text:
\section{Introduction}
 The reduced Hamiltonian of $q$-state Potts model\cite{potts,wfypotts} can be written as
 \begin{eqnarray}
 -{\mathcal H}/k_BT=K\sum\limits_{<i,j>}\delta_{\sigma_i\sigma_j},\label{pottsmodel}
 \end{eqnarray}
 where $\sigma_i$, $\sigma_j$ are the Potts spins on sites $i$ and $j$. The Potts spin can take $q$ values, namely $\sigma_i=0,1,\cdots,q-1$.
The sum takes over all nearest-neighboring sites $<i,j>$. $K$ is the coupling constant between spin $\sigma_i$ and $\sigma_j$.
 This model can be mapped to the random-cluster model\cite{FK1,FK2} with partition sum
 \begin{eqnarray}
 Z_{\rm rc}=\sum\limits_{\mathcal G} v^{N_b} q^{N_c}, \label{rcmodel}
 \end{eqnarray}
where ${\mathcal G}$ is a graph on lattice ${\mathcal L}$ that consists of $N_b$ random occupied bonds and $N_c$ clusters. $v=\exp K-1$ is the bond weight.
$q$ is the cluster weight, which is not restricted to be integer in (\ref{rcmodel}).
 An edge is set to be occupied by a `bond' with probability  $p=v/(1+v)$ or vacant with probability $1-p$.
A cluster is defined as a group of sites connected by the bonds.

When $q=1$, the random-cluster model (\ref{rcmodel}) reduces to the bond percolation model\cite{Stauffer},
 in which the physicists have special interests.
A central problem of the research in percolation model is the determination of the thresholds of the percolation transition on different lattices.
After decades of investigation, the percolation thresholds on a lots of lattices have been determined. Most of the results are numerical ones,
exact results are scarce. An important exact result for bond percolation threshold is the one for the triangular-type lattice (see Fig. \ref{tri&kag}(a)).
 As is summarized by Ziff {\it et al.}\cite{Ziffperco,Chayes2006},
the percolation threshold of the lattice is determined by
\begin{eqnarray}
P(i,j,k)=P(\bar{i},\bar{j},\bar{k}), \label{ABC}
\end{eqnarray}
where $P(i,j,k)$ is the probability that the three apexes $i$, $j$ and $k$ of the triangular cell (Fig. \ref{tri&kag}(b)) are connected in a same cluster
via the bonds in the cell, while
$P(\bar{i},\bar{j},\bar{k})$ is the probability that the three apexes are isolated in three different clusters. There are other expressions to
determine the percolation thresholds of the triangular-type lattices\cite{Ziffperco}, but they are in fact equivalent to (\ref{ABC}).
(\ref{ABC}) was used to determine the exact percolation thresholds of a series of lattices, such as the martini lattices\cite{martinisite,Ziffperco}
and the stack-of-triangle lattices\cite{voidsperco}. All of these lattices belong to the triangular-type lattice, with different internal structure in the triangular cells. Ziff and Scullard\cite{Ziffexact} extend (\ref{ABC}) to a series of lattices which
they call bowtie lattices, as shown in Fig. \ref{bowtie}. The threshold of bond percolation
on the bowtie-A lattice was first determined by Wierman\cite{st-tri2}, using the star-triangle transformation\cite{st-tri1}.

\begin{figure}[htpb]
\includegraphics[scale=0.25]{threesite.eps}
\caption{(a) the triangular-type lattice, (b) the triangular cell, (c) the kagome-type lattice. The hatched triangular cells can have internal structures.}
\label{tri&kag}
\end{figure}

\begin{figure}[htpb]
\includegraphics[scale=0.16]{bowtie.eps}
\caption{Bowtie lattices: (a) bowtie-A lattice, (b) bowtie-B lattice, (c) bowtie-C lattice, (d) bowtie-D lattice.}
\label{bowtie}
\end{figure}

 When $q\ne1$, the random-cluster model (\ref{rcmodel}) can be considered as a general percolation model with cluster weight in the partition sum.
The determination of the critical points of this model on different lattices is also a challenge in statistical physics.
 Similar to the problem of percolation threshold, exact critical points of the random-cluster model are also scarce.
 Up to now, exact critical points were found for only a small number of two-dimensional lattices, such as the square lattice, the honeycomb lattice, the martini lattices\cite{pottsmartini},
and the stack-of-triangle lattices\cite{paperI,paperII}. In these lattices, the honeycomb lattice and the martini lattices belong to the triangular-type
lattice, with complex internal structure in the triangular cells.

For the Potts model on the triangular-type lattice, the partition sum can be written as
\begin{eqnarray}
Z_{\rm tri}=\sum\limits_{\{\sigma\}} \prod\limits_{\Delta} W_{\Delta}(i,j,k), \label{Ztri}
\end{eqnarray}
where $i,j,k$ are the three apexes of the cell, and the sum and product take over all the cells.
$W_{\Delta}(i,j,k)$ is the Boltzman weight of a hatched triangular cell, it can be written in the form of
\begin{eqnarray}
W_{\Delta}(i,j,k)=A+B(\delta_{\sigma_i\sigma_j}+\delta_{\sigma_j\sigma_k}+\delta_{\sigma_k\sigma_i})+C\delta_{\sigma_i\sigma_j\sigma_k}.
\end{eqnarray}
For example, if the internal structure of the cells are simple triangles, then
\begin{eqnarray}
W_{\Delta}(i,j,k)&=&\exp[K(\delta_{\sigma_i\sigma_j}+\delta_{\sigma_j\sigma_k}+\delta_{\sigma_k\sigma_i})]\nonumber\\
&=&(1+v\delta_{\sigma_i\sigma_j})(1+v\delta_{\sigma_j\sigma_k})(1+v\delta_{\sigma_k\sigma_i})\nonumber\\
&=&1+v(\delta_{\sigma_i\sigma_j}+\delta_{\sigma_j\sigma_k}+\delta_{\sigma_k\sigma_i})+(v^3+3v^2)\delta_{\sigma_i\sigma_j\sigma_k},
\end{eqnarray}
where $v=\exp K-1$ and $\delta_{\sigma_i\sigma_j\sigma_k}=\delta_{\sigma_i\sigma_j}\delta_{\sigma_j\sigma_k}$.
This gives $A=1, B=v$, and $C=v^3+3v^2$.

 Kelland\cite{Kelland} showed that the partition sum (\ref{Ztri}) is a self-dual one with self-dual point
\begin{eqnarray}
qA=C. \label{qA}
\end{eqnarray}
Wu and Lin\cite{WuandLin} established rigorously that this point is
indeed the critical frontier in the ferromagnetic regime
\begin{eqnarray}
3B+C>0,\quad 2B+C>0.
\end{eqnarray}
This is a very important result  in determining the critical point of Potts model.
For example, substituting the expressions $A=1$ and $C=v^3+3v^2$ to (\ref{qA}), we get the critical frontier of the Potts model on
the triangular lattice,
which writes $v^3+3v^2=q$. Substituting $p=v/(1+v)$ to this equation with $q=1$,
we get the threshold of bond percolation on the triangular lattice, i.e., $p_c=2\sin(\pi/18)$, which is a famous result\cite{st-tri1}.

Basing on (\ref{qA}), Wu gives the  critical points
of the kagome-type lattices (Fig. \ref{tri&kag}(c)) through a homogeneous conjecture\cite{paperI}
\begin{eqnarray}
(q^2A+3qB+C)^2-3(qB+C)^2-(q-2)C^2=0.
\end{eqnarray}
 This conjecture was checked by a lot of accurate numerical
results given by the finite-size scaling analysis\cite{paperII}, which shows that Wu's conjecture was not
exact but an excellent approximation.

In this paper, we will show that (\ref{qA}) can be extend to the bowtie lattices for determining the exact critical points of
the random-cluster model on these lattices.
This is very similar to Ziff's extension of (\ref{ABC}) to this type of lattices for determining the bond percolation thresholds.
Furthermore, we do Monte Carlo simulations of the random-cluster model on the bowtie-A lattice, the numerical results of the critical
points are consistent with the theoretical predictions.
 In the procedure of simulations, we find that the shape-dependent properties\cite{shapeuniv1,shapeuniv} of the percolation model on the
bowtie-A lattice are somewhat different from those of an isotropic lattice, by comparing some numerical results of bond percolation and site percolation.

\section{Critical frontier on bowtie lattices}
\label{bowtie_cri}
In fact, (\ref{ABC}) is exactly equivalent to the $q=1$ case of (\ref{qA}). $A$ is the sum of the weights
of sub-graphs with the three apexes $i,j,k$ are isolated in three different clusters, corresponding to the
probability $P(\bar{i},\bar{j},\bar{k})$. $C$ is the sum of the weights of sub-graphs with the three apexes connected in a same cluster,
corresponding to the probability $P(i,j,k)$. For example, in a simple triangle, $P(\bar{i},\bar{j},\bar{k})=(1-p)^3$ and $P(i,j,k)=p^3+3p^2(1-p)$.
Substituting $p=v/(1+v)$ to these expressions, one can easily find that $P(\bar{i},\bar{j},\bar{k})$ and  $P(i,j,k)$ are different to $A$ and $C$ respectively with
a constant $[1/(1+v)]^3$.  This property holds for any triangular cell (but the constant may be different for different internal structure).
In Ref. \cite{paperII}, a computer program was used to calculate the expressions of $A,B,C$ of $n\times n$ stack-of-triangle cell up to $n=4$.
 In Ref. \cite{voidsperco}, Haji-Akbari and Ziff give the expressions of $P(\bar{i}, \bar{j},\bar{k})$ and $P(i,j,k)$. Comparing these
expressions, we can see more clearly the equivalence of (\ref{ABC}) and the $q=1$ case of (\ref{qA}).

(\ref{ABC}) follows the duality arguments\cite{dual0,dual}, it
can be used to calculate the threshold of bond percolation on a lattice composed of triangular cells, and the lattice must
be self-dual under the transformation shown in Fig. \ref{bowtie_generate}(a).  The triangular-type lattice obviously meets this condition. Another type of lattice
that meets this condition is the lattice shown in Fig. \ref{bowtie_generate}(b), and the four bowtie lattices can be generated by inserting different internal
structures in the hatched triangular cells of this lattice. This lattice is first given by Ziff and Scullard\cite{Ziffexact}, we call this lattice the
`bowtie-generating lattice', because it `generates' the four bowtie lattices. If simple triangles are inserted in the cells of the bowtie-generating lattice, it
generates a lattice with double bonds as shown in Fig. \ref{pre_bA}. The bond percolation with uniform probability $p$ on the bowtie-A lattice can be got by
 setting the double bonds with probability $1-\sqrt{1-p}$, which is the Wierman's trick\cite{st-tri2}.

\begin{figure}[htpb]
\includegraphics[scale=0.20]{triB.eps}
\caption{(a) duality transformation of a triangular cell, (b) the bowtie-generating lattice, which is self-dual under the duality transformations of
the cells. Four bowtie lattices as shown in Fig. \ref{bowtie} can be generated by this lattice by inserting different internal structures in the
hatched triangular cells.}
\label{bowtie_generate}
\end{figure}

\begin{figure}[htpb]
\includegraphics[scale=0.20]{pre_bA.eps}
\caption{A lattice constructed by inserting simple triangles in the cells of bowtie-generating lattice and its bond weights of the random-cluster model (top).}
\label{pre_bA}
\end{figure}

Similar to (\ref{ABC}), (\ref{qA}) also follows the duality arguments\cite{Kelland, WuandLin}, this inspires us to extend (\ref{qA}) to calculate the
critical points of random-cluster model on other lattice which meets the condition of self-duality. As an example, we calculate the critical points of
the bowtie lattices in this paper.

\subsection{bowtie-A lattice}
For generality, we set the coupling constants of the three bonds of the triangle in the double-bond lattice with $K,M,N$. The bond weights
 are $v=\exp{K}-1, u=\exp{M}-1$ and $w=\exp{N}-1$, which are shown in Fig. \ref{pre_bA}. The expressions of $A$ and $C$ are
\begin{eqnarray}
A&=&1,\\
C&=&uvw+uv+vw+wu.
\end{eqnarray}
Substituting $A$ and $C$ to (\ref{qA}), we get the critical frontier of the Potts model on the double-bond lattice
\begin{eqnarray}
uvw+uv+vw+wu=q. \label{double_bond}
\end{eqnarray}
Setting
\begin{eqnarray}
M=K, N=K/2,
\end{eqnarray}
namely
\begin{eqnarray}
u=v, w=\sqrt{1+v}-1,\label{uw}
\end{eqnarray}
we get a Potts model with uniform coupling constant $K$ or random-cluster model with uniform bond
weight $v$ on the bowtie-A lattice. Substituting (\ref{uw}) to (\ref{double_bond}), we get the critical
frontier of the Potts model on the bowtie-A lattice, which writes
\begin{eqnarray}
(v^2+2v)(\sqrt{1+v}-1)+v^2=q, \quad \mbox{bowtie-A lattice.}\label{bowtieAcri}
\end{eqnarray}
Solving (\ref{bowtieAcri}) with $q=1$, we get $v_c=0.679~312~786$ or $K_c=0.518~384~653$. This yields the threshold of bond percolation on the
bowtie-A lattice, i.e., $p_c=v_c/(1+v_c)=1-\exp(-K_c)=0.404~518~319$, which is exactly equal to the result that given by Wierman\cite{st-tri2}
or Ziff {\it et al.}\cite{Ziffexact}.

The case of $q=2$, which is the Ising model, has been exactly solved by Str$\check{\rm e}$cka and $\check{\rm C}$anov$\acute{\rm a}$\cite{bowtieIsing} through
mapping the model into a free-fermion eight-vertex model, and the critical points is given by
\begin{eqnarray}
[\cosh(4K_I)-1]e^{2K_I}=2, \label{bAIsing}
\end{eqnarray}
where $K_I=K/2$ is the coupling constant of nearest-neighboring Ising spins. Solving (\ref{bAIsing}), we get $K_{Ic}=0.333~135~959$.
Furthermore, if we solve (\ref{bowtieAcri}) with $q=2$, we get $K_c=0.666~271~918$.
The two results are exactly consistent with each other.

The cases of $q=1$ and 2 show the correctness of (\ref{bowtieAcri}) for the bowtie-A lattice. For the other case, especially the case that $q$
 is not an integer, the correctness of (\ref{bowtieAcri}) is not verified. Thus we do extensive Monte Carlo simulation and finite-size scaling analysis
for the Potts model on the bowtie-A lattice, the numerical results will be given in Sec. \ref{MC}.

Setting the double-bond weights $w=0$ (namely $N=0$), (\ref{double_bond}) gives the critical frontier
of Potts model on the square lattice
\begin{eqnarray}
uv=q,
\end{eqnarray}
which is a famous result\cite{Baxter}.

\subsection{bowtie-B, and bowtie-D lattices}
The critical frontier of the Potts model on the bowtie-B, the bowtie-C and the bowtie-D lattices can be obtained by the same way that we used for the bowtie-A lattice.
In this paper, we only pay attention to the random-cluster model on these lattices with uniform bond weight $v$.
\begin{figure}[htpb]
\includegraphics[scale=0.35]{subnets.eps}
\caption{(a) internal structure for constructing the bowtie-B and bowtie-D lattices, (b) internal structure for constructing the lattice as shown
in Fig. \ref{bowtiec}(a). $v=\exp K-1, v_i=\exp K_i -1, w_i=\exp L_i-1$ are bond weights, $M$ is the 3-site coupling
 constant between the three spins of the triangle in center.}
\label{subnets}
\end{figure}

In order to get the critical frontier of the Potts model on the bowtie-B and the bowtie-D lattices, we insert the network as shown in Fig. \ref{subnets}(a) to the
 cells of the bowtie-generating lattice, this generates a lattice that also has double bonds. The expression of $A$ and $C$ are
\begin{eqnarray}
 A&=&q^2+5qv+8v^2+v^3,\\
 C&=&3v^4+v^5+v_1(2qv^2+8v^3+5v^4+v^5),
\end{eqnarray}
where $v=\exp K-1$ and $v_1=\exp K_1-1$ are the bond weights (see Fig. \ref{subnets}(a)). Substituting them to (\ref{qA}), we get the critical frontier
\begin{eqnarray}
 3v^4+v^5+v_1(2qv^2+8v^3+5v^4+v^5)-q(q^2+5qv+8v^2+v^3)=0.
\end{eqnarray}
Setting $K_1=K/2$, namely $v_1=\sqrt{1+v}-1$, we get the critical frontier of the Potts model on the bowtie-B lattice with uniform coupling constant $K$
\begin{eqnarray}
3v^4+v^5+(\sqrt{1+v}-1)(2qv^2+8v^3+5v^4+v^5)-q(q^2+5qv+8v^2+v^3)=0,&&\nonumber\\
 \mbox{bowtie-B lattice.}&& \label{bowtieB}
\end{eqnarray}
Solving this formula with $q=1$, we get the critical point $v_c=1.142~305~296$ or $K_c=0.761~882~490$. This yields the threshold of bond percolation
$p_c=0.533~213~122$, which is exactly consistent with the value given in Ref. \cite{Ziffexact}.

Setting $K=0$, namely $v_1=0$, we get the critical frontier
of the Potts model on the bowtie-D lattice with uniform $K$
\begin{eqnarray}
 3v^4+v^5-q(q^2+5qv+8v^2+v^3)=0,
\quad \mbox{bowtie-D lattice.} \label{bowtieD}
\end{eqnarray}
Solving this formula with $q=1$, we get the critical point $v_c=1.669~919~079$ or $K_c=0.982~048~164$. This yields the bond percolation threshold
$p_c=0.625~456~813$, which is exactly consistent with the value given in Ref. \cite{Ziffexact}. (\ref{bowtieD}) is also the critical frontier of the Potts model
on the martini-A lattice, see Ref. \cite{pottsmartini} for details.

\subsection{bowtie-C lattice}
\begin{figure}[htbp]
\includegraphics[scale=0.16]{bowtiec.eps}
\caption{After the decimation of the dotted spins, the lattice (a) becomes the lattice (b), namely the bowtie-C lattice.}
\label{bowtiec}
\end{figure}
By inserting the network as shown in Fig. \ref{subnets}(b) to the triangular cells of the bowtie-generating lattice,
 we get a lattice as shown in Fig. \ref{bowtiec}(a). The expressions of $A, C$ for the network as shown in Fig. \ref{subnets}(b) have been
given by Wu\cite{pottsmartini}:
\begin{eqnarray}
A&=&v_1v_2v_3+v_1v_2(q+w_1+w_2)+v_2v_3(q+w_2+w_3)+v_3v_1(q+w_3+w_1)\nonumber\\
&&+(q+v_1+v_2+v_3)\times [q^2+q(w_1+w_2+w_3)+h]\\
C&=&v_1v_2v_3h
\end{eqnarray}
where
\begin{eqnarray}
v_i&=&e^{K_i} -1, w_i=e^{L_i} -1\nonumber\\
h&=&e^{M+L_1+L_2+L_3}-e^{L_1}-e^{L_2}-e^{L_3}+2.
\end{eqnarray}
$K_i,L_i$ are the 2-site coupling constant, $M$ is the 3-site coupling constant.
Setting the bond weights $w_1=w_2=w_3=v_2=v_3=v$ (i.e., $L_1=L_2=L_3=K_2=K_3=K$) and the 3-site coupling constant $M=0$,
we get the critical frontier of the Potts model on the lattice as shown in Fig. \ref{bowtiec}(a):
\begin{eqnarray}
q(q+2v)[q^2+3qv+v^2(4+v)]+[q^3+5q^2v-v^4(3+v)+qv^2(8+v)]v_1=0. \label{bowtieC_cri}
\end{eqnarray}
Decimating the dotted spins in the lattice as shown in Fig. \ref{bowtiec}(a), we get the lattice as shown in Fig. \ref{bowtiec}(b), which is
the bowtie-C lattice. If the bond weight $v_1$ is set as $v_1=v+\sqrt{v^2+vq}$, we get the critical frontier of the Potts model on the
bowtie-C lattice with uniform bond weight $v$:
\begin{eqnarray}
q(q+2v)[q^2+3qv+v^2(4+v)]+[q^3+5q^2v-v^4(3+v)+qv^2(8+v)](v+\sqrt{v^2+qv})=0. &&\nonumber\\
\mbox{bowtie-C lattice.}\quad&& \label{bowtieC}
\end{eqnarray}
Solving this formula with $q=1$, we get the critical point $v_c=2.057~439~254$ or $K_c=1.117~577~720$. This yields the bond percolation threshold
$p_c=0.672~928~906$, which is exactly consistent with the result that given in Ref. \cite{Ziffexact}.

We also solve (\ref{bowtieAcri}), (\ref{bowtieB}), (\ref{bowtieD}), and (\ref{bowtieC}) with other value of $q$, some results are listed in Table. \ref{bowtie_exact}.
In the table, we only give the results with $q\le 4$, because the phase transition of the Potts model becomes discontinuous when $q>4$\cite{Baxter}.

\begin{table}[htbp]
\caption{Exact critical points of the Potts model on the bowtie lattices, A = bowtie-A lattice, B = bowtie-B lattice, C = bowtie-C lattice,
D = bowtie-D lattice.}
 \begin{tabular}{c|c|c|c|c}
    \hline
      $q$   &$K_c$(A)     &$K_c$(B)    &$K_c$(C) &$K_c$(D)\\
    \hline
    $1.0$   &0.518 384 653 &0.761 882 490 &1.117 577 720 &0.982 048 164           \\
    $1.5$   &0.601 796 427 &0.880 501 865 &1.263 365 870 &1.116 885 906 \\
    $2.0$   &0.666 271 918 &0.971 526 940 &1.372 908 989 &1.218 755 726 \\
    $2.5$   &0.719 275 592 &1.045 947 678 &1.461 235 663 &1.301 189 204 \\
    $3.0$   &0.764 490 171 &1.109 155 997 &1.535 511 667 &1.370 686 799  \\
    $3.5$   &0.804 030 933 &1.164 234 324 &1.599 748 425 &1.430 907 221  \\
    $4.0$   &0.839 235 249 &1.213 123 525 &1.656 430 741 &1.484 126 262 \\
    \hline
\end{tabular}
\label{bowtie_exact}
\end{table}

\subsection{bowtie-dual lattices}
The critical frontier of the Potts model on the dual of the bowtie lattice, namely the bowtie-dual lattice (see Fig. 5 of Ref. \cite{Ziffexact}),
 can be obtained by the dual relation\cite{dual}
\begin{eqnarray}
v^*\cdot v=q,
\end{eqnarray}
or
\begin{eqnarray}
(e^{K^*}-1)(e^K-1)=q,
\end{eqnarray}
where $K$ is the coupling constant on the bowtie lattice, and $K^*$ is the coupling constant on the dual lattice. $v=\exp K-1$ and $v^*=\exp K^*-1$
 are the bond weights on the bowtie lattice and the bowtie-dual lattice, respectively.
Thus the critical frontier of the Potts model on the bowtie-dual lattice is given by
\begin{eqnarray}
(e^{K^*_c}-1)(e^{K_c}-1)=q.
\end{eqnarray}

\section{Monte Carlo simulation and numerical results}
\label{MC}
\subsection{Simulation of the Potts model on the bowtie-A lattice}
Using the cluster algorithm\cite{Qian}, we do Monte Carlo simulation of the Potts model on the bowtie-A lattice with $q\ge1$.
This cluster algorithm is a combination of Swendsen-Wang algorithm\cite{SW} and `coloring trick'\cite{color1,color2}, which drastically reduces
 the critical slowing down problem.
This type of algorithm has been developed and used to simulate O($n$) loop model\cite{DengAlg,Ding2009}, Eulerian bond-cubic model\cite{DingCubic} and so forth.

In the random-cluster model (\ref{rcmodel}), the behaviors of the clusters are similar to those of percolation model.
There are small clusters on the configurations when the bond weight $v$ is small, but the clusters will growth as the value of $v$ increases.
The largest cluster diverges at the critical point $v_c$ (or $K_c$) in the thermodynamic limit, which is called a `percolating cluster'.
However, for a finite system, the largest cluster may be different to the `percolating cluster'.
There are different rules to define the `percolating cluster' for a finite system with periodic boundary condition. Here we use
the `wrapping cluster'\cite{wrapping}, which
is defined as a cluster that connects itself along at least one of the periodic directions.

Basing on the definition of the `wrapping cluster', the `wrapping probabilities' on a rectangular lattice are defined as
\begin{eqnarray}
R_x&=&\langle \mathcal{R}_x \rangle,\nonumber\\
R_y&=&\langle \mathcal{R}_y \rangle, \nonumber\\
R_{xy}&=&\langle \mathcal{R}_{xy}\rangle,
\end{eqnarray}
where $\langle \cdots \rangle$ stands for ensemble average. $\mathcal{R}_x$ is 1 (zero) if there is a (no) cluster that wraps along the $x$ direction,
 whether or not the cluster wraps along $y$ direction.
The definition of $\mathcal{R}_y$ is similar to $\mathcal{R}_x$. $\mathcal{R}_{xy}$ is defined as
$\mathcal{R}_{xy}=\mathcal{R}_x\cdot\mathcal{R}_y$, whose value is 1 if and only if there exists a cluster that wraps along $x$ and $y$ directions at the same time.
\begin{figure}[htbp]
\includegraphics[scale=0.2]{bowtie_R.eps}
\caption{Definition of wrapping probabilities on the bowtie-A lattice.}
\label{bowtie_R}
\end{figure}

The bowtie-A lattice that we simulated in the paper can be viewed as a `rectangular lattice' with internal structure in the rectangular cells,
see Fig. \ref{bowtie_R}. This
rectangular lattice is obviously not symmetry between $x$ and $y$ directions; therefore the values of $R_x$ and $R_y$ will be not the same,
which will be shown concretely later.
The wrapping probabilities $R_x$, $R_y$ and $R_{xy}$ have the similar finite-size scaling behavior\cite{fssa,fssb}
\begin{eqnarray}
R=R_0+a_1(K-K_c)L^{y_t}+a_2(K-K_c)L^{2y_t}+\cdots+b_1L^{y_1}+b_2L^{y_2}+\cdots, \label{Rfss}
\end{eqnarray}
where $R$=$R_x$, $R_y$ or $R_{xy}$. $R_0$ is the value of $R$ at the critical point $K_c$. In present paper, we call it `critical wrapping probability'.
$y_t$ is the thermal exponent of the Potts model, and $y_i$ the correction-to-scaling exponent, which has negative value. $a_i$ and $b_i$
are unknown parameters. The value of $y_t$ can be derived by Coulomb gas method\cite{CG} or conformal invariance\cite{conformal}
\begin{eqnarray}
 y_t&=&3-\frac{3}{2g}, \label{yt}\\
{\rm with} \quad \sqrt{q}&=&-2\cos(\pi g), \quad 1/2\le g \le 1,
\end{eqnarray}
where $g$ is the coupling constant of the Coulomb gas.

In order to illustrate our numerical procedure, we take the $q=1.5$ Potts model as an example.
The cluster algorithm easily allows us to do meaningful simulations for system
 with  linear size up to $L=256$. After equilibrating the system, $10^9$ samples were taken for each
value of $K$ for $8\le L\le 64$, and $1.4\times 10^8$ samples for $128\le L\le 256$.
 Figure. \ref{Ry15} is an illustration of $R_y$ versus $K$ with different $L$ for $q=1.5$ Potts model.
\begin{figure}[htpb]
\includegraphics[scale=1.0]{Ry1.5.eps}
\caption{$R_y$ versus $K$ for various system size of $q=1.5$ Potts model. All the error bars are much smaller than the data points, the lines between points are added for illustration purpose.}
\label{Ry15}
\end{figure}
Using the Levenberg-Marquardt least squares algorithm, we fit the data according to the finite-size-scaling formula (\ref{Rfss}). The fitting yields
$R_{0y}=0.7070(2)$, $R_{0x}=0.4363(2)$, $R_{0xy}=0.3780(1)$, $K_c=0.6017964(6)$ and $y_t=0.887(1)$. In these results,
we can see that the critical point $K_c$ is consistent with our theoretical
prediction $K_c=0.60179643$, and the thermal exponent is consistent with the Coulomb gas prediction $y_t=0.8867$.

We also simulate the cases of $q=1, 2.5, 3, 3.5$ and 4.  All the numerical results and the theoretical predictions
are summarized in Table. \ref{bowtieA_num}.
In the fitting of the data for $q=4$ Potts model, logarithmic corrections\cite{log,log1,log2} should be included. Instead of (\ref{Rfss}), we fit the data
according to
\begin{eqnarray}
R=R_0+a_1(K-K_c)L^{y_t}(\log L)^{y_1}+a_2(K-K_c)^2L^{2y_t}(\log L)^{2y_1}+\cdots+b_1\frac{\log \log L}{\log L}+\frac{b_2}{\log L},
\end{eqnarray}
with $y_1<0$. However, we can see (from the table) that the
results for $q=4$ Potts model are relatively inaccurate. This is because of the limited system size.
For an accurate fitting with logarithmic corrections,  data with larger system sizes are necessary.

\begin{table}[htbp]
\caption{Critical properties of the Potts model on the bowtie-A lattice, T=Theoretical results, N=Numerical results. The value of $K_c$ (N) can be obtained by fitting
the data of $R_y$, $R_x$, or $R_{xy}$, here we list the best estimation.}
 \begin{tabular}{c|l|l|l|l|l|l|l}
\hline
     $q$      & $K_c$ (T)  & $K_c$ (N)    &$y_t$ (T) & $y_t$ (N) & $R_{0y}$ & $R_{0x}$ & $R_{0xy}$\\
    \hline
    $1$     &0.518 384 653 &0.5183847(7)  &0.7500    &0.750(1)   &0.6490(1)&0.3838(1)&0.3161(1)  \\
    $1.5$   &0.601 796 427 &0.6017964(6)  &0.8867    &0.887(1)   &0.7070(2)&0.4363(2)&0.3780(1)\\
    $2.5$   &0.719 275 592 &0.7192756(11) &1.1018    &1.101(2)   &0.7750(1)&0.5151(2)&0.4729(2)\\
    $3$     &0.764 490 171 &0.7644906(9)  &1.2000    &1.197(6)   &0.7975(1)&0.550(1) &0.515(1)\\
    $3.5$   &0.804 030 933 &0.8040314(25) &1.3050    &1.296(12)  &0.8156(2)&0.585(1) &0.557(1)\\
    $4$     &0.839 235 249 &0.839235(2)   &1.5000    &1.50(7)    &0.829(5)  &0.62(1) &0.60(1)\\
    \hline
\end{tabular}
\label{bowtieA_num}
\end{table}

\subsection{Simulation of site percolation}
We also simulate the site percolation on the bowtie-A lattice, the finite-size scaling analysis gives $s_c=0.5479148(7)$,
$y_t=0.750(1)$, $R_{0y}=0.6090(1)$, $R_{0x}=0.4286(1)$ and  $R_{0xy}=0.3349(1)$.
 The value of $s_c$ coincides with 0.5475(8), given by van der Marck\cite{Marck}, and our result is much more accurate.

\section{Shape-dependent properties of the bowtie-A lattice}
We also do Monte Carlo simulations of the bond and site percolation models
on the square lattice. For the percolation thresholds, the numerical estimations give $p_c=0.5000000(4)$ (bond), and $s_c=0.5927460(5)$ (site). The accuracy
of $s_c$ is somewhat lower than the result given by Feng {\it et al}\cite{Feng}. The numerical results of the critical wrapping probabilities are listed in
Table \ref{bowtieA_shapeuniv}, these results give $R_{0e}$=$R_{0x}+R_{0y}-R_{0xy}$=0.6902(5) for the bond percolation and 0.6905(5) for the site percolation,
which are consistent with the literature value $R_{0e}=0.6904737$\cite{wrapvalue,shapeuniv}. $R_{0e}$ is the critical value of the probability that
there is a cluster wrapping along one or both of the two coordinate directions.

\begin{figure}[htpb]
\includegraphics[scale=0.25]{shape.eps}
\caption{(a) A rectangular lattice (on macroscopic level) with square symmetry on microscopic level, with aspect ratio $r=3/4$;
 (b) A rectangular lattice (on macroscopic level) with triangular symmetry on microscopic level, with aspect ratio $r=\sqrt{3}/2$.}
\label{shape}
\end{figure}

For a lattice that is isotropic on microscopic level (equivalent in the coordinate directions, or with the triangular or hexagonal symmetry, etc.),
the values of the wrapping probabilities are functions of the system shape and boundary conditions\cite{shapeuniv,wrapvalue}, which are independent of the
percolation type.
In saying the shape, it is on the macroscopic level. For example, a macroscopic rectangular lattice may be a lattice with square or triangular symmetry on
microscopic level, see Fig. \ref{shape} for examples or see Ref. \onlinecite{shapeuniv1,shapeuniv} for details.
From now on, in saying `isotropic' or `anisotropic' in the text, it is on the microscopic level.
For an isotropic rectangular lattice with periodic boundary conditions (not twisted),
the value of the wrapping probability $R_{0e}$ is determined by the aspect ratio of the lattice\cite{shapeuniv}
\begin{eqnarray}
R_{0e}(r)=1-\frac{1}{2}[Z_c(8/3,r)-Z_c(2/3,r)], \label{R0}
\end{eqnarray}
where $r$ is the aspect ratio and $Z_c(h,r)$ is
\begin{eqnarray}
Z_c(h)=\frac{\sqrt{h/r}}{\eta^2(w)} \sum\limits_{n=-\infty}^{\infty}\sum\limits_{n^\prime=-\infty}^\infty \exp \big\{-\frac{\pi h}{r}[n^{\prime2}+n^2r^2]  \big\},
\end{eqnarray}
$\eta(w)=w^{1/24}\prod\limits_{k=1}^{\infty}(1-w^k)$ is the Dedekind eta function and $w=e^{-2\pi r}$. When $r=1$, (\ref{R0}) gives $R_{0e}=0.6904737$ for the square lattice.

In our Monte Carlo simulations, the bowtie-A lattice is viewed as a rectangular lattice (with periodic b.c.), as shown in Fig. \ref{bowtie_R},
and the aspect ratio is $r=1/\sqrt{3}$. However,
the wrapping probability of the bowtie-A lattice is not determined by this aspect ratio.
(\ref{R0}) gives $R_{0e}(1/\sqrt{3})=0.7439918$, this is inconsistent with
our numerical values $R_{0e}=0.7167(1)$ (bond) and 0.7027(1) (site). Furthermore,
we can see that the values of $R_{0}$ (include $R_{0e}, R_{0x}, R_{0y}$, and $R_{0xy}$, see Table. \ref{bowtieA_shapeuniv}.)
for the bond percolation are different from those for the site percolation. These results show the difference between the bowtie-A lattice
and an isotropic lattice in the aspect of shape-dependent properties.

The difference may be caused by the anisotropy of the bowtie-A lattice,
 and (\ref{R0}) is only valid for an isotropic lattice.
 After mapping the percolation model on the bowtie-A lattice onto the  Gaussian model\cite{wrapvalue},  it will be an anisotropic one.
Thus a rescaling on $x$ or $y$ direction is required  in order to obtain an
isotropic Gaussian model. After the rescaling, the isotropic model would have an aspect ratio that is not equal to $1/\sqrt{3}$ (Unfortunately, we
don't know the value of the effective aspect ratio). This is the reason why our numerical value of $R_{0e}$ for the bowtie-A lattice
doesn't coincide with $R_{0e}(1/\sqrt{3})$.

More important, there is no reason why the bond and site percolation models on the bowtie-A lattice should map onto the same
(anisotropic) Gaussian model, and after rescaling to the isotropic Gaussian model, there would be different aspect ratios for
the bond and site percolation models. Thus, there would be different values of the wrapping probabilities for the bond and site percolation models,
 which are verified by our numerical results.

Another quantity that we simulate for investigating the shape-dependent properties is the average
density of clusters, which is defined as the average number of clusters per site.
We find that on such an anisotropic lattice, the average density of clusters also behaves as\cite{shapeuniv1}
\begin{eqnarray}
n=n_c+b/N+\cdots,
\end{eqnarray}
when the system is at the critical point. $n_c$ is the value of $n$ for the infinite system, and $N$ the total number of sites of a finite system.
$b$ is the excess number of clusters over the bulk value of clusters $n_cN$. For an isotropic lattice with periodic boundary conditions (not twisted),
the value of $b$ is also a function of system
shape, which has the same value for the bond and site percolation models\cite{shapeuniv1}
\begin{eqnarray}
b(r)=\frac{5\sqrt{3}r}{24}+w^{5/4}(2\sqrt{3}r-\frac{1}{2})+w^2(\sqrt{3}r-1)+w^{5/48}+2w^{53/48}-w^{23/16}+w^{77/48}+\cdots,
\end{eqnarray}
where $w=e^{-2\pi r}$. This equation gives $b(1/\sqrt{3})=0.943713$, which is obviously inconsistent with our numerical results for the bowtie-A lattice:
 $b^{\rm B}=0.9120(6)$ and $b^{\rm S}=0.8957(4)$ (B denotes bond percolation, S denotes site percolation).
More important, the value of $b^{\rm B}$ is different from that of $b^{\rm S}$.
The physical essence of these results for $b$ is the same to that for the wrapping probabilities, it may also be caused by the anisotropy of the bowtie-A lattice.

In conclusion (Table \ref{bowtieA_shapeuniv}), the shape-dependent properties of the percolation model on the bowtie-A lattice are somewhat different from
those of an isotropic lattice, this may be caused by the anisotropy of the bowtie-A lattice.
The bond and site percolation models on the bowtie-A lattice correspond
with different anisotropies after mapping onto Gaussian models, and thus with different aspect ratios after rescaling to isotropic models,
which lead to the difference of shape-dependent properties between the bond and site percolation models.

\begin{table}[htbp]
\caption{Critical values of the wrapping probability $R_0$, the average density of clusters $n_c$ and the
excess number of clusters $b$ on the square and bowtie-A lattices, Site=Site percolation, Bond=Bond percolation. $^c$=Reference\cite{shapeuniv1},
$^d$=Reference\cite{shapeuniv},$^e$=Reference\cite{wrapvalue},$^f$=Reference\cite{Kleban}}
 \begin{tabular}{l|l|l|l|l|l|l}
\hline
     System           & $R_{0y}$  & $R_{0x}$ &$R_{0xy}$& $R_{0e}$ &$b$  & $n_c$                      \\
    \hline
    Bond (square)     &0.5208(5)  &0.5210(5) &0.3516(5) &0.6902(5) &0.883(1)&0.098076(1) \\
                      &0.5210583$^{d,e}$  &0.5210583$^{d,e}$&0.3516429$^{d,e}$&0.6904737$^{d,e}$&0.8838(5)$^c$&0.0980763(8)$^c$ \\
                      & & & &&0.883756 $^f$ & $\frac{3\sqrt{3}-5}{2}$$^c$\\
    Site (square)     &0.5211(5)(1)&0.5209(5)&0.3514(4) &0.6905(5)&0.8834(11)&0.027598(1)\\
                      &0.5210583$^{d,e}$&0.5210583$^{d,e}$&0.3516429$^{d,e}$&0.6904737$^{d,e}$&0.8832(3)$^c$&0.0275981(3)$^c$ \\
    Bond (bowtie-A)   &0.6490(1)  &0.3838(1) &0.3161(1) &0.7167(1) &0.9120(6)&0.119212(1)  \\
    Site (bowtie-A)   &0.6090(1)  &0.4286(1) &0.3349(1) &0.7027(1) &0.8957(4)&0.023990(1)  \\
    \hline
\end{tabular}
\label{bowtieA_shapeuniv}
\end{table}

\section{Conclusion}
\label{con}

In conclusion, we have given the exact critical frontier of the Potts model on the bowtie lattices.
For $q=1$, the critical frontier yields the thresholds of bond percolation on these lattice, which are consistent with the results given by
Ziff {\it et al}. For the $q=2$ Potts model on the bowtie-A lattice, the critical point coincides with the exactly solved critical point
of the Ising model on this lattice.
Furthermore, we do Monte Carlo simulations of the Potts model on the bowtie-A lattice, the numerical results, which includes the cases with
noninteger $q$, are in agreement with the exact critical points in a high precision.

Furthermore, by comparing the numerical results of the critical wrapping probability $R_0$ and the excess number of clusters $b$ for the
bond and site percolation models, we find that the shape-dependent properties of the percolation model on the bowtie-A lattice are somewhat different from
those of an isotropic lattice, which may be caused by the anisotropy of the bowtie-A lattice.
\acknowledgments
We are much indebted to Prof. R. M. Ziff for valuable discussions. We thanks Prof. Wenan Guo for a critical reading of the manuscript.
 This research is supported by the National Science Foundation of
China  under Grant \#11175018, and by the High Performance Scientific Computing Center (HSCC) of the Beijing Normal University.

redpajama_set_name: RedPajamaArXiv


text:
Cardiology/
Cardiac Rehab Program
At the ribbon cutting for Saint Joseph's Cardiac Rehab and Physical Therapy program, Yonkers City Council Member Shanae Williams and Council President Mike Khader presented a proclamation on behalf of Mayor Mike Spano. Pictured above are Shanae Williams, Saint Joseph's President and CEO Michael J. Spicer, Mike Khader, cardiologists Stuart Moser, M.D., Richard Greif, M.D., Jose Tavaras, M.D. and Richard Feldman, M.D.
Saint Joseph's has opened a physician-supervised Cardiac Rehab Program at its Cardiovascular Center at 530 Yonkers Avenue in Yonkers. The program offers two levels of service, one for cardiac rehab patients within a year of discharge from a hospital after a heart attack who require a more intensive level of monitoring, and a regular rehab program for musculoskeletal and orthopedic cases. For information, please call (914) 308-7350.
Saint Joseph's
Cardiac Rehab and
530 Yonkers Avenue
Yonkers 10704

redpajama_set_name: RedPajamaCommonCrawl


You know that a regular exercise routine can help your heart stay healthy, keep diseases like diabetes at bay and make your muscles strong.
To leverage the power of the mind-body connection, it's helpful to understand how that connection works. This guide will show you how exercise changes your brain, how it improves both your physical and mental health, and what exercises are best for boosting brain function.
Memory is a complex and incredible function of the mind. Many people describe having a fully functioning memory as "staying sharp"—a primary concern for many aging adults.
Sandra Bond Chapman, PhD, founder and chief director at the UT Dallas Center for BrainHealth, has found that aerobic exercise can help address this worry. In a 2013 study that Chapman led, researchers looked at adults from 57 to 75 years old over the course of 12 weeks. One group of participants exercised on treadmills and stationary bikes for one hour, three days each week. The other group did not exercise.
The researchers found that the active participants had improved memory function and increased blood flow to the hippocampus, the part of the brain that controls memory.
Further, exercise can produce the proteins that the brain needs to retain and create new cells in the hippocampus, researcher David DiSalvo explains. "Research has discovered that exercise stimulates the production of a protein called FNDC5 that is released into the bloodstream while we're breaking a sweat," DiSalvo writes at Forbes.
In other words, exercise stimulates the growth and the preservation of the region of the brain responsible for storing memories. And science shows this exercise-hippocampus connection has several benefits for our minds.
High-interface memory is an important component in memory. It's how our brains distinguish patterns in what we see. For example, high-interface memory helps you recognize that a dog of an unfamiliar breed is still a dog.
Cameron Evans at Psychology Today reports on some of the surprising research that looked into exercise's effect on this type of memory. For this study, researchers at McMaster University in Canada sorted participants into three groups. One group performed exercise, one group performed exercise and cognitive training, and the final group did neither. It seems obvious that that the second group—which received both cognitive and physical training—would see the most improvement. After all, they were the only ones receiving cognitive training—and cognitive training is the conventional way to improve a person's mind.
At first, the researchers saw exactly that. "The results showed that, unlike the control group, participants who underwent the cognitive and exercise training sessions improved on the high-interference memory task," Evans writes.
However, something unexpected happened when they saw the results from the group who simply exercised.
As the results from the second group showed memory improvement, "so did those who only received exercise training," Evans continues. The research seems to suggest that exercise alone can help people with cognitive abilities. While this study focused on young adults, the researchers are currently working on a similar study with older adults and hypothesize the findings will be similar.
Memory consolidation is another important function of the brain. It's what helps people form long-term memories.
Shikha Snigdha and fellow researchers at the Institute for Memory Impairments and Neurological Disorders found that exercise can help keep this memory function in tact. The researchers explain that even if you feel you are losing your ability to retain long-term memories, exercise can have an immediate effect on mitigating decline—an effect that continues over time.
Not only does exercise help improve your memory functions, it can also play defense against lifestyle factors that would otherwise negatively impact your memory.
New York Times science reporter Gretchen Reynolds explains that while certain lifestyle factors—namely stress—can prevent our synapses from firing off properly, research suggests that exercise can counter that effect.
A team of researchers from Brigham Young University's Department of Physiology and Developmental Biology and Neuroscience Center recently published the results of a study that explored the relationships among stress, memory and the synapses in the hippocampuses of mice.
They found that mice that experienced stress showed signs of impaired memories, whereas the mice that exercised after experiencing stress showed normal memory function. "Collectively, exercise can mitigate some of the negative impact stress has on hippocampal function when both occur concurrently," the researchers conclude.
Helping you remember things isn't the only way exercise can impact your grey matter. Working out regularly can help you stave off depression, which is particularly common in the aging population.
Dr. Robert Madan at the University of Toronto has made it his mission to end depression among the elderly. He explains that memory diseases like Alzheimer's and dementia are closely related. By taking care of your memory, you can help fight off depression.
So, exactly how helpful can exercise be in the fight against depression? Ellen Goldstein at the University of Wisconsin-Madison and fellow researchers found that exercise is as effective as practicing mindfulness. "Findings suggest that mindfulness and exercise training share similar mechanisms that can improve global mental health, including adaptive responses to stress," the researchers write.
We have already shown how meditation is a key component of a healthy lifestyle. Adding exercise to the mix will only reinforce your other memory-boosting healthy habits.
If you read these studies and find yourself wishing you'd focused more on exercise in your youth, don't feel discouraged. Research has found that it's never too late to see the benefits of exercise on your cognitive abilities.
Even after 50, however, starting on a regular exercise routine is still a great idea, even if it's never been a key aspect of your life to that point.
Ronald Petersen, MD, PhD, director of the Alzheimer's Disease Research Center at the Mayo Clinic, is the lead author on a study that shows people who are experiencing mild cognitive impairment can still see benefits from exercising.
"We need not look at aging as a passive process; we can do something about the course of our aging," Petersen says. "So if I'm destined to become cognitively impaired at age 72, I can exercise and push that back to 75 or 78. That's a big deal."
The most important part of creating an exercise routine is to do something you can stick to.
Walking and riding a stationary bike are the main types of exercise in each of these studies cited. Interestingly, Beth Howard of AARP also recommends trying yoga. You don't have to pay a small fortune to join a yoga studio either. Apps like Gaia, YogaGlo and Yoga Studio allow you to practice yoga anywhere.
If yoga and running aren't your style, consider resistance exercises such as weight lifting. Carrie Cuttler and fellow researchers report in the journal Sport, Exercise, and Performance Psychology that resistance training bolsters prospective memory, which is what helps you remember your to-do list.
Memory loss doesn't have to be a given as you age. Understanding how working out affects your brain can help you retain your memory, fight mental illness and live a happier life for longer.
Exercise is an important factor in how our brains age, and VitaYears™ Memory Support Supplement can take those health benefits up another level. Whether you prefer aerobic exercises, resistance training or yoga, you can see significant improvement in your cognitive abilities just by getting your body moving.


text:
Parochlus duséni är en tvåvingeart som beskrevs av Lars Zakarias Brundin 1966. Parochlus duséni ingår i släktet Parochlus och familjen fjädermyggor. Inga underarter finns listade i Catalogue of Life.

Källor

Fjädermyggor
duséni

redpajama_set_name: RedPajamaWikipedia


I just pocketed $40 from Amazon! How much have you saved today?
Tomorrow is Prime Day on Amazon. What does that mean? Basically, it's their own "Christmas in July" (or to be more specific, "Black Friday in July.").
In other words, it's a deal extravaganza, and it starts Tuesday, July 12, at 3 am EDT. All of the deals are for Prime members only, but in case you don't have Prime, you can sign up for a free trial here.
While I'm not planning on staying up for the kickoff, I was able to score a few great deals in advance.
This one is super easy. If you've got Alexa, use her to shop on Amazon and get $10 off any item over $20. Offer expires at 11:59 p.m. (PST) July 12, 2016.
Stock up on something you always buy – this is an easy $10!
Another easy stock-up situation. Use code PRIMEDAY25 for a $25 future credit towards beauty products.
This one is a little confusing because the discount doesn't show up in your total. However, as long as the words "promotion applied" appear (see the gray box below), you're good. You'll get the credit after your order ships.
Um, wow. This is super first-world, but I went ahead and did it because a) I love a deal and b) I am super curious about this! It's like a "panic button"!
I'm excited that I just saved $40! Did you take advantage of any of these deals? Did you find any others? I'd love to know!
Yes! I used Alexa to buy sunscreen for my upcoming "glamping" trip organized by my favorite camp buddy and 21 of our friends and family! Plus I used my Discover It card which has 5% back on Amazon this month, and that will double at the end of year! Stack those deals!
Wait, I just saved $40 more! I bought a Peet's card ($20 for $25) a Spafinder card I will use towards a yoga retreat ($100 for $80) and a $50 Amazon gift card that will get me a $10 credit. And I will get $15 cashback from Discover! So far I have gotten $200 worth of stuff for $148,50. Not bad.
Hottie – Put that $50 towards something fun (like a big b-day trip)!
My husband did the shopping, so I'm not sure how much he saved, but an adult sized unicorn onsie is apparently on its way.


Wound Management Products Market are generally as categorized as traditional and advanced. Advanced wound care products are very important in the treatment of the diabetic ulcers. Standard treatment for this includes infection control, local ulcer care and education on foot cares. The advanced wound care products market, in terms of revenue, was valued at US$ 8,400 Mn in 2017 and is expected to reach US$ 11,500 Mn by 2021.
Some of the key participants of global advanced wound care products market are Smith & Nephew, 3M Healthcare, Mölnlycke Healthcare AB, ConvaTech, Hollister Incorporated, Lohmann & Rauscher, Nitto Denko, Advanced Medical Solutions, DeRoyal Industries, Genewel Co., Ltd., Organogenesis Inc., Laboratoires URGO, BSN Medical, Medtronic, B.Braun (B. Braun Melsungen AG), Coloplast A/S, Paul Hartmann, Kinetic Concepts, Inc., Medline Industries, Inc. and Winner Medical Co., Ltd. amongst others.
Surgical wound is the wound made by a cutting instrument and made in sterile environments where many variables such location, size, bacteria and nature wounds can be controlled. Demand for the wound care needs has been increasing from the last decade. In order to satisfy them, many manufactures are developing application specific products which can be used in the acute wounds, chronic wounds and surgical wounds.
There are more than 4.5 million pressure ulcers cases in the world which requires treatment every year. Wound management products are also used for the treatment such as melanomas, carcinomas, amputations and other complicated cancer treatments.
High expenses in advanced wound care can critically impact the market. It has been reported that wounds accounts for approximately 4% of total health systems cost. The breakdown of wound costs in the acute setting is the material cost which is 15 to 20%, nurse time of about 30 to 35% and hospitalization which ranges above 50%.
Advanced technologies to provide many opportunities in the advanced wound care products market. For instance., electrical stimulation is the application which involves use of electric current to help encourage wound closure and activate cell growth. It has been observed that, patients with spinal cord injuries took 13 days to heal with electrical stimulation in comparison with 31.5 days for standard treatment.
Asia pacific in the advanced wound care products accounted for US$ 1,770 Mn in the 2017 and to represent significant growth opportunities in the future. Rising number of surgeries due to the chronic wounds has led to high demand for the technologically advanced services which will lead more usage of advanced wound care products. Along with this, rising disposable income in this region has also made the advanced medical treatments more affordable to the people.
The advanced wound care products market report provides both, qualitative and quantitative research of the market, as well as it helps to gain important insights of current and future market and trending technologies adopted by the key players. The report also offers extensive research on the key players in this market and detailed insights on the competitiveness of these players. The key business strategies such as mergers and acquisitions(M&A), affiliations, collaborations, and contracts adopted by the major players are also recognized and analyzed in the report. For each company, the report recognizes their manufacturing base, competitors, product type, application and specification, pricing, and gross margin.


Summer is such a great time to fill your diet with fresh fruit. It's also a great option to have at a party or shower. Here's a great dip to include with your next fruit tray or just to have on hand for you & your family!
It's low fat too so it's a guilt free treat!
Prepare Dream Whip according to package directions. In a separate bowl, prepare the jello according to directions, but don't allow it to completely set. Only refrigerate for 45 minutes to an hour - until thickened, but still a liquid. Fold jello & strawberry yogurt into the whipped topping. Refrigerate for 2 more hours, until thickened.
Note: If it gets runny, you can whip it for 2 minutes on high & it will bring it right back to life!
This looks so good! I could probably eat it with just a spoon and no fruit haha.
YUM! Thanks for sharing - I'm always looking for skinny recipes. Pinned it!


The Nissan Safari (Patrol) is considered one of the toughest SUVs around the world and is used for a variety of applications from mudding to military. This one comes with a variety of power options including sunroof to go with an adjustable 4WD drive-train. This is a great option for RSMCs (Rural and Suburban Mail Carriers) who need a tougher or larger vehicle for those bigger routes.
This one's currently on it's way fresh from Japan, reserve it before someone else does!
At RightDrive we accept trade-ins of all types.
Ask about our RSMC buy back guarantee, and extended service plans.


Kerkyra 2 L -1.5, -2.0 Under 9.0 N.A.
Panionios 3 W 0.0, +0.5 Under 9.0 N.A.
Olympiakos 1 L +1.5, +2.0 Under 9.0 N.A.
Iraklis 1 L 0.0, -0.5 Over 9.0 N.A.
Panathinaikos 2 L -1.0, -1.5 Under 9.5 N.A.
Panathinaikos 1 D +0.5 Draw 9.0 N.A.
Athletic Bilbao 3 L +0.5 Over 9.5 N.A.
Apollon Smyrnis 3 L -1.5 Under N.A.
Panathinaikos 1 W 0.0, +0.5 Over 9.0 N.A.
FK Qabala 1 W -1.0 Under 9.5 N.A.
AZ W 0.0, -0.5 Over 9.5 N.A.


text:
Biden Is Losing It, Media Knows It And Sheltering Him. Kamala To Step-Up
Posted on September 23, 2021 September 23, 2021 Author ThePatriotSignal Comments(2)
President Biden declined to answer any questions from news reporters in the Oval Office yesterday while holding a meeting with UK Prime Minister Boris Johnson.
The pair had a brief conversation which discussed climate change, transport infrastructure as well as lifting the ban on British beef. At the end of their conversation Johnson asked, "Would it be okay if we have just a couple of questions, just a couple?", looking over to Biden, who replied, "Good luck."
A reporter from the UK then pressed the president on whether the UK would be at "the back of the queue" for a post-Brexit trade deal, as former President Obama said on a state visit to the UK in 2016 which took heavy criticism from Brits, to which the president did not answer.
MOMENTS AGO: Staff orders news media to leave The White House as UK Prime Minister Boris Johnson speaks mid-sentence pic.twitter.com/MXg2m0o1hq
— Breaking911 (@Breaking911) September 21, 2021
The UK leader then called on reporters from his own press pool to which he answered several questions, to which Biden sat and watched on. White House Staff interrupted Johnson mid-sentence to rush all reporters out of the Oval Office, to the angry of the reporters themselves.
The president has come under fire for consistently implying that he is not in charge of when and where he is able to take questions from news reporters. This is something that has never been seen before under any previous administration or president.
Biden's refusal to take questions came after he addressed the UN General Assembly, in which he called for the world to come together to tackle the many issues facing the world in the coming years, such as climate change, hunger and terrorism. He also said it is vital that the world works together to defeat Covid-19 and bring an end to the pandemic.
Democrats and news media alike both criticized former President Trump for not taking any media questions at appearances during his four years in office. Let's see if they begin holding Joe Biden and his administration to the same standards.
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Biden FURIOUS Over Movie About Hunter Biden
Nonips
Her dad said in an interview that she was NOT born in America so she cannot be president and to many say she is incompetent to be in charge of anything as she said I Fuk, and sucked my way the whole biden administration need arrested and prosecuted for Terrorist, Treasonous. Law breaking criminals much less stealing from Taxpayers
Reply to Nonips
For the love of God, We all know all that's said about her is TRUE. We might as well keep Biden at least he is known to be an idiot ! She started off early in her career having sex with whoever could help her advance. The entire country is in a death spiral
because of these two idiots. COME BACK TRUMP!!