text:
University of Surrey makes key appointment for new School of Veterinary Medicine
Monday, June 3, 2013 - Mark Humph
The University of Surrey is delighted to announce the appointment of Professor Christopher Proudman, MA, Vet MB, PhD, Cert EO, FRCVS, as the founding Head of its new School of Veterinary Medicine.
Professor Proudman joins the University of Surrey from the University of Liverpool where he has held numerous managerial and leadership roles in the School of Veterinary Science, the wider University and in external partnerships. Previous roles include Head of Department of Veterinary Clinical Science, membership of the Senior Management Teams of the Faculty of Veterinary Science and the Institute of Translational Medicine, and Chair of Cheshire West and Chester's Rural Regeneration Stratgy Board. Professor Proudman has been an equine clinician and clinical teacher for over 20 years. His current research focus is in the area of equine gastrointestinal health and disease, and he has recently undertaken research on behalf of the Horseracing Betting Levy Board, veterinary charitable trusts and the Egyptian Government.
In announcing Professor Proudman's appointment Dean of the Faculty of Health and Medical Sciences, Professor Lisa Roberts commented: "I'm delighted to welcome Chris to the University of Surrey to lead the delivery of our exciting new vision for our School of Veterinary Medicine. He brings with him a wealth of knowledge and a strong track record of research and leadership, and I very much look forward to working closely with him in the development of the new School."
Professor Proudman is expected to commence work at Surrey in early September.
(Source: surrey.ac.uk)
Major veterinary players unite in commitment to evidence-based veterinary medicine
RCVS clarifies role on new veterinary schools
Joint RCVS and University of Surrey news: University of Surrey's veterinary degree receives formal approval from veterinary regulator
RCVS news: RCVS makes changes to new CPD policy following feedback from professions
RCVS news: College releases new position statement on complementary and alternative medicines


On behalf of Sharisa Battle of Schimmel & Parks, APLC on Thursday, October 18, 2018.
Perhaps your employer does not include you in company emails because of your nationality. Perhaps you were refused a promotion because you practice a specific religion. Maybe you receive less pay, for the same amount of work, due to your gender. In any of these discriminatory cases, you have the authority to file a complaint with the Equal Employment Opportunity Commission (EEOC).
On behalf of Sharisa Battle of Schimmel & Parks, APLC on Thursday, October 4, 2018.
A growing number of young couples are opting to not have children. Several have speculated to the reason for the shift. It could be the increased cost of living the country has experienced recently, a greater desire to focus on personal interests over raising a family or simply a change in social norms.


text:
Q: Why do the following products of Pauli operators commute? I've read in Mermin (https://doi.org/10.1063/1.2810588) that the operators $\hat{\sigma}_{y1} \hat{\sigma}_{x2} \hat{\sigma}_{x3}$, $\hat{\sigma}_{x1} \hat{\sigma}_{y2} \hat{\sigma}_{x3}$ and $\hat{\sigma}_{x1} \hat{\sigma}_{x2} \hat{\sigma}_{y3}$ all commute as "the product of any two of them differs from the product in the reverse order by an even number of such anti-commutations" (page 9).
Take for instance $[\hat{\sigma}_{y1} \hat{\sigma}_{x2} \hat{\sigma}_{x3}, \hat{\sigma}_{x1} \hat{\sigma}_{y2} \hat{\sigma}_{x3}]$; I think that I must not understand the implementation of the tensor products as my guess would be that $\hat{\sigma}_{y1} \hat{\sigma}_{x2} \hat{\sigma}_{x3} \hat{\sigma}_{x1} \hat{\sigma}_{y2} \hat{\sigma}_{x3} - \hat{\sigma}_{x1} \hat{\sigma}_{y2} \hat{\sigma}_{x3} \hat{\sigma}_{y1} \hat{\sigma}_{x2} \hat{\sigma}_{x3} = [\hat{\sigma}_{y1}, \hat{\sigma}_{x1}][\hat{\sigma}_{x2}, \hat{\sigma}_{y2}] \hat{\sigma}_{x3} \neq 0$. Any help would be much appreciated. Thank you!

A: $$(_^1 _^2_x^3)( _^1\sigma_^2_^3)\\=-\sigma_x^1_^1 _^2_x^3 \sigma_^2_^3\quad (x1,y1\, {\rm anticommute})\\
=+ \sigma_x^1\sigma_y^2_^1 _^2_x^3 _^3\quad (x2,y2 \,{\rm anticommute})\\
=+(_^1\sigma_^2_^3)(_^1 _^2_x^3)\quad  (x3,x3 \,{\rm commute})
$$
Everything else commutes.

redpajama_set_name: RedPajamaStackExchange


Sabah, is a one of the of NECC Al Shajaea clinic beneficiaries, she used to attend our clinic , she is 37 years old, married with 8 family members , lived in Gaza city, Al-Shajaea area, her husband is unemployed, we lived in very hard economic situation, some day we can't offer a piece of bread for my children, we can't a save for them their simplest needs.	"These reasons let me suffer of some psychological problems, I became constantly angry, nervous, worry, depressed and no stop for sad thoughts in my mind".	She complained to the psychosocial counselor in the clinic and registered in the PSS program in the NECC family care center.	The psychosocial counselor calmed her down, assuring that she is not a psycho patient and she told Sabah that knowing your disease is the first step of being healthy. Mrs. Sabah is joining "mind and body skills" sessions conducted in the clinic with the psychosocial counselor to overcome her problems, she applied 6 sessions with the counselor, showing that her health status is much important that anything else in the world, also she should have faith in God. After my experience she said, I advise all women to take care of their mental health which reflects both mental and physical health."I would like to thank NECC Shajaeaa Clinic and its unique staff for the great support they gave to me and to others. I wish the continuous good work for the psycho- social support program because it is so important and essential in every individual's life.


On this page, we suggest you different documents which will help you to prepare your staying at Dar Hamza. We hope they will bring you all the necessary informations to guide you on your travel. However, don't hesitate to contact us for any supplementary requests or further informations.


And over the past few months it has been refreshing, but somewhat painful, to hear his honest assessments of Chelsea's poor displays.
Speaking to Sky Sports ahead of Chelsea's Saturday clash against Huddersfield (Watch live, 10 a.m. ET on NBCSN and online via NBCSports.com), Rudiger lambasted his team for their 4-0 defeat at Bournemouth a few days ago.
"Everyone needs to be ashamed of what happened in the last game," Rudiger said. "Everyone needs to think very deeply about that because you can win, lose or draw, but it's the manner which is important. In the second half, I can't really describe it. We went out and they scored after two minutes then things went very badly.
When you add in Maurizio Sarri's comments that his underestimated the task of being Chelsea's new manager and he finds it hard to motivate this group of players, it appears a reality check has occurred rather quickly for the Blues.
Chelsea have lost their place in the top four on goal difference and have only won one of their last four Premier League games.
This honesty from Sarri and Rudiger will be welcomed by Chelsea's fans, but it will do little to lift and gloom around the west London club at a pivotal stage of the season.
They've reached the League Cup final and latter stages of the Europa League, but finishing in the top four should be the bare minimum requirement for Sarri in his first season in charge of Chelsea. The style he is asking them to play, with Jorginho at the center of it and Kante asked to play a more attacking role, has been criticized but Sarri says he will not change his philosophy.
That may be the biggest problem, as Chelsea's players seem to have decided amongst themselves that they don't want to play "Sarri-ball" and would rather play their own way.
Right now a top four finish looks anything but guaranteed as Eden Hazard has lost his superb early-season form and the team which opened up the Premier League season without a defeat in their first 11 games has now lost three of their last seven.
Chelsea are all over the place defensively, but it appears that keeping focused and staying motivated is their biggest problem right now.
That is a damning indictment on these players and Sarri.


text:
Iraq violence: Islamic State 'behind Baghdad attacks'
Image source, Reuters
There are fears IS would step up attacks during the Muslim fasting month of Ramadan
Islamic State (IS) militants have said they were behind a spate of bomb attacks in the Iraqi capital, Baghdad, which killed at least 28 people.
In a statement posted on social media, IS described Sunday night's bombings as the "pounce of the monotheists on the chests of the apostates".
Several of the blasts were carried out by suicide bombers in Shia areas of the capital.
The series of attacks was one of the worst to hit Baghdad in recent weeks.
There have been fears IS would step up its bombings campaign during the Muslim fasting month of Ramadan.
A series of car bombs exploded across the Iraqi capital on Sunday
Image source, AP
Debris from the blasts was still visible on Monday
A car bomb in the northern Shaab district on Sunday killed at least 10 people, and a suicide bomber struck at a nearby market, claiming more lives.
Another car bomb exploded in the Bunouk district killing at least nine people. Scores of others were wounded in the attacks, medics said.
IS has occupied parts of Iraq and is battling government forces in the north and west of the country.
The government announced on Monday that it had begun a long-awaited, large-scale military offensive to dislodge IS from Anbar province,
Meanwhile, the UN has said that about 15,000 civilians have been killed in Iraq since the start of 2014.
It says that it recorded at least 44,136 civilian casualties (14,947 killed and 29,189 wounded) as a result of "non-international armed conflict in Iraq".
The report says the figure, which runs up to the end of April 2015, only takes account of verifiable casualties and that the real toll may be much higher.
Face-to-face with 'IS bombers'
Virtuoso performance - at a bomb site
Can Iraq's army dislodge IS?
The war against 'Islamic State' in maps and charts
In depth: Islamic State conflict
Islamic State group

redpajama_set_name: RedPajamaCommonCrawl


69/250 is part of the HW City series and the 2015 Treasure Hunt set. The metalflake copper car boasts blue stripes with white stars, red and black graphics, and a large football graphic on the hood.
The sides also have a football graphic as well as the low production symbol. It has tinted windows, a tan interior, black PR5 wheels with red chrome rims, and a tan plastic Malaysia base.
This vehicle can be found in mix "D" of 2015 Hot Wheels mainline cases.
The Fast Felion is a unique design that made its debut in the 2009 New Models series.


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text:
\section{Introduction}\label{sec:intro}

Optimization problems have spread in wide area of science, for example, information science and statistical physics.
Properties of optimization problems can be summarized as follows:
\begin{itemize}
 \item There are many elements.
 \item A cost function can be defined.
       The best solution is the state where the cost function takes the maximum value or the minimum value depending on the definition of the given problem.
\end{itemize}
An well-known example of optimization problem is the so-called ``traveling salesman problem''.
The traveling salesman problem is to find the shortest path covering all the cities.
Here, we can path through each city only once.
In this problem, the cost function is the length of path.
In general, it is difficult to obtain the best solution by naive approach since the number of candidates of a solution is very huge.
Then, in the optimization problem, it is a central issue how to find the best (or better) solution.
There are many types of optimization problems.
Depending on an optimization problem, there are specific algorithms to solve the problem.

To treat optimization problem in a general way, on the other hand, a couple of approaches have been developed from a viewpoint of statistical physics.
The cost function can be regarded as the internal energy in statistical physics.
The cost function of the best solution corresponds to the internal energy of the ground state.
Kirkpatrick {\it et al.} proposed a pioneering generic algorithm -- simulated annealing\cite{Kirkpatrick-1983,Kirkpatrick-1984}.
Simulated annealing is a method to obtain not so bad solution of optimization problem by decreasing temperature.
In high temperature, since the probability distribution at equilibrium state is almost flat, the state can be changed easily.
As we decrease temperature gradually, generated probability distribution expects to approach the equilibrium probability distribution at each temperature.
Then we can find not so bad solution of problem by simulated annealing.
Because of the Geman-Geman's argument\cite{Geman-1984}, we can succeed to obtain the best solution with the probability unity if we decrease the temperature slow enough.
Although there could be better algorithms specific to each optimization problem, simulated annealing is regarded as a stable and generic method because of easy implementation and independency from problems.
The purpose of our study is to establish a good general method for optimization problems from a viewpoint of quantum statistical physics.

The organization of this paper is as follows.
We first review on quantum annealing which has been believed a good general method to obtain not so bad solution of optimization problems.
In section 3, we will review how to implement Monte Carlo simulation.
In section 4, we will introduce a model which can treat clustering problems.
We also consider quantum effect on this model in this section.
In section 5, some remarks on this problem will be shown.
In section 6, we will show results obtained by our proposed method.
In section 7, we will summarize our study.
We will review on the concepts of ``invisible fluctuation'' in the appendix A.

\section{Quantum annealing}\label{sec:qa}

Quantum annealing is expected that it succeeds to obtain the better solution of optimization problems than simulated annealing\cite{Kadowaki-1998,Kadowaki-2002,Farhi-2001,Santoro-2002,Charkrabarti-book1,Charkrabarti-book2,Das-2008,Ohzeki-2011}.
This method is based on quantum statistical physics.
There have been a couple of realization methods of quantum annealing: (i) stochastic method, (ii) deterministic method, and (iii) experiment on artificial lattice.

The stochastic method is realized by quantum Monte Carlo method in many circumstances\cite{Santoro-2002,Kurihara-2009}.
The quantum Monte Carlo method is an established method to obtain equilibrium properties of strongly correlated quantum systems.
Efficient algorithms for quantum annealing have been developed such as the cluster algorithm\cite{Nakamura-2008,Morita-2009}.
Owing to these masterly methods, quantum Monte Carlo simulation can be adopted for large-scale systems.

There are a couple of deterministic methods for quantum annealing.
The first one is based on the time-dependent Schr\"odinger equation\cite{Kadowaki-1998}.
This method can trace the real-time evolution which can be observed in real experiments.
This method is called as ``quantum adiabatic evolution'' in quantum information science\cite{Farhi-2001}.
This method cannot treat large-scale systems because of the limitation of size of memory in computer.
The second one is based on a time-dependent density matrix renormalization group\cite{Suzuki-2007,Laguna-2007}.
By this method, we can study time-evolution of one-dimensional quantum systems.
However it is difficult to treat two-dimensional or three-dimensional quantum systems by this method.
Both of two methods are based on the principle of quantum mechanics.
However, it is not necessary to treat the Schr\"odinger equation directly if we adopt the deterministic method as quantum annealing.
This is because our purpose is to obtain not so bad solution of given problem.
One of the examples is mean-field calculation\cite{Tanaka-2002,Sato-2009}.
This method can treat large-scale systems as well as the quantum Monte Carlo simulation.
Then, this method has been widely adopted for optimization problems.

Next there are a couple of proposals for experiments on artificial lattice for quantum annealing.
For example, we can generate quantum state by optical lattice\cite{Jane-2003,Porras-2004,Lewenstein-2007,Friedenauer-2008}.
In addition, the Ising model with transverse field can be realized by superconducting flux qubits\cite{Harris-2009,Berkley-2010,Harris-2010,Amin-2011}.
These methods are expected as a new type of quantum computer.

There have been a number of studies on quantum annealing from a viewpoint of theoretical physics.
Convergence theorem for quantum annealing was proved as well as that for simulated annealing\cite{Morita-2006,Morita-2007,Morita-2008}.
According to this theorem, the permitted upper bound of sweeping speed of quantum field in quantum annealing is larger than that of temperature in simulated annealing.
From this theorem, quantum annealing seems better than simulated annealing in principle.
However, we often sweep temperature and/or quantum field faster than that upper bound in practice.
Then, it is nontrivial whether quantum annealing is better than simulated annealing from a viewpoint of practical situation.
Microscopic behavior in quantum annealing has been also investigated.
It is well-known that frustrated systems have some interesting static behavior induced by thermal fluctuation and quantum fluctuation.
Dynamical properties of frustrated systems have been studied from a viewpoint of quantum annealing\cite{shu-t-2007a,Matsuda-2009,shu-t-2009a,shu-t-2010a,shu-t-2010b,shu-t-2010d}.
Novel type of implementation of quantum annealing itself is an important topic.
One of the examples is quantum annealing based on the Jarzynski equality\cite{Ohzeki-2010a}.
This method is expected an efficient method since it uses both merits which come from thermal fluctuation and quantum fluctuation.

In this paper, we adopt the first strategy -- the quantum Monte Carlo method as the realization of quantum annealing.

\section{Monte Carlo simulation}\label{sec:mc}

In this section, we review how to implement Monte Carlo simulation.
It has been often used in order to obtain the equilibrium properties of strongly correlated systems such as magnetic systems and bosonic systems.
Equilibrium physical quantities of the system which is expressed by the Hamiltonian ${\cal H}$ at finite temperature $T$ is given as
\begin{eqnarray}
 \langle {\cal O} \rangle_{\rm eq}^{(T)} =
  \frac{{\rm Tr}\, {\cal O} {\rm e}^{-\beta {\cal H}}}{{\rm Tr}\, {\rm e}^{-\beta {\cal H}}},
\end{eqnarray}
where $\beta$ denotes the inverse temperature $1/T$ and here the Boltzmann constant $k_{\rm B}$ is set to be unity.
If we consider a small system, we can obtain all of the equilibrium physical quantities by naive method.
If we consider large scale systems, however, we cannot obtain the equilibrium properties by naive method in practice.
Monte Carlo simulation enables us to calculate equilibrium physical quantities with high accuracy by the following relation.
\begin{eqnarray}
 \frac{\sum_\Sigma {\cal O}(\Sigma) {\rm e}^{-\beta {\cal H}(\Sigma)}}{\sum_\Sigma {\rm e}^{-\beta {\cal H}(\Sigma)}}
  \to \langle {\cal O} \rangle_{\rm eq}^{(T)},
\end{eqnarray}
where $\Sigma$ denotes sample, in other words, state.
Physical quantity converges to the equilibrium value as the number of samples increases.
In fact, the above calculation is inefficient if states are generated by uniform distribution.
In order to make the method more efficient, we generate states according to equilibrium probability distribution which is proportional to the Boltzmann factor ${\rm e}^{-\beta E(\Sigma)}$.
This method is called the importance sampling.
We can obtain the equilibrium value as follows:
\begin{eqnarray}
 \frac{\sum_\Sigma {\cal O}(\Sigma)}{\sum_\Sigma} \to \langle {\cal O} \rangle_{\rm eq}^{(T)}.
\end{eqnarray}
In order to generate a state from the equilibrium distribution, we just have to use Markov chain Monte Carlo method.
Time evolution of probability distribution is given as the master equation:
\begin{eqnarray}
 \nonumber
 P(\Sigma_i,t+\Delta t) =
  &&-\sum_{j \neq i} P(\Sigma_i,t) w(\Sigma_j|\Sigma_i) \Delta t \\
  &&+ \sum_{j \neq i} P(\Sigma_j, t) w (\Sigma_i|\Sigma_j) \Delta t
  + P(\Sigma_i,t) w(\Sigma_i|\Sigma_i) \Delta t,
\end{eqnarray}
where $P(\Sigma_i,t)$ denotes probability of the state $\Sigma_i$ at time $t$ and $w(\Sigma_j|\Sigma_i)$ represents transition probability from the state $\Sigma_i$ to the state $\Sigma_j$ in unit time.
Transition probability $w(\Sigma_j|\Sigma_i)$ obeys
\begin{eqnarray}
 \sum_{\Sigma_j} w(\Sigma_j|\Sigma_i) = 1 \,\,\,\, (\forall \Sigma_i).
\end{eqnarray}
The master equation can be represented as
\begin{eqnarray}
 {\bf P}(t+\Delta t) = {\cal L} {\bf P}(t),
\end{eqnarray}
where ${\bf P}(t)$ is a vector-representation of probability distribution $\{P(\Sigma_i,t)\}$ and ${\cal L}$ is called the transition matrix whose elements are expressed as
\begin{eqnarray}
 {\cal L}_{ji} = w(\Sigma_j|\Sigma_i) \Delta t,
  \,\,\,\,\,
 {\cal L}_{ii} = 1- \sum_{j\neq i}{\cal L}_{ji} =
 1 - \sum_{j\neq i} w(\Sigma_j|\Sigma_i) \Delta t.
\end{eqnarray}
It should be noted that ${\cal L}$ is a non-negative matrix by the definition.
This time evolution is the Markovian since the time-evolution operator ${\cal L}$ does not depend on time.
If the time-evolution operator ${\cal L}$ obeys (i) detailed balance condition and (ii) ergordicity, we can obtain the equilibrium probability distribution in the limit of $t \to \infty$ because of the Perron-Frobenius theorem.

\section{Model}\label{sec:model}

Clustering problem is one of the important problems in information science and engineering.
Since it is difficult to obtain the best solution of the clustering problem by naive method, development of a new method which can obtain the best (or not so bad) solution is an important issue.
We can obtain not so bad solution by using simulated annealing as we mentioned in the section 1.
We propose a new type of quantum annealing and succeed to obtain better solution by proposed quantum annealing method as will be mentioned.

In the beginning of this section, we will explain a model to consider clustering problems.
After that, we will introduce new kind of fluctuation -- quantum fluctuation -- into this model.
Next, we will review on implementation method of the quantum annealing method.

\subsection{Clustering problem}

In clustering problems, there are $N$ elements in the space, which is depicted in Fig.~\ref{fig:clustering_problem}(a).

\begin{figure}[ht]
 \begin{center}
  $\begin{array}{ccc}
  \includegraphics[scale=0.8]{./clusteringproblem_a.eps}&
   \hspace{5mm}&
  \includegraphics[scale=0.8]{./clusteringproblem_b.eps} \\
    ({\rm a}) & \hspace{5mm} & ({\rm b}) \\
  \includegraphics[scale=0.8]{./clusteringproblem_c.eps}&
   \hspace{5mm}&
  \includegraphics[scale=0.8]{./clusteringproblem_d.eps} \\
    ({\rm c}) & \hspace{5mm} & ({\rm d})
  \end{array}$
  \caption{
  The dots represent elements.
  The dotted boxes denote clusters.
  In these figures, the number of elements $N=69$ and the number of clusters $Q=4$.
  (a) There are $N$ elements in the space.
  (b) Not so good solution.
  (c) Not so bad solution.
  (d) The best solution.
  }
  \label{fig:clustering_problem}
 \end{center}
\end{figure}

Clustering problems are to decide which the best partition of these elements.
In other words, clustering problems are to search the best division of $N$ elements into $Q$ sub-categories.
Clustering problems have been applied for not only natural science but also social science.
For instance, divisions of the articles in newspaper in terms of contents and analysis of questionnaire can be considered as clustering problems.
Figure \ref{fig:clustering_problem} (b), (c), and (d) represent not so good, not so bad, and the best solution, respectively.
In practice, it is difficult to obtain the best solution by direct method for huge number $N$, since the number of total states is $Q^N$.
Then we often use simulated annealing to solve clustering problems as one of useful generic methods.
However, even if we use simulated annealing, it is difficult to obtain the best solution.
The energy landscape of clustering problems is very complicated such as random spin systems and frustrated systems.
Then we should develop more efficient algorithm than the simulated annealing method.
Actually, there is a pioneering method to obtain more better solution than simulated annealing method.
This method is called exchange method\cite{Hukushima-1996}.
In the exchange method, we prepare some independent layers where the temperatures are different.
We sometimes exchange the states between two layers according to the Boltzmann weight.
In this paper, we adopt another strategy, quantum annealing method as more efficient algorithm.
We introduce quantum term into this model to implement the quantum annealing in the next section.

\subsection{Quantum fluctuation}

Before introducing a quantum fluctuation, we review on a classical Hamiltonian of the original clustering problem.
The classical Hamiltonian is given by
\begin{eqnarray}
 {\cal H}_{\rm c} = {\rm diag}\left( E(\Sigma_1),E(\Sigma_2),\cdots,E(\Sigma_{Q^N})\right),
\end{eqnarray}
where $E(\Sigma_i)$ denotes the eigenenergy of $i$-th state $\Sigma_i$.
Suppose we consider the case for $Q=3$ and $N=2$ as an example.
The classical Hamiltonian is given as
\begin{eqnarray}
 {\cal H}_{\rm c} =
  \left(
   \begin{array}{ccccccccc}
    E(\Sigma_1) & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & E(\Sigma_2) & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & E(\Sigma_3) & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & E(\Sigma_4) & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & E(\Sigma_5) & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & E(\Sigma_6) & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & E(\Sigma_7) & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & E(\Sigma_8) & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & E(\Sigma_9)
   \end{array}
  \right),
\end{eqnarray}
where the definitions of the states from $\Sigma_1$ to $\Sigma_9$ are summarized in Table \ref{table:def_sigma_q3n2}.
$\sigma_1$ and $\sigma_2$ in Table \ref{table:def_sigma_q3n2} represent the states of the first element and the second element, respectively.

\begin{table}[h]
 \tbl{Definitions of the states from $\Sigma_1$ to $\Sigma_9$.}
  {\begin{tabular}{cccccccccc}
    \toprule
    & $\Sigma_1$ & $\Sigma_2$ & $\Sigma_3$ & $\Sigma_4$ & $\Sigma_5$ & $\Sigma_6$ & $\Sigma_7$ & $\Sigma_8$ & $\Sigma_9$ \\
    \hline
    $\sigma_1$ & $1$ & $2$ & $3$ & $1$ & $2$ & $3$ & $1$ & $2$ & $3$ \\
    $\sigma_2$ & $1$ & $1$ & $1$ & $2$ & $2$ & $2$ & $3$ & $3$ & $3$ \\
    \botrule
  \end{tabular}}
 \label{table:def_sigma_q3n2}
\end{table}

Next we introduce a quantum fluctuation into this model.
Since when $Q=2$, the classical Hamiltonian is equivalent to that of the Ising model.
It is natural to extend transverse field in the Ising spin system as a quantum fluctuation.
Then we adopt the following definition as a quantum fluctuation.
\begin{eqnarray}
 {\cal H}_{\rm q} = - \Gamma \sum_{i=1}^N \sigma_i^x
  = -\Gamma \sum_{i=1}^N (\hat{1}_Q - \mathbb{E}_Q),
\end{eqnarray}
where $\hat{1}_Q$ denotes the matrix whose all elements are unity and $\mathbb{E}_Q$ represents identity matrix.
Both $\hat{1}_Q$ and $\mathbb{E}_Q$ are $Q\times Q$ matrices.
It should be noted that $\sigma_i^x$ denotes the $x$-component of $s=1/2$ Pauli matrix at the site $i$ when $Q=2$.
Suppose we consider the case for $Q=3$ and $N=2$ as the previous example, the quantum part of the Hamiltonian ${\cal H}_{\rm q}$ is given as
\begin{eqnarray}
 {\cal H}_{\rm q} =
 \left(
  \begin{array}{ccccccccc}
   0 & -\Gamma & -\Gamma & -\Gamma & 0 & 0 & -\Gamma & 0 & 0 \\
   -\Gamma & 0 & -\Gamma & 0 & -\Gamma & 0 & 0 & -\Gamma & 0 \\
   -\Gamma & -\Gamma & 0 & 0 & 0 & -\Gamma & 0 & 0 & -\Gamma \\
   -\Gamma & 0 & 0 & 0 & -\Gamma & -\Gamma & -\Gamma & 0 & 0 \\
   0 & -\Gamma & 0 & -\Gamma & 0 & -\Gamma & 0 & -\Gamma & 0 \\
   0 & 0 & -\Gamma & -\Gamma & -\Gamma & 0 & 0 & 0 & -\Gamma \\
   -\Gamma & 0 & 0 & -\Gamma & 0 & 0 & 0 & -\Gamma & -\Gamma \\
   0 & -\Gamma & 0 & 0 & -\Gamma & 0 & -\Gamma & 0 & -\Gamma \\
   0 & 0 & -\Gamma & 0 & 0 & -\Gamma & -\Gamma & -\Gamma & 0
  \end{array}
  \right).
\end{eqnarray}
When we use quantum Monte Carlo method, all we have to do is to calculate the probability of the state $\Sigma$.
In the next section, we will show how to calculate the probability of the state $\Sigma$ by path-integral representation\cite{Trotter-1959,Suzuki-1976}.

\subsection{Path integral representation}

We consider the following Hamiltonian
\begin{eqnarray}
 {\cal H} = {\cal H}_{\rm c} + {\cal H}_{\rm q}.
\end{eqnarray}
When the Hamiltonian ${\cal H}$ is a diagonal matrix which corresponds to simulated annealing, {\it i.e.} $\Gamma = 0$, the probability of the state $\Sigma$ at finite temperature $T$ is given as
\begin{eqnarray}
 \label{eq:p_sa}
 p_{\rm SA}(\Sigma;\beta) =
  \frac{{\rm e}^{-\beta E(\Sigma)}}{{\rm Tr}\, {\rm e}^{-\beta {\cal H_{\rm c}}}}
  = \frac{1}{Z}\langle \Sigma | {\rm e}^{- \beta {\cal H}_{\rm c}} | \Sigma \rangle,
\end{eqnarray}
where the denominator is called the partition function in statistical physics.
The partition function $Z$ is calculated as
\begin{eqnarray}
 Z = {\rm Tr}\, {\rm e}^{-\beta {\cal H}_{\rm c}}
  = \sum_\Sigma \langle \Sigma | {\rm e}^{-\beta {\cal H}_{\rm c}} | \Sigma \rangle
		 = \sum_\Sigma {\rm e}^{-\beta E(\Sigma)}.
\end{eqnarray}
We can change the state by using the ``single-spin-flip'' type of heat bath method,
\begin{eqnarray}
 \label{eq:sa-update}
 p_{\rm SA}^{\rm update} (\sigma_i = s | \Sigma \backslash \sigma_i)
  = \frac{{\rm e}^{-\beta E(\sigma_i=s,\Sigma \backslash \sigma_i)}}{\sum_{s'=1}^Q {\rm e}^{-\beta E(s',\Sigma \backslash \sigma_i)}},
\end{eqnarray}
where $\Sigma \backslash \sigma_i$ means $\{\sigma_j|j\neq i\}$ and $p(A|B)$ denotes a conditional probability of $A$ given $B$.
The denominator of Eq.~(\ref{eq:sa-update}) can be calculated where the computational cost is ${\cal O}(Q)$.

In similar with the classical case, the probability of the state $\Sigma$ at finite temperature $T$ and finite quantum field $\Gamma$ is given as
\begin{eqnarray}
 \label{eq:prob_qa_direct}
 p_{\rm QA}(\Sigma;\beta,\Gamma) =
  \frac{\langle \Sigma | {\rm e}^{-\beta {\cal H}} | \Sigma \rangle}
  {\sum_{\Sigma'} \langle \Sigma' | {\rm e}^{-\beta {\cal H}} | \Sigma' \rangle}
  = \frac{\langle \Sigma | {\rm e}^{-\beta {\cal H}} | \Sigma \rangle}{Z}.
\end{eqnarray}
Note that it is difficult to calculate $\langle \Sigma | {\rm e}^{-\beta {\cal H}} | \Sigma \rangle$, since the Hamiltonian including quantum field has off-diagonal elements.
For small systems, we can exactly calculate all of the elements ${\rm e}^{-\beta {\cal H}}$ by using the unitary transform.
However, we can not obtain them for large system in practice.
In order to calculate the probability given by Eq.~(\ref{eq:prob_qa_direct}), we should rewrite the numerator of it by path-integral representation.
Then we obtain the probability
\begin{eqnarray}
 \nonumber
 &&p_{\rm QA}(\Sigma;\beta,\Gamma) =
  \frac{1}{Z}\langle \Sigma |
  \left( {\rm e}^{-\frac{\beta}{m}{\cal H}_{\rm c}} {\rm e}^{-\frac{\beta}{m}{\cal H}_{\rm q}} \right)^m | \Sigma \rangle + {\cal O}(\frac{1}{m})\\
  &&= \frac{1}{Z} \sum_{\Sigma^{(1)'}} \sum_{\Sigma^{(2)}} \cdots \sum_{\Sigma^{(m)'}}
   \prod_{j=1}^m
   \langle \Sigma^{(j)} | {\rm e}^{-\frac{\beta}{m} {\cal H}_{\rm c}} | \Sigma^{(j)'} \rangle
   \langle \Sigma^{(j)'} | {\rm e}^{-\frac{\beta}{m} {\cal H}_{\rm q}} | \Sigma^{(j+1)} \rangle,
\end{eqnarray}
where $m$ is called as the Trotter number and $\Sigma^{(1)}= \Sigma^{(m+1)} = \Sigma$ which corresponds to periodic boundary condition along the Trotter axis.

Here we define
\begin{eqnarray}
 \label{eq:correlation_function}
 &&s(\Sigma^{(j)},\Sigma^{(j+1)}) :=
  \frac{1}{N} \sum_{i=1}^N \delta \left( \sigma_i^{(j)},\sigma_i^{(j+1)}\right),\\
 \label{eq:def_f_beta_gamma}
 &&f(\beta,\Gamma) := N \log \left(1+\frac{Q}{{\rm e}^{\frac{Q\beta\Gamma}{m}}-1} \right),
\end{eqnarray}
where $\sigma_i^{(j)}$ denotes the $i$-th element on the $j$-th Trotter layer.
Then we can obtain
\begin{eqnarray}
 &&\langle \Sigma^{(j)} | {\rm e}^{-\frac{\beta}{m}{\cal H}_{\rm c}} | \Sigma^{(j)'} \rangle
  \propto p_{\rm SA}\left( \Sigma^{(j)};\frac{\beta}{m}\right)
  \delta\left( \Sigma^{(j)},\Sigma^{(j)'}\right),\\
 &&\langle \Sigma^{(j)'} | {\rm e}^{-\frac{\beta}{m}{\cal H}_{\rm q}} | \Sigma^{(j+1)} \rangle
  \propto {\rm e}^{s(\Sigma^{(j)'},\Sigma^{(j+1)}) f(\beta,\Gamma)},
\end{eqnarray}
since ${\cal H}_{\rm c}$ is a diagonal matrix and the equation
\begin{eqnarray}
 \left( \sigma_i^x \right)^l = (\hat{1}_Q - \mathbb{E}_Q)^l =
  \frac{1}{Q}\left[ (1-Q)^l - 1 \right] \hat{1}_Q + (-1)^l \mathbb{E}_Q
\end{eqnarray}
is satisfied.
Then we obtain
\begin{eqnarray}
 p_{\rm QA}(\Sigma;\beta,\Gamma) =
  \frac{1}{Z} \sum_{\Sigma^{(2)}} \cdots \sum_{\Sigma^{(m)}}
  \prod_{j=1}^m p_{\rm SA}(\Sigma^{(j)};\frac{\beta}{m})
  {\rm e}^{s(\Sigma^{(j)},\Sigma^{(j+1)}) f(\beta,\Gamma)}.
\end{eqnarray}
By using path-integral representation, the probability of the state $\Sigma$ of the $d$-dimensional quantum system can be represented by that of $(d+1)$-dimensional classical system approximately (see Fig.\ref{fig:concept_pathintegral}).

\begin{figure}[h]
 \begin{center}
  \includegraphics[scale=0.7]{./concept_pathintegral.eps}
  \caption{
  Conceptual diagram of path integral.
  In order to calculate the probability of the state $\Sigma$ of quantum system, we add an extra dimension which is called the Trotter axis.
  The probability $p_{\rm QA}(\Sigma;\beta,\Gamma)$ can be calculated by taking sum of configurations of path depicted by the bold line.
  }
  \label{fig:concept_pathintegral}
 \end{center}
\end{figure}

Here it should be noted that the function $f(\beta,\Gamma)$ is a monotonic decreasing function of inverse temperature $\beta$ and quantum field $\Gamma$ (see Fig.\ref{fig:graph_f}).

\begin{figure}[h]
 \begin{center}
  \includegraphics[scale=0.8]{./f_beta.eps}
  \caption{
  $f(\beta,\Gamma)$ as a function of $\Gamma$ for $N=5000$, $Q=30$, and $m=50$.
  }
  \label{fig:graph_f}
 \end{center}
\end{figure}

In the quantum Monte Carlo simulation by using path-integral representation, we update the state according to the probability as follows:
\begin{eqnarray}
 \label{eq:p_qast_update}
&& p_{\rm QA-ST}^{\rm update}(\sigma_i^{(j)}=s|\Sigma^{(j)} \backslash \sigma_i^{(j)},\Sigma^{(j-1)},\Sigma^{(j+1)};\beta,\Gamma) = \frac{P(s)}{\sum_{s'=1}^Q P(s')},\\
&& P(s) = {\rm e}^{\left\{
 -\frac{\beta}{m}E(\Sigma^{(j)}_s)
 + [ s(\Sigma^{(j-1)},\Sigma^{(j)}_s) + s(\Sigma^{(j)}_s,\Sigma^{(j+1)}) ]
 f(\beta,\Gamma)
\right\}},
\end{eqnarray}
where $\Sigma^{(j)}_s$ represents the state $\sigma_i^{(j)}=s$ given $\Sigma^{(j)}\backslash \sigma_i^{(j)}$.

\section{Some remarks}

In the section 4, we introduced the quantum fluctuation and obtained the probability of the quantum state by the path-integral representation.
Some remarks will be shown in this section.

\subsection{Labels of the clusters}

The value $s(\Sigma^{(j)},\Sigma^{(j+1)})$ given by Eq.~(\ref{eq:correlation_function}) expresses the correlation function between the state on the $j$-th Trotter layer and that on the $j+1$-th Trotter layer.
When the divisions are the same but the labels of the clusters are completely different as shown in Fig.~\ref{fig:samedivision}, $s(\Sigma^{(j)},\Sigma^{(j+1)})$ becomes zero.

\begin{figure}[h]
 \begin{center}
  $\begin{array}{ccc}
  \includegraphics[scale=0.8]{./purity_a.eps}&
   \hspace{5mm}&
  \includegraphics[scale=0.8]{./purity_b.eps} \\
    ({\rm a}) \quad \Sigma^A & \hspace{5mm} & ({\rm b}) \quad \Sigma^B
  \end{array}$
  \caption{Both (a) and (b) are the same divisions but the name of the clusters are completely different.}
  \label{fig:samedivision}
 \end{center}
\end{figure}

When we decrease the quantum field $\Gamma$ slow enough, not only division but also the labels of the clusters should become the same.
If the state on the $j$-th layer is $\Sigma^{(j)} = \Sigma_A$ and that on the $j+1$-th layer is $\Sigma^{(j+1)}= \Sigma_B$ such as Fig.~\ref{fig:samedivision} by accident, we cannot only gain a benefit but also are faced with a problem by introducing path-integral representation.
This is similar situation with the domain wall problem in the ferromagnetic Ising model.
In the ferromagnetic Ising model without external magnetic field, the ground state is that all spins are up or down.
Suppose we consider the case that the spins in the left half are up and the spins in the right half are down.
If we execute the standard Monte Carlo simulation, it is difficult to obtain the stable state ({\it i.e.} the ground state) from such an initial state for a short time.
For clustering problems, the domain wall problem gets more seriously comparing with the standard ferromagnetic Ising model since the number of the same divisions is $Q!$.
In order to avoid the domain wall problem, we introduce a new parameter ``modified correlation function'' and approximate the probability distribution.
The definition of the modified correlation function $\tilde{s}(\Sigma^{(j)},\Sigma^{(j+1)})$ is
\begin{eqnarray}
 &&\tilde{s}(\Sigma^{(j)},\Sigma^{(j+1)})
  := \frac{1}{N} \sum_{c=1}^k
  {\rm max}_{c'=1,\cdots,k} [Y(\Sigma^{(j)})Y^T(\Sigma^{(j+1)})]_{c,c'},\\
 &&Y(\Sigma^{(j)}) := (\sigma^{(j)}_1,\sigma^{(j)}_2,\cdots,\sigma^{(j)}_N),
\end{eqnarray}
where $Y(\Sigma^{(j)})$ denotes $N \times Q$ matrix.
For $Q=3$ and $N=2$, $Y(\Sigma_3)$, where $\Sigma_3$ was defined in Table \ref{table:def_sigma_q3n2}, is given as
\begin{eqnarray}
 Y(\Sigma_3) =
  \left(
   \begin{array}{cc}
   0 & 1\\
   0 & 0\\
   1 & 0\\
   \end{array}
  \right).
\end{eqnarray}
It should be noted that the correlation function $s(\Sigma^{(j)},\Sigma^{(j+1)})$ can be expressed in a similar way such as
\begin{eqnarray}
 s(\Sigma^{(j)},\Sigma^{(j+1)}) =
  \frac{1}{N} {\rm Tr} [Y(\Sigma^{(j)})Y^T(\Sigma^{(j+1)})].
\end{eqnarray}

Let us show some properties of the modified correlation function comparing $s(\Sigma^{(j)},\Sigma^{(j+1)})$ in the next section.
Suppose we consider the case for $Q=3$ and $N=7$ in the section 5.1.1 and 5.1.2.

\subsubsection{Example A}

We consider a case that the states on the $j$-th layer and that on the $j+1$-th layer are shown in Table~\ref{table:example1}.

\begin{table}[h]
 \tbl{Example A.}{
  \begin{tabular}{cccccccc}
   \toprule
    & $\sigma_1$ & $\sigma_2$ & $\sigma_3$ & $\sigma_4$ & $\sigma_5$ & $\sigma_6$ & $\sigma_7$ \\
   \hline
   $\Sigma^{(j)}$ & $1$ & $1$ & $1$ & $2$ & $2$ & $3$ & $3$ \\
   $\Sigma^{(j+1)}$ & $3$ & $3$ & $2$ & $3$ & $1$ & $1$ & $2$ \\
   \botrule
  \end{tabular}}
 \label{table:example1}
\end{table}

Here, $Y(\Sigma^{(j)})$ and $Y(\Sigma^{(j+1)})$ are given as
\begin{eqnarray}
 Y(\Sigma^{(j)}) =
  \left(
   \begin{array}{ccccccc}
    1 & 1 & 1 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & 1 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 1 & 1
   \end{array}
  \right),
  \,\,\,\,\,
  Y(\Sigma^{(j+1)}) =
  \left(
   \begin{array}{ccccccc}
    0 & 0 & 0 & 0 & 1 & 1 & 0 \\
    0 & 0 & 1 & 0 & 0 & 0 & 1 \\
    1 & 1 & 0 & 1 & 0 & 0 & 0
   \end{array}
  \right).
\end{eqnarray}
Then, the modified correlation function is calculated as
\begin{eqnarray}
 \tilde{s}(\Sigma^{(j)},\Sigma^{(j+1)}) = \frac{1}{7} \sum_{c=1}^3
  {\rm max}_{c'=1,\cdots,3}
  \left(
   \begin{array}{ccc}
    0 & 1 & 2 \\
    1 & 0 & 1 \\
    1 & 1 & 0
   \end{array}
  \right)_{c,c'} = \frac{4}{7}.
\end{eqnarray}
Note that $s(\Sigma^{(j)},\Sigma^{(j+1)}) = 0$ by the definition.
Next we fix the labels of $\Sigma^{(j)}$ and rename the labels of $\Sigma^{(j+1)}$.
The values of $s(\Sigma^{(j)},{\cal P}_\pi\Sigma^{(j+1)})$ are shown in Table \ref{table:example1_2}, where ${\cal P}_\pi$ ($\pi = 1, \cdots, Q!$) denotes label permutation operator.

\begin{table}[h]
 \tbl{The value of correlation function by applying label permutation operator ${\cal P}_\pi$ for example A.}{
  \begin{tabular}{ccccccccc}
   \toprule
    & $\sigma_1$ & $\sigma_2$ & $\sigma_3$ & $\sigma_4$ & $\sigma_5$ & $\sigma_6$ & $\sigma_7$ & $s(\Sigma^{(j)},{\cal P}_\pi\Sigma^{(j+1)})$ \\
   \hline
   ${\cal P}_1\Sigma^{(j+1)}$ & $3$ & $3$ & $2$ & $3$ & $1$ & $1$ & $2$ & $0$  \\
   ${\cal P}_2\Sigma^{(j+1)}$ & $3$ & $3$ & $1$ & $3$ & $2$ & $2$ & $1$ & $2/7$ \\
   ${\cal P}_3\Sigma^{(j+1)}$ & $1$ & $1$ & $3$ & $1$ & $2$ & $2$ & $3$ & $4/7$ \\
   ${\cal P}_4\Sigma^{(j+1)}$ & $1$ & $1$ & $2$ & $1$ & $3$ & $3$ & $2$ & $3/7$ \\
   ${\cal P}_5\Sigma^{(j+1)}$ & $2$ & $2$ & $3$ & $2$ & $1$ & $1$ & $3$ & $2/7$ \\
   ${\cal P}_6\Sigma^{(j+1)}$ & $2$ & $2$ & $1$ & $2$ & $3$ & $3$ & $1$ & $3/7$ \\
   \botrule
  \end{tabular}}
 \label{table:example1_2}
\end{table}

In this case, the maximum value of $s(\Sigma^{(j)},{\cal P}_\pi\Sigma^{(j+1)})$ is the same as the value of modified correlation function $\tilde{s}(\Sigma^{(j)},\Sigma^{(j+1)})$.

\subsubsection{Example B}

We also consider another case that the states on the $j$-th layer and that on the $j+1$-th layer are shown in Table~\ref{table:example2} as an another example.

\begin{table}[h]
 \tbl{Example B.}{
  \begin{tabular}{cccccccc}
   \toprule
    & $\sigma_1$ & $\sigma_2$ & $\sigma_3$ & $\sigma_4$ & $\sigma_5$ & $\sigma_6$ & $\sigma_7$ \\
   \hline
   $\Sigma^{(j)}$ & $1$ & $2$ & $2$ & $2$ & $2$ & $3$ & $3$ \\
   $\Sigma^{(j+1)}$ & $2$ & $1$ & $1$ & $1$ & $3$ & $1$ & $1$ \\
   \botrule
  \end{tabular}}
 \label{table:example2}
\end{table}

Here, $Y(\Sigma^{(j)})$ and $Y(\Sigma^{(j+1)})$ are given as
\begin{eqnarray}
 Y(\Sigma^{(j)}) =
  \left(
   \begin{array}{ccccccc}
    1 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 1 & 1 & 1 & 1 & 0 & 0 \\
    0 & 0 & 0 & 0 & 0 & 1 & 1
   \end{array}
  \right),
  \,\,\,\,\,
  Y(\Sigma^{(j+1)}) =
  \left(
   \begin{array}{ccccccc}
    0 & 1 & 1 & 1 & 0 & 1 & 1 \\
    1 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 & 1 & 0 & 0
   \end{array}
  \right).
\end{eqnarray}
Then, the modified correlation function is calculated as
\begin{eqnarray}
 \tilde{s}(\Sigma^{(j)},\Sigma^{(j+1)}) = \frac{1}{7} \sum_{c=1}^3
  {\rm max}_{c'=1,\cdots,3}
  \left(
   \begin{array}{ccc}
    0 & 1 & 0 \\
    3 & 0 & 1 \\
    2 & 1 & 1
   \end{array}
  \right)_{c,c'} = \frac{6}{7}.
\end{eqnarray}
Note that $s(\Sigma^{(j)},\Sigma^{(j+1)}) = 0$ by the definition.
As in the section 5.1.1, when we fix the labels of $\Sigma^{(j)}$ and rename the labels of $\Sigma^{(j+1)}$, the values of $s(\Sigma^{(j)},{\cal P}_\pi\Sigma^{(j+1)})$ are shown in Table \ref{table:example2_2}.

\begin{table}[h]
 \tbl{The value of correlation function by applying label permutation operator ${\cal P}_\pi$ for example B.}{
  \begin{tabular}{ccccccccc}
   \toprule
   & $\sigma_1$ & $\sigma_2$ & $\sigma_3$ & $\sigma_4$ & $\sigma_5$ & $\sigma_6$ & $\sigma_7$ & $s(\Sigma^{(j)},{\cal P}_\pi\Sigma^{(j+1)})$ \\
   \hline
   ${\cal P}_1\Sigma^{(j+1)}$ & $2$ & $1$ & $1$ & $1$ & $3$ & $1$ & $1$ & $0$ \\
   ${\cal P}_2\Sigma^{(j+1)}$ & $3$ & $1$ & $1$ & $1$ & $2$ & $1$ & $1$ & $1/7$ \\
   ${\cal P}_3\Sigma^{(j+1)}$ & $1$ & $2$ & $2$ & $2$ & $3$ & $2$ & $2$ & $4/7$ \\
   ${\cal P}_4\Sigma^{(j+1)}$ & $3$ & $2$ & $2$ & $2$ & $1$ & $2$ & $2$ & $3/7$ \\
   ${\cal P}_5\Sigma^{(j+1)}$ & $1$ & $3$ & $3$ & $3$ & $2$ & $3$ & $3$ & $4/7$ \\
   ${\cal P}_6\Sigma^{(j+1)}$ & $2$ & $3$ & $3$ & $3$ & $1$ & $3$ & $3$ & $2/7$ \\
   \botrule
  \end{tabular}}
  \label{table:example2_2}
\end{table}

In this case, the maximum value of $s(\Sigma^{(j)},{\cal P}_\pi\Sigma^{(j+1)})$ is not the same as the modified correlation function $\tilde{s}(\Sigma^{(j)},\Sigma^{(j+1)})$.
Although the modified correlation function expresses some kinds of similarity between the states $\Sigma^{(j)}$ and $\Sigma^{(j+1)}$, it is not necessary commutative {\it i.e.} $\tilde{s}(\Sigma^{(j)},\Sigma^{(j+1)}) \neq \tilde{s}(\Sigma^{(j+1)},\Sigma^{(j)})$.
In the case of example A,
\begin{eqnarray}
 \tilde{s}(\Sigma^{(j)},\Sigma^{(j+1)}) = \tilde{s}(\Sigma^{(j+1)},\Sigma^{(j)}) = \frac{4}{7}.
\end{eqnarray}
On the other hand, in the case of example B,
\begin{eqnarray}
 \tilde{s}(\Sigma^{(j)},\Sigma^{(j+1)}) = \frac{6}{7},
  \,\,\,\,\,
  \tilde{s}(\Sigma^{(j+1)},\Sigma^{(j)}) = \frac{5}{7}.
\end{eqnarray}
Here we summarize the properties of the modified correlation function:
\begin{itemize}
 \item The modified correlation function does not necessary commute.
       \begin{eqnarray}
	\tilde{s}(\Sigma^{(j)},\Sigma^{(j+1)}) \neq \tilde{s}(\Sigma^{(j+1)},\Sigma^{(j)}).
       \end{eqnarray}
 \item For any ${\cal P}_\pi$ ($\pi=1,\cdots,Q!$), the inequality
       \begin{eqnarray}
	\nonumber
	0 \le s(\Sigma^{(j)},{\cal P}_\pi\Sigma^{(j+1)})
	 = {\rm Tr} \left[ Y(\Sigma^{(j)})Y({\cal P}_\pi\Sigma^{(j+1)})^T \right]
	 \le \tilde{s}(\Sigma^{(j)},\Sigma^{(j+1)}) \le 1
       \end{eqnarray}
       is satisfied.
\end{itemize}
Although the modified correlation function is not necessary the same as the maximum value of $s(\Sigma^{(j)},{\cal P}_\pi\Sigma^{(j+1)})$, we use this value in our proposed algorithm.
This is because our aim is to find not so bad solution of given problem as mentioned before.

\subsection{Speed-up by Using Modified Correlation Function}

When we adopt the quantum Monte Carlo simulation by the path-integral representation, it is enough to update the state according to Eq.~(\ref{eq:p_qast_update}) in principle.
In the previous section, we introduced the new parameter ``modified correlation function'' to avoid the so-called domain wall problem.
Instead of Eq.~(\ref{eq:p_qast_update}), we adopt the following the transition probability based on modified correlation function:
\begin{eqnarray}
\nonumber
&&p_{\rm QA-ST+modify}^{\rm update}(\sigma_i^{(j)}=s|\Sigma^{(j)}\backslash \sigma_i^{(j)},\Sigma^{(j-1)},\Sigma^{(j+1)};\beta,\Gamma)\\
\label{eq:prob_qastpurity}
&&=\frac
  {
  \exp\left[
       -\frac{\beta}{m}E(\Sigma^{(j)}_s)+\tilde{S}(\Sigma^{(j-1)},\Sigma^{(j)}_s,\Sigma^{(j+1)})f(\beta,\Gamma)
      \right]
}{
\sum_{t=1}^Q \exp\left[
       -\frac{\beta}{m}E(\Sigma^{(j)}_t)+\tilde{S}(\Sigma^{(j-1)},\Sigma^{(j)}_t,\Sigma^{(j+1)})f(\beta,\Gamma)
      \right]
},\\
 &&\tilde{S}(\Sigma^{(j-1)},\Sigma^{(j)}_s,\Sigma^{(j+1)})
  := \tilde{s}(\Sigma^{(j-1)},\Sigma^{(j)}_s) + \tilde{s}(\Sigma^{(j)}_s,\Sigma^{(j+1)}).
\end{eqnarray}
Since $\tilde{s}(\Sigma^{(j-1)},\Sigma^{(j)})$ is not necessary commutative, there are four possibilities of the definition of $\tilde{S}(\Sigma^{(j-1)},\Sigma^{(j)},\Sigma^{(j+1)})$ as follows:
\begin{eqnarray}
 \label{eq:def_purity_oneoffour}
 &&\tilde{S}(\Sigma^{(j-1)},\Sigma^{(j)}_s,\Sigma^{(j+1)})
  = \tilde{s}(\Sigma^{(j-1)},\Sigma^{(j)}_s) + \tilde{s}(\Sigma^{(j)}_s,\Sigma^{(j+1)}),\\
 &&\tilde{S}(\Sigma^{(j-1)},\Sigma^{(j)}_s,\Sigma^{(j+1)})
  = \tilde{s}(\Sigma^{(j-1)},\Sigma^{(j)}_s) + \tilde{s}(\Sigma^{(j+1)},\Sigma^{(j)}_s),\\
 &&\tilde{S}(\Sigma^{(j-1)},\Sigma^{(j)}_s,\Sigma^{(j+1)})
  = \tilde{s}(\Sigma^{(j)}_s,\Sigma^{(j-1)}) + \tilde{s}(\Sigma^{(j)}_s,\Sigma^{(j+1)}),\\
 &&\tilde{S}(\Sigma^{(j-1)},\Sigma^{(j)}_s,\Sigma^{(j+1)})
  = \tilde{s}(\Sigma^{(j)}_s,\Sigma^{(j-1)}) + \tilde{s}(\Sigma^{(j+1)},\Sigma^{(j)}_s).
\end{eqnarray}
For small systems we confirm that when we adopt Eq.~(\ref{eq:def_purity_oneoffour}) as the definition of $\tilde{S}(\Sigma^{(j-1)},\Sigma^{(j)}_s,\Sigma^{(j+1)})$, we can obtain better solution than the other choices.
Then, we adopt the relation given by Eq.~(\ref{eq:def_purity_oneoffour}) as the definition of $\tilde{S}(\Sigma^{(j-1)},\Sigma^{(j)}_s,\Sigma^{(j+1)})$.

\subsection{Simultaneous Control of Thermal Fluctuation and Quantum Fluctuation}

In our algorithm given by Eq.~(\ref{eq:prob_qastpurity}), thermal fluctuation $\beta/m$ and quantum fluctuation $f(\beta,\Gamma)$ coexist.
Then, it is expected that we can obtain the better solution than simulated annealing by controlling the thermal fluctuation and quantum fluctuation simultaneously with ingenuity.

In order to investigate how to control both of the thermal fluctuation and quantum fluctuation, we first consider limiting cases.
When the temperature is very low than the quantum field, {\it i.e.} $\beta/m \gg f(\beta,\Gamma)$, the probability distributions in each layer obey Eq.~(\ref{eq:p_sa}).
On the other hand, if the quantum field is very weak, {\it i.e.} $\beta/m \ll f(\beta,\Gamma)$, the state in each layers becomes the same state such as $\Sigma = \Sigma^{(j)}$ for all $j$.
By calculating small systems, we find how to control the thermal fluctuation and quantum fluctuation simultaneously.
The best condition is as follows:
(i) At the first step, to find the metastable state on each layers, we set the temperature $\beta/m \gg f(\beta,\Gamma)$.
(ii)After that, $f(\beta,\Gamma)$ overstrides thermal fluctuation.
The curve $f^*$ in Fig.~\ref{fig:proposed_algorithm} shows a conceptual diagram for our proposed schedule.

\begin{figure}[h]
 \begin{center}
  \includegraphics[scale=0.5]{./schedule.eps}
  \caption{
  The dotted curve denotes the schedule of inverse temperature.
  $f_1$ is too rapid for decreasing the quantum field.
  On the other hand, $f_2$ is too slow for decreasing the quantum field.
  $f^*$ denotes the best schedule.
  }
  \label{fig:proposed_algorithm}
 \end{center}
\end{figure}

The dotted curve in Fig.~\ref{fig:proposed_algorithm} indicates the schedule of cooling temperature which corresponds to the simulated annealing.
The curve depicted $f^*$ in Fig.~\ref{fig:proposed_algorithm} is the best schedule for obtaining better solution than the simulated annealing.
In the schedule depicted $f_1$ in Fig.~\ref{fig:proposed_algorithm}, quantum mixing effect does not make sense.
In the schedule depicted $f_2$ in Fig.~\ref{fig:proposed_algorithm}, on the other hand, the states on each layers behave independently.
The quantum fluctuation effect is strong than the thermal fluctuation effect in the schedule depicted $f_2$.
It is essentially the same as the simulated annealing.

Here we assume the scheduling functions of temperature and quantum field as follows:
\begin{eqnarray}
 &&\beta(t) = \beta_0 r_\beta^t,\\
 &&\Gamma(t) = \infty,  \,\,\,\,\, (t < \tau),  \,\,\,\,\,\,\,\,
  \Gamma(t) = \Gamma_0 \exp(-r_\Gamma^{t-\tau}) \,\,\,\,\, (t \ge \tau),
\end{eqnarray}
where $\tau$ corresponds to the time $\beta(\tau) = m$.
When $Q\beta\Gamma/m \ll 1$, the interaction along the Trotter axis $f(\beta,\Gamma)$ given by Eq.~(\ref{eq:def_f_beta_gamma}) is approximately given as
\begin{eqnarray}
 f(\beta,\Gamma) \sim -N \log(\frac{\beta\Gamma}{m})
  = N r_\Gamma^t - N \log (\frac{\beta\Gamma_0}{m}).
\end{eqnarray}
From this equation, it is enough to set large enough $\Gamma_0$ and $r_\beta < r_\Gamma$ in order to prepare the schedule $f^*$.
In other words, independent simulated annealing is performed until $\tau$ and after that we decrease quantum field to obtain the better solution than the conventional simulated annealing.

\section{Results}\label{sec:results}

We perform numerical experiment for the following three problems by proposed quantum annealing and simulated annealing for comparison.
We prepare the number of replicas $m = 50$ in the whole experiments.
The initial inverse temperature and the initial quantum field are set to be $\beta_0 = 0.2m$ and $\Gamma_0 = {\rm e}^{1/2}$, respectively.
Here the ratio of changing temperature is set to be $r_\beta = 1.05$.
In the simulated annealing, we prepare independent $55$ samples and use the same initial temperature and the ratio of changing temperature as the case of quantum annealing.
We study three problems as follows:
\begin{description}
 \item[(a)] Evaluation of mixture of Gaussian by using MNIST data\cite{MNIST} \\
	    ($Q=30$,$N=5000$)
 \item[(b)] Evaluation of latent Dirichlet allocation\cite{Blei-2003} by using Reuters data\cite{REUTERS} \\
	    ($Q=20$,$N=2000$)
 \item[(c)] Evaluation of latent Dirichlet allocation by using NIPS corpus\cite{NIPS} \\
	    ($Q=20$,$N=1000$)
\end{description}
In all of the experiments, we fix the ratio of changing the inverse temperature $r_\beta = 1.05$.

\begin{figure}[h]
 \begin{center}
  $\begin{array}{ccccc}
   \includegraphics[scale=0.4]{./mnist.eps}&
    \hspace{3mm}&
   \includegraphics[scale=0.4]{./reuters.eps}&
   \hspace{3mm}&
   \includegraphics[scale=0.4]{./nips.eps}\\
    ({\rm a}) & \hspace{3mm} & ({\rm b}) & \hspace{3mm} & ({\rm c})
   \end{array}$
  \caption{
  Time development of the minimum energy in all of the layers.
  The circles represent results obtained by simulated annealing for comparison.
  The triangles, asterisks, and squares denote obtained results by proposed quantum annealing for $r_\Gamma = 1.02$, $r_\Gamma = 1.10$, and $r_\Gamma = 1.20$, respectively.
  (a) Evaluation of mixture of Gaussian by using MNIST data ($Q=30$,$N=5000$).
  (b) Evaluation of latent Dirichlet allocation by using Reuters data ($Q=20$,$N=2000$).
  (c) Evaluation of latent Dirichlet allocation by using NIPS corpus ($Q=20$,$N=1000$).
  }
  \label{fig:results}
 \end{center}
\end{figure}

Figure \ref{fig:results} denotes the time development of the minimum energy in all of the layers for each problems.
The circles in Fig.~\ref{fig:results} represent results obtained by simulated annealing for comparison.
The triangles, asterisks, and squares denote obtained results by proposed quantum annealing for $r_\Gamma = 1.02$, $r_\Gamma = 1.10$, and $r_\Gamma = 1.20$, respectively.
The schedule for $r_\Gamma = 1.02$ corresponds to schedule depicted $f_2$ in Fig.~\ref{fig:proposed_algorithm} whereas the schedules for $r_\Gamma = 1.10$ and $r_\Gamma = 1.20$ correspond to schedule depicted $f^*$ in Fig.~\ref{fig:proposed_algorithm}.
From Fig.~\ref{fig:results} it is clear that we can obtain the better solution than simulated annealing if we adopt the schedule $f^*$.

\section{Conclusion}\label{sec:conclusion}

In this paper, we developed hybrid type of the quantum annealing for clustering problems.
When we apply the standard quantum annealing for these problems, we face on the difficulty of the domain wall problem.
Since the number of the same division is $Q!$, the domain wall problem gets more serious for large $Q$.
To avoid such a problem, we introduced new parameter ``modified correlation function'' instead of the standard correlation function.
We investigated the best schedule of changing the thermal fluctuation and the quantum fluctuation simultaneously.
We first apply strong quantum field and then we obtain the metastable states.
At the second step, we can obtain the better solution than the simulated annealing by decreasing the quantum field.
Actually, we succeeded to obtain the better solution for clustering problems than the simulated annealing.
We expect that the proposed schedule of changing the thermal fluctuation and the quantum fluctuation is generally efficient for other type of problems.
However, it is an open problem when to use the quantum annealing.
To solve this problem, we should study the efficiency of quantum annealing for the problems where the {\it difficulty} of the problem can be controlled.

The authors are grateful to Naoki Kawashima, Seiji Miyashita, Hiroshi Nakagawa, Daiji Suzuki, and member of T-PRIMAL for their valuable comments.
S.T. is partly supported by Grant-in-Aid for Young Scientists Start-up (21840021) from the JSPS, MEXT Grant-in-Aid for Scientific Research (B) (22340111), and the ``Open Research Center'' Project for Private Universities: matching fund subsidy from MEXT.
R.T. is partly supported by Global COE Program ``the Physical Sciences Frontier'', MEXT, Japan.
The computation in the present work was performed on computers at the Supercomputer Center,
Institute for Solid State Physics and University of Tokyo and at Taisuke Sato's group, Tokyo Institute of Technology.

redpajama_set_name: RedPajamaArXiv


Europa League – Follow the Football match between FC Banants and Omonia Nicosia live with Eurosport. The match starts at 17:00 on 30 June 2016. Our live coverage lets you follow all the key moments as they happen.
Who will come out on top in the battle of the managers Aram Voskanyan or Pambos Christodoulou? Find out by following our live matchcast.
Have your say by voting on who will win between FC Banants and Omonia Nicosia? Enjoy some pre-match reading with related articles about these two Enjoy some pre-match reading with related articles about these two Football teams.
Head-to-head: see historical stats and visit our detailed profiles for FC Banants vs Omonia Nicosia. Get all the latest on Football: fixtures, results and tables.


text:
Stine Østergaard Jørgensen (Dronninglund, 3 september 1990) is een Deens handbalspeelster, die uitkomt voor het Deens nationale handbalteam.

Jørgensen begon op vijfjarige leeftijd met handbal bij de lokale club Dronninglund IF. Op zestienjarige leeftijd ging ze naar Aalborg DH als verdedigster.

Met het Deense nationale team won ze in 2008 een bronzen medaille bij het Wereldkampioenschap U18. In 2011 werd ze vierde op het Wereldkampioenschap, waar ze negen keer scoorde voor het Deense nationale team. Twee jaar later werd ze derde op het Wereldkampioenschap.

Privé
Stine Jørgensen heeft een relatie met badmintonspeler Jan Ø. Jørgensen.

Deens handballer

redpajama_set_name: RedPajamaWikipedia


If you take an ultrasound scan of a person with diabetes, we see the presence of hardened viral structures. Like a very slow growing cancer cells.
If we apply high intensity ultrasound, these cells pop. Clearing the disease. The cheapest source of HIUS, is an 8 W one MHZ ultrasound massage device. Medically licensed for safe home use.
Applying to the bottom right of the chest for ½ a minute, it clears type two diabetes. Like all my stepfather and mother. And every diabetic who is in my close circle of friends.
Annoyingly type one diabetes, only minutes after three days. 2016 I can found 1 minute of HIUS clears all diabetes totally in ten seconds.
Medics all bought and validated a HIUS device cleared all cancers in 2002. As 23 medical professors from the Moffitt cancer centre published a Medical Paper on this subject.
Registered doctors had to confirm the HIUS work, or they ceased to be registered doctors 16 years ago. All subsequent medical practice criminal.
High intensity ultrasound 150 W 40 kHz or 8W 1MHz applied externally, from a massage device, just once will cure all the diseases of age!
They are legally required to use HIUS to clear all cancers. At one appointment. No cancer drug prescription is legal. Neither is and radiotherapy or the massively defective cancer surgery.
When a patient presents with diabetes, they are also legally required to use HIUS to clear the diabetes. I have a 33 person, 100% success rate in clearing both type one and type two diabetes around the world.
A 20 per cent success rate, is medically significant. So doctors all have a required HIUS device. Its use does not affect any drugs. It is just totally effective at clearing cancers and diabetes.
At high school (Bolton School, Greater Manchester England) I sat next to the table for diabetic students. I have lost three non-blood relatives to type two diabetes.
It transpires that ½ minute of HIUS totally clears the disease. Medically proved all around the world. Prescription of metformin or insulin is defective and criminal medicine.
Since I've published this work on the Internet 2013. No registered Dr. Has ethically been allowed to prescribe the diabetic medicine. And yet they have!
Which has resulted in diabetes the eighth largest killer of people on earth.
So from 2013 around 1000 people have died from type two diabetes in the UK. Each death warrants a fine of 10,000,000 UK pounds, it involved Dr. Gets 25 years in high-security prison without parole.
We are talking about the total fine of one trillion UK pounds. The involved doctors will die in jail.
At least at my high school pupils on the diabetic table, can now have their diabetes cures. In under 1 minute.


text:
Home News IAAF Diamond League: Okagbare wins in Rabat
IAAF Diamond League: Okagbare wins in Rabat
Africa's Queen of the Tracks, Blessing Okagbare – Ighoteguonor, has secured her first victory in the 2019 IAAF Diamond League series.
Okagbare coasted to victory in the women's 100m event on Sunday at the Prince Moulay Abdellah Stadium in Rabat, Morocco.
She won Sunday's 100m race with a Season Best Time of 11.05s ahead of her Ivorian rival and current African champion, Marie Josee Ta Lou, who posted a time of 11.09s while Crystal Emmanuel of Canada was third in 11.30s.
Sunday's race was Okagbare's 60th appearance in the prestigious Diamond League and she marked it with her first-ever win in Rabat.
In her previous outings this year in the Diamond League, Okagbare in Doha finished in the third position in the 200m while it was slightly better at the Shanghai Diamond League where she finished second in the 100m.
Before finally breaking the jinx on Sunday, Okagbare had failed to win any of her previous Diamond League races in Rabat.
In her debut Diamond League race in Rabat in 2016, Okagbare ran 11.11 seconds to place second behind Jamaica's Elaine Thompson
Okagbare was fourth in 2017 (11.03 seconds)was fourth in 2017 (11.03 seconds) behind Thompson again who ran a blistering 10.87 seconds while it was worse in 2018 for the University of Texas graduate as she ran a time of 23.42 seconds to place 7th in the 200m.
Having finally conquered in Rabat where the athletics events of the 2019 All-Africa Games will be held, it is yet to be seen if Okagbare will be available to compete for Nigeria in the 'African Olympics' later this year.
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A human child is raised by a wolf pack in India.
Recently released ex-con & brother are forced on the run.
A woman is sent to a mental hospital in Ireland.
Boy must form a band after wooing a girl.
Film adaptation of Shakespeare's play about General Macbeth.
Teen princesses celebrate V-E Day with a London night out.
The Autobots and Decepticons continue their war.
Ordinary man learns that sperm donation led to 533 kids.


text:
package org.apache.camel.model.language;

import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlRootElement;

import org.apache.camel.spi.Metadata;

/**
 * For PHP expressions and predicates
 *
 * @version
 */
@Metadata(label = "language", title = "PHP")
@XmlRootElement(name = "php")
@XmlAccessorType(XmlAccessType.FIELD)
public class PhpExpression extends ExpressionDefinition {

    public PhpExpression() {
    }

    public PhpExpression(String expression) {
        super(expression);
    }

    public String getLanguage() {
        return "php";
    }
}

redpajama_set_name: RedPajamaGithub


text:
Obituaries » Gordon "Connie" Constantine Martin
Gordon "Connie" Constantine Martin
November 4, 1933 - January 20, 2021
Due to COVID restrictions and his own personal wishes, there will be no service. The family will have a private gathering at a later date.
Subscribe to updates for Gordon "Connie" Martin
Gordon "Connie" Constantine Martin, 87, was born on November 4, 1933, in Martinsville, Va., later moving to Lynchburg with his parents, Thomas Elton and Nellie Mills Martin, who predecease him. He graduated from Madison Heights High School.
Connie began working at the age of 13 delivering for a drug store on his bicycle. During high school he worked for Leggett Department Store and loved decorating for Christmas, a favorite holiday, especially the Santa and reindeer in front of Leggett. He developed a fond friendship with Harold Leggett. Connie's love of math and building led him to Montague-Betts Steel Fabricating Company where he learned the steel business and rose to the position of vice-president of sales. He made many life-long friends there. He was especially proud of his work on The World Trade Center. At the closing of Montague-Betts, he tried retirement but was not ready. He went to work for Wexco as an Account Executive and finally retired at the age of 79.
Connie loved his relationship with the E. C. Glass Boosters organization and proudly served as the president of the board. He developed many friendships with the coaches. He was a member of Boonsboro Country Club for over 60 years where he played many rounds of golf with his friends. He was an avid Hokie fan and treasured his round of golf with Frank Beamer. Connie was a long-time member of the Rivermont Avenue Baptist Church, now Gospel Community Church, where he served as a Deacon.
Connie is survived by his loving wife, Lauralyn, of 52 years and his four adored children – RoxAnne Frey (Mike), Steve (Lisa), Peter (Julie), and Lantz (Kristi). He was so proud of his grandchildren, Dawn Hoyle (Mickey), Jacob Martin (Cianne), Joy Landa (Sam), Kayla Lutes (Jason), Cameron, Carter, and Christian Martin, Alexis Gossom, Lantz and Lyncoln Martin, and five great-grandchildren.
He was predeceased by his grandson, Joshua Steven Martin. He was also predeceased by his brother, Thomas Martin, and his sister, Judith Bordeaux.
He loved his dogs, Katie, Trudy, and Luna, and often said "if dogs don't go to Heaven, I'm not going." We know Connie is in Heaven.
If you choose, please remember Connie with a memorable contribution to The Lynchburg Humane Society, 1211 Old Graves Mill Road, Lynchburg, Va, 24502: or Gospel Community Church, 1301 Rivermont Avenue, Lynchburg, Va 24504. Most of all, Connie wants his friends to know how much he loved them.
Tharp Funeral Home & Crematory, Lynchburg, is assisting the family. To send condolences, please visit tharpfuneralhome.com.


Sometimes the folks over at Screen Junkies have to do a bit of reaching to really come up with funny things to comment on in their Honest Trailers. However, with "Aquaman," you can't help but think that the writers had plenty of extra jokes that they couldn't fit into the video's relatively short five-minute run-time.
Let's be honest, despite making over $1 billion at the box office and making fans' heads explode with superhero bliss, "Aquaman" is not a great film.
Continue reading 'Aquaman' Honest Trailer: From Fish-Talking Zero To Hunky Hero at The Playlist.


I've tracked the issue to the use of java.net.HttpURLConnection without setting the contentLength (in some cases) and not using 'fixed length streaming mode' when contentLength is available.
I'm attaching a patch that works for me with the DeployTask. I know that org.apache.catalina.ant.AbstractCatalinaTask is subclassed by many tasks, but I haven't tested all of those.
I can probably provide some code to reproduce if needed.
"When output streaming is enabled, authentication and redirection cannot be handled automatically."
This has been fixed in 6.0.x and will be included in 6.0.25 onwards. Thanks again for the patch.


text:
A gun crew from the Vicksburg National Military Park fires a cannon during a demonstration in 2018 for July 4. The park will hold artillery training for volunteers Saturday and Sunday. (File photo)
Military Park puts out call for volunteers
By John Surratt
Published 7:01 pm Friday, May 3, 2019
Area residents have the opportunity Saturday and Sunday to help tell the story of the Siege of Vicksburg.
The Vicksburg National Military Park is looking for volunteers interested in becoming involved in the park's Living History Program. The park will be conducting artillery training for new artillery volunteers Saturday and Sunday from 8:30 a.m. to 3 p.m. to prepare them for the park's summer cannon program.
In case of rain, the training will be held at the park's maintenance area.
"We're recruiting more folks to help out with the summer program," Ranger Lela White said, adding the firing demonstrations are held each weekend during the summer and July 4. "We start the weekend before Memorial Day," White said. "We don't fire on Memorial Day."
White is the park's black powder expert. To qualify, she spent two weeks of historic weapons training at an Army National Guard base in Alabama.
She said the park's training program includes teaching safety procedures, providing information on the park, and drilling on the cannon. During the safety training, she said, the volunteers "are taught about the explosive nature of black powder and the hazards they need to be aware with the cannons.
"They are also taught misfire procedures, the procedures to deal with any misfires, and procedures to fire the cannon."
She said a cannon crew usually includes seven people, but can include up to eight or nine.
People ages 16 and older interested in participating in the training can call White at 601-636-0583, or by email at lela_white@nps.gov.
"Sixteen can only do a couple of positions, but 18 can do everything," she said.
About John Surratt
John Surratt is a graduate of Louisiana State University with a degree in general studies. He has worked as an editor, reporter and photographer for newspapers in Louisiana, Mississippi and Alabama. He has been a member of The Vicksburg Post staff since 2011 and covers city government. He and his wife attend St. Paul Catholic Church and he is a member of the Port City Kiwanis Club.
More by John
Truck wreck snarls I-20 traffic for hours
A Jackson man received minor injuries when the 18-wheeler he was driving left Interstate 20 near the Clay Street exit... read more


text:
Q: Queries with complex filtering LINQ to XML c# I want to understand the complex filtering with LINQ XML.
I have created a simple XML example (DataBaseCities.xml):
<?xml version="1.0" encoding="utf-8"?>
<DataBase>
    <DocumentInfo version="1.0" schemaVersion="1.0"/>
    <ListOfMegaCities>
        <MegaCities city="Moscow" continent="Europe">
            <VariantConstraint>
                <LanguageRef LanguageId="russian">
                    <LanguageDialectsRef DialectsId="north"/>
                </LanguageRef>
            </VariantConstraint>
            <Districts>
                <CityDistrict District="Arbat"/>
                <CityDistrict District="Basmanny"/>
            </Districts>
        </MegaCities>
        <MegaCities city="New York" continent="North America">
            <VariantConstraint>
                <LanguageRef LanguageId="english">
                    <LanguageDialectsRef DialectsId="west"/>
                </LanguageRef>
                <LanguageRef LanguageId="spanish">
                    <LanguageDialectsRef DialectsId="cental"/>
                </LanguageRef>
            </VariantConstraint>
            <Districts>
                <CityDistrict District="Queens"/>
                <CityDistrict District="Bronx"/>
            </Districts>
        </MegaCities>
        <MegaCities city="London" continent="Europe">
            <VariantConstraint>
                <LanguageRef LanguageId="english">
                    <LanguageDialectsRef DialectsId="west"/>
                </LanguageRef>
                <LanguageRef LanguageId="spanish">
                    <LanguageDialectsRef DialectsId="central"/>
                </LanguageRef>
                <LanguageRef LanguageId="french">
                    <LanguageDialectsRef DialectsId="central"/>
                </LanguageRef>
            </VariantConstraint>
            <Districts>
                <CityDistrict District="Greenwich"/>
                <CityDistrict District="Westminster"/>
            </Districts>
        </MegaCities>
    </ListOfMegaCities>
</DataBase>

And I try to filter like:
        XElement root = XElement.Load(@"DataBaseCities.xml");
        IEnumerable<XElement> ListOfMegaCities =
            from el in root.Descendants("MegaCities")
            where
                (from add in el.Descendants("LanguageRef")
                 where
                      (string)add.Attribute("LanguageId") == "english"
                 select add)
            .Any()
            select el;

        foreach (XElement el in ListOfMegaCities)
        {
            Console.WriteLine((string)el.Attribute("city"));
        }

So the Output is:
New York
London

But I want to filter more then one Attribute.

*

*If I try to filter with these lines:
(string)add.Attribute("LanguageId") == "english" &&
(string)add.Attribute("LanguageId") == "spanish"
Why it dont work?

*How I can filter "DialectsId" too?

Example: I want to get "New York" with exactly this filtering:
LanguageId="english"
DialectsId="west"
LanguageId="spanish"
DialectsId="cental"

Maybe someone has a good source for complex filtering? I found this one https://learn.microsoft.com/de-de/dotnet/standard/linq/write-queries-complex-filtering but that helped me only partially...

A: I have used All instead of Any so that it filters only those cities which satisfy all the conditions. I am used to writing LINQ Expressions instead of LINQ Query.
List<XElement> ListOfMegaCities = root.Descendants("MegaCities")
                                 .Where(lr => lr.Descendants("LanguageRef")
                                 .All(
                                      li => (li.Attribute("LanguageId").Value == "english"
                                          && li.Element("LanguageDialectsRef").Attribute("DialectsId").Value == "west")
                                          || (li.Attribute("LanguageId").Value == "spanish"
                                          && li.Element("LanguageDialectsRef").Attribute("DialectsId").Value == "cental")
                                            )
                                 ).ToList();

A: First, see @zaggler comment. Deserialize and manipulate object will be more easily.
Second, if you want the cities where only english and spanish is spoken, see @AkshayGaonkar answer. If you want cities where english and spanish (and maybe other languages) is spoken, see my answer.

To write complex Linq query, I begin to write simulacrum SQL :
-- I want all cities
select * from MegaCities as el
where
    -- that speak english with west dialect
    exists (
        select * from el.LanguageRef as add
        where add.LanguageId = 'english' and
            add.LanguageDialectsRef.DialectsId = 'west'
    )
    -- and also speak spanish with central dialect
    and exists (
        select * from el.LanguageRef as add
        where add.LanguageId == 'spanish' and
            add.LanguageDialectsRef.DialectsId = 'cental'
    )

Then I can transpose to Linq :
IEnumerable<XElement> ListOfMegaCities =
    from el in root.Descendants("MegaCities")
    where
        (from add in el.Descendants("LanguageRef")
         where add.Attribute("LanguageId").Value == "english" &&
            add.Element("LanguageDialectsRef").Attribute("DialectsId").Value == "west"
         select add)
        .Any()
        &&
        (from add in el.Descendants("LanguageRef")
         where add.Attribute("LanguageId").Value == "spanish" &&
            add.Element("LanguageDialectsRef").Attribute("DialectsId").Value == "cental"
         select add)
        .Any()
    select el;

Maybe not the best trick, but can help sometimes.


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BMW is sticking with plans to open a new factory in Mexico in 2019 despite warnings from President-elect Donald Trump that he will slap a border tax on cars imported from the country.
The German automaker's head of sales and marketing, Ian Robertson, told Richard Quest of CNN that BMW is committed to the factory in Mexico.
Robertson explained that the company needs to expand manufacturing beyond its U.S. factory in Spartanburg, South Carolina, already the biggest BMW plant in the world.
"We need additional capacity, so we have chosen to build a plant in Mexico," he said. The Spartanburg plant produces up to 450,000 cars per year, he said.
"In a global setup, this is normal business," he said. "At the end of the day, our commitment to the U.S. is strong."
Trump has publicly threatened automakers with a tariff on Mexican-made cars that are sold in the U.S. In recent weeks, he has specifically called out Ford (F), GM (GM) and Toyota (TM).
Last week, Ford scrapped plans to build a new $1.6 billion Mexican plant. CEO Mark Fields said the decision was made "independently" and it had not made a special deal with Trump.
Trump responded with a "thank you" tweet to Ford. "This is just the beginning - much more to follow," he said.
BMW has already started work on the new factory in San Luis Potosi, Mexico. The facility will open in 2019 and produce the BMW 3 Series Sedan.
BMW says the factory will create at least 1,500 new jobs.
"The cars which will be produced in Mexico are for the global market, not only North America," the company said.
The German automaker has more than 30 factories in 14 countries around the world.
The South Carolina plant employs 8,800 people and is investing $1 billion to increase production capacity. About 70% of cars made there are exported globally.
BMW sold nearly 370,000 cars in the U.S. last year.