text:
\section{Introduction}\label{sec:secIntro}
Recent developments in quantum information have resulted in an
increasing number of applications: for instance, quantum
teleportation, quantum dense coding, quantum cryptography, quantum
logic gates, quantum algorithms and
etc~\cite{Nielsen2000,Galindo2002,Gisin and Thew,Gisin and
Ribordy,rev}. Many tasks in quantum information processing (QIP)
possess different properties from their classical
counterparts. One such case is quantum NOT gate.
Classically, we can use the NOT gate to invert (complement) a bit, by
changing the value of a bit, from 0 to 1 and vice versa.
Complementing a qubit, however, is another matter. The complement of
a state $\ket{\psi}$ is the state $\ket{\psi^{\bot}}$ that is
orthogonal to it. In the quantum case, as shown by Bu\v{z}ek,
Hillery and Werner~\cite{BHW-UNG1999}, it is impossible to build a
device that  transforms an unknown quantum state into the state
orthogonal to it. That is to say, we cannot design a perfect
universal-NOT (U-NOT) gate. This difference between
classical and quantum information processing is closely
related to the no-cloning theorem~\cite{Wootters1982}. However, such
no-go theorem does not forbid imperfect quantum quantum
cloning~\cite{Buzek1996, Gisin1997, Brub1998, Gisin1998, Werner1998,
Keyl1999, Brub2000, phase1, phase2, buc, phase3, phase4, phase5,
Zou2005, Cerf2002, Fan2003, Fan2001, Cerf and Durt2002, Durt2003,
Hu2008}. Also, approximate quantum NOT gates do
exist~\cite{BHW-UNG1999} and it is interesting to know how well we
can orthogonalize an unknown quantum state. Bu\v{z}ek, Hillery and
Werner~\cite{BHW-UNG1999} have introduced a U-NOT gate that implements an
approximate NOT operation to an unknown quantum state $\ket{\psi}$
on the Bloch's sphere and generates an output that is as close as
possible to $\ket{\psi^{\bot}}$, which is orthogonal to
$\ket{\psi}$.

In many real applications of the quantum information system, we
often have partial information about a  2-level quantum state,
i.e., the state is distributed on a specific area on the Bloch
sphere. Such partial information as in phase covariant
$1$-to-$1$ NOT gate allows us to orthogonalize such states by
transforming $\ket{0}$ to $\ket{1}$ and $\ket{1}$ to $-\ket{0}$.
Thus, any phase covariant states can be orthogonalized
perfectly. In this work, we study the problem of 1-to-$M$ quantum NOT gate where
the input state is uniformly distributed between two latitudes of
the Bloch sphere rather than the whole Bloch sphere. By considering the case in which the
two latitudes are brought to the poles, we obtain the U-NOT gate with the
optimal fidelity $F=2/3$~\cite{BHW-UNG1999}. However, if the two latitudes
collapse into a geodesic circle of the Bloch sphere, we
obtain the phase covariant NOT gate.

Taking qubit $\ket{\psi}=\cos\frac{\theta}{2}\ket{0}+
\sin\frac{\theta}{2} e^{i\phi}\ket{1}$ with $\phi\in[0,2\pi]$ and
$\theta_{1}\leq\theta\leq\theta_2$ as input, and the outputs of our
$1$-to-$M$ NOT gate will always be some multipartite entangled
states. As a result, the controllable generation of these entangled
states becomes very important. But in general, it is extremely
difficult to generate experimentally multipartite entangled states
through a single global unitary operation. For this purpose, the
sequential generation of the entangled states appears to be more promising
and a lot of effort has been made in recent years in this direction.
The general
sequential generation of entangled multiqubit states in the realm of
cavity QED have been systematically studied
in~\cite{SchonPRL2005,SchonPRA2007}. It is pointed out that the
$1$-to-$M$ sequential quantum cloning is
possible~\cite{DelgadoPRL2007}. Dang and Fan~\cite{DangJPA2008}
extended the sequential quantum cloning to the general $N$-to-$M$ case
and considered also $d$-level systems.

To this end, we consider the following state:
\begin{equation}
  \ket{\psi}=\cos{\frac{\theta}{2}}\ket{0}+
  \sin{\frac{\theta}{2}}e^{i\phi}\ket{1}
\end{equation}
where $\phi\in[0,2\pi]$ and $\theta_{1}\leq\theta\leq\theta_2$ with
constants $\theta_1,\theta_2$. The states we considered here are
distributed uniformly between two latitudes on the Bloch sphere.
When $\theta_{1}=0$ and $\theta_{2}=\pi$, we get the situation of
the U-NOT gate. In this way, the result of U-NOT gate is recovered as
special cases of our NOT gate. When $\theta_{1}=\pi/2$ and
$\theta_{2}=\pi/2$, we obtain the NOT gate for phase covariant
states.

This paper is arranged as follows: We formulate our problem and
present analytical results to our situation in the next section. In
Sec.\uppercase\expandafter{\romannumeral3}, we analyze the
$1$-to-$M$ NOT gate within a sequential generation scheme and
express the sequential NOT gate in explicit form. We end the paper
with some concluding remarks.

\section{Quantum NOT gate for qubits between two latitudes on the Bloch
sphere  }\label{sec:NG} The state we wish to orthogonalize can be
written as
\begin{equation}\label{eq:State}
  \ket{\psi}=\cos{\frac{\theta}{2}}\ket{0}+
  \sin{\frac{\theta}{2}}e^{i\phi}\ket{1}
\end{equation}
where $\phi\in[0,2\pi]$ and $\theta_1\leq\theta\leq\theta_2$, i.e.,
the states we considered here are distributed uniformly within a region enclosed by
two latitudes on the Bloch sphere. We assume
the following unitary transformation for our NOT gate:
\begin{eqnarray}\label{eq:eqNG}
  U: &&
  \ket{0}\ket{X}\rightarrow
  \sum_{k=0}^{M}{\ket{(M-k)0,k1}\otimes\ket{A_{k}}} \nonumber \\
  &&
  \ket{1}\ket{X}\rightarrow
  \sum_{k=0}^{M}{\ket{k0,(M-k)1}\otimes\ket{A_{M+k+1}}}
\end{eqnarray}
where $\ket{(M-k)0,k1}$ denotes symmetric and normalized states with
$M-k$ qubits in $\ket{0}$ and $k$ qubits in $\ket{1}$. This ensures a
symmetric NOT gate and that all the
first $M$ qubits at the output of the NOT gate are the same.
$\ket{A_{l}}(l=0,1,\cdots,2M+1)$ are unormalized states. Let
$a_{k,l}=\iprod{A_l}{A_k}$ and denote $a_k=\iprod{A_k}{A_k}$ for
short where $a_{k,l}$ are the parameters that we want to determine.

After applying the unitary operation $U$, we can get the following
state with the input qubit $\ket{\psi}$ described by
Eq.~\eqref{eq:State}:
\begin{eqnarray}\label{eq:Out}
  \ket{\psi_{\textrm{out}}} &&
    =\cos\frac{\theta}{2}\sum_{k=0}^{M}{\ket{(M-k)0,k1}\otimes\ket{A_{k}}}
    \nonumber\\
    & &+\sin\frac{\theta}{2}e^{i\psi}\sum_{k=0}^{M}{\ket{k0,(M-k)1}\otimes\ket{A_{M+k+1}}}
\end{eqnarray}
By taking partial trace, we obtain the reduced density matrix
$\rho_{k}$ for the $k$-th output qubit, and all the reduced density
matrix are the same for $k=1,2,\cdots,M$. With the reduced density
matrix $\rho_{k}$, we can calculate the fidelity:
\begin{eqnarray}\label{eq:Fidelity}
  & &
    F=\bra{\psi^{\perp}}\rho_{k}\ket{\psi^{\perp}}\nonumber\\
  &=&
    \sin^{2}\frac{\theta}{2}\cos^{2}\frac{\theta}{2}\left[
    \sum_{k=0}^{M-1}{\frac{M-k}{M}(a_{k}+a_{M+k+1})}\right.
    \nonumber \\
  & &
    \left.+2\sum_{k=0}^{M-1}{\frac{\sqrt{(M-k)(K+1)}}{M}
    \sqrt{|a_{M+k+1,M-k-1}|^{2}}}\right] \nonumber\\
  & &
    +\sin^{4}\frac{\theta}{2}\sum_{k=0}^{M-1}
    {\frac{\binom{M-1}{M-k-1}}{\binom{M}{k+1}}a_{M+k+2}} \nonumber\\
  & &
    +\cos^{4}\frac{\theta}{2}\sum_{k=0}^{M-1}
    {\frac{\binom{M-1}{k}}{\binom{M}{k+1}}a_{k+1}} \nonumber\\
  & &
    +e^{i\phi}C_{1}+e^{-i\phi}C_{1}^{*}+e^{2i\phi}C_{2}+e^{-2i\phi}C_{2}^{*}
\end{eqnarray}
where $C_{1}^{*}$ is the complex conjugation of $C_{1}$ and the same
for $C_{2}^{*}$. sincee the input state $\ket{\psi}$ given by
Eq.~\eqref{eq:State} is arbitrary, the parameters $\phi\in [0,2\pi]$
and $\theta\in[ \theta_1, \theta_2]$ are unknown and are
distributed uniformly on a belt of the Bloch sphere.
We need to average the
fidelity over all possible cases. The last four terms in
Eq.~\eqref{eq:Fidelity} disappear as a consequence of the averaging
over all
possible angles $\phi$. Moreover, we obtain the same optimal NOT gate
by assuming that $C_{1}$ and $C_{2}$ are equal to zero. On the other
hand, by using the definitions of $a_{k,l}$ we can easily get that
$a_{k,l}=a_{l,k}^{*}$, $|a_{k,l}|^{2}= a_{k,l}*a_{l,k}\leq
a_{k}a_{l}$. Equality is obtained if and only if $a_{k,l}=a_{l,k}$ are
real numbers.  Using Eq.~\eqref{eq:Fidelity}, the fidelity becomes:
\begin{eqnarray}\label{eq:SimpleF}
    F
  &=&
    \sin^{2}\frac{\theta}{2}\cos^{2}\frac{\theta}{2}\left[
    \sum_{k=0}^{M-1}{\frac{M-k}{M}(a_{k}+a_{M+k+1})}\right.
    \nonumber \\
  & &
    \left.+2\sum_{k=0}^{M-1}{\frac{\sqrt{(M-k)(K+1)}}{M}
    \sqrt{a_{M+k+1}a_{M-k-1}}}\right] \nonumber\\
  & &
    +\sin^{4}\frac{\theta}{2}\sum_{k=0}^{M-1}
    {\frac{\binom{M-1}{M-k-1}}{\binom{M}{k+1}}a_{M+k+2}} \nonumber\\
  & &
    +\cos^{4}\frac{\theta}{2}\sum_{k=0}^{M-1}
    {\frac{\binom{M-1}{k}}{\binom{M}{k+1}}a_{k+1}}
\end{eqnarray}
Averaging the fidelity over all possible angles
$\theta$~\cite{Gisin1997}, and observing that $\sum_{k=0}^{M}{a_k}=
\sum_{k=0}^{M}{a_{M+k+1}}=1$, we have
\begin{eqnarray}\label{eq:MF}
  \bar{F}
  &=&
    \frac{\int_{\theta_{1}}^{\theta_{2}}{F\sin{\theta}d{\theta}}}
    {\int_{\theta_{1}}^{\theta_{2}}{\sin{\theta}d{\theta}}}
    \nonumber\\
  &=&
    \frac{1}{2}+\frac{1}{6}K
    \nonumber\\
  & &
    +P\sum_{k=0}^{M-1}
    {\frac{\sqrt{(M-k)(k+1)}}{M}\sqrt{a_{M+k+1}a_{M-k-1}}} \nonumber\\
  & &
    -Q\sum_{k=0}^{M-1}{\frac{M-k}{M}a_{k}}
    -R\sum_{k=0}^{M-1}{\frac{M-k}{M}a_{M+k+1}}
\end{eqnarray}
where
$K=\cos^{2}{\theta_{1}}+\cos{\theta_1}\cos{\theta_2}+\cos^{2}{\theta_2}$,
$P=\frac{3-K}{6}$,
$Q=\frac{K}{6}+\frac{1}{4}\left(\cos{\theta_1}+\cos{\theta_2}\right)$,
$R=\frac{K}{6}-\frac{1}{4}\left(\cos{\theta_1}+\cos{\theta_2}\right)$,
and $K,P,Q,R$ are constants with given $\theta_1$ and $\theta_2$. In
order to get the optimal quantum NOT gate, we should maximize
$\bar{F}$ with respect to $a_k(k=0,1,\cdots,M-1,M+1,M+2,\cdots,2M)$.

We now seek a solution of $a_k$ with maximum $\bar{F}$. It is interesting to note that
if the state lies somehwere  on the whole Bloch sphere, $\theta_{1}=0$ and $\theta_2=\pi$, and we have
$K=1$, $P=\frac{1}{3}$ and $Q=R=\frac{1}{6}$. The optimal fidelity
is $\bar{F}=\frac{2}{3}$ with $a_{M+k+1}=a_{M-k-1}
(k=0,1,\cdots,M-1)$, recovering the well known result for the
$1$-to-$M$ U-NOT gate in~\cite{BHW-UNG1999}. In this case, the
fidelity is constant and the optimal Universal NOT gate can be
realized via a "measurement + re-preparation"
scheme~\cite{BHW-UNG1999}. In the general situation, with states
uniformly distributed in a belt on the Bloch sphere, the fidelity is
dependent on the number of output qubits $M$ as shown below in this
article. So we can not realized via a "measurement + re-preparation"
scheme for the general case.

For the case in which the state is phase covariant,
$\theta_1=\theta_2=\frac{\pi}{2}$, and we have
$K=0$, $P=\frac{1}{2}$ and $Q=R=0$. The optimal fidelity is
$\bar{F}=\frac{1}{2}+\frac{\sqrt{M(M+2)}}{4M}$ for even $M$, and
$\bar{F}=\frac{1}{2}+\frac{M+1}{4M}$ for odd $M$. This fidelity is
just equal to the fidelity of optimal $1$ to $M$ phase-covariant
quantum cloning machine~\cite{Brub2000,Fan2001,Scarani2005}. As
mentioned above, the $1$-to-$1$ phase-covariant NOT gate can be
constructed perfectly. So we can achieve the $1$-to-$M$ optimal
phase-covariant NOT gate by combining the $1$-to-$1$ perfect NOT
gate with the $1$-to-$M$ optimal phase-covariant cloning machine.
The fidelities of the $1$-to-$M$ optimal phase-covariant NOT
gate and QCM must be the same as analyzed before.

In the general situation, we need to optimize the fidelity in
Eq.~\eqref{eq:MF} under the restrictions $0\leq a_{k}\leq 1
(k=0,1,\cdots,M-1,M+1,\cdots,2M),$  $\sum_{k=0}^{M-1}{a_k}\leq 1,$
and $\sum_{k=0}^{M-1}{a_{M+k+1}}\leq 1.$ By considering the smoothness
of $\bar{F}$, the maximum value should be achieved at the extremal
points or on the boundary. We analyze the optimization problem with
restrictions and get the following optimal NOT gate for the following
situations:

\begin{enumerate}
  \item When $|\theta_{1}-\frac{\pi}{2}|\geq
  |\theta_{2}-\frac{\pi}{2}|$ and $M$ is odd, we have
  $a_{\frac{M-1}{2}}=\min((\frac{P}{2Q})^2,1),
  a_{M}=1-a_{\frac{M-1}{2}}, a_{\frac{3M+1}{2}}=1,
  a_{\frac{3M+1}{2},\frac{M-1}{2}}=a_{\frac{M-1}{2},\frac{3M+1}{2}}=
  -\sqrt{a_{\frac{M-1}{2}}}$, and $a_{k,l}=0$ otherwise. The
  fidelity is $\bar{F}=\frac{1}{2}+\frac{K}{6}+\frac{M+1}{2M}(
  \frac{P^{2}}{4Q}-R)$ for
  $a_{\frac{M-1}{2}}=(\frac{P}{2Q})^{2}$, and
  $\bar{F}=\frac{1}{2}+\frac{K}{6}+\frac{M+1}{2M}(P-Q-R)$ for
  $a_{\frac{M-1}{2}}=1$.
  \item When $|\theta_{1}-\frac{\pi}{2}|<
  |\theta_{2}-\frac{\pi}{2}|$ and $M$ is odd, we have
  $a_{\frac{M-1}{2}}=1,
  a_{\frac{3M+1}{2}}=\min((\frac{P}{2R})^2,1),
  a_{2M+1}=1-a_{\frac{3M+1}{2}},
  a_{\frac{3M+1}{2},\frac{M-1}{2}}=a_{\frac{M-1}{2},\frac{3M+1}{2}}=
  -\sqrt{a_{\frac{3M+1}{2}}}$, and $a_{k,l}=0$ otherwise. The
  fidelity is $\bar{F}=\frac{1}{2}+\frac{K}{6}+\frac{M+1}{2M}(
  \frac{P^{2}}{4R}-Q)$ for
  $a_{\frac{3M+1}{2}}=(\frac{P}{2R})^{2}$, and
  $\bar{F}=\frac{1}{2}+\frac{K}{6}+\frac{M+1}{2M}(P-Q-R)$ for
  $a_{\frac{3M+1}{2}}=1$.
  \item When $|\theta_{1}-\frac{\pi}{2}|\geq
  |\theta_{2}-\frac{\pi}{2}|$ and $M$ is even, we have
  $a_{\scriptstyle{\frac{M}{2}}}=\min(\frac{P^{2}(M+2)}{4Q^{2}M},1),
  a_{\scriptstyle{M}}=1-a_{\scriptstyle{\frac{M}{2}}},
  a_{\scriptstyle{\frac{3M}{2}}}=1,
  a_{\scriptstyle{\frac{3M}{2},\frac{M}{2}}}=
  a_{\scriptstyle{\frac{M}{2},\frac{3M}{2}}}=
  -\sqrt{a_{\scriptstyle{\frac{M}{2}}}}$, and $a_{k,l}=0$ otherwise.
  The fidelity is $\bar{F}=\frac{1}{2}+\frac{K}{6}+
  \frac{M+2}{2M}(\frac{P^{2}}{4Q}-R)$ for
  $a_{\scriptstyle{\frac{M}{2}}}=\frac{P^{2}(M+2)}{4Q^{2}M}$, and
  $\bar{F}=\frac{1}{2}+\frac{K}{6}+
  P\frac{\sqrt{\frac{M}{2}(1+\frac{M}{2})}}{M}-
  R\frac{M+2}{2M}-\frac{1}{2}Q$ for
  $a_{\scriptstyle{\frac{M}{2}}}=1$.
  \item When $|\theta_{1}-\frac{\pi}{2}|<
  |\theta_{2}-\frac{\pi}{2}|$ and $M$ is even, we have
  $a_{\frac{M}{2}-1}=1,
  a_{\frac{3M}{2}+1}=\min(\frac{P^{2}(M+2)}{4R^{2}M},1),
  a_{2M+1}=1-a_{\frac{3M}{2}+1},
  a_{\frac{M}{2}-1,\frac{3M}{2}+1}=
  a_{\frac{3M}{2}+1,\frac{M}{2}-1}= -\sqrt{a_{\frac{3M}{2}+1}}$, and
  $a_{k,l}=0$ otherwise. The fidelity is
  $\bar{F}=\frac{1}{2}+\frac{K}{6}+\frac{M+2}{2M}(\frac{P^{2}}{4R}-Q)$
  for $a_{\frac{3M}{2}+1}=\frac{P^{2}(M+2)}{4R^{2}M}$, and
  $\bar{F}=\frac{1}{2}+\frac{K}{6}+
  P\frac{\sqrt{\frac{M}{2}(1+\frac{M}{2})}}{M}-
  Q\frac{M+2}{2M}-\frac{1}{2}R$ for $a_{\frac{3M}{2}+1}=1$.
\end{enumerate}
The explicit NOT gate transformations have already been
presented in Eq.~\eqref{eq:eqNG} by letting
$\ket{A_{[\frac{M}{2}]}}=-\sqrt{a_{[\frac{M}{2}]}}\ket{\uparrow},
\ket{A_{M}}=\sqrt{1-a_{[\frac{M}{2}]}}\ket{\downarrow},
\ket{A_{[\frac{3M+1}{2}]}}=\ket{\uparrow}$, and $\ket{A_{k}}=0$
otherwise for case 1 and 3; by letting
$\ket{A_{[\frac{M-1}{2}]}}=\ket{\uparrow},
\ket{A_{[\frac{3M}{2}+1]}}=-\sqrt{a_{[\frac{3M}{2}+1]}}\ket{\uparrow},
\ket{A_{2M+1}}=\sqrt{1-a_{[\frac{3M}{2}+1]}}\ket{\downarrow}$, and
$\ket{A_{k}}=0$ otherwise for case 2 and 4.

It is interesting to note that the output states of the NOT gate given
by Eq.~\eqref{eq:eqNG} are always entangled. As shown by Bu\v{z}ek,
Hillery and Werner~\cite{BHW-UNG1999}, the optimal U-NOT gate can be
realized via a ``measurement + repreparation" scheme. Moreover for each
measurement result obtained, the prepared state can be taken to be a
product one. In this case, it is an ``easy" operations, and requires
no generation of entanglement, nor unitary operations - just
measurement and preparation of product states. Unfortunately, there
is no ``measurement + repreparation" scheme in the general situation,
as the optimal fidelity is dependent on the number of output qubits
$M$. So the generation of entanglement is unavoidable. As a result,
the controlled generation of these entangled states becomes very
important. In the next section, we consider the generation of these
entangled states and present the sequential quantum NOT gate.

\section{The $1$-to-$M$ Sequential Quantum NOT gate}\label{sec:SNG}

As shown in~\cite{SchonPRL2005,DelgadoPRL2007,DangJPA2008}, the
sequential generation of a multiqubit state is as follows. Let
$\mathcal{H}_{A}\simeq\mathbb{C}^{D}$ and $\mathcal{H}_{B}\simeq
\mathbb{C}^{2}$ be the Hilbert spaces characterizing a
$D$-dimensional ancillary system and a single qubit respectively. At
every step of the sequential generation of a multiqubit state, a
unitary time evolution will be acting on the joint system
$\mathcal{H}_{A}\otimes\mathcal{H}_{B}$. Assuming that each qubit is
initially in the state $\ket{0}$, we disregard the qubit at the
input and write the evolution in the form of an isometry
$V:\mathcal{H}_{A}\to \mathcal{H}_{A}\otimes\mathcal{H}_{B}$, where
$V=\sum_{i,\alpha,\beta}{V^{i}_{\alpha,\beta}\ket{\alpha,i}\bra{\beta}}$,
each $V^{i}$ is a $D\times D$ matrix and the isometry condition
takes the form $\sum_{i=0}^{1}{[V^{i}]^{\dagger}V^{i}}=1$. If we
apply successively $n$ operations of this form to an initial state
$\ket{\varphi_{I}}\in\mathcal{H}_{A}$, we obtain the state
$\ket{\Phi}=V^{[n]}\cdots V^{[2]}V^{[1]}\ket{\varphi_{I}}$. The $n$
generated qubits are in general entangled. Assuming in the last step
the ancilla decouples from the system, such that
$\ket{\Phi}=\ket{\varphi_{F}}\otimes\ket{\varphi}$, and we are left with
the $n$-qubit state
\begin{equation}\label{eq:eqMPS}
  \ket{\varphi}=\sum_{i_{1},\cdots,i_{n}=0}^{1}{\bra{\varphi_{F}}V^{[n]i_{n}}\cdots
  V^{[1]i_{1}}\ket{\varphi_{I}}\ket{i_{n}\cdots i_{1}}},
\end{equation}
where $\ket{\varphi_{F}}$ is the final state of the ancilla. The
state~\eqref{eq:eqMPS} is a matrix-product state (MPS) (cf., e.g.,
\cite{PerezQIC2007}, and references therein), already
comprehensively studied in~\cite{SchonPRL2005, SchonPRA2007,
DelgadoPRL2007, DangJPA2008, PerezQIC2007, AffleckPRL1987,
VidalPRL2003, VerstraetePRL2004}. Moreover, it was proven that any
multiqubit MPS can be sequentially generated~\cite{SchonPRL2005}.

The NOT gate given by Eq.~\eqref{eq:eqNG} can approximately
orthogonalize one input state to $M$ copies. Next we show that this
general $1$-to-$M$ NOT gate can be generated through a sequential
procedure. The basic idea is to show that the final states
$\ket{\Psi_{1M}^{k}}$ in Eq.~\eqref{eq:eqNG} can be expressed in its
MPS form. As presented in~\cite{SchonPRL2005}, any MPS can be
sequentially generated. We shall follow the method as
in~\cite{DelgadoPRL2007, DangJPA2008, VidalPRL2003}.

Taking one output entangled state in case 1 for example. We have
\begin{equation}\label{eq:OutExample}
  \ket{\Psi_{1M}^{0}}=-\sqrt{\gamma}
  \ket{\frac{M+1}{2}0,\frac{M-1}{2}1}\ket{0}+\sqrt{1-\gamma}\ket{M1}\ket{1}
\end{equation}
where $\gamma=a_{\frac{M-1}{2}}$. By Schmidt decomposition, we first
express the quantum state $\ket{\Psi_{1M}^{0}}$ as a bipartite pure
state in $\mathcal{H}_{A_{1}}\otimes\mathcal{H}_{B_{1}}$ with two
particle sets $A_{1}=\{1\}$ and $B_{1}=\{2,3,\cdots,M+1\}$.
\begin{eqnarray}\label{eq:eqSD1}
  \ket{\Psi_{1M}^{0}}
  &=&
    \lambda_{1}^{[1]}\ket{0}\ket{\psi_{1}^{[2,\cdots,M+1]}}+
    \lambda_{2}^{[1]}\ket{1}\ket{\psi_{2}^{[2,\cdots,M+1]}}\nonumber
    \\
  &=&
    \sum_{\alpha_{1},i_{1}}{\Gamma_{\alpha_{1}}^{[1]i_{1}}
    \lambda_{\alpha_{1}}^{[1]}\ket{i_{1}}\ket{\psi_{\alpha_{1}}^{[2,\cdots,M+1]}}},
\end{eqnarray}
where $\lambda_{\alpha_{1}}^{[1]}$ are eigenvalues of the first
qubit reduced density operator and we find
$\lambda_{1}^{[1]}=\sqrt{\gamma\frac{M+1}{2M}}$,
$\lambda_{2}^{[1]}=\sqrt{1-\gamma\frac{M+1}{2M}}$. Matching indices
in Eq~\eqref{eq:eqSD1}, we have
$\Gamma_{\alpha_{1}}^{[1]0}=\delta_{\alpha_{1},1}$ and
$\Gamma_{\alpha_{1}}^{[1]1}=\delta_{\alpha_{1},2}$. To correspond
with the MPS form in Eq.~\eqref{eq:eqMPS}, we define
\begin{equation}\label{eq:eqV1}
  V_{\alpha_{1}}^{[1]i_{1}}=\Gamma_{\alpha_{1}}^{[1]i_{1}}\lambda_{\alpha_{1}}^{[1]}.
\end{equation}

By successive Schmidt decomposition, the quantum state
$\ket{\Psi_{1M}^{0}}$ in Eq.~\eqref{eq:OutExample} can be considered
as a bipartite pure state in
$\mathcal{H}_{A_{n}}\otimes\mathcal{H}_{B_{n}}$ with particle sets
$A_{n}=\{1,2,\cdots,n\}$ and $B_{n}=\{n+1,n+2,\cdots,M+1\}$, where
$1<n\leq M$. We have
\begin{equation}\label{eq:eqSDn}
  \ket{\Psi_{1M}^{0}}= \sum_{l=0}^{n}{ \lambda_{l+1}^{[n]}
  \ket{(n-l)0,l1}\ket{\psi_{l+1}^{[n+1,\cdots,M+1]}}},
\end{equation}
where $\lambda_{l+1}^{[n]}$ are eigenvalues of the first $n$ qubits
reduced density operator of $\ket{\Psi_{1M}^{0}}$. We can obtain
\begin{widetext}
\begin{equation}\label{eq:eqLambdan}
  \left\{
  \begin{array}{ll}
    \lambda_{l+1}^{[n]}=\sqrt{\gamma\frac{\binom{n}{l}
    \binom{M-n}{\frac{M-1}{2}-l}}{\binom{M}{\frac{M-1}{2}}}}, &
    1<n\leq\frac{M-1}{2}, l=0,1,\cdots,n-1 ;\quad
    \frac{M+1}{2}\leq n\leq M,
    l=n-\frac{M+1}{2},\cdots,\frac{M-1}{2}.\\
    \lambda_{n+1}^{[n]}=\sqrt{1-\gamma+\gamma
    \frac{\binom{M-n}{\frac{M-1}{2}-n}}{\binom{M}{\frac{M-1}{2}}}},
    & \begin{array}{lrl}
      1<n\leq \frac{M-1}{2}. &
      \hspace{5mm}\lambda_{n+1}^{[n]}=\sqrt{1-\gamma}, &
      \frac{M+1}{2}\leq n\leq M.
    \end{array} \\
    \lambda_{l+1}^{[n]}=0, & \textrm{otherwise}.
  \end{array}
  \right.
\end{equation}
and
\begin{equation}\label{eq:eqReducedState}
  \left\{
  \begin{array}{ll}
    \ket{\psi_{l+1}^{[n+1,\cdots,M+1]}}=
    -\ket{(\frac{M+1}{2}-n+l)0,(\frac{M-1}{2}-l)1}\ket{0}, &
    \hspace{-40mm} 1<
    n\leq \frac{M-1}{2},l=0,1,\cdots,n-1. \\
    \ket{\psi_{n+1}^{[n+1,\cdots,M+1]}}=
    -\frac{\sqrt{\gamma}}{\lambda_{n+1}^{[n]}}
    \sqrt{\frac{\binom{M-n}{\frac{M-1}{2}-n}}{\binom{M}{\frac{M-1}{2}}}}
    \ket{\frac{M+1}{2}0,(\frac{M-1}{2}-n)1}\ket{0}+
    \frac{\sqrt{1-\gamma}}{\lambda_{n+1}^{[n]}}\ket{(M-n)1}\ket{1},
    & 1<n\leq \frac{M-1}{2}. \\
    \ket{\psi_{l+1}^{[n+1,\cdots,M+1]}}=
    -\ket{(\frac{M+1}{2}-n+l)0,(\frac{M-1}{2}-l)1}\ket{0}, &
    \hspace{-40mm} \frac{M+1}{2}\leq n\leq M,
    l=n-\frac{M+1}{2},\cdots,\frac{M-1}{2}. \\
    \ket{\psi_{n+1}^{[n+1,\cdots,M+1]}}= \ket{(M-n)1}\ket{1}, &
    \hspace{-40mm} \frac{M+1}{2}\leq n\leq M.\\
    \ket{\psi_{l+1}^{[n+1,\cdots,M+1]}}=0, & \hspace{-40mm}
    \textrm{otherwise}.
  \end{array}
  \right.
\end{equation}
\end{widetext}
According to the results in Eq.~\eqref{eq:eqLambdan} and
\eqref{eq:eqReducedState}, we get the following recursion formula

\begin{eqnarray}\label{eq:eqRecursion}
  & &
    \ket{\psi_{l+1}^{[n,n+1,\cdots,M+1]}}\nonumber\\
  &=&
    \frac{\sqrt{\binom{n-1}{l}}}{\lambda_{l+1}^{[n-1]}}
    \left[\frac{\lambda_{l+1}^{[n]}}{\sqrt{\binom{n}{l}}}
    \ket{0}\ket{\psi_{l+1}^{[n+1,\cdots,M+1]}}\right. \nonumber\\
  & &
    \left.+
    \frac{\lambda_{l+2}^{[n]}}{\sqrt{\binom{n}{l+1}}}
    \ket{1}\ket{\psi_{l+2}^{[n+1,\cdots,M+1]}}\right]
\end{eqnarray}
Comparing Eq.~\eqref{eq:eqRecursion} with the following relation
\begin{eqnarray*}\label{eq:eqRecursionPar}
  & &
    \ket{\psi_{l+1}^{[n,n+1,\cdots,M+1]}}\nonumber \\
  &=&
    \sum_{\alpha_{n},i_{n}}{
    \Gamma_{(l+1)\alpha_{n}}^{[n]i_{n}}\lambda_{\alpha_{n}}^{[n]}
    \ket{i_{n}}\ket{\psi_{\alpha_{n}}^{[n+1,\cdots,M+1]}}},
\end{eqnarray*}
we have
\begin{equation}\label{eq:eqGamma1}
  \Gamma_{(l+1)\alpha_{n}}^{[n]0}=\delta_{(l+1)\alpha_{n}}
  \frac{\sqrt{\binom{n-1}{l}}}
  {\lambda_{l+1}^{[n-1]}\sqrt{\binom{n}{l}}},
\end{equation}
\begin{equation}\label{eq:eqGamma2}
  \Gamma_{(l+1)\alpha_{n}}^{[n]1}=\delta_{(l+2)\alpha_{n}}
  \frac{\sqrt{\binom{n-1}{l}}}
  {\lambda_{l+1}^{[n-1]}\sqrt{\binom{n}{l+1}}}.
\end{equation}
In order to get the MPS form in Eq.~\eqref{eq:eqMPS}, we define that
\begin{equation}\label{eq:eqVn}
  V_{\alpha_{n}\alpha_{n-1}}^{[n]i_{n}}=
  \Gamma_{\alpha_{n-1}\alpha_{n}}^{[n]i_{n}}\lambda_{\alpha_{n}}^{[n]},
  \quad (1<n\leq M).
\end{equation}

After performing $M$ sequential Schmidt decompositions, the states
on the rhs in Eq.~\eqref{eq:eqSDn} can be written as
$\ket{\psi_{\frac{M+1}{2}}^{[M+1]}}=-\ket{0}$ and
$\ket{\psi_{M+1}^{[M+1]}}=\ket{1}$. Checking the above-defined $V$,
we find that the isometry condition
$\sum_{i_{n}}{[V^{[n]i_{n}}]^{\dagger}V^{[n]i_{n}}}=1$ is satisfied.

Until now, we have found that the output state of the general
quantum NOT gate can be expressed as a MPS as in
form~\eqref{eq:eqMPS}. So the sequential quantum NOT gate is
obtainable.

\section{Concluding Remark}

In summary, by applying quantum orthogonalizing transformations
for the state uniformly distributed between two latitudes on the
Bloch sphere, we present a general $1$-to-$M$ quantum NOT gate. The
usual U-NOT gate is a special case and we find out that the optimal
fidelity of the U-NOT gate is consistent with the one studied
in~\cite{BHW-UNG1999}. For another special case, we point out the
relation between the phase-covariant $1$-to-$M$ NOT gate and the
phase-covariant QCM. In the general situation, there is no
``measurement + repreparation" scheme as the U-NOT gate can be
realized via it. Consequently, the generation of entanglement is
unavoidable and the controlled generation of entangled states
becomes very important. To this end, we analyze the NOT gate within a
sequential generation scheme and show that the sequential quantum
NOT gate is feasible.

{\bf Acknowledgement:} This work was supported in part by the
National Basic Research Program of China grant No. 2007CB907900 and
2007CB807901, NSFC grant No. 60725416 and China Hi-Tech program
grant No. 2006AA01Z420.  KLC acknowledges support by the National
Research Foundation and Ministry of Education, Singapore

redpajama_set_name: RedPajamaArXiv


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text:
Az A26-os autópálya egy 197,1 km hosszúságú autópálya Olaszországban. Az út Piemont és Liguria régiókon halad keresztül. Fenntartója az .

Útvonal

Források

Olaszország autópályái
Genova közlekedése
Piemont közlekedése

redpajama_set_name: RedPajamaWikipedia


* The installation requires Microsoft Internet Explorer and .NET Framework (version 3.5). The provision on the platform Microsoft Windows 8 is described on this page.
* In order to execute the CHM file its properties have to be changed after saving it to the local hard drive, see also Microsoft Knowledge Base. In the menu Properties/General under the last menu item Security access must be permitted. This change has to be saved.


text:
DeSantis signs father engagement bill into law
by Caden DeLisa | Apr 11, 2022
Gov. Ron DeSantis on Monday signed House Bill 7065 (HB 7065) into law, which includes educational programs, mentorship programs, and one-on-one support to encourage responsible and involved fatherhood in Florida. The bill is tied to nearly $70 million in funding to provide a wide spectrum of family and youth support through the Department of Juvenile Justice (DJJ) and the Department of Children and Families (DCF).
A chief effort of the bill would direct the department to contract for the production of the "Responsible Fatherhood Initiative," partnering with a yet-to-be-decided company to launch a website and circulate materials that would inform fathers on effective parenting. The fatherhood initiative would include a print, television, and social media campaign to serve as public service announcements.
"There are more than 18 million children in our country who live without a father in their home," said DeSantis. "This has a severe impact on children and often leads to dropping out of school, crime, and substance abuse. Incredibly, there are those who diminish the importance of fatherhood and the nuclear family – we will not let that happen in our state. I am proud to say we are doing everything we can to support involved fatherhood in Florida."
Under the bill, DCF would additionally be required to award grants to non-profit organizations to address the needs of fathers. To exemplify, the bill would require that grants be geared toward helping fathers in seeking employment, managing child support obligations, transitioning from a period of incarceration, accessing health care, understanding child development, and enhancing parenting skills.
According to the U.S. Department of Justice, children from fatherless homes account for:
63 percent of youth suicides
90 percent of all homeless and runaway youths
85 percent of all children that exhibit behavioral disorders 71 percent of all high school dropouts
70 percent of juveniles in state-operated institutions
75 percent of adolescent patients in substance abuse centers
Senate President Wilton Simpson placed child welfare as one of his principal concerns for the ongoing 2022 Legislative Session, making a vow in his opening remarks to amend the risk children living in homes without fathers face.
"The earlier in life we can give a child a safe permanent home, the better opportunities that child will have for the rest of their lives," said Simpson, who was adopted as a child and grew up in a house with foster children," said Simpson. "Costs of early childhood care are consistently identified as one of the biggest barriers for would-be foster families. We will address the gap between what the voucher pays and the actual costs of early childhood care. We need to make sure more children raised by their foster relatives have access to the waiver."
Improving child welfare has long been an issue facing Florida's lawmakers, gauged as far back as 2019 by House Speaker Chris Sprowls in a speech.
"We cannot legislate fatherhood, accountability, or character, but we can provide supports for fathers to equip and encourage them to take an active role in the lives of their children," said Sprowls. "By addressing the root cause of so many of the challenges facing our sons and daughters, we will see the positive impact of this legislation send shockwaves for generations to come."

redpajama_set_name: RedPajamaCommonCrawl


text:
Hotels in Ilfracombe, England
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Royal Britannia Hotel
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About Ilfracombe
Set around a picturesque harbour on the North Devon coast, Ilfracombe is clustered with Victorian architecture. It's home to one of England's oldest working lighthouses and a striking modern theatre and also offers easy access to the legendary South West Coast Path.
Things to do in Ilfracombe
If you're interested in the local history, visit St Nicholas Chapel, which was built in the 14th century overlooking the harbour on Lantern Hill. It's one of the oldest working lighthouses in the country, having guided ships since the middle ages, and exhibits old documents and photographs detailing Ilfracombe's cultural heritage.
If you're interested in Devon's navigational history, visit the Bull Point Lighthouse, which overlooks the rugged coast to the north of Mortehoe village. The fully-automated modern lighthouse replaced an original 19th-century landmark and offers magnificent views along the coast from its lighthouse keeper's cottages.
A short drive from Ilfracombe takes you to the Heritage-listed Watermouth Castle, which was built in the mid-19th century as a country estate. In addition to its Victorian antiques and vintage amusement machines, the castle complex boasts themed rides in the Adventure, Merry-go and Gnome Land zones.
One of the most prominent buildings in Ilfracombe is the award-winning Landmark Theatre, referred to locally as "Madonna's Bra". Its double conical design hosts regular musical, dance and comedy performance, having replaced the Victorian-era Pavilion Theatre that once stood on Ilfracombe's coastal promenade.
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Limited amount of blue-colored vinyl available.
With Soon Away, GRMLN veers from the pop-punk-rooted road the band started on with last year's Empire. It's an aggressive album, darker and heavier than what's come before. While it carries these characteristics, there is also a certain peace to Soon Away, thanks to songwriter Yoodoo Park's personal growth.
The album grapples with letting go and getting used to good-byes. "Find a way to be wanted and here for someone else/ Break away from the notion, the poison in my head," Park sings on the cathartic "Faux." The singer-guitarist sees the constant changes of life allowing people to embrace the true nature of living. The teachings of Krishna were an inspiration to Park while writing the record and it's a force that helped define Park's perspective in these songs. Still, there's an intensity to these guitar-driven tracks. "Go, but you're breaking without me/ When days can crawl around/ To waste myself away," Park sings on opener "Jaded," which is built around one of GRMLN's catchiest riffs.
After writing the songs—some in Japan, some while on tour in the U.S.—Park went back to Different Fur Studios in San Francisco with engineers Patrick Brown and Sean Paulson, who worked on Empire. On Soon Away, Park is joined by his brother, Tae San, who assisted with bass parts and some song structures; and longtime drummer Keith Frerichs.


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Irvine Ranch Conservancy, a pioneering non-profit managing more than 30,000 acres of open space parks in Orange County, is celebrating its 10th anniversary.
Over the last decade, the organization has partnered with public agencies such as OC Parks, City of Irvine and City of Newport Beach to connect people to globally-significant local wildlands.
"We started out with one trail and eight volunteers, and today we manage more than 150 miles of trails and support more than 800 volunteers who help provide nearly 2,000 public activities each year," said Michael O'Connell, the Conservancy's founding and current Executive Director.
To celebrate their history, the Conservancy is hosting a special Wilderness Access Day on November 21 in Bommer Canyon.
This canyon is part of the Irvine Open Space Preserve, and the location recognizes and celebrates the Conservancy's first partner, the City of Irvine.
During this special event in Bommer Canyon, visitors will be able to explore the entire trail system as well as learn about the local flora and fauna through a pop-up nature center and ranching history with an educational nature walk.
The Conservancy was founded in 2005 with initial funding from Irvine Company Chairman Donald Bren to help care for tens of thousands of public open space acres preserved in Orange County.
At the time, Irvine Company owned more than 20,000 acres of open space, all of which has since been transferred to the County of Orange.
The land is part of the historic Irvine Ranch, and about half of the Irvine family's original 93,000 acres is permanently-preserved public open space.
With the Conservancy's stewardship assistance, nearly 40,000 acres of these public lands have been designated both a National Natural Landmark by the U.S. National Park Service and California Natural Landmark for their biological and geological significance.
"Everything we do is about partnership," said O'Connell.
The Conservancy doesn't itself own any land, but assists public owners with all aspects of stewardship.
An independent board of directors oversees the organization's strategy and operations, and more than 40 employees manage day-today functions including habitat restoration, scientific research, trail planning and maintenance, public education and recreation programs, as well as volunteer administration for park facilities across the county.
In the last five years, the Conservancy has provided more than 7,000 public activities on the land, and highly-trained Conservancy volunteers have contributed more than 100,000 hours.
Community members have also given back to the land with more than 16,000 hours of public volunteer service given to Conservancy projects since 2010.
All visitors attending the November 21 event will receive a commemorative 10th anniversary pin and Irvine Ranch Natural Landmarks keychain.
The event begins at 8 a.m., and ends at 2 p.m., with a special naturalist-led nature walk at 9 a.m.
Pre-registration for both the Wilderness Access Day and nature walk is recommended to guarantee a spot. In addition, the City of Irvine's new Bommer Nature Garden will be open during the event, along with the pop-up nature center.
Visit http://www.LetsGoOutside.org/activities to register for this event, or to find other outdoor activities to enjoy.


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text:
Mudéjar es un término que deriva de la palabra árabe  (transcrita mudaʒʒan, que significa «aquel a quien se ha permitido quedarse»), que se utiliza para designar a los musulmanes que permanecieron viviendo en territorio conquistado por los cristianos, aunque segregados en barrios llamados morerías y menos específicamente aljamas y bajo su control político, durante el periodo de la Reconquista, que se desarrolló a lo largo de la Edad Media en la península ibérica.

Historia
Al principio se les permitió seguir practicando el islam, utilizar su lengua y mantener sus costumbres. Se organizaban en comunidades denominadas aljamas o morerías con diversos grados de autogobierno, según las condiciones de rendición, o de subordinación: en el caso de Baleares, la total esclavitud, en otros casos, la vinculación en condiciones de servidumbre feudal. En Valencia existían los denominados moros paliers (protegidos del rey) y otros de menor grado de protección (decimati y quintati).

En su gran mayoría, de condición social humilde, eran campesinos con una especial vinculación a la agricultura de regadío (huertas y vegas, terrazas en las laderas) o artesanos especializados (albañilería, oficios textiles -cordobanes, sedas-). Con el tiempo, las condiciones de convivencia y tolerancia se fueron endureciendo, restringiendo los contactos sociales y económicos entre comunidades; a la tradicional separación de carnicerías (basada en el especial método de sacrificio) se añadió la prohibición de contactos profesionales y matrimonios mixtos.

Las revueltas mudéjares fueron numerosas a partir del  y provocaron la despoblación de algunas zonas (valle del Guadalquivir, norte de Alicante), aunque se mantuvieron en otras, sobre todo en el Levante, tanto castellano (Murcia), como aragonés (resto del reino de Valencia -Denia, Játiva, Segorbe-, e incluso en el valle del Ebro aragonés -Borja, Tarazona, Huesca, Teruel, Zaragoza, Calatayud-, y el Bajo Ebro y Bajo Segre catalanes). A finales de la Edad Media representaban el 11% de la población de la Corona de Aragón.

La Guerra de Granada (1482-1492) proporcionó la definitiva ampliación del concepto de mudéjares a todos los musulmanes peninsulares. En principio las condiciones de la rendición les permitían su continuidad y el ejercicio de la religión islámica; sin embargo, el incumplimiento de las condiciones pactadas por parte de los cristianos originó los primeros conflictos. A partir de la revuelta del Albaicín y la de los mudéjares granadinos en 1499, por el decreto de 1502 fueron obligados a convertirse al cristianismo, pasando así a denominarse moriscos, que a pesar de ello continuaron con sus costumbres diferenciadas y el ejercicio clandestino de su religión. La rebelión de las Alpujarras (1568) llevó a la dispersión de los moriscos granadinos por el interior de Castilla (no así los de Levante). La imposibilidad de integración y el recelo a su connivencia con los piratas berberiscos y el Imperio turco llevaron a la decisión de la expulsión de los moriscos en 1609, aunque sus efectos reales fueron escasos, excepto en los reinos de Aragón y Valencia, donde la expulsión fue completa.

Arte mudéjar

El estilo de arte que se empezó a llamar mudéjar en el , desarrollado particularmente en arquitectura, consistió en la aplicación a los edificios cristianos de influencias de estilo hispano-musulmán debidas a la albañilería de tradición andalusí. Es un fenómeno autóctono y exclusivamente hispánico, que se manifestó tanto en elementos arquitectónicos estructurales como decorativos (arco de herradura, artesonados) y en la preferencia por el uso de ciertos materiales (yeso, ladrillo -simple o vitrificado en azulejos-, madera -vista en los artesonados-, etc.). A pesar de su nombre, no hay una relación directa entre este estilo y la población mudéjar de la Edad Media.

Desde el  se desarrolló una arquitectura historicista denominada neomudéjar.

Véase también
 Aljamía
 Literatura aljamiada
 Mozárabe
 Mudéjar aragonés (declarado Patrimonio de la Humanidad)
 Arquitectura mudéjar de Cuéllar, mayor foco en Castilla y León
 Arquitectura religiosa de Sevilla
 Morisco

Referencias

Enlaces externos

 Arte Mudéjar en la provincia de Zaragoza

Arabismos


text:
Arrest of Indian Comedians Sparks Free Speech Debate
Mumbai-based comic Munawar Faruqui, four others, were arrested for allegedly insulting Hindu gods.
By Prachi Sibal January 2, 2021
Hind Rakshak chief Aklavya Singh Gaur and comic Munawar Faruqui at the show on Jan. 1. (Prachi Sibal)
Indore, 2 January 2020 — The arrest of a stand-up comedian in the central Indian city of Indore on Jan. 1 for allegedly hurting religious sentiments has yet again revealed the diminishing space of dissent in the country, claim free speech activists and lawyers.
Mumbai-based comic Munawar Faruqui took the stage at Café Monroe in Indore on the evening of Jan. 1 for a show when he was stopped by Aklavya Singh Gaur, the convenor of Hind Rakshak. The organization claims to be a defender of Hindu culture and society.
A video shared with Zenger News by an audience member shows Gaur haranguing Faruqui about his choice of jokes, accusing him of hurting the sentiments of Hindus while remaining tight-lipped about Islam.
During the altercation, which went on for 10 minutes, Faruqui also mentions that he had issued an apology on Instagram. The show begins only after he declares that there will be no political jokes, eliciting disappointment from the audience.
It begins, but only to be stopped 15 minutes later when police from the Tukoganj police station arrives at the venue.
Indore, Madhya Pradesh. (Map by Urvashi Makwana)
The police took Faruqui into custody along with two other comedians — Prakhar Vyas and Nalin Yadav.
Anshuman Shrivastav, a former advocate general of the state of Madhya Pradesh, is representing the arrested comedians and confirmed to Zenger News that a first information report or police complaint had been filed against the comedians as well as two others — Priyam Vyas and Edwin Anthony.
"Sections 269 and 188 of the Indian Penal Code, under which the accused have been charged, relate to measures for controlling Covid-19," said Shrivastav. "They have also been charged with making filthy and derogatory comments against Hindu gods under Section 295A." Zenger News has a copy of the FIR.
"I have nothing against political jokes and am a patron of arts," Singh Gaur told Zenger News.
"However, if someone crosses limits, it is necessary to do something about it. If someone hurts religious sentiments, is it wrong to take legal action?"
"Let him make jokes on Islam, not Hinduism. Can they make jokes on Allah or Paigambar the same way? Why are Hindus soft targets then?"
This isn't Fauruqi's first brush with the law. In April last year, Ramesh Solanki, a former member of the Shiv Sena party, filed a complaint with Mumbai police alleging that the comedian had insulted Hindu gods Ram and Sita.
After being produced in front of the magistrate, all five have been remanded to judicial custody till January 13. A bail plea will be filed on Monday, Shrivastav confirms.
Comedians in India have been under the heat for several months.
In December, the Supreme Court of India initiated contempt proceedings against popular comedian Kunal Kamra for his tweets allegedly challenging the court's objectivity in granting bail to TV news anchor Arnab Goswami, a self-professed supporter of prime minister Narendra Modi's Bharatiya Janata Party.
Kamra refused to retract his tweets and issued a statement saying he will not apologize.
Another Mumbai-based comedian, Agrima Joshua, was embroiled in a controversy for allegedly insulting Chhatrapati Shivaji, a 17th century king who had established the Maratha dynasty, which ruled over large swathes of central and south India in the 18th and 19th centuries. Shivaji is also the subject of hero-worship for many right-wing Hindu organizations, especially in the western state of Maharashtra.
Anil Deshmukh, Maharashtra's home minister, directed police action against her while members of the Maharashtra Navnirman Sena damaged a venue where she had performed.
The Jan. 1 event from which Faruqui and the others were arrested was organized by New Delhi-based events company, Curtain Calls, and locally produced by Vyas and Yadav.
"I put up shows all over the country, and we tie with smaller local producers," said Vishesh Arora, owner, Curtain Calls and Faruqui's tour manager. "The producers, who are also comedians, were picked up by the cops along with Munawar. It seemed pre-planned,"
Videos of the altercation and of the subsequent arrest have made their way to social media and the comedy circuit is abuzz with discussions and tweets in support of Faruqui.
Singh Gour said he had filed the police complaint.
"Munawar Faruqui is a serial offender," he added. "They have insulted Lord Shiva, Ganesh, Kartikeya and even brought up Home Minister Amit Shah. Three stand-up comedians and two event organizers have been named in the complaint."
"We had watched his earlier videos and had heard about the show a few days ago. We have always raised our voice in such matters. But this particular time, it was done in a planned manner. We had bought tickets for five people, the rest remained outside. We have shared videos of the show yesterday with the police."
But Singh Gour denied that he was present at the venue during the disturbances that several audience videos captured.
"I had got out of the show and some of my people were outside with me," he said. "I tried to pacify them. There were some others who were still inside. They were all from Hind Rakshak."
Indore-resident Sadakat Liyakat Khan told Zenger News that 10-15 people had entered the venue while the show was in progress.
"Singh Gour went on stage and then left after their confrontation," he added. "He was in front of the camera and polite while others who came with him were rude and threatening. Faruqui had performed no more than 15 minutes, when they began hitting audience members."
(Edited by Uttaran Das Gupta and Anindita Ghosh.)
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text:
Владимир Ивич () е бивш сръбски футболист, играл като полузащитник, и настоящ треньор по футбол.

Кариера

Кариера като футболист
Родом от Зренянин, Ивич дебютира във футбола с екипа на местния Пролетер на 17 години. Четири сезона по-късно е закупен от Партизан. С "Гробарите" се състезава в продължение на шест сезона, след което преминава в Борусия Мьонхенгладбах. В немския отбор обаче не успява да се наложи и записва участие в едва 4 мача за целия сезон. Вследствие от това през лятото на 2005 г. приема трансфер в посока Гърция, обличайки екипа на АЕК. С екипа на атинския клуб се подвизава до края на сезон 2006/07, когато става част от Арис Солун. След един успешен сезон с Арис, Ивич става играч на кръвния ПАОК. С екипа на "Черно-белите" играе футбол до края на сезон 2011/12, когато прекратява активната си спортна кариера.

За националния отбор на Сърбия и Черна гора записва 8 мача, но никога не е първи избор.

Кариера като треньор
Една година след като обявява края на кариерата си, Ивич се завръща в последния си отбор – ПАОК, и поема младежката формация до 21 години.

През март 2016 г. наследява Игор Тудор начело на първия тим. Под негово ръководство резултатите се подобряват, и успява да го изведе до второто място в първенството.

През 2017 г. печели Купата на Гърция.

Успехи

Като футболист
 Партизан Белград
 Шампион на Югославия (3): 1998/99, 2001/02, 2002/03
 Носител на Купата на Югославия (1): 2000/01

Като треньор
 ПАОК
 Носител на Купата на Гърция (1): 2016/17

Външни препратки
 Профил  в superleaguegrrece.net
 Профил в fussballdaten.de
 Профил в reprezentacija.rs
 Статистика в national-football-teams.com

Сръбски футболисти
Сръбски треньори по футбол
Футболисти на Партизан
Футболисти на Борусия Мьонхенгладбах
Футболисти на АЕК Атина
Родени в Зренянин


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In the previous two chapters of the "Infinity Multiplication" series for PHP, I had demonstrated the methods and the programming codes for multiplying together two numbers that were stored as string type data, and the equation was processed at the rate of one digit at a time. If you need a reference on the procedures for the multiplication equation, please refer to chapter one and two of the "Infinity Multiplication" series. The chapters' names are "Infinity Multiplication – Beyond integer" and "Infinity Multiplication – Decimal, Precise Float Calculation".
The previous two chapters' programming codes are error proof. This is because the programming codes for the previous chapters were built to process the equation at the rate of only one digit at a time. Besides that, we were also storing the answer for each cycle of multiplication on a separate temporary answer string before joining the temporary answer string to the final output string. By having a temporary answer string, it is easier to inspect if there are any errors in the formula or in the code. Nonetheless, the programming codes of the previous chapters are inefficient and should not be used in a production environment.
In this chapter, I will demonstrate what in my opinion consider to be the most optimized way for optimizing the previous chapters' programming codes and methods. By optimizing the methods and the codes, our program can execute efficiently, and therefore, we can utilize the program for the production environment. Same as the previous chapters, the source code of this chapter was tested against hundred of thousand plus test cases. With this article, I also included the source code for testing all the previous chapters' programming codes and including the code of this chapter. I also did a benchmark test for this production version of infiX versus the previous versions of infiX. Note that infiX is the name of the program that is the programming code examples for this chapter and the all the previous chapters.
Before further discussion in regard to how to optimize the program to operate more efficiently, I will first present the results of the benchmark test for this version of infiX versus the previous version of infiX. When testing for how fast the program executes, I produced two strings of digits, each string contained forty digits. The strings are then passed into each version of infiX 100,000 times, and the time it took for each version of infiX to complete the calculation was recorded as the benchmark score. This test ran twice and the average result between the two tests is used as the final benchmark score. Note that both versions of infiX were calculating the same two strings of digits 100,000 times.
This version of infiX for PHP took 14.333 seconds to compute the results for 100,000 cases, and the test strings didn't have a fractional part. With the decimal, the program took 14.801 seconds to compute the results 100,000 times. For a comparison of the execution time, the previous version of infiX took 228.092 seconds to compute the results for 100,000 cases, and the test string didn't have a fractional part. When there is a fractional part, the previous version took 250.223 seconds to compute the result for 100,000 test cases.
What are the changes in the methods and in the programming code that give the program the ability to operate more than ten times faster? Instead of processing one digit at a time, the program was converted to process the equation at the rate of five digits at a time. For a summary of each multiplication cycle, the previous version of infiX had to convert a digit that was stored as string type data to an integer type data twice, convert the result value to string type data, stores the result value of the multiplication procedure to a temporary answer string, add the temporary answer string to the final output string, and calculate the carry-over values for both, the addition and the multiplication procedures. By converting the program to compute the equation at the rate of five digits at a time, we would still have the same procedures for each multiplication cycle. Nonetheless, instead of only consuming a single digit from both numbers per multiplication cycle, each of our multiplication cycles would now consume five digits. Thus, any single process in the multiplication cycle would be applied to a block of five digits and not just one. By processing more digits per cycle, the program saves processing time from all its sub-process of a multiplication cycle.
Besides processing more digits per cycle, the temporary answer string is now removed and the program would only use the single output string for storing and processing all the result values. By removing the temporary answer string, the program would also save memory usage, and therefore, requires far less memory to operate.
Besides the above changes, the "for" loop was converted into the "while" loop. The "while" loop was benchmarked to execute faster than the "for" loop. The substr() function was chosen due to its speed and execution time. For replacing a value in a string or searching for a value in a string, where it is possible, instead of using the regex engine, I used the native built-in function that can manipulate strings. For example, I used the ltrim() and rtrim() function for simple trimming left and right of the strings instead of the preg_replace() function.
Native built-in functions that involve modifying or searching for a value within a string was tested and shown to execute faster than if to use the regex engine for the same purpose. Nonetheless, the native built-in functions might not be as flexible when it's come to complex conditions. I can only assume that the regex engine took longer to compute a result is because the regex engine needs to evaluate the pattern's information to understand the conditions within pattern before applying the conditions onto a string.
Although this version of infiX only processes five digits at a time, this number may have to be configurated to a lower amount if we are truly working in a 32 bits environment. This is because the maximum value that a 32 bits integer can hold is 2,147,483,647. Which is about 10 digits in length. For a multiplication equation, five digits multiply by another five digits have the potential to produce a maximum result value that is ten digits in length and that value is 9,999,999,999. This is a value that is a little bit larger than the maximum value that a 32 bits integer can hold. Thus, if truly working in a 32 bits environment, the number of digits to be processed per cycle should be configurated at the rate of four digits at a time.
When it comes to the formula for multiplying five digits at a time, the formula is similar to the formula of multiplying one digit at a time. Nonetheless, when writing the result to the output string it can be a complicated task. This simply because the program would now operate without a temporary answer string. Before going into the examples and the explanations of the programming code, let's first look at the new formula for the multiplication procedures.
*** Carry-overM = the carry-over value for the multiplication procedure.
Carry-overA = the carry-over value for the addition procedure.
With the formula above in context, we can see that we are multiplying all the digits in A string to the digits of B string in a block of five digits at a time. Each block of five digits in A string is multiplied by the first block of five digits in B string, then to the second block, then to third and so on if there is. For each time when we multiplied a block of five digits from A to the first block of five digits in B, starting from right to left of the answer string, we would only store a block of five digits at a time to the answer string. When there are more than five digits in the result value then all the digits after the fifth digits is to be kept as the carry-over value for the next equation. When storing the result into the answer string, the first five digits that are the rightmost in the result value would be the digits that are to be stored into the answer string. The sixth digits and beyond would be kept as the carry-over value.
When it comes to the procedure of when we are adding the carry-over value to the result value of the equation. Similar to the above, the first five most digits that are the rightmost in the result is the result value, digits that are beyond that is the carry-over value.
After multiplying all the digits in A string to the first block of five digits in B string, we would then multiply all the digits in A string to the next block of five digits in B string. When we are moving over to the next block of five digits in B string, our starting position for adding the equation's result value into the answer string would move over to the left by one block that is five digits in length. Every time when the equation is moving to another block of five digits in B string, the starting position in the answer string would also move over to the left by one block of digits. For the procedure of adding the result value of an equation into the answer string, the rules from above that are in regard to the carry-over values, and how many digits are the result and how many digits are the carry-over value would also apply to this procedure.
When it comes to the variables that we need in the program, the first variable our program needs is a variable that store a value that represents how many digits that the program is processing per multiplication cycle. This value can extend the flexibility of the program. If we were to work in an environment where the integer data type is stored in larger bit blocks then we can increase this value so that the program can process more digits per cycle of calculation. On the other hand, in the event where we have to work in a more restricted environment and integer values are stored in smaller bit blocks then we can decrease this value to adapt the program to the environment. For the naming context, the mentioned variable is named "digit".
The second variable we would need is a variable that will store a value that is the index count of how many times we had read from B string. In the circumstance of this article, we know that each time we read a string, we are reading a block of five digits. When it comes to the ending of the strings, the amount of how many digits that are left for the program to read from can vary. Nonetheless, this program was built to considers the variation on the number of digits in the final block of digits for both strings. For the naming context, the variable for storing the value of the read count for B string is named "posr".
Besides the above variables, the program would also need a variable that stores a value that represents the position that the program is working on in the answer string. That variable is named "posw". For a fourth new variable, the variable is named "prepos". When utilizing the substr() function, the behavior of the substr() function is that when a negative value is passed into the substr() function, then the substr() function will treat that value as a position in the string counting from right to left, with negative one being the rightmost position. This is the opposite of when a positive value is passed into the substr() function. Since we are working from right to left in the strings, for some procedures, it is easier to use negative values. Therefore, a negative version of the value that is stored in the "digit" variable is assigned to the "prepos" variable to enhance the simplicity for the equation. This below is the variables that we need to declare outside of the calculation loop.
For every time the program process a position in B string, the program would have to process all the positions in A string. Thus, the outer loop will stay at a position in B string until all the position in A string is read, and after each cycle, the outer loop will move the read point for B string by five positions. For the inner loop, after processing a position in A string, the loop will also move the read point for A string by five positions.
To get five digits from B string, we can use the substr() function. For the usage of the substr() function, the function can take three arguments as input. The first argument is the string itself. The second argument should be an integer value, and the substr() function treats that value as the position of where to get the first character from. If the second argument's value is a value of zero or is a positive value then the substr() function treats that value as a position that is counting from left to right. If the second argument's value is a negative value then the substr() function will treat that value as a position that is counting from right to left with -1 being the leftmost position.
For the third argument, the substr() function will take an integer value, and the third argument is an optional one. If the third argument is omitted then the substr() function will extract all the characters from the position that was defined by the value of the second argument to the end of the string. If the third argument's value is a negative value then the substr() function will treat that value as a position that is counting from right to left. The substr() function will then extract all the characters from and including the position that was defined by the second argument to the character that is just before the position that was defined by the third argument. Counting from left to right of the string, if the position that was defined by the third argument somehow become before the position that was defined by the second argument, then the substr() function will return an empty string. If the third argument's value is a positive value then substr() will treat that value as the number of characters that the function should extract from the string, starting from the position that was defined by the second argument. If the string contains fewer characters than the number of characters that the substr() function should extract, then the function will extract all that is available.
When it comes to utilizing the substr() function, The first value that we are passing as the second argument is the current position of where we are at in B string subtract to the number of digits that the program is processing per cycle. Notice that we do not subtract the string's length to a value of one to get the current position. This is because if we were to extract five rightmost digits in a string that is 20 positions long, although the last position is position 19 if to extract five digits we would have to start at position 15. Therefore, in the example, we can just utilize the length of the string, which is a value of 20, and subtract that value to a value of 5 to get the the starting position, which would be position 15. Also, we do not want a negative value as the second argument for the substr() function. This event occurs when the last block of digits contains less than the number of digits that the program is processing per cycle. Thus, if the value that we are going to pass as the second argument is a negative value then we would convert that value to a value of zero, which is the leftmost and is the first position in a string.
The value that we want to pass as the third argument for the substr() function is how many digits we want to get. This value is defined by the "digit" variable. Nonetheless, when comes to the last block of digits in either string A or B and the digits block has fewer digits than the amount that was defined by the "digit" variable, then we would extract the digits that we had already extracted in the previous loop cycle, and therefore, the equation would produce an erroneous result. Thus, if the position that we are in subtract to a value of zero is smaller than the value of the "digit" variable, then we would use the position that we are in and subtract that value to a value of zero as the value that we will pass as the substr() function's third argument. The digits that we are extracting from the string would then be converted into an integer value. Nonetheless, if the substr() function did not get anything from the string then the intval() function will produce an integer value of zero for the empty value that was returned by the substr() function.
When we are executing the inner loop for the calculating procedures and when the answer string did not yet contain a value, we would simply place the result value of the procedures into the output string bases on the following conditions. When it comes to processing the result value of the current procedure, one of the values that we have to evaluate for is the carry-over value. To get the carry-over value from the result value, we can utilize the substr() function on the result value to get the character from the zero position to five positions away from the rightmost of the result value. To do this, we pass a value of zero as the second argument for the substr() function, then we pass the "prepos" variable's value as the third argument for the substr() function. If the previous statement fails to produce any digits then the intval() will return an integer value of zero. When it comes to storing the carry-over value, the carry-over value for the multiplication procedure is assigned to a variable that is named "carryOverM".
When it comes to appending the result value to the answer string, we would always append the result value in the format of a digits block that is five digits in length. Therefore, we would have to append to the result a number of zeroes for the event of when the result value contains less than five digits. The following examples will demonstrate the circumstance of how the result value can have less than five digits. For the first example, we have an equation of 1 00125 x 1 00001. In that equation, when executing the equation, the first block of five digits from A multiply by the first block of five digits of B would translate into an equation of 1 x 125. The result value of the equation would be a value of 125. If we do not append zeroes to the beginning of the result value to make up for the missing length then when it comes to the next calculation, there will only be a value of one that is going to be added to the beginning of the result value. Which will give us a value of 1125 as the result value for the equation instead of the correct answer of 1 00125.
To be able to always consistently produce a result value that is five digits in length, we add four zeroes to the left side of the result value. We then utilize the substr() function to extract the five rightmost digits from the just modified result value. For adding zeroes to the result value, we would only execute this procedure after when we had completed the evaluation of the result value for any carry-over value.
After multiplying all the digits block of A string to the first block of digits in B string, we then append any carry-over value to the leftmost of the answer string. The principle of a consistent number of digits in a block of digits would also apply to this procedure. Take note here that when I subtract the index count for A string, of which the value is held by a variable that named "ai", it can also be done by subtracting "ai" to the value of the "digit" variable as mentioned earlier. This code here is just for the demonstration purpose of which showed that we are moving by five positions per cycle. Nonetheless, for easy customization, this can be changed to "ai – digit", and when modifying the value of "bi", it can also be changed to "bi – digit".
This code block below demonstrate the mentioned procedures in PHP.
The previous procedures are for when the answer string did not yet contain a value. However, it can get a bit complicated for when the answer string does contain a value. Each time we read a digits block in A string, we would have to calculate the positions of where we are working on in the answer string. Besides the working on position, the starting position for the answer string would move over to the left by one digits block for each digits block that we read in B string. Also, the position that we are working on in the answer string would also move over to the left by one digits block after we process a digits block from A string.
To calculate the position that we are working on in the answer string, we would first calculate the starting position of the answer string. To get the starting position of the answer string, we multiply how many times we had read from B string to the value of the "prepos" variable, which is just the negative version of the "digit" variable. After getting the starting position of the answer string, we then multiply the number of times we had read from A string to the value of the "prepos" variable. Then we add the result value of that equation to the value of the starting position of the answer string.
For this time of multiplying a digits block from A string to the digits block in B string, we have to add the result value of the equation to a value that exists in the position of where we are working on in the answer string. The result value for this equation is stored in a temporary placeholder, the temporary placeholder is the variable with the name "tempadd". To get the digits from the answer string, we will utilize the substr() function. For the substr() function, we use a value that is the value of the current position that we are working on in the answer string in negative and add that value to the value of the "prepos" variable. The previous value is then passed to the substr() function as the second argument. For the third argument, we use the value that is the position of where we are currently at in the answer string.
If the tempadd variable holds a value that is more than five digits in length then we would produce a carry-over value for the addition procedure. The carry-over value for our addition procedure is held by a variable that is named carryOverA. For an addition equation, the carry-over value can never be more than a value of one if we are only adding two sets of digits together. Thus, our carry-over value our addition procedure can only be a value of one when the result value contains a value that is more than five digits in length. Else, we will assign a zero value to the carry-over value. The value of the carryOverM and the carryOverA variable will be added to the result value of the equation of next loop execution.
After producing a value for the tempadd variable, we would then place the value of the tempadd variable into our answer string. When it comes to placing the tempadd's value into the answer string, we first get all the digits from the beginning of the answer string to the position that is before the position of where we grabbed the first digit from the answer string. We then place our tempadd's value to the right side of the previous value. Then we get the digits from the position that is after the position of the last digit that we grabbed from the answer string to the end of the answer string. We place this value on the right side of where we placed the tempadd's value. All the digits that we pieced up together with the tempadd's value are assigned as a new value to the answer string.
After completely multiplying all the digits in A string to a set of digits in B string, we would add the value of the carryOverM and carryOverA variable together, and place this value to the left of the answer string. The below code block demonstrate the above procedures into programming statements for PHP.
As for testing and debugging the infiX program, the code block below is what I used to test the previous version of InfiX against the native built-in multiplication feature of the PHP's platform. For the previous version of infiX, the program was built to process the multiplication equation at the rate of only one digit at a time. Therefore, if the program always correctly produces result values for equations that involve smaller strings of digits, then the program would also be able to consistently produces correct result values for larger strings of digits. I also tested the infiX program against Ruby's native multiplication feature. As it seems, Ruby's native multiplication feature can also calculate very large numbers.
When it comes to numbers that do have a fractional part, do take into consideration that the native built-in calculating feature would not be able to provide the actual precise floating point calculation. The tester code below will produce two random numbers for each test case. The tester will then multiplying the numbers together using the native multiplication feature of PHP. The tester will then pass the numbers as string type data into the infiX function. If there is a mismatch between the two result values, then the tester will print the test case information.
To use this tester, simply look for version two of infiX, which included in the chapter that is called "Decimal, Precise Float Calculation" for PHP. Then place this tester code with the program. The number of test cases that the tester will execute was preset at a value of 5000. This value can be increased or decreased as necessary. For PHP, it is safe to set the number of cases to run at less than or equal to 100,000 cases at a time. Otherwise, it might take sometimes to get back the results value. This code block below is the tester code for the version two of infiX.
When it comes to testing this version of infiX, I tested the result values of this version of infiX to the result values of the previous version of infiX. Since the previous version of infiX was already thoroughly tested, this version's result value would be correct if matches the result values of the previous version. The previous version of infiX does take a long time to produce result values. Nonetheless, as a prototype, I built the previous version as a demonstrator for the articles that I wrote and also as a tester for testing future releases of the infiX program.
To use the tester code below, simply place this version of infiX and the version two of infiX in the same file and with the tester code. Rename one of the infiX function to infiX2. This tester will produce five blocks of random digits that are before the decimal for each string of digits. The random blocks of digits can be anywhere between one or five digits. The tester will then pad 0 to 15 zeroes behind the decimal, and then the tester will place three blocks of random numbers to the right of the zeroes that were just added to the string. After that, the tester will place another 0 to 15 zeroes and then the last two blocks of random digits to the right side of the previous 3 block of random digits. It is a recommendation that this tester should not be set to execute more than 50,000 cases at a time. It can take a very long time to produce the results. This is because the previous version of InfiX does take a long time to produce an answer.
The below code block is the tester code that was written in PHP.
The below code snippet is the infiX program. This version is usable for a production environment.
// Notice: This is the normal usage version of infinity multiplication written for PHP and is designed for a production environment.
//         The first limitation of this version is how much memory PHP allow the string to be store.
//         The second limitation is how much memory PHP can assign to the program.
//         3100 Mb of memory is requires for multiplying two strings of digits where each string is 500MB in size.
Original Post Name: "Working With Number – Infinity Multiplication – Optimised The Code – PHP".


Georges is known for foregoing traditional thick sauces for oils and broths, which are lighter and capable of more beguiling scents. Dishes involve pairings so perfectly calibrated that one feels even the slightest variation could cause the delicately erected structure to collapse. Bizarre combinations somehow attain a feeling of inevitability. One senses behind every dish the hands of a master who is at the height of his powers. Dishes like the Egg Caviar, a dish much imitated (see, e.g., Mélisse) but seldom equaled, and the turbot with chateau chalon sauce are a testament to his virtuosity. The classic tasting menu is often better than the seasonal tasting menu, but both are always worth trying. The gracefully choreographed service is formal, competent and often tableside. The vast wine list is magnificent, with both trophy wines and thoughtfully selected midrange wines. We would be remiss if we did not mention the adjacent bar-restaurant Nougatine, which is worth its own trip, as well as the renowned affordability of the popular lunchtime prix fixe menu. Some detractors claim that Jean Georges is merely recycling proven hits from its youth rather than expanding the boundaries of culinary possibility. The continued execution of timeless dishes at this level of craftsmanship should be cause for celebration rather than criticism.


A programme initiated by NEPAD SANBio, supported by the Finnish - Southern African Partnership Programme, BioFISA II, the FemBioBiz programme attracted over 500 female entrepreneurs and students doing (or envisioning) business with biotechnologies and other technologies in health or nutrition from 8 SADC countries.
With the aim to create a network of women bioentrepreneurs in the region by providing support to entrepreneurs for acceleration of their businesses, the programme is run with a 3-phased approach.
In South Africa, from the 15 participants that were trained during phase 1, only 10 were selected to participate in a two further boot camps focusing on key aspects of business development.
The businesses that emerged victorious were: vertical farming products and consultancy company Sibayeni Metrofarming run by Samkelisiwe Chunda, Karen Vosloo's clay skin product manufacturing company Skin Clays, Lesotho native fruit and vegetables preserve company Limomonane Products by Lipuo Nkholi, and fish scale biopolymer manufacturer Nomahlubi Nazo's start-up.
Facilitated by the Africa Women Innovation and Entrepreneurship Forum (AWIEF), the four winners each will receive marketing consultancy services from AfricanRedhead; business case consulting from Tinodiwanashe Kujenga; professional image, digital and pitch consulting from Kudakwashe Kuzviwanza as well as Florian Akinbiyi; and coaching from International Coach Federation's Trish Neill. Lastly, participants will also get to study four world class online courses of their choice. What is more, following the win, one of the winners is already in discussions with an angel investor.
For updates and results from the other participating countries as they become available, visit our FemBioBiz Season 2 pages.