text:
\section{Introduction}

Alday-Maldacena hypothesis \cite{AM1}-\cite{AMlast},
an artfully-motivated corollary of the AdS/CFT realization \cite{AdS/CFT}
of string-gauge duality \cite{SGT},
is one of the most spectacular ideas of recent years.
Our understanding of the subject is described in
\cite{MMT1}-\cite{M} and we do not go into any details here.
For us

\bigskip

\centerline{
\framebox{\vbox{\centerline{
the Alday-Maldacena hypothesis is that}\centerline{
the
AdS Plateau problem in its weak form is explicitly resolvable}}}}

\bigskip
\noindent
Moreover, the hypothesis {\it per se} is even more explicit:
the (regularized) area of the minimal surface in AdS space,
\beq\new\begin{array}{c}
A_\Pi = \int \sqrt{\det_{ab} G_{MN}^{AdS}(Y_{min})\partial_a Y_{min}^M
\partial_b Y_{min}^N}
\label{SNG}
\end{array}\eeq
with a given boundary $\Pi$ at the flat infinity (absolute)
of AdS
and
the Abelian Wilson average, the (regularized) double integral along the boundary
\beq\new\begin{array}{c}\label{DLI}
D_\Pi = \frac{1}{4}{\oint\oint}_{\Pi}
\frac{d\vec Y d\vec Y'}{(\vec Y-\vec Y')^2}
\end{array}\eeq
are proportional to each other
\beq\new\begin{array}{c}
\framebox{$
A_\Pi^{reg} \ \stackrel{?}{=}\ \kappa_\Pi D_\Pi^{reg}$}
\label{AMh}
\end{array}\eeq
where $\kappa_\Pi$ depends only on the angles (non-smooth points)
of $\Pi$, not on its smooth deformations.
It is a {\it weak} form of the Plateau problem, because only
the {\it area} of the minimal surface, not its shape
is claimed to be explicitly evaluated.

\bigskip

A physical motivation for this hypothesis comes from string-gauge
duality, supplemented by the BDS conjecture \cite{BDS}:
explicit (hypothetical) formulas for multi-loop diagrams
in the planar limit of $N=4$ Super-Yang-Mills
theory.
The BDS conjecture, adapted in \cite{AM1,dks,BHT} with the help of
the KT-duality \cite{KT}, was that the perturbation
theory at weak coupling provides an answer for the MHV amplitudes
in terms of the Abelian double-loop integral:
\beq\new\begin{array}{c}
{\cal A}_{BDS} \sim \exp (\tilde\kappa D_\Pi)
\label{BDS}
\end{array}\eeq
what allowed to extend it straightforwardly to the strong coupling
regime, by simply changing the coefficient $\tilde \kappa$
to some other $\kappa$.
The $\tilde\kappa$-dependence of the coupling constant is pretty
sophisticated and, as any parameter in the effective action \cite{UFN3},
is controlled by a hidden integrable structure \cite{instr}.
The $\kappa$-$\tilde\kappa$ relation can be also dictated
by this hidden structure, but this still remains to be revealed.

Since at strong coupling the stringy description should be
relevant (with non-critical string described by
the AdS/CFT-correspondence \cite{AdS/CFT}),
the same quantity should be given by $\exp (A_\Pi)$,
and, therefore, one concludes that $A_\Pi = \kappa D_\Pi$,
as conjectured in (\ref{AMh}).

\bigskip

Unfortunately, a closer examination of the BDS conjecture (\ref{BDS})
leads to its serious modification \cite{Vo,Koanti}: as originally
anticipated in \cite{dhks3},
the loop diagrams are actually summed into a {\it non}-Abelian
Wilson average, thus, the {\it answers}  can not be explicitly
exponentiated and continued to the strong coupling
region, which makes the Alday-Maldacena hypothesis (\ref{AMh})
groundless.
Worst of all, the lack of a simple formula like (\ref{BDS})
with momentum-dependence in $\Pi$ and $D_\Pi$
separated from the coupling-constant dependence
in $\tilde \kappa_\Pi$ makes it impossible
to use the ordinary perturbative (diagram) calculations
for analysis of the strong coupling regime:
there is no straightforward way to extrapolate between the
weak and strong coupling regions.
Thus, the relation (\ref{AMh}) and its possible modifications
have to be analyzed directly, without references to diagram
calculations in SYM theory.

This is a well-defined problem, because both sides of
(\ref{AMh}) are pure geometric and contain no reference to
quantum field theory.
However, it is a difficult problem,
because it is difficult to explicitly
construct the minimal AdS surface with a given boundary:
the AdS Plateau problem does not look much simpler than the Euclidean
one, an old "hard problem" in fundamental mathematics.
Currently the number of explicit examples, where the AdS
Plateau problem is {\it fully} resolved
is too small to draw far-going conclusions.
Approximate methods are also difficult to develop, they
are even more involved than in the Euclidean case
because of the need to regularize the area and extract
finite contributions to divergent expressions
(see \cite{IMM1,IM8} for a possible approach, which still
needs to be applied to particular examples).

A way out is provided by consideration of wavy lines \cite{wavy,AM3}:
of small smooth deformations of exactly solvable examples,
i.e. of $\Pi$ deviating slightly and smoothly from a $\Pi_0$,
for which the AdS minimal surface is explicitly known.
This idea is not quite consistent with the original physical
setting, where $\Pi$ were rather polygons with the light-like
sides, and, therefore, are not smooth.
The wavy lines can get relevant in the case of scattering of
$n\rightarrow\infty$ particles with tiny momenta $\sim 1/n$ \cite{AM3}.
Though in this way we loose any direct contact with the finite-$n$
calculations (the only ones that can be made diagrammatically
in the weak coupling regime),
we get instead an infinite-dimensional family of curves $\Pi$
to check if and how eq.(\ref{AMh}) is actually modified.

\section{Alday-Maldacena hypothesis for the wavy circle}

In \cite{IMM2,M} we proposed to do this calculation for
$\Pi_0=$circle,\footnote{
Another possible choice for exactly solvable $\Pi_0$
is a pair of circles \cite{concc}.
This generalization of (\ref{circ}) is very interesting,
because there is a phase transition when the distance
between the two circles large enough as compared with their radii.
}
where the AdS Plateau problem has the explicit solution
\beq\new\begin{array}{c}
r^2 = 1-z\bar z, \ \ \ \ z= y_1+iy_2, \\
y_0=y_3=0
\label{circ}
\end{array}\eeq
Here $y_0,y_1,y_2,y_3,r$ are the standard Poincar\'e coordinates
on $ADS_5$ with the metric
\beq\new\begin{array}{c}
ds^2 = \frac{-dy_0^2 + d\vec y\,^2 + dr^2}{r^2}
\end{array}\eeq
and we refer to \cite{IMM2,M}
for further details about notations and other peculiarities
of our approach.

\bigskip

If the unit circle \ $\Pi_0:\ z=e^{i\phi}$\ is substituted by
any other smooth contour\footnote{
According to Riemann's theorem we can actually parameterize
in this way an {\it arbitrary} smooth planar curve,
not only an {\it infinitesimal} deformation of a circle $\Pi_0$.
}, lying in the plane $y_0=y_3=0$,
\beq\new\begin{array}{c}\label{RiemPar}
\Pi:\ z=H\left(e^{i\phi}\right),\ \ \ \
H(\zeta) = \zeta + \sum_{k=0}^\infty h_k\zeta^k,
\end{array}\eeq
then both sides of eq.(\ref{AMh}) can be represented
as formal series in powers of $h$ and $\bar h$:
\beq\new\begin{array}{c}
\begin{array}{ccccc}
A_\Pi^{reg} &=& A_\Pi - \frac{\pi L_\Pi}{2\mu} + 2\pi &=&
- 3\pi\left(\sum_{m,n} (-)^{m+n}
A^{(m|n)}_{k_1\ldots k_m|l_1\ldots l_n}
h_{k_1+1}\ldots h_{k_m+1}\bar h_{l_1+1}\ldots \bar h_{l_n+1}\right),
\\
D_\Pi^{reg} &=& D_\Pi - \frac{\pi L_\Pi}{4\lambda}
+ \frac{\pi^2}{2} &=&
- \pi^2\left(\sum_{m,n} (-)^{m+n}
D^{(m|n)}_{k_1\ldots k_m| l_1\ldots l_n}
h_{k_1+1}\ldots h_{k_m+1}\bar h_{l_1+1}\ldots h_{l_n+1}\right)
\end{array}
\label{ADwavy}
\end{array}\eeq
In these terms, the Alday-Maldacena hypothesis (\ref{AMh}) is that all
the coefficients $A^{(m|n)}$ and $D^{(m|n)}$ coincide,
$A^{(m|n)} = D^{(m|n)}$, while
\beq\new\begin{array}{c}
\kappa_{smooth}
= \left(\frac{\pi}{3}\right)^{-1}
\end{array}\eeq
Intermediate equations in (\ref{ADwavy}) show
how divergencies are subtracted from the area and the contour integral,
where $\mu$ and $\lambda$ are the corresponding
regularization parameters \cite{IMM2} and
\beq\new\begin{array}{c}
L_\Pi = 2\pi\left(1 + \sum_{m,n} L^{(m|n)}_{k_1\ldots k_m|l_1\ldots l_n}
h_{k_1+1}\ldots h_{k_m+1}\bar h_{l_1+1}\ldots \bar h_{l_n+1}\right)
\label{Lwavy}
\end{array}\eeq
is the length of the contour $\Pi$.
We explicitly write only sums over expansion orders $m$ and $n$,
summations are of course performed also over the indices
$k_i$ and $l_j$.
The unit shift of indices in (\ref{ADwavy}) and (\ref{Lwavy}),
$h_{k+1}$ instead of $h_k$,
is convenient because with such a definition
\beq\new\begin{array}{c}
\sum_{i=1}^m k_i = \sum_{j=1}^n l_j
\label{seru}
\end{array}\eeq
in all the sums (such a shift was not made in \cite{IMM2,M}).

\bigskip

In \cite{IMM2} the first two coefficients were calculated, and they
turned out to be
\beq\new\begin{array}{c}
A^{(1|1)}_{k|k} = \frac{(k+1)k(k-1)}{6} = D^{(1|1)}_{k|k}, \\
A^{(2|1)}_{k_1,k_2|k_1+k_2} = \frac{(k_1+1)(k_2+1)}{12}
\Big(k_1^2+k_2^2+3k_1k_2-k_1-k_2\Big) = D^{(2|1)}_{k_1,k_2|k_1+k_2}
\label{AD1121}
\end{array}\eeq
We explicitly took into account the selection rule (\ref{seru})
and corrected a trivial, but misleading error: in \cite{IMM2,M}
we got a slightly different expression for
$A^{(2|1)}_{k,k|2k}$, what made it different from
$D^{(2|1)}_{k,k|2k}$,
and what we called an {\it anomaly}.
In fact, the entire $A^{(2|1)}$ is what it is in
(\ref{AD1121}),\footnote{
An error in \cite{IMM2} is clear from the line
"$h:=z->h[K]*z^K + h[l]*z^L;$"  in Appendix III
(the very last line at page 21 in the arXiv version of that paper).
It is clear that at $K=L$ there is a double-counting
and the "diagonal" terms with $h_K^2$, $h_K^3$ should be
actually divided by $2^2=4$, $2^3=8$ and so on:
this correction converts $A_\Pi^{reg}$ of \cite{IMM2,M}
into that in (\ref{AD1121}) of the present paper.}
thus $A^{(2|1)} = D^{(2|1)}$
and for the first time {\it the anomaly} shows up in the $(2|2)$ terms,
see below.

Actually (\ref{AD1121}) looks like a serious evidence
{\it in support} of the Alday-Maldacena hypothesis (\ref{AMh}):
it demonstrates the coincidence of {\it infinitely many}
terms at both sides.
As we see below, it looks probable that this
coincidence is not quite accidental: we now believe that it
is extended to all other terms with $m=1$ or $n=1$:
\beq\new\begin{array}{c}
A^{(m|1)}_{k_1,\ldots,k_m|k_1+\ldots+k_m} =
D^{(m|1)}_{k_1,\ldots,k_m|k_1+\ldots+k_m}, \\
A^{(1|n)}_{l_1+\ldots+l_m|l_1,\ldots,l_n} =
D^{(1|n)}_{l_1+\ldots+l_m|l_1,\ldots,l_n}
\label{ADm1}
\end{array}\eeq
(the second relation is of course obtained from the first
one by complex conjugation),
though we check it only up to $m=3$ in this paper.
Moreover,
with additional assumption of polynomial dependence on indices,
(\ref{ADm1}) can be
considered as implication of conformal invariance
\cite{dks,dhks1,dhks3,Komar,IMM2} and already for this reason has good
chances to be true.
At the same time, the conformal invariance does not seem to
control the terms with $min(m,n)\geq 2$ with
indices $k_i+1$ and $l_i+1$ exceeding $2$
(since the invariance algebra contains
only three generators: deformations of $\partial/\partial h_0$,
$\partial/\partial h_1$ and $\partial/\partial h_2$).
Moreover, the index dependence of the area fails to be polynomial in this case
and, therefore, with our present intuition
one can expect deviations from (\ref{AMh}) in the $(2|2)$ terms.
We see below that this expectation turns out to be true:
indeed, already $A^{(2|2)} \neq D^{(2|2)}$ and

\bigskip

\centerline{
\framebox{\vbox{\centerline{
The Alday-Maldacena hypothesis (\ref{AMh}) {\it fails}
for the wavy circle}}}}

\bigskip

\section{The difference between smooth and non-smooth $\Pi$
and non-uni\-ver\-sa\-li\-ty of the $\kappa$-factor}

Of course we already learned from \cite{AM3}
that it fails, but in a different, much weaker sense.
What we know from \cite{AM3} is
that the coefficients $\kappa_\Pi$ can not be fully universal (see the Appendix):
they are different in the $n=4$ case, $\kappa_{\Box} = 1$ \cite{AM1},
and in the $n=\infty$ case with $\Pi =$infinitely long
strip \cite{AM3}, $\kappa_{strip}=4\frac{(2\pi)^2}{\left(\Gamma\left(1/4\right)\right)^4}
\approx 1$.
Moreover, (\ref{ADwavy}) implies \cite{IMM2,M} that
$\kappa_{smooth} = 3/\pi\approx 1$ is different from both
$\kappa_{\Box}$ and $\kappa_{strip}$:
that was another ingredient of {\it the anomaly} of ref.\cite{IMM2}.
However, from (\ref{AD1121}) one can see a possible origin for
this non-universality of the relative coefficient:
the coefficients $A^{(m|n)}$ and $D^{(m|n)}$ are
rational in $k$ and $l$, thus the series
(\ref{ADwavy}) are nicely convergent for smooth
curves $\Pi$ (when $h_k$ falls faster than any negative power
of $k$ at large $k$), but they diverge when $h_k \sim 1/k^2$,
and this is exactly the behavior when angles are present
in $\Pi$.

Indeed, the periodic step function (Fig.\ref{step})
has the Fourier expansion
$$\frac{qh}{\pi^2}\ {\rm Im} \sum_{m=0}^\infty
\frac{e^{iq(2m+1)\phi}}{2m+1},$$
which means that the behavior $h_k \sim \frac{1}{k}$ of
Fourier coefficients can lead to jumps.
The $\phi$-integral of this step function, a saw,
Fig.\ref{pila} has the expansion
$$\frac{h}{\pi ^2}\ {\rm Re} \sum_{m=0}^\infty
\frac{e^{iq(2m+1)\phi}}{(2m+1)^2},$$
thus, $h_k \sim \frac{1}{k^2}$ can provide jumps in the derivative,
i.e. angles in the curve itself.

\begin{figure}
\begin{center}
\begin{picture}(200,40)(50,-20)
\put(10,10){\line(1,0){40}}
\put(50,-10){\line(1,0){40}}
\put(90,10){\line(1,0){40}}
\put(130,-10){\line(1,0){20}}
\put(190,-10){\line(1,0){20}}
\put(210,10){\line(1,0){40}}
\put(250,-10){\line(1,0){40}}
\put(0,0){\line(1,0){150}}
\put(190,0){\vector(1,0){120}}
\put(10,-15){\vector(0,1){40}}
\put(0,-10){\line(1,0){10}}
\put(170,0){\makebox(0,0)[cc]{$\ldots$}}
\put(50,4){\makebox(0,0)[cc]{$\pi/q$}}
\put(90,4){\makebox(0,0)[cc]{$2\pi/q$}}
\put(133,4){\makebox(0,0)[cc]{$3\pi/q$}}
\put(247,4){\makebox(0,0)[cc]{$2\pi-\pi/q$}}
\put(290,4){\makebox(0,0)[cc]{$2\pi$}}
\put(3,11){\makebox(0,0)[cc]{$\frac{qh}{2\pi}$}}
\end{picture}
\caption{{\footnotesize The periodic step function with period $2q$.
The amplitude is adjusted so that the amplitude of saw in
Fig.\ref{pila} is exactly $h$. }}
\label{step}
\end{center}
\end{figure}
\begin{figure}
\begin{center}
\begin{picture}(200,40)(50,-20)
\put(10,-10){\line(2,1){40}}
\put(50,10){\line(2,-1){40}}
\put(90,-10){\line(2,1){40}}
\put(130,10){\line(2,-1){20}}
\put(190,0){\line(2,-1){20}}
\put(210,-10){\line(2,1){40}}
\put(250,10){\line(2,-1){40}}
\put(0,0){\line(1,0){150}}
\put(190,0){\vector(1,0){120}}
\put(10,-15){\vector(0,1){40}}
\put(0,-5){\line(2,-1){10}}
\put(170,0){\makebox(0,0)[cc]{$\ldots$}}
\put(50,-5){\makebox(0,0)[cc]{$\pi/q$}}
\put(90,4){\makebox(0,0)[cc]{$2\pi/q$}}
\put(133,-5){\makebox(0,0)[cc]{$3\pi/q$}}
\put(247,-5){\makebox(0,0)[cc]{$2\pi-\pi/q$}}
\put(290,4){\makebox(0,0)[cc]{$2\pi$}}
\put(6,11){\makebox(0,0)[cc]{$h$}}
\end{picture}
\caption{{\footnotesize The periodic saw function of period
$2q$ and amplitude $h$.}}
\label{pila}
\end{center}
\end{figure}

Looking already at
the first line in (\ref{AD1121}), one sees that $s=2$
in $h_k \sim \frac{1}{k^s}$ is exactly the convergency boundary
for these series. For the infinite strip of \cite{AM3} with
$$H_{strip}(\zeta) = \log \frac{1-\zeta}{1+\zeta} =
-2\sum_{m=0}^\infty \frac{\zeta^{2m+1}}{2m+1}$$
we have even $h_k \sim \frac{1}{k}$ so that even the length of
$\Pi$,
\beq\new\begin{array}{c}
L_\Pi = 2\pi \left(1 + \frac{1}{4}\sum_{k=0}^\infty k^2|h_k|^2 +
O(h^3)\right)
\end{array}\eeq
diverges, not only $A$ and $D$.

Of course, the regularization of integrals $A_\Pi$ and $D_\Pi$
eliminates these divergencies at the angles and at spatial infinities,
as it does with the divergencies near the boundaries,but the
renormalizations are different for finite terms
(with rapidly decreasing $h_k$), for logarithmic
($h_k\sim 1/k^2$) and for quadratic ($h_k\sim 1/k$) divergencies.
Thus, it is not a big surprise that, if (\ref{AMh}) would even hold
for non-smooth curves, the coefficient $\kappa_\Pi$
can depend on the number of angles.\footnote{
Moreover, one can even imagine that both $A_\Pi^{reg}$ and $D_\Pi^{reg}$
split into two items, one for a smooth, another one for an angle-containing
constituents of the curve $\Pi$, and the two items coincide pairwise but
with different coefficients.}
Evaluation of anomalous ratio $\frac{\kappa_{angles}}{\kappa_{smooth}}
\neq 1$ remains an interesting open problem.

\section{Abelian double integral \label{Dcalc}}

We now proceed to detailed examination of the two sides
of eq.(\ref{AMh}).
The double integral at the r.h.s. in the formula
is actually much simpler than the minimal area in its l.h.s.
In this section, we explicitly represent the {\it generic}
coefficient $D^{(m|n)}$ as a simple multiple sum,
thus, generalizing eq.(1.13) of \cite{IMM2} from $n=1$ or $m=1$
to arbitrary $(m|n)$:
\beq\new\begin{array}{c}
D^{(m;n)}_{k_1,\ldots,k_m| l_1,\ldots,l_n} =
{\rm symmetrized}\left(
\sum_{i_1=0}^{k_1}\ldots \sum_{i_m=0}^{k_m}\
\sum_{j_1=0}^{l_1}\ldots \sum_{j_n=0}^{l_n}\
(k_m-i_m)(l_n-j_n)\right)
= \\ = \frac{1}{p!}\sum_{p=1}^m
\frac{1}{q!}\sum_{q=1}^n \left(
\sum_{i_1=0}^{k_1}\ldots \sum_{i_m=0}^{k_m}\
\sum_{j_1=0}^{l_1}\ldots \sum_{j_n=0}^{l_n}\
(k_p-i_p)(l_q-j_q)\right)
\label{Dgen}
\end{array}\eeq
with the constraint
\beq\new\begin{array}{c}
\sum_{p=1}^m i_p + \sum_{q=1}^n j_q\ =\
\sum_{p=1}^m k_m\ =\ \sum_{q=1}^n l_q
\label{conDgen}
\end{array}\eeq
We derive this formula in subsection
\ref{derivD}, but consider a number
of examples and applications before.

\subsection{Examples}

In particular, (\ref{AD1121}) is an immediate
corollary of (\ref{Dgen}):
\beq\new\begin{array}{c}
D^{(2)} \equiv \sum_{\stackrel{k,l=0}{k=l}}^\infty
D^{(1|1)}_{k|l} h_{k+1}\bar h_{l+1}
= \sum_{k=0}^\infty |h_{k+1}|^2
\left.\sum_{i=0}^k (k-i)(k-j)\right|_{i+j=k}
= \sum_{k=0}^\infty \frac{(k-1)k(k+1)}{6}|h_{k+1}|^2
\end{array}\eeq
\beq\new\begin{array}{c}
D^{(3)} \equiv \sum_{\stackrel{k_1,k_2,l=0}{k_1+k_2=l}}^\infty
D^{(2|1)}_{k_1,k_2|l} h_{k_1+1}h_{k_2+1}\bar h_{l+1}
+ \sum_{\stackrel{k,l_1,l_2=0}{k=l_1+l_2}}^\infty
D^{(1|2)}_{k|l_1,l_2} h_{k+1}\bar h_{l_1+1}\bar h_{l_2+1}
\end{array}\eeq
with
\beq\new\begin{array}{c}
D^{(2|1)}_{k_1,k_2|k_1+k_2}  = {\rm symm}_{k_1\leftrightarrow k_2}
\left(
\left.\sum_{i_1=0}^{k_1}\sum_{i_2=0}^{k_2}
\sum_{j=0}^{k_1+k_2} (k_2-i_2)(k_1+k_2-j)\right|_{i_1+i_2+j=k_1+k_2}
\right) = \\ = \frac{1}{12}(k_1+1)(k_2+1)
(k_1^2+3k_1k_2+k_2^2-k_1-k_2)
= D^{(1;2)}_{k_1+k_2;k_1,k_2}
\end{array}\eeq
Further, one easily reproduces from (\ref{Dgen}) the result of \cite{IMM2} for
more terms
\beq\new\begin{array}{c}
D^{(4)} \equiv
\sum_{\stackrel{k_1,k_2,k_3,l=0}{k_1+k_2+k_3=l}}^\infty
D^{(3|1)}_{k_1,k_2,k_3|l} h_{k_1+1}h_{k_2+1}h_{k_3+1}\bar h_{l+1}
+ \\
+ \sum_{\stackrel{k,l_1,l_2=0}{k_1+k_2=l_1+l_2}}^\infty
D^{(2|2)}_{k_1,k_2|l_1,l_2} h_{k_1+1}h_{k_2+1}
\bar h_{l_1+1}\bar h_{l_2+1}
+ \sum_{\stackrel{k,l_1,l_2,l_3=0}{k=l_1+l_2+l_3}}^\infty
D^{(1|3)}_{k|l_1,l_2,l_3} h_{k+1}\bar h_{l_1+1}\bar h_{l_2+1}
\bar h_{l_3+1}
\end{array}\eeq
with
\beq\new\begin{array}{c}
D^{(3|1)}_{k_1,k_2,k_3|k_1+k_2+k_3}  =
{\rm symm}_{k_1,k_2,k_3}
\left(
\left.\sum_{i_1=0}^{k_1}\sum_{i_2=0}^{k_2}\sum_{i_3=0}^{k_3}
\sum_{j=0}^{k_1+k_2+k_3} (k_3-i_3)(k_1+k_2+k_3-j)
\right|_{i_1+i_2+i_3+j=k_1+k_2+k_3}
\right) = \\ = \frac{1}{18}(k_1+1)(k_2+1)(k_3+1)
\Big(k_1^2+k_2^2+k_3^2 + 3(k_1k_2 + k_2k_3 + k_3k_1)-k_1-k_2-k_3\Big)
= D^{(1|3)}_{k_1+k_2+k_3|k_1,k_2,k_3}
\end{array}\eeq
and
\beq\new\begin{array}{c}
D^{(2;2)}_{k_1,k_2|l_1,l_2} =
{\rm symm}_{k_1\leftrightarrow k_2}
\ {\rm symm}_{l_1\leftrightarrow l_2}
\left(
\left.\sum_{i_1=0}^{k_1}\sum_{i_2=0}^{k_2}\sum_{j_1=0}^{l_1}
\sum_{j_2=0}^{l_2} (k_2-i_2)(l_2-j_2)
\right|_{i_1+i_2+j_1 + j_2=k_1+k_2=l_1+l_2}
\right)
\end{array}\eeq

\subsection{Generating functions}

To more effectively deal with various double integral formulas, one can deal, instead of the
coefficients $D$'s, with the generating functions. To this end, one introduces the
following generating functions:
\beq\new\begin{array}{c}
D(x_1...x_m;y_1...y_n)\equiv \sum D^{(m|n)}_{k_1...k_m|l_1...l_n}x_1^{k_1}...x_m^{k_m}
y_1^{l_1}...y_m^{l_m}\\
h(x)\equiv \sum h_nx^n\\
\bar h(x)\equiv\sum\bar h_nx^n
\end{array}\eeq
Then, for instance\footnote{Note that with another natural definition of generating functions
$D$'s,
\beq\new\begin{array}{c}
D(x_1...x_m;y_1...y_n)\equiv \sum {D^{(m|n)}_{k_1...k_m|l_1...l_n}\over x_1^{k_1+2}...x_m^{k_m+2}
y_1^{l_1+2}...y_m^{l_m+2}}
\end{array}\eeq
formula (\ref{gfDh}) looks even simpler:
\beq\new\begin{array}{c}
D^{(m,1)}=\sum D^{(m,1)}_{i_1\ldots i_m}h_{i_1+1}\ldots h_{i_m+1} \bar h_{i_1+\ldots+i_m+1}=
\oint\ldots\oint D^{(m,1)}(x_1,\ldots,x_m;y)h(x_1)\ldots h(x_m)
\bar h(y)\frac{dx_1}{2\pi i}\ldots \frac{dx_m}{2\pi i}\frac{dy}{2\pi i}
\end{array}\eeq
with $h(x)$ instead of $h(\bar x)$ and the integration contours encircling zero, but not
obligatory unit circles.},
\beq\new\begin{array}{c}\label{gfDh}
D^{(m,1)}=\sum D^{(m,1)}_{i_1\ldots i_m}h_{i_1+1}\ldots h_{i_m+1} \bar h_{i_1+\ldots+i_m+1}=\\
=\oint\ldots\oint D^{(m,1)}(x_1,\ldots,x_m;y)h(\bar x_1)\ldots h(\bar x_m)
\bar h(\bar y)\frac{dx_1}{2\pi i}\ldots \frac{dx_m}{2\pi i}\frac{dy}{2\pi i}
\end{array}\eeq
where integration goes over unit circles, and similarly for other $D$'s.

\subsubsection{The simplest generating functions}

The simplest generating functions can be immediately obtained from explicit formulas in indices:
\beq\new\begin{array}{c}
D^{(1,1)}_{i} = \frac{i+1}{6}(i^2-i), \\
D^{(2,1)}_{ij} = \frac{(i+1)(j+1)}{12}(i^2+j^2+3ij - i - j),\\
D^{(3,1)}_{ijk} = \frac{(i+1)(j+1)(k+1)}{18}
\Big(i^2+j^2+k^2 + 3(ij+jk+ki) - i - j - k\Big), \\
\ldots  \\
D^{(m,1)}_{i_1\ldots i_m} =
\frac{(i_1+1)\ldots(i_m+1)}{6m}\left(
\sum_{p=1}^m i_p^2 + 3\sum_{p<q} i_pi_q - \sum_{p=1} i_p\right)
\end{array}\eeq
These series can be summed up to give the following expressions
\beq\new\begin{array}{c}
D^{(1;1)}(x,y) = \frac{(xy)^2}{(1-xy)^4}
= \Big(\xi(1 -\xi)\Big)^2,  \\
D^{(2,1)}(x_1,x_2;y) = \frac{(x_1y+x_2y - 2x_1x_2y^2)^2}
{2(1-x_1y)^4(1-x_2y)^4} =
\frac{1}{2}\Big(\xi_1\xi_2(2 - \xi_1-\xi_2)\Big)^2,  \\
\ldots \\
D^{(m,1)}(x_1,\ldots,x_m;y) =
\frac{1}{m}\Big(\xi_1\ldots \xi_m(m - \xi_1 - \ldots - \xi_m)\Big)^2
\end{array}\eeq
where $\xi_i = \frac{1}{1-x_iy}$.

\subsubsection{Generating functions and Miwa variables}

Another possibility, which allows one to look at the generating functions differently, is
to introduce instead of the variables $h_k$ and $\bar h_k$ (we omit from these sets
$h_0$ and $\bar h_0$ since they anyway do not enter the results due to the conformal
invariance) the two infinite sets of new
Miwa-like variables $\{x_i\}$ and $\{y_i\}$ \cite{Miwa,UFN3}
\beq\new\begin{array}{c}
h_n\equiv\sum_i \alpha_i x_i^{n-1}\ ,\ \ \ \ \ \ \ \ \ \ \bar h_n\equiv\sum_i \bar\alpha_i y_i^n
\end{array}\eeq
where $\alpha_i$ and $\bar\alpha_i$ are multiplicities, i.e. the number of coinciding
$x_i$ and $y_i$ correspondingly. Then,
\beq\new\begin{array}{c}
z(\zeta)=\zeta+\sum_{k=2} h_k\zeta^k=\zeta+\sum\limits_{i,k}\alpha_ix_i^k \zeta^{k+1}=
\zeta+\sum_i\frac{\alpha_i\zeta}{1-x_i\zeta}
\end{array}\eeq
and similarly
\beq\new\begin{array}{c}
\bar z(\zeta)=\bar\zeta+\sum\limits_j\frac{\bar\alpha_j\bar\zeta}{1-y_i\bar\zeta}\\
\end{array}\eeq
In fact, one can generically put all the multiplicities equal to
unity. However, one can instead preserve non-trivial multiplicities in order to immediately obtain
the following formula that allows one to produce the generating functions from the double loop
integral $D[z,\bar z]$
\beq\new\begin{array}{c}
D^{(n|m)}(x|y)=\left.\left[\left(\prod\limits_{i=1}^n\frac{\partial}{\partial \alpha_i}
\prod\limits_{j=1}^m\frac{\partial}{\partial \bar\alpha_j}\right)D[z,\bar z]\right]\right|_{\alpha_i,\beta_j=0}
\end{array}\eeq
The derivatives here should be taken w.r.t. different $\alpha_i$'s and, similarly, w.r.t.
different $\bar\alpha_i$'s.

\subsubsection{Generating functions and Schwarzian}

It turns out that all the terms $D^{(m|1)}$ can be explicitly summed up
\cite{IMM2} or derived by the above described technique, and are expressed through the
Schwarzian derivative:
\beq\new\begin{array}{c}
D^{(\cdot|1)} \equiv
\sum_{m=1}^\infty D^{(m|1)}_{k_1\ldots k_m|k_1+\ldots+k_m}
h_{k_1+1}\ldots h_{k_m+1}\bar h_{k_1+\ldots+k_m+1} =-\frac{1}{6}
\oint \bar h(\bar \zeta)S_\zeta(z)\zeta^2 d\zeta
\end{array}\eeq
where the contour integral is taken along the unit circle
$\zeta = e^{i\phi}$, $z = \zeta + \sum_k h_k\zeta^k$ and the Schwarzian derivative is
\beq\new\begin{array}{c}
S_\zeta(z) = \frac{z'''}{z'} - \frac{3}{2}\left(\frac{z''}{z'}\right)^2
\end{array}\eeq
Similarly one can simply derive that
\beq\new\begin{array}{c}
D(h,\bar h)=\pi^2\sum\limits_{n=1}^{\infty}(-1)^{n}\left\{\frac{1}{2\pi i}
\oint d\zeta_k \bar h(\bar \zeta_k)\right\}_{k=1}^n
\left(\frac{1}{6}\sum_{i}\frac{\zeta_i^{2n}}{\prod\limits_{j\neq i}(\zeta_i-\zeta_j)^2}S(z_i)-\sum\limits_{i\neq j}\frac{2\zeta_i^n \zeta_j^n \sigma(z_1,z_2)}{(\zeta_i-\zeta_j)^2\prod\limits_{p\neq i,j}(\zeta_i-\zeta_p)(\zeta_j-\zeta_p)}\right)
\label{hexp}
\end{array}\eeq
where
\beq\new\begin{array}{c}
\sigma(z_1,z_2)=\frac{z_1' z_2'}{(z_1-z_2)^2}
\end{array}\eeq
Advantage of the representation is that $s$ and $\sigma$ are explicit invariants of the conformal
symmetry, see s.6.3 below.

\subsection{Derivation of (\ref{Dgen}) \label{derivD}}

To complete this section, we now return to the formal derivation of (\ref{Dgen}):

$$
D_\Pi = \oint_\Pi\oint_\Pi \frac{\frac{1}{2}(dzd\bar z' + d\bar zdz')}
{|z-z'|^2}
= \frac{1}{2} \oint_{|\zeta|^2=1} \oint_{|\zeta'|^2=1}
PQ\bar Q\ \frac{d\zeta d\bar\zeta'}
{|\zeta-\zeta'|^2} + (\zeta\leftrightarrow \zeta',
\bar\zeta\leftrightarrow \bar\zeta'),
$$
where
$$
P = 1 + \frac{\partial h(\zeta)}{\partial\zeta} +
\frac{\overline{\partial h(\zeta')}}{\partial\bar\zeta'} +
\frac{\partial h(\zeta)}{\partial\zeta}
\frac{\overline{\partial h(\zeta')}}{\partial\bar\zeta'} =(1 + p_{1|0})(1 + p_{0|1})=
1 + p_{1|0} + p_{0|1} + p_{1|1}
$$
with $p_{1|1} = p_{1|0}p_{0|1}$, while
$$
Q = \left(1+\frac{h(\zeta)-h(\zeta')}{\zeta-\zeta'}\right)^{-1}
= (1+q_{1|0})^{-1}
$$
and $\bar Q = (1+ q_{0|1})^{-1}$ with $q_{0|1}=\overline{q_{1|0}}$.
The subscripts here count the numbers of $h$ and $\bar h$
in the corresponding expressions.

It is easy to check\footnote{
Indeed, the direct computation gives:
$$
(PQ\bar Q)_{0|0} = 1;
$$
$$
(PQ\bar Q)_{1|0} = p_{1|0} - q_{1|0},\ \ \ \
(PQ\bar Q)_{0|1} = p_{0|1} - q_{0|1};
$$
$$
(PQ\bar Q)_{2|0} = p_{1|0}(-q_{1|0}) + (-q_{1|0})^2 =
(-q_{1|0})(p_{1|0} - q_{1|0}), \ \ \ \
(PQ\bar Q)_{0|2} = p_{0|1}(-q_{0|1}) + (-q_{0|1})^2 =
(-q_{0|1})(p_{0|1} - q_{0|1}),
$$
$$
(PQ\bar Q)_{1|1} = p_{1|1} - p_{1|0}q_{1|0} - p_{0|1}q_{0|1} +
q_{1|0}q_{0|1} = (p_{1|0}-q_{1|0})(p_{0|1}-q_{0|1});
$$
and so on. Eq.(\ref{PQQ}) is an obvious generalization of
elementary formulas, obtained in this straightforward way.
}
that for all $m,n\geq 1$
\beq\new\begin{array}{c}
(PQ\bar Q)_{m|n} = (-q_{1|0})^{m-1}(-q_{0|1})^{n-1}
(p_{1|0}-q_{1|0})(p_{0|1}-q_{0|1})
\label{PQQ}
\end{array}\eeq
Now,
$$
p_{1|0}-q_{1|0} = \frac{\partial h(\zeta)}{\partial\zeta} -
\frac{h(\zeta)-h(\zeta')}{\zeta-\zeta'} =
\sum_{k=1}^\infty h_k\zeta^{k-1}
\left(k - \frac{1-(\zeta'/\zeta)^k}{1-\zeta'/\zeta}\right)
$$
If it is further divided by $(1-\zeta'/\zeta)$, then
\beq\new\begin{array}{c}
\frac{p_{1|0}-q_{1|0}}{1-\zeta/\zeta'}\ =\
\sum_{k=1}^\infty h_k\zeta^{k-1}\,
\frac{k-1-x-\ldots - x^{k-1}}{1-x}\ =  \\ =
\sum_{k=1}^\infty h_k\zeta^{k-1}\Big(
1 + (1+x) + (1+x+x^2) + \ldots + (1+x+\ldots +x^{k-2})\Big) = \\
= \sum_{k =1}^\infty\sum_{i=0}^{k-1} (k-i-1) h_k\zeta^{k-1}
(\zeta'/\zeta)^i
\end{array}\eeq
where $x=\zeta'/\zeta$.
Similarly,
\beq\new\begin{array}{c}
\frac{p_{0|1}-q_{0|1}}{1-\bar\zeta/\bar\zeta'} =
\sum_{l=1}^\infty \sum_{j=0}^{l-1} (l-j-1)\bar h_l(\bar\zeta')^{l-1}
(\bar\zeta/\bar\zeta')^j
\end{array}\eeq
At the same time
\beq\new\begin{array}{c}
q_{1|0} = \frac{h(\zeta)-h(\zeta')}{\zeta-\zeta'} =
\sum_{k=1}^\infty h_k\zeta^{k-1} \sum_{i=0}^{k-1} (\zeta'/\zeta)^i
\end{array}\eeq
and
\beq\new\begin{array}{c}
q_{0|1} = \frac{h(\bar\zeta)-h(\bar\zeta')}{\bar\zeta-\bar\zeta'} =
\sum_{l=1}^\infty \bar h_l(\bar\zeta')^{l-1} \sum_{j=0}^{l-1}
(\bar\zeta/\bar\zeta')^j
\end{array}\eeq
It now follows from (\ref{PQQ}) that
$$
\frac{(PQ\bar Q)_{m|n}}{(1-\zeta'/\zeta)(1-\bar\zeta/\bar\zeta')}
= (-)^{m+n}\left(\sum_{k=1}^\infty h_k\zeta^{k-1}
\sum_{i=0}^{k-1} (\zeta'/\zeta)^i\right)^{m-1}\!\!
\left(\sum_{k =1}^\infty\sum_{i=0}^k (k-i) h_k\zeta^{k-1}
(\zeta'/\zeta)^i\right)
\cdot $$ $$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdot
\left(\sum_{l=1}^\infty \bar h_l(\bar\zeta')^{l-1} \sum_{j=0}^{l-1}
(\bar\zeta/\bar\zeta')^j\right)^{n-1}\!\!
\left(\sum_{l=1}^\infty \sum_{j=0}^l (l-j)\bar h_l(\bar\zeta')^{l-1}
(\bar\zeta/\bar\zeta')^j\right) =
$$ $$
= (-)^{m+n}\sum_{k_1,\ldots,k_m=0}^\infty\
\sum_{l_1,\ldots,l_n=0}^\infty
h_{k_1+1}\ldots h_{k_m+1}\bar h_{l_1+1}\ldots \bar h_{l_n+1}
\cdot $$ $$ \cdot
\zeta^{\sum_{p=1}^m k_p}(\bar\zeta')^{\sum_{q=1}^n l_q}
\prod_{p=1}^{m-1} \left(\sum_{i=0}^{k_p} (\zeta'/\zeta)^i\right)
\left(\sum_{i=0}^{k_m} (k_m\!-i) (\zeta'/\zeta)^i\right)
\prod_{q=1}^{n-1} \left(\sum_{j=0}^{l_q} (\bar\zeta/\bar\zeta')^i\right)
\left(\sum_{j=0}^{l_n} (l_n\!-j) (\bar\zeta/\bar\zeta')^j\right)
= $$ $$ = (-)^{m+n}\sum_{k_1,\ldots,k_m=0}^\infty\
\sum_{l_1,\ldots,l_n=0}^\infty
h_{k_1+1}\ldots h_{k_m+1}\bar h_{l_1+1}\ldots \bar h_{l_n+1}
\cdot $$ $$ \cdot
\zeta^{\sum_{p=1}^m k_p}(\bar\zeta')^{\sum_{q=1}^n l_q}
\sum_{i_1=0}^{k_1}\ldots \sum_{i_m=0}^{k_m}
\sum_{j_1=0}^{l_1}\ldots \sum_{j_n=0}^{l_n}
(\zeta'/\zeta)^{\sum_{p=1}^m i_p}
(\bar\zeta/\bar\zeta')^{\sum_{q=1}^n j_p}
(k_m\!-i_m)(l_n\!-j_n)
$$
Once again one can see convenience of the shift of indices $k\rightarrow k+1$
and $l\rightarrow l+1$ in $h_k$ and $\bar h_l$.

Integration over $d\zeta/\zeta$ and $d\bar\zeta'/\bar\zeta'$
along the unit circle now imposes the constraints (\ref{conDgen})
\beq\new\begin{array}{c}
\sum_{p=1}^m i_p + \sum_{q=1}^n j_q = \sum_{p=1}^m k_p +
\sum_{q=1}^n l_q
\end{array}\eeq
and provides an expression (\ref{Dgen}) for generic
\beq\new\begin{array}{c}
D^{(m|n)}_{k_1\ldots k_m|l_1\ldots l_n} =
{\rm symmetrized}\left( \sum_{i_1=0}^{k_1}\ldots \sum_{i_m=0}^{k_m}
\sum_{j_1=0}^{l_1}\ldots \sum_{j_n=0}^{l_n}(k_m\!-i_m)(l_n\!-j_n)\right)
\end{array}\eeq

\section{Minimal area \label{Acalc}}

Turning to the l.h.s. of eq.(\ref{AMh}),
we begin with briefly reminding the perturbation theory
for AdS minimal surfaces, suggested in
\cite{IMM1,IM8} and adjusted to the wavy lines problem
in \cite{IMM1}.

\subsection{Perturbative evaluation of the minimal surface}

After substitution of
\beq\new\begin{array}{c}
y_0 = 0, \\
z= y_1+iy_2 = \zeta + H(\zeta) = \zeta + \sum_{k\geq 0}^\infty
h_k\zeta^k, \\
r(\zeta,\bar\zeta) = \sqrt{1-\zeta\bar\zeta + a(\zeta,\bar\zeta)}
\end{array}\eeq
the AdS Nambu-Goto action (\ref{SNG})
becomes
\beq\new\begin{array}{c}\label{plan-NGA}
A_\Pi = \int\frac{\sqrt{|\partial H|^2
\Big(|\partial H|^2r^2 + |\partial r^2|^2\Big)}}
{r(r^2+\mu^2)}\ d^2\zeta \ \stackrel{\mu = 0}{\longrightarrow} \
\int \frac{|1+\partial h|^2 \sqrt{1-\zeta\bar\zeta + a +
\frac{(\partial a - \bar\zeta)(\bar\partial a - \zeta)}{|1+\partial h|^2}}}
{(1-\zeta\bar\zeta +a)^{3/2}}\ d^2\zeta
\equiv \int {\cal L}d^2\zeta
\end{array}\eeq
where solution $a(\zeta,\bar\zeta)$ for the equations of motion,
i.e. the shape of the minimal surface should be substituted,
subjected to the boundary $a(e^{i\phi},e^{-i\phi}) = 0$
at $\Pi$, and $\mu$ is a small IR-regularization parameter
(one can ignore it in solving the equations of motion, but
it is important for evaluating the minimal area, i.e. the action
itself).

This action implies the equation of motion in the form
\beq\new\begin{array}{c}
0 =
\partial\left(\frac{\partial {\cal L}}{\partial (\partial a)}\right)
+ \bar\partial\left(\frac{\partial {\cal L}}
{\partial (\bar\partial a)}\right) -
\frac{\partial{\cal L}}{\partial a} =
\frac{1}{4(1-\zeta\bar\zeta)^{3/2}}\left\{
\Delta_{NG}\Big(a(\zeta,\bar\zeta)\Big)
+ R(a;h,\bar h)\right\}
\end{array}\eeq
with the Nambu-Goto operator
\beq\new\begin{array}{c}
\Delta_{NG} = 4\partial\bar\partial
- \zeta^2\partial^2-2\zeta\bar\zeta\partial\bar\partial -
\bar\zeta^2\bar\partial^2
\label{NGO}
\end{array}\eeq
(it depends on the AdS metric and on the choice of the
circle solution $r_0^2 = 1-\zeta\bar\zeta$)
and with a somewhat sophisticated,
but straightforwardly derivable
expression $R(a;h,\bar h)$.

The equation of motion can be now iteratively expanded in powers
of $h$ and $\bar h$, as we did before in study of the double
contour integral:
\beq\new\begin{array}{c}\label{NGE}
\Delta_{NG}\Big(a^{(1)}(\zeta,\bar\zeta)
+ \bar\zeta h(\zeta) + \zeta\bar h(\bar\zeta)\Big) = 0, \\
\Delta_{NG}\Big(a^{(k)}(\zeta,\bar\zeta)
\Big) =
- R^{(k)}(a;h,\bar h) \ \ \ \ {\rm for} \ k\geq 2.
\end{array}\eeq
The superscripts here refer to the order in $h$, and
we used here the fact \cite{IMM2} that $R^{(0)} = R^{(1)} = 0$.
Since $R(a;h,\bar h)$ is a non-linear function
of all its arguments, all the components
$a^{(j)}$ with $1\leq j \leq k-1$ contribute to $R^{(k)}$.

The $h^k$-component $a^{(k)}$  contributes only to the
$h^{2k}$-component of the action, $A^{(2k)}$, since
$a^{(1)}+\ldots+a^{(k-1)}$ satisfies the equation of motion:
because of this, the $a^{(k)}$-linear term in
$$A\Big(a^{(1)}+\ldots+a^{(k-1)}+a^{(k)}\Big) =
A\left(a^{(1)}+\ldots+a^{(k-1)}\right)  +
a^{(k)}\frac{\delta A}{\delta a}\left(a^{(1)}+\ldots+a^{(k-1)}\right)
+ \ldots$$
is vanishing, and $a^{(k)}$ can contribute only
quadratically, i.e. only to $A^{(2k)}$.
In particular, $A^{(2)}$ and $A^{(3)}$ depend only on $a^{(1)}$
\cite{IMM2}, into $A^{(4)}$ and $A^{(5)}$ only $a^{(1)}$ and $a^{(2)}$
are contributing, $a^{(3)}$ first contributes to $A^{(6)}$ and so on.

It can be instructive to look at the polynomial example here.
Let us take for the action
$$
S(a,h) = -ah + \frac{1}{2}a^2\Big(1-\alpha h - \beta h^2 - \gamma h^3
+ \ldots\Big)
$$
Then, if one substitutes a solution of the equations of motion
expanded in powers of $h$,
$$
a = a_1 + a_2 + a_3 + a_4 + \ldots,
$$
one obtains
$$
S_{min} = \left(\frac{1}{2}a_1^2-a_1h\right)
+ \left(a_1a_2-\alpha a_1^2 h - a_2h\right)
+ \left(\frac{1}{2}a_2^2 + a_1a_3 - \alpha a_1a_2h
-\frac{1}{2}\beta a_1^2 h^2 - a_3h\right) + $$ $$
+ \left(a_2a_3 + a_1a_4 - \frac{1}{2}a_2^2 h - \alpha a_1a_2h
- \beta a_1a_2 h^2 - \frac{1}{2}\gamma a_1^2h^3 - a_4h\right)
+ O(h^6)
$$
The coefficients in front of $a_3$ and $a_4$ in this expressions
are respectively $(a_1-h) + (a_2-\alpha a_1h)$ and $(a_1-h)$.
However, these brackets are vanishing by definition of $a_1$ and $a_2$
which are the iterative solutions to the equations of motion
$$
a = \frac{h}{1-\alpha h - \beta h^2 - \gamma h^3 + \ldots}
= h + \alpha h^2 + (\beta+\alpha^2)h^3
+ (\gamma+2\alpha\beta+\alpha^3)h^4 + \ldots
$$

\subsection{Inverting Nambu-Goto operator}

Now we need a systematic way to solve the Nambu-Goto equations
with the non-vanishing right hand sides and with the
vanishing (Dirichlet) boundary conditions.

The zero modes of Nambu-Goto operator $\Delta_{NG}$, eq.(\ref{NGO})
are defined through
\beq\new\begin{array}{c}
g_k(\eta) = \frac{1+k\sqrt{1-\eta}}{(1+\sqrt{1-\eta})^k},
\ \ \ \ \ \
\tilde g_k(\eta) = \frac{1-k\sqrt{1-\eta}}{(1-\sqrt{1-\eta})^k}
= \frac{(1-k\sqrt{1-\eta})(1+\sqrt{1-\eta})^k}{\eta^k},\\
G_k(u) = g_k(1-u^2) = \frac{1+ku}{(1+u)^k}, \ \ \ \
\tilde G_k(u) = \tilde g_k(1-u^2) = \frac{1-ku}{(1-u)^k},\ \ \ \
k\geq 2
\end{array}\eeq
with $\eta=\zeta\bar\zeta$ and $u=\sqrt{1-\eta} = \sqrt{1-\zeta\bar\zeta}$
as follows:
\beq\new\begin{array}{c}
\Delta_{NG} \Big(\zeta^k g_k(\eta)\Big) = 0, \ \ \ \
\Delta_{NG} \Big(\zeta^k \tilde g_k(\eta)\Big) = 0
\end{array}\eeq
The solution to
\beq\new\begin{array}{c}
\Delta_{NG}\Big(\zeta^k F_k(\eta)\Big) = \zeta^k m_k(\eta),
\end{array}\eeq
which is finite at $\eta=0$ and satisfies $F_k(\eta=1)=0$,
is provided by the usual "variation-of-constants" method
\beq\new\begin{array}{c}
F_k(\eta) = -\tilde g_k(\eta) \int^{1}_{\sqrt{1-\eta}}
\frac{M_k(u)d\mu_k}{\tilde G_k(u)} +
g_k(\eta) \left\{\int^{1}_{\sqrt{1-\eta}}
\frac{M_k(u)d\mu_k}{G_k(u)} +
\int_0^1 \left(\frac{1}{\tilde G_k(u)}-\frac{1}{G_k(u)}\right)
M_k(u)d\mu_k\right\},
\end{array}\eeq
where $M_k(u) = m_k(1-u^2)$,
\beq\new\begin{array}{c}
d\mu_k=\frac{(1-k^2 u^2)du}{2k(k^2-1)u^2}
\end{array}\eeq
and each of the three integrals is convergent for
non-singular $M_k(\eta)$.
In the particular cases of $k=0$ and $k=1$,
these formulas are not directly applicable,
we have instead
\beq\new\begin{array}{c}
F_0(\eta) = \int^{\sqrt{1-\eta}}_{0}
\frac{u^2du}{1-u^2}\int_u^1\frac{M_0(u)du}{u^2}
\end{array}\eeq
and
\beq\new\begin{array}{c}
F_1(\eta) = \int^{\sqrt{1-\eta}}_{0}\frac{u^2du}{(1-u^2)^2}
\int^u_1\frac{M_1(u)(1-u^2)du}{u^2}
\end{array}\eeq

\subsection{The first terms of the $a(h)$ expansion}

Now one can easily solve (\ref{NGE}) and get
\beq\new\begin{array}{c}
a(\zeta,\bar\zeta) = \sum_{k=0}^\infty
\Big(h_{k+1}\zeta^{k} + \bar h_{k+1}\bar\zeta^{k}\Big)
\left(\frac{\Big(1+k\sqrt{1-\zeta\bar\zeta}\Big)
\Big(1-\sqrt{1-\zeta\bar\zeta}\Big)^{k}}
{(\zeta\bar\zeta)^{k}}
-\zeta\bar\zeta\right)
+  \\
+ \frac{1}{4}\sum_{k,l=0}^\infty \Big(h_{k+1}h_{l+1}\zeta^{k+l} +
\bar h_{k+1}\bar h_{l+1} \bar\zeta^{k+l}\Big)
\left(\frac{k(k+1)}{(1+\sqrt{1-\zeta\bar\zeta})^{k-1}}
+ \frac{l(l+1)}{(1+\sqrt{1-\zeta\bar\zeta})^{l-1}}
\ -\right. \\ \left.
-\ \frac{k(k+1)+l(l+1)+(k+1)(l+1)(k+l)\sqrt{1-\zeta\bar\zeta}
}
{(1+\sqrt{1-\zeta\bar\zeta})^{k+l}}\right)
+  \\
+ \sum_{k>l}^\infty (h_{k+1}\bar h_{l+1} \zeta^{k-l}
+ \bar h_{k+1} h_{l+1}\bar\zeta^{k-l})b_{kl}(z\bar z) +
\sum_k |h_{k+1}|^2 b_{kk}(z\bar z)
+ O(h^3,h^2\bar h,h\bar h^2,\bar h^3)
\end{array}\eeq
where
\beq\new\begin{array}{c}
b_{kl}(\eta) =
\left(\frac{k(k-1)}{2(1+u)^{k+1}}+\frac{l(l-1)}{2(1+u)^{l+1}}
-1\right)\eta^{l+1}
- \frac{k^2+l^2-2+(k+l)(kl-1)u}{2(1+u)^{k+l}}\,\eta^{l} + \\
+ \frac{(k+l)\big(1+(k-l)u\big)}{2(1+u)^{k-l}}
\end{array}\eeq
for $k\geq l$ and $\eta=1-u^2$.
In particular,
\beq\new\begin{array}{c}
b_{k0}(\eta) = g_{k}(\eta)-\eta = \frac{1+ku}{(1+u)^{k}}-\eta
\end{array}\eeq

\subsection{Expression for $A^{(1|1)}$, $A^{(2|1)}$ and $A^{(3|1)}$}

Now we are ready to present a few first expressions for the minimal area and the corresponding
double loop integral. As it was stated by the initial hypothesis, they coincide:

\beq\new\begin{array}{c}
A^{(1|1)}_{i} = \frac{i(i^2-1)}{6} = D^{(1|1)}_{i}\\
A^{(2|1)}_{ij} = \frac{(i+1)(j+1)}{12}\Big(i^2+j^2 + 3ij -
(i+j)\Big) = D^{(2|1)}_{ij} \label{A31} \end{array}\eeq
\beq\new\begin{array}{c}
A^{(3|1)}_{ijk} =
\frac{(i+1)(j+1)(k+1)}{18}\Big(i^2+j^2+k^2 + 3(ij+jk+ik) -
(i+j+k)\Big) = D^{(3|1)}_{ijk} \label{Vco} \end{array}\eeq

\subsection{Expression for $A^{(2|2)}$}

On the other hand, in this (fourth) order we already have an expression which is {\it different}
for the area and the double integral. Indeed, for $k\leq i,j\leq l$
\beq\new\begin{array}{c}
A_{ij|kl}^{(2|2)} = \delta_{i+j,k+l}
\frac{k+1}{48(i+j-1)(i+j+1)} \times  \\ \times
\Big\{2ij(i^4+5i^3j+8i^2j^2+5ij^3+j^4)+2(i+j)^5
-2(k^2-k+1)i^2j^2 +  \\
+k^2(k^2+k-2)(i^2-ij+j^2)
+(3k^4+3k^3-10k^2+4k-2)ij-k^2(k^2+k-2)-  \\
-\frac{1}{i+j}\Big(
2(k^3+k^2-2k+2)(i^4+j^4)
+(7k^3+9k^2-16k+16)ij(i^2+j^2)
+(9k^3+15k^2-24k+24)i^2j^2 -  \\
-2(k^3+k^2-2k+1)(i^2+j^2)
-(5k^3+3k^2-8k+4)ij\Big) \Big\}
\ \ \times h_{i+1}h_{j+1}\bar h_{k+1}\bar h_{l+1}
\end{array}\eeq
while
\beq\new\begin{array}{c}
D^{(2|2)}_{ij|kl} = \delta_{i+j,k+l}
\frac{1}{24}\Big((i+1)(j+1)(k+1)(i^2+3ij+j^2-i-j) -  \\
- (i+j+2)(k+2)(k+1)k(k-1)
+ \frac{3}{5}(k+3)(k+2)(k+1)k(k-1)\Big)
\ \ \times h_{i+1}h_{j+1}\bar h_{k+1}\bar h_{l+1}
\end{array}\eeq
These expressions do not coincide unless one of the indices is equal to 1, i.e. the expansion
contains $h_2$ or $\bar h_2$. In this latter case $A$ and $D$ coincide, as a corollary of the
conformal invariance we discuss in the next section.

Note that these formulas at particular values of indices read as
\beq\new\begin{array}{c}
A_{ij|ij}^{(2|2)} =
\frac{i+1}{48(i+j-1)(i+j)(i+j+1)}\Big(
i(-i^6 -2i^5j +4i^4j^2 +18i^3j^3 +24i^2j^4 +12ij^5 +2j^6) +\\
+(i^6 +6i^5j +20i^4j^2 +34i^3j^3 +28i^2j^4 +12ij^5 +2j^6)
+i(3i^4 +10i^3j +14i^2j^2 +12ij^3 +4j^4)-  \\
-(3i^4 +10i^3j +18i^2j^2 +16ij^3 +4j^4)
-2i(i^2  +4ij + +3j^2)
+2(i+j)^2
\Big)\\
\ \ \times |h_{i+1}|^2|h_{j+1}|^2,\ \ \ \ i\leq j
\end{array}\eeq
and
\beq\new\begin{array}{c}
A^{(2|2)}_{ii|ii} =
\frac{i(i+1)(57i^5+103i^4+43i^3-51i^2-16i+8)}
{96(2i-1)(2i+1)} \ \  \times |h_{i+1}|^4
\end{array}\eeq
while
\beq\new\begin{array}{c}
D^{(2|2)}_{ii|ii} =
\frac{i(i+1)(9i^3-16i^2+9i-4)}{60} \ \  \times |h_{i+1}|^4
\end{array}\eeq
In the following section we are going to describe how the conformal invariance governs this discrepancy.

\section{Conformal invariance}
The structure of $AdS_5$ implies the conformal invariance with generators that form the $SL(2)$-algebra.
In terms of the $h$-parameters calculated in \cite{IMM2} this generators represent a set of
first Virasoro-like constraints
\beq\new\begin{array}{c}
\hat{J}_{-}=\frac{\partial}{\partial h_0}\\
\hat{J}_0=\frac{\partial}{\partial h_1}+\sum\limits_{k=0}^{\infty}h_k\frac{\partial }{\partial h_k}\\
\hat{J}_{+}=\frac{\partial}{\partial h_2}+2\sum_{k=0}^{\infty} h_k \frac{\partial}{\partial h_{k+1}}+\sum_{k,m=0}^{\infty}h_k h_m \frac{\partial}{\partial h_{k+m}}
\end{array}\eeq
Since the minimal area and the double contour integral are equivalent up to the third order,
it deserves investigating the both sides from this point of view.

\subsection{Invariance of the double contour integral}

Consider the action of the global transformation on the double loop integral.
Under the fractional linear $SL(2,R)$ transformation
$z \rightarrow \frac{az+b}{cz+d}$, the ingredients of the
{\it planar} double-loop integral
change as follows
$$
dz_1 d\bar z_2 \rightarrow \frac{dz_1d\bar z_2}
{(cz_1+d)^2(c\bar z_2+d)^2}, \ \ \ \ \
d\bar z_1 dz_2 \rightarrow \frac{d\bar z_1dz_2}
{(c\bar z_1+d)^2(cz_2+d)^2}, \ \ \ \ \
|z_1-z_2|^{-2} \rightarrow |cz_1+d|^2|cz_2+d|^2|z_1-z_2|^{-2},
$$
so that
\beq\new\begin{array}{c}
D \ \stackrel{(\ref{DLI})}{=} \ \frac{1}{2}
\oint\oint \frac{dz_1 d\bar z_2 + d\bar z_1 dz_2}{|z_1-z_2|^2}
\ \longrightarrow \   \\ \longrightarrow \
\frac{1}{2}
\oint\oint \frac{dz_1 d\bar z_2}{|z_1-z_2|^2}\cdot
\frac{c\bar z_1+d}{cz_1+d}\cdot\frac{cz_2+d}{c\bar z_2+d} +
\frac{1}{2}
\oint\oint \frac{d\bar z_1 dz_2}{|z_1-z_2|^2}\cdot
\frac{cz_1+d}{c\bar z_1+d}\cdot\frac{c\bar z_2+d}{cz_2+d}
\end{array}\eeq
It is not very obvious from this formula that $D$
is invariant. However, it actually is.
The simplest way to see this is to look at an infinitesimal
$SL(2,R)$ transformation, picking up the contribution
to the variation of $D$, which is linear in  $d-1$ and $c$.
Then, the variation of $D$ is
\beq\new\begin{array}{c}
\delta D = -\frac{c}{2}\oint\oint
\left(\frac{1}{z_1-z_2} - \frac{1}{\overline{z_1-z_2}}\right)
\Big(d\bar z_1 dz_2 - dz_1d\bar z_2\Big) \
 = c \oint\oint  \frac{y\left(dydX-dxdY\right)}{x^2 +y^2},
\end{array}\eeq
where we denoted $z_1, z_2$ as $z_{1,2} = x_{1,2}+iy_{1,2}$
and $x=x_1-x_2$, $y=y_1-y_2$, $X=x_1+x_2$, $Y=y_1+y_2$.
The integrand is then a total derivative.
In fact,  the above integrals are divergent and
formal manipulations make sense only after regularization,
which, however, breaks the invariance and {\it an anomaly}
 shows up when the regularization is removed:
\beq\new\begin{array}{c}
\delta D = \delta \left(\frac{1}{2}
\oint\oint \frac{dz_1 d\bar z_2 + d\bar z_1 dz_2}
{|z_1-z_2|^2+\lambda^2} \right) =
\frac{\pi}{4\lambda} \delta L
\end{array}\eeq
where $L=L_\Pi$ is the length of the curve $\Pi$.
However, $D^{reg} = D - \pi\frac{L}{4\lambda}$ remains fully
invariant \cite{IMM2}.

\subsection{Invariance of the minimal area}
The initial expression for the minimal area (\ref{SNG}) is obviously invariant under the action
of such generators because of its construction. Introducing the $\mu$-regularization seems to
break this symmetry. Actually, if one considers the action of generators which transforms a
planar contour into another planar contour
\beq\new\begin{array}{c}
r\rightarrow r(1-\gamma \bar z -\bar \gamma z)\\
z\rightarrow z(1-\bar \gamma z)+\gamma r^2\\
\bar z\rightarrow \bar z(1-\gamma \bar z)+\bar \gamma r^2
\end{array}\eeq
the $\mu$-regularization obviously breaks the invariance (in terms of (\ref{plan-NGA}))
\beq\new\begin{array}{c}
\int d^2z \mathcal{L}(z,\bar z,r)\rightarrow \int d^2 z \mathcal{L}(z,\bar z,r)
\left(1-\mu^2\frac{\gamma \bar z+\bar\gamma z}{r^2+\mu^2}\right)
\end{array}\eeq
After regularization, the status of the area invariance is similar to the status of the double
loop integral: on one hand, the expression for the area behaves as $\sim L/\mu+\ldots$,
the violation is proportional to $\mu^2$ and it does not contribute to the regularized part
of the area; on the other hand, the additional divergency arises in the denominator, and this can
cause some troubles.

\bigskip

An explicit calculation of the minimal area in h-terms shows that the conformal invariance
persists. Moreover, it seems to be deeply related to the equality of terms that contain
$h_2$ or $\bar h_2$ at the fourth order of the both sides of (\ref{AMh}).
The invariance arguments can be directly applied to the fifth order contributions into the area,
the corresponding term can be written as follows
\beq\new\begin{array}{c}
A^{(5)}=\frac{3}{2}\sum\limits_{i,j,k,l,m}\left(X_{ijk;lm} h_{i+1} h_{j+1} h_{k+1} \bar h_{l+1} \bar h_{m+1}+Y_{ijkl;m} h_{i+1} h_{j+1} h_{k+1} h_{l+1}\bar h_{m+1}\right)+h.c.
\end{array}\eeq
with the additional condition
\beq\new\begin{array}{c}
X_{ijk;lm}\sim \delta_{i+j+k,l+m}\qquad Y_{ijkl;m}\sim \delta_{i+j+k+l,m}
\end{array}\eeq
One can easily construct a relation which must be satisfied
\beq\new\begin{array}{c}
3!A^{(5)}_{1jk;lm}-2(2! A^{(4)}_{j(k+1);lm}+2!A^{(4)}_{(j+1)k;lm})+2A^{(3)}_{j+k+1;lm}=0
\end{array}\eeq
This relation between coefficients is not universal, it depends on number of coinciding indices in
various terms. To present a dramatic example, consider one of such relations
\beq\new\begin{array}{c}
3A^{(5)}_{112;22}-2(A^{(4)}_{13;22}+A^{(4)}_{22;22})+A^{(3)}_{4;22}=0
\end{array}\eeq
(note that $A^{(5)}_{22;22}$ does not coincide with $D^{(5)}_{22;22}$.)
This relation allows one to anticipate, e.g., the coefficient
$A^{(5)}_{112;22}=\frac{419}{30}$. One can check that this coefficient is, indeed, given by this
number.

\bigskip

In fact, the conformal invariance allows one to fix numerous terms at the both sides.
Consider, for instance, formula (\ref{Vco}). One can expect it is a symmetric
polynomial of degree three in all indices. Actually one knows
it is zero if any one of the indices is
equal to minus one (which corresponds to presence of $h_0$), just since the generator $\hat J_-$
should cancel it due to the conformal invariance.
Therefore, a generic form of this expression is
\beq\new\begin{array}{c}
A^{(3|1)}=\alpha(i+1)(j+1)(k+1)(i^2+j^2+k^2+\beta(ij+jk+ik)+\gamma(i+j+k)+\delta)
\end{array}\eeq
Further, the conformal invariance gives two more equations for $k=0$ and $k=1$:
\beq\new\begin{array}{c}
\alpha(i^2+j^2+\beta ij+\gamma(i+j)+\delta)=\frac{1}{18}(i^2+j^2+3ij-(i+j))\\
2\alpha(i^2+j^2+1+\beta(ij+j+i)+\gamma(i+j+1)+\delta)=\frac{2}{18}(i^2+j^2+3ij+2(i+j))
\end{array}\eeq
These equations allow one to unambiguously define all the coefficients and to restore
the expression not only for $A^{(1...3|1)}$, but for all $A^{(n|1)}$ as well.
Hence, the assumptions of the
polynomial structure and of the conformal invariance of the
regularized minimal area and the double loop integral allow one to claim that the
$h$-linear terms (or $\bar h$-linear ones) in the both sides of (\ref{AMh}) coincide with
each other. Note, however, that the assumption of polynomial structure fails for $A^{(n|m)}$
with min$(m,n)\ge 2$ (though remains true for $D^{(n|m)}$).

\subsection{Towards an explicit "non-perturbative" derivation of the area}

Since the Alday-Maldacena conjecture fails, one has to correct it in some way.
At the moment, one can foresee, at least, what could be elementary building blocks for corrections.

Indeed, the consideration above establishes that $A^{(\cdot|1)}=D^{(\cdot|1)}$.
Then, as in section 4.2.3, one can sum up the $\bar h$-linear terms of the {\it minimal area}
\beq\new\begin{array}{c}
A^{(\cdot|1)} \equiv
\sum_{m=1}^\infty A^{(m|1)}_{k_1\ldots k_m|k_1+\ldots+k_m}
h_{k_1+1}\ldots h_{k_m+1}\bar h_{k_1+\ldots+k_m+1} =\frac{1}{6}
\oint \bar h(\bar \zeta)S_\zeta(z)\zeta^2 d\zeta
\end{array}\eeq
This does not come as a surprise that the result is expressed through the conformal
invariant Schwarzian derivative. Moreover, if one anticipates that it enters linearly, the answer
is completely fixed up to a coefficient. This explains why $A^{(\cdot|1)}=D^{(\cdot|1)}$.

To construct
further $A$, one has to introduce higher conformal structures into the game.
One can directly make certain that the following structures are invariant under the rational
transformation, which reflects action of the conformal $SL(2)$-algebra:
\beq\new\begin{array}{c}
S(z) = \frac{z'''}{z'} - \frac{3}{2}\left(\frac{z''}{z'}\right)^2
\end{array}\eeq
\beq\new\begin{array}{c}
S_2(z_1,z_2)\sim\frac{dz_1 dz_2}{(z_1-z_2)^2} =
\frac{d\zeta_1d\zeta_2}{(\zeta_1-\zeta_2)^2}
\times\\ \times
\left(1 + \sum_k kh_k\zeta_1^{k-1}\right)
\left(1 + \sum_k kh_k\zeta_2^{k-1}\right)
\left\{1 + \sum_{m=1}^\infty (-)^m(m+1)
\left(h_k\frac{\zeta_1^k-\zeta_2^k}{\zeta_1-\zeta_2}\right)^m\right\}
\end{array}\eeq
\beq\new\begin{array}{c}
S_3(z_1,z_2,z_3)\sim\frac{dz_1 dz_2 dz_3}{(z_1-z_2)(z_2-z_3)(z_3-z_1)} =
\frac{d\zeta_1d\zeta_2d\zeta_3}
{(\zeta_1-\zeta_2)(\zeta_2-\zeta_3)(\zeta_3-\zeta_1)}\times\\
\times\left(1 + \sum_k kh_k\zeta_1^{k-1}\right)
\left(1 + \sum_k kh_k\zeta_2^{k-1}\right)
\left(1 + \sum_k kh_k\zeta_3^{k-1}\right)
\cdot
\end{array}\eeq
These expressions should serve as building blocks for higher $A$, which becomes
different from $D$ already in the $\bar h^2$-term, i.e. at the
level of non-perturbative $A^{(\cdot|2)}$. However, a close inspection of this
difference as well as of the invariants is beyond the scope of the present paper.

\section{Non-planar case}
\subsection{Equations}

To go beyond the planar case, i.e. to lift the last requirement in (\ref{circ}),
we switch on one of the coordinate fields
that was put zero before (for the sake of definiteness, we choose $y_0$).
Thus, we add its contributions to the area and to the loop integral
\beq\new\begin{array}{c}
A=\int \frac{\sqrt{-8\partial r \bar \partial r \partial y_0 \bar \partial y_0 +4 \partial r \bar \partial r \partial H \bar \partial \bar H-4 \partial y_0\bar \partial y_0\partial H\bar \partial\bar H+4(\partial r)^2(\bar \partial y_0)^2+4(\bar \partial r)^2(\partial y_0)^2+(\partial H \bar \partial \bar H)^2}}{r^2+\mu^2}\\
D=\frac{1}{4}\oint\oint \frac{\frac{1}{2}(dz d\bar z'+dz'd\bar z)-dy_0 dy_0'}{(z-z')(\bar z-\bar z')-(y_0-y_0')^2}
\end{array}\eeq
As previously, we use the gauge invariance to fix the fields $y_1$ and $y_2$
as in (\ref{RiemPar}). The corresponding Euler-Lagrange equations is a set of two non-linear
equations for the fields $r$ and $y_0$. In contrast to the boundary condition
$\left.r\right|_{\zeta\bar\zeta=1}=0$, we consider for $y_0$
\beq\new\begin{array}{c}\label{npbc}
\left.y_0(\zeta,\bar \zeta)\right|_{\zeta\bar\zeta=1}=\sum\limits_{k=0}^{\infty} q_k \zeta^k+\sum\limits_{k=1}^{\infty}q_{-k}\bar\zeta^{k}
\end{array}\eeq
which is nothing but the Fourier expansion of an arbitrary boundary condition.
We again develop an iterative procedure, looking now for formal expansion in powers of both the
$h$- and $q$-parameters. Beyond the boundary (in the bulk) one has
\beq\new\begin{array}{c}
r(\zeta,\bar\zeta)=\sqrt{1-\zeta\bar\zeta+a^{(1)}(\zeta,\bar\zeta)+a^{(2)}(\zeta,\bar\zeta)+\ldots}\\
y_0(\zeta,\bar\zeta)=b^{(1)}(\zeta,\bar\zeta)+b^{(2)}(\zeta,\bar\zeta)+\ldots
\end{array}\eeq
and the NG equations convert into
\beq\new\begin{array}{c}
\Delta_{NG}(a^{(1)}(\zeta,\bar \zeta)+h(\zeta)\bar\zeta+\bar h(\bar\zeta)\zeta)=0\\
\label{b-eq}\Delta_{NG}b^{(1)}(\zeta,\bar \zeta)=0\\
\Delta_{NG}a^{(k)}(\zeta,\bar \zeta)=-R^{(k)}(a,b;h,\bar h)\\
\Delta_{NG}b^{(k)}(\zeta,\bar \zeta)=-Q^{(k)}(a,b;h,\bar h)
\end{array}\eeq
One can easily construct the solution of the second equation in (\ref{b-eq})
which satisfies the boundary conditions
(\ref{npbc})
\beq\new\begin{array}{c}
b^{(1)}(\zeta,\bar\zeta)=\sum\limits_{k=0}^{\infty} q_k \zeta^k g_{k}(\zeta\bar\zeta)+\sum\limits_{k=1}^{\infty}q_{-k}\bar\zeta^{k}g_k(\zeta\bar\zeta)
\end{array}\eeq
Then, all other iterations must satisfy the trivial boundary condition
$\left.b^{(k)}\right|_{\zeta\bar\zeta=1}=0,\quad k\geq 2$.

\subsection{Explicit results}

The first $q$-dependent contributions to the regularized area and to the double loop integral
are
\beq\new\begin{array}{c}
\frac{A}{3\pi}=\frac{D}{\pi^2}=\sum\limits_{k=1}^{\infty}\frac{2}{3}k(k^2-1)q_kq_{-k}+\sum\limits_{i,j\geq 1}\frac{ij(i^2+j^2+3(i+j)-2)}{6}\left(q_i q_j\bar h_{i+j+1}+q_{-i}q_{-j}h_{i+j+1}\right)-\\
-\frac{(i+1)j(i^2+2j^2+3ij+2i-2)}{3}\left(h_{i+1} q_j q_{-i-j}+\bar h_{i+1} q_{-j}q_{i+j}\right)
\end{array}\eeq
Therefore, up to the third order, $A$ and $D$ coincide.

In the fourth order, the situation is much similar to the planar case.
For instance, the both expressions are of the form
\beq\new\begin{array}{c}
A,D\sim T_{ij;kl}q_i q_j q_{-k} q_{-l} \delta_{i+j,k+l}+Y_{ijk} (q_i q_j q_k q_{-i-j-k}+q_{-i}q_{-j}q_{-k}q_{i+j+k})
\end{array}\eeq
where the coefficients $T_{ij;kl}$ are generally different, while
\beq\new\begin{array}{c}
Y_{ijk}\sim ijk\left(\frac{2}{3}(i^2+j^2+k^2)+(ij+kj+jk)-2\right)
\end{array}\eeq
coincide for the both sides. This expression must be fixed by the conformal invariance as
in the planar case.

\subsection{Conformal invariance in the non-planar case}

$AdS_{d+1}$ space can be considered as a quadric (hyperboloid)
in $R^{d+2}$, defined by
\beq\new\begin{array}{c}
Y_+Y_- + \vec Y^2 = 1,
\label{embed}
\end{array}\eeq
which can be conveniently parameterized by
\beq\new\begin{array}{c}
\vec y = \frac{\vec Y}{Y_+}, \ \ \ r=\frac{1}{Y_+},  \\
\vec Y = \frac{\vec y}{r}, \ \ \ Y_+ = \frac{1}{r}, \ \ \
Y_- = \frac{r^2-\vec y^2}{r}
\end{array}\eeq
The metric on $AdS_{d+1}$, induced by this embedding from the
flat one, has the Poincar\'e form
\beq\new\begin{array}{c}
dY_+dY_- + d\vec Y^2 = \frac{d\vec y^2 - dr^2}{r^2}
\label{AdSmet}
\end{array}\eeq
The symmetry transformations from $SO(d,2)$ act linearly
on the $(d+2)$-component vector $(Y_+,Y_-,\vec Y)$,
\beq\new\begin{array}{c}
\delta\left(\begin{array}{c} Y_- \\ Y_+ \\ \vec Y \end{array}\right)=
\left(\begin{array}{ccc} b & 0 & 2\vec\beta\\ 0 & -b& 2\vec\gamma
\\ -\vec\gamma & -\vec\beta & B \end{array}\right)
\left(\begin{array}{c} Y_- \\ Y_+ \\ \vec Y \end{array}\right)
\end{array}\eeq
Consider the action of the $\vec{\gamma}$-generator on $AdS_5$:
\beq\new\begin{array}{c}
\delta Y_-=0, \quad \delta Y_+=2(\vec{\gamma}\,\vec{Y}),\quad \delta \vec{Y}=-\vec{\gamma}\,Y_-
\end{array}\eeq
Expressing $Y$'s via the coordinates on $AdS_5$, one can immediately derive the transformations
of these latter:
\beq\new\begin{array}{c}
\delta r=-2(\vec{\gamma}\,\vec{y})r\\
\delta \vec{y}=-2\vec{y}\,(\vec{\gamma}\,\vec{y})+\vec{\gamma}\,\vec{y}\,^2-\vec{\gamma}r^2
\end{array}\eeq
This means that the two planar generators can be realized at the boundary ($r=0$) as
\beq\new\begin{array}{c}
\hat{\gamma}_1=2y_0y_1\frac{\partial}{\partial y_0}+(y_1^2+y_0^2-y_2^2)\frac{\partial}{\partial y_1}+2y_1 y_2\frac{\partial}{\partial y_2}\\
\hat{\gamma}_2=2y_0y_2\frac{\partial}{\partial y_0}+2y_1 y_2\frac{\partial}{\partial y_1}+(y_2^2+y_0^2-y_1^2)\frac{\partial}{\partial y_2}
\end{array}\eeq
or, equivalently,
\beq\new\begin{array}{c}
\hat{\bar{\gamma}}=\hat{\gamma}_1+i\hat{\gamma}_2=2y_0(y_1+i y_2)\frac{\partial}{\partial y_0}+y_0^2\left(\frac{\partial}{\partial y_1}+i\frac{\partial}{\partial y_2}\right)+\left(y_1^2-y_2^2+2iy_1y_2\right)\frac{\partial}{\partial y_1}+
\\+\left(2y_1y_2+i y_2^2-i y_1^2\right)\frac{\partial}{\partial y_2}=2y_0z \frac{\partial}{\partial y_0}+2y_0^2\frac{\partial}{\partial \bar z}+z^2\left(\frac{\partial}{\partial y_1}-i\frac{\partial}{\partial y_2}\right)=\\
=2\left(y_0 z\frac{\partial}{\partial y_0}+y_0^2\frac{\partial}{\partial \bar z}+z^2\frac{\partial }{\partial z}\right)
\end{array}\eeq

Let us now check the invariance of the double contour integral \beq\new\begin{array}{c}
D=\oint\oint\frac{(dy dy')}{(y-y')^2+\lambda^2} \end{array}\eeq w.r.t. the
action of the generator $\vec{\hat{\gamma}}$: \beq\new\begin{array}{c} \delta
D=2\oint\oint\frac{(dy d\delta
y')}{(y-y')^2+\lambda^2}-2\oint\oint\frac{(dy d
y')(y-y',\delta(y-y'))}{\left[(y-y')^2+\lambda^2\right]^2}=
\\=
2\oint\oint\frac{dy(-2dy'(\gamma y')-2y'(\gamma dy')+2\gamma(y'\,dy'))}{(y-y')^2+\lambda^2}+2\oint\oint\frac{(y-y')^2(y+y',\gamma)}{\left[(y-y')^2+\lambda^2\right]^2}=\\
=-4\lambda^2\oint\oint\frac{(dy\,dy')(y'\gamma)}{\left[(y-y')^2+\lambda^2\right]^2}+4\oint\oint\frac{(\gamma dy')((y-y')dy)}{(y-y')^2+\lambda^2}=\\
=-4\lambda^2\oint\oint\frac{(dy\,dy')(y'\gamma)}{\left[(y-y')^2+\lambda^2\right]^2}+2\oint
(\gamma dy')\oint d\log \left[(y-y')^2+\lambda^2\right] \end{array}\eeq The
second term is a total derivative and can be neglected. The first
term seems to tend to zero since $\lambda\rightarrow 0$, on the
other hand there is an additional divergency in the denominator and
the invariance can not be claimed as obvious.

Similarly, other generators of the conformal transformations are (the $SO(d)$ rotations
omitted):
\beq\new\begin{array}{c}\label{generators}
\hat\beta_0: \hfill\frac{\partial}{\partial y_0}\ , \\
\hat\beta:  \hfill\frac{\partial}{\partial z}\ , \\
\hat{\bar\beta}: \hfill \frac{\partial}{\partial\bar z}\ ,  \\
\hat b: \hfill y_0\frac{\partial}{\partial y_0} + z\frac{\partial}{\partial z}
+ \bar z\frac{\partial}{\partial\bar z}\ , \\
\hat\gamma_0: \hfill \phantom{acd}(z\bar z+y_0^2)\frac{\partial}{\partial y_0}
+2y_0\left(z\frac{\partial}{\partial z} +
\bar z\frac{\partial}{\partial\bar z}\right), \\
\hat\gamma: \hfill y_0\bar z\frac{\partial}{\partial y_0} +
y_0^2\frac{\partial}{\partial z}
+ \bar z^2\frac{\partial}{\partial\bar z}\ , \\
\hat{\bar\gamma}: \hfill y_0z\frac{\partial}{\partial y_0} +
z^2\frac{\partial}{\partial z} +
\underline{y_0^2\frac{\partial}{\partial \bar z}} \end{array}\eeq

One can represent these generators in the $h$,$q$-variables using
the following relations
\beq\new\begin{array}{c}
\frac{\partial \mathcal{F}}{\partial h_k}=\oint ds\,\zeta^k \frac{\delta \mathcal{F}}{\delta z},\quad
\ \ \ \ \ \ \frac{\partial \mathcal{F}}{\partial q_k}=\oint ds\, \zeta^k \frac{\delta \mathcal{F}}{\delta y_0}
\end{array}\eeq
Then, one obtains
\beq\new\begin{array}{c}\label{gen_ind}
\hat b=\frac{\partial}{\partial h_1}+\frac{\partial }{\partial \bar h_1}+\sum\limits_{k=1}^{\infty} \bar h_k \frac{\partial}{\partial \bar h_k}+\sum\limits_{k=-\infty}^{\infty}q_k \frac{\partial }{\partial q_k}\\
\hat {\bar \gamma}=\frac{\partial}{\partial h_2}+2\sum\limits_{k=1}^{\infty}h_k\frac{\partial}{\partial h_{k+1}}+\sum\limits_{k,m\geq 1}h_k h_m\frac{\partial}{\partial h_{k+m}}+\sum\limits_{k=-\infty}^{\infty}q_k \frac{\partial}{\partial q_{k+1}}+\sum\limits_{k,m}q_k h_m \frac{\partial}{\partial q_{k+m}}+\\
+\sum\limits_{k,m\geq 0} q_{-k}q_{-m}\frac{\partial}{\partial \bar h_{k+m}}+2\sum\limits_{k\geq m}q_mq_{-k}\frac{\partial}{\partial \bar h_{k-m}}
\end{array}\eeq

Now let us see how the conformal invariance works in the non-planar
case. To this end, consider the following terms
\beq\new\begin{array}{c}\label{MR} \sum_k
M_k q_kq_{-k} + \sum_{i,j}R_{ij}(h_{i+1}q_jq_{-i-j} + \bar
h_{i+1}q_{-j}q_{i+j})+ \sum A_k h_{k+1}\bar h_{k+1} + \sum_{k,m}
B_{k,m}q_{-k-m-1}q_{m}h_{k+1}h_2
\end{array}\eeq
They should (and do) satisfy
the non-planar-non-planar relations
\beq\new\begin{array}{c}\label{104}
2R_{0j} + 2M_{j}=0,  \\
R_{1j} + M_{j+1} + M_j=0
\end{array}\eeq
where 2 at the l.h.s. of the first formula comes from
the action of $\partial/\partial h_1$ and $\partial/\partial\bar h_1$,
which are both present in ${\hat b}$.

Less trivial example arises when one acts with the generators onto (\ref{MR}) and compare coefficients
in front of $q_{-k-m}q_{m}h_k$. This leads to the following relation for the coefficients
\beq\new\begin{array}{c}\label{confexp}
B_{k-1,m}+2 R_{k,m}+R_{k-1,m}+R_{k-1,m+1}+M_{k+m}+M_m+2A_{k-1}=0
\end{array}\eeq
The expression for $B$ can be explicitly found and reads
\beq\new\begin{array}{c}\label{106}
B_{k,m}=\frac{2}{3}m(k+1)(2k^2+6km+7k+4m^2+6m+2)
\end{array}\eeq
Formula (\ref{confexp}) is also checked to be valid.

Note that this derivation of formulas for the conformal generators
is, in fact, a bit naive. In principle, one should take into account
that that the conformal transformations change the gauge-fixing
condition, as one can already see from (\ref{generators}). Indeed,
the generators $\hat \gamma$'s explicitly mix purely holomorphic
$z$-coordinates with $y_0$ which contains anti-holomorphic parts.
Still, eqs.(\ref{104})-(\ref{106}) demonstrate that the generators (\ref{gen_ind})
are adequate, at least, in these examples.

\subsection{Non-planar generalization of Schwarzian}

Now we are going to make an attempt to construct the same relation in invariant terms and
introduce a kind of "non-planar" Schwarzian. To this end, we add to $(z,\bar z)$ the holomorphic
and anti-holomorphic parts of $y_0$, $(y,\bar y)$ so that under the conformal transformation
\beq\new\begin{array}{c}\label{pro}
z\rightarrow \frac{az+b}{cz+d},\ \ \ \ \ \ \  y \rightarrow \frac{y}{cz+d}
\end{array}\eeq
Then
\beq\new\begin{array}{c}
z' \rightarrow \frac{z'}{cz+d},  \\
z'' \rightarrow \frac{z''}{(cz+d)^2} - \frac{2c(z')^2}{(cz+d)^3}, \\
z''' \rightarrow \frac{z'''}{(cz+d)^2}
-\frac{6cz'z''}{(cz+d)^3} + \frac{6c^2(z')^3}{(cz+d)^4}
\end{array}\eeq
and
\beq\new\begin{array}{c}
y' \rightarrow \frac{y'}{cz+d} - \frac{cyz'}{(cz+d)^2}, \\
y'' \rightarrow \frac{y''}{cz+d} - \frac{2cy'z'}{(cz+d)^2}
- \frac{cyz''}{(cz+d)^2} + \frac{2c^2y(z')^2}{(cz+d)^3},  \\
y''' \rightarrow \frac{y'''}{cz+d} - \frac{3cy''z'}{(cz+d)^2}
-\frac{3cy'z''}{(cz+d)^2} + \frac{6c^2y'(z')^2}{(cz+d)^3}
-\frac{cyz'''}{(cz+d)^2} + \frac{6c^2yz'z''}{(cz+d)^3}
-\frac{6c^3y(z')^3}{(cz+d)^4}
\end{array}\eeq
It follows that
\beq\new\begin{array}{c}
S(z) = \frac{z'''}{z'} - \frac{3}{2}\left(\frac{z''}{z'}\right)^2
\rightarrow S(z)
\end{array}\eeq
is invariant, while
\beq\new\begin{array}{c}
\sigma_2 (y,z) = \left(\frac{y''}{z'} - \frac{y'z''}{(z')^2}\right)
\rightarrow
(cz+d)\sigma_2 (y,z)
\end{array}\eeq
is projective invariant, as well as
\beq\new\begin{array}{c}\label{boxed}
\sigma_3 (y,z) = \left(\frac{y'''}{z'} - \frac{3}{2}\frac{y''z''}{(z')^2}
- \boxed{1}\frac{y'z'''}{(z')^2} + \boxed{\frac{3}{2}}\frac{y'(z'')^2}{(z')^3}\right)
\rightarrow (cz+d) \sigma_3(y,z)
\end{array}\eeq
In a little more detail,
\beq\new\begin{array}{c}
\frac{y'''}{z'} - \frac{3}{2}\frac{y''z''}{(z')^2}
\rightarrow (cz+d)
\left(\frac{y'''}{z'} - \frac{3}{2}\frac{y''z''}{(z')^2}\right)
- cy S(z)
\end{array}\eeq
and this is combined into invariant with $-\frac{y'}{z'}S(z)$
with Schwarzian $S(z)$ invariant and $\frac{y'}{z'} \rightarrow
(cz+d)\frac{y'}{z'} - cy$.

This could imply that the $\bar y$-linear part of the exact answer
is just
\beq\new\begin{array}{c}\label{ndi}
\oint \bar y \Big( \sigma_3(y,z)\zeta^2 + 3 \sigma_2(y,z)\zeta\Big)d\zeta
\end{array}\eeq
However, this is an invariant of transformations (\ref{pro}), i.e. of
\beq\new\begin{array}{c}\label{cgen}
\frac{\partial}{\partial z},  \\
z\frac{\partial}{\partial z} + \frac{1}{2}y\frac{\partial}{\partial y},\\
z^2\frac{\partial}{\partial z} + yz\frac{\partial}{\partial y}
\end{array}\eeq
which are different from (\ref{generators}). Especially important is the lack of the underlined
term from (\ref{generators}) in (\ref{cgen}). Therefore, (\ref{ndi}) is not equal to the
double integral. However, the difference is minor (it is rather simple to verify that (\ref{116})
simulates equation (\ref{confexp})):
\beq\new\begin{array}{c}\label{116}
D=\frac{2}{3}\pi^2\frac{1}{2\pi i}\oint dx \bar y(\bar x)\left(x^2\left(\frac{y'''}{z'}-\frac{3}{2}\frac{y''z''}{z'^2}+\frac{3}{4}\frac{y' z''^2}{z'^3}-\frac{1}{2}\frac{y' z'''}{z'^2}\right)+3x\left(\frac{y''}{z'}-\frac{y'z''}{z'^2}\right)\right)-
\\-\frac{\pi^2}{6}\frac{1}{2\pi i}\oint x^2 dx \bar z(\bar x) S(z)+\ldots
\end{array}\eeq
i.e. the two boxed coefficients in (\ref{boxed}) are actually modified by a factor of $1/2$.
This of course violates the invariance under (\ref{cgen}), but, in fact, this is exactly
what is necessary to ensure the invariance under (\ref{generators}): the change of
coefficients is fully compensated by the change of $\bar z$ by $y^2$ in the term
$\oint\bar z S(z)\zeta^2 d\zeta$ (the missed underlined term in (\ref{cgen})).

\section{Conclusion}

In this paper we applied the constructive approach of
\cite{IMM2},\cite{IMM1}-\cite{M}
and evaluated the regularized area of AdS minimal surface
bounded by a wavy circle at the AdS infinity up to the 4-th
order in deviations $h,\bar h$ from the ideal circle.
We confirm the hypothesis of \cite{IMM2,M} that infinitely many
terms of the form $h^m\bar h$ and $h\bar h^n$ coincide
at the l.h.s. and the r.h.s. of (\ref{AMh}).
However, the $4$-th order involves also the terms $h^2\bar h^2$,
the first ones where the real violation of Alday-Maldacena
hypothesis (\ref{AMh}) can be expected, since these
terms are not controlled by the conformal invariance
\cite{dks,dhks1,dhks3,Komar,IMM2}, supplemented by assumptions about
polynomial dependence of expansion coefficients on their indices.
The deviation from (\ref{AMh}) is indeed observed, moreover, the
assumption of polynomial dependence also turns false for these
terms.

This seems to be in parallel with existing observations
about violation of the BDS conjecture (\ref{BDS}) at $n=6$
\cite{Vo,Koanti}, where the terms, which are not controlled by
the conformal invariance break (\ref{AMh}) and look far more
sophisticated.
An advantage of the wavy lines case is a constructive approach
to the evaluation of minimal areas (still not quite developed
for the finite-$n$ case, for this one should further proceed
along the lines of \cite{IMM1,IM8}).
The disadvantage is the lack of direct contact with
Feynman diagram calculations and with the non-Abelian
Wilson line conjecture at weak coupling.

\bigskip

The wavy line calculation has both beauties and puzzles
of its own. The general structure of the $D^{(m|n)}$
coefficients is fully revealed in s.\ref{Dcalc},
but equally explicit formulas for generic $A^{(m|n)}$
remain to be found, despite a clear recursive procedure
to invert the peculiar Nambu-Goto Laplacian $\Delta_{NG}$
is formulated and applied in s.\ref{Acalc}.
This procedure can be easily computerized,
but, unless properly optimized the calculations
can require a lot of computer time.

On the other hand,
even formulas (\ref{Dgen}) do not possess any
simple representation in terms of differential geometry:
they are not representable as local integrals of
the curvatures of $\Pi$ and its derivatives \cite{IMM2},
thus, their differential geometry meaning remains an open
question.

The same calculation can be (but is not yet) performed for
a wavy deformation of the other solvable example of planar curves:
the two concentric circles \cite{concc}.
Instead we extended the analysis of \cite{IMM2} to
non-planar deformations of the circle.

\bigskip

To conclude, the failure of the Alday-Maldacena conjecture
in its most naive form (\ref{AMh}) is now confirmed both
at finite and infinite $n$.
However, this does not look like an end of the story.
While working out the ways to analyze (\ref{AMh}) we
found that the AdS Plateau problem is not as hopeless as
it looked from the very beginning: one can try and get
concrete and explicit formulas about minimal surfaces and
their regularized areas, which do not at all look
structureless.
This opens a way to systematically investigate the deviations
from (\ref{AMh}), to find hidden symmetries of the both sides
and, finally, to find a modification of the Abelian formula
at the r.h.s. which should serve as a strong-coupling
counterpart of the non-Abelian Wilson average in the weak-coupling
regime and, thus, provide an explicit formulation
of the string/gauge duality for scattering amplitudes.

\section*{Acknowledgements}

Our work is partly supported by the Grant-in-Aid for
Scientific Research 2054278 from the Ministry of Education,
Science and Culture, Japan (H.I.), by NWO project 047.011.2004.026,
by the joint grants 09-01-92440-CE (A.M.'s), 09-02-91005-ANF,
by the Russian President's Grant of
Support for the Scientific Schools NSh-3035.2008.2 and
by Russian Federal Nuclear Energy Agency
(D.G. and A.M.'s), by RFBR grants
07-02-00878 (D.G. and A.Mir.) and 07-02-00645 (A.Mor.).

\newpage

\section*{Appendix: $\kappa$-coefficients in different examples}

We briefly calculate here the coefficient $\kappa$ of proportionality between the
area and the double integral, (\ref{AMh}) and demonstrate that it is typically very close to 1.

For smooth curves the minimal area and the double loop integral are the expressions
of the similar form
\beq\new\begin{array}{c}
A=\kappa_A L+A_{reg}+const_A\\
D=\kappa_D L+D_{reg}+const_B
\end{array}\eeq
Here $L$ is the corresponding length of the boundary where the divergency accumulates.
As we discussed in the paper, $A$ and $D$ can be presented for a smooth curve in the form
\beq\new\begin{array}{c}
A_{\Pi}=\frac{\pi L_{\Pi}}{2\mu} -2\pi-3\pi \frac{1}{6}\frac{1}{2\pi i}\oint dx x^2 \bar h(x) h'''(x)+\ldots \\
D_{\Pi}=\frac{\pi L_{\Pi}}{4\lambda}-\frac{\pi^2}{2}-\pi^2 \frac{1}{6}\frac{1}{2\pi i}\oint dx x^2 \bar h(x) h'''(x) +\ldots
\end{array}\eeq
Therefore, $\kappa_A/\kappa_D=2\lambda/\mu$, while $\kappa_{smooth}$ defined to be
the ratio $A_{reg}/D_{reg}$, is $3/\pi=0.955$ (by modulo difference of $A_{reg}$ by $D_{reg}$
in higher orders and up to inessential numerical additive constants). In fact, to
find $\kappa_A/\kappa_D$, one suffices to consider the case of unit circle. Then,
\beq\new\begin{array}{c}
A=\int\limits_0^1 r dr \int\limits_{0}^{2\pi} d\varphi \frac{1}{\sqrt{1-r^2}(1-r^2+\mu^2)}=2\pi\frac{\arctan\left(\frac{1}{\mu}\right)}{\mu}=\frac{\pi}{2\mu}(2\pi)+\ldots\\
D=\frac{1}{8}\int\int \frac{d\zeta d\bar\zeta'+d\zeta'd\bar\zeta}{(\zeta-\zeta')(\bar \zeta-\bar\zeta')+\lambda^2}=\frac{\pi}{4\lambda}(2\pi)+\ldots
\end{array}\eeq

Another instance is the boundary presented in \cite{AM3}, an infinite strip (two parallel infinite
lines).
We just briefly remind this example following \cite{IMM2}.
The form of the surface can be explicitly found and reads
\beq\new\begin{array}{c}
y(r)=\pm \int\limits_{0}^{r_{max}}\frac{\xi^2 d\xi}{\sqrt{r_{max}^4-\xi^4}}
\end{array}\eeq
A width of the strip (a distance between lines), $a$ is related to $r_{max}$ via
\beq\new\begin{array}{c}
a=\frac{\pi r_{max}}{\sqrt{2}K\left(\frac{1}{\sqrt{2}}\right)}
\end{array}\eeq
where $K$ is the complete elliptic integral of
the second kind. With the $\mu$-regularization the minimal area reads
\beq\new\begin{array}{c}
A=2L r_{max}^2\int\limits_{0}^{r_{max}}\frac{dr}{(r^2+\mu^2)\sqrt{r_{max}^4-r^4}}=
\frac{\sqrt{2}L}{r_{max}}\frac{1}{1+\mu^2}\Pi\left(-\frac{1}{1+\mu^2},\frac{1}{\sqrt{2}}\right)
\end{array}\eeq
where $L$ is the length of the strip, $L\rightarrow\infty$. Thus,
\beq\new\begin{array}{c}
A=\frac{\pi^2 L}{\sqrt{2}K\left(\frac{1}{\sqrt{2}}\right)a}\left(\frac{1}{\mu}-\frac{1}{\sqrt{2}K\left(\frac{1}{\sqrt{2}}\right)}\right)
\end{array}\eeq
The corresponding double loop integral in this case reads
\beq\new\begin{array}{c}
D=\frac{L\pi}{2\lambda}-\pi \frac{L}{2a}
\end{array}\eeq
Therefore, the ration of $A_{reg}$ and $D_{reg}$ is similar
$\kappa_{strip}=4\frac{(2\pi)^2}{\left(\Gamma\left(1/4\right)\right)^4}\approx 0.914$.
Note that in this case $L_{\Pi}=2L+2a\approx 2L$.

At last, consider an example of the rectangular with light-like edges.
The expressions for the minimal area and for the double loop integral are of a slightly different
form in this case so that the divergency is basically due to unsmoothness of the curve.
Then, the dimensional regularization better suits the problem. The minimal area in this
regularization is (up to inessential numerical factors)
\beq\new\begin{array}{c}
A=\frac{2}{\epsilon^2}\left((1+b)^{\epsilon}+(1-b)^{\epsilon}\right)-\frac{1}{4}\log^2 \frac{s}{t}
\end{array}\eeq
where $b$ is the rectangular parameter related to the variables $s,t$ \cite{AM1} via
\beq\new\begin{array}{c}
\frac{(1+b)^2}{(1-b)^2}=\frac{s}{t}
\end{array}\eeq
In this case the ratio of finite parts of the area and the loop integral, $\kappa_{\Box}=1$.
Being a dramatic example, the expression for the area has been calculated in numerical papers
(\cite{AM1,MMT1,Popol} etc). Here we just remind the calculation of the double loop integral
in dimensional regularization (see \cite{MMT1}),
\beq\new\begin{array}{c}
D=\frac{1}{4}\oint\oint\frac{dy^{\mu}dy'_{\mu}}{(y-y')^{(2+\epsilon)}}
\end{array}\eeq
First consider the contribution from angles of the rectangle. We parameterize the two adjacent
edges as
\beq\new\begin{array}{c}
y_0=\frac{a\sqrt{1+b^2}\tanh u}{1+b\tanh u}, \quad y_0=-\frac{a\sqrt{1+b^2}\tanh u'}{1-b\tanh u'}\\
y_1=\frac{a}{1+b\tanh u},\quad y_1=-\frac{a\tanh u'}{1-b\tanh u'}\\
y_2=\frac{a\tanh u}{1+b\tanh u},\quad y_2=\frac{a}{1-b\tanh u'}
\end{array}\eeq
Here $a$ is an arbitrary parameter.
After substitution and change of variables, one can immediately derive
\beq\new\begin{array}{c}
D_{angle}=\frac{1}{4}(b+1)^2\left(\int\limits_{-1}^1\frac{dt}{(1+b t)^{1-\epsilon/2}(1-t)^{1+\epsilon/2}}\right)^2=2^{\epsilon}\frac{(1-b)^{\epsilon}}{\epsilon^2}
\end{array}\eeq
The factor $-2$ is since we change the direction of integration in the second integral
and the contour is passed twice in different integrals. The contribution from the
opposite angles is evidently the same, while the other two angles contribute with the opposite
sign of $b$. The contribution of the opposite sides is convergent and no regularization is needed:
\beq\new\begin{array}{c}
D_{opposite}=-4\frac{1}{8}(1+b^2)\int\limits_{-1}^{1}dt\int\limits_{-1}^{1}dt'\frac{1}{(1+b t)(1-b t')(1-tt')}=
\\=-\frac{\pi^2}{4}-\log^2\left(\frac{1+b}{1-b}\right)=-\frac{\pi^2}{4}-\frac{1}{4}\log^2\frac{s}{t}
\end{array}\eeq
The factor $4$ arises because of 4 opposite edges and corresponding integrations.

redpajama_set_name: RedPajamaArXiv


text:
That Ole Devil Called Love - Alison Moyet
"That Ole Devil Called Love" by Alison Moyet
ALISON MOYET - THAT OLE DEVIL CALLED LOVE ( LYRICS ) VINYL 1985
Billie Holiday - That ole devil called love
That Ole Devil Called Love (2009 Remastered)
Alison Moyet - That ole devil called love 1985 Top of The Pops
Rebecca Ferguson - That Ole Devil Called Love (Audio)
Other popular songs similar to "That Ole Devil Called Love" by Alison Moyet.
Alison Moyet - Love Letters
Alison Moyet - Wishing You Were Here
Gloria Estefan - Don't Wanna Lose You
Joan Armatrading - Drop The Pilot
Five Star - Rain or Shine
Fairground Attraction - Find My Love
Randy Crawford - Rainy Night In Georgia
The Pointer Sisters - Slow Hand
Everything But the Girl - I Don't Want To Talk About It
Gloria Estefan - Anything for You
Wet Wet Wet - Wishing I Was Lucky
Wet Wet Wet - Sweet Little Mystery

redpajama_set_name: RedPajamaCommonCrawl


This easy to make, healthy family meal was sent to us by Stephanie from Neenah, Wisconsin. Her Slow Cooker Vegan Chili Recipe features meatless crumbles of a plant based and protein rich meat substitute. Stephanie opted to use a Morning Star product but tells us that Beyond Beef is another brand of meatless crumbles that she recommends for this Slow Cooker Vegan Chili Recipe.
Stephanie tells us she improvises when creating this Slow Cooker Vegan Chili Recipe, using whatever she may have on hand that keeps it vegan, healthy and most of all delicious. We love this Reader's Circle Recipe for it's simplicity, convenience and broad appeal to the whole family. Besides tasting great, this Slow Cooker Vegan Chili Recipe is perfect for busy parents who are on the go. Thank you, Stephanie, for sharing this great recipe with the Platter Talk's Reader's Circle.
Share your recipes with Platter Talk! Send us a pic (doesn't have to be fancy – a phone picture works just fine!) along with the recipe and we will share it with our readers.
Email to [email protected]dotcom and include a photo. Any photo will do, even a phone pic!
Place all ingredients in a slow cooker on low for 6-8 hours.
This looks amazing! Trying this for dinner tonight! Thank you!!
This chili looks fantastic! I just sent this to my vegan cousin hoping she will make it and invite me over LOL!
I like how you think, Beeta; I think I may want to stop by as well! Thanks so much for dropping by and leaving a comment!
Who needs the meat in this fabulous looking chili. Yum!
Thank so much, Nancy. And agreed!
I have never tried a vegan chili but I bet I could make this and my husband not even know the difference. Thanks.
I haven't cooked with morning star crumbles in a while, but they are quite good. Excellent option in a recipe like this comforting bowl of chili! This looks so delicious and perfect for chilly weather.
Too funny! We are thinking alike here… I am attending a vegan pot luck this weekend and just picked up everything I need to make a began friendly chili! This one sounds great Dan!!!
We love a good chili and one that we can put into a slow cooker and walk away from it- is even better! Thanks for an easy recipe!
Why spend a small fortune on morningstar products or others like it when you can just use generic TVP? Dry TVP doubles in volume and looks, tastes just like hamburger. I make tacos with it all the time and no one knows the difference. I use NOW brand 12 oz TVP, which makes almost 4 pounds of product. $3. Can you get 4 pounds of beef for $3? Morningstar is even more than beef.
Great comments, Ben. And a good reminder that cooking like everything else, often boils down to personal preferences and more times than not, budget Thanks so much for stopping Platter Talk.


text:
package com.prafaelo.lecs.collection;

import java.util.ArrayList;
import java.util.Collection;
import java.util.HashMap;
import java.util.Map;

import com.prafaelo.lecs.LibraryUtil;
import com.prafaelo.lecs.system.user.DefaultUser;
import com.prafaelo.lecs.system.user.LibraryUser;

public class Book {

	private final int code;

	private static int count=0;

	private static final Map<Genre, Collection<Book>> books = new HashMap<Genre, Collection<Book>>();

	private final Collection<BookCopy> copies = new ArrayList<BookCopy>();

	private Genre genre;

	private String title;

	private Collection<String> authors;

	private int year;

	private String local;

	private String observation;

	private LibraryUser reserveUser;

	public Book(int qtyCopies, Genre genre, String title, Collection<String> authors, int year, String local, String observations) {
		setGenre(genre);
		setTitle(title);
		setAuthors(authors);
		setYear(year);
		setLocal(local);
		setObservation(observations);
		this.code = ++count;
		if(!insertInBooks()){
			System.err.println("(Error) This book is already created!");
			return;
		}
		setInitialCopies(qtyCopies);
	}

	public Book(){
		setGenreTyped();
		setTitleTyped();
		setAuthorsTyped();
		setYearTyped();
		setLocalTyped();
		setObservationTyped();
		this.code = ++count;
		if(!insertInBooks()){
			System.err.println("(Error) This book is already created!");
			return;
		}
		setInitialCopiesTyped();
		System.out.println("\nSuccessfully Registered!");
	}

	public static void updateTyped(){
		Book book = Book.getBookTyped();

		if(book != null){
			String isUpdate = LibraryUtil.getTypedValue("Do you want to update this book?");
			if(Boolean.parseBoolean(isUpdate)){

				StringBuffer updatedValues = new StringBuffer();

				boolean isUpdateGenre = Boolean.parseBoolean(LibraryUtil.getTypedValue("Update Genre?"));
				if(isUpdateGenre){
					book.setGenreTyped();
					updatedValues.append("|Genre=" + book.getGenre()+"|");
				}

				boolean isUpdateTitle = Boolean.parseBoolean(LibraryUtil.getTypedValue("Update Title?"));
				if(isUpdateTitle){
					book.setTitleTyped();
					updatedValues.append("|Title=" + book.getTitle()+"|");
				}

				boolean isUpdateAuthors = Boolean.parseBoolean(LibraryUtil.getTypedValue("Update Authors?"));
				if(isUpdateAuthors){
					book.setAuthorsTyped();
					updatedValues.append("|Authors="+book.getAuthors()+"|");
				}

				boolean isUpdateYear = Boolean.parseBoolean(LibraryUtil.getTypedValue("Update Year?"));
				if(isUpdateYear){
					book.setYearTyped();
					updatedValues.append("|Year="+book.getYear()+"|");
				}

				boolean isUpdateLocal = Boolean.parseBoolean(LibraryUtil.getTypedValue("Update Local?"));
				if(isUpdateLocal){
					book.setLocalTyped();
					updatedValues.append("|Local="+book.getLocal()+"|");
				}

				boolean isUpdateObservation = Boolean.parseBoolean(LibraryUtil.getTypedValue("Update Observation?"));
				if(isUpdateObservation){
					book.setObservationTyped();
					updatedValues.append("|Observation="+book.getObservation()+"|");
				}
				System.out.println("Successfully updated ["+updatedValues+"]");
				return;
			}
		}

		System.out.println("(Error) This book was not updated!");
	}

	public static Collection<Book> getBooksByGenreTyped(){
		Collection<Book> booksFound=new ArrayList<Book>();
		String genreTyped = LibraryUtil.getTypedValue("Enter genre:");

		Collection<Book> books = Book.getBooks().get(new Genre(genreTyped));
		if(books!=null){
			booksFound.addAll(books);
			if(booksFound.size()>1)
				System.out.println("Found " + booksFound.size() + ":");
			else
				System.out.println("Found " + booksFound.size() + ":");

			for(Book book : booksFound){
				System.out.println(book);
			}
			return booksFound;
		}
		System.err.println("(Error) Genre not found!");

		return booksFound;
	}

	public static Book getBookTyped(){
		Book book=null;
		Collection<Book> booksFound=new ArrayList<Book>();

		String genreTyped = LibraryUtil.getTypedValue("Enter genre:");
		String titleTyped = LibraryUtil.getTypedValue("Enter title:");

		if(genreTyped.equals("")){
			for(Collection<Book> books : Book.getBooks().values()){
				for(Book bookRead : books){
					if(bookRead.title.equalsIgnoreCase(titleTyped))
					booksFound.add(bookRead);
				}
			}
		} else {
			Collection<Book> books= Book.getBooks().get(new Genre(genreTyped));
			if(books!=null){
				for(Book bookRead : books){
					if(bookRead.title.equalsIgnoreCase(titleTyped))
						booksFound.add(bookRead);
				}

			} else
				System.err.println("(Error) Genre not found!");
		}

		if(booksFound.size()>1){
			System.out.println("Found " + booksFound.size() + ":");
			for(Book bookFound : booksFound){
				System.out.println(bookFound);
			}
			String code = LibraryUtil.getTypedValue("Enter book code:");
			for(Book bookFound : booksFound){
				if(bookFound.code == Integer.parseInt(code)) {
					book = bookFound;
					break;
				}
			}
		} else if(booksFound.size() > 0){
			System.out.println("Found " + booksFound.size() + ":");
			book = booksFound.iterator().next();
			System.out.println(book);
		} else
			System.err.println("(Error) Book not found!");

		return book;
	}

	public void delete(){
		Book.books.get(getGenre()).remove(this);
	}

	public static void reserve(){
		Book book = Book.getBookTyped();

		if(book!=null){

			System.out.println("Found " + book.getCopies().size() + " copies:");
			for(BookCopy bookCopy : book.getCopies()){
				System.out.println(bookCopy);
			}

			if(book.getAvailableCopiesSize()>0){
				System.err.println("(Error) It is not possible to reserve! Copies available.");
				return;
			}

			if(book.getReserveUser()!=null){
				System.err.println("(Error) It is already reserved: " + book.getReserveUser());
				return;
			}

			int libraryCode = Integer.parseInt(LibraryUtil.getTypedValue("Enter library code:"));
			LibraryUser libraryUser = LibraryUser.LIBRARYUSERS.get(libraryCode);

			if(libraryUser==null){
				System.out.println("User not found! Register:");
				libraryUser = new DefaultUser();
			}

			book.setReserveUser(libraryUser);
		}
	}

	public int getAvailableCopiesSize(){
		int countAvailableBookCopy=0;
		if(getCopies().size() > 1){
			for(BookCopy bookCopy : getCopies()){
				if(bookCopy.isAvailable()){
					countAvailableBookCopy++;
				}
			}
		}

		if(countAvailableBookCopy>0){
			countAvailableBookCopy--;
		}

		if(getReserveUser()!=null){
			if(countAvailableBookCopy>0){
				countAvailableBookCopy--;
			}
		}
		return countAvailableBookCopy;
	}

	public static void deleteBookCopyTyped(){

		Book book = Book.getBookTyped();

		if(book != null){

			if(book.getCopies().size()>0){

				System.out.println("Found " + book.getCopies().size() + " copies:");
				for(BookCopy bookCopy : book.getCopies()){
					System.out.println(bookCopy);
				}
				if(book.getAvailableCopiesSize()>0){
					String isDelete = LibraryUtil.getTypedValue("Do you want to delete this book Copy?");
					if(Boolean.parseBoolean(isDelete)){

						String codeBookCopy = LibraryUtil.getTypedValue("Enter book Copy code:");
						for(BookCopy bookCopy : book.getCopies()){
							if(bookCopy.getCode() == Integer.parseInt(codeBookCopy)){
								if(!bookCopy.delete()){
									System.err.println("(Error) This book copy was not deleted!");
								}
								break;
							}
						}
					}
				} else {
					System.err.println("(Error) There is no copies available!");
				}
			} else {
				System.err.println("(Error) Book Copy not found!");
			}

		}
	}

	private boolean insertInBooks(){
		Collection<Book> booksByGenre = Book.getBooks().get(getGenre());
		if(booksByGenre != null){
			if(booksByGenre.contains(this)){
				return false;
			} else {
				booksByGenre.add(this);
			}
		} else {
			booksByGenre = new ArrayList<Book>();
			booksByGenre.add(this);
			Book.getBooks().put(getGenre(), booksByGenre);
		}
		return true;
	}

	public static Collection<Book> getAllBooks(){
		Collection<Book> booksFound = new ArrayList<Book>();
		System.out.println("Found " + Book.getBooks().size() + ":");
		for (Collection <Book> books : Book.getBooks().values()){
			for(Book book : books){
				booksFound.add(book);
				System.out.println(book);
			}
		}
		return booksFound;
	}

	public static void listAllBooks(){
		Book.getAllBooks();
	}

	public static void listBookByTitle(){
		Book.getBookTyped();
	}

	public static void listBookByGenreTyped(){
		Book.getBooksByGenreTyped();
	}

	public void setGenre(Genre genre) {
		this.genre = genre;
	}

	public void setGenreTyped(){
		String genreName = LibraryUtil.getTypedValue("Enter genre:");
		Genre genre = new Genre(genreName);
		setGenre(genre);
	}

	public String getTitle() {
		return title;
	}

	public void setTitle(String title) {
		this.title = title;
	}

	public void setTitleTyped(){
		String title = LibraryUtil.getTypedValue("Enter title:");
		setTitle(title);
	}

	public Collection<String> getAuthors() {
		return authors;
	}

	public void setAuthors(Collection<String> authors) {
		this.authors = authors;
	}

	public void setAuthorsTyped(){
		Collection<String> authors= new ArrayList<String>();
		boolean hasAnotherauthors=false;
		do {
			String author = LibraryUtil.getTypedValue("Enter author:");
			authors.add(author);
			hasAnotherauthors = Boolean.parseBoolean(LibraryUtil.getTypedValue("Another authors?"));
		} while(hasAnotherauthors);
		setAuthors(authors);
	}

	public int getYear() {
		return year;
	}

	public void setYear(int year) {
		this.year = year;
	}

	public void setYearTyped(){
		int year = Integer.parseInt(LibraryUtil.getTypedValue("Enter year:"));
		setYear(year);
	}

	public String getObservation() {
		return observation;
	}

	public void setObservation(String observation) {
		this.observation = observation;
	}

	public void setObservationTyped(){
		String observation = LibraryUtil.getTypedValue("Enter observation:");
		setObservation(observation);
	}

	public String getLocal() {
		return local;
	}

	public void setLocal(String local) {
		this.local = local;
	}

	public void setLocalTyped(){
		String local = LibraryUtil.getTypedValue("Enter local:");
		setLocal(local);
	}

	public Genre getGenre() {
		return genre;
	}

	public static Map<Genre, Collection<Book>> getBooks() {
		return books;
	}

	public Collection<BookCopy> getCopies() {
		return copies;
	}

	public void setInitialCopiesTyped(){
		int qtyCopies = Integer.parseInt(LibraryUtil.getTypedValue("Enter quantity copies:"));
		setInitialCopies(qtyCopies);
	}

	public void setInitialCopies(int qtyCopies){
		for(int i = 0; i< qtyCopies; i++){
			new BookCopy(this);
		}
	}
	public int getCode() {
		return code;
	}

	public static int getCount() {
		return count;
	}

	public LibraryUser getReserveUser() {
		return reserveUser;
	}

	public void setReserveUser(LibraryUser reserveUser) {
		this.reserveUser = reserveUser;
	}

	@Override
	public String toString() {
		StringBuilder builder = new StringBuilder();
		builder.append("Book [code=").append(code)
		.append(", genre=").append(genre)
		.append(", title=").append(title)
		.append(", copies=").append(copies.size())
		.append(", availableCopies=").append(getAvailableCopiesSize())
		.append(", authors=").append(authors)
		.append(", year=").append(year)
		.append(", local=").append(local)
		.append(", observation=")
		.append(observation).append("]")
		.append(", reserveUser=")
		.append(reserveUser).append("]");
		return builder.toString();
	}

	@Override
	public int hashCode() {
		final int prime = 31;
		int result = 1;
		result = prime * result + ((genre == null) ? 0 : genre.hashCode());
		result = prime * result + ((title == null) ? 0 : title.toLowerCase().hashCode());
		return result;
	}

	@Override
	public boolean equals(Object obj) {
		if (this == obj)
			return true;
		if (obj == null)
			return false;
		if (getClass() != obj.getClass())
			return false;
		Book other = (Book) obj;
		if (genre == null) {
			if (other.genre != null)
				return false;
		} else if (!genre.equals(other.genre))
			return false;
		if (title == null) {
			if (other.title != null)
				return false;
		} else if (!title.equalsIgnoreCase(other.title))
			return false;
		return true;
	}
}

redpajama_set_name: RedPajamaGithub


This is an event recording of the November 2, 2016 webinar with presenters Paul Connor and Connie McCutcheon. 23 participants joined this event from 14 different organizations across Niagara.
During this webinar presenters described described what an FTZ Point is (and isn't), reviewed how an FTZ Point works and potential benefits to local business, and explored why Niagara is primed for this designation.


text:
Asha Srinivasan
Mizzou faculty ensemble ZouM to perform Friday, September 8 in Columbia and Saturday, September 9 in St. Louis
The Mizzou faculty collective ZouM will showcase new music from six living composers with performances next weekend in Columbia and St. Louis. They'll play at 7:30 p.m. Friday, September 8 at Whitmore Recital Hall on the MU campus, then travel to St. Louis for a concert at 7:30 p.m. Saturday, September 9 at Washington University's…
MICF resident composers in the news
Here's the latest news and updates from some of the resident composers at past Mizzou International Composers Festivals: * Jeremy Podgursky, MICF 2010, is one of six winners this year of a Charles Ives scholarship from the American Academy of Arts and Letters. The $7,500 scholarships are awarded to composition students of great promise, as…
"Conversations with Composers" podcast – Episode 103
Here's the third installment in our series of "Conversations With Composers" podcasts recorded at this year's Mizzou New Music Summer Festival. Episode 103 is the second and final part of a conversation including guest composer Donnacha Dennehy and resident composers Stephanie Berg, Brian Ciach, Charlie Piper and Asha Srinivasan. Give it a listen using the…
Here's another installment in our series of "Conversations With Composers" podcasts. Recorded during the 2012 Mizzou New Music Summer Festival, episode 102 is the first part of a conversation among guest composer Donnacha Dennehy and resident composers Stephanie Berg, Brian Ciach, Charlie Piper and Asha Srinivasan. You can hear it in the embedded player below:
Summer Festival Spotlight: Asha Srinivasan
Born in Utah and brought up in India and the USA, Asha Srinivasan creates music that combines elements from two different cultures. An assistant professor of music at Lawrence University in Appleton, Wisconsin, Srinivasan (pictured) is one of eight resident composers for the 2012 Mizzou New Music Summer Festival. She is part of an musical…
Mizzou New Music Summer Festival's resident composers will bring diversity to 2012 event
From a London-based composer pursuing a doctorate at the Royal Academy of Music, to a Greek studying music in upstate New York, to a Kansas City native working towards her Master's degree at Mizzou, the resident composers selected for this year's Mizzou New Music Summer Festival are perhaps the most diverse group yet in the…


text:
Q: Java: Waiting for mouse input (click) I am making some sort of turn-based battle system in a JFrame where the player clicks a button when it's his turn. The problem is: how can the program wait for a mouse click on the button? It goes like this:
while it is the player's turn {

*

*wait for mouse input

*if input == attack (for example)
   -> attack

else if input == item
   -> use item

and so on

A: You don't wait for a click; you let Swing do that for you. Instead, you put whatever you want to do into an ActionListener and attach it to the button, so it gets executed when the button is clicked.
As far as the turns go, you just need a member variable someplace that keeps track of whose turn it is; the button handler then has to look at that variable to know what to do.
One good way to structure things, by the way, might be to have a Player class, and a Game class, and a member in Game called currentPlayer. Then the ActionListener (which keeps the Game object as a member variable of its own) could look at currentPlayer in the Game and simply invoke makeMove() on the appropriate Player object.

redpajama_set_name: RedPajamaStackExchange


In preparation for the upcoming Create tour for startups, I've been challenging our speakers to do something a bit unusual for most of them: Draft their talks out longhand.
Don't get me wrong, we've got some great speakers who can show up to an event cold and improve a presentation better than most people can do a talk that's been worked on for months. And, they speak so often, they all have several slide decks ready to go for their areas of expertise and can mix or match as needed. For this tour, however, we're asking many of them to bring their knowledge about their topic area, focus it for the needs of early- to mid-phase startups, and to do it in a way that's useful for our audience of startup developers and technical founders.
For example, Phoummala Schmitt is giving a talk about governance at our first two stops in New York and Toronto next week. This isn't a typical topic for a startup conference. Having worked at startups who grow up and end up as (or part of) bigger companies, however, I've run into many topics around governance and compliance that not only did we not think about, we made decisions that were counter to being able to handle our requirements later on. So, we're going to try to bring a bit of this kind of perspective back to startups.
No, we're not going to tell you to do things the way Microsoft does it. That would be patently ridiculous. What Phoummala will do, however, is give the audience some tools and techniques that can help determine what the starting guardrails for a startup should be. Ones that can develop over time as needed.
Tricky, right? It'd be straightforward for us to come in talking about ISO-standard this and compliance that and totally lose the audience in less than two minutes flat.
That's where drafting the talk out longhand comes in. As we've worked the last few weeks together over Phoummala's words, we've talked about examples we can use, what concepts feel right for the audience we expect, and what we should cut because it would distract from the core takeaways we want to leave with our audience.
It's not an easy thing to do, especially when a speaker is used to throwing some slides together, showing up, and pushing play on themselves. But the effort is worth it. In this case, I think we've now got a talk that will be useful to developers at a startup. One that will leave them with some of the things I wish I'd known a while ago.
If you want to see how it turned out, along with the rest of the presentations we've been working on, please join us in New York or Toronto next week.


text:
Napoleon může být:

Jméno osoby
 Napoleon Bonaparte (1769–1821) – francouzský vojevůdce a císař
 Napoleon II. (1811–1832) – syn císaře Napoleona I., vévoda zákupský
 Napoleon III. (1808–1873) – francouzský císař
 Napoleon Evžen Ludvík Bonaparte (1856–1879) – syn císaře Napoleona III., po smrti svého otce byl bonapartistickou frakcí prohlášen jako císař Napoleon IV.
 Napoleon Murphy Brock (* 1945) – americký hudebník
 Napoleon Hill (1883–1970) – americký spisovatel
 Napoleon Prato (1822–???) – rakouský politik z Tyrolska

Jiné významy
 Napoleon (automobilka) – francouzský výrobce automobilů
 Napoleon (koňak) – druh alkoholického nápoje
 Napoléon (1850) – francouzská pancéřová loď
 Jméno vládnoucího kance v Orwellově románu Farma zvířat
 Napoleon (dělo) – dělo francouzské konstrukce užívané v americké občanské válce

Externí odkazy

redpajama_set_name: RedPajamaWikipedia


You can decide to use whichever 3 word address appeals to you most, for aesthetic or practical reasons. People will often choose a combination of words that they particularly like, or the address of the 3m x 3m square covering a specific useful point, e.g. the front door, side gate or parking space.