing Haden to a long-term deal was a priority for Cleveland this off-season. During the draft, the team selected Oklahoma State cornerback Justin Gilbert early in the first round to play opposite Haden. Last season, Haden had four interceptions, recorded 60 tackles and was credited with 21 passes defenced. He had his first career multiple-interception game and scored his first touchdown, returning an interception 29 yards at Cincinnati on Nov. 17. According to STATS, Hadens 21 passes defenced in 2013 tied for third in the NFL and his 67 since joining the Browns four years ago are second in the league during that span. Haden averages 1.19 passes defenced per game, the second-highest average among players with 50-plus PBUs since at least 1994. For the most part, Haden has been an exemplary player and citizen since joining the Browns. His only major misstep was the four-game suspension he received in 2012 for violating the leagues substance abuse policy after he tested positive for the stimulant Adderall. First-year Browns coach Mike Pettine is thrilled to have Haden for several years. "Its important for us to extend our core players and continue to grow with guys that represent everything we want this team to be: tough, hard-working and passionate," Pettine said. "Joe obviously loves his teammates, and he loves this city. I know he wants to keep growing and improving as a player. Thats what we need, and thats what we want." With Hadens contract situation resolved, the Browns are expected to focus on getting Pro Bowl tight end Jordan Cameron signed to a long-term deal. Cameron had a breakout season in 2013, making 80 catches for 917 yards and seven TDs. Cameron is in the final season of his contract and has reportedly turned down two offers from the Browns. At Clevelands recent voluntary minicamp, Cameron said hes blocking out any discussion about negotiations to improve. "Ive got a lot of stuff to work on," he said. "Im not anywhere near where I can be. I still have a lot to prove to myself, and Im willing to work." Ryan Rua Jersey . In a pregame tribute commemorating his final contest at Coors Field on Wednesday night, Helton caught the ceremonial first pitch from his daughter with his wife, younger daughter and good friend Peyton Manning watching from the field. Jon Niese Jersey . The Marlies surrendered two power-play goals and failed to score on six man-advantage opportunities en route to a 4-1 defeat in American Hockey League action on Saturday. http://www.officialrangersgearshop.com/R...er-Kids-Jersey/. Bozak and Clarkson suffered the injuries in pre-training camp workouts, according to Carlyle, who said they made "small amounts of improvement" from Friday to Saturday. After the players were deemed less than 100 per cent by doctors after Thursdays medical testing, the Leafs held them out of on-ice testing and are being cautious. Elvis Andrus Jersey . Each day, TSN.ca provides the latest rumours, reports and speculation from around the NHL beat. Duck Calling With several marquee centres available via trade and expected to be available via free agency, the Anaheim Ducks could be primed to be big players in the coming days. Drew Robinson Jersey . The Heat centre scored 10 of his 30 points early in the first quarter to silence the Toronto fans as Miami defeated the Raptors 113-101 on Friday night.New York, NY (SportsNetwork.com) - MLS expansion club New York City FC announced Tuesday that it has signed United States international Mix Diskerud on a free transfer. Diskerud made his debut with the U.S. national team in a friendly match against South Africa in 2010. He has accumulated a total of 25 caps and represented the U.S. at the 2014 World Cup. New York City FC have made my wish as an American a reality by giving me the opportunity to play and live in the U.S., Diskerud said. Everybody I have spoken to on the mens national team about (NYCFC head coach) Jason Kreis has spoken highly of him to me. I felt eexactly the same way after meeting him.dddddddddddd Im sure he will help me to develop my game, to succeed with our team here in New York and to stay involved with the national team. Coming to play for New York City FC was an easy decision for me. A native of Oslo, Norway, the 24-year-old midfielder began his club career with Stabaek in 2008, logging a total of 10 goals in 81 league appearances. Diskerud enjoyed a brief loan spell at Genk before making the switch to Rosenborg in 2012. 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CPSA Column: June Industry News and Update
By EditorJuly 3, 2017Connecticut, Top News, Association Talk
By Carrol J. Hughes, Executive Director, Connecticut Package Store Association
STATE BUDGET SOLUTION NOT REACHED BEFORE DEADLINE
Carroll J. Hughes, Executive Director, Connecticut Package Store Association
The legislature will now begin a special session to arrive at a solution to the 2018-2019 biennial budget, which is facing a $5 billion deficit.
One reason a budget resolution was not reached by the June 7th deadline was that state employee concessions negotiated by the Governor could not be voted on for approval until early to mid-July. Several unions have to approve the contract in order for the agreement to be ratified. If a certain number of them do not approve, the agreements would be invalid and the savings could not be factored into the budget. Republicans have indicated that they do not feel the concessions negotiated with the unions garner enough savings. They were looking for close to $1.1 billion per year, instead of the approximate $700 million next fiscal year and $850 million in the Governor's 2018-2109 concession package.
Negotiations between Republicans, Democrats and the Governor are expected to take place over the period from June 12 through mid-July. There is strong speculation that the actual budget vote could extend into September, as vacation schedules and other factors make a budget deal more elusive than it has been in more than 25 years.
In the meantime, if a budget is not passed by the beginning of the new fiscal year on July 1st, the Governor is required to run the state under the existing budget, which is experiencing an $800 million revenue shortfall due to a dramatic decline in income tax receipts. This will cause many additional problems as the administration will hold back payments to providers and reduce agency expenses in an effort to keep the state operating within existing resources.
MINIMUM BOTTLE NOT EXPECTED TO BE IN BUDGET
Although it is has been aggressively pushed by the Governor, the elimination of minimum bottle pricing is not expected to be included in the budget as a "revenue item." This is due to the fact that the inclusion would result in a major loss of stores and their respective sales, income and local property taxes. A bill to eliminate minimum bottle pricing was offered by Governor Malloy in the General Law Committee, but the committee declined to move the bill forward after the outpouring of opposition at the public hearing.
Based on projections provided by the Governor, CPSA has estimated a $6-9 million sales tax revenue drop, due to the decrease in the average bottle price. It could also result in the loss of more than 3,000 employees from up to 600 stores, which would mean more than $4 million in lost income tax.
A trickle down effect would occur if these stores were to close, causing employee losses in retail, wholesale, sales, and delivery and warehouse workers. The indirect loss of service employees to the stores would further reduce income and sales tax receipts for the state. Closing stores also means a loss of rent payments and mortgages, affecting shopping centers already suffering from the rampant closures of retail chain stores as online retail marketplaces expand.
These factors and other points were presented by CPSA members and their lobbying team to legislators, who heard our concerns and acted in our favor. We thank all those who came to support retailers, wholesalers and other entities that would be affected by this potential loss at the public hearings in Hartford and other events. We will remain vigilant as budget negotiations continue.
TOTAL WINE LAWSUIT DISMISSED BY FEDERAL COURT
On June 6th, the U. S District Court for the District of Connecticut dismissed the lawsuit by Total Wine & More seeking to end the state's liquor pricing system, including minimum bottle pricing. Chief U.S. District Judge Janet C. Hall ruled that Connecticut's liquor pricing system cannot be pre-empted by a federal antitrust law that prohibits activities restricting interstate commerce and competition in the marketplace. She said this falls under the purview of Connecticut's executive and legislative branches of government.
As I stated in an interview with the Hartford Courant: We think this is a good decision in properly interpreting the way the alcoholic beverage laws are crafted across the country, not just in Connecticut. With the repeal of Prohibition, the federal government gave states the power to regulate alcoholic beverages. Some states chose to control it themselves … Other states set up a system whereby they would strongly regulate all aspects of alcohol sales, which is what Connecticut did. Our state regulations were upheld by the judge.
We thank the Wine & Spirit Wholesalers of Connecticut, the Connecticut Beer Wholesalers Association and the Connecticut Restaurant Association for their efforts in fighting this lawsuit alongside the Connecticut Package Stores Association.
For more information on any of these topics and more visit CPSA online.
Carroll J. HughesConnecticut Package Store AssociationCT Liquor Law
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Eyeball sculpture damaged in Dallas protest being repaired
Posted: Dec 2, 2020 / 02:09 PM CST / Updated: Dec 2, 2020 / 03:08 PM CST
DALLAS (AP) — A large sculpture of an eyeball in downtown Dallas is in the process of being repaired after it was tagged with graffiti during a protest following George Floyd's death.
The 30-foot-high sculpture that has caught the attention of passersby since it was installed in 2013 was tagged with graffiti on May 29. The message "NOW UC US," along with Floyd's initials and his words, "I can't breathe," were spray-painted on the sculpture.
Floyd died after a white Minneapolis police officer pressed his knee against the handcuffed Black man's neck for several minutes while Floyd said he couldn't breathe. His death sparked mass demonstrations around the world.
Sculptor Tony Tasset told The Dallas Morning News that a lot of the grafiti has been cleaned from the eyeball he created, but that there are some chips and cracks that will also be repaired. He said the repairs will cost "north of $100,000."
"It feels terrible, of course, to have your work damaged. But I'm supportive of Black Lives Matter. I'm sympathetic not to the vandalism but certainly to the anger," Tasset said.
Tasset, who lives in Sawyer, Michigan, plans to fly to Dallas for that last phase of repairs, which are expected in March.
The eyeball was originally created for a temporary exhibit in Chicago. It was placed in a garden near The Joule, a luxury hotel in Dallas, in 2013 after being purchased by the hotel's owner, Tim Headington.
Jeny Bania, a spokeswoman for The Joule, said the eyeball has required "comprehensive cleaning and repair." She added that moving forward, they'll "be enhancing our electronic and personal oversight of the art."
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Wonderful Opportunity In Prime Leslieville*3 Bedroom Semi Detached Home*Large Principal Rooms. Eat-In Kitchen With W/O To Sunroom & Outdoor Deck*Laneway Access In Back For 1 Parking In Garage*Best Location In Leslieville-Walk Down The Street To Some Amazing Shops/Cafes/Restaurants-The Heart Of Leslieville-Also Ttc To Downtown-Jimmy Simpson Rec Centre/Park-Easy Access To The Lakeshore.
2 Fridges, 2 Stoves, 2 Sets Washer Dryer.Currently Set Up As Income Generating Property With Leases In Place Until June 2019. Tenants Must Be Assumed. Can Be Easily Converted Back To Single Family.


Since watching Great British Sewing Bee, I've been really inspired by one contestant - Tilly - and her inspiring sewing blog. Her picnic blanket skirt tutorial caught my eye, and I set out to make one myself.
Without further ado... my picnic blanket skirt.
I followed her tutorial step by step, using a mint pique fabric and my own fabric covered buttons. In the process, I somehow grossly overestimated how large the waist should be, so I actually made it a wrap skirt (thought you can't tell in the photos). The top button is functional, along with a hidden button on the inside of the waistband, but the rest of the buttons are just for looks. Next time I make it I'll use correct measurements and make all the buttons functional. For now, though, I loved this first attempt. I do have one correction to make - the top button is sitting glaringly off-centered because, for some reason, I put the buttonhole in horizontally instead of vertically. I'll fix it soon and it'll be officially finished.
Thanks, Tilly, for such a great tutorial!
So cute! You're posts are making me want to clear out enough room to set up my sewing table!
FYI buttons that can have strain, like at waist or down shirt front, are usually placed horizontal, so there is a little "give" in the button hole, rather than just poppint open. That said, cute skirt! And marking button holes is very difficult!
How cute are you?! Love it!
Its beautiful, already visited Tilly and the buttons. Making me one soon.
Much better and confident shots than last time.
Jokes apart u look so beautiful n delicate..i love the colour of the skirt..
But I still miss Liam.
You look darling!!! I love the sense of accomplishment on your face ~ you should be proud of your first attempt, you did a great job!
Mint green is a winner any day- but you found the ultimate perfect color and style! Let me know if you ever wanna sell it to me;) haha... But really.
This is beautiful...you are beautiful!! I can't believe all the creative things you're good at... sewing, painting, re-purposing, jewelry-making... seriously! You are talented.
Whatsapp on Android is the most effective up until now seen. whatsuply Whatsapp is not readily available for iPad and also iPod touch.


text:
Q: C# File lock in Windows Service I wrote one windows service to do the following.

*

*Call Hand Punch machine's vendor .exe to generate a text file to get data from Hand Punch machine. Store in a temporary folder. The .exe called using System.Diagnostics.Process.Start

*Attach and send email the file
SmtpClient client = new SmtpClient();
MailMessage mail = new MailMessage("f@abc.com", "t@abc.com");
string[] fileEntries = Directory.GetFiles(...);
foreach (string fileName in fileEntries)
{
    Attachment attachment = new Attachment(fileName);
    mail.Attachments.Add(attachment);
}
client.Send(mail);

*Move the file from temporary folder to Sent folder
I could accomplish step 1 and 2, but when I try step 3 ( to move the file), the file is locked by my service ( but I am sure it is not writing anything to it).
How can I move the file?

A: SmtpClient.Send is locking files. When I used 'using', solved my issue.


Squire has introduced a new range of Stronghold® Half Euro body only padlocks that accept 45mm half euro cylinders, and they don't need any extra modification.
Designed to the highest specification, these cylinder operated padlock bodies are based on Squire's ultra-high security Stronghold® range of CEN graded padlocks.
They are available in two body widths of 55mm (CEN grade 4) and 65mm (CEN grade 5), with open shackle and long shackle options.
Any manufacturer 45mm half euro cylinders can be used, but they require the same CEN grades as the padlocks to meet CEN standards.
Padlock features also include solid hardened steel lock body, electrophoretic anti-corrosion finish and hardened boron alloy steel shackle.
They offer anti-drill protection and are highly corrosion resistant with over 400-hours salt spray test, making them ideal for harsh environments.
All padlocks come with Squire's personal 10-year guarantee as standard.
"We're proud to introduce these very high specification padlocks using our Stronghold® range of CEN grade locks.
Designed to reflect Squire's premium brand status for 'toughness guaranteed,' Stronghold® range of Half Euro padlocks is supported by the independent crime fighting charity Crimestoppers.


Adults should choose a small to medium-sized brush head. This should have soft to medium, multi-tufted, round-ended nylon bristles or 'filaments'.
The head should be small enough to reach into all parts of the mouth, especially the back of the mouth; where it can be difficult to reach.
You can also use more specialized toothbrushes. For instance, people with sensitive teeth can use softer-bristled brushes, while people with crooked or irregular teeth can use smaller-headed toothbrushes.
Kids toothbrushes vary in size and shape, depending on their need and ability.
Choose a toothbrush with soft bristles, that is made of a round-ended or "polished" material. These types of brush heads, clean kids' teeth well and are not rough on their gums.
A toddler's toothbrush has a small head and large handle with soft grip, that are easy for small hands to hold.
Powered toothbrushes are also available for this age group and older children, which can be engaging and fun for them to use.


text:
Dubai plans world's largest solar project
UAE Environment
Concentrated Solar Power (CSP) expected to be operational within five years in Dubai
Published: June 02, 2016 16:13 Derek Baldwin, Chief Reporter
Saeed Al Tayer and Waleed Salman, executive vice-president, strategy and business development, address media at the announcement of the new project yesterday. Image Credit: Virendra Saklani/Gulf News
Dubai: The largest Concentrated Solar Power (CPS) project to be built on a single site in the world will begin power generation in Dubai within the next five years, officials announced on Thursday.
The long view is that the new CPS site will generate 1,000 megawatts (MW) of power by 2030 as part of the Dubai Clean Energy Strategy to generate 75 per cent of Dubai's power from clean energy by 2050.
Concentrated solar power is generated by circular rings of solar mirrors called heliostats that direct sunlight into a central collection tower where the sun's rays power a steam turbine to generate electricity.
The project will surpass the existing world's largest CPS tower in Morocco that has a power generating capacity of 150MW, senior energy officials said.
Saeed Mohammad Al Tayer, managing director and CEO of Dubai Electricity and Water Authority (Dewa), announced the project at a press conference at Burj Khalifa, pledging a cleaner, self-sufficient future for Dubai through renewable energy.
The new CPS project, Al Tayer said, will deliver inexpensive power at less than 8 US cents per kilowatt hour as it is generated from the site to be located at the existing Mohammad Bin Rashid Al Maktoum Solar Park.
The solar project and solar park when completed is expected to slash carbon emissions in Dubai by more than 6.5 million tonnes of harmful CO2, helping Dubai and the UAE meets its commitment to the Paris Agreement to keep global warming temperature increases below 2 degrees Celsius this century.
Al Tayer said that Dewa has issued a call for tenders inviting international solar consultants to submit proposals on the first 200MW phase of the project.
He told Gulf News in an interview after the announcement that the benefit of using concentrated solar power is that it can be stored for eight to 12 hours after generation which can help power the emirate through the night hours.
The first phase of the new CPS project should be operational by 2021, Al Tayer said, and will "contribute to the happiness of society by making Dubai the city with the lowest carbon footprint in the world by 2050".
"Today's announcement of the first CSP project will be followed by other major projects to diversify Dubai's energy mix. This will contribute to achieving the vision or our wise leadship to make Dubai number one globally, and our vision to become a sustainable, innovative world-class utility," Al Tayer told those gathered at the launch.
Hadi Tahboub, vice-president — Programmes Director of Middle East Solar Industry Association, told Gulf News said the new project puts Dubai one step closer to becoming the world leader in solar power generation.
"Dubai is taking the gutsy step of fully embracing renewable energy," said Tahtoub. "Dubai is showing people how it is done."
Dubai Clean Energy Strategy 2050
The aim of the clean energy strategy is to increase the amount of renewable energies used to power Dubai's electricity demand while reducing fossil fuel use by 2050:
• 7 per cent of total power output from clean energy by 2020
• 25 per cent of total power output from clean energy by 2030
Source: Dewa
More From Environment
Fake news: No earthquake in Dubai
Fujairah: Endangered lizard found in shipping container
Meet the amazing Zayed Sustainable Prize winners
UAE's National Experts Programme Fellows shine at ADSW
Caicedo pleads with Brighton for Arsenal move


text:
var chai = require("chai");
var chaiAsPromised = require("chai-as-promised");
chai.use(chaiAsPromised);

var errorComponents = require("../js/components/error.js")

var should = require('chai').should(),
        expect = require('chai').expect,
        assert = require('chai').assert;

var Promise = require('bluebird');
var _ = require('lodash');

var express = require('express');
var request = require('supertest')

describe('Error components', function() {
    it('should "errorComponents" has good signature', function() {
        return assert.typeOf(errorComponents, 'function', 'errorComponents is not a function')
    });

    it('should "errorComponents" has mandatory method', function() {

        var c = new errorComponents();

        expect(c).to.respondTo('NotFoundError');
        expect(c).to.respondTo('AccessDeniedError');
        expect(c).to.respondTo('handleError');
    });

})

describe('#NotFoundError', function() {

    it("should return an Error object", function() {

        var c = new errorComponents();

        var error = new c.NotFoundError("MESSAGE", "type", "id");
        expect(error.item_id).equal("id");
        expect(error.type).equal("type");
        expect(error.message).equal("MESSAGE");
        expect(error.name).equal("NotFoundError");

    })

    it("should return an Error object", function() {

        var c = new errorComponents();

        var error = new c.NotFoundError("", "type", "id");
        expect(error.item_id).equal("id");
        expect(error.type).equal("type");
        expect(error.message).equal("NOT_FOUND");
        expect(error.name).equal("NotFoundError");
    })
});

describe('#AccessDeniedError', function() {

    it("should return an Error object", function() {

        var c = new errorComponents();

        var error = new c.AccessDeniedError("MESSAGE", "type");
        expect(error.reason).equal("type");
        expect(error.message).equal("MESSAGE");
        expect(error.name).equal("AccessDeniedError");

    })

    it("should return an Error object", function() {

        var c = new errorComponents();

        var error = new c.AccessDeniedError("", "type");
        expect(error.reason).equal("type");
        expect(error.message).equal("NOT_ALLOWED");
        expect(error.name).equal("AccessDeniedError");
    })
});

describe('#handleError', function() {

    it("should handle NotFoundError default error", function(done) {

        var app = express();
        app.use(function(req, res) {

            var c = new errorComponents({debug: false});

            var e = new c.NotFoundError("", "user", "1");
            c.handleError(e, req, res);

            return res;
        });
        request(app)
                .post('/')
                .expect(function(res) {

                    if (!('date' in res.body))
                        return "missing date key";

                    if (!('details' in res.body))
                        return "missing details key";

                    if (!('error' in res.body))
                        return "missing error key";

                    if (res.body.error != "NOT_FOUND")
                        return "error key invalid";

                    if (res.body.statusCode != "404")
                        return "status code key invalid";
                })
                .expect(404, done);
    })

    it("should handle NotFoundError with custom message", function(done) {

        var app = express();
        app.use(function(req, res) {

            var c = new errorComponents({debug: false});

            var e = new c.NotFoundError("CUSTOM", "user", "1");
            c.handleError(e, req, res);

            return res;
        });
        request(app)
                .post('/')
                .expect(function(res) {

                    if (!('date' in res.body))
                        return "missing date key";

                    if (!('details' in res.body))
                        return "missing details key";

                    if (!('error' in res.body))
                        return "missing error key";

                    if (res.body.error != "CUSTOM")
                        return "error key invalid";

                    if (res.body.statusCode != "404")
                        return "status code key invalid";
                })
                .expect(404, done);
    })

    it("should handle AccessDeniedError default error", function(done) {

        var app = express();
        app.use(function(req, res) {

            var c = new errorComponents({debug: false});

            var e = new c.AccessDeniedError("", "you can not make this action");
            c.handleError(e, req, res);

            return res;
        });
        request(app)
                .post('/')
                .expect(function(res) {

                    if (!('date' in res.body))
                        return "missing date key";

                    if (!('details' in res.body))
                        return "missing details key";

                    if (!('error' in res.body))
                        return "missing error key";

                    if (res.body.error != "NOT_ALLOWED")
                        return "error key invalid";

                    if (res.body.statusCode != "403")
                        return "status code key invalid";
                })
                .expect(403, done);
    })

    it("should handle AccessDeniedError with custom message", function(done) {

        var app = express();
        app.use(function(req, res) {

            var c = new errorComponents({debug: false});

            var e = new c.AccessDeniedError("CUSTOM", "you can not make this action");
            c.handleError(e, req, res);

            return res;
        });
        request(app)
                .post('/')
                .expect(function(res) {

                    if (!('date' in res.body))
                        return "missing date key";

                    if (!('details' in res.body))
                        return "missing details key";

                    if (!('error' in res.body))
                        return "missing error key";

                    if (res.body.error != "CUSTOM")
                        return "error key invalid";

                    if (res.body.statusCode != "403")
                        return "status code key invalid";
                })
                .expect(403, done);
    })

    it("should handle default Error", function(done) {

        var app = express();
        app.use(function(req, res) {

            var c = new errorComponents({debug: false});

            var e = new Error("foo");
            c.handleError(e, req, res);

            return res;
        });
        request(app)
                .post('/')
                .expect(function(res) {

                    if (!('date' in res.body))
                        return "missing date key";

                    if (!('details' in res.body))
                        return "missing details key";

                    if (!('error' in res.body))
                        return "missing error key";

                    if (res.body.error != "foo")
                        return "error key invalid";

                    if (res.body.statusCode != "500")
                        return "status code key invalid";
                })
                .expect(500, done);
    })

    it("should handle default Error with custom statusCode", function(done) {

        var app = express();
        app.use(function(req, res) {

            var c = new errorComponents({debug: false});

            var e = new Error("foo");
            e.statusCode = 408;
            c.handleError(e, req, res);

            return res;
        });
        request(app)
                .post('/')
                .expect(function(res) {

                    if (!('date' in res.body))
                        return "missing date key";

                    if (!('details' in res.body))
                        return "missing details key";

                    if (!('error' in res.body))
                        return "missing error key";

                    if (res.body.error != "foo")
                        return "error key invalid";

                    if (res.body.statusCode != "408")
                        return "status code key invalid";
                })
                .expect(408, done);
    })

    it("should handle default Error with custom statusCode and details", function(done) {

        var app = express();
        app.use(function(req, res) {

            var c = new errorComponents({debug: false});

            var e = new Error("foo");
            e.statusCode = 409;
            e.details = "bar"
            c.handleError(e, req, res);

            return res;
        });
        request(app)
                .post('/')
                .expect(function(res) {

                    if (!('date' in res.body))
                        return "missing date key";

                    if (!('details' in res.body))
                        return "missing details key";

                    if (!('error' in res.body))
                        return "missing error key";

                    if (res.body.error != "foo")
                        return "error key invalid";

                    if (res.body.statusCode != "409")
                        return "status code key invalid";

                    if (res.body.details != "bar")
                        return "details invalid";
                })
                .expect(409, done);
    })

    it("should handle ReferenceError", function(done) {

        var app = express();
        app.use(function(req, res) {

            var c = new errorComponents({debug: false});
            try {
                console.log(vars / 0);
            } catch (e) {
                c.handleError(e, req, res);
            }

            return res;
        });
        request(app)
                .post('/')
                .expect(function(res) {

                    if (!('date' in res.body))
                        return "missing date key";

                    if (!('details' in res.body))
                        return "missing details key";

                    if (!('error' in res.body))
                        return "error is not defined";

                    if (res.body.error != "vars is not defined")
                        return "error key invalid";

                    if (res.body.statusCode != "500")
                        return "status code key invalid";
                })
                .expect(500, done);
    })
});

redpajama_set_name: RedPajamaGithub