text:
\section{Introduction}

The coefficient inverse problem of identifying the unknown leading coefficient of an elliptic equation is the basis of electrical impedance tomography (EIT) and electrical resistivity tomography (ERT) non-invasive technique for investigation of the internal structure of bodies from voltage and current boundary measurements. The inverse problem for elliptic equation  has numerous applications in geophysics \cite{Chambers1999, Chambers2006}, medical imaging \cite{Bodenstein2009a, Putensen2019} and nondestructive testing \cite{McCarter1989, Karhunen2010}.

The coefficient inverse problem for elliptic equation  was first posed in \cite{Calderon1980}. It is known to be nonlinear and ill-posed and, therefore, special algorithms are needed to solve such problems numerically. Also, the issues of stability and uniqueness are crucial for the development of robust numerical algorithms. The uniqueness of a large class of isotropic coefficients was considered in many works \cite{Sylvester1986, Novikov1988, Nachman1996, Brown1997, Astala2006}. In \cite{Astala2006}, the optimal regularity condition for coefficient in two dimensions was obtained. The stability of the inverse problem was investigated in \cite{Alessandrini1988, Allers1991, Barcelo2001a}.
To achieve stability, it is required to place some constraints on the coefficient.
For example, in \cite{Alessandrini1988}, a logarithmic stability estimate was obtained.

There are many reconstruction methods and procedures for solving the inverse problem \cite{Cheney1999, Borcea2002}. These methods can be divided into two main groups: non-iterative and iterative methods.
The non-iterative methods include  factorization \cite{Bruhl2000, Gebauer2007},  layer stripping \cite{Somersalo1991, Sylvester1992},  D-bar methods \cite{Siltanen2001, Knudsen2009}, NOSER \cite{Cheney1990}.
The factorization method is a method based on an explicit criterion for detecting inhomogeneities inside bodies \cite{Bruhl2001}.
In layer-stripping algorithms, one needs first to find the unknown coefficient on the boundary of the body and then progress inside layer by layer.
The D-bar method is based on evaluating a nonlinear Fourier transform of the coefficient from EIT data and the inversion of the transform.
NOSER (Newton's One-Step Error Reconstructor) is an algorithm based on the minimization of an error functional, but it takes only one step of Newton's method with constant coefficient as an initial guess. Most of the calculations, including the gradient of the functional, can be done analytically.

Iterative methods for solving the inverse problems include minimization algorithms based on either least squares \cite{Yorkey1987, Dobson1992, Borcea2001} or equation-error \cite{Wexler1985, Kohn1990} formulations.
Least squares methods usually iteratively minimize the norm of the difference between electrical potential due to the applied current and the measured potential on the boundary.
Equation-error approaches are also known as variational methods, derived from Dirichlet and Thompson variational principles \cite{Borcea2002}.
The convergence of iterative methods can be ensured using some regularization technique, which typically means that a specific regularization term is added to a functional.
Iterative methods require calculating first and, in some optimization methods, second derivatives of the objective functionals.
The adjoint method can be used to calculate both derivatives \cite{Dorn2000} efficiently.

In many applications of the coefficient inverse problem, one can assume that objects under investigation contain several materials with piecewise constant coefficients
This assumption allows us to reduce the inverse problem of recovering the distribution of  coefficient inside the body to the problem of reconstructing shapes of the materials.
Among the shape reconstruction methods, the level set method is known to be the most powerful one.

The level set method was first proposed in \cite{Osher1988} to track evolving interfaces.
There, an evolving domain is represented by a  continuous level set function. Then, the motion of this domain is expressed via a Hamilton-Jacobi equation for the level set function.
In \cite{Santosa1996}, this approach was first used to solve inverse problems where the desired unknown is a  characteristic function of some geometry.
Next, the unknown is represented by a level set function.
The evolution of the level set function minimizes a functional and leads to a solution of the inverse problem.

 In \cite{Osher2001}, the level set method was used to solve the inverse problem associated with shape optimization for the eigenvalue problem for the Laplace equation. This approach was used to solve the inverse problem for electrical impedance tomography in \cite{Ito2001, Chan2004, Chung2005}.
More recently, the level set based methods were applied to elliptic inverse problems in \cite{Agnelli2018, Lin2018, Liu2019}. Surveys on level set methods for solving inverse problems can be found in \cite{Tai2004, Burger2005a, Gibou2018}.

In this paper, we present a numerical algorithm based on the level set idea for solving the inverse problem associated with electrical impedance tomography.
We assume that the leading coefficient is a piecewise constant function, and the values of coefficients are known.
The main idea is to implicitly represent the interface between regions with known coefficients as the zero value of an auxiliary elliptic equation's solution.
The algorithm is based on minimizing the squared norm of the difference between potentials due to applied currents and measured potentials. We previously used this approach to successfully recover a piecewise constant lower coefficient  \cite{Kolesov2019} and the right-hand side \cite{Ivanov2019} of an elliptic equation.

The paper is organized as follows. In the next section, we introduce the inverse problem. The details of the proposed reconstruction algorithm, the cost functional, and the calculation of its derivative using the adjoint method are discussed in section 3. Then, we present a series of numerical experiments to show our algorithm's ability to recover the unknown interface from noisy data.
The final section is the conclusion.

\section{Problem statement}

Let $\Omega \subset \mathbb{R}^2=d$ ($d=2,3$) be the bounded domain  with sufficiently smooth boundary $\partial\Omega$ and $\sigma(\bm x) \in L^{\infty}(\Omega)$ is the coefficient such that $\sigma(\bm x) \geq \sigma_0 > 0$.
We consider the boundary value problem for elliptic equation
\begin{equation}
\label{eq:forward}
- \nabla \cdot \sigma (\bm x) \nabla u = 0, \quad \bm x \in \Omega,
\end{equation}
\begin{equation}
\label{eq:g}
\sigma (\bm x)  \pdn{u} = g (\bm x), \quad \bm x \in \partial\Omega,
\end{equation}
where $u(\bm x)$ is the electrical potential, $\bm n$ is  the unit outward normal to $\partial \Omega$ and $g(\bm x)$ is the applied current density. In addition, both $u(\bm x)$ and  $g(\bm x)$ must satisfy the following constraints
\begin{equation}
\label{eq:uds}
\int_{\partial \Omega} u(\bm x) ds = 0,
\end{equation}
\begin{equation}
\label{eq:gds}
\int_{\partial \Omega} g(\bm x) ds = 0.
\end{equation}
The problem (\ref{eq:forward})--(\ref{eq:gds}) correspond to the continuum model for electrical impedance tomography.
The inverse problem is to determine the distribution of  coefficient $\sigma(\bm x)$ inside $\Omega$ using a set of given values of applied current density $g(\bm x)$ on $\partial \Omega$ and the corresponding measured values of potential $u( \bm x)$ on $\partial \Omega$ . The set of $g(\bm x)$ and $u(\bm x)$ on $\partial \Omega$  is also known as the Neumann-to-Dirichlet or current-to-voltage map in problems of electrical impedance tomography.  It is well known that this inverse problem does not have unique solution. Therefore, we need to narrow the class of admissible solutions.

In many applications, such as medical imaging, geophysics, and nondestructive testing, it is a-priori known that the object to be imaged contains several materials with piecewise constant conductivities. Let $N$ be the number of materials, $D_i$ be the subdomain containing material with  $\sigma_i$, ($i = 1, \dots, N$),  $\Omega = \cup_{i=1}^N D_i$.  Then,  the distribution of  coefficient $\sigma (\bm x)$ can be represented as
\begin{equation*}
\sigma (\bm x) = \sum_{i=1}^N  \sigma_i \chi_{D_i}(\bm x),
\end{equation*}
where $\chi_{D_i}(\bm x)$  ($i=1,\dots, N$) is the characteristic function of the subdomain $D_i$:
\begin{equation}\label{eq:chi}
\chi_i(\bm x) = \begin{cases}
1, & \bm x \in D_i,\\
0, & \bm x \in \Omega \backslash D_i.
\end{cases}
\end{equation}
In this work, for simplicity  we restrict ourselves with only two materials with conductivities $1$ and $2$ an denote by $D$ the subdomain with coefficient $2$. Therefore, the coefficient distribution $\sigma (\bm x)$ can be written as
\begin{equation} \label{eq:coef}
\sigma (\bm x) = 1 + \chi_{D}(\bm x)),
\end{equation}
with $\chi_{D} (\bm x)$ is the characteristic function of the subdomain $D$.

In this case, the inverse problem of determining the distribution of coefficient $\sigma (\bm x)$ reduces to the reconstruction of the shape of the subdomain $D$.

\section{Reconstruction algorithm}

In this section, we propose a new algorithm for reconstruction of the coefficient $\sigma (\bm x)$.
The algorithm is based on minimizing  a cost functional  using a gradient method.

\subsection{Cost functional}

We introduce a level set function $q(\bm x)$, which describes subdomain  $D$ as follows
\begin{equation}\label{eq:levelset}
\begin{cases}
q(\bm x) \geq 0, & \bm x \in D,\\
q(\bm x) < 0, & \bm x \in \Omega \backslash D,
\end{cases}
\end{equation}
and the Heaviside function $H(q)$
\begin{equation}
\label{eq:heaviside}
H(q(\bm x)) =
\begin{cases}
1, & q(\bm x) \geq 0, \\
0, & q(\bm x) < 0.
\end{cases}
\end{equation}
Then, the coefficient $\sigma (\bm x)$ (\ref{eq:coef}) can be defined as
\begin{equation}\label{eq:c_new}
\sigma (\bm x) = 1 + H(q(\bm x)).
\end{equation}
Clearly, to determine  $\sigma (\bm x)$  it is sufficient to identify the level set function $q(\bm x)$. Many different level set functions were used to solve the inverse problem of electrical impedance tomography \cite{Dorn2006}. Commonly, the level set function can be determined using a signed distance function, which can be updated by solving the Hamilton-Jacobi equation.

In this work, the level set function is the solution of the following elliptic equation:
\begin{equation} \label{eq:aux}
- \gamma \Delta q + q = f(\bm x), \quad \bm x \in \Omega,
\end{equation}
\begin{equation} \label{eq:aux_bc}
q = 0, \quad \bm x \in \partial\Omega,
\end{equation}
where $\gamma = \mbox{const} > 0$ is the parameter.  The key of our approach is to determine the right hand side $f(\bm x)$ such as the solution $q(\bm x)$ of problem (\ref{eq:aux}), (\ref{eq:aux_bc}) describes the desired coefficient $\sigma (\bm x)$ (\ref{eq:c_new}).
In fact, the parameter $\gamma$ can be seen as a smoothing parameter for function $f(\bm x)$ and $q(\bm x)$ is smoothed out function. For $\gamma = 0$, we have $q(\bm x) = f(\bm x)$.  Note that we can employ the different differential operator instead of $-\Delta$.

Let $M$ be the number of measurements. For $1 \leq j \leq M$, let $g_j(\bm x)$ be a given applied current density on $\partial \Omega$ and $m_j(\bm x)$ be the corresponding measurement of the potential on $\partial \Omega$.
To find $f(\bm x)$, we minimize the following least-squares cost functional
\begin{equation} \label{eq:J}
J(f) = \frac{1}{2} \sum_{j=1}^M \int_{\partial \Omega} \left|u_j(\bm x; f) - m_j(\bm x) \right|^2 \, \mathrm{d}s.
\end{equation}
where $u_j(\bm x; f)$, $j = 1, \dots, M$ are the solutions of the problems
\[
- \nabla \cdot \sigma (\bm x) \nabla u_j = 0, \quad \bm x \in \Omega,
\]
\[
\sigma(\bm x)  \pdn{u_j} = g_j (\bm x), \quad \bm x \in \partial\Omega.
\]
Here, $\sigma(\bm x) $ is a coefficient corresponding to $f(\bm x)$ via (\ref{eq:c_new})--(\ref{eq:aux_bc}).
Note that in (\ref{eq:J}) the functions $m_j(\bm x)$ correspond to the solution of above problem for the desired coefficient.
Both $u_j(\bm x; f)$ and $m_j(\bm x)$ must satisfy (\ref{eq:uds}).

\subsection{Variational formulations}

For discretization in space we use the finite element method, so we need to obtain variational forms of boundary value problems  (\ref{eq:forward})--(\ref{eq:gds}) and   (\ref{eq:aux}), (\ref{eq:aux_bc}).
First, we define the functional spaces:
\[
V = \left\{v \in H^1(\Omega): \int_{\partial \Omega} v \, d\bm s = 0\right \},  \quad Q = \left\{v \in H^1(\Omega): v(\bm x ) = 0,  \, \bm x \in \partial \Omega  \right \},
\]
where $H^1(\Omega)$ is Sobolev space.
We multiply equation (\ref{eq:forward}) by a test function $v \in V$, integrate the resulting equation over $\Omega$ and perform integration by parts to eliminate second-order derivative of $u$:
\begin{equation*}
\int_{\Omega} \sigma \nabla u \,  \nabla v \, d\bm x - \int_{\partial \Omega} \sigma \pdn{u} \, v \, d\bm s = 0.
\end{equation*}
Taking into account the boundary condition (\ref{eq:g}), yields
\begin{equation*}
\int_{\Omega} \sigma \nabla u \,  \nabla v \, d\bm x = \int_{\partial \Omega} g \, v \, d\bm s.
\end{equation*}
Next, we define the following bilinear form
\begin{equation*}
a(u,v) = \int_{\Omega} \sigma \nabla u \,  \nabla v \, d\bm x,
\end{equation*}
and linear form
\begin{equation*}
L_a(v) = \int_{\partial \Omega} g \, v \, d\bm s.
\end{equation*}
Then the variational formulation of problem (\ref{eq:forward})--(\ref{eq:gds}) read as: find $u \in V$ such as
\begin{equation}
\label{eq:forward_var}
a(u,v) = L_a(v),  \quad \forall v \in V.
\end{equation}

The variational formulation of problem  (\ref{eq:aux}), (\ref{eq:aux_bc}) is derived similarly: find $q \in Q$ such that
\begin{equation}
\label{eq:aux_var}
b(q, w) = L_b(w),  \quad \forall w \in Q,
\end{equation}
where
\begin{equation*}
b(q, w) = \int_{\Omega} \gamma \nabla q \,  \nabla w \, d\bm x +  \int_{\Omega}  q \, w \, d\bm x,
\end{equation*}
\begin{equation*}
L_b(w) = \int_{\Omega}  f \, w \, d\bm x,
\end{equation*}

Note that both bilinear forms $a(u,v) $ and $b(q, w)$ are symmetric
\begin{equation*}
a(u,v) = a(v,u), \quad b(q, w) = b(w, q).
\end{equation*}
This property is useful for calculating the gradient of $J(f)$.

\subsection{Gradient of functional}

To minimize the functional $J(f)$ we use a gradient based method.Thus, we need to calculate the gradient of $J(f)$ with respect to the function $f(\bm x)$. First, note that  $J(f)$ is the functional of  the functions $u_j(\bm x)$, $j = 1, \dots, M,$ which depend on  coefficient $\sigma(\bm x)$. In turn, $\sigma(\bm x)$ is the function of the level set $q(\bm x)$ via (\ref{eq:c_new}). Finally, the function $q(\bm x)$  also depends on the objective function $f(\bm x)$ by (\ref{eq:aux}). Therefore, by the chain rule, to compute the gradient of $J$ we use the following equation
\begin{equation}
\label{eq:dJdf1}
\d{J}{f} = \sum_{j=1}^{M} \pd{J}{u_j} \pd{u_j}{\sigma} \pd{\sigma}{q} \pd{q}{f}.
\end{equation}
The terms $\partial J / \partial u_j$ and $\partial \sigma / \partial q$ are straightforward to compute using the following equations:
\begin{equation}
\label{eq:djdu}
\pd{J}{u_j}  = u_j - m_j,
\end{equation}
and
\begin{equation}
\label{eq:dsigmadq}
\pd{\sigma}{q} = \delta(q(\bm x)),
\end{equation}
where $\delta (\cdot)$ is the Dirac delta function.
 By contrast, $\partial u_j / \partial \sigma$ and $\partial q / \partial f$ are rather difficult to compute.
So, we use the adjoint method to compute the gradient of $dJ/df$.

First, by taking the derivative of (\ref{eq:forward_var}) with respect to $q$, we get
\[
\pd{a}{u_j} \pd{u_j}{\sigma}  + \pd{a}{\sigma} = 0, \quad j = 1, \dots, M.
\]
Since $\partial a / \partial u_j$ is invertible, the following equation for $\partial u_j/\partial\sigma$ is obtained
\begin{equation}
\label{eq:dudsigma}
\pd{u_j}{\sigma} = - \left( \pd{a}{u_j} \right)^{-1} \pd{a}{\sigma} .
\end{equation}
By analog, we can derive the equation for $\partial q/\partial f$ from (\ref{eq:aux_var})
\begin{equation}
\label{eq:dqdf}
 \pd{q}{f}  = \left( \pd{b}{q} \right)^{-1} \pd{L_b}{f}
\end{equation}
Substituting  (\ref{eq:dudsigma}) and (\ref{eq:dqdf}) into (\ref{eq:dJdf1}) yields
\begin{equation*}
\d{J}{f} =- \sum_{j=1}^{M} \pd{J}{u_j} \left( \pd{a}{u_j} \right)^{-1} \pd{a}{\sigma}  \pd{\sigma}{q} \left( \pd{b}{q} \right)^{-1} \pd{L_b}{f}.
\end{equation*}

Now we take adjoint of the above equation
\begin{equation}
\label{eq:dJdf2}
\d{J}{f}^*=\pd{L_b}{f}^* \left( \pd{b}{q} \right)^{-1*}   \pd{\sigma}{q}^* \sum_{j=1}^{M} \pd{a}{\sigma}^*  \left(  \pd{a}{u_j} \right)^{-1*} \d{J}{u_j}^*
\end{equation}
Next, we define new variables $z_j$ as follows
\[
z_j = \left( \pd{a}{u_j} \right)^{-1*} \, \d{J}{u_j}^*, \quad j = 1, \dots, M,
\]
and obtain the following adjoint equations associated with (\ref{eq:forward_var})
\begin{equation*}
\pd{a}{u_j}^{*} \, z_j = \d{J}{u_j}^*, \quad j = 1, \dots, M.
\end{equation*}
Taking into account that the bilinear form $a$ is symmetric and  (\ref{eq:djdu}) we get
\begin{equation}
\label{eq:forward_adjoint}
a(z_j,v) = \int_{\partial \Omega} (u_j - m_j) \, v \, ds, \quad j = 1, \dots, M.
\end{equation}
Substituting the adjoint solutions $z_j$ into (\ref{eq:dJdf2}) leads to
\begin{equation}
\label{eq:dJdf3}
\d{J}{f}^*=\pd{L_b}{f}^* \left( \pd{b}{q} \right)^{-1*}   \pd{\sigma}{q}^* \sum_{j=1}^{M} \pd{a}{\sigma}^*   z_j
\end{equation}

Next, we define another adjoint variable $\lambda$ as follows
\begin{equation*}
\lambda = \left( \pd{b}{q} \right)^{-1*}   \pd{\sigma}{q}^* \sum_{j=1}^{M} \pd{a}{\sigma}^*   z_j
\end{equation*}
and the adjoint equation associated with (\ref{eq:aux_var})
\begin{equation*}
\pd{b}{q}^{*}  \lambda = \pd{\sigma}{q}^* \sum_{j=1}^{M} \pd{a}{\sigma}^*   z_j
\end{equation*}
The bilinear form $b$ is also symmetric, so we have the adjoint problem associated with (\ref{eq:aux_var})
\begin{equation}
\label{eq:aux_adjoint}
b(\lambda, w) =  \delta(q) \sum_{j=1}^M \int_{\Omega} \mathop{\rm \nabla }\nolimits u_j \cdot \mathop{\rm \nabla }\nolimits z_j \, w \, dx
\end{equation}

Finally, we substitute the adjoint solution $\lambda$ into (\ref{eq:dJdf3})
\begin{equation*}
\d{J}{f}^* = \pd{L_b}{f}^* \, \lambda
\end{equation*}
which yields the final equation for computing the gradient of functional $J$ with respect to $f$:
\begin{equation}
\label{eq:dJdf4}
\d{J}{f} = \int_{\Omega} \lambda \, w \, dx.
\end{equation}
To compute the gradient $dJ/df$ (\ref{eq:dJdf4})  we need to solve the adjoint problems (\ref{eq:forward_adjoint}) and (\ref{eq:aux_adjoint}).

\subsection{Algorithm}

In this section, we propose the new algorithm for solving our coefficient inverse problem. It is based on finding $f(\bm x)$ such that the solution  $q(\bm x)$ of problem (\ref{eq:aux}), (\ref{eq:aux_bc}) describe the desired coefficient  $\sigma (\bm x)$ via (\ref{eq:c_new}). To find $f(\bm x)$ we minimize the functional $J(f)$(\ref{eq:J}) using a gradient based iterative procedure:

\begin{algorithm}
\caption*{\textbf{Reconstruction algorithm}}
\begin{algorithmic}[1]
\STATE{
\textbf{input} $M$, $g_j$, $m_j$, $j=1,\dots,M$, $f^0$, $ \gamma, \alpha, tol, K$ \COMMENT{$K$ is a large number}
}
\STATE{
\textbf{output} $\sigma(\bm x)$
}
\FOR{$k=0$ \TO $K$}
\STATE{
$q^k(\bm x) \leftarrow$ the solution of  (\ref{eq:aux_var})
}
\STATE{
$\sigma^k(\bm x) \leftarrow 1 + H(q^k(\bm x))$
}
\FOR{$j=1$ \TO M}
\STATE{
$u_j^k \leftarrow$ the solution of  (\ref{eq:forward_var}) for given $g_j(\bm x)$
}
\ENDFOR
\STATE{
$J^k(f^k) \leftarrow \frac{1}{2} {\displaystyle\sum_{j=1}^M \int_{\partial \Omega} \left|u_j(f;\bm x) - m_j(\bm x) \right|^2 \, \mathrm{d}s }$
}
\FOR{$j=1$ \TO M}
\STATE{
$z_j^k(\bm x) \leftarrow$ the solution of adjoint problem  (\ref{eq:forward_adjoint}) for given $m_j(\bm x)$
}
\ENDFOR
\STATE{
$\lambda^k(\bm x) \leftarrow$ the solution of adjoint problem  (\ref{eq:aux_adjoint})
}
\STATE{
$\cfrac{dJ^k}{df^k}  \leftarrow  \int_{\Omega} \lambda \, w \, dx.$
}

\IF{ $\left\|\cfrac{dJ^k}{df^k} \right \|_{L^{\infty}(\Omega)}  < tol$}
\STATE{
$\sigma(\bm x) \leftarrow \sigma^k(\bm x)$
}
\STATE{\textbf{stop}}
\ENDIF
\STATE{
$f^{k+1} \leftarrow f^k - \beta^k \cfrac{dJ^k}{df^k} $
}
\COMMENT{step size $\beta^k $ is calculated by line search}
\ENDFOR
\STATE{
$\sigma(\bm x) \leftarrow \sigma^k(\bm x)$
}
\end{algorithmic}
\end{algorithm}

The stopping criteria $tol$ will be discussed later. The numerical implementation of the reconstruction is performed in Python. For solving partial differential equations, we use an open-source computing platform FEniCS \cite{AlnaesBlechta2015a}.

\section{Numerical experiments}

In this section, we examine the effectiveness of the proposed algorithm. A series of numerical experiments have been performed to confirm that our algorithm can recover the interface between two materials with piecewise constant coefficients from computationally simulated data.

\subsection{Implementation details}

For numerical computation, we consider unit circle domain $\Omega$.
To generate boundary data for potential $m$ we employ sets of equidistant electrodes along the boundary with width $\theta = \pi / 20$. See Figure \ref{fig:electrodes} for a scheme of distribution of 2, 4, and 6 electrodes. The following current injection pattern is used for measurements: we set $g(\bm x) = 1$ on one electrode, $g(\bm x) = -1$ on the opposite electrode, and $g(\bm x) = 0$ elsewhere.
 In this manner, we obtain one independent measurement for two electrodes, two measurements for four electrodes, etc.

\begin{figure
\begin{center}
\subfloat[\label{fig:2elecrodes}]{
\begin{tikzpicture}[scale=0.55]
\draw [line width=1.0,fill=black!5!white] (0,0) circle (3);
\draw [line width=3.0, red] (-0.23,2.99) -- (0.24,2.99);
\draw [line width=3.0, red] (0.23,-2.99) -- (-0.23,-2.99);
\end{tikzpicture}
}\hspace{1em}
\subfloat[\label{fig:4elecrodes}]{
\begin{tikzpicture}[scale=0.55]
\draw [line width=1.0,fill=black!5!white] (0,0) circle (3);
\draw [line width=3.0, red] (-0.23,2.99) -- (0.24,2.99);
\draw [line width=3.0, red] (2.99,0.23) -- (2.99,-0.23);
\draw [line width=3.0, red] (0.23,-2.99) -- (-0.23,-2.99);
\draw [line width=3.0, red] (-2.99,0.23) -- (-2.99,-0.23);
\end{tikzpicture}
}\hspace{1em}
\subfloat[\label{fig:6elecrodes}]{
\begin{tikzpicture}[scale=0.55]
\draw [line width=1.0,fill=black!5!white] (0,0) circle (3);
\draw [line width=3.0, red] (-0.23,2.99) -- (0.24,2.99);
\draw [line width=3.0, red] (2.47,1.7) -- (2.71,1.29);
\draw [line width=3.0, red] (2.71,-1.29) -- (2.47,-1.7);
\draw [line width=3.0, red] (0.23,-2.99) -- (-0.23,-2.99);
\draw [line width=3.0, red] (-2.47,-1.71) -- (-2.71,-1.29);
\draw [line width=3.0, red] (-2.71,1.29) -- (-2.47,1.7);
\end{tikzpicture}
}
\end{center}
\caption{Distribution of equidistant electrodes (in red) along the boundary of the domain: a) 2 electrodes, b) 4 electrodes, c) 6 electrodes}
\label{fig:electrodes}
\end{figure}

In our experiments, we reconstruct two images: ellipse and two circles with different diameters.
To generate boundary data $m_j(\bm x)$, $j = 1, \dots, M$, $\bm x \in \partial \Omega$ for each test case we solve the forward problem (\ref{eq:forward})--(\ref{eq:gds}) using an unstructured mesh adapted to represent internal structure of imaged materials. For reconstruction purposes, we use a different mesh with no knowledge of imaged materials. All meshes are created using the open source software for mesh generation \texttt{Gmsh}.
The computational meshes used for data generation and reconstruction are shown in Figure \ref{fig:meshes}. Note that the mesh in Figure \ref{fig:mesh} is used to reconstruct the images.

\begin{figure}
\begin{center}
\subfloat[\label{fig:mesh_ellipse}]{\includegraphics[width=0.33\textwidth]{mesh_ellipse}}
\subfloat[\label{fig:mesh_circles}]{\includegraphics[width=0.33\textwidth]{mesh_circles}}
\subfloat[\label{fig:mesh}]{\includegraphics[width=0.33\textwidth]{mesh}}
\end{center}
\caption{Computational meshes used for: (a) data generation for ellipse,  (b) data generation for circles,  (d) reconstruction}
\label{fig:meshes}
\end{figure}

For numerical solution we need to approximate the Heaviside function $H(q(\bm x))$ (\ref{eq:heaviside}) and the delta function $\delta (q(\bm x))$ by the following smooth functions $H_{\alpha} (q(\bm x))$ and $\delta_{\alpha} (q(\bm x))$
\begin{equation}
\label{eq:heaviside_alpha}
H_{\alpha}(q(\bm x)) =
\begin{cases}
0, & q(\bm x) < 0, \\
\displaystyle{\frac{1}{2} - \frac{1}{2} \cos \left( \pi \frac{ q}{\alpha} \right)}, & 0 \leq q(\bm x) < \alpha \\
1, & q(\bm x) \geq \alpha,
\end{cases}
\end{equation}
and
\begin{equation}
\label{eq:delta_alpha}
\delta_{\alpha}(q(\bm x)) =
\begin{cases}
0, & q(\bm x) < 0, \\
\displaystyle{\frac{\pi}{2 \alpha} \sin \left( \pi \frac{ q}{\alpha} \right)}, & 0 \leq q(\bm x) < \alpha \\
0, & q(\bm x) \geq \alpha,
\end{cases}
\end{equation}
where $\alpha > 0$  is smoothing parameter.

We add some uniform noise to the generated data.  More precisely, the noisy data $\tilde{m}_j(\bm x)$ is obtained by adding to $m_j(\bm x)$ a uniform  noise as follows:
\[
\tilde{m}_j(\bm x) = m_j(\bm x)  + \epsilon \|m_j(\bm x) \|_{L^2(\partial \Omega)} \frac{\theta_j(\bm x) }{\|\theta_j(\bm x) \|_{L^2(\partial \Omega)} }, \, j = 1, \dots, M, \, \bm x \in \partial \Omega,
\]
where $\epsilon$ is the noise level, $\theta_j (\bm x)$ is random numbers uniformly distributed on the interval $(-1.0, 1.0)$. Note that the noise is added only on the boundary $\partial \Omega$. The norms $ \|m_j(\bm x) \|_{L^2(\partial \Omega)}$ and $\|\theta_j(\bm x) \|_{L^2(\partial \Omega)}$ are used to scale $\theta_j (\bm x)$. Figure \ref{fig:noisy_data} shows noiseless and noisy data with $\epsilon=0.1$ on the perimeter of the computational domain $\partial \Omega$. Noiseless data is illustrated by blue dashed line, noisy data  by orange solid line. The data is generated for ellipse with 2 electrodes (1 measurement).

\begin{figure}
\begin{center}
\includegraphics[width=0.8\textwidth]{noisy_data}
\end{center}
\caption{Noiseless data $m_1(\bm x)$ (blue dashed line) and noisy data $\tilde{m}_1(\bm x)$ with $\epsilon=0.1$ (orange solid line)}
\label{fig:noisy_data}
\end{figure}

The initial guess for the control function $f(\bm x)$ is chosen as follows
\begin{equation*}
f^0(\bm x) =
\begin{cases}
1, & \bm x \in D^0 , \\
0, & \bm x \in \Omega \backslash D^0,
\end{cases}
\end{equation*}
where $D^0$ is some initial guess. It is taken as a circle  with the radius $r = 0.2$ and the center $\bm x_0 = (0.0, 0.0)$ of the domain $\Omega$. This initial control function $f^0(\bm x)$ is shown in Figure \ref{fig:f0}.
In Figure \ref{fig:q0} we show the functions $q^0(\bm x)$ obtained by solving (\ref{eq:aux_var}) with $\gamma = 0.001$.
The initial guess for the smoothed Heaviside function $H^0_{\alpha}(q(\bm x))$ (\ref{eq:heaviside_alpha}) with $\alpha = 0.01$ corresponding to $q^0(\bm x)$ in Figure \ref{fig:q0} is displayed in Figure \ref{fig:c0}. Note that $H^0_{\alpha}(\bm x)$ is shown in red (equal to 1)  for $q(\bm x) > 0.01$.

\begin{figure
\begin{center}
\subfloat[\label{fig:f0}]{\includegraphics[width=0.33\textwidth]{f0}}
\subfloat[\label{fig:q0}]{\includegraphics[width=0.33\textwidth]{q0_1}} %
\subfloat[\label{fig:c0}]{\includegraphics[width=0.33\textwidth]{c0_1}}
\end{center}
\caption{ The initial guess for: (a) the control function $f^0(\bm x)$,  (b) the function $q^0(\bm x)$,  (c) the smoothed Heaviside function $H^0_{\alpha}(q(\bm x))$  }
\label{fig:initial_fq}
\end{figure}

Figure \ref{fig:fq_gammas} shows the dependence of the level set function $q(\bm x)$ on the parameter $\gamma$. Here, the blue solid line plots  the values of $f(\bm x)$ over a line $(-0.5,0.0)$--$(0.5,0.0)$, the orange dashed  and green dash-dotted lines plot the values of $q(\bm x)$, corresponding to $\gamma=0.0005$, $\gamma=0.001$, and $\gamma=0.005$, respectively, over the same line. We can see that  as the value of $\gamma$ increases, the level set functions $q(\bm x)$ is more smoothed out.

\begin{figure}
\begin{center}
\includegraphics[width=0.8\textwidth]{fq_gammas}
\end{center}
\caption{Dependence of the level set function $q(\bm x)$ on the parameter $\gamma$}
\label{fig:fq_gammas}
\end{figure}

For minimization, we use the steepest descent method.
The stopping criterion depends on the noise level $\epsilon$ as follows:
\[
\left \| \frac{dJ}{df}^k \right \|_{L^{\infty}(\Omega)} \leq \epsilon \beta,
\]
where $k$ is iteration number, $\beta$ is equal to $10^{-9}$.

\subsection{Reconstruction of ellipse}

First, we consider the ellipse object  (Figure \ref{fig:mesh_ellipse}).
We want to examine the dependence of reconstruction results on the number of measurements $M$ and the parameter $\gamma$.
For this, we reconstruct the piecewise constant coefficient from computationally simulated data with uniform noise with $0.01$ noise level.
In Table \ref{tab:ellipse_Je} the first and second columns show the number of measurements $M$ and the parameter $\gamma$, respectively. In the next two columns we show the number of iterations until convergence and the corresponding value of the functional $J$. The last column represent the reconstruction error $\varepsilon$ calculated as follows:
\begin{equation}
\label{eq:err}
\varepsilon = \frac{\| \chi_D(\bm x) - H_{\alpha}(q(\bm x)) \|_{L^2(\Omega)}}{\|\chi_D(\bm x) \|_{L^2(\Omega)}},
\end{equation}
where $\chi_D (\bm x)$ is the characteristic function of ellipse (\ref{eq:chi}) and $H_{\alpha} (q(\bm x))$ is the reconstructed value of the smoothed Heaviside function (\ref{eq:heaviside_alpha}).
As can be seen in Table \ref{tab:ellipse_Je}, the number of measurements $M$ is crucial for reconstruction accuracy. Using only one measurement $M=1$ is clearly not enough to accurately recover the desired coefficient and the number of iteration until convergence, in this case, is significantly larger. In this case, three measurements are sufficient for reconstruction, since using four measurements leads to slightly worse results. As for the parameter $\gamma$, we need to use $\gamma = 0.001$ for the best results. Using less than or greater than $0.001$ leads to increased error $\varepsilon$ and more iterations are needed for convergence of minimization.  Note that in these calculations we use $\alpha = 0.01$.

In Figures \ref{fig:j_err_ellipse_gammas} and \ref{fig:j_err_ellipse_Ms}, we show the evolution of functional $J^k$ and error $\varepsilon^k$ with iterations $k$ for different values of $\gamma$ and $M$, respectively.
In Figure \ref{fig:j_err_ellipse_gammas}, the blue solid, orange dashed, and green dash-dotted lines shows the values of $J^k$ (Figure \ref{fig:j_ellipse_gammas}) and $\varepsilon^k$   (Figure \ref{fig:e_ellipse_gammas}) for  $\gamma = 0.0005$,  $ 0.001$,  and $0.002$, respectively.
Similarly, in Figure \ref{fig:j_err_ellipse_Ms}, the blue solid, orange dashed, and green dash-dotted lines shows the values of $J^k$ (Figure \ref{fig:j_ellipse_Ms}) and $\varepsilon^k$   (Figure \ref{fig:e_ellipse_Ms}) for  $M = 2$,  $3$,  and $4$, respectively.
Reconstructions with different  $\gamma$ are performed with 3 measurements ($M = 3$).
While, $\gamma = 0.001$ is used for reconstructions with different $M$.
In Figure \ref{fig:j_err_ellipse_gammas}, we see that the larger the value of $\gamma$, the smaller the initial values of the functional $J$ and the error $\varepsilon$.
Although after a certain amount of iterations, the reconstruction error $\varepsilon$ for $\gamma=0.002$ becomes larger than the error $\varepsilon$ for $\gamma=0.001$ (see  Figure \ref{fig:e_ellipse_gammas}).
This is due to the fact that the larger $\gamma$,  the smoother the function $q(\bm x)$, and this results in loss of detail in the smoothed Heaviside function $H_{\alpha}(\bm x)$. Similar situation  can be seen in Figure \ref{fig:j_err_ellipse_Ms}. When we use more measurements,  the initial error $\varepsilon$ increases. But again, in the process of minimization, the error $\varepsilon$ for $M=4$ becomes larger than the error $\varepsilon$ for $M=3$. We can also see that the larger the number of measurements $M$, the minimization process takes more iterations to converge.

Figure \ref{fig:ellipse_iters} shows the reconstructed smoothed Heaviside functions $H_{\alpha}(q(\bm x))$ after 1, 10, 15, 20, 30 and 50 iterations. These results are obtained from 3 measurements with noise $\epsilon=0.01$ using $\gamma = 0.001$ and $\alpha = 0.01$. The shape of the true object (ellipse) is outlined with the solid white line. Note that in Figures  \ref{fig:j_err_ellipse_gammas} and \ref{fig:j_err_ellipse_Ms} the reconstruction results correspond to the orange dashed line. We can see that during the first 10 iterations, the recovered coefficient does not change much, and starting from 20 iterations, we already recovered the ellipse shape. Subsequent iterations are needed to increase the quality of recovery.

\begin{table}
\begin{center}
\begin{tabular}{c|c|c|c|c}
\hline
$M$ & $\gamma$ & Iterations & $J\cdot 10^{-7}$ & $\varepsilon$ \\ \hline
	& 0.0005	& 308	& 1.203	  & 0.225	 \\
1	& 0.0010	& 328	& 1.204	  & 0.211	 \\
	& 0.0020	& 351	& 1.205	  & 0.232	  \\
\hline
 	& 0.0005	& 43	& 2.478 & 0.138	  \\
2	& 0.0010	& 42	& 2.465 & 0.114	  \\
	& 0.0020	& 40	& 2.484  & 0.127	  \\
\hline
	& 0.0005	& 52	& 4.582	  & 0.102	  \\
3	& 0.0010	& 54	& 4.580     & 0.101	  \\
	& 0.0020	& 54	& 4.603	  & 0.120	  \\
\hline
	& 0.0005	& 46	& 5.839	  & 0.156	  \\
4	& 0.0010	& 56	& 5.808	  & 0.116 \\
	& 0.0020	& 63	& 5.826	  & 0.125	 \\
\hline
\end{tabular}
\end{center}
\caption{Dependence of $J$, $\varepsilon$, and number of iterations on the number of measurements $M$ and the parameter $\gamma$ for reconstruction of ellipse}
\label{tab:ellipse_Je}
\end{table}

\begin{figure
\begin{center}
\subfloat[\label{fig:j_ellipse_gammas}]{\includegraphics[width=0.5\textwidth]{j_ellipse_gammas}}
\subfloat[\label{fig:e_ellipse_gammas}]{\includegraphics[width=0.5\textwidth]{e_ellipse_gammas}}
\end{center}
\caption{The evolution of functional $J^k$ and error $\varepsilon^k$ with iterations $k$ for different values of $\gamma$: (a)  -- $J^k$,  (b)   -- $\varepsilon^k$}
\label{fig:j_err_ellipse_gammas}
\end{figure}

\begin{figure
\begin{center}
\subfloat[\label{fig:j_ellipse_Ms}]{\includegraphics[width=0.5\textwidth]{j_ellipse_Ms}}
\subfloat[\label{fig:e_ellipse_Ms}]{\includegraphics[width=0.5\textwidth]{e_ellipse_Ms}}
\end{center}
\caption{The evolution of functional $J^k$ and error $\varepsilon^k$ with iterations $k$ for different values of $M$:  (a) -- $J^k$,  (b)  -- $\varepsilon^k$}
\label{fig:j_err_ellipse_Ms}
\end{figure}

\begin{figure
\begin{center}
\subfloat[\label{fig:ellipse_iter1}]{\includegraphics[width=0.33\textwidth]{ellipse_iter1}} %
\subfloat[\label{fig:ellipse_iter10}]{\includegraphics[width=0.33\textwidth]{ellipse_iter10}}
\subfloat[\label{fig:ellipse_iter15}]{\includegraphics[width=0.33\textwidth]{ellipse_iter15}} \\
\subfloat[\label{fig:ellipse_iter20}]{\includegraphics[width=0.33\textwidth]{ellipse_iter20}}
\subfloat[\label{fig:ellipse_iter30}]{\includegraphics[width=0.33\textwidth]{ellipse_iter30}}
\subfloat[\label{fig:ellipse_iter50}]{\includegraphics[width=0.33\textwidth]{ellipse_iter50}}
\end{center}
\caption{Reconstructed smoothed Heaviside function $H_{\alpha}(q(\bm x))$  after: (a) --  1, (b) -- 10,  (c) -- 15,  (d) -- 20,  (e) -- 30, and (f) -- 50  iterations }
\label{fig:ellipse_iters}
\end{figure}

\subsection{Reconstruction of circles}

Next, we consider two circles  in Figure \ref{fig:mesh_circles}.
We perform the same experiments as with the ellipse using computationally simulated data with uniform noise with noise level.

In Table \ref{tab:circles_Je} we list the number of measurements $M$, the parameter $\gamma$, the number of iterations until convergence, the final value of the functional $J$, the reconstruction error $\varepsilon$ (\ref{eq:err}).
We can confirm that the number of measurements $M$ is essential for good accuracy.
In this case, we need to use more than two measurements.
When we use only one measurements $M=1$,  we get poor quality results with big errors.  Also,  the algorithm did not converged in 1000 iterations and was terminated.
We can see that $\gamma = 0.006$ is the optimal value for better results and using less than or greater than it leads to increased error $\varepsilon$.
Note that when we increase the value of $\gamma$, more iterations are needed for convergence of the algorithm.
Here, in all calculations we use $\alpha = 0.01$.

Similar to Figures \ref{fig:j_err_ellipse_gammas} and \ref{fig:j_err_ellipse_Ms} for the reconstruction of ellipse,
Figures \ref{fig:j_err_circles_gammas} and \ref{fig:j_err_circles_Ms} illustrate the evolution of functional $J^k$ and error $\varepsilon^k$ with iterations $k$ for different values of $\gamma$ and $M$, respectively.
In Figure \ref{fig:j_err_circles_gammas},
the blue solid, orange dashed, and green dash-dotted lines shows
the values of $J^k$ (Figure \ref{fig:j_circles_gammas}) and $\varepsilon^k$   (Figure \ref{fig:e_circles_gammas}) for  $\gamma = 0.004$,  $ 0.006$,  and $0.008$, respectively.
Similarly, in Figure \ref{fig:j_err_circles_Ms}, the blue solid, orange dashed, and green dash-dotted lines shows the values of $J^k$ (Figure \ref{fig:j_circles_Ms}) and $\varepsilon^k$   (Figure \ref{fig:e_circles_Ms}) for  $M = 3$,  $4$,  and $5$, respectively.
Reconstructions with different  $\gamma$ are performed with $M = 3$ and reconstructions with different $M$ are performed with $\gamma = 0.006$.

Figure \ref{fig:circles_iters} shows the reconstructed smoothed Heaviside functions $H_{\alpha}(q(\bm x))$ after 5, 25, 50, 150, 300 and 650 iterations. These images are obtained from 5 measurements with noise $\epsilon=0.01$ using $\gamma = 0.006$ and $\alpha = 0.01$. The shape of true objects is outlined with the solid white line. We can see that after 150 iterations, the circles are almost recovered.  The location and shape of the bigger circle are already reconstructed. Further iterations are required to recover the true location of the second circle.

\begin{table}
\begin{center}
\begin{tabular}{c|c|c|c|c}
\hline
$M$ & $\gamma$ & Iterations & $J\cdot 10^{-7}$ & $\varepsilon$ \\ \hline
	& 0.004	& 1000	& 1.154 & 	  0.853	\\
1	& 0.006	& 1000	& 1.385 &  1.543 \\
	& 0.008	& 1000	& 1.702 &   1.645	\\
\hline
	& 0.004	& 92	    & 6.119 & 	  0.883	\\
2	& 0.006	& 112	& 6.113 & 	  0.863	 \\
	& 0.008	& 123	& 6.121 & 	  0.837	\\
\hline
	& 0.004	& 429	& 4.324	& 0.145	\\
3	& 0.006	& 702	& 4.307	& 0.103	\\
	& 0.008	& 923	& 4.306	& 0.097	\\
\hline
	& 0.004	& 300	& 5.546	& 0.149	\\
4	& 0.006	& 673	& 5.473	& 0.085	\\
	& 0.008	& 938	& 5.471		& 0.086	\\
\hline
	& 0.004	& 325	& 7.537		& 0.106	\\
5	& 0.006	& 654	& 7.503	& 0.082	\\
	& 0.008	& 956	& 7.496	& 0.084	\\
\hline
\end{tabular}
\end{center}
\caption{Dependence of $J$, $\varepsilon$, and number of iterations on the number of measurements $M$ and the parameter $\gamma$ for reconstruction of circles}
\label{tab:circles_Je}
\end{table}

\begin{figure
\begin{center}
\subfloat[\label{fig:j_circles_gammas}]{\includegraphics[width=0.5\textwidth]{j_circle_circle_gammas}}
\subfloat[\label{fig:e_circles_gammas}]{\includegraphics[width=0.5\textwidth]{e_circle_circle_gammas}}
\end{center}
\caption{The evolution of functional $J^k$ and error $\varepsilon^k$ with iterations $k$ for different values of $\gamma$: (a) -- $J^k$, (b)   -- $\varepsilon^k$}
\label{fig:j_err_circles_gammas}
\end{figure}

\begin{figure
\begin{center}
\subfloat[\label{fig:j_circles_Ms}]{\includegraphics[width=0.5\textwidth]{j_circle_circle_Ms}}
\subfloat[\label{fig:e_circles_Ms}]{\includegraphics[width=0.5\textwidth]{e_circle_circle_Ms}}
\end{center}
\caption{The evolution of functional $J^k$ and error $\varepsilon^k$ with iterations $k$ for different values of $M$: (a) -- $J^k$,  (b)  -- $\varepsilon^k$}
\label{fig:j_err_circles_Ms}
\end{figure}

\begin{figure
\begin{center}
\subfloat[\label{fig:circles_iter5}]{\includegraphics[width=0.33\textwidth]{circles_iter5}} %
\subfloat[\label{fig:circles_iter25}]{\includegraphics[width=0.33\textwidth]{circles_iter25}}
\subfloat[\label{fig:circles_iter50}]{\includegraphics[width=0.33\textwidth]{circles_iter50}} \\
\subfloat[\label{fig:circles_iter150}]{\includegraphics[width=0.33\textwidth]{circles_iter150}}
\subfloat[\label{fig:circles_iter200}]{\includegraphics[width=0.33\textwidth]{circles_iter300}}
\subfloat[\label{fig:circles_iter650}]{\includegraphics[width=0.33\textwidth]{circles_iter650}}
\end{center}
\caption{Reconstructed smoothed Heaviside function $H_{\alpha}(q(\bm x))$  after: (a)  -- 5,  (b) -- 25, (c)  -- 50,  (d) -- 150,  (e) -- 300, and (f) -- 650  iterations }
\label{fig:circles_iters}
\end{figure}

Finally, we want to see how our algorithm can handle the different values of noise. We consider the two circles.
In Table \ref{tab:circles_err} the first and second columns show the number of measurements $M$ and the parameter $\gamma$, respectively. In the next four columns we show the reconstruction error $\varepsilon$ (\ref{eq:err}) for different values of noise level $\epsilon = 0.01$,  $0.02$, $0.03$ and $0.04$.
As can be seen in Table \ref{tab:circles_err}, we need to increase the number of measurements $M$ to handle big noise in our data.
Also, we observe that when $ M $ is less than 8, the effect of the value of $ \gamma $
That means that the effect of $M$ is much more significant for better reconstruction.
When $M$ is equal or greater than 8, $\gamma = 0.006$ gives better accuracy.
Figure \ref{fig:circles_err} illustrates the reconstructed smoothed Heaviside functions $H_{\alpha}(q(\bm x))$ from data with noise $\epsilon = 0.02$, $0.03$ and $0.04$. These results are obtained from 10 measurements using $\gamma = 0.006$.

\begin{table}
\begin{center}
\begin{tabular}{c|c|c|c|c|c}
\hline
$M$ & $\gamma$  & $\epsilon=0.01$ & $\epsilon=0.02$ & $\epsilon=0.03$ & $\epsilon=0.04$  \\
		\hline
	& 0.004		& 0.149			& 0.142		& 0.235	&	0.394	\\
4	& 0.006		& 0.085		& 0.156		& 0.257	&	0.314	\\
	& 0.008		& 0.086		& 0.148		& 0.231		&	0.326  \\
	\hline
	& 0.004		& 0.106			& 0.168		& 0.224	&	0.259	\\
5	& 0.006		& 0.082		& 0.149		& 0.216		&	0.272	\\
	& 0.008		& 0.084		& 0.130		& 0.196		&	0.287	\\
	\hline
	& 0.004		& 0.128			& 0.227	& 0.220	&	0.252	\\
6	& 0.006		& 0.099		& 0.176		& 0.213		&	0.289	\\
	& 0.008		& 0.086		& 0.141		& 0.216		&	0.265	\\
	\hline
	& 0.004		& 0.211			& 0.135		& 0.321		&	0.265	\\
7	& 0.006		& 0.086		& 0.146		& 0.259 	&	0.288	\\
	& 0.008		& 0.095		& 0.130		& 0.217		&	0.302	\\
	\hline
	& 0.004		& 0.101			& 0.151		& 0.233	&	0.246	\\
8	& 0.006		& 0.086		& 0.123		& 0.170		&	0.234	\\
	& 0.008		& 0.089		& 0.144		& 0.218		&	0.251	\\
	\hline
	& 0.004		& 0.101			& 0.145		& 0.194		&	0.249	\\
9	& 0.006		& 0.085		& 0.134		& 0.166		&	0.226	\\
	& 0.008		& 0.090		& 0.135		& 0.205	&	0.238	\\
	\hline
	& 0.004		& 0.131			& 0.139		& 0.205	&	0.222	\\
10	& 0.006		& 0.081			& 0.119		& 0.177		&	0.223	\\
	& 0.008		& 0.090		& 0.133		& 0.187		&	0.224	\\
	\hline
\end{tabular}
\end{center}
\caption{Dependence of $\varepsilon$ on the number of measurements $M$ and the parameter $\gamma$ for reconstruction of circles with noisy data $\epsilon = 0.01$, $0.02$, $0.03$ and $0.04$}
\label{tab:circles_err}
\end{table}

\begin{figure
\begin{center}
\subfloat[\label{fig:circles_err2}]{\includegraphics[width=0.33\textwidth]{circles_err2}} %
\subfloat[\label{fig:circles_err3}]{\includegraphics[width=0.33\textwidth]{circles_err3}}
\subfloat[\label{fig:circles_err4}]{\includegraphics[width=0.33\textwidth]{circles_err4}}
\end{center}
\caption{Reconstructed smoothed Heaviside function $H_{\alpha}(q(\bm x))$  from data with noise: (a)  $\epsilon=0.02$,  (b) $\epsilon=0.03$,  (c) $\epsilon=0.04$, }
\label{fig:circles_err}
\end{figure}

\section{Conclusion}

This paper presents the numerical algorithm for the reconstruction of the piecewise constant  coefficient. The performance of the algorithm is validated on numerical experiments. We have determined that the number of measurements is a crucial parameter for achieving good accuracy.  The smoothing parameter  $\gamma$ also affects the reconstruction results. The optimal values of these parameters can be determined experimentally.

In this work,  we assumed that an object to be imaged contains only two materials with known piecewise constant coefficients. More general cases with several unknown coefficients will be a subject of our future work. It is well known that the complete electrode model is better suited for electrical impedance tomography than the continuum model considered here. We will extend the proposed algorithm to work with the complete electrode model in the future.

redpajama_set_name: RedPajamaArXiv


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Before joining The Intern Group in 2014, John held senior positions in the investment operations field, including Senior Manager for Investment Application Services at Liberty Mutual (one of the USA's largest insurance companies), and AVP at Bank of New York-Mellon. John holds a Bachelor of Liberal Arts degree from Harvard University with a field of study in Economics, which he earned while working full-time. A travel enthusiast, John has visited over 30 countries, and believes deeply in the value of international experiences as a lever for educational, professional, and personal growth.
I am a second year student at the University of Chichester studying Film and Television Studies. I am interested in having an insight into the workings of a television company in London for the summer period as I live near London.
I am very intrigued to see how a company that specialises in television operates and am particularly interested in meeting a producer. This was a role I enjoyed taking on during a university module called Filming Performance. I have very good communication skills, can meet deadlines and am used to working as part of a team on modules at University. I have had some experience on technical software such as Final Cut Pro through my course and other technical hardware through working with a free-lance Video Producer for a couple of days.
I would very much like to have an interview please to discuss what options there may be. Thank you very much for taking the time for reading this email.
Thanks for your message and your interest in our TV & Film program. I'll send you an email today with more information and how you can apply to have an interview with our admissions department. Make sure to keep an eye out for it!
I am Himani Tewary from India. Currently, I am in my senior year of my Master's degree in Journalism and Mass Communication from Gujarat University, Ahmedabad. I will complete my degree by April 2018. I have interned in many media house including radio and print as well as UNICEF (India.) I am interested in 2 months internship programme in media house of London to boost up my CV.
I would like to know if I am eligible for internship programs offered The Intern Group and how to register myself for the same. Thank you.
Thanks for your message and your interest in our media programs in London. We have placed interns in many different types of roles in the field of journalism and media, including content creation, publishing, writing articles, print media, online media and more. I'll send you an email today with some more information on how you can get started with an application.
Thanks for your message and your interest in our international internship program in London. Internships are a great way for recent graduates to gain hands-on experience in their career field right after university. I'll send you some more information today regarding our London program.
Hi, I just finished my Degree in Journalism at Multimedia University of Kenya in February 2018. I am expecting to graduate in December 2018. I am looking for an opportunity to intern in London before I graduate to boost up my CV, gain international exposure and perfect the skills I have acquired in my studies. I have already interned in several Media Stations in Kenya(my home country).Kindly assist me with any opportunities that I am eligible to apply.Thank you in advance.


Endodontics - root canal therapy can help save a smile and stop the pain.
If you have a tooth that is cracked, damaged or decayed, a root canal can stop further damage and save your tooth.
You want to keep your natural teeth for as long as possible, and root canal therapy can help you do just that. Each year, millions of people receive this common treatment, which is performed by a dental specialist, called an endodontist, who has received two extra years of specialized training.
If a tooth is painful, discolored, sensitive to cold and hot, or your gums are very tender, these are signs a tooth may be infected. This can happen if a cavity has grown very deep or the tooth is cracked or broken. When that happens, bacteria infect the soft tissue inside your tooth, where antibiotics can't reach.
An endodontist is highly skilled in diagnosing and treating the soft tissue, or pulp, inside a tooth, which consists of nerves and blood vessels. This tissue extends down to the tooth's roots.
Your dentist and root canal specialist will work together to restore your natural tooth. The root canal specialist will carefully examine your teeth, jaw, gums, and mouth to determine the extent of the problem and recommend what treatment is needed. He or she will explain the treatment and answer any questions you might have. Root canal therapy is usually performed under a local anesthetic. In general, this involves removing the infected pulp inside of your tooth chamber and root canal. The area is then carefully cleaned, medicated and permanently sealed. Your dentist may need to restore your tooth with a permanent crown so that it looks and feels completely natural.
Problems that require root canal therapy can't be put off, so if you are experiencing the symptoms described earlier, call us today for an appointment.
To make sure you can afford this procedure, we offer a number of flexible payment options. We also accept nearly all dental insurance plans, and special programs are available for patients without insurance.


Shower shampoo dispenser – Use the dispenser shower instead as easily as you press the last bit of shampoo bottles and can be distracting. It is very easy for you to get your shower cubicles cluttered with bottles especially if you use some type of liquid gel. The solution to this is to simply get the dispenser wall mounted shower easily accessible help to free up space. The good thing about this is that you can easily pull the proper amount of shampoo or conditioner or without wasting. Through the pump mechanism operated normally with just a push button that gives the right amount. The most common types are those that have two separate containers along but you can easily buy them with up to four.
This is very suitable for obvious reasons as you can get quick and easy access to your favorite shower shampoo dispenser or other liquids. Another added benefit is that you can do away with all the bottles that may be scattered around the shower floor until you have more free space. If you purchase such large quantities of liquid, you can contribute to a clean environment so don't always throw the bottle.
Before you make a purchase, make sure you get transparent so you know exactly when it needs to be recharged. While you can go with the model that the focus will be only on the table, they are not very stable and tend to move around a lot shower shampoo dispenser. When you've made the decision to buy, make sure you check out the online retail stores for good deals and discounts.
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Find Out Reclaimed Barn Door in Here!


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Zarif's special aide visits Moscow for regional consultations
TEHRAN – Ali Asghar Khaji, who acts as senior assistant to Foreign Minister Mohammad Javad Zarif for political affairs, visited Moscow on Monday for talks on regional issues with Russian officials, IRNA reported.
Khaji was slated to hold talks with Russian Deputy Foreign Minister Sergey Vershinin and President Vladimir Putin's special envoy for Middle East affairs Mikhail Bogdanov.
Both sides were supposed to review important regional developments, especially those related to Syria and Yemen.
Recently, Russian Foreign Minister Sergei Lavrov said Moscow backs Iran's proposal for a non-aggression pact with Persian Gulf Arab states.
During his visit to Iraq on May 25-26, Foreign Minister Mohammad Javad Zarif offered a proposal to leaders of the Persian Gulf states to sign the pact as a prelude to de-escalation of conflicts in the Persian Gulf region.
Lavrov said that the initiative was discussed with the [Persian] Gulf Cooperation Council.
The Israeli regime and the United States have been trying hard to demonize Iran in the eyes of Saudi Arabia, the United Arab Emirates and Bahrain.
Iran has amicable relationship with Oman, Qatar and Kuwait, three members of the regional Arab bloc.
MH/PA
Ali Asghar Khaji

redpajama_set_name: RedPajamaCommonCrawl


A bill soon to be introduced in the state Senate will use an increase in gasoline taxes to improve safety and create jobs, according to Senate officials.
Cheryl Hicks, executive director of the Senate Democratic Transportation Committee, said the bill is a three-year $2.5 billion spending plan. Sen. John Wozniak, D-Westmont, is chairman of the committee.
She said the bill does not outline any specific project, but it will make sure funds are available for work across the state.
The bill calls for an increase in the gasoline tax. Pennsylvania has a 12-cent flat tax — and oil companies have a franchise tax based on the wholesale price of gasoline. Under current law there is a $1.25 per gallon cap.
While the tax is levied on companies, not the consumer, there is no measure to stop companies from passing the increase to the consumer. The gasoline tax in considered a user fee for those who travel on the roads, Hicks said.
Hicks said the motor license fund, which is the fund that holds money for transportation projects, has remained stagnant since 1997, when the last gas tax increase was approved.
There has been a 20-percent increase in road project costs since 1997, she said.
She said Pennsylvania roads and bridges are in need of repair and funding is needed to complete those projects.
Hicks said that if the bill reaches the governor's desk, lawmakers believe he will sign it, but there are concerns that some lawmakers may not want to raise the gasoline tax.
State Rep. Carl Walker Metzgar, R-Allegheny Township, said that he is familiar with the proposed Senate plan.
Metzgar said that some of his colleagues are in favor of raising the tax.
Somerset County Commissioner John Vatavuk said improvements to the state's transportation systems are needed so Pennsylvania can compete with other states. He said Maryland, Ohio and Virginia have all passed new transportation bills to address their needs. Vatavuk said one Pennsylvania company relocated to Virginia because their bridges are rated for their larger tanker trucks.
Vatavuk said that the bill being passed may improve the state's chances of completing Route 219 from Meyersdale to Maryland.


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Мартин Вагнер (, роден на 24 февруари 1968 в Офенбург) е бивш германски футболист. Започва кариерата си в родния си клуб "Офенбург", където е открит за професионалния футбол.

В Първа Бундеслига халфът играе за Нюрнберг, Кайзерслаутерн и Волфсбург. С "червените дяволи" той печели Купата на Германия през 1996 г. и шампионската титла през 1998 г. Във финалната среща през 1996 г. Мартин Вагнер отбелязва единственото и победното попадение срещу Карлсруе. Неприятното в случая е, че лаутерите изпадат от Първа Бундеслига само няколко дни преди това. С екипа на Кайзерслаутерн Вагнер изиграва 138 мача (16 гола) в Първа Бундеслига и 31 мача (7 гола) във втора лига и има решаваща роля за завръщането на клуба в германския елит.

В националния отбор по Футбол на Германия Вагнер играе 6 пъти през 1993 и 1994 г. Той е част от разширения състав на бундестима за Световното първенство в САЩ'94.

Полузащитникът спира с професионалния футбол през 2001 г. с екипа на Волфсбург.

Вагнер е мениджър на футболисти, като представя играчи в и извън Германия. Също така той е и младежки треньор в клуба "Еберсвай".

Външни препратки
 Уебсайт на мениджърската агенция на Мартин Вагнер

Германски футболисти
Футболисти на ФК Нюрнберг
Футболисти на Кайзерслаутерн
Футболисти на ФФЛ Волфсбург

redpajama_set_name: RedPajamaWikipedia


TUBAS (Ma'an) — Israeli forces on Thursday destroyed over 40 Palestinian homes and structures in the Tubas district of the occupied West Bank as ongoing Israeli policies in the Jordan Valley continue to drive Palestinians out of their homes.
A spokesperson for Israel's Coordination of Government Activities in the Territories (COGAT) did not have immediate information on the demolitions.
Threats of displacement for the thousands of Palestinian Bedouins living in the Jordan Valley in particular have reportedly increased dramatically since 2012.
Rights groups argue that Israel aims to fully annex the strategic area of land and is unlikely to return the occupied area to Palestinians.
Palestinians are frequently ordered to evacuate their villages during military trainings, and B'Tselem reported that military exercises carried out last month caused severe damage to crops belonging to Palestinian communities.


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The Globes
Here's absolutely everything you need to know about tonight's Golden Globes, and how to watch it in Ireland
The booziest ceremony takes place tonight.
By Nicola Byrne Sunday 6 Jan 2019, 9:00 AM
Jan 6th 2019, 9:01 AM 4,095 Views No Comments
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THE GOLDEN GLOBES celebrates the best in film and television, and kicks off awards season tonight.
Source: Sipa USA/SIPA USA/PA Images
Unfortunately, it takes place in LA, so if you're planning on tuning in, it's gonna be a late one. But as attested from previous years, it's the booziest of ceremonies and usually throws up a few gems.
Here's everything you need to know.
Any Irish nominations this year?
Source: Hfpa Photographer/Zuma Press/PA Images
Monaghan actress Caitriona Balfe is up for Best Actress for the TV show Outlander, for the fourth consecutive year.
Source: Twentieth Century Fox/Zuma Press/PA Images
Irish co-production The Favourite also got several nods, including Best Motion Picture Musical or Comedy, Best Screenplay, Best Actress for Olivia Colman, and Best Supporting Actress for Rachel Weisz and Emma Stone.
Who are the faves?
On the drama side, it looks like A Star Is Born will dominate. It's only a matter of how many speeches Bradley Cooper will give, with the movie up for best drama, as well as the man himself up for best actor and director. Gaga will probably take best actress in a drama, and is definitely a safe bet for original song, too.
Source: Gerber Pictures/Zuma Press/PA Images
Comedy categories indicate a sweep for The Favourite being the eh, Favourite.
Source: Amazon Prime
In TV, it's likely that Julia Roberts will take an acting gong for Homecoming, while Killing Eve, The Kominsky Method, and The Marvelous Mrs Maisel are all shoo ins for consideration.
What's the ceremony gonna be like?
Well hosting duties are in the hands of unlikely duo Sandra Oh of Killing Eve and Grey's Anatomy fame, and Andy Samberg, of SNL and Brooklyn 99. You can expect them to perform some musical numbers, and don't expect too much Trump jokes. They've already stated they'll be avoiding any reference to him.
Source: Willy Sanjuan/AP/Press Association Images
Presenters include our own Saoirse Ronan (where is her nom, eh?), Ben Stiller, Idris Elba, Catherine Zita-Jones, Dick Van Dyke, and Chrissy Metz.
Source: Matt Crossick/Matt Crossick/Empics Entertainment
Jeff Bridges will be accepting the Cecil B DeMille Award, an accolade that produced memorable speeches from Oprah Winfrey and Meryl Streep over the last two years.
Great, so how do I watch it in Ireland?
The red carpet will be covered on E! as usual from 11pm until about 1am. The ceremony itself will be broadcast on NBC at around 1am, so if you can source a link up to that, brilliant.
It won't be over until about 4am.
If you can't stay up that late, RTE 2 will be airing the ceremony highlights at 9.30pm on Monday night.
We'll have full coverage on DailyEdge.ie in the morning. Of course.
Follow us: dailyedge
Nicola Byrne
nicola@dailyedge.ie
See more articles by Nicola Byrne
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We don't get a lot of snow around here, but when there's a threat of an inch, schools and businesses close in precaution. (Don't make fun. We don't know how to drive in that stuff.) Unfortunately for Jeremy and Jackson, the Miller School for Boys is open every day, snow or no snow.
Still, we make little allowances for the delightful weather. It's nice to not have to go anywhere. We sleep in a little later, take our time getting to school work, and this morning was no different. Jeremy, who protested that one to three inches of snow was no big deal, was the first one to throw on some layers and take the dog out. It's Major, after all, who loves snow the most.
The meteorologist called for one to three inches in our specific area, and sure enough, that's what fell. It was beautiful.
I tried to get pictures of our neighbors, but the snowflakes were so fast and fat that I couldn't get the horses in focus, and it was too cold to cross the road to get closer.
The temperatures will continue to drop overnight, which gives me pause when I think of the surrounding trees and the heavy snow covering all the limbs.
Major really is the happiest boy in the snow.
Once the snow stopped and the clouds cleared, we had blue skies once again and the sun melted all the remaining ice on the roads.


Within the cosmetics industry there is certainly no shortage of beauty products which boast to contain a new-fangled natural ingredient, however when we heard of the Escargot Collection it peaked our interest.
The Escargot Collection is a snail secretion filtrate product, meaning it contains the slime which you'll notice on the common garden snail. Sounds icky right? But while the west has been slow on the uptake, surprisingly in Asia, one product is sold every 6 seconds.
Despite us referring to this as a new-fangled idea earlier in the article, we were in fact able to find evidence of snails being used for medicinal purposes dating back to ancient Greece. Throughout history people have harnessed such a common garden pest for everything from medicine to skincare.
While we've researched many quotes and studies around the product, we thought it was best to just try the product out for ourselves.
We received both the Escargot Collection Revitalising Moisturising Emulsion (pictured) and Escargot Collection Revitalising Lifting Cream. After getting over the mental barrier of applying snail slime to my face, I found that the product absorbed very well into my skin. I'm always weary of applying excessive amounts of moisturiser to my face from the fear of becoming too oily and greasy from the application. With the escargot collection, only a small application was needed to adequately moisturise my skin without having to reapply in a short amount of time.
After a month of 2 applications a day of the Lifting Cream (in the morning and before bed at night) my skin did feel more smooth. I was surprised that in a short amount of time how big of a difference I saw. The redness in my face also reduced which I can only imagine is thanks to the medicinal qualities of the lotion. For us this was the better lotion of the two.
On the second, we weren't as impressed with its absorption. That's not to say it was particularly "bad" , however it did not absorb into my skin as great as the Lifting Cream and we found it slightly more slimey. If you're only after a single face-cream, I would highly recommend the Lighting cream over the emulsion.
All in all, both products did what they said they would do. My skin was clearer, softer and less tired. So, should we all throw out our synthetic creams and switch over to the naturally produced snail slime that is the Escargot Collection? Maybe! With all the over perfumed and chemical infused creams on the market, there's nothing wrong with a return to nature and a detox from the overly chemical infused products which appear on TV. FAULT would definitely recommend you keep an open mind and get at least the lifting cream to see how much it can improve your beauty regiment!


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TimeMachiner
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Ask Me Questions At My Funeral
Post author By Aaron
Mourning the dead is a process that differs by culture, geographic location, customs, and a myriad of other factors. One interesting turn of how technology may augment this process is the work of StoryFile. This company whose purpose is to memorialize Holocaust survivors has been employed at times to let the deceased "attend" and "answer" questions posed to them at their own funerals. Victor Tangermann at Futurism has more.
Marina Smith who passed away in June, was able to address the mourners at her own funeral. StoryFile used 20 cameras to film her answering around 250 questions prior to her death. This data was then fed into a software tool that was able to virtually recreate her after she passed.
The illusion must have been pretty shocking to funeral goers. Stephen Smith was able to converse with the virtual representation of his mother in real time, and even gave attendants the opportunity to ask questions as well.
"Mum answered questions from grieving relatives after they had watched her cremation," Smith told The Telegraph. "The extraordinary thing was that she answered their questions with new details and honesty," he added. "People feel emboldened when recording their data. Mourners might get a freer, truer version of their lost loved one."
I'd heard of StoryFile's work to preserve the stories and memories of Holocaust survivors earlier in 2022. It's an ambitious and important effort to keep history from being erased or rewritten.
For people like Smith who decided to use this for her own final planning, it feels both comforting and creepy. I am sure she felt a sense of relief to share anything she wanted with the utmost candor. On the attendee side, I could completely understand feeling creeped out by it, especially when you're in the throes of grieving that person's loss.
Sometimes truth is stranger than fiction but this gets us closer to the plot of Upload in a small way.
TimeMachiner is my one-person project I run in my off time when I'm not working my day job in IT. If you enjoy my work, consider leaving a tip or becoming a member. Your support is appreciated and goes a long way to keep my work going.
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Hire Clubhouse
Head Race
Learn2Row
Junior Squad continue to make club history
Leave a Comment	/ News
Newark Rowing Club's Junior Women's squad secure a qualifying spot to race for the Diamond Jubilee Challenge Cup at Henley Royal Regatta – Newark's first women's crew to qualify for the prestigious regatta.
Henley Royal Regatta 2021 qualifiers saw Newark Rowing Club history with the first women's crew to qualify for the regatta. 17 year old Connie Ferrer, Neve Woodman and 18 year old Maisie Archer-Dytch and Emily McCrery raced for a qualifying spot for the Diamond Jubilee Challenge Cup, Junior Quads. The girls will progress to side-by-side racing at Henley Royal Regatta in Henley-on-Thames on 12 August 2021.
In order to qualify for the Regatta, crews must race a 1,900m time trial and make a qualifying time. A total of 14 crews raced for a spot in the Diamond Jubilee Challenge Cup with only 7 crews making it to race side-by-side at the Regatta itself.
"Henley Royal Regatta is one of the biggest, and most prestigious regattas in the world, and as a club, we are so incredibly proud of the girls securing a qualifying spot," comments Club Captain, Andy Joy. "This is a testament to the hard work and dedication shown by the athletes and their coach over the past months during a period of unprecedented uncertainty."
We wish the girls the best of luck when they head to Henley for side-by-side racing on Thursday 11 August.
Established in 1839, Henley Royal Regatta is one of the most prestigious rowing events in the world. Composed of over 300 races, the Regatta is not one to be missed in the sporting and social calendar. Attracting Olympic and club athletes alike, the Regatta promises five days of dramatic knock-out style racing.
OPEN DAYS DURING THE OLYMPICS
Newark Rowing Club are hosting 3 open days during the Olympics. When we see sport on such big occasions like the Olympics it can inspire…
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⭐⭐ Star New Year Head ⭐⭐ Wow
#throwbackthursday Throwback to our 2011 Henley
HAPPY NEW YEAR🥂🍾 Hope there aren't too m
🎉MEMBER OF THE YEAR🎉 Our 2021 Member of th
MERRY CHRISTMAS 🥳🎄 Wishing all our members
As the year is drawing to a close, we would like t
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Trang chủ Chưa được phân loại Vietjet Air to launch direct Hanoi-New Delhi flight
Vietjet Air to launch direct Hanoi-New Delhi flight
Vietjet Air will become the first Vietnamese carrier to launch a direct flight to India after receiving permission this week.
Vietjet Air will become the first in Vietnam to open Hanoi-New Delhi direct service. Photo by Shutterstock/Jimmy Tran.
The Civil Aviation Authority of Vietnam has approved the budget carrier's request based on an air transport agreement between Vietnam and India signed in 2013.
The airline has not announced when it would launch the first flight.
The number of Indian tourists to Vietnam remains low compared to other Asian countries like China and South Korea, with only 130,000 arrivals in 2018. But the year-on-year growth was a whopping 30 percent.
China and South Korea are Vietnam's top feeder markets with 4.9 million and 3.4 million visitors last year respectively.
India is one of Vietnam's top 10 trading partners. Bilateral trade reached $10.69 billion last year, a two-fold increase against 2016, according to Vietnam's Ministry of Industry and Trade.
Vietjet Air had last year stated that it planned to launch a direct air route connecting the country's business hub Ho Chi Minh City to New Delhi to meet the growing overseas travel demand of Vietnamese holiday-goers. However, no flights have taken off so far.
Indian budget carrier Indigo Airlines had recently stated that it plans to launch daily direct flights between India's Kolkata City and Hanoi from October 3.
Vietnam first allowed foreign tourists to apply for a visa online in February 2017, starting with 40 countries including China, South Korea and the U.S. In December that year six countries were added to the list, including Australia and India.
Vietnam's aviation industry has been growing at the third fastest pace in the Asia-Pacific, according to the International Air Transport Association (IATA). It welcomed 12.5 million air passengers last year, up 14.4 percent from 2017, the General Statistics Office reported.
Vienam's five airlines are Vietnam Airlines, its low-cost carrier Jetstar Pacific, budget airline Vietjet Aviation, Bamboo Airways and Vietnam Air Services Co.
According to Mastercard, Vietnam has the second fastest growing outbound market in the Asia Pacific region after Myanmar, with a projected annual growth of 9.5 percent between 2016 and 2021.
Mastercard has forecast that 7.5 million Vietnamese travelers will travel outside the country in 2021.
The capital city Hanoi was recently named among the top 15 destinations in the world by TripAdvisor travelers. It is also on top of the list of most affordable destinations in the world, with backpackers needing just $18.29 a day, according to the 2019 Backpacker Index formulated by Price of Travel, the world's largest database of travel costs.
Bài trướcElation first, depression next: pitfalls of studying abroad
Bài tiếp theoFive Google Doodles that have depicted Vietnamese culture
Mastercard lists Hanoi, HCMC among top 20 Asia-Pacific travel destinations
Vietnam's telco giants get green light for 5G tech tests
Vietjet, Masan buy 77 pct of bonds issued by Viet Capital
US embargo on Russian contractor delays key power project in Vietnam
'Moving' Georgian love monument finds new home in central Vietnam
Inside Southeast Asia's largest bear rescue center
Five Google Doodles that have depicted Vietnamese culture
Lonely Planet lists central Vietnam in top 10 Asia-Pacific destinations
Vietnam budget carrier to fly from Nha Trang to China's Yinchuan


As today (19th November) is World Toilet Day we're dedicating our latest blog entry to toilets and their importance in human life and what we should be doing in order to make sure that a place that we spend an average of 13 months of our lives sitting on is as clean as it needs to be.
Having access to a toilet is a basic human right and is something that most of us will take for granted but, alarmingly, 4.5 billion people on our planet are currently living without access to a safe toilet – that's a staggering 62.5% of the world living without access to safe sanitation. That may seem crazy for us that live in first-world countries, but unfortunately for those who don't this is a very real and serious problem. As well as this, 892 million people worldwide are practising open defecation, which is essentially turning our beautiful planet into an open sewer, which doesn't take much to realise that this is going to cause problems for the environment and its ecosystems.
This is a global issue as toilets play a hugely important role for all of us, not only for a safe space to relieve ourselves, but they are also important for the sustainable, inclusive development of the human race. In a nutshell, toilets are vital for the healthy development of people, especially children. Safe sanitation in general is just as important. Having a safe area to relieve of urine and faeces helps to provide a healthy living environment for everyone, protect natural resources such as surface water, groundwater and soil, and provide safety, security and dignity for everyone when they use the facility.
So what needs to be done in order to tackle these issues and reduce the number of people who do not have access to safe sanitation? The short answer is investment. In 2012 the World Health Organisation estimated that for every dollar that the USA invested into global sanitation, the global economic return on sanitation spending is $5.50. This number alone proves that investment in the area is the answer to the problem. Nearly half of global malnutrition instances are linked to chronic diarrhoea caused by a lack of clean water, decent sanitation and good hygiene, showing that the global population's lack of access to clean and safe toilets is having a huge impact on the future of millions of children worldwide suffering from malnutrition. Over half of the world are struggling to develop, stay healthy and earn themselves a better future, all for the want of a toilet. It's time that changed.
As well as the global need for improved sanitation, it is also important that we ensure our toilets at home and work are clean in order to maintain a healthy lifestyle. Bathrooms can be germ breeding grounds if they are not maintained correctly, resulting in illness and potential damage to bathroom surfaces and fixtures. To help the prevention of bacteria and other germs we would recommend cleaning your toilet twice a week, your sink weekly and your shower and bath once every couple of weeks, but there is nothing wrong with cleaning them more often if you use them more frequently. If you're looking for new products to tackle the germs in your bathroom, our Everfresh Apple is great for cleaning toilets and Clean Fast is great for cleaning all surfaces in your bathroom.


Claude Puel said it was time for everyone associated with Leicester City to put the Riyad Mahrez saga behind them and move on.
Mahrez ended his 10-day exile by returning to training on Friday and a he made his comeback as a second-half substitute against Manchester City after missing the previous two games following the collapse of his desired move to the Etihad Stadium.
Mahrez has held clear the air talks with his team-mates and Puel said there is no issue in the dressing room after Mahrez's walk out, and said it was time to draw a line under the whole saga.
"I said before in my last press conference it was important he came back quickly," said Puel after City's 5-1 defeat ended a tough week for the club.
"It was important he maintained good fitness alone and it was important to come back quickly into the squad.
"We have lost time and now it is important to look forward. The window is finished and it is important now to move on.
"Of course we are stronger with our best player.
"It is important to move on together with solidarity, a good understanding and forget this last result, and have a good focus and concentration on our future games.
Puel defended his decision to throw Mahrez back into the fray at the expense of other players like Demarai Gray.
"Since of the beginning of the window we took the decision with the club and all the time to protect the team and the squad," he added.
"Riyad came back and he made perhaps a mistake, but it is important to have all the best players with us.
"Riyad is a valuable player for the team, like Demarai and all the players.
"All the players welcomed him with a good feeling. Now we have to move on together and prepare for the next game.


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Police to Investigate Asylum Seeker's Lynching in Be'er Sheva
Videos show people kicking him, throwing a bench at him and pinning him to the ground with a chair, after he was shot by police after Sunday's terrorist attack.
Yaniv Kubovich
Almog Ben Zikri
Security forces at the Be'er Sheva Central Bus Station after a shooting attack, October 18, 2015.Eliyahu Hershkovitz
Police will conduct an investigation into the circumstances surrounding the shooting and death of an asylum seeker, Haftom Zarhum, 29, of Eritrea, who was killed in Sunday's Be'er Sheva shooting attack.
Be'er Sheva attack indicates calm is still far away
Gunman behind Be'er Sheva shooting attack identified as Bedouin man
Two dead, 11 wounded in shooting attack in southern Israel
In a hearing at the Southern District Police Monday morning, police decided that the central unit of the Negev District Police would undertake the specific investigation.
An IDF soldier was killed and 10 other people were wounded in a shooting attack at the Central Bus Station in the Southern Israeli city of Be'er Sheva. The terrorist, a Bedouin Israeli citizen, was also shot and killed. Zarhum was shot following the initial gunfire by a security guard, who evidently thought he was linked to the gunman. Zarhum traveled to Be'er Sheva to obtain a visa and was on his way home when he was shot.
In videos captured at the scene, Zarhum is seen attacked by the people around him, including a soldier, after being shot. People are seen kicking him, throwing a bench at him and pinning him to the ground with a chair. Some of the witnesses made efforts to stop the attackers.
Police stressed that this is not a manslaughter investigation; it is investigating the entire event, as he had sustained injuries before the attack.
Police decided to open an investigation as this was an attack on a defenseless person after the terrorist was killed—there was no more threat. Police explain that they will proceed with this investigation "cautiously:" On the one hand, they do not want to deter citizens from acting during terrorist attacks, but on the other, they must hold each individual accountable for their actions.
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Israel police


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These Airbnb hosts earned more than $15,000 on Thursday after the company let them buy IPO shares
A woman talks on the phone at the Airbnb office headquarters in the SOMA district of San Francisco, California.
Gabrielle Lurie | Reuters
Since starting an Airbnb host in 2016, Travis Schurr has become a mainstay on the site, listing more than 20 houses in the Las Vegas area. He got started when one of the properties in his home-flipping company didn't sell, so he turned it into a short-term rental.
Now, Schurr and his wife, Janie, are among Airbnb hosts making money from a huge first-day pop in the company's IPO. In its public debut this week, Airbnb set aside up to 3.5 million non-voting shares for hosts, accounting for as much as 7% of the total offering.
Schurr, 45, said he's never been able to buy into an IPO before and called the decision a "no-brainer" given what he knows about how they typically perform. He gladly bought 200 shares, the maximum available to him, for $68 apiece, expecting a significant pop when trading started on Thursday.
"All the fundamentals are out the window," Schurr said, citing IPOs like DoorDash, which surged 86% in its debut on Wednesday. "When you're allocated shares in an IPO and you can get in, nine times out of 10 everybody's getting in if you have the money."
Schurr's $13,600 initial investment is worth $28,942 after Airbnb spiked 113% to close at $144.71. He had pre-registered to buy 275 shares, the most allowed, but was informed Wednesday night that he was only allocated 200. Other hosts told CNBC the same thing happened to them.
At the close of trading on Thursday, Airbnb had a market cap of $86.5 billion on a non-diluted basis, more than hotel chains Marriott, Hilton and Wyndham Hotels combined. Airbnb raised at least $3.4 billion in the IPO, among the most-ever for a U.S. tech company.
Airbnb didn't say how many hosts participated in the IPO or if it fully allocated the 3.5 million shares. Assuming 200 shares was the maximum allotment, the company could have issued that amount to as many 17,500 of its more than 4 million hosts.
To be eligible, hosts had to be based in the U.S. and active on the platform in 2019 or 2020, accepting a reservation by Nov. 1. Participants told CNBC they were informed of the program on Nov. 16, and had until Nov. 20 to pre-register. Seniority helped determine who got into the offering.
"If demand for the program exceeds capacity, we may invite hosts to participate based on tenure, as determined by the year they first hosted on Airbnb," the company said in its prospectus.
Getting beyond Wall Street
By opening up a piece of the IPO to hosts, Airbnb added a twist to the first-day IPO pop, an event that most often benefits large money managers who have close relationships to underwriters like Morgan Stanley and Goldman Sachs, the two firms that led Airbnb's offering.
Airbnb stuck to the general IPO playbook, but made some tweaks around the edges. It mimicked a strategy used earlier this year by Unity Software and also DoorDash, holding a hybrid auction that let executives solicit bids from investors and choose a price based on where they landed. It also allowed employees with vested shares to sell a portion of their equity in the IPO rather than having to wait for the lock-up expiration.
Airbnb isn't the first company to open its IPO to partner businesses. In Uber's IPO last year, the company set aside up to 3% of the 180 million shares for drivers. The investment has so far not paid off very well: Uber debuted at $45 in May 2019, and then traded below the IPO price until last month. It's now at $54.35, representing a 21% increase from the IPO, while the S&P 500 is up 27% over that stretch.
Uber vs. S&P 500
Even with the company's blockbuster IPO, Airbnb hosts have had a tough year in the pandemic. Because of the plunge in tourism, Airbnb's gross booking volume tumbled 39% in the first three quarters of 2020 from a year earlier to $18 billion, and the decline is expected to continue through the year.
Airbnb didn't help itself with the way it responded to cancellations. In the initial weeks after Covid-19 hit the U.S., it gave travelers refunds, leaving hosts at a loss. In March, Airbnb set aside $250 million to compensate property owners for missed stays, but by the following month hundreds of hosts were complaining in online forums that they hadn't received any compensation or even follow-up communications.
While an IPO allotment to a small percentage of hosts won't make up for a year of financial struggles experienced by so many people who relied on the site for income, Airbnb optimists like Richard Fertig say the IPO reflects brighter days ahead for the business.
Richard Fertig
Source: Richard Fertig
A former hedge fund manager, Fertig lives in Jackson Hole, Wyoming, and is now a full-time real estate owner, with boutique hotels in Costa Rica, Palm Beach, Florida, and Palm Springs, California. He's been listing properties on Airbnb for about eight years, and says he has millions of dollars in real estate tied to short-term rentals.
Like Schurr, Fertig signed up to purchase 275 shares and was ultimately awarded 200.
"I think it will be a very popular and successful IPO," Fertig said, in an interview on Wednesday, before the offering. "We'll have a very nice initial bump but I also think the business has very strong legs going forward because of the rate of adoption and growth and use."
'Experience trumps luxury'
Fertig, who teaches YouTube courses about Airbnb and short-term rentals, said his business slumped from March through early June, but picked back up in the summer and hasn't let up since. Travelers have flocked to beach towns and rural getaways for family vacations and to work remotely, playing right into Fertig's sweet spot.
"The world today is more about experiences," Fertig said, from his Costa Rica hotel. "Experience trumps luxury."
Property owners weren't the only ones who got to participate in the IPO. Daniel Chan, a Bay Area magician who regularly entertains techies, performs shows as part of Airbnb Experiences. He's done magic for C-level executives at the company and has been booking shows through Airbnb for about four years.
Daniel Chan is among a number of small-business owners who rely on the tech industry's parties and conferences and are now staring at blank calendars with no idea of when they will be able to return to their livelihoods.
Courtesy of Daniel Chan
Speaking on Thursday, just before starting a show on Zoom, Chan, who promotes himself as "the billionaires' magician," said he was allotted 200 shares. He, too, originally registered for 275 when the price range was $44 to $50.
Chan said he got nervous when the range ticked up and said he was disappointed when it jumped to $68 because he thought it would be much harder to generate gains. He's surprised how wrong he was.
Now, the question for Chan is how long to hold. He doesn't have a lot of experience with stock investing, though he said he's purchased one share of Google every time he's performed for the company, and now owns over $100,000 worth of Alphabet stock.
During the pandemic, much of Chan's business has come through Airbnb — he said he has 22 virtual shows lined up this month. While Chan likes the idea of backing the company that's supporting his business, he's been around Silicon Valley long enough to see some ugly implosions.
"I might sell slowly, but my wife is saying don't sell at all," said Chan, adding that the capital gains tax he faces is a reason to stand pat. "I don't think it will ever be Pets.com, but you've seen stocks go down so it's a really iffy thing."
WATCH: Airbnb opens at $146