text:
\section{Introduction}

AdS/CFT correspondence \cite{AdSCFT1,AdSCFT2,AdSCFT3} is a leading
formalism providing a consistent relation between gauge and
gravitation theories and has given us a rather deep insight on
possible unification of them. Such a correspondence captures an
equivalence between the conformal field theory and its dual theory
of gravity in AdS space with a strong/weak duality point of view.
It is believed that it provides a non-perturbative formulation
access to strongly coupled regimes of quantum field theories which
cannot be accessed by the traditional perturbation approach.
Indeed, AdS/CFT correspondence is one of the important examples of
entanglement between gravity and quantum information
\cite{AdSCFT4,AdSCFT5,AdSCFT6,AdSCFT7}. Complexity has recently
been introduced as a complement to entanglement providing
additional information on the quantum properties of black holes.

In order to calculate the holographic duality of quantum computational
complexity one may follow two proposals. The $complexity = volume$ (CV) \cite%
{A1,A2} and complexity = action (CA) \cite{B1,B2} conjectures. The
former one states that the complexity of the boundary state is
proportional to the maximum volume of codimension hypersurface
bounded by the CFT slices and the latter conjecture specifies that
complexity on the boundary CFT is proportional to the on-shell
action of the Wheeler-DeWitt(WDW) patch in the bulk. According to
these conjectures, Lloyd showed that the growth rate of complexity
for the Schwarzschild like black holes is bounded by the total
energy of the system \cite{Lloyd2000}. Cai, et.al proved that the
so-called
Lloyd bound should be modified for charges or rotating black holes \cite%
{Cai2016}. Although there are many cases of black holes that
confirm the mentioned upper bound \cite{L1,L2,L3,L4,L5,L6,L7}, its
validity is an open question since there are different types of
black hole violating the Lloyd bound \cite{V1,V2,V3}. Accordingly,
the validity of CV and CA conjectures and their possible
modifications deserve further study.

In this direction, here, we focus on the holographic complexity of
nonlinearly charged regular black holes in $\mathcal{P}$ frame.
Indeed, the motivation for studying this black hole is three
folds: regularity, nonlinearity and complexity. The first item is
the regularity of the black hole with known action, exactly. The
second one is nonlinearity which has the key role to avoid the
singularity. Although such a nonlinear electrodynamics is an ad
hoc model near the black hole, it behaves like the Maxwell field
far from the horizon radius. The third one is the complexity of
black holes which is reasonable for regular black holes since the
singularity may violate the interpretation of physical quantities
such as entangled entropy and complexity.

The subject of singularity itself and its possible solution
(leading to regularity) are highly interesting from different
viewpoints: mathematical and geometrical points of view, classical
gravity and its quantum standpoints, cosmological side, effective
field theory and AdS/CFT correspondence features, string theory
and supergravity aspects, etc. Indeed there are two important
singularities that their natures are not yet fully understood: the
singularity that is covered by an event horizon of black holes and
the big bang singularity. So one of the substantial questions is
how one can prevent the singularity. Although it is believed that
quantum gravity can be able to smooth out singularities, there is
no consistent theory of quantum gravity overcoming this issue
despite many attempts. Thus, the question should again be asked
how we can preclude the singularity in an alternative theory.

Regarding the Einstein theory of gravity coupling minimally with
an appropriate nonlinear electrodynamics theory, various regular
black hole metrics are constructed (for an incomplete list, please
take a look at \cite{Balart2014,Ayon-Beato1998,Bronnikov2001,Arellano2006,Hassain2008,Hollenstein2008,Balart2009}%
). The exact regular solution of $f(R)$ gravity has been studied in \cite%
{Hollenstein2008}. Gravitational lensing \cite{Eiroa:2010wm}, dynamical \cite%
{Fernando:2012yw} and thermodynamical stability
\cite{Hendi:2020knv} of regular black hole has been studied. A new
Smarr-type formula for the black hole in nonlinear electrodynamics
has been obtained in \cite{kastor09,Balart:2017dzt}. In this
paper, we concentrate on the complexity of a simple static and
rotating regular black hole with nonlinear electrodynamics as a
source. The rest of the paper is organized as follows. In Sec.
\ref{sec2}, we briefly review the complexity and thermodynamics of
regular black hole, especially deriving its Smarr formula. Section
\ref{sec3} is devoted to study the complexity of
large-extremal-static regular black hole in the framework of CA
and CV conjectures and comparing the results in two frameworks. In
Sec. \ref{sec4}, we repeated the same computation of complexity
for the large-extremal rotating regular black holes. After that,
conclusions are given. In the appendix, we also investigate two
other regular cases and calculated thermodynamic quantities,
complexity and complexity of formation numerically.

\section{basic formalism \label{sec2}}

Electrically charged black hole solutions can be studied through
an alternative form of nonlinear electrodynamics obtained by the
Legendre transformation. The action of nonlinear electrodynamics
in $\mathcal{P}$ frame minimally coupled to the Einstein gravity
with the York-Gibbons-Hawking surface term and the two-dimensional
joint term is given by
\begin{eqnarray}
\mathcal{A}&=&\frac{1}{4\pi}\int d^{4} x \sqrt{-g}\left[ \frac{1}{4} (%
\mathcal{R}-2\Lambda) -\left(\frac{1}{2} P^{\mu\nu} F_{\mu\nu}- \mathcal{H} (%
\mathcal{P})\right)\right]  \notag \\
&&+\frac{1}{8\pi}\int d^{3}x \sqrt{-h} \mathcal{K}+\frac{1}{8\pi}\int d^{2}x
\sqrt{-\gamma}\; a ,  \label{eqI}
\end{eqnarray}
where $g$ is the determinant of the metric, $\mathcal{R} $ is the Ricci
scalar and the negative cosmological constant is denoted by $\Lambda=-\frac{3%
}{l^2}$. In addition, the anti-symmetric tensor $P_{\mu\nu}= \mathcal{L}%
_{F}F_{\mu \nu}$ in which $\mathcal{L}_{\mathcal{F}}\equiv\frac{\partial
\mathcal{L}}{\partial \mathcal{F}} $, $\mathcal{F}=\frac{1}{4}%
F_{\mu\nu}F^{\mu\nu } $ and $F_{\mu \nu} $ is the Faraday tensor.
Furthermore, the structure function is $\mathcal{H(\mathcal{P})}=2\mathcal{F}
\mathcal{L}_{\mathcal{F}}-\mathcal{L}$ with $\mathcal{P} \equiv\frac{1}{4}%
P_{\mu\nu}P^{\mu\nu }=\mathcal{F} \mathcal{L}_{\mathcal{F}}^2$ since $d%
\mathcal{H}=\mathcal{L}_{\mathcal{F}}^{-1}d\left(\mathcal{F}\mathcal{L}_{%
\mathcal{F}}^2\right)=\mathcal{H}_{\mathcal{P}}d\mathcal{P}$, where $%
\mathcal{H}_{\mathcal{P}}=\frac{d\mathcal{H}}{d\mathcal{P}}=\mathcal{L}_{%
\mathcal{F}}^{-1}$. Moreover, $h$ and $\gamma$ are, respectively, the
determinant of the induced metrics $h_{\mu \nu}$ in three-dimensions and
two-dimensional $\gamma_{\mu \nu}$. Also, $\mathcal{K}$ is the trace of the
extrinsic curvature of the induced metric $h_{\mu \nu}$
\begin{equation}  \label{extcur}
\mathcal{K}=n^{\mu}{}_{;\mu}=\dfrac{1}{\sqrt{-g}}\partial_{\mu}(\sqrt{-g}%
n^{\mu}),
\end{equation}
where $n^{\mu}$ is the normal vector. The integrant $a$ is defined as
\begin{equation*}
a=\ln \left(-\frac{1}{2}N.\bar{N}\right),
\end{equation*}
in which $N$ is the future-directed null normal to the left-moving null
surface and $\bar{N}$ denotes the future-directed null normal to the
right-moving null surface.

The first integral of Eq. (\ref{eqI}) represents the bulk action while the
second and third ones stand for the boundary and joint parts of the WDW
patch \cite{Lehner:2016vdi}. The WDW patch is used to obtain the rate of
temporal change of the action. A typical WDW patch of a black hole with two
horizons is shown in Fig. \ref{wdw} which is evolved in time from $t_{0}$ to
$t_{0}+\delta t$. According to this figure, the contribution of different
parts is bulk region $V_{1}$ and $V_{2}$, and null-null surface joints $A$, $%
B$, $C$ and $D$ (for more details, see \cite{Lehner:2016vdi}).
Hence, we can write
\begin{eqnarray}
\partial \mathcal{A} &=&\dfrac{1}{4 \pi G}\int_{V_{1}} \sqrt{-g}~\mathcal{L}%
\; dt dr d\theta d\phi -\dfrac{1}{4 \pi G}\int_{V_{2}} \sqrt{-g}~\mathcal{L}%
\; dt dr d\theta d\phi  \notag \\
&&+\frac{1}{8\pi G} \int_{B}\sqrt{-\gamma }\ a_{B}\ d\theta d\phi - \frac{1}{%
8\pi G}\int_{A} \sqrt{-\gamma}\ a_{A}\ d\theta d\phi  \notag \\
&&+\frac{1}{8\pi G}\int_{D} \sqrt{-\gamma}\ a_{D}\ d\theta d\phi -\frac{1}{%
8\pi G}\int_{C}\ \sqrt{-\gamma}\ a_{C}\ d\theta d\phi \; ,  \label{action 2}
\end{eqnarray}
where $\mathcal{L}$ is the Lagrangian of Einstein-nonlinear electrodynamics.
In order to compute the contribution of joints, we use the following
transformation of $N $ and $\bar{N}$
\begin{equation}
N_{\alpha }=-b_{1}\partial _{\alpha }(t-r^{\ast }),\ \ \ \ \ \ \bar{N}
=b_{2}\partial _{\alpha }(t+r^{\ast }),
\end{equation}
in which $b_1$ and $b_2$ are two arbitrary positive constants and the
tortoise coordinate $r^{\ast}$ is defined as $r^{\ast }=\int \frac{dr}{f(r)}$%
. So, the contribution of joints is calculated as
\begin{equation}
S_{B}-S_{A}=\frac{\delta t}{4}\ \left(2 r f(r) \left[ \ln \left(\frac{f(r)}{%
b_{1}b_{2}}\right) + \frac{r f^{\prime }(r)}{2 f(r)} \right]\right) ^{r_{B}}
_{r_{A}},  \label{joint1}
\end{equation}
\begin{equation}
S_{D}-S_{C}=\frac{\delta t}{4}\ \left(2 r f(r) \left[ \ln \left(\frac{f(r)}{%
b_{1}b_{2}}\right) + \frac{r f^{\prime }(r)}{2 f(r)}
\right]\right)_{r_{A}},  \label{joint12}
\end{equation}
where prime denotes derivative with respect to $r$. In Fig.
\ref{fig1}, we assume close cylindrical hypersurfaces consisting
of a past spacelike surface $B$, a truncated null cone $N$, a
future spacelike surface $A$ and boundaries for intersection of
spacelike and null surfaces $x$ and $y$.

\begin{center}
\begin{figure}[tbp]
\hspace{0.2cm} \centering
\subfigure[]{\includegraphics[width=0.4\columnwidth]{pic10.eps}} %
\subfigure[]{\includegraphics[width=0.4\columnwidth]{pic11.eps}}
\caption{Closed hypersurfaces consisting of a past spacelike
surface $B$, a truncated null cone $N$ and a future spacelike
surface $A$. (a) Same direction of normal vectors, (b) Opposite
direction of normal vectors. } \label{fig1}
\end{figure}
\end{center}

The contribution of surfaces $A$ and $B$ canceled each other in
Fig. (\ref{fig1}a) due to the same direction of normal vectors.
While for Fig. (\ref{fig1}b) according to opposite direction of
normal vectors, contributions of boundaries of intersection of
surfaces, $x$ and $y$ cancel each other. Therefore depending on
what we choose for direction of normal vector, boundary terms or
joint contributions will be vanished \cite{Lehner:2016vdi}.

It is notable that applying the variational principle to the
action (\ref{eqI}), one finds the field equations as
\begin{equation}  \label{gg}
G_{\mu}^\nu+\Lambda\delta_{\mu}^\nu=T_{\mu}^\nu= \frac{1}{2}\left[\mathcal{H}%
_{\mathcal{P}}P_{\mu\lambda}P^{\nu\lambda}- \delta_{\mu}^\nu\left(2\mathcal{P%
}\mathcal{H}_{\mathcal{P}} -\mathcal{H}\right)\right],
\end{equation}
\begin{equation}  \label{nn}
\nabla_\nu P^{\mu\nu}=0,
\end{equation}
where $G_{\mu \nu} $ is the Einstein tensor. By integrating equation (\ref%
{nn}) with the assumption of spherical symmetry spacetime, we obtain the
following static solution
\begin{equation}  \label{eqmetric1}
ds^2=-f(r)dt^2+\frac{dr^2}{f(r)}+r^2d\Omega_{2}^2,\hspace{0.5cm}f(r)=1-\frac{%
1}{r}\int \rho(r)r^2 dr,
\end{equation}
\begin{equation}  \label{eqn:p}
P_{\mu \nu}=2\delta^{t}_{[\mu}\delta^{r}_{\nu]} \frac{q}{r^{2}},\hspace{0.5cm%
} \mathrm{or} \hspace{0.5cm} \mathcal{P}=-\frac{q^{2}}{2r^{4}},
\end{equation}
where $q$ is an integration constant and $\mathcal{H}=2\rho(r)$.

In the following, for the importance of Smarr relation, we will
obtain it for the case of spherically symmetric static charged
regular black holes. In order to do that we use the Komar formula
for the mass of black holes as follows
\begin{equation}
M=-\dfrac{1}{8\pi }\oint_{\infty }\nabla ^{\alpha }\xi ^{\beta }dS_{\alpha
\beta },  \label{eqkomar}
\end{equation}%
where $\xi $ is a timelike Killing vector which satisfies Killing equation
and $dS_{\alpha \beta }$ is a two-dimensional surface element of
the boundary at infinity. Since the spacetime has two boundaries,
we will write the Komar integral for the mass as a sum of an
integral over a closed null surface at the horizon $H$ and an
integral on the spacelike hypersurface $\Sigma $ which is bounded
by the horizon and infinity as
\begin{equation}
M=-\dfrac{1}{8\pi }\oint_{H}\nabla ^{\alpha }\xi ^{\beta }dS_{\alpha \beta }-%
\dfrac{1}{8\pi }\int_{\Sigma }\nabla_{\beta} \nabla ^{\alpha }\xi
^{\beta }d\Sigma _{\alpha }.  \label{eqkomarr}
\end{equation}%
For the first term, we have
\begin{equation}
-\dfrac{1}{8\pi }\oint_{H}\nabla ^{\alpha }\xi ^{\beta }dS_{\alpha \beta }=%
\dfrac{\kappa A}{4\pi },
\end{equation}%
where $A$ is the horizons surface area, $\kappa $ denotes the
surface gravity that is constant at the event horizon and
satisfies $\xi ^{\alpha }{}_{;\beta }\xi _{\alpha }=\kappa \xi
_{\beta }$. In order to evaluate the second term in equation
(\ref{eqkomarr}), we consider the Stokes theorem for an
antisymmetric tensor field $B^{\alpha \beta }$ as follows
\begin{equation}
\oint_{S}B^{\alpha \beta }dS_{\alpha \beta }=2\int_{\Sigma }B^{\alpha \beta
}{}_{;\beta }d\Sigma _{\alpha }.
\end{equation}%
For the antisymmetric tensor $B^{\alpha \beta }=\nabla ^{\alpha }\xi ^{\beta
}$ of the bulk term $\Sigma $, we have
\begin{equation}
B^{\alpha \beta }{}_{;\beta }=(\nabla ^{\alpha }\xi ^{\beta })_{;\beta
}=-(\nabla ^{\beta }\xi ^{\alpha })_{;\beta }=-\square \xi ^{\alpha },
\end{equation}%
where $\square =\nabla _{\alpha }\nabla ^{\alpha }$. Recalling that $\nabla
_{\rho }\nabla _{\mu }\xi _{\nu }=R_{\nu \mu \rho }{}^{\sigma }\xi _{\sigma
} $, we get
\begin{equation}
\oint_{S}\nabla ^{\alpha }\xi ^{\beta }dS_{\alpha \beta
}=-2\int_{\Sigma }R^{\alpha }{}_{\beta }\xi ^{\beta }d\Sigma
_{\alpha }.
\end{equation}%
Now, by using the Einstein field equation $R_{\alpha \beta }=4\pi \left(
2T_{\alpha \beta }-g_{\alpha \beta }T\right) $, we obtain
\begin{equation}
\oint_{S}\nabla ^{\alpha }\xi ^{\beta }dS_{\alpha \beta }=-16\pi
\int_{\Sigma }\left( T_{\beta }^{\alpha }-\dfrac{1}{2}\delta _{\beta
}^{\alpha }T\right) \xi ^{\beta }d\Sigma _{\alpha }.  \label{eqmass}
\end{equation}%
On the other hand, considering equation (\ref{gg}), the nonzero components
of energy-momentum tensor are
\begin{equation}
T^{t}{}_{t}=T^{r}{}_{r}=-\dfrac{\mathcal{H}}{8\pi },\hspace{0.5cm}T^{\theta
}{}_{\theta }=T^{\phi }{}_{\phi }=\dfrac{1}{8\pi }\left( 2\mathcal{P}%
\mathcal{H_{P}}-\mathcal{H}\right).
\end{equation}%
By rewriting the RHS of Eq. (\ref{eqmass}), one can obtain
\begin{eqnarray}
-16\pi \int_{\Sigma }\left( T_{\beta }^{\alpha }-\dfrac{1}{2}\delta _{\beta
}^{\alpha }T\right) \xi ^{\beta }d\Sigma _{\alpha } &=&  \notag \\
-16\pi \int_{\Sigma }\left( T_{\beta }^{\alpha }-\dfrac{1}{4}\delta _{\beta
}^{\alpha }T\right) \xi ^{\beta }d\Sigma _{\alpha }+4\pi \int_{\Sigma
}T\delta ^{\alpha }{}_{\beta }\xi ^{\beta }d\Sigma _{\alpha } &=&  \notag \\
-2\int_{\Sigma }\dfrac{q}{r^{2}}Ed\Sigma _{\alpha }+4\pi \int T\sqrt{-h}%
drd\theta d\phi .&&
\end{eqnarray}%
since the timelike Killing vector is $\xi ^{\beta }=\delta
_{t}^{\beta }$ and $E^{2}=-2\mathcal{P}\mathcal{H_{P}}^{2}$ with
$E=q/r^{2}\mathcal{H_{P}}$ identically. Finally, by inserting
above results in Eq. (\ref{eqkomarr}), we obtain
\begin{equation}
M=\dfrac{\kappa A}{4\pi }+q\Phi -\dfrac{1}{2}\int
T\sqrt{-h}drd\theta d\phi,
\end{equation}%
where $h$ is the trace of the induced metric of spacelike hypersurface.

\section{The complexity growth of regular electric black holes \label{sec3}}

Here, we consider a class of known regular electric black holes in
AdS spacetime and calculate its complexity growth rate. For the
sake of completeness, we will point out other regular black hole
solutions in the appendix.

\begin{center}
\begin{figure}[tbp]
\hspace{0.4cm} \centering
\subfigure{\includegraphics[width=0.8\columnwidth]{r2vsqthree.eps}}
\caption{The behavior of q in terms of $r_{+}$ for $M=l=1$. dotted
line (first case), dashed line (second case) and black line (third
case).} \label{r2}
\end{figure}
\end{center}

The motivations of considering the forthcoming kind of
$\mathcal{H}$ function are its simplicity and its agreement to the
correspondence which indicates such a modified nonlinear theory
should reduce to the usual Maxwell theory in the weak-field limit.
The ansatz of structure function which is inspired by the
exponential distribution is defined as \cite{Balart:2014cga}
\begin{equation}  \label{eqhamilton}
\mathcal{H}=\mathcal{P}e^{-\left( \frac{\gamma^{\frac{3}{2}}(-\mathcal{P})^{%
\frac{1}{4}}}{2^{\frac{5}{4}} \chi}\right) }+\frac{6}{l^2},
\end{equation}
in which for the weak field limit ($\mathcal{P} \ll 1$), it
reduces to the Maxwell field as
\begin{equation}
\mathcal{H}\approx \mathcal{P}+\frac{6}{l^2}+\mathcal{O}(\mathcal{P}^{\frac{5%
}{4}}).
\end{equation}

Considering the mentioned structure function with Eqs. (\ref{eqn:p}) and (%
\ref{eqmetric1}), we can find the following spherically symmetric
exponentially black hole solution \cite{Balart:2014cga,
Balart:2014jia, Culetu:2014lca, Hendi:2020knv, Ghosh:2014pba}
\begin{equation}  \label{eqmetric0}
-g_{tt}=f(r)=1+\frac{r^{2}}{l^{2}}-\frac{2\chi}{r}e^{\left(-\frac{\gamma^{2}%
}{2\chi r}\right) }.
\end{equation}

Using the series expansion of this solution for large values of
$r$, it is noticeable that its asymptotical behavior can be found
by the following expression
\begin{equation}
-g_{tt}\approx 1+\frac{r^{2}}{l^{2}}-\frac{2\chi}{r}+\frac{\gamma^{2}}{r^{2}%
}+\mathcal{O}\left(\frac{1}{r^{4}}\right),
\end{equation}
which is the Reissner-Nordstr\"{o}m-AdS metric function provided
$\chi$ and $\gamma$ are associated with the mass and electric
charge of the system, respectively. In other words, the
asymptotical behavior of the obtained solution is completely
matched to the Reissner-Nordstr\"{o}m-AdS black hole.

In order to check the first law and the Smarr formula in the
extended phase space, it is convenient to consider the
cosmological constant as a varying
thermodynamic quantity and interpret it as a thermodynamic pressure, as $P=%
\frac{3}{8\pi l^{2}}$. In addition, its conjugate variable of the introduced
pressure is the thermodynamic volume
\begin{equation}  \label{eqvol}
V=\frac{4\pi}{3}r_{+}^{3},
\end{equation}
where $r_{+}$ is the event horizon radius obtained via $f(r_{+}) = 0$. The
temperature would be found straightforwardly through the use of surface
gravity ($\kappa$) interpretation with the following explicit relation
\begin{equation}
T=\frac{\kappa}{2 \pi}=\left. \frac{1}{4 \pi}\frac{df(r)}{dr}%
\right|_{r=r_{+}}= \left(m-\frac{q^{2}}{2r_{+}}\right) \frac{e^{-\left(
\frac{q^2}{2mr_{+}}\right)}} {2\pi r_{+}^{2}}+\frac{r_{+}}{2\pi l^{2}},
\end{equation}
in which for the large black hole event horizon ($r_{+} \gg l$) becomes
\begin{equation}
T\approx \dfrac{3r_{+}}{4\pi l^2}+\mathcal{O}\left(\dfrac{1}{r_{+}}\right).
\end{equation}
Since we are working in the Einstein gravity, the black hole
entropy $S$ pursues the area law, yielding
\begin{equation}
S=\frac{A}{4}=\pi r_{+}^{2}.
\end{equation}
In addition, the electrostatic potential $\Phi$ can be obtained at the event
horizon versus spatial infinity as the reference
\begin{equation}
\Phi = \int_{r_{+}}^{\infty}E dr=\frac{3m}{2q}-\frac{(6mr_{+}-q^{2})e^{-%
\left(\frac{q^2}{2mr_{+}}\right)}}{4qr_{+}},
\end{equation}
and the asymptotic limit becomes
\begin{equation}
\Phi \approx \dfrac{q}{r_{+}}+\mathcal{O}\left(\dfrac{1}{r_{+}^{5}}\right).
\end{equation}
For a black hole embedded in AdS spacetime, employing the relation
between the cosmological constant and thermodynamic pressure would
result to interpret the mass of black hole as the enthalpy. The
enthalpy can be written in terms of thermodynamic quantities as
\begin{equation}
H=m=\frac{q^2}{2r_{+}\mathcal{W}\left(\frac{q^2 l^2}{r_{+}^2(l^2+r_{+}^2)}%
\right)},
\end{equation}
where $\mathcal{W}$ is the $Lambert\ W$ function. In the case of large black
hole event horizon, we have
\begin{equation}  \label{eqappH}
H\approx \dfrac{r_{+}^{3}}{2 l^{2}}+\dfrac{r_{+}}{2}+\mathcal{O}\left(\dfrac{%
1}{r_{+}}\right).
\end{equation}

It is straightforward to check that these thermodynamic quantities satisfy
the first law of black hole thermodynamics in the enthalpy representation
\begin{equation}
dH=TdS+VdP+\Phi dq,
\end{equation}
and the Smarr relation is given by
\begin{equation}  \label{smarr1}
\frac{H}{2}+PV-TS-\frac{q\Phi}{2}+\frac{1}{4}\int w dv=0,
\end{equation}
where the last term of Eq. (\ref{smarr1}) comes from the fact that the
energy-momentum tensor is not traceless, as
\begin{equation}  \label{eqtrace}
\int w dv=\frac{1}{2}\int_{r_{+}}^{\infty}T^{\mu}_{\mu}\ r^{2}dr = m-\left(m+%
\frac{q^{2}}{2r_{+}}\right)e^{-\left(\frac{q^2}{2mr_{+}}\right)},
\end{equation}
and in the case of $r_{+} \gg l$ becomes
\begin{equation}
\int w dv \approx \dfrac{q^4 l^2}{4 r_{+}^{5}}+\mathcal{O}\left(\dfrac{1}{%
r_{+}^{7}}\right).
\end{equation}

\begin{center}
\begin{figure}[tbp]
\centering
\par
\includegraphics[width=8cm]{dif.eps}
\caption{The left hand side of Eq. (\protect\ref{smarr1}) versus
$r$ for $M=0.8, q=0.1, l=1$ .} \label{smarr}
\end{figure}
\end{center}

Now, we want to calculate the complexity growth rate in the above
background. For this aim, we should calculate the Ricci scalar for the
solution (\ref{eqmetric0}) as
\begin{equation}  \label{ricciscalar}
\mathcal{R}=-\frac{12}{l^{2}}+\frac{q^{4}e^{-\left( \frac{q^{2}}{2mr}\right)
}}{2mr^{5}},
\end{equation}
whereas by using of Eq. (\ref{eqhamilton}), the Lagrangian of nonlinear
electrodynamics is given by
\begin{equation}  \label{lagrangian}
\mathcal{L}=\frac{-q^2(4mr-q^2)e^{-\left( \frac{q^2}{2mr}\right) }}{2mr^{5}}-%
\frac{6}{l^{2}}.
\end{equation}

Using Eqs. (\ref{ricciscalar}) and (\ref{lagrangian}), we can calculate the
bulk action growth as
\begin{equation}
\frac{d\mathcal{A}_{bulk}}{dt} = \frac{1}{4}\int_{r_{-}}^{r_{+}}r^{2} \left(%
\mathcal{R}+ \frac{6}{l^{2}} -\mathcal{L}\right)dr =m\left[e^{\left( -\frac{%
q^{2}}{2mr}\right) }\right]_{r_{-}}^{r_{+}},
\end{equation}
while the growth rate of the surface term (Eq. (\ref{eqI})) is given by \cite%
{ElMoumni:2020nse}
\begin{eqnarray}
\frac{d\mathcal{A}_{boundary}}{dt}&=&\frac{1}{8\pi}\int_{\partial \mathcal{M}%
}(\sqrt{-h}\mathcal{K})d\Omega_{2}= \frac{1}{2}\left[ \sqrt{f(r)}\frac{%
\partial}{\partial r}(r^{2}\sqrt{f(r)})\right]_{\partial \mathcal{M}} \\
&=&\left[ \frac{3r^{3}}{2l^{2}} +r-\left(\frac{3m}{2} +\frac{q^{2}}{4r}%
\right) e^{\left(-\frac{q^{2}}{2mr}\right) }\right]_{r_{-}}^{r_{+}},  \notag
\end{eqnarray}
where we should note that in the calculation of boundary term we have used $%
\sqrt{-h}=\sqrt{f}r^{2}\sin(\theta)$ and $n^{\mu}=(0,\sqrt{f},0,0)$.

Besides, the contribution of joint terms can be calculated as
\begin{equation}
S_{D}-S_{C}=\frac{\delta t}{4}\ \left(2 r f(r) \left[ \ln \left(\frac{f(r)}{%
b_{1}b_{2}}\right) + \frac{r f^{\prime }(r)}{2 f(r)}
\right]\right) _{r_{C}}.  \label{joint2}
\end{equation}
After some manipulations, we can find
\begin{align}
S_{B}-S_{A}&=  \notag \\
&\frac{\delta t}{4} \left[2me^{-(\frac{q^2}{2mr})} -\frac{q^{2}e^{-(\frac{q^2%
}{2mr})}}{r} +\frac{2r^3}{l^2}+\left(2r-4m e^{-(\frac{q^2}{2mr})} +\frac{2r^3%
}{l^2}\right) \ln\left(\frac{1-\frac{2m}{r} e^{-(\frac{q^2}{2mr})}+ \frac{r^2%
}{l^2}}{b_{1} b_{2}}\right) \right]_{r_A},  \label{eqjoint1} \\
S_{D}-S_{C}&=  \notag \\
&\frac{\delta t}{4}\left[2me^{-(\frac{q^2}{2mr})} -\frac{q^{2} e^{-(\frac{q^2%
}{2mr})}}{r} +\frac{2r^3}{l^2}+ \left(2r-4m e^{-(\frac{q^2}{2mr})} +\frac{%
2r^3}{l^2}\right) \ln\left(\frac{1-\frac{2m}{r} e^{-(\frac{q^2}{2mr})} +%
\frac{r^2}{l^2}}{b_{1}b_{2}} \right)\right]_{r_C}.
\label{eqjoint2}
\end{align}
Since at the late time $r_{A}$ and $r_{C}$ approach to $r_{-}$ and $r_{+}$,
respectively, while $f(r)$ vanishes, one finds Eqs. (\ref{eqjoint1}) and (%
\ref{eqjoint2}) reduce to
\begin{eqnarray}
S_{B}-S_{A}&=&\frac{\delta t}{4}\left[2me^{-\left(\frac{q^2}{2mr}\right)} -%
\frac{q^{2}e^{-\left(\frac{q^2}{2mr}\right)}}{r}+\frac{2r^3}{l^2}\right]%
_{r_A}, \\
S_{D}-S_{C}&=&\frac{\delta t}{4}\left[2me^{-\left(\frac{q^2}{2mr}\right)} -%
\frac{q^{2}e^{-\left(\frac{q^2}{2mr}\right)}}{r}+\frac{2r^3}{l^2}\right]%
_{r_C}.
\end{eqnarray}

So, the total growth rate of the action for such a black hole configuration
within WDW patch at late time approximation is simplified as
\begin{equation}
\frac{d\mathcal{A}}{dt}=(r_{+}-r_{-})\left[ 1+\frac{3}{2l^{2}}%
(r_{+}^{2}+r_{-}^{2}+r_{+}r_{-})\right] +\left( \frac{q^{2}}{4r_{-}}+\frac{m%
}{2}\right) e^{\left( -\frac{q^{2}}{2mr_{-}}\right) }-\left( \frac{q^{2}}{%
4r_{+}}+\frac{m}{2}\right) e^{\left( -\frac{q^{2}}{2mr_{+}}\right) }.
\label{eqtotalaction}
\end{equation}

In order to describe Eq. (\ref{eqtotalaction}) in terms of $r_{-}$ and $%
r_{+} $, we can use the redefinitions $m$ and $q^{2}$ in terms of $r_{-}$
and $r_{+}$ as%
\begin{equation}
m =\frac{r_{-}^{2}(r_{-}^{2}+l^{2})\ln (A)A^{\left( \frac{r_{+}}{%
r_{+}-r_{-}}\right) }}{2(r_{+}-r_{-})l^{2}\mathcal{W}\left( \frac{\ln
(A)A^{\left( \frac{r_{+}}{r_{+}-r_{-}}\right) }r_{-}^{2}(l^{2}+r_{-}^{2})}{%
r_{+}(r_{+}-r_{-})(l^{2}+r_{+}^{2})}\right) },\hspace{0.8cm}
q^{2} =\frac{\ln (A)r_{-}^{2}(l^{2}+r_{-}^{2})r_{+}}{(r_{+}-r_{-})l^{2}e^{%
\frac{r_{+}\ln (A)}{r_{-}-r_{+}}}},
\end{equation}%
where
\begin{equation}
A=\frac{r_{+}(l^{2}+r_{+}^{2})}{r_{-}(l^{2}+r_{-}^{2})}.
\end{equation}%
Although it is straightforward to rewrite Eq. (\ref{eqtotalaction}) in terms
of $r_{-}$ and $r_{+}$, we ignore its explicit relation for the sake of
brevity. Finally, by using the above calculated thermodynamic quantities, it
is easy to show that%
\begin{equation}
\frac{d\mathcal{A}}{dt}\leq 2\left(m-q\phi +\frac{1}{2}\int wdv\right)_{+}-2\left(m-q\phi +%
\frac{1}{2}\int wdv\right)_{-},  \label{rate}
\end{equation}%
indicating that the action growth rate of the black hole in the WDW patch
has been bounded. For instance, for the values $m=0.8,q=0.1,l=1$ this bound
is about $1.5761$, and the differences between left and right (right mines
left) is about $0.0396$ which is clearly non-negative. It is worth
mentioning that although there are some extra terms in the right hand side
of Eq. (\ref{rate}) due to considering the nonlinear electrodynamics as a
source, the action growth rate is bounded.

\subsection{Extreme case: $r_{+}\approx r_{-}$}

Now, we focus on the extremal solutions. By introducing $\alpha=l/r_{+}$, $%
\epsilon=1-r_{-}/r_{+}$ and in the case of extremal, large black hole,
equation (\ref{eqtotalaction}) becomes
\begin{equation}
\dfrac{d \mathcal{A}}{dt}\approx\left (\dfrac{9 r_{+}} {4 \alpha^2}%
-r_{+}\alpha^{2}+\dfrac{7r_{+}}{4}\right)\epsilon + \mathcal{O}(\epsilon^2)+%
\mathcal{O}(\alpha^4),
\end{equation}
in terms of $r_\pm$ leads to
\begin{equation}
\dfrac{d \mathcal{A}}{dt}\approx \dfrac{9 }{4 l^{2}}
(r_{+}^{3}-r_{-}r_{+}^{2})+\dfrac{7}{4}(r_{+}-r_{-})+ \mathcal{O}\left(%
\dfrac{1}{r_{+}}\right).
\end{equation}
By using of equations (\ref{eqvol}), (\ref{eqappH}) and ($r_{+}\approx r_{-}$%
) one can rewrite above equation as follows \cite{Couch:2016exn}
\begin{equation}
\dfrac{d \mathcal{A}}{dt}\approx \dfrac{9}{2}P (V_{+}-V_{-}),\hspace{1cm}%
r_{+}\gg l,
\end{equation}
which shows that in the large black holes the rate of complexity is
proportional to $P\Delta V$, i.e, is controlled by thermodynamical volume.
Besides, it is notable that one can obtain such a relation in terms of
entropy for the static black holes since in the static case the
thermodynamic volume is not independent of entropy ($V=4/(3\sqrt{\pi}%
)S^{3/2} $).

In the case of extremal black hole with small degenerate horizon radius, we
have
\begin{equation}
\dfrac{d \mathcal{A}}{dt}\approx \left(\dfrac{13}{\alpha^{2}} +3\right)%
\dfrac{\epsilon}{4}+\mathcal{O} \left(\dfrac{1}{\alpha^{3}}\right)+\mathcal{O%
}(\epsilon^{2}).
\end{equation}
By using of CV conjecture, the rate of complexity is \cite{An:2018dbz}
\begin{equation}
\dfrac{d\mathcal{C}_{\mathcal{V}}}{dt}=\dfrac{4\pi}{Gl}\sqrt{-f(r_{min})}%
r_{min}^{2},
\end{equation}
where $r_{min}$ is the turning point of maximal surface. The late time limit
of $d\mathcal{C}_{\mathcal{V}}/dt$ becomes
\begin{equation}
\dfrac{d\mathcal{C}_{\mathcal{V}}}{dt}=\dfrac{4\pi}{Gl} \sqrt{-f(\hat{r}%
_{min})}\hat{r}^{2}_{min}, \hspace{1cm}t\longrightarrow\infty,
\end{equation}
where $\hat{r}_{min}$ is the extreme value point of $\sqrt{-f(r_{min})}%
r_{min}^{2}$. For the present case we need to obtain the zeros of the
following equation
\begin{equation}
l^2(6 M \hat{r}_{min}+q^2)\exp\left(-\dfrac{q^2}{2 M \hat{r}_{min}}\right)-2
\hat{r}^{2}_{min}(2 l^2+3 \hat{r}^{ 2}_{min})=0.
\end{equation}
in the case of small $q$, it becomes
\begin{equation}
\hat{r}_{min}\approx r_{0 min}+\varepsilon r_{1 min}=r_{0}+ \dfrac{q^2
l^2\varepsilon }{3M l^{2}-12 r_{0}^{3}-4r_{0}l^{2}},
\end{equation}
where
\begin{equation}
r_{0}=\dfrac{(108Ml^2+4\sqrt{32l^4+729l^4M^2})^{\frac{1}{3}}}{6}- \dfrac{%
4l^2}{(108Ml^2+4\sqrt{32l^4+729l^4M^2})^{\frac{1}{3}}}.
\end{equation}

In the case of extremal black hole with large horizon radius, we can write
\begin{equation}
\dfrac{\hat{r}_{min}}{r_{+}} \approx 1.5-0.5 \varepsilon+ \mathcal{O}%
(\varepsilon,\alpha^{2}),
\end{equation}
and correspondingly, for the late time of growth of complexity, we obtain
\begin{equation}
\dfrac{d\mathcal{C}_{\mathcal{V}}}{dt}\approx \dfrac{40\pi}{\alpha^2}%
(r_{+}-r_{-})\approx 20 P\Delta V.
\end{equation}
The comparison of the complexity growth from the CV and CA dualities are
\begin{equation}\label{Rrate}
\mathcal{R}_{rate}=\dfrac{d\mathcal{C}_{\mathcal{A}}/dt}{d\mathcal{C} _{%
\mathcal{V}}/dt}=\dfrac{9 P \Delta V}{40 P \Delta V}=0.225.
\end{equation}

\subsection{Complexity of formation}

The complexity of formation is the difference between complexity
in the process of forming the entangled TFD state and preparing
two individual copies of the vacuum state of the left and right
boundary CFTs \cite{Chapman:2016hwi,Cottrell:2017ayj}
\begin{equation}
\Delta \mathcal{C}_{\mathcal{A}}=\dfrac{1}{\pi }(\mathcal{A}_{BH}-2\mathcal{A%
}_{AdS}).
\end{equation}

\begin{figure}[tbp]
\centering
\includegraphics[width=6cm] {WDWpatch.eps}
\caption{Penrose diagram for charged black holes.}
\label{wdw}
\end{figure}
For the case of charged black hole with two horizons, the complexity of
formation in CA conjecture becomes \cite{Chapman:2016hwi}
\begin{equation}\label{eqdeltaca}
\Delta \mathcal{C}_{\mathcal{A}}=\dfrac{1}{\pi }(\Delta \mathcal{A}_{bulk}+%
\mathcal{A}_{joint,meet}).
\end{equation}%
Now, we evaluate the action for the obtained black hole solutions. The
tortoise coordinate for the both horizons of the solution is%
\begin{equation}
r^{\ast }(r)=\dfrac{l^{2}r_{+}\ln \left( \dfrac{|r-r_{+}|}{r+r_{+}}\right) }{%
3r_{+}^{2}+l^{2}-(l^{2}+r_{+}^{2})\mathcal{W}\left( \dfrac{l^{2}q^{2}}{%
r_{+}^{2}(l^{2}+r_{+}^{2})}\right) }+\dfrac{l^{2}r_{-}\ln \left( \dfrac{%
|r-r_{-}|}{r+r_{-}}\right) }{3r_{-}^{2}+l^{2}-(l^{2}+r_{-}^{2})\mathcal{W}%
\left( \dfrac{l^{2}q^{2}}{r_{-}^{2}(l^{2}+r_{-}^{2})}\right) },
\label{eqtor1}
\end{equation}%
The point where the ingoing null rays from the two asymptotic
regions meet inside the black hole between the two horizons
($r_{-}<r_{meet}<r_{+}$) can be calculated numerically using Eqs.
(\ref{eqtor1})-(\ref{eqtor3}) which reads
\begin{equation}
r^{\ast }(r_{meet})=0.
\end{equation}%
\begin{center}
\begin{figure}[tbp]
\hspace{0.4cm} \centering
\subfigure{\includegraphics[width=0.4\columnwidth]{rmeetrp.eps}} %
\subfigure{\includegraphics[width=0.4\columnwidth]{rmeetq.eps}}
\caption{The behavior of $r_{meet}$ in terms of $r_{+}$ (left) and $q$
(right) for $M=l=1$. red line (first case), black line (second case) and
blue line (third case).}
\label{fig5}
\end{figure}
\end{center}
We show the results for $r_{meet}$ in Fig. \ref{fig5} for the considered
solution. According to the Fig. \ref{fig5}a, by increasing event horizon
radius, $r_{meet}$ decreases from extremal value to $r_{-}$. Then, by
increasing the electric charge, $r_{meet}$ increases (Fig. \ref{fig5}b). By
using of numerical curve fitting and choosing the appropriate branches of
the logarithms, one can obtain the following relation for $r_{meet}(r_{+})$
\begin{equation}
r_{meet}\approx 0.6825r_{+}^{2}-2.0259r_{+}+1.336,
\end{equation}
The above approximation functions are necessary to obtain the $A_{bulk}$ and
$A_{joint}$ in the following.\\
The bulk action of the black hole is
\begin{align}
A_{bulk,BH}=& 4\dfrac{4\pi }{16\pi G}\int_{r_{meet}}^{r_{max}}\left(
\mathcal{R}+\dfrac{6}{l^{2}}-\mathcal{L}\right) dr\int_{0}^{v_{\infty
}-r^{\ast }(r)}dt  \notag \\
=& \dfrac{2q^{2}}{G}\int_{r_{meet}}^{r_{max}}dr\left( \dfrac{\exp \left( -%
\dfrac{q^{2}}{2Mr}\right) }{r^{4}}\right) (v_{\infty }-r^{\ast }(r))\approx
\dfrac{8.27 r_{+}^2}{G l^4}-\dfrac{0.005 r_{+}}{G l^4}+\dfrac{0.00003 r_{+}}{%
G \delta l^2}+\mathcal{O}(1),
\end{align}%
where $v_{\infty }=\lim_{r\rightarrow \infty }r^{\ast }(r)$ is the
constant defining the future null boundary, $r_{max}=\frac{l^{2}}{\delta }-%
\frac{\delta }{4}+\frac{M\delta ^{2}}{3l^{2}}-\frac{q^{2}\delta ^{3}}{8l^{4}}
$ and $\delta $ is a some cutoff surface associated with a UV divergence. In
order to obtain the action of AdS spacetime, one uses%
\begin{equation}\label{eqads0}
f_{AdS}=1+\dfrac{r^{2}}{l^{2}},
\end{equation}%
in which the tortoise coordinate for AdS spacetime is%
\begin{equation}\label{eqads1}
r_{0}^{\ast }{}(r)=l\arctan \left( \dfrac{r}{l}\right) ,\hspace{0.5cm}%
v_{0\infty }=\dfrac{\pi l}{2}.
\end{equation}%
Thus the bulk action of AdS vacuum is%
\begin{align}\label{eqads2}
A_{AdS}& =4\dfrac{4\pi }{16\pi G}\int_{0}^{r_{0max}}dr\left( \mathcal{R}+%
\dfrac{6}{l^{2}}\right) \int_{0}^{v_{0\infty }-r_{0}^{\ast }{}(r)}dt=-\dfrac{%
6}{Gl^{2}}\int_{0}^{r_{0max}}dr(v_{0\infty }-r_{0}^{\ast }{}(r))=  \notag \\
& \dfrac{-1}{G}\left[ -l^{2}\ln \left( 1+\dfrac{r_{0max}^{2}}{l^{2}}\right)
+r_{0max}^{2}\left( 1+\dfrac{\pi r_{0max}}{l}-\dfrac{2r_{0max}}{l}\arctan (%
\dfrac{r_{0max}}{l})\right) \right] ,
\end{align}%
where $\mathcal{R}_{ads}=-12/l^{2}$ and $r_{0max}=\frac{l^{2}}{\delta }-%
\frac{\delta }{4}$. Now, we should calculate the following relation for the
solution
\begin{equation}
\Delta A_{bulk}=A_{bulk,BH}-A_{AdS}.  \label{eqact}
\end{equation}%
The action at the meet point $r_{meet}$ is%
\begin{equation}
\mathcal{A}_{joint}=\frac{2}{G}r_{meet}^{2}\ln [f(r_{meet})],  \label{eqjoin}
\end{equation}%
which in the case of large black hole becomes%
\begin{equation*}
\mathcal{A}_{joint}\approx 0.0186\ln \left( \dfrac{0.0682r_{+}^{2}}{l}%
\right) r_{+}^{4}+\mathcal{O}(r_{+}^{3}).
\end{equation*}%
The total action is the sum of the bulk (\ref{eqact}) and joint (\ref{eqjoin}%
) terms. Substituting the numerical solution for $r_{meet}$, we
obtain the interesting result plotted in Fig. \ref{CDV1}.
According to these figures, one finds the bulk action of black
hole and complexity of formation are linear functions of entropy.
For the large black hole, we have
\begin{equation}
\Delta \mathcal{C}_{\mathcal{A}}\approx \dfrac{8.27 C_{T}S}{l^{2}}+\mathcal{O%
}(S^{1/2}),
\end{equation}%
where $C_{T}=3l^{2}/\pi ^{3}G$ is the boundary central charge.

\begin{figure}[tbp]
\hspace{0.2cm} \subfigure{\includegraphics[width=0.7\columnwidth]{deltacvfirstt.eps}} %
\subfigure{\includegraphics[width=0.7\columnwidth]{deltacafirstt.eps}}
\caption{Plot of $\Delta \mathcal{C}_{\mathcal{V}} $ (up) and
$\Delta \mathcal{C}_{\mathcal{A}}$ (down) in terms of $r_{+}$ for
static exponential solution.. } \label{CDV1}
\end{figure}

Here, we want to obtain the complexity of formation by using of CV
conjecture. To do that, we consider the maximal volume functional for the $%
t=0$ timeslice (the straight line connecting the two boundaries through the
bifurcation surface in the Penrose diagram) as follows \cite%
{Chapman:2016hwi,Balushi:2020wkt}%
\begin{equation}
\mathcal{V}=8\pi \int_{r_{+}}^{r_{max}}dr\dfrac{r^{2}}{\sqrt{f}}=4\pi l
r_{+}^2+2\pi l^3+\dfrac{\pi l \delta^2}{4}+8\pi l^3 \ln\left(\dfrac{l}{\sqrt{%
\delta r_{+}}}\right)+\mathcal{O}\left(\dfrac{l^5}{r_{+}^5}\right),
\label{eqvolumfun}
\end{equation}%
where the corresponding contribution from two copies of the vacuum AdS
background is%
\begin{equation}
\mathcal{V}_{AdS}=8\pi \int_{0}^{r_{max}}dr\dfrac{r^{2}}{\sqrt{f_{AdS}}}%
=4\pi l\left[ r_{max}\sqrt{l^{2}+r_{max}^{2}}-l^{2}\ln \left( \dfrac{r_{max}+%
\sqrt{l^{2}+r_{max}^{2}}}{l}\right) \right] =\dfrac{4\pi l^{5}}{\delta ^{2}}%
-4\pi l^{3}\ln \left( \dfrac{2l}{\delta }\right) -\dfrac{\pi
l\delta ^{2}}{4}.
\end{equation}%
The complexity of formation is obtained by subtracting from (\ref{eqvolumfun}%
) the volume of the AdS vacuum
\begin{equation}\label{eqdeltacvform}
\Delta \mathcal{C}_{\mathcal{V}}=\dfrac{\mathcal{V}-2\mathcal{V}_{AdS}}{%
G_{N}R}=\dfrac{8\pi }{G_{N}R}\left[ \int_{r_{+}}^{r_{max}}\dfrac{r^{2}dr}{%
\sqrt{f(r)}}-\int_{0}^{r_{max}}\dfrac{r^{2}dr}{\sqrt{f_{AdS}}}\right] =%
\dfrac{41.3C_{T}S}{l^{3}}+\mathcal{O}(S^{1/2}).
\end{equation}%
To understand the behaviour of complexity of formation, we obtain
the plot of $\Delta \mathcal{C}_{\mathcal{V}}$ as a function of
$r_{+}$ in Fig. \ref{CDV1}.

It is interesting, of course, to compare the results of CA and CV
duality. The ratio of the complexity of formation for large black
holes $r_{+}\gg l$ is
\begin{equation}\label{Rform}
\mathcal{R}_{form}=\dfrac{\Delta \mathcal{C}_{\mathcal{A}}}{\Delta \mathcal{C%
}_{\mathcal{V}}}\approx 0.2,
\end{equation}%
and by using (\ref{Rrate}) and (\ref{Rform}) one gets
\begin{equation}
\mathcal{R}_{rate}-\mathcal{R}_{form}=0.02.
\end{equation}
As can be seen, the differences between two ratios is small. So, the two
ratios agree very well and this comparison suggests that the two holographic
approaches to complexity are consistent.

\section{Generalization to rotating solution:}\label{sec4}

The metric of rotating charged regular black hole in the Boyer-Lindquist
coordinates is obtained as
\begin{equation}
dS^{2}=-\dfrac{\Delta _{r}}{\Sigma }\left( dt-\dfrac{a\sin ^{2}(\theta )}{%
\Xi }d\phi \right) ^{2}+\dfrac{\Sigma }{\Delta _{r}}dr^{2}+\dfrac{\Sigma }{%
\Delta _{\theta }}d\theta ^{2}+\dfrac{\Delta _{\theta }\sin ^{2}(\theta )}{%
\Sigma }\left( adt-\dfrac{r^{2}+a^{2}}{\Xi }d\phi \right) ^{2},
\label{metric}
\end{equation}%
where
\begin{align}
\Delta _{r}& =(r^{2}+a^{2})(1+\dfrac{r^{2}}{l^{2}})-2f,\;\;\;\;\Sigma
=r^{2}+a^{2}\cos ^{2}(\theta ),\;\;\;\;\Xi =1-\dfrac{a^{2}}{l^{2}},  \notag
\\
\Delta _{\theta }& =1-\dfrac{a^{2}}{l^{2}}\cos ^{2}(\theta
),\;\;\;\;f(r)=Mr\exp \left( -\dfrac{q^{2}}{2Mr}\right) .
\end{align}%
The outer and inner horizons are determined by the equation $\Delta (r_{\pm
})=0$, respectively. The required thermodynamical quantities to check the
first law are \cite{Hendi:2020knv}%
\begin{align}
P=& \dfrac{3}{8\pi l^{2}},\hspace{0.5cm}V=\dfrac{r_{+}A}{3}+\dfrac{4\pi J^{2}%
}{3\mathcal{M}},\hspace{0.5cm}A=\dfrac{4\pi (r_{+}^{2}+a^{2})}{\Xi },\hspace{%
0.5cm}J=\dfrac{Ma}{\Xi ^{2}},\hspace{0.5cm}\mathcal{M}=\dfrac{M}{\Xi ^{2}},%
\hspace{0.5cm}Q=\dfrac{q}{\Xi },  \notag \\
T=& -\dfrac{(2Mr_{+}+q^{2})e^{-\left( \dfrac{q^{2}}{2Mr_{+}}\right) }}{4\pi
(r_{+}^{2}+a^{2})r_{+}}+\dfrac{r_{+}(r_{+}^{2}+2r_{+}^{2}+a^{2})}{2\pi
l^{2}(r_{+}^{2}+a^{2})},\hspace{0.5cm}\Phi =\dfrac{qr_{+}}{r_{+}^{2}+a^{2}},%
\hspace{0.5cm}\Omega =\dfrac{a}{r_{+}^{2}+a^{2}}\left( 1+\dfrac{r_{+}^{2}}{%
l^{2}}\right) .
\end{align}%
where
\begin{equation}
M=\dfrac{q^{2}}{2r_{+}W\left( \dfrac{l^{2}q^{2}}{%
(r_{+}^{2}+a^{2})(r_{+}^{2}+l^{2})}\right) }.
\end{equation}%
So, the first law of thermodynamic is satisfied as \cite{Hendi:2020knv}%
\begin{equation}
d\mathcal{M}=TdS+PdV+\Phi dQ+\Omega dJ,
\end{equation}%
Integrating the on-shell Einstein-Hilbert bulk action, directly,
in the case of $\frac{q}{M}\ll 1$ and $\frac{a}{l} \ll 1$ leads to
\begin{align}
\frac{d\mathcal{A}_{bulk}}{dt}& =\frac{1}{8}\int_{r_{-}}^{r_{+}}\int_{0}^{%
\pi }\sqrt{-g}\left( \mathcal{R}+\frac{6}{l^{2}}-\mathcal{L}\right)
drd\theta =  \notag \\
& -\left[ \dfrac{%
24Mr^{7}-24Ml^{2}q^{2}r^{3}+9l^{2}q^{4}r^{2}+24Ma^{2}r^{5}+24Ma^{2}q^{2}l^{2}r-11a^{2}l^{2}q^{4}%
}{48\Xi ml^{2}r^{4}}\right] _{r_{-}}^{r_{+}},
\end{align}%
since $\sqrt{-g}=\dfrac{\sin\theta }{\Xi}\Sigma$ and%
\begin{equation}
\mathcal{L}=\dfrac{2q^{2}e^{-\dfrac{q^{2}}{2Mr}}((Mr+q^{2})a^{4}\cos
^{4}\theta +(-6Mr+q^{2})a^{2}r^{2}\cos ^{2}\theta +Mr^{5})}{%
Mr(r^{2}+a^{2}\cos ^{2}\theta )^{4}}.
\end{equation}%
The growth rate of the surface term in the case of $\frac{q}{M}\ll
1$ and $\frac{a}{l}\ll 1$
(Eq. (\ref{eqI})) is%
\begin{equation}
\dfrac{d\mathcal{A}_{boundary}}{dt} =\frac{1}{8\pi }\int_{\partial \mathcal{%
M}}(\sqrt{-h}\mathcal{K})d\Omega _{2}=\left[ \dfrac{\Delta _{r}^{^{\prime }}%
}{4\Xi }\right] _{r_{-}}^{r_{+}}=
\left[ \dfrac{-8M^{2}r^{2}l^{2}+8Mr^{3}(l^{2}+a^{2}+2r^{2})+q^{4}l^{2}}{%
16Ml^{2}\Xi r^{2}}\right] _{r_{-}}^{r_{+}},
\end{equation}%
since the determinant of the induced metric on the null
hypersurface $r_{\pm }$ is $h=-\dfrac{\sin ^{2}\theta \Sigma
\Delta _{r}}{\Xi ^{2}}$ and the trace of extrinsic curvature can
be calculated from Eq. (\ref{extcur}) and the normal vector is
$n_{\mu }=(0,\sqrt{\frac{\Delta _{r}}{\Sigma }},0,0)$. Combining
the bulk action and boundary term, we can obtain the growth rate
of total action as follows
\begin{equation}
\dfrac{d\mathcal{A}}{dt}=\left[ \dfrac{%
24Mq^{2}l^{2}r^{3}-6q^{4}l^{2}r^{2}-24Ma^{2}q^{2}l^{2}r+11a^{2}q^{4}l^{2}-24M^{2}l^{2}r^{4}+24Ml^{2}r^{5}+24Mr^{7}%
}{48\Xi Ml^{2}r^{4}}\right] _{r_{-}}^{r_{+}},
\end{equation}%
where
\begin{equation*}
q^{2}=\dfrac{r_{+}(a^{2}l^{2}+a^{2}r_{-}^{2}+r_{-}^{2}l^{2}+r_{-}^{4})\ln
\left( \dfrac{r_{-}(l^{2}+r_{+}^{2})(a^{2}+r_{+}^{2})}{%
r_{+}(a^{2}+r_{-}^{2})(l^{2}+r_{-}^{2})}\right) }{(r_{+}-r_{-})l^{2}\left(
\dfrac{r_{-}(l^{2}+r_{+}^{2})(a^{2}+r_{+}^{2})}{%
r_{+}(a^{2}+r_{-}^{2})(l^{2}+r_{-}^{2})}\right) ^{\frac{r_{+}}{r_{-}-r_{+}}}}.
\end{equation*}
In the case of $\alpha \ll 1$, $\beta \ll 1$ and $\epsilon \ll 1$, one finds%
\begin{eqnarray}
q^{2}/r_{+}^{2} &=&\left( \dfrac{60.26}{\alpha ^{2}}-100.43+\left( -\dfrac{%
100.43}{\alpha ^{2}}+140.6\right) \beta ^{2}\right) +  \notag \\
&&\left( -\dfrac{120.51}{\alpha ^{2}}+100.43+\dfrac{100.43\beta ^{2}}{\alpha
^{2}}\right) \epsilon +\mathcal{O}(\epsilon ^{2})+\mathcal{O}(\alpha ^{2})+%
\mathcal{O}(\beta ^{4}),
\end{eqnarray}%
while for $\alpha \gg 1$, $\beta \ll 1$ and $\epsilon \rightarrow 0$, we have%
\begin{eqnarray}
q^{2}/r_{+}^{2} &=&2.72+\dfrac{13.59}{\alpha ^{2}}-\left( 8.15+\dfrac{29.9}{%
\alpha ^{2}}\right) \beta ^{2}-  \notag \\
&&\left( 2.72+\dfrac{27.2}{\alpha ^{2}}-\dfrac{29.9\beta ^{2}}{\alpha ^{2}}%
\right) \epsilon +\mathcal{O}(\epsilon )+\mathcal{O}(\beta ^{4})+\mathcal{O}%
(1/\alpha ^{4}).
\end{eqnarray}

In addition, in the case of $\alpha \ll 1$, $\beta \ll 1$ and $\epsilon
\rightarrow 0$, we obtain%
\begin{equation}
\dfrac{d\mathcal{A}}{dt}\approx \dfrac{22.3 r_{+}^3}{l^{2}}-\dfrac{9.32
r_{-}r_{+}^2}{l^2} +\mathcal{O}(r_{+}),
\end{equation}%
and for $\alpha \ll 1$, $\beta \ll 1$ and $\epsilon \rightarrow 1$, one
achieves%
\begin{equation}
\dfrac{d\mathcal{A}}{dt}\approx \dfrac{0.21r_{+}\alpha ^{4}\beta ^{6}}{%
\epsilon ^{12}}+\mathcal{O}(\epsilon ^{-11}),
\end{equation}%
since

\begin{align}
V/r_{+}^{3}\approx & \dfrac{0.8\times 10^{-9}\alpha ^{6}\beta ^{6}}{\epsilon
^{12}},\hspace{1cm}\epsilon \rightarrow 1,  \notag \\
V/r_{+}^{3}\approx & 84.13+560.9\alpha ^{6},\hspace{1cm}\epsilon \rightarrow
0.
\end{align}
As can be seen in both cases $\epsilon \rightarrow 1,0$, the rate of
complexity proportional to the volume of black holes instead of entropy of
black holes.

\subsection{Complexity of formation}

In this section complexity of formation for rotating black holes
in the limit $\frac{a}{l}\ll1$ and $\frac{q}{M}\ll1$ and in the
conjectures Complexity-Action and Complexity-Volume is calculated.

To do so, we calculated Lagrangian of the bulk and vacuum AdS in
the limit $ \frac{a}{l}\ll1$ and $\frac{q}{M}\ll1$, which in the
second-order approximation are written the following form
\begin{equation*}
L_{bulk}=\frac{-4r^{3}l^{2}-4r^{3}a^{2}-4a^{2}l^{2}r+\frac{%
4q^{2}l^{2}(l^{2}+a^{2})}{r}-\frac{4a^{2}q^{2}l^{4}}{r^{3}}}{8l^{4}},\;\;\;\;\;\;L_{AdS}=-\frac{2r^{3}}{l^{2}}.
\end{equation*}
For obtaining the complexity of formation we should compute the following relation:
\begin{equation*}
\pi \Delta C_{A}=A_{bulk}-A_{Ads}+A_{joints}=\int \left( \int \sqrt{-g}
\pounds _{bulk}(r_{\infty }^{\ast}-r^{\ast })dr-\int \sqrt{-g} \pounds %
_{AdS}(r_{\infty }^{\ast }-r^{\ast })dr\right) d\theta
+A_{joints},
\end{equation*}
where
\begin{equation*}
\pounds _{bulk}=R+\frac{6}{l^{2}}-L,\;\;\;\;\pounds
_{AdS}=R+\frac{6}{l^{2}},\;\;\;\;r^{\ast }=\int
\sqrt{\frac{g_{rr}}{g_{tt}}}dr.
\end{equation*}
Since $\lim_{r \rightarrow \infty }r^{\ast }=0\ $and$\ r^{\ast }$
in the limit $\frac{q}{M}\ll1$ and $\frac{a}{l}\ll 1$ can be
calculated analytically, but the integral of the first term with
respect to $r$ can not be calculated analytically thus it is
actually more convenient to use integration\ by parts to eliminate
the appearance of $r^{\ast }$ inside this expression, because
after that we must integrating with respect to $\theta $ therefore
the first integral must be calculated analytically,

\begin{equation*}
A_{bulk}=\int\int\sqrt{-g}\pounds _{bulk}(-r^{\ast})drd\theta =\int\left(
(-r^{\ast})\int\sqrt{-g}\pounds _{bulk} dr+\int\left(\sqrt{-\frac{g_{rr}}{%
g_{tt}}}\int \sqrt{-g}\pounds _{bulk}dr\right)dr \right) d\theta.
\end{equation*}

$A_{Ads}$ is obtained in the same way. After some calculations
complexity of formation for slowly rotating black holes in the
limit $\frac{q}{M}\ll 1$ within Complexity-Action conjecture is
obtained which is shown in Fig. \ref{dcasr}.
\begin{center}
\begin{figure}[tbp]
\centering
\includegraphics[width=14cm]{deltacaslowrot}
\caption{Complexity of formation in the CA conjecture versus
$\frac{r_-}{r_+} $ for rotating black hole.} \label{dcasr}
\end{figure}
\end{center}

The behavior of  $\frac{r_{meet}}{r_{+}}$ ($r_{meet}$) versus
$r_{-}/r_{+}$ ($r_{+}$) is plotted in Fig. \ref{meet12}. We find
that $r_{meet}$ is approximately a linear function of event
horizon and the best fitting curve relation is $r_{meet}=-2.5
r_{+} + 2.6$.

\begin{figure}[tbp]
\hspace{0.2cm}
\subfigure{\includegraphics[width=0.7\columnwidth]{rmeetr2versusr1r2.eps}}
\subfigure{\includegraphics[width=0.7\columnwidth]{rmeetversusr2.eps}}
\caption{ The behavior of $\frac{r_{meet}}{r_{+}}$ versus
$\frac{r_{-}}{r_{+}}$ (up) and $r_{meet}$ versus $r_{+}$ (down).}
\label{meet12}
\end{figure}

\textbf{Complexity of formation in CV conjecture}

For calculating the complexity in CV conjecture, we must find the
surface with maximal volume. For this purpose we should write the
line element in the tortoise coordinate
\begin{eqnarray*}
ds^{2} &=&d\upsilon ^{2}\left( -\frac{\Delta _{r}}{\Sigma }+\frac{%
a^{2}\Delta _{\theta }\sin ^{2}(\theta )}{\Sigma }\right) +dr^{2}\left( -%
\frac{\Delta _{r}}{\Sigma }\frac{g_{rr}}{g_{tt}}+\frac{a^{2}\Delta _{\theta
}\sin ^{2}(\theta )}{\Sigma }\frac{g_{rr}}{g_{tt}}+\frac{\Sigma }{\Delta _{r}%
}\right) \\
&&+2drd\upsilon \sqrt{-\frac{g_{rr}}{g_{tt}}}\left( \frac{\Delta _{r}}{%
\Sigma }-\frac{a^{2}\Delta _{\theta }\sin ^{2}(\theta )}{\Sigma }\right) +%
\frac{\Sigma }{\Delta _{\theta }}d\theta ^{2}+d\phi ^{2}\left( -\frac{\Delta
_{r}}{\Sigma }\frac{a^{2}\sin ^{4}(\theta )}{\Xi ^{2}}+\frac{%
(a^{2}+r^{2})^{2}\Delta _{\theta }\sin ^{2}(\theta )}{\Xi ^{2}\Sigma }\right)
\\
&&-d\phi d\upsilon \left( \frac{a\sin ^{2}(\theta )}{\Sigma \Xi }\right)
\left( 2\Delta _{r}-(a^{2}+r^{2})\Delta _{\theta }\right) +drd\phi \sqrt{-%
\frac{g_{rr}}{g_{tt}}}\left( -2\frac{\Delta _{r}}{\Sigma }+\frac{%
a(a^{2}+r^{2})\Delta _{\theta }\sin ^{2}(\theta )}{\Xi \Sigma
}\right),
\end{eqnarray*}
where
\begin{equation*}
\upsilon =t+r^{\ast },\;\;\;\;r^{\ast }=\int \sqrt{-\frac{g_{rr}}{g_{tt}}}dr=\int \sqrt{\frac{\Sigma ^{2}}{%
\Delta _{r}^{2}-a^{2}\Delta _{r}\Delta _{\theta }\sin ^{2}(\theta )}}dr.
\end{equation*}
To find the maximal surface we describe the surface with the
parametric relations $r=r(\lambda )$ and $\upsilon =\upsilon
(\lambda )$ with some parameter $\lambda$, then the volume of the
surface becomes in the following form
\begin{equation}\label{eqmaxvol}
V=\int \sqrt{\sigma }d^{2}x=4\pi \int \int dr d\theta r^{2}\left( \sqrt{\left[-%
\frac{\Delta _{r}}{\Sigma }+\frac{\Delta _{\theta }a^{2}\sin ^{2}(\theta )}{%
\Sigma }\right]\dot{\upsilon} ^{2}+2\dot{r}\dot{\upsilon} \sqrt{1-\frac{\Delta _{\theta }a^{2}\sin
^{2}(\theta )}{\Delta _{r}}}}\right),
\end{equation}
where dots indicate derivatives with respect to $\lambda $ and the
factor 2 is originated from this fact that the surface is composed
from two equivalent parts. We are free to choose $\lambda $ to
keep the radial volume element fixed as follow
\begin{equation}\label{eq415}
r^{2}\sqrt{\left[-\frac{\Delta _{r}}{\Sigma }+\frac{\Delta_{\theta }a^{2}\sin^{2}%
}{\Sigma}\right]\dot{\upsilon} ^{2}+2\dot{r}\dot{\upsilon}\sqrt{1-\frac{\Delta _{\theta }a^{2}\sin
^{2}(\theta) }{\Delta _{r}}}}=1,
\end{equation}
this Lagrangian is independent of $\upsilon $ and hence there is a
conserved quantity given by
\begin{equation}\label{eq416}
E=-\frac{\partial\pounds }{\partial \upsilon}=\left(\frac{\Delta_{r}}{ \Sigma}-%
\frac{\Delta_{\theta} a^{2}\sin (\theta)^{2}}{\Sigma}\right)\dot{\upsilon}-\sqrt{%
1-\frac{\Delta_{\theta} a^{2}\sin (\theta)^{2}}{\Delta_{r}}}\dot{r}.
\end{equation}

By substituting $\dot{\upsilon}$ from (\ref{eq415}) into the
(\ref{eq416}), one can obtain
\begin{equation*}
E^{2}=r^{4}\left(\frac{\Delta _{r}}{\Sigma }-\frac{\Delta _{\theta
}a^{2}\sin ^{2}(\theta )}{\Sigma
}\right)+r^{8}\dot{r}^{2}\sqrt{1-\frac{\Delta _{\theta }a^{2}\sin
^{2}(\theta )}{\Delta _{r}}},
\end{equation*}
with substituting above equations into the relation
(\ref{eqmaxvol}) for maximal surface volume one can find that
\begin{equation*}
V=4\pi \int_{r_{\min }}^{r_{\max}} \int_{0}^{\pi}\frac{dr}{\dot{r}}=2\Omega
\int_{r_{\min }}^{r_{\max }}\int_{0}^{\pi }\frac{r^{4}\sqrt[4]{1-\frac{%
\Delta _{\theta }a^{2}\sin ^{2}(\theta )}{\Delta _{r}}}d\theta dr}{\sqrt{%
r^{4}(\frac{\Delta _{r}}{\Sigma }-\frac{\Delta _{\theta }a^{2}\sin
^{2}(\theta )}{\Sigma })+E^{2}}}.
\end{equation*}

Here we wish to take $r_{\max }$ to be infinity, but this will
yield a divergent result in general. A finite result can be
obtained by performing a carefully matched subtraction of the AdS
vacuum. Here $r_{min}$\ is the turning point of the surface which
determined by the condition $\dot{r}=0$
\begin{equation*}
E^{2}+r_{\min }^{4}\left(-\frac{\Delta _{r}}{\Sigma }+\frac{\Delta
_{\theta }\sin ^{2}(\theta )}{\Sigma }\right)_{r_{\min }}=0.
\end{equation*}

A simple calculation shows that $r_{min}$ will be on or inside the (outer)
horizon, and so we have that, $g_{tt}(r_{\min })< 0$ , $\upsilon
(\lambda _{\min })> 0$ $\Longrightarrow $ E$< 0$ and we recall that $%
g_{tt}(r_{\min })< 0$ in the region between the inner and event
horizon. We are interested in maximal slice which $\tau =0$\ in
this case $r_{\min }=r_{+}$\ which gives $E=0$. Thus, the
complexity of formation becomes
\begin{eqnarray*}
\Delta C_{V} =\frac{V_{bulk}-2V_{ads}}{G_{N}L}=\frac{4\pi }{G_{N}L}\left(
\int_{r_{+}}^{r_{\max }}\int_{0}^{\pi }\frac{r^{2}\sqrt{\frac{\Sigma }{%
\Delta _{r}}}}{(1-\frac{\Delta _{\theta }a^{2}\sin ^{2}(\theta )}{\Delta _{r}%
})^{\frac{1}{4}}}d\theta dr-
 \int_{0}^{r_{0\max }}\int_{0}^{\pi }2r^{2}\sqrt{\frac{\Sigma }{%
r^{2}(1+\frac{r^{2}}{l^{2}})-2Mr}}d\theta dr\right),
\end{eqnarray*}
integrating with respect to $\theta$ gives us
\begin{eqnarray*}
\Delta C_{Vbulk} &=&\frac{8l\pi ^{2}}{G_{N}L}\int_{r_{+}}^{r_{\max }}\frac{dr}{%
\left[ r^{3}-(-r^{2}+2m)l^{2}\right] ^{\frac{5}{2}}}\left[ \left(\frac{r^{7}}{2}%
-r^{6}m+\left(-\frac{q^{2}}{8}-\frac{3}{32}a^{2}\right)r^{5}+\frac{5a^{2}q^{2}r^{3}}{32}%
\right)l^{2}+\frac{r^{9}}{4}-\frac{a^{2}r^{7}}{16}+\right.  \\
&&\left. \left(\frac{1}{4}r^{5}-mr^{4}+\left(-\frac{q^{2}}{8}+m^{2}-\frac{a^{2}}{32}%
\right)r^{3}+\left(\frac{q^{2}}{4}m-\frac{a^{2}m}{16}\right)r^{2}+\frac{a^{2}}{4}\left(m^{2}+\frac{%
7q^{2}}{16}\right)r+\frac{a^{2}q^{2}m}{16}\right)l^{4}\right]-\frac{8l\pi ^{2}}{G_{N}L}\times  \\
&& \int_{0}^{r_{0\max }}dr\frac{l\sqrt{r}}{%
\sqrt{(r-2m)l^{2}+r^{3}}}.
\end{eqnarray*}

As the next step, we have to integrate the above relation with
respect to $r$. In this case the integral cannot be calculated
analytically, and therefore, we should calculate it numerically
which the results are shown in Fig. \ref{dcvsr}.
\begin{figure}[tbp]
\hspace{0.4cm} \centering
\includegraphics[width=14cm]{deltacvslowrot.eps}
\caption{Complexity of formation in the CV conjecture versus $\frac{r_-}{r_+}
$.}
\label{dcvsr}
\end{figure}

According to Figs. \ref{dcasr} and \ref{dcvsr}, it is obvious that
complexity of formation in two mentioned conjectures has same
behavior and it is bounded without divergency. In the limit
$\frac{r_-}{r_+}\rightarrow 0$ the value of the complexity of
formation (calculated via CV conjecture) is reduced and tends to
$4\times 10^{21}$ while at the large value of $\frac{r_-}{r_+}\sim
0.25$, it's value approximately tends to a constant. Since
increasing the electric charge leads decreasing (increasing) inner
horizon (event horizon), it is easy to find that $\frac{r_-}{r_+}$
is an increasing function of electric charge, $q$, and therefore,
the complexity of formation in both conjectures increases as one
increases the electric charge.

\section{Conclusion \label{sec5}}

In this paper, we studied the complexity of regular static and
rotating black holes in the presence of nonlinear electrodynamics
. For the case of the static black
hole, by looking at the action growth rates of the Wheeler-De Witt
patch, we showed the Lloyd bound is satisfied
and its form remains unaltered
. Then, we have obtained the rate of complexity and complexity of
formation by using both CA and CV conjectures. From the comparison
of the result, we concluded that the two approaches to the
complexity are consistent. In the case of slowly rotating one and
with assumption $\frac{q}{M}\ll 1$, we obtained the rate of
complexity proportional to the volume of the black hole (in the
limit $\alpha\ll 1$ and $\beta \ll1$). So, one can conclude that
the rate of complexity for both static and rotating black holes is
proportional to thermodynamical volume instead of entropy.

The complexity of formation is numerically obtained by using both
conjectures. Its behavior is reflected in Figs. \ref{dcasr} and
\ref{dcvsr} which shows that for both conjectures, the complexity
of formation has the same behavior and is bounded.

In the appendix, we investigated two other regular cases and
obtained the rate of complexity and complexity of formation by
using of both conjectures for them. According to Figs. \ref{dcast}
and \ref{dcvst}, it is obvious that for each case, the complexity
of formation obtained from two approaches are relatively
consistent and on the other hand behavior of complexity of
formation for these two regular cases have interesting
similarities with each other.

\textbf{Acknowledgements}\\
The authors thank Shiraz University Research Council.

\section{Appendix}
For the sake of completeness, here, we investigate complexity
conjecture of two more examples of static regular black holes in
$\mathcal{P}$.

\subsection{Second Case:}

Here, we consider another structure function inspired by the
log-logistic distribution as the second class of solution which is
given by \cite{Balart:2014cga,Dymnikova:2004zc}
\begin{equation}
\mathcal{H}=\frac{\mathcal{P}}{(1+\gamma_{0} \sqrt{-\mathcal{P}})^{2}}+\frac{6}{%
l^{2}},  \label{H2}
\end{equation}%
where $\gamma_{0} =\frac{r_{0}^{2}}{\sqrt{2\gamma ^{2}}}$, and
$r_{0}$ and $\gamma
$ are two parameters. In the weak field limit ($\mathcal{P}\ll 1$), Eq. (\ref%
{H2}) reduces to
\begin{equation}
\mathcal{H}\approx \mathcal{P}+\frac{6}{l^{2}}+\mathcal{O}(\mathcal{P}^{%
\frac{3}{2}}).
\end{equation}

Taking such a structure function into account with
$\mathcal{H}=2\rho (r)$ and Eqs. (\ref{eqn:p}) and
(\ref{eqmetric1}), we can, directly, achieve the following
$arctan$ form of the solution
\cite{Dymnikova:2004zc,Sajadi:2017glu}
\begin{equation}
f(r)=1-\frac{4\chi }{\pi r}\left[ arctan(x)-\frac{x}{1+x^{2}}\right] +%
\frac{r^{2}}{l^{2}},\;\;\;r_{0}=\frac{\pi \gamma ^{2}}{8\chi },\;\;\;x=%
\frac{r}{r_{0}},
\end{equation}
with the following asymptotical behavior for large $r$
\begin{equation}
-g_{tt}\approx 1+\frac{r^{2}}{l^{2}}-\frac{2\chi }{r}+\frac{\gamma ^{2}}{%
r^{2}}+\mathcal{O}\left( \frac{1}{r^{4}}\right).
\end{equation}%
As a result, the obtained solution behaves like
Reissner-Nordstr\"{o}m-AdS, asymptotically, when we adjust $\chi $
and $\gamma $ as the mass and electric charge of the system,
respectively.

Such as before, we can obtain the following thermodynamic
quantities for the recent $arctan$ solution
\begin{eqnarray}
P &=&\frac{3}{8\pi l^{2}}, \\
V &=&\frac{4\pi r_{+}^{3}}{3}, \\
S &=&\pi r_{+}^{2}, \\
T &=&\frac{m\left( (1+x_{+}^{2})^{2}\arctan
(x_{+})-x_{+}-3x_{+}^{3}\right)
}{\pi ^{2}r_{+}^{2}(1+x_{+}^{2})^{2}}+\frac{r_{+}}{2\pi l^{2}}, \\
\Phi &=&\int_{r_{+}}^{\infty }Edr=\frac{3m}{\pi q}\left( \frac{\pi }{2}%
-\arctan (x_{+})+\frac{(1+\frac{5}{3}x_{+}^{2})x_{+}}{(1+x_{+}^{2})^{2}}%
\right) ,
\end{eqnarray}%
where $x_{+}=x|_{r=r_{+}}$. Having the mentioned thermodynamic
quantities, one can directly examine the first law of black hole
thermodynamics in the enthalpy representation
\begin{equation}
dH=TdS+VdP+\Phi dq,
\end{equation}
where
\begin{equation}
H=m=\frac{\pi r_{+}(1+\frac{r_{+}^{2}}{l^{2}})}{4\left( \arctan (x_{+})-%
\frac{x_{+}}{1+x_{+}^{2}}\right) }.
\end{equation}

Moreover, the Smarr relation is given by
\begin{equation}
\frac{H}{2}+PV-TS-\frac{q\Phi }{2}+\frac{1}{4}\int wdv=0,
\end{equation}%
where
\begin{equation}
\int wdv=\frac{1}{2}\int_{r_{+}}^{\infty }T_{\mu }^{\mu }\ r^{2}dr=m-\frac{%
2mx_{+}(x_{+}^{2}-1)}{\pi (1+x_{+}^{2})^{2}}-\frac{2m}{\pi
}\arctan (x_{+}).
\end{equation}

Now, we calculate joint terms according to joint contribution
which is described in the first class. After simplification, we
can write
\begin{eqnarray}
{S_{B^{\prime }B}} &=&\frac{1}{4}\delta t\ \left[ {\frac{{4m}}{{\pi {r^{2}}}}%
\left( {\arctan (x)-\frac{{x}}{1+x^{2}}}\right) -\frac{{4m}}{{\pi r}}\left( {%
\frac{{2x{^{2}}}}{{{{{r_{0}}(1+x^{2})}^{2}}}}}\right) +\frac{{2r}}{{{l^{2}}}}%
}\right] _{r_B},  \notag \\
{S_{C^{\prime }C}} &=&\frac{1}{4}\delta t\left[ {\frac{{4m}}{{\pi {r^{2}}}}%
\left( {\arctan (x)-\frac{{x}}{1+x^{2}}}\right) -\frac{{4m}}{{\pi r}}\left( {%
\frac{{2x{^{2}}}}{{{{{r_{0}}(1+x^{2})}^{2}}}}}\right) +\frac{{2r}}{{{l^{2}}}}%
}\right] _{r_C}.
\end{eqnarray}

Considering the metric function obtained here, it is easy to show
that the Ricci scalar is calculated as
\begin{equation}
\mathcal{R}=\frac{4q^{2}}{r_{0}^{4}(1+x^{2})^{3}}-\frac{12}{l^{2}},
\end{equation}%
and also, the Lagrangian of nonlinear electrodynamics can be
calculated at the event horizon with the following explicit form
\begin{equation}
\mathcal{L}=\frac{16m(1-x_{+}^{2})}{\pi r_{0}^{3}(1+x_{+}^{2})^{3}}-\frac{6}{%
l^{2}}.
\end{equation}

Consequently, the growth rate of bulk action and surface term are,
respectively,
\begin{equation}
\frac{d\mathcal{A}_{bulk}}{dt}=\left[ \frac{2m}{\pi }\left( \arctan (x)-{%
\frac{{x}}{1+x^{2}}}\right) \right] _{r_{-}}^{r_{+}},
\end{equation}
and
\begin{equation}
\frac{d\mathcal{A}_{boundray}}{dt}=\left[ r-\frac{3m}{\pi }\arctan (x)+\frac{%
mx(3+x^{2})}{\pi (1+x^{2})^{2}}+\frac{3r^{3}}{2l^{2}}\right]
_{r_{-}}^{r_{+}}.
\end{equation}

Finally, we find that the total growth rate of the action for such
black hole configuration within WDW patch at late time
approximation can be
collected as%
\begin{equation}
\frac{d\mathcal{A}}{dt}=\left[ r-\frac{m}{\pi }\arctan (x)-\frac{mx(1-x^{2})%
}{\pi (1+x^{2})^{2}}+\frac{3r^{3}}{2l^{2}}\right]
_{r_{-}}^{r_{+}}.
\end{equation}%
Thus, by using the above calculated thermodynamic quantities, it
is easy to
show that the Lloyd bound is satisfied. As an example, for the values $%
m=0.8,q=0.1,l=1$ this bound is about $1.6872$, and the differences
between left and right (right mines left of Eq. (\ref{rate})) is
about $0.0666$.

\subsection{Third Case:}

As the third case, we consider the following structure function
which is inspired by the Fermi-Dirac distribution
function\cite{AyonBeato:1999rg,Balart:2014cga}
\begin{equation}
\mathcal{H}=\frac{\mathcal{P}}{\cosh ^{2}\left( \frac{\gamma ^{\frac{3}{2}}(-%
\mathcal{P})^{\frac{1}{4}}}{2^{\frac{5}{4}}\chi }\right)
}+\frac{6}{l^{2}}, \label{H3}
\end{equation}%
with the following weak field limit ($\mathcal{P}\ll 1$)%
\begin{equation}
\mathcal{H}\approx \mathcal{P}+\frac{6}{l^{2}}+\mathcal{O}(\mathcal{P})^{%
\frac{3}{2}}.
\end{equation}

Considering the spherically symmetric spacetime with previous relations for $%
\mathcal{H}$, the third class of solution is%
\begin{equation}
f(r)=1-\frac{2\chi }{r}\left[ 1-\tanh \left( \frac{\gamma ^{2}}{2\chi r}%
\right) \right] +\frac{r^{2}}{l^{2}}. \label{f3}
\end{equation}%
According to the series expansion of the metric function for large
$r$, one can find this solution behaves like the
Reissner-Nordstr\"{o}m-AdS, asymptotically, as
\begin{equation}
-g_{tt}\approx 1+\frac{r^{2}}{l^{2}}-\frac{2\chi }{r}+\frac{\gamma ^{2}}{%
r^{2}}+\mathcal{O}\left( \frac{1}{r^{4}}\right),
\end{equation}%
whereas $\chi $ and $\gamma $ will be associated with the mass and
electric charge of the system, respectively.

Using the previous approach in the extended phase space, the
following thermodynamical quantities can be obtained, directly,
\begin{eqnarray}
P &=&\frac{3}{8\pi l^{2}}, \\
V &=&\frac{4\pi r_{+}^{3}}{3}, \\
S &=&\pi r_{+}^{2}, \\
T &=&\frac{m}{2\pi r_{+}^{2}}\left[ 1-\tanh \left( \frac{q^{2}}{2mr_{+}}%
\right) \right] -\frac{q^{2}}{4\pi r_{+}^{3}}\left[ 1-\tanh ^{2}\left( \frac{%
q^{2}}{2mr_{+}}\right) \right] +\frac{r_{+}}{2\pi l^{2}}, \\
\Phi &=&\int_{r_{+}}^{\infty }Edr=\frac{3m}{2q}-\frac{3mr_{+}\left( 1+e^{%
\frac{q^{2}}{mr_{+}}}\right)
-q^{2}e^{\frac{q^{2}}{mr_{+}}}}{qr_{+}\left(
1+e^{\frac{q^{2}}{mr_{+}}}\right) ^{2}}.
\end{eqnarray}%
It is evident that these thermodynamic quantities satisfy the
first law of black hole thermodynamics as
\begin{equation}
dH=TdS+VdP+\Phi dq,
\end{equation}%
where in the enthalpy representation, we have%
\begin{equation}
H=m.
\end{equation}%
In addition, the related Smarr formula is given by%
\begin{equation}
\frac{H}{2}+PV-TS-\frac{q\Phi }{2}+\frac{1}{4}\int wdv=0,
\end{equation}%
where
\begin{equation}
\int wdv=\frac{1}{2}\int_{r_{+}}^{\infty }T_{\mu }^{\mu }r^{2}dr=m-\frac{%
2mr_{+}\left( 1+e^{\frac{q^{2}}{mr_{+}}}\right) +2q^{2}e^{\frac{q^{2}}{mr_{+}%
}}}{r_{+}\left( 1+e^{\frac{q^{2}}{mr_{+}}}\right) ^{2}}.
\end{equation}%
According to the previous section, joint contribution can be
calculated as
below%
\begin{equation}
S_{B^{\prime }B}=\frac{1}{4}\delta t\left( {\frac{{2m}}{{r^{2}}}}\left[ {{%
1-\tanh }\left( {\frac{{q^{2}}}{{2mr}}}\right) }\right] {-\frac{{q^{2}}}{{%
r^{3}}}}\left[ {{1-{{{\tanh }^{2}\left( {\frac{{q^{2}}}{{2mr}}}\right) }}}}%
\right] {+\frac{{2r}}{{l^{2}}}}\right) _{r_B},
\end{equation}%
\begin{equation}
{S_{C^{\prime }C}}=\frac{1}{4}\delta t\left( {\frac{{2m}}{{r^{2}}}}\left[ {{%
1-\tanh }\left( {\frac{{q^{2}}}{{2mr}}}\right) }\right] {-\frac{{q^{2}}}{{%
r^{3}}}}\left[ {{1-{{{\tanh }^{2}\left( {\frac{{q^{2}}}{{2mr}}}\right) }}}}%
\right] {+\frac{{2r}}{{l^{2}}}}\right) _{r_C}.
\end{equation}%
Moreover, the Ricci scalar for this solution is simplified as%
\begin{equation}
\mathcal{R}=\frac{q^{4}\sinh \left( \frac{q^{2}}{2mr}\right)
}{mr^{5}\cosh ^{3}\left( \frac{q^{2}}{2mr}\right)
}-\frac{12}{l^{2}},
\end{equation}%
and the nonlinear Lagrangian is as follows
\begin{equation}
\mathcal{L}=\frac{q^{2}\left( q^{2}\tanh \left(
\frac{q^{2}}{2mr}\right)
-2mr\right) }{mr^{5}\cosh ^{2}\left( \frac{q^{2}}{2mr}\right) }-\frac{6}{%
l^{2}}.
\end{equation}%
It is notable that the growth rate of the bulk action and its
related
surface term can be written as%
\begin{eqnarray}
\frac{d\mathcal{A}_{bulk}}{dt} &=&\left[ \frac{2m}{1+e^{\left( \frac{q^{2}}{%
mr}\right) }}\right] _{r_{-}}^{r_{+}}, \\
\frac{d\mathcal{A}_{boundray}}{dt} &=&\left[ r-(\frac{3m}{2}+\frac{q^{2}}{4r}%
)+\frac{3m}{2}\tanh \left( \frac{q^{2}}{2mr}\right)
+\frac{q^{2}}{4r}\tanh ^{2}\left( \frac{q^{2}}{2mr}\right)
+\frac{3r^{3}}{2l^{2}}\right] _{r_{-}}^{r_{+}}.
\end{eqnarray}%
As a result, the total growth rate of action for such a black hole
configuration within WDW patch at late time approximation is%
\begin{equation}
\frac{d\mathcal{A}}{dt}=\left(
r+\frac{3r^{3}}{2l^{2}}-\frac{m}{2}\left[ 1-\tanh \left(
\frac{q^{2}}{2mr}\right) \right] -\frac{q^{2}}{4r}\left[ 1-\tanh
^{2}\left( \frac{q^{2}}{2mr}\right) \right] \right)
_{r_{-}}^{r_{+}}.
\end{equation}

For instance, for the values $m=0.8,q=0.1,l=1$ this bound is about
$8.5783$,
and the differences between left and right (right mines left of Eq. (\ref%
{rate})) is $7.0399$.

\subsection{Complexity of formation}

Similar to the complexity of formation of the first case we would
like to study the complexity of formation of second and third
cases. In this way, $r^{\ast }(r)$ and $r_{meet}$ of the second
and the third solutions can be obtained as follows:
\begin{align}
r^{\ast }(r)=& \dfrac{\pi l^{2}r_{+}^{2}\ln \left( \dfrac{|r-r_{+}|}{r+r_{+}}%
\right) }{4ml^{2}\arctan (\dfrac{r_{+}}{r_{0}})+\dfrac{2\pi r_{+}^{5}}{%
r_{+}^{2}+r_{0}^{2}}+\dfrac{2r_{0}r_{+}(\pi
r_{0}^{3}r_{+}^{2}-6ml^{2}r_{+}^{2}-4ml^{2}r_{0}^{2})}{%
(r_{+}^{2}+r_{0}^{2})^{2}}}+  \notag \\
& \dfrac{\pi l^{2}r_{-}^{2}\ln \left( \dfrac{|r-r_{-}|}{r+r_{-}}\right) }{%
4ml^{2}\arctan (\dfrac{r_{-}}{r_{0}})+\dfrac{2\pi r_{-}^{5}}{%
r_{-}^{2}+r_{0}^{2}}+\dfrac{2r_{0}r_{+}(\pi
r_{0}^{3}r_{-}^{2}-6ml^{2}r_{-}^{2}-4ml^{2}r_{0}^{2})}{%
(r_{-}^{2}+r_{0}^{2})^{2}}}\hspace{2.7cm}second\;case  \label{eqtor2} \\
r^{\ast }(r)=& \dfrac{l^{2}r_{+}^{3}\ln \left( \dfrac{|r-r_{+}|}{r+r_{+}}%
\right) }{2mr_{+}l^{2}\left( 1-\tanh \left(
\dfrac{q^{2}}{2mr_{+}}\right) \right) -q^{2}l^{2}\left( 1-\tanh
^{2}\left( \dfrac{q^{2}}{2mr_{+}}\right)
\right) +2r_{+}^{4}}+  \notag \\
& \dfrac{l^{2}r_{-}^{3}\ln \left( \dfrac{|r-r_{-}|}{r+r_{-}}\right) }{%
2mr_{-}l^{2}\left( 1-\tanh \left( \dfrac{q^{2}}{2mr_{-}}\right)
\right) -q^{2}l^{2}\left( 1-\tanh ^{2}\left(
\dfrac{q^{2}}{2mr_{-}}\right) \right)
+2r_{-}^{4}}\hspace{2.5cm}third\;case  \label{eqtor3}
\end{align}
and
\begin{align}
r_{meet}& \approx 0.068r_{+}^{2}-1.23r_{+}+1.1433,\hspace{1.5cm}%
second\;\;case \\
r_{meet}& \approx -0.35387r_{+}^{2}-0.64296r_{+}+0.9821.\hspace{0.5cm}%
third\;\;case
\end{align}
Accordingly, the bulk action of the black holes are
\begin{align}
A_{bulk,BH}=& \dfrac{16Mr_{0}}{\pi
G}\int_{r_{meet}}^{r_{max}}dr\left(
\dfrac{1}{(r^{2}+r_{0}^{2})^{2}}\right) (v_{\infty }-r^{\ast }(r))\hspace{1.5cm}second\;case \\
A_{bulk,BH}=& \dfrac{2q^{2}}{G}\int_{r_{meet}}^{r_{max}}dr\left( \dfrac{%
1-\tanh ^{2}\left( \dfrac{q^{2}}{2Mr}\right) }{r^{4}}\right)
(v_{\infty 5}-r^{\ast }(r))\hspace{0.5cm}third\;case
\end{align}
here $r_{max}$ is the same as the first case. The relation for
$A_{joint}$ for large black holes are written in the following
form:
\begin{align}
\mathcal{A}_{joint}\approx & 0.0185\ln \left(
\dfrac{0.068r_{+}^{2}}{l}\right)
r_{+}^{4}+\mathcal{O}(r_{+}^{3}).\;\;\;second\;case \\
\mathcal{A}_{joint}\approx & 0.1252\ln \left( \dfrac{-0.35387r_{+}^{2}}{l}%
\right) r_{+}^{4}+\mathcal{O}(r_{+}^{3}).\;\;\;third\;case
\end{align}

\begin{center}
\begin{figure}[tbp]
\centering
\includegraphics[width=12cm] {ajoint.eps}
\caption{Plot of $A_{joint}$ for second and third cases which are
shown in black and dotted line respectively.} \label{jointt}
\end{figure}
\end{center}

The figure of $A_{joint}$ for second and third cases are plotted
in Fig. \ref{jointt}. According to Fig. \ref{jointt}, it is
obvious that for small black hole the contribution of the joints
to the action is large for large black holes, this contribution
becomes smaller and goes to zero.

Finally according to above relations and the results related to
the $AdS$ spacetime (\ref{eqads0})-(\ref{eqads2}), $\Delta C_{A} $
from (\ref{eqdeltaca}) can be obtained. The plot of $\Delta C_{A}$
for second and third cases are shown in Fig. \ref{dcast}.

Figure \ref{dcast} describes that complexity of formation for
second and third cases increases for large black holes. Finally
$\Delta C_{V}$ according to Eq. (\ref{eqdeltacvform}) for these
two cases can be obtained.

Figure \ref{dcvst} shows that $\Delta C_{V}$ for both cases has
the same behavior as with $\Delta C_{A}$, increasing with radius
of the black hole. To summarize the behavior of $\Delta C_{A}$ and
$\Delta C_{V}$ for three cases versus $r_{+}$ are plotted in Fig.
\ref{three}.

\begin{center}
\begin{figure}[tbp]
\centering
\includegraphics[width=12cm] {deltacvast.eps}
\caption{Plots of $\Delta C_{A}$ for second and third cases are
shown in the right and left panels respectively.} \label{dcast}
\end{figure}
\end{center}
\begin{center}
\begin{figure}[tbp]
\centering
\includegraphics[width=12cm] {dcvast2.eps}
\caption{Plots of $\Delta C_{V}$ for second and third cases are
shown in the right and left panels respectively.} \label{dcvst}
\end{figure}
\end{center}

\begin{center}
\begin{figure}[tbp]
\centering
\includegraphics[width=16cm] {threecases.eps}
\caption{Plot of Complexity of formation for the different
geometries. For first, second and third cases respectively are
shown from up to down, and left panels describe $\Delta C_{A}$ and
right ones indicate $\Delta C_{V} $.} \label{three}
\end{figure}
\end{center}

redpajama_set_name: RedPajamaArXiv


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Silk was made by Yuri Vishnevsky, with music and sound crafted by Mat Jarvis.
Art shared with Silk is licensed under Creative Commons.
If you enjoyed Silk, please let me know!
Note — Silk has sound. Muted. Mute?


Will not rust, rot or corrode and ships for significantly less than traditional tanks.
Made in the US with 22 oz. extruded coated thermoplastic material, reinforced with polyester and nylon scrims.
RF (radio frequency) welding in the fabrication process ensures fail-save welds and a contaminate-free seal.


text:
Free schools: freedom for whom?
Andrew Dolan looks at the problems posed by free schools
April 10, 2014 · 6 min read
Photo: Policy Exchange/Flickr
Funded by the state yet set up and controlled by external parties, free schools have since their introduction in 2010 divided opinion and in some cases even communities. Their most vocal detractor, the National Union of Teachers (NUT), claim free schools acquire a disproportionate share of funding, undermine teachers' professional status, pay and conditions and threaten democratic local accountability of schools. For their supporters, no one more so than the Secretary of State for Education, Michael Gove, free schools are a solution to educational underachievement, providing autonomy, flexibility and an unprecedented opportunity for parents to influence the learning experience of their children. Indeed, 'power to the parents' was the dominant discourse accompanying the policy announcement. Yet, as a recent report by the National Audit Office revealed, only 29 per cent of free schools currently open are ran by parent and community groups and just 18 per cent by teacher-led groups. Such statistics, striking as they are, beg the question, who else is running free schools, how and for what purpose?
The second largest group running free schools are faith organisations (26 per cent), yet there are also a significant number ran by not-for-profit companies such as E-ACT, which operates 24 academies, including free schools. The involvement of such companies has raised the spectre of the privatisation of compulsory and further education in the UK. According to the NUT, companies and academy chains bidding to run free schools are lobbying the Government to allow them to do so on a for-profit basis. Although for this to happen a change in law is required, leaked details of discussions held in the Department for Education advocating profit-making in free schools and academies—and Gove's failure to rule this out—have exacerbated concerns.
Yet even without a change in law the so-called 'free schools revolution' offers a number of avenues for financial gain. For senior employees of the charities running free schools and academies the rewards can be considerable. As E-ACT's former director general, Sir Bruce Liddington earned almost £300,000 in 2010/11, making him one of the highest paid people in education. Similarly, when one considers that for heads of free schools pay is not determined by a national framework and is typically higher than in schools under local council control, it is perhaps unsurprising that so many applications have been led by would-be head teachers—almost 25 per cent in 2011, according to the Institute of Education—many of whom have no teaching experience.
More importantly, it is already legal for charitable trusts to outsource the day-to-day running of the school to for-profit Educational Management Organisations (EMO's). Opened in 2012, IES Breckland in Sussex is the first free school to operate on this basis after its parent-led governing trust signed a ten-year contract worth £21 million with the EMO Internationella Engleska Skolan (IES) to provide for the entire running of the school, including teachers' salaries. IES has since admitted that under its management the education provision at IES Breckland has been sub-standard, strengthening claims that education provision and profit are fundamentally incompatible.
Even when profit is not an immediate possibility, however, there is enough evidence to suggest that free schools are already colonised by the logic of the private sector, primarily a fixation with cost reduction and efficiency maximisation. The contract dispute at STEM 6 College in Islington provides a case in point. Established as a free school in 2013 by the Skills and Development Agency—a not-for-profit company with a for-profit subsidiary—STEM 6 recently offered teachers what amounted to zero-hour contracts. The contracts stipulated that 'the school reserves the right to temporarily lay [staff] off from work without normal contractual pay or to reduce . . . normal working hours and reduce . . . pay proportionately'. Staff were also threatened with 'legal consequences' should they not sign the new contracts and STEM 6 refused to grant union recognition.
Following a two-day strike by teachers at STEM 6 in early February—the first strike at a free school—and the threat of a second, management eventually agreed to grant union recognition and enter into contractual negotiations. According to its assistant secretary, Ken Muller, Islington NUT is hopeful that a suitable conclusion will soon be reached and further strike action avoided.
Although the partial victory of the STEM 6 teachers sets an encouraging precedent as regards union organising and zero-hour contracts in free schools, the attempts of management to drive down costs by attacking labour conditions and their initial refusal to grant union recognition suggest that far from simply providing autonomy for parents, free schools can also provide carte blanche for employers to exploit staff. As with the private sector, the absence of regulation, in this instance a nationally negotiated pay scale, means very different things for those at the top than it does for those at the bottom, for whom precarisation is a very real possibility and in some cases already a reality.
Alongside the STEM 6 dispute, revelations of serious financial mismanagement at Barnfield Federation, which runs a number of academies including one free school, the stripping of ten academies from E-ACT for poor performance, and an increasing number of failing free schools point to an educational sub-sector facing serious problems, not least of all the prospect of future privatisation.
What happens if a university fails?
David Ridley reflects on the Augar Review
Governments fixate on STEM education – so why won't they listen to scientists on climate change?
Governments could do well to learn from school students, writes 17-year-old Climate Striker Cate Davies
Let's harness the power of political education
We need political education to build a confident, fighting movement writes Isaac Kneebone-Hopkins, an organiser for Bristol Transformed.
The rise of the student worker
The student population today is unrecognisable from that of a generation or more ago, writes Matt Myers. And it is central to any socialist project for the future.
What can academics learn from the Lucas Plan?
With the right organising and the right plan, UCU workers can transform universities from within. By David Ridley
'Could do better': Racism in schools
Remi Joseph-Salisbury writes that institutional racism is not just about individual teachers, but a lack of clear school-wide or nationwide policy.

redpajama_set_name: RedPajamaCommonCrawl


text:
SET Edition: U.S. Sports l Global Sports
U.S SPORTS
Minnesota Wild Place Jason Zucker On Injured Reserve With Concussion
By Matt Jussim
Feb 25, 2016 07:40 AM EST
NHL, NHL News, Jason Zucker, Minnesota Wild, Minnesota Wild News, NHL Injury Update, Chicago Blackhawks, Stadium Series
Jason Zucker #16 of the Minnesota Wild is helped off of the ice after taking a hard hit during the 2016 Coors Light Stadium Series game against the Chicago Blackhawks at TCF Bank Stadium on February 21, 2016 in Minneapolis, Minnesota."
The Minnesota Wild have placed Jason Zucker on injured reserve after the forward sustained a concussion over the weekend.
Zucker was placed on the injured list on Wednesday after being injured on Sunday in the win against the Chicago Blackhawks, according to Chad Graff at the St. Paul Pioneer Press. Zucker took an elbow high on his head from Blackhawks defenseman Michal Rozsival. Zucker hit his head on the ice after falling down in the second period. Rozsival was given a penalty for the play and was ejected from the game.
At first, it appeared that Zucker may have avoided a concussion. Wild coach John Torchetti said that Zucker did not have a concussion, but the diagnosis changed on Wednesday, leading to the IR move. Zucker did not play on Tuesday against the Islanders after sustaining the injury in the win against the Blackhawks on Sunday.
The Wild made a corresponding move by adding Jordan Schroeder to the roster from their AHL affiliate in Iowa, according to Fox Sports. The Wild have two road games coming up against the Flyers no Thursday and the Capitals on Friday. Zucker is not expected to travel with the team for either of those games, and he could be out for longer than that, the Minneapolis Star Tribune reports.
Zucker was replaced in the lineup on Tuesday against the Islanders with Chris Porter. Zucker has 12 goals and 10 assists in 58 games this season for Minnesota. Schroeder now will get the chance to play, and he previously has appeared in 13 games for Minnesota this season. The next chance that Zucker will have to play for Minnesota after the two-game road trip is on Sunday against the Florida Panthers at home.
The Wild are currently trying to stay in the playoff race in the Western Conference. The team has a 27-23-10 record this season with 64 points, and they are two points back of the Colorado Avalanche for the second wild card spot. The Nashville Predators currently are in the first wild card spot with 69 points.
• Washington Capitals Trade For Mike Weber, Key Move For Playoff Run?
• NHL Rookie Of The Year Race Between Jack Eichel And Artemi Panarin?
• Shea Weber Sits Out For Nashville Predators, Returns Against Chicago Blackhawks?
Get the Most Popular Stories in a Weekly Newsletter
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Joey Bosa Dallas Cowboys' Top NFL Draft Target If Available At Fourth Pick?
Bears Trying To Sign Alshon Jeffery To Long-Term Contract
Peyton Manning Retires From NFL After Super Bowl Win
MLB News and Notes
Jacksonville Jaguars Have Most Cap Space In NFL; What's The Plan?
Colin Kaepernick 'More Than Open' To 49ers Return
Steelers couldn't care less that they're the nastiest group in the NFL playoffs Jan 09, 2017 11:17 AM EST
Who Will Win Super Bowl in 2017, Patriots or Cowboys? Jan 03, 2017 11:01 AM EST
Will Sean McVay become the youngest coach in NFL history? Jan 02, 2017 09:22 PM EST
Top 10 Best NFL players who should've made Pro Bowl Dec 21, 2016 11:37 AM EST
Atlanta Falcons Looking To Sign Sanu As A Replacement For Roddy White Mar 01, 2016 03:15 PM EST
LeBron James vs Michael Jordan: Who is the Greatest of All Time?
NBA's Eastern Conference teams concered whether they will ever be able to beat LeBron James
Does LeBron James play better when Kyrie Irving and Kevin Love are on the bench?
Will Russell Westbrook Become the MVP This Year?
Michael Jordan Not Too Impressed with Warrior's 73 Wins?
Does Lebron James Want Carmelo Over Kevin Love?
@sportswr
© Copyright 2019 Sports World Report. All Rights Reserved.


My 2001 911 Cab top, intermittently will not open. When I push the top buttton, AT TIMES, it won't open and the light on the dash flashes. Sometimes the top operates normallly. Does anyone know why this occurs?


Did everyone have a great holiday weekend? Ours was totally relaxed and just right. Lots of vegetables and watermelon in the mix, and just a smidgin' of gettin'er done in terms of outside work. We generally have to stay home on "fireworks holiday" nights because our dog is terrified of them, so this comic rings especially true. I know that this is an issue for a lot of families, and seeing "Lost Dog" signs in my neighbourhood after a holiday weekend always gives me a heavy heart. I know for sure that my girl just isn't herself when she hears those booms, so we do everything we can to make her comfortable. I've kept my eye on those thundershirts as an option for a while, too.
Aside from the regular snuggles, treats, toys, antlers, playtime, and walks (our girl is spoooooiled–I think a few of you can relate haha), I like to give her something with peanut butter when she's been really good or when I know she's gonna have a tough day. As soon as I slide the lid off of a jar of PB (which is often cuz PB is life), she comes prancing into the kitchen, making her silly little grunt noises the whole way.
I really love this book by Tuulia and Virpi though (amazing Instagram page with great quotes about ice cream if you're interested in seeing more). As if a book of all vegan and healthy ice cream wasn't enough temptation for me, the photos are so enticing. They're all dreamy light, just-right ice cream drips, and perfectly-imperfect styling to a tee. Some other recipes from the book that I've been eyeing up: the choco-vanilla bomb, the $5 milkshake (anything with a Pulp Fiction reference is thumbs up), strawberry basil creamsicles, white chocolate sauce, raw chocolate-covered cashew ice cream, and the dreamy blackberry ice cream cake.
NOTES: I added a tiny squeeze of fresh lemon juice to both the ice cream and the caramel sauce. Just personal preference really. In the book, the recommendation for the cake crumble is ground cardamom, but I went full-on vanilla with ground vanilla beans AND extract. Also note that you will have to refrigerate a can of coconut milk overnight for this recipe.
Make the crumble: Combine the nuts, dates, ground vanilla, vanilla extract, sea salt, and maple syrup in a blender. Blend on high until you have a moist, crumbly mixture. Set aside.
Make the caramel: In a small bowl, vigorously whisk together the peanut butter, maple syrup, plant-based milk, sea salt, and lemon juice. Once you have a smooth sauce, set aside.
Make the ice cream: Rinse out the blender pitcher. Open the can of coconut milk and scoop the top layer of cream, being careful to avoid the water at the bottom of the can. Transfer the coconut cream to the blender. Add the frozen bananas, vanilla, maple syrup, sea salt, and lemon juice to the blender as well. Blend the mixture on high, using a tamper to push the mixture down into the blades. Once you have a smooth mixture, stop the motor.
At this point, I had to freeze my ice cream for a good 45 minutes, stirring it up every 15 minutes. Once you have that slightly firm/soft-serve like texture, serve the ice cream with the peanut butter caramel and vanilla cake crumble.
I HAVE to get my hands on this book asap. Have seen some many incredible recipes from it around the blogosphere already, and it really feels like a must-have this summer.
So nice to have a little pal in the kitchen licking the spoons as well ;-) <3 I have to show these photos to my dog crazy girlfriend, she's gonna die :'D (and make us get a dog asap).
Sounds like you're having a super relaxed summer so far! I hope it continues in that vibe.
HOLY MOLY!! Is there anything better than being able to eat ice cream and have it actually be GOOD for you!
Love that this recipe doesn't require a ice cream maker (I think buying one would be like opening pandoras box), and it's pretty much all the ingredients I can put into a smoothie, which means that it's healthy. I am sure of it! Thank you for sharing… making tonight.
Lovely as always, Laura! I've been seeing the new-ish N'Ice Cream book cropping up around the interwebs too … it's great to get your take on it! Thank you.
This literally made me drool.


From the 12th to the 14th November the Internet Governance Forum (IGF) annual meeting took place in Paris. On the theme of Internet of Trust, this meeting gathered various attendees from different nationalities and with different backgrounds. It was an opportunity for France to present its commitment and vision for a secured and stable cyberspace. The Agence Nationale de la Sécurité des Systèmes d'Information (ANSSI) shares and develops this vision in its cyberdefense strategy, on both national and international side.
In cyberspace, some actions, even if restricted, can cause damages on a large scale. Facing growing threats, the cooperation between private, public sector and the civil society must be strenghtened.
To this aim, national and international regulation must define the responsibility of each actors in order to ensure the stability of cyberspace.
President Mr. Emmanuel MACRON and Minister for Europe and Foreign Affairs Mr. Jean-Yves LE DRIAN spoke aside from the IGF meeting to present the Paris Call for Cyberspace trust and security.
The Agence Nationale de la Sécurité des Systèmes d'Information (ANSSI) promotes the development of an open, safe and stable cyberspace. On the international scene, ANSSI is actively participating both at political and normative level.
"Hack back", "gray zone", "active cyber defense", "passive cyber defense", etc. The multiplicity of this terminology, but also the ambiguity of these concepts prevent most actors from clearly understanding the underlying challenges they trigger. It is paramount for all stakeholders, in order to exchange properly, to discuss with a shared and immediately comprehensible terms. This could be achieved through a categorization of the actions discussed and a refocusing of the debate on the risk they imply.
In order for Cyberspace to remain an area of opportunity for States, citizens and companies, the regulations should evolve via standards and rules internationally shared, based on international laws, in accordance with the principles of freedom and human rights.
The Paris call for cyberspace trust and security, launched by President on Monday, November 12th 2018, during the Internet Governance Forum meets these concerns.
This call reinforces the international cooperation and invites states, private sector and civil society to commit for a more stable cyberspace. France will develop this ambition through OECD's Global Forum, which will take place on 13th and 14th December 2018.
For many years, the stability of cyberspace has become a priority for France who try to share this preoccupation on an international level.
In its Strategic review for cyber defense, published on 12 February 2018, France makes a cyber threats overview, expresses many proposals to improve the cyber defense and cyber security of the French society. Particularly, the review puts forward the need to prevent private actors from conducting offensive cyber operations, and more specifically as a retaliation – from hacking back.
In April 2017, the Secretary General for Defense and National Security and The Agence Nationale de la Sécurité des Systèmes d'Information (ANSSI) organized a meeting dedicated to build peace and international security of the digital society (« Construire la paix et la sécurité internationales de la société numérique »). This meeting gathered many actors with various backgrounds in order to share academic thoughts for the preservation of peace and international security in the cyberspace.
With the Paris Digital Week, France pursues its path and wishes to generate adequate conditions for dialogue and collaboration among all interested parties involved in the stability of cyberspace.
It is paramount for such a shared responsibility to unit us through a collective commitment.


The TruCourse Rudder is a radical new kayak rudder system with unique practical and high-performance features. Kit includes TruCourse™ Rudder, Kayak Adjustable Footbrace with Rudder Control and most materials needed to attach a rudder to a kayak.
The Sea-Lect Designs TruCourse Rudder Assembly is a radical new kayak rudder system with unique practical and high-performance features. The multi-position blade enables kayakers to adjust rudder depth to suit conditions. The foil-shaped blade slices through the water with minimum drag. It is very lightweight. The rudder weighs only 1.4 pounds. The rudder pulls up and slips forward when not in use. Quiet when pulled up. No loud clunk to chase away the fish and wildlife when coming into shallow water. Tough, durable engineered injection molded polymer and stainless steel components ensure top performance and reliability. Can be attached to most kayaks designed for rudder with the use of the proper mounting gudgeon.


text:
package org.buildmlearn.practicehandwriting.activities.selection;

import android.app.Activity;
import android.content.Intent;
import android.content.pm.ActivityInfo;
import android.graphics.Point;
import android.os.Bundle;
import android.view.Display;
import android.view.View;
import android.widget.Button;
import android.widget.LinearLayout;

import org.buildmlearn.practicehandwriting.R;
import org.buildmlearn.practicehandwriting.activities.practice.FreehandActivity;
import org.buildmlearn.practicehandwriting.activities.practice.PracticeActivity;
import org.buildmlearn.practicehandwriting.activities.information.SplashActivity;

import java.util.Random;

import info.hoang8f.widget.FButton;

/**
 * Activity to let the user choose a character
 */
public class CharacterSelectionActivity extends Activity {

    /**
     *  colours for the buttons
     */
    private int[] colours = {R.color.Green,
            R.color.Orange,
            R.color.Pink,
            R.color.Red,
            R.color.Yellow
    };

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setRequestedOrientation(ActivityInfo.SCREEN_ORIENTATION_NOSENSOR);
        setContentView(R.layout.activity_character_selection);
        LinearLayout root = (LinearLayout) findViewById(R.id.CharacterLL);
        LinearLayout child = new LinearLayout(this);

        Display display = getWindowManager().getDefaultDisplay();
        Point size = new Point();
        display.getSize(size);

        int horizontal_padding = (int) getResources().getDimension(R.dimen.activity_horizontal_margin);
        int vertical_padding = (int) getResources().getDimension(R.dimen.activity_vertical_margin);
        //width and height to fit 4 buttons on the screen.
        LinearLayout.LayoutParams params = new LinearLayout.LayoutParams((size.x-8*horizontal_padding)/4,(size.x-8*horizontal_padding)/4);
        params.setMargins(horizontal_padding, vertical_padding, horizontal_padding, vertical_padding);
        int count = 0;
        for(int i=0;i< SplashActivity.CHARACTER_LIST.length;i++) {
            String s = SplashActivity.CHARACTER_LIST[i];
            FButton fb = new FButton(this);
            fb.setText(s);
            fb.setTextSize(((size.x - 8 * horizontal_padding) / 4) * 1.0f/4.0f);
            fb.setAllCaps(false);
            fb.setButtonColor(getResources().getColor(colours[new Random().nextInt(colours.length)])); //setting a random colour
            fb.setShadowEnabled(false);
            fb.setCornerRadius((size.x - 8 * horizontal_padding) / 4); //radius to fit 4 buttons on the screen
            fb.setLayoutParams(params);
            fb.setOnClickListener(new View.OnClickListener() {
                @Override
                public void onClick(View view) {
                    Intent intent;
                    if (getIntent().getStringExtra(getResources().getString(R.string.practice_mode)).equals(getResources().getString(R.string.practice)))
                        intent = new Intent(CharacterSelectionActivity.this, PracticeActivity.class);
                    else //only else and not else if as the value of getIntent().getStringExtra(getResources().getString(R.string.practice_mode)) could take only 2 values
                        intent = new Intent(CharacterSelectionActivity.this, FreehandActivity.class);
                    intent.putExtra(getResources().getString(R.string.practice_string), ((Button) view).getText());
                    startActivity(intent);
                }
            });
            child.addView(fb);

            count++;
            if(count==4) {
                root.addView(child);
                child = new LinearLayout(this);
                count=0;
            }
        }
        root.addView(child);
    }
}

redpajama_set_name: RedPajamaGithub


Start the new year off right with a hike in your favorite state park. Fort Macon will be offering two ranger led hikes this year, both starting from the Visitor Center at 2pm. One will be a short nature hike on the Yarrow's Loop Trail and inlet beach covering about 3/4 of a mile. The second hike will be along the 3.2 mile Elliot Caues Trail.


This espace is a place for sharing information about DIRAC at IHEP computing center. DIRAC is a software package for creating distributed computing infrastructures. Detailed information can be found in the DIRAC project home page.
Here we present our work on several projects related to the DIRAC software itself. DIRAC is also being used at IHEP for building a computing grid for processing the data of the BES experiment, which is documented here.
Exploring the suitability of unstructured data stores for recording and exploiting DIRAC accounting data.