text:
\section{Introduction} \label{sec:intro}
\subsection{Background and Motivation}
Weighted voting games are a classical family of cooperative games that have been extensively studied
in the game theory and social choice literature. Such games model a common voting scenario where each agent (player),
associated with a weight, casts a ``YES'' (for) or ``NO'' (against) vote: if the weighted sum of the ``YES'' votes exceeds
a threshold, then the voting outcome is ``YES'', otherwise the outcome is ``NO''. Examples of such practical scenarios
include the voting system of the European Union, stockholder companies, and resource allocation in multi-agent systems \cite{EGGW08,KKZ11}.

Although having a larger weight might help an agent affect the voting outcome,
her influence on the result of the game is not always proportional to her weight.
Thus, instead of using agents' weights, the power of an agent over the outcome
is usually measured in a systematic way by a {\em power index}.
Over the years, many power indices have been proposed and studied,
such as the Shapley value~\cite{shapley} (also known as Shapley-Shubik index
for weighted voting games~\cite{SS}), the Banzhaf index~\cite{ban},
the Deegan-Packel index~\cite{DP}, and the Holler index~\cite{Hol}.
The problem of computing the agents' power indices in a given game
has received ample attention and its computational complexity
is well-understood for many game representations and power index functions
(see, e.g., ~\cite{PK90,DP94,A08}).

\subsection{Our Contributions}
In this work, we focus on the {\em inverse power index problem}
--- that is, the problem of {\em designing} a weighted voting game  with a given set of power indices.
As we will explain in detail below, the inverse problem has been extensively studied in various fields,
including game theory, social choice theory, and learning theory. Various works have provided
heuristic methods, exponential time algorithms, or polynomial time approximation algorithms
with provable performance guarantees for this problem.

Despite this wealth of prior work on the algorithmic version of the inverse problem,
prior to this work its computational complexity was not well-understood, even for the most popular
power indices (Shapley values, Banzhaf indices). {\em In this paper, we study and essentially resolve the computational complexity of the inverse power index problem for weighted voting games, with respect to a broad and extensively studied
family of power indices. Specifically, we show that the inverse problem is computationally intractable
under standard complexity assumptions.}
More specifically, we prove that for a large class of power indices --- that includes the popular Banzhaf index, Shapley values,
and the class of semivalues~\cite{W79} --- the inverse problem cannot be in the polynomial hierarchy (PH),
unless the polynomial hierarchy collapses. Prior to this work, it was conceivable that there
exists an exact polynomial time algorithm for this problem. It follows from our hardness result
that the existence of such an algorithm is unlikely.

\subsection{Related work}
Several heuristic algorithms for the inverse Banzhaf index problem have been proposed
in the social choice and game theory literature.
Aziz, Paterson, and
Leech \cite{APL07} give an approximation algorithm that on input target Banzhaf indices and a desired $\ell_2$-distance bound
outputs a weighted voting game with integer weights that has Banzhaf indices within the desired distance bound.
Unfortunately, no theoretical guarantees are provided regarding the convergence rate of this method,
and it is not known whether it converges to an approximately optimal solution. Two related heuristic algorithms
that return weighted voting games with Banzhaf indices within a given distance from the target indices are proposed
by  Laruelle and Widgren~\cite{LW98} and Leech~\cite{L03}.
Similarly to~\cite{APL07}, Fatima, Wooldridge, and Jennings  \cite{FWJ08}
give an iterative approximation algorithm for the inverse Shapley value problem that on input target Shapley values,
a quota, and a desired average percentage difference, outputs a weighted voting game with the given quota
that has Shapley values within the desired distance.
It is shown that each iteration runs in quadratic time and that the algorithm eventually converges,
but no theoretical guarantees are given regarding the convergence rate, i.e., the time
until a desired approximation is achieved

An exact algorithm for both the inverse Banzhaf index and inverse Shapley value problems is given by Kurz~\cite{K11}.
The proposed method relies on integer linear programming and returns a weighted voting game
that minimizes the $\ell_1$-distance from the target power indices.
This method has running time exponential in the number of players.
Another exact, but exponential-time, algorithm for the inverse power index problem
is given by De Keijzer, Klos, and Zhang~\cite{KKZ11}.
The~\cite{KKZ11} algorithm outputs a weighted voting game
in which the power indices of the players are as close to the target vector as possible.
The presented algorithm is based on an enumeration of all weighted voting games:
for each weighted voting game, the algorithm computes the power indices,
their distance from the target ones and it then outputs the game with the smallest distance.
Since there exist $2^{\Omega(n^2)}$ weighted voting games with $n$ players,
this algorithm also runs in exponential time.

In addition to the several heuristics and exponential time algorithms that have been proposed,
a line of recent works in theoretical computer science~\cite{OS11, DDFS12,DDS12} have obtained
polynomial time approximation algorithms with provable performance guarantees
for the inverse problem with respect to both the Banzhaf indices and the Shapley values.
These algorithms output a weighted voting game whose power indices have
small $\ell_2$-distance from the target indices. These algorithmic results were recently
extended to more general classes of functions~\cite{DK18-degd}.

The inverse power index problem is also of significant interest in various other fields,
such as circuit complexity and computational learning theory. The reader is referred to~\cite{OS11}
for a detailed summary of this connection. In the fields of computational complexity and learning theory,
linear threshold functions (LTFs),
which are equivalent to weighted voting games if we allow negative weights~\cite{KKZ11},
have been of great significance and have been studied for several decades~\cite{Rosenblatt:58, chow,
MTT:61, Dertouzos:65, MinskyPapert:68}.
A fundamental result of C. K. Chow from the the early 1960s~\cite{chow}
shows that linear threshold functions are characterized by their degree-$0$ and degree-$1$ ``Fourier coefficients'',
now known as {\em Chow parameters}. Given this structural result, the following natural computational question
--- now known as ``the Chow parameters problem''~\cite{OS11}  --- arises:
Given the Chow parameters of an LTF,
reconstruct a weights-based representation of the function. Interestingly, the Chow parameters
are essentially equivalent to the non-normalized Banzhaf indices~\cite{DPS79}, and therefore
the inverse Banzhaf index problem is tantamount to the Chow parameters problem.

In addition to the aforementioned algorithmic results, a number of
complexity results have been established concerning weighted voting games.
Aziz \cite{A08}, Elkind {\em et al.}~\cite{EGGW07}, and Elkind {\em et al.}~\cite{EGGW08}
study the computational complexity of various problems related to weighted voting games.
Aziz~\cite{A08} studies the complexity of computing various indices such as the Shapley values,
the Banzhaf, and the Deegan-Packel indices for a given simple game when the game is given in different forms.
Finally, Faliszewski and Hemaspaandra \cite{FH08} study the complexity of the power index {\em comparison problem}:
given two weighted voting games and a player, decide on which game the given player
has higher influence as it is computed by a specific power index. They show that this problem is intractable,
namely  PP-complete, for both the Shapley values and Banzhaf index. To achieve this, they extend the
\#P-metric-completeness of computing the Shapley values, proved by Deng and Papadimitriou~\cite{DP94}.
They prove that, whereas computing the Banzhaf indices of a weighted voting game is \#P-parsimonious-complete \cite{PK90}, computing the Shapley values is \#P-many-one complete and it cannot be strengthened to \#P-parsimonious-complete.
Gopalan, Nisan, and Roughgarden \cite{GNR15} study the convex polytope
consisting of the Chow parameters of all Boolean functions.
They show that the linear optimization problem over this polytope is \#P-hard;
a result that indicates, but does not logically imply, that the inverse Banzhaf index problem may be intractable.

\section{Preliminaries}
\paragraph{Notation}
We write $wt(x)$ to denote the {\em weight} of a Boolean vector $x \in \{-1,1\}^n$,
i.e., the number of 1's in $x$. For any $i \in \{1, \dots, n\}$ and $x \in \{-1,1\}^n$ such that $x_i=-1$,
we write $x^i$ to denote the vector obtained when we flip the $i^{th}$ coordinate of $x$.
We denote by $\textbf{1}$ (resp. $\textbf{-1}$) the vector in $\{-1, 1\}^n$ with all coordinates equal to $1$ (resp. $-1$).
We will denote by $\mathrm{Conv}(S)$ the convex hull of the set $S$.
We will use $\mathrm{sign}: \mathbb{R} \rightarrow \mathbb{R}$ for the function
that takes value $1$ if $z \geq 0$ and value $-1$ if $z<0$.

\smallskip

\noindent Our basic object of study is the family of linear threshold functions (LTFs) over $\{-1,1\}^n$:

\begin{definition}[Linear Threshold Function] \label{def:ltf}
A {\em linear threshold function (LTF)} is any function
$f_{w, \theta}: \{-1,1\}^n \rightarrow \{-1, 1\}$ such that
$f_{w, \theta}(x)=\mathrm{sign}(w \cdot x - \theta)$ for some weight vector
$w \in \mathbb{R}^n$ and  threshold $\theta \in \mathbb{R}$.
\end{definition}

Note that weighted voting games are equivalent to LTFs with {\em non-negative weights}.
We leverage this equivalence throughout this paper. At various points, we
may refer to a weighted voting game as an LTF without further elaboration.

\paragraph{Semivalues}
We mainly focus on power indices that belong to {\em the class of semivalues}.
Semivalues are a fundamental family of power indices, introduced by Weber \cite{W79}
as generalizations of the Shapley value that do not satisfy the efficiency axiom \cite{DNW81}.
Since their introduction, semivalues have received considerable attention, see, e.g., \cite{E87,CFP03,CF08}.

We start by providing the definition of semivalues \cite{roth} in terms of weighting coefficients,
as they were characterized by Dubey, Neyman, and Weber \cite{DNW81}:
\begin{definition}[Semivalues] \label{def:semiv}
For a positive integer $n$, a probability vector $p^n=(p_0^n, \dots, p_{n-1}^n) \in \mathbb{R}^n$ is
a vector such that $\sum_{t=0}^{n-1}\binom{n-1}{t}p_{t}^n=1$ and $p_t^n \geq 0,\text{ for } t \in \{0, \dots, n-1\}$.
The $i$-th semivalue corresponding to the probability vector $p^n$ of a Boolean function
$f:\{-1,1\}^n \rightarrow \{-1,1\}$ is defined to be
\begin{align*}
\widetilde{f}^{p^n}(i)=\sum_{x \in \{-1,1\}^n: x_i=-1}p^n_{wt(x)}f(x)x_i+\sum_{x \in \{-1,1\}^n: x_i=1}p^n_{wt(x)-1}f(x)x_i \;,
\end{align*}
for $i \in \{1, \dots, n\}$.
\end{definition}

Intuitively, we can interpret $p^n$ as the vector of probabilities that a given player
will join a coalition of size $t$~\cite{CF08}, $0 \leq t \leq n-1$. With this interpretation,
the $i$-th semivalue computes the probability of the event that player $i$ is a {\em pivot},
i.e., the probability that the output of the game would change from $1$ to $-1$
if the $i$-th player (the $i$-th variable) were to change her vote from $1$ to $-1$.

We call a semivalue {\em regular} if it is defined by strictly
positive probability vectors~\cite{CF08}.

\paragraph{Remark}
The Shapley values and Banzhaf indices are the semivalues defined
by $p^n_t=\frac{(n-t-1)! t!}{n!}$ \cite{SS} and $p^n_t=\frac{1}{2^{n-1}}$ \cite{DPS79}, respectively.
Note that both indices are regular semivalues.

\paragraph{Reformulation of Semivalues}
The following equivalent way to express a set of semivalues will
be useful throughout this paper.
Setting $p_{-1}^n=p_{n}^n=0$, we observe that we can rewrite the semivalues vector defined by the probability vector
$p^n$ as follows.
For $i \in \{1, \dots, n\}$, we have:
\begin{align}\label{sem}
\widetilde{f}^{p^n}(i)&= \frac{1}{2}\sum_{x \in \{-1,1\}^n }f(x)x_i \left(p^n_{wt(x)}+p^n_{wt(x)-1}\right)
+\frac{1}{2}\sum_{x \in \{-1,1\}^n}f(x) \left(p^n_{wt(x)-1}-p^n_{wt(x)}\right) \;.
\end{align}
From this representation, a probability distribution $\mu_{p^n}$ over $\{-1,1\}^n$ emerges,
defined as follows. For $x \in \{-1,1\}^n$, we have:
$$\mu_{p^n}(x): = \frac{\mu'_{p^n}(x)}{\Lambda(p^n)} \;,$$
where $\mu'_{p^n}(x): =p^n_{wt(x)}+p^n_{wt(x)-1}$ and
$\Lambda(p^n):=\sum_{x \in \{-1,1\}^n}\mu'_{p^n}(x)$ is the normalizing factor.

For a Boolean function $f: \{-1,1\}^n \rightarrow \{-1, 1\}$, we will write
$\widehat{f}^{p^n}(i): =\sum_{x \in \{-1,1\}^n} \mu'_{p^n}(x)f(x)x_i$ for the first term of (\ref{sem})
and $C_f^{p^n} := \sum_{x \in \{-1,1\}^n}f(x)(p^n_{wt(x)-1}-p^n_{wt(x)})$ for the second term of (\ref{sem}).
One can view the first term, $\widehat{f}^{p^n}(i)$, as the expectation $\mathbb{E}_{x \sim  \mu_{p^n}} [f(x)x_i]$
(up to the normalizing factor $\Lambda(p^n)$).

\medskip

We now define the notion of a {\em reasonable} probability vector to describe
the family of semivalues for which our computational hardness results apply:

\begin{definition}[Reasonable Probability Vector] \label{def:reas}
A probability vector $p^n \in \mathbb{R}^{n}$ is called {\em reasonable} if there exists a $t$ with
$t = \Omega(n)$ and $n-t = \Omega(n)$ such that $p^n_t > 0$.
\end{definition}

The intuition behind the above definition is that the distribution $\mu_{p^n}$
has support $2^{\Omega(n)}$. Note that this happens when there exists
a $t$ with $\binom{n}{t} = 2^{\Omega(n)}$ such that $p^n_t > 0$.
We recall that all regular semivalues (including the Banzhaf indices and Shapley values)
satisfy this property.

\begin{rem}
{\em For computational purposes, throughout this paper, we will assume that
each value $p^n_t$ defining our probability vector is a rational number
that can be described as a ratio of integers with $\mathrm{poly}(n)$ bits.}
\end{rem}

\paragraph{Inverse Semivalues Problem} We are ready to define the inverse semivalues problem.
Let $n$ denote the number of players and consider the semivalues defined by a known
probability vector $p^n$. Given a vector $c = (c_1, \ldots, c_n)$ of target semivalues,
we want to either find a weighted voting game with these target
semivalues or decide that there does not exist any weighted voting game with
semivalues vector $c$.

\smallskip

\medskip
\fbox{\parbox{6in}{
\noindent \textbf{Name:} SV$_{p^{n}}$-Inverse Problem

\medskip

\textbf{Input:} A vector $(c_1, \dots, c_{n}) \in \mathbb{Q}^{n}$ and $\theta \in \mathbb{Q}$.

\smallskip

\textbf{Question:} Output $w \in \mathbb{Q}_{+}^n$ with $\sum_{i=1}^n w_i=1$ such that $\widetilde{f}_{w, \theta}^{p^n}(i)=c_i$,
for $i \in \{1,\dots, n\}$, or ``NO'' if no such $w$ exists.

}}

\section{Main Result: Computational Intractability of Inverse Power-Index Problem} \label{sec:main-thm}

\subsection{Statement of Main Result and Proof Overview}

The main result of this paper is the following:

\begin{theorima}[Main Result, Informal Statement] \label{thm:main}\label{main}
For semivalues defined by the probability vector $p^n$, if $p^n$ is a reasonable probability vector,
then the SV$_{p^n}$-Inverse problem is not in the polynomial hierarchy, unless the polynomial hierarchy collapses.
\end{theorima}

As an immediate corollary of Theorem~\ref{thm:main}, we obtain that
the inverse power index problem is similarly intractable for the class of regular semivalues,
which includes the Shapley values and the Banzhaf indices.

\paragraph{Proof Overview}
We start with a brief overview of our proof establishing Theorem~\ref{main}.
To prove hardness of the inverse problem, we examine the convex polytope $\mathcal{C}_{p^{n+2}}$ consisting of the convex combinations of the semivalues of linear threshold functions with zero threshold and weight vectors
of a specific form described below (Definition~\ref{poly}).
We prove (Theorem~\ref{thm:opt-hard}) that if the probability vector $p^n$ defining
the semivalues is {\em reasonable}, then the  linear optimization over $\mathcal{C}_{p^{n+2}}$
is \#P-hard (under Turing reductions). Then, we proceed to show (Theorem~\ref{thm:svr-hard})
that the optimization problem can be solved using an oracle for the semivalues {\em verification problem},
i.e., the problem of verifying that the given target semivalues are the actual semivalues of a given linear threshold function,
or an oracle for the {\em inverse problem} for weight vectors of the aforementioned specific form.
We thus conclude that the verification and the inverse semivalues problem for linear threshold functions with this
specific weight structure cannot be in the polynomial hierarchy. Finally, using a lemma that shows
that semivalues characterize the space of linear threshold functions with the same threshold (Lemma~\ref{equiv}),
we show hardness of the inverse and verification problems for linear threshold functions with positive weights,
i.e., for weighted voting games, as desired.

Our proof strategy bears some similarities to the approach by Gopalan, Nisan, and Roughgarden \cite{GNR15}
(also exploited by Dughmi and Xu \cite{DH16}). In particular, Gopalan, Nisan, and Roughgarden \cite{GNR15} show
that the linear optimization problem over the polytope consisting of the Chow parameters of all Boolean functions is \#P-complete,
and therefore there cannot exist an efficient membership oracle for this polytope.  We note that our results include the results of \cite{GNR15} regarding Chow parameters as a very special case: as previously mentioned, the non-normalized Banzhaf indices of a linear threshold function are equal to its Chow parameters and they are semivalues.

Despite this similarity, our proof involves a number of novel ideas that seem necessary
in order to handle a broad range of probability distributions that could define a semivalue.
One of the difficulties comes from the fact that we want to prove hardness for the class
of weighted voting games, i.e., LTFs with {\em positive} weights. While this requirement is easy
to handle for the Banzhaf indices (Chow parameters), it poses non-trivial difficulties for more
general semivalues. To handle this, we propose a generalization of the definition of the Khintchine constant
from the uniform distribution to any probability distribution and establish that it is hard to compute
under a restricted set of weights that is crucial for our proof (Theorem~\ref{thm:KK-hard}).
Another crucial ingredient of our proof is a new structural result (Lemma~\ref{equiv})
establishing that the set of semivalues uniquely determines a weighted voting game.

\subsection{Proof of Main Result}
In this subsection, we proceed with the detailed proof of
Theorem~\ref{thm:main}.

\paragraph{Semivalues Polytope} Our analysis makes essential use of the
convex polytope defined as the convex hull of the set of semivalues for
all linear threshold functions whose weights-based representation is of a specific form:
Namely, their threshold $\theta=0$ and their weight vectors consist of $n$ positive coordinates
and two coordinates each of whose weights is equal to minus a half times the sum of the first $n$ coordinates.
Formally, we introduce the following definition:

\begin{definition} \label{poly}
For a positive integer $n$, define $\mathcal{C}_{p^{n+2}} := \mathrm{Conv}(\mathcal{A}_{p^{n+2}})$,
where
$$\mathcal{A}_{p^{n+2}} := \left\{c \in \mathbb{R}^{n+2}: \exists w \in \mathbb{R}^{n+2}, \; w_i > 0, \;1 \leq i \leq n,\; w_{n+1}=w_{n+2}=-\sum_{i=1}^nw_i/2,\; c_i=\widetilde{f}^{p^{n+2}}_{w,0}(i), \; 1 \leq i \leq n+2 \right\} \;.$$
\end{definition}

The first main step of our proof involves showing that the linear optimization problem
over the above defined polytope is computationally hard.

\paragraph{Linear Optimization over $\mathcal{C}_{p^{n+2}}$}
We firstly prove that if the semivalues' probability distribution defined by $p^{n+2}$ has a sufficiently large  support
(in particular, if $p^{n+2}$ is a reasonable vector as in Definition~\ref{def:reas}),
then the linear optimization problem over the polytope $\mathcal{C}_{p^{n+2}}$ is \#P-hard.

The linear optimization problem for semivalues defined by any probability vector $p^{n+2}$
is captured by the following family of problems:

\medskip

\fbox{\parbox{6in}{

\noindent \textbf{Name:} SV$_{p^{n+2}}$-Optimization Problem

\smallskip

\textbf{Input:} A vector $a=(a_1, \dots, a_{n+2}) \in \mathbb{Q}^{n+2}$.\\
\textbf{Question:} Compute $\max_{c \in \mathcal{C}_{p^{n+2}}} a \cdot c$.

}}

\medskip

The main result of this subsection is the following:

\begin{theorima} \label{thm:opt-hard}
If $p^{n+2}$ is a {\em reasonable} probability vector, the SV$_{p^{n+2}}$-Optimization Problem is \#P-hard.
\end{theorima}

\smallskip

We prove Theorem~\ref{thm:opt-hard} by reducing from an intermediate problem --- that of
computing the Khintchine constant of a vector {\em with respect to the probability distribution $\mu_{p^n}$}:

\medskip

\fbox{\parbox{6in}{
\noindent \textbf{Name:} Khintchine$_{\mu_{p^n}}$

\smallskip

\textbf{Input:} A vector $a = (a_1, \dots, a_n) \in \mathbb{Q}^{n}$.\\
\textbf{Question:} Compute $K_{\mu_{p^n}}(a)=\mathbb{E}_{x \sim \mu_{p^n}}[|a \cdot x|]$.

}}

\medskip

The Khintchine constant has been extensively studied with respect to the uniform distribution
on the Boolean hypercube (see, e.g.,~\cite{Szarek:76, DeDS16} and references therein). We note that
\cite{GNR15} established the intractability of computing this quantity under the uniform distribution.
We show:

\begin{theorima} \label{thm:KK-hard}
If $p^{n+2}$ is a reasonable probability vector, the Khintchine$_{\mu_{p^{n+2}}}$ problem is \#P-hard,
even restricted to inputs $(a_1, \dots, a_n, -A/2, -A/2)$, where $A=\sum_{i=1}^n a_i$
and $a_i>0$ for $i \in \{1, \dots, n\}$.
\end{theorima}
\begin{proof}

We start by showing that the \#Partition problem for the distribution $\mu_{p^n}$ is hard
and then reduce the latter problem to the former.

\medskip

\noindent \textbf{Name:} \#Partition$_{\mu_{p^n}}$

\smallskip

\noindent \textbf{Input:} A vector $w=(w_1, \dots, w_n) \in \mathbb{Z}^{n}$.\\
\textbf{Question:} Compute $\text{Pr}_{x \sim \mu_{p^n}}[w \cdot x = 0]$.

\medskip

We start with the following proposition:

\begin{prop} \label{thm:partition}
If $p^{n+2}$ is a reasonable probability vector, \#Partition$_{\mu_{p^{n+2}}}$ is \#P-hard,
even restricted to inputs $(w_1, \dots, w_n, -W/2, -W/2)$,
where $W=\sum_{i=1}^n w_i$ and $w_i>0$ for $1 \leq i \leq n$.
\end{prop}
\begin{proof} We reduce from the following problem:

\medskip

\noindent \textbf{Name:} \#R-Partition \\
\textbf{Input:} Positive integers $c_j,1 \leq j \leq n$ and a positive integer $b_1n \leq k \leq b_2n$,
where $0 < b_1 \leq b_2 < 1$, such that if there is a subset $S$ of $\{1,\dots,n\}$
with $\sum_{i \in S} c_i=\sum_{i=1}^{n} c_i/2$, then $|S|=k$ or $|S|=n-k$.\\
\textbf{Question:} Compute the number of subsets $S$ of $\{1,\dots,n\}$ such that
$\sum_{i \in S} c_i=\sum_{i=1}^{n} c_i/2$.

\medskip

Given an instance $c_1, \dots, c_{n}$ of the \#P-complete \#R-Partition \cite{DP94},
we construct an instance of \#Partition$_{\mu_{p^{n+2}}}$ as follows:
We set $w_i = c_i$, for $1 \leq i \leq n$, and define $w= (w_1, \dots, w_{n}, -W/2, -W/2)$,
where $W=\sum_{i=1}^n w_i$. We have:

\begin{align*}
\text{Pr}_{x \sim \mu_{p^{n+2}}} [w \cdot x =0]
&=\sum_{x \in \{-1,1\}^{n+2}: w \cdot x =0}\mu_{p^{n+2}}(x)=
\frac{|\{x:w \cdot x=0, x \not \in \{\textbf{-1},\textbf{1}\}\}|}{2\Lambda(p^{n+2})}(p^{n+2}_{n-k+1}+p^{n+2}_{n-k})\\
&+\frac{|\{x:w \cdot x=0, x \not \in \{\textbf{-1},\textbf{1}\}\}|}{2\Lambda(p^{n+2})}(p^{n+2}_{k+1}+p^{n+2}_{k})
+\frac{1}{\Lambda(p^{n+2})}(p^{n+2}_{n+1}+p^{n+2}_{0}) \;,
\end{align*}
as the \#R-Partition problem guarantees that every solution has size $k$ or $n-k$
and the number of solutions with size $k$ are equal to the number of solutions with size $n-k$.
That is, every $x \in \{-1,1\}^{n+2} \setminus \{\textbf{-1},\textbf{1}\}$ such that $w \cdot x=0$
is guaranteed to have weight $k+1$ or weight $n-k+1$ as $x_{n+1}$ has to be different
than $x_{n+2}$, otherwise the only solutions are $\{\textbf{-1},\textbf{1}\}$.
So, for every solution of the \#R-partition problem we have two
$x \in \{-1,1\}^{n+2} \setminus \{\textbf{1}, \textbf{-1}\}$ such that $w \cdot x=0$.
Thus,
\begin{align*}
\left| \left\{ S \subset \{1,\dots,n\}:\sum_{i \in S} c_i =\sum_{i=1}^{n}c_i/2 \right\} \right| =
\frac{\Lambda(p^{n+2})\text{Pr}_{x \sim \mu_{p^{n+2}}} [w \cdot x =0]-(p^{n+2}_{n+1}+p^{n+2}_{0})}{(p^{n+2}_{n-k+1}+p^{n+2}_{n-k}+p^{n+2}_{k+1}+p^{n+2}_{k})} \;.
\end{align*}
This completes the proof of Proposition~\ref{thm:partition}.
\end{proof}

Given an instance of \#Partition$_{\mu_{p^{n+2}}}$, i.e.,
$a=(a_1,\dots,a_{n}, -A/2, -A/2)$, where $A=\sum_{i=1}^na_i$ and $a_i > 0$ for $1 \leq i \leq n$,
we construct the following three Khintchine$_{\mu_{p^{n+2}}}$ instances:
\begin{align*}
    c&=2(a_1, a_2,\dots, a_{n},-A/2,-A/2), \\
    d&=(a_1-y, a_2, \dots, a_{n},-A/2+y/2,-A/2+y/2),\\
    e&=(a_1+y,a_2, \dots, a_{n},-A/2-y/2,-A/2-y/2),
\end{align*}
where $0<y< 1/2 < a_1$. We will show that solving the above instances of Khintchine$_{\mu_{p^{n+2}}}$ suffices
to solve our given instance of \#Partition$_{\mu_{p^{n+2}}}$. This will complete the proof of Theorem~\ref{thm:KK-hard}.
To do so, we require some case analysis and explicit calculations.

\smallskip

\noindent For any $x \in \{-1,1\}^{n+2}$, we have that:
\begin{align*}
 &|d \cdot x|= \begin{cases}
       |\sum_{i=1}^na_ix_i +A-2y|, & \textrm{ if  } x_1=1, x_{n+1}=x_{n+2}=-1\\
	  |\sum_{i=1}^na_ix_i+A|, & \textrm{ if  } x_1=-1, x_{n+1}=x_{n+2}=-1\\
      |\sum_{i=1}^na_ix_i-A|, &\textrm{ if  } x_1=1,x_{n+1}=x_{n+2}=1 \\
     |\sum_{i=1}^na_ix_i-A+2y|, &\textrm{ if  } x_1=-1, x_{n+1}=x_{n+2}=1\\
  |\sum_{i=1}^na_ix_i-y|, &\textrm{ if  } x_1=1,x_{n+1}\neq x_{n+2}\\
|\sum_{i=1}^na_ix_i+y|, &\textrm{ if  } x_1=-1,x_{n+1}\neq x_{n+2}\\
   \end{cases}
\end{align*}
and similarly
\begin{align*}
 |e \cdot x|= \begin{cases}
       |\sum_{i=1}^na_ix_i+A+2y|, &\textrm{ if  } x_1=1,x_{n+1}=x_{n+2}=-1\\
	  |\sum_{i=1}^na_ix_i+A|, &\textrm{ if  } x_1=-1, x_{n+1}=x_{n+2}=-1\\
      |\sum_{i=1}^na_ix_i-A|, &\textrm{ if  } x_1=1, x_{n+1}=x_{n+2}=1 \\
     |\sum_{i=1}^na_ix_i-A-2y|, &\textrm{ if  } x_1=-1, x_{n+1}=x_{n+2}=1\\
  |\sum_{i=1}^na_ix_i+y|, &\textrm{ if  } x_1=1, x_{n+1}\neq x_{n+2}\\
|\sum_{i=1}^na_ix_i-y|, &\textrm{ if  } x_1=-1,x_{n+1}\neq x_{n+2}\\
   \end{cases}
\end{align*}
Given the above, we observe that the following hold:
\begin{itemize}
\item For $x \in \{-1,1\}^{n+2}$, $wt(x) \in \{0, n+2\}$,
\[|c \cdot x|= |e \cdot x|=|d \cdot x|=0 \;.\]
\item For $x \in \{-1,1\}^{n+2}$, $x_{n+1} = x_{n+2}=x_1$,
\[|d \cdot x| +|e \cdot x|=|c \cdot x| \;.\]
\item For $x \in \{-1,1\}^{n+2}$, $x_{n+1} \neq x_{n+2}$:
\begin{itemize}
\item If $c \cdot x \neq 0$,
\[|d \cdot x|+|e \cdot x|=|c \cdot x| \;,\]
as it holds that
$|\sum_{i=1}^n a_ix_i-y|+|\sum_{i=1}^n a_i x_i+y|=2\max(|\sum_{i=1}^n a_i x_i|,|y|)$
and $|\sum_{i=1}^n a_i x_i| \geq 1$.
\item If $c \cdot x=0$, then
\[ |d \cdot x|+|e \cdot x|=2y \;. \]
\end{itemize}
\item For $x \in \{-1,1\}^{n+2}$, $x_{n+1}=x_{n+2}$, $x_1 \neq x_{n+1}$
\[|d \cdot x|+|e \cdot x|=|c \cdot x| \;, \]
as similarly with the above case, we have that
$|\sum_{i=1}^{n+2}a_ix_i-2y|+|\sum_{i=1}^{n+2}a_i x_i+2y|=2\max(|\sum_{i=1}^{n+2}a_i x_i|,|2y|)$
and $|\sum_{i=1}^{n+2} a_i x_i| \geq 1 $.
\end{itemize}
Hence, we have:
\begin{dmath*}
    K_{\mu_{p^{n+2}}}(d)+K_{\mu_{p^{n+2}}}(e)-K_{\mu_{p^{n+2}}}(c) =
    \mathbb{E}_{x \sim \mu_{p^{n+2}}}\left[ |d \cdot x| \right]+
    \mathbb{E}_{x \sim \mu_{p^{n+2}}}\left[|e \cdot x|\right]-
     \mathbb{E}_{x \sim \mu_{p^{n+2}}}\left[|c \cdot x|\right] =
    \sum_{x \in \{-1,1\}^{n+2}} 2y \cdot \mu_{p^{n+2}}(x) \cdot \mathds{1}_{c \cdot x=0, x \not \in \{\bf{-1},\bf{1}\}} \;.
\end{dmath*}
So, we get:
\begin{align*}
\mathrm{Pr}_{x \sim \mu_{p^{n+2}}}[a \cdot x=0]=& \frac{K_{\mu_{p^{n+2}}}(d)+K_{\mu_{p^{n+2}}}(e)-K_{\mu_{p^{n+2}}}(c)}{2y}+\mathrm{Pr}_{x \sim \mu_{p^{n+2}}}[x=\textbf{-1}]+\mathrm{Pr}_{x \sim \mu_{p^{n+2}}}[x=\bf{1}]\\&=\frac{K_{\mu_{p^{n+2}}}(d)+K_{\mu_{p^{n+2}}}(e)-K_{\mu_{p^{n+2}}}(c)}{2y}+\frac{p_0+p_{n+1}}{\Lambda(p^{n+2})} \;.
\end{align*}
The proof of Theorem~\ref{thm:KK-hard} is now complete.
\end{proof}

We are now ready to prove Theorem~\ref{thm:opt-hard}.
\begin{proof}[Proof of Theorem~\ref{thm:opt-hard}]
We reduce from the Khintchine$_{\mu_{p^{n+2}}}$ problem: given the vector
$a=(a_1, \dots, a_n, a_{n+1}=-A/2, a_{n+2}= -A/2) \in \mathbb{Q}^{n+2}$,
where $A=\sum_{i=1}^n a_i$ and $a_i >0$ for $0 \leq i \leq n$, we want to compute
$\max_{c \in \mathcal{C}_{p^{n+2}}}a \cdot c$.
For any $c \in \mathcal{C}_{p^{n+2}}$, using our reformulation of semivalues (\ref{sem}), we have:
\begin{align*}
a \cdot c&= \frac{1}{2}\sum_{i=1}^{n+2}a_i \widehat{f}^{p^{n+2}}(i)+\frac{1}{2}\sum_{i=1}^{n+2}a_iC_f^{p^{n+2}}
\\&=\frac{\Lambda(p^{n+2})}{2}\sum_{x \in \{-1,1\}^{n+2}}f(x)\mu_{p^{n+2}}(x)\sum_{i=1}^{n+2}a_ix_i
 \\&\leq \frac{\Lambda(p^{n+2})}{2}\sum_{x \in \{-1,1\}^{n+2}}\mu_{p^{n+2}}(x) \left| \sum_{i=1}^{n+2}a_ix_i \right|\;,
\end{align*}
where $f$ is a linear threshold function.

Reducing from a weight vector that has the sum of its coordinates equal to zero was essential for this step:
the term $C_f^{p^{n+2}}$ vanishes and we end up with the term
$\sum_{x \in \{-1,1\}^{n+2}}f(x)\mu_{p^{n+2}}(x)\sum_{i=1}^{n+2}a_ix_i$
that is upper bounded by  $\sum_{x \in \{-1,1\}^{n+2}}\mu_{p^{n+2}}(x)|\sum_{i=1}^{n+2}a_ix_i|$.

This upper bound is tight, as we show below, which is crucial for our argument.
If one were to reduce from a vector with sum different than $0$ or include in the polytope only linear threshold functions with positive weights,
it would not have been possible to obtain a tight upper bound.

Observe that
\[a \cdot c =\frac{\Lambda(p^{n+2})}{2} \sum_{x \in \{-1,1\}^{n+2}}\mu_{p^{n+2}}(x)\left|\sum_{i=1}^{n+2}a_ix_i \right|\;,\]
for $c=(\widetilde{f}^{p^{n+2}}(1),\dots, \widetilde{f}^{p^{n+2}}(n+2))$,
where $f(x)=\mathrm{sign}(\sum_{i=1}^{n+2}a_ix_i).$
So,
\[\mathbb{E}_{x \sim \mu_{p^{n+2}}}\left[\left|\sum_{i=1}^{n+2}a_ix_i \right|\right]=
\frac{2\max_{c \in \mathcal{C}_{p^{n+2}}}a \cdot c}{\Lambda(p^{n+2})} \;.\]
That is, a solution to our instance of linear optimization over our polytope
gives a solution to the initial instance of the Khintchine$_{\mu_{p^{n+2}}}$ problem,
and the proof of Theorem~\ref{thm:opt-hard} is complete.
\end{proof}

\paragraph{Linear Optimization Using a Verification Oracle}
We prove that the {\em restricted verification problem}, defined below, is computationally hard, where
the input linear threshold functions are defined by weight vectors of the specific
form described in Definition \ref{poly}.

\medskip

\fbox{\parbox{6in}{
\noindent \textbf{Name:} SVR$_{p^{n+2}}$-Verification Problem

\smallskip

\textbf{Input:} A vector $(c_1, \dots, c_{n+2}) \in \mathbb{Q}^{n+2}$ and a vector $w=(w_1, \dots, w_{n+2})\in \mathbb{Q}^{n+2}$ such that $w_{n+1}=w_{n+2}=-\sum_{i=1}^nw_i/2$ and $w_i>0, i \in \{1, \dots, n\}$.\\
\textbf{Question:} Does it hold that $\widetilde{f}^{p^{n+2}}_{w, 0}(i)=c_i$ for $i \in \{1,\dots, {n+2}\}$?

}}

\medskip

\begin{theorima} \label{thm:svr-hard} \label{hi}
If $p^{n+2}$ is a reasonable probability vector, the SVR$_{p^{n+2}}$-Verification problem
is not in the $k$-th level of the polynomial hierarchy,
unless \#P is contained in the $(k+2)$-level.
\end{theorima}

The main idea behind the proof is that one can solve the linear optimization problem using a membership oracle
of the polytope which can be obtained if we have an efficient algorithm for the restricted verification problem.
Since the vertices of the polytope correspond to semivalues of linear threshold functions, if we have an efficient
algorithm for the verification problem, then we can efficiently verify that a vector is a vertex of the polytope,
and. using Caratheodory's theorem, we can get a membership oracle. In this way, we obtain a contradiction,
unless the polynomial hierarchy collapses: if the verification problem is in the polynomial hierarchy,
then a \#P-hard problem lies in the polynomial hierarchy.

The {\em restricted membership problem}
for any probability vector $p^{n+2}$ is defined below:

\medskip

\fbox{\parbox{6in}{

\noindent \textbf{Name:} SVR$_{p^{n+2}}$-Membership Problem

\smallskip

\textbf{Input:} A vector $c=(c_1, \dots, c_{n+2}) \in \mathbb{Q}^{n+2}$.\\
\textbf{Question:} Is $c$ in $\mathcal{C}_{p^{n+2}}$ ?

}}

\medskip

\begin{proof}[Proof of Theorem~\ref{thm:svr-hard}]
We first prove the following lemma that shows how an efficient oracle
for the restricted verification problem
can be used to obtain a membership oracle:

\begin{lemma}
If the SVR$_{p^{n+2}}$-Verification problem is in the $k$-th level of PH,
then the SVR$_{p^{n+2}}$-Membership problem is in the $(k+1)$-level.
\end{lemma}

\begin{proof}
Assume that the SVR$_{p^{n+2}}$-Verification problem is in the $k$-th level of PH.
By Caratheodory's theorem a point $c$ is in $\mathcal{C}_{p^{n+2}}$ iff it is a convex combination
of at most $n+3$ vertices of $\mathcal{C}_{p^{n+2}}$, i.e., $c=\sum_{i=1}^m \lambda_i x^{(i)}$,
where $x^{(i)}$ is a vertex, $\lambda_i \geq 0$ for $1 \leq i \leq m \leq n+3$,
and $\sum_{i=1}^m \lambda_i=1$. So, it can be certified that a given point $c$ is in
$\mathcal{C}_{p^{n+2}}$ by finding the $m \leq n+3$ vertices $x^{(i)}$ and computing the $m$
scaling factors $\lambda_i$. Given the $x^{(i)}$, one can verify that $x^{(i)}$ is a vertex of $\mathcal{C}_{p^{n+2}}$
by finding a weight vector $w^{(i)}$ of the form described in Definition 4 such that $x^{(i)}$
is the $\widetilde{f}^{p^{n+2}}_{w^{(i)},0}$ vector, as the vertices of $\mathcal{C}_{p^{n+2}}$
correspond to linear threshold functions with weights of this specific form. So, if we are given
the $x^{(i)}$ and the corresponding $w^{(i)}$, we can verify in polynomial time with a $k$-th level
oracle that $x^{(i)}$ is the $\widetilde{f}^{p^{n+2}}_{w^{(i)},0}$ vector
as we assumed that the SVR$_p^{n+2}$-Verification problem is in the $k$-th level of PH.
Thus, there is a polynomial-size certificate that can be checked in polynomial time with a $k$-th level oracle
when $c$ is in $\mathcal{C}_{p^{n+2}}$: the $m \leq n+3$ vertices $x^{(i)}$, where each $x^{(i)}$ can be
represented by poly($n$) bits by assumption; and the $m$ corresponding $w^{(i)}$ vectors,
where each $w^{(i)}$ can be represented by poly($n$) bits,
as every linear threshold function can be represented with weight $w=(w_1, \dots, w_{n+2})$
such that each $w_i$ is an integer that satisfies $|w_i| \leq 2^{(O(n\log n))}$ \cite{Mur61}.
Given the $x^{(i)}$ and the $w^{(i)}$, it can be verified in polynomial time with a $k$-th level
oracle that the $x^{(i)}$ are vertices and then we can compute in polynomial time the $\lambda_i$
coefficients by solving the linear system $c=\sum_{i=1}^m \lambda_i x^{(i)}$. Thus, if the
SVR$_{p^{n+2}}$-Verification problem is in the $k$-th level of PH,
the SVR$_{p^{n+2}}$-Membership problem is in the $(k+1)$-level.
\end{proof}

As $\mathcal{C}_{p^{n+2}}$ has non-empty interior, if the SVR$_{p^{n+2}}$-Membership problem is in the $(k+1)$-level of PH,
then using the ellipsoid algorithm, we could solve the optimization problem using a polynomial number
of membership-oracle calls (page 189, \cite{ShL82}). Hence, we would have that the
SV$_{p^{n+2}}$-Optimization problem, which by Theorem~\ref{thm:opt-hard} is $\#$P-hard,
is in the $(k+2)$-level of PH. This completes the proof of Theorem~\ref{thm:svr-hard}.
\end{proof}

\paragraph{Hardness of the Verification Problem for Weighted Voting Games}
One important issue is that the computational problems we have considered so far
involve linear threshold functions some of whose weights can be negative.
This seemed necessary to some extent for our arguments, as it is crucially exploited
in the proof of Theorem~\ref{thm:opt-hard}.

We now show how to switch to weighted voting games (i.e., LTFs with non-negative weights),
which was our initial goal. Using a bijection between the semivalues of a linear threshold function
with weight vector $w = (w_1, \ldots, w_n)$ of the form described in Definition \ref{poly}
and a linear threshold function with weight vector $w' = (|w_1|, \ldots, |w_n|)$,
we show the equivalence between the restricted verification problem and the verification problem
for linear threshold functions defined by positive weights.

\medskip

\fbox{\parbox{6in}{

\noindent \textbf{Name:} SV$_{p^{n}}$-Verification Problem

\smallskip

\textbf{Input:} A vector $(c_1, \dots, c_{n}) \in \mathbb{Q}^{n}$, a vector $w=(w_1,\dots, w_{n}) \in \mathbb{Q}_{+}^n$
and $\theta \in \mathbb{Q}$.\\
\textbf{Question:} Does it hold that $\widetilde{f}^{p^{n}}_{w, \theta}(i)=c_i$ for $i \in \{1,\dots, {n}\}$?

}}

\medskip

\begin{theorima} \label{thm:verification-equiv}
If the SV$_{p^{n+2}}$-Verification problem  is in the $k$-th level of PH,
then the SVR$_{p^{n+2}}$-Verification problem is in the $k$-th level of PH.
\end{theorima}
\begin{proof}
We use the following lemma that shows how one can compute the semivalues
of a linear threshold function with weight vector $w$ of the form
described in Definition \ref{poly} given the semivalues of the linear threshold function
with weight vector the absolute values of $w$, and vice-versa.

\begin{lemma} \label{pton}
For a positive integer $n$, fix $a_i > 0, 1 \leq i \leq n$, and $A=\sum_{i=1}^na_i$.
Consider the LTFs $f(x)=\mathrm{sign}(\sum_{i=1}^n a_i x_i - (A/2) x_{n+1} -(A/2) x_{n+2})$ and
$g(x)=\mathrm{sign}(\sum_{i=1}^n a_i x_i+(A/2) x_{n+1}+(A/2) x_{n+2})$. Then,
we have the following:
\begin{enumerate}
\item[(i)] For $1 \leq i \leq n $, it holds
$\widetilde{g}^{p^{n+2}}(i)=\widetilde{f}^{p^{n+2}}(i)-2(p^{n+2}_{n+1}-p^{n+2}_{n-1})$, and
\item[(ii)] For $n+1 \leq i \leq n+2$, it holds
$\widetilde{g}^{p^{n+2}}(i)=\widetilde{f}^{p^{n+2}}(i)+2\sum_{t=0}^{n-1}{\binom{n}{t}}(p^{n+2}_t+p^{n+2}_{t+1})$.
\end{enumerate}
\end{lemma}
\begin{proof}
For $i \in \{1, \dots, n\}$, from the definitions of semivalues and functions $f$, $g$ we get:
\begin{dmath*}
\ \widetilde{g}^{p^{n+2}}(i)
=\sum_{x:x_i=-1}p_{wt(x)}^{n+2}(g(x^i)-g(x))
=\sum_{\substack{x:x_i=-1,\\x_{n+1}=x_{n+2}=-1}}p_{wt(x)}^{n+2}\left(\mathrm{sign}\left(\sum_{j\leq n:j \neq i}a_jx_j+a_i-A\right)-\mathrm{sign}\left(\sum_{j\leq n:j \neq i}a_jx_j-a_i-A\right)\right)+
\sum_{\substack{x:x_i=-1,\\x_{n+1}=x_{n+2}=1}}p_{wt(x)}^{n+2}\left(\mathrm{sign}\left(\sum_{j\leq n:j \neq i}a_jx_j+a_i+A\right)-\mathrm{sign}\left(\sum_{j\leq n:j \neq i}a_jx_j-a_i+A\right)\right)
\end{dmath*}
\begin{dmath*}
+\sum_{\substack{x:x_i=-1,\\x_{n+1}\neq x_{n+2}}}p_{wt(x)}^{n+2}\left(\mathrm{sign}\left(\sum_{j\leq n:j \neq i}a_jx_j+a_i\right)-\mathrm{sign}\left(\sum_{j\leq n:j \neq i}a_jx_j-a_i\right)\right)
\end{dmath*}
and
\begin{dmath*}
\widetilde{f}^{p^{n+2}}(i)
=\sum_{x:x_i=-1}p_{wt(x)}^{n+2}(f(x^i)-f(x))
=\sum_{\substack{x:x_i=-1,\\x_{n+1}=x_{n+2}=1}}p^{n+2}_{wt(x)}\left(\mathrm{sign}\left(\sum_{j\leq n:j \neq i}a_jx_j+a_i-A\right)-\mathrm{sign}\left(\sum_{j\leq n:j \neq i}a_jx_j-a_i-A\right)\right)
+\sum_{\substack{x:x_i=-1,\\x_{n+1}=x_{n+2}=-1}}p^{n+2}_{wt(x)}\left(\mathrm{sign}\left(\sum_{j\leq n:j \neq i}a_jx_j+a_i+A\right)-\mathrm{sign}\left(\sum_{j\leq n:j \neq i}a_jx_j-a_i+A\right)\right)
+\sum_{\substack{x:x_i=-1,\\x_{n+1}\neq x_{n+2}}}p^{n+2}_{wt(x)}\left(\mathrm{sign}\left(\sum_{j\leq n:j \neq i}a_jx_j+a_i\right)-\mathrm{sign}\left(\sum_{j\leq n:j \neq i}a_jx_j-a_i\right)\right) \;.\end{dmath*}
As $f(x)=-1$ for any $x \neq \bf{1}$ with $x_{n+1}=x_{n+2}=1$,
$f(x)=1$ for any $x$ with $x_{n+1}=x_{n+2}=-1$,
$g(x)=1$ for any $x$ with $x_{n+1}=x_{n+2}=1$, and
$g(x)=-1$ for any $x\neq(1,\dots,1,-1,-1)$ with $x_{n+1}=x_{n+2}=-1$,
we get that
\[\widetilde{f}^{p^{n+2}}(i)-\widetilde{g}^{p^{n+2}}(i)=2(p^{n+2}_{n+1}-p^{n+2}_{n-1}) \;.\]
For $i = n+1$, we can write:
\begin{dmath*}
\widetilde{g}^{p^{n+2}}(n+1)
=\sum_{x:x_{n+1}=-1}p_{wt(x)}^{n+2}(g(x^{n+1})-g(x))
=\sum_{\substack{x:x_{n+1}=-1,\\x_{n+2}=-1}}p^{n+2}_{wt(x)}\left(\mathrm{sign}\left(\sum_{j\leq n}a_jx_j\right)-
\mathrm{sign}\left(\sum_{j\leq n}a_jx_j-A\right)\right)
+ \sum_{\substack{x:x_{n+1}=-1,\\x_{n+2}=1}}p^{n+2}_{wt(x)}\left(\mathrm{sign}\left(\sum_{j\leq n}a_jx_j+A\right)
-\mathrm{sign}\left(\sum_{j\leq n}a_jx_j\right)\right)\;.
\end{dmath*}
and
\begin{dmath*}
\widetilde{f}^{p^{n+2}}(n+1)
=\sum_{x:x_{n+1}=-1}p_{wt(x)}^{n+2}(f(x^{n+1})-f(x))
= \sum_{\substack{x:x_{n+1}=-1,\\x_{n+2}=1}}p^{n+2}_{wt(x)}\left(\mathrm{sign}\left(\sum_{j\leq n}a_jx_j-A\right)
-\mathrm{sign}\left(\sum_{j\leq n}a_jx_j\right)\right)
+\sum_{\substack{x:x_{n+1}=-1,\\x_{n+2}=-1}}p^{n+2}_{wt(x)}\left(\mathrm{sign}\left(\sum_{j\leq n}a_jx_j\right)
-\mathrm{sign}\left(\sum_{j\leq n}a_jx_j+A \right)\right) \;.
\end{dmath*}
We thus have
\begin{dmath*}
\widetilde{f}^{p^{n+2}}(n+1)-\widetilde{g}^{p^{n+2}}(n+1)
=\sum_{\substack{x:x_{n+1}=-1,\\x_{n+2}=-1}}p^{n+2}_{wt(x)}\left(\mathrm{sign}\left(\sum_{j\leq n}a_jx_j-A\right)
-\mathrm{sign}\left(\sum_{j\leq n}a_jx_j+A\right)\right)
+\sum_{\substack{x:x_{n+1}=-1,\\x_{n+2}=1}}p^{n+2}_{wt(x)}\left(\mathrm{sign}\left(\sum_{j\leq n}a_jx_j-A\right)
-\mathrm{sign}\left(\sum_{j\leq n}a_jx_j+A\right)\right)
=-2\sum_{t=0}^{n-1}{n \choose t}(p^{n+2}_t+p^{n+2}_{t+1}) \;.
\end{dmath*}
In the same way, we get that
\[\widetilde{f}^{p^{n+2}}(n+2)-\widetilde{g}^{p^{n+2}}(n+2)
=-2\sum_{t=0}^{n-1}{n \choose t}(p^{n+2}_t+p^{n+2}_{t+1}) \;.
\]
This completes the proof of Lemma~\ref{pton}.
\end{proof}

Given an instance $(a_1, \dots, a_n, -A/2, -A/2)$, $(c_1, \dots, c_{n+2})$ of the SVR$_{p^{n+2}}$-Verification problem,
we construct the following instance of SV$_{p^{n+2}}$-Verification problem:
$(a_1, \dots, a_n, +A/2, +A/2)$, $\theta=0$,
\begin{align*}
&\Big(c_1-2(p^{n+2}_{n+1}-p^{n+2}_{n-1}), \dots, c_n-2(p^{n+2}_{n+1}-p^{n+2}_{n-1}), c_{n+1}+2\sum_{t=0}^{n-1}{\binom{n}{t}}(p^{n+2}_t+p^{n+2}_{t+1}),\\
&c_{n+2}+2\sum_{t=0}^{n-1}{\binom{n}{t}}(p^{n+2}_t+p^{n+2}_{t+1})\Big) \; .
\end{align*}
By Lemma \ref{pton}, we have that the SVR$_{p^{n+2}}$-Verification instance is a ``YES''-instance
iff the SV$_{p^{n+2}}$-Verification instance is a ``YES''-instance.
This completes the proof of Theorem~\ref{thm:verification-equiv}.
\end{proof}

\paragraph{Verification Using Inverse Oracle}
The final step of our proof is to show that the inverse problem for semivalues
is at least as hard as the verification problem. While this is intuitively obvious,
the proof requires the following non-trivial structural result:
The semivalues of a weighted voting game characterize the game within the space of weighted voting games.

\begin{theorima} \label{thm:inverse-vs-ver}
If the SV$_{p^{n}}$-Inverse problem is in the $k$-th level of PH,
then the SV$_{p^{n}}$-Verification problem is in the $(k+1)$-level.
\end{theorima}
\begin{proof}
The proof makes essential use of the following lemma that shows
that if two LTFs with normalized weights and
the same threshold have the same semivalues, then they are equal
on all points $x$ that are given positive probability by the distribution $\mu'_{p^n}$.
This lemma is qualitatively similar to (and inspired by) Chow's Theorem~\cite{chow},
that shows that a linear threshold function is uniquely determined by its
Chow parameters:

\begin{lemma} \label{equiv}
Let $f(x) = \mathrm{sign}(w \cdot x - \theta)$ and $g(x) = \mathrm{sign}(v \cdot x - \theta)$
where $\sum_{i=1}^n w_i = \sum_{i=1}^n v_i$. If $\widetilde{f}^{p^{n}}(i)=\widetilde{g}^{p^{n}}(i)$ for $i \in \{1, \dots, n\}$,
then $f(x)=g(x)$ for all $x \in \{-1,1\}^n$ such that $p^n_{wt(x)}+p^n_{wt(x)-1} \neq 0$
and $|w \cdot x-\theta|+|v \cdot x-\theta| \neq 0$.
\end{lemma}
\begin{proof}
By our assumption that $\widetilde{f}^{p^{n}}(i)=\widetilde{g}^{p^{n}}(i)$ for $i \in \{1, \dots, n\}$
it follows that:
\begin{align*}
&2\sum_{i=1}^n w_i (\widetilde{f}^{p^n}(i)-\widetilde{g}^{p^{n}}(i))+
2\sum_{i=1}^n v_i(\widetilde{g}^{p^{n}}(i)-\widetilde{f}^{p^n}(i))
-\theta \sum_{x \in \{-1,1\}^n}(f(x)-g(x))\mu'_{p^n}(x) \\&
- \theta\sum_{x \in \{-1,1\}^n}(g(x)-f(x))\mu'_{p^n}(x)=0
\end{align*}
Recalling our reformulation of the semivalues (\ref{sem}), we equivalently have:
\begin{align*}
&\sum_{i=1}^n w_i \left(\widehat{f}^{p^n}(i)-\widehat{g}^{p^n}(i)+C_f^{p^n}-C^{p^n}_g\right)
+ \sum_{i=1}^n v_i \left(\widehat{g}^{p^n}(i)-\widehat{f}^{p^n}(i)+C_g^{p^n}-C^{p^n}_f\right)\\
&-\theta \sum_{x \in \{-1,1\}^n}(f(x)-g(x)) \mu'_{p^n}(x)
- \theta \sum_{x \in \{-1,1\}^n}(g(x)-f(x)) \mu'_{p^n}(x)=0\\
& \Leftrightarrow
\sum_{x \in \{-1,1\}^n}(f(x)-g(x))\mu'_{p^n}(x) (w \cdot x)
+\sum_{x \in \{-1,1\}^n}(g(x)-f(x))\mu'_{p^n}(x) (v \cdot x) \\
&-\theta\sum_{x \in \{-1,1\}^n}(f(x)-g(x))\mu'_{p^n}(x)
-\theta\sum_{x \in \{-1,1\}^n}(g(x)-f(x))\mu'_{p^n}(x)=0 \\
& \Leftrightarrow \sum_{x \in \{-1,1\}^n} (f(x)-g(x)) \mu'_{p^n}(x) (w\cdot x-\theta)
+\sum_{x \in \{-1,1\}^n}(g(x)-f(x))\mu'_{p^n}(x)(v \cdot x-\theta)=0\\
&\Leftrightarrow \sum_{x \in \{-1,1\}^n} \mu'_{p^n}(x)|f(x)-g(x)| |w \cdot x-\theta|
+\sum_{x \in \{-1,1\}^n}\mu'_{p^n}(x)|g(x)-f(x)| |v \cdot x-\theta|=0 \\
& \Leftrightarrow \sum_{x \in \{-1,1\}^n} \mu'_{p^n}(x) |f(x)-g(x)| (|w \cdot x-\theta|+|v \cdot x-\theta|)=0 \;.
\end{align*}
Hence, for any $x \in \{-1,1\}^n$ such that $|w \cdot x-\theta|+|v \cdot x-\theta| \neq 0$
and $\mu'_{p^{n}}(x) \neq 0$, we have that $f(x)=g(x)$.
This completes the proof of Lemma~\ref{equiv}.
\end{proof}

We are now ready to complete the proof.
Given an SV$_{p^{n}}$-Verification instance
$a=(a_1, \dots, a_n)$, $\theta$, $(c_1, \dots, c_{n})$,
we create the following instance of the SV$_{p^n}$-Inverse problem:
$\theta' = \theta/\sum_{i=1}^n{a_i}$, $(c_1, \dots, c_{n})$. Then, if the SV$_{p^{n}}$-Inverse instance
is a ``NO''-instance, we have a ``NO''-instance of the SV$_{p^{n}}$-Verification problem.
If the SV$_{p^n}$-Inverse output is a weight vector $w=(w_1, \dots, w_n)$,
we can check with a co-NP oracle if the functions $f_{w,\theta'}$
and $f_{a,\theta}$ have the same semivalues:
By Lemma \ref{equiv}, they have the same semivalues iff
there is no $x \in \{-1, 1\}^n$ such that $\mu'_{p^n}(x) > 0$ and
$f_{w,\theta'}(x) \neq f_{a,\theta}(x)=f_{\frac{a}{\sum_{i=1}^n{a_i}}, \frac{\theta}{\sum_{i=1}^n{a_i}}}(x)$.
This completes the proof of Theorem~\ref{thm:inverse-vs-ver}.
\end{proof}

Theorem~\ref{main} now follows by combining Theorems~\ref{thm:svr-hard},~\ref{thm:verification-equiv},
and~\ref{thm:inverse-vs-ver}.

\section{Conclusions}
The inverse power index problem has received considerable attention in game theory and social choice,
and the inverse Banzhaf index problem has been relevant in other fields as well, such as circuit complexity
and computational learning. In this paper, we proved that the inverse semivalue problem, for {\em reasonable}
probability distributions, is computationally intractable. As special cases, we deduce that the inverse Banzhaf
index and inverse Shapley value problems are also intractable.
A number of interesting open questions remain: Can we design efficient approximation algorithms for the inverse problem
in the case of more general semivalues? Can we characterize the computational complexity of the inverse power index problem
for power indices that do not belong in the semivalues class?

\paragraph{Acknowledgements.} We thank Shaddin Dughmi for his contributions to the early stages of this work.
We are grateful to Kousha Etessami for numerous helpful discussions.

\bibliographystyle{alpha}

redpajama_set_name: RedPajamaArXiv


Signed by author with gift inscription on title page. Previous owner bookplate on blank front page. Slight wear to some front page edges. Text and illustrations unmarked and interior otherwise clean. Dust jacket price clipped. Otherwise clean and intact. First edition, first printing. Please see image if available. 190 pages.


text:
Ignacy Kwarta (ur. 25 lipca 1898 w Poczapińcach, zm. 23 maja 1945 w Krakowie) – doktor nauk medycznych, więzień Auschwitz-Birkenau, działacz społeczny.

Młodość
Urodzony w rodzinie chłopskiej, był synem Jakuba. Wczesne dzieciństwo spędził w Poczapińcach (Ukraina); po ukończeniu szkoły powszechnej we wcześniej wspomnianej miejscowości kontynuował naukę w gimnazjum w Tarnopolu, aż do wybuchu I wojny światowej. W 1916 udał się wraz z rodziną do Wiednia, gdzie 26 lutego 1917 zdobył świadectwo dojrzałości, po uprzednim zdaniu egzaminu w języku polskim. Od kwietnia 1916 (do końca wojny) pozostawał w służbie wojskowej w armii Austro-Węgier, dochodząc do szarży podporucznika. Po odzyskaniu przez Polskę niepodległości wstąpił do polskiej armii. 20 października 1919 rozpoczął studia medyczne na wydziale lekarskim Uniwersytetu Jana Kazimierza we Lwowie. Jako stypendysta Ministerstwa Obrony Narodowej 4 grudnia 1924 uzyskał tytuł doktora wszech nauk lekarskich.

Praca po studiach
Po ukończeniu studiów został skierowany został do wojskowej służby medycznej w Korpusie Ochrony Pogranicza w Batalionie "Ludwikowo", pełniąc tam funkcję lekarza. W późniejszym czasie obłożono go zarzutami o współpracę z komunistami, a także odmowę wypisania nieprawdziwego aktu zgonu członka bandy przemytniczej zastrzelonego w czasie potyczki granicznej, dając tym samym dowód lekarskiej niezawisłości (według niektórych źródeł mówi się o zastrzelonym członku bojówek ukraińskich, co jest błędne zarówno z uwagi na odcinek granicy, jak i czas zdarzenia), co w ostateczności przyczyniło się do zwolnienia ze służby przed 1 listopada 1926 r., ponieważ dokumenty wykazujące liczbę oficerów ze wspomnianego dnia, nie zawierają jego nazwiska.

Niedługo po odbyciu przeszkolenia dla lekarzy powiatowych w maju 1929 podjął pracę w Świeciu koło Grudziądza. Oprócz pracy zawodowej, zorganizował też młodzieżowe koło "Wici" i oddział Towarzystwa Uniwersytetów Robotniczych. Uczestniczył też w dyskusjach w Kole Wolnomyślicieli.

Po oskarżeniu o działalność komunistyczną został zwolniony z pracy i w 1931 przeniósł się do Warszawy, a władze rozwiązały lokalną organizację Towarzystwa Uniwersytetów Robotniczych w Świeciu. W Warszawie Kwarta został referentem w Ministerstwie Opieki Społecznej, gdzie jednak już w październiku 1933 został zawieszony w obowiązkach, a w styczniu kolejnego roku zwolniony.

Przed II wojną światową w Jędrzejowie
Ostatnim miastem pobytu Ignacego Kwarty był Jędrzejów. Znalazł się tutaj za pośrednictwem Feliksa Przypkowskiego, który pragnął, aby Kwarta został jego następcą. Od 15 sierpnia 1934 sprawował funkcję lekarza powiatowego w Jędrzejowie.

W Jędrzejowie kontynuował swoją działalność polityczną, współpracując z komunistami, socjalistami i ludowcami, a także z organizacją Związku Młodzieży Wiejskiej "Wici". Angażował się w działalność społeczną, był prezesem lokalnego oddziału Polskiego Czerwonego Krzyża, współtworzył oddział Towarzystwa Uniwersytetu Robotniczego TUR i prowadził szkolenia drużyn ratowniczych. Wygłaszał wykłady propagandowe w Lidze Ochrony Powietrznej i Przeciwgazowej, w Powiatowym Komitecie Przysposobienia Wojskowego i Wychowania Fizycznego oraz w Komitecie do Walki z Gruźlicą. Miejscowa ludność ceniła go za humanitarne gesty, jakie czynił w stosunku do najuboższych, udzielając bezpłatnych porad, a nawet przekazując pieniądze na zakup najpotrzebniejszych lekarstw.

Wygłaszane przez niego lewicowe poglądy oraz kontakty z działaczami PPS (m.in. ze Stanisławem Dubois) spowodowały, że 2 listopada 1934 został aresztowany pod zarzutem publicznej krytyki władz sanacyjnych i osadzony na Zamku w Kielcach. W więzieniu podjął dwutygodniowy strajk głodowy. Na skutek licznych interwencji opinii publicznej, prasy PPS, protestów komunistów z Francji i innych krajów, a także staraniem żony doktora Kwarty, Marii, która dotarła do prezydenta Ignacego Mościckiego z prośbą o uwolnienie, opuścił więzienie 22 stycznia 1935. Jednocześnie wojewoda kielecki zwolnił go ze stanowiska lekarza powiatowego i zawnioskował o przeniesienie na emeryturę.

W ostatnich latach przed wojną Kwarta pozostawał zaangażowany w lewicową działalność społeczną, przewodniczył oddziałowi TUR w Jędrzejowie, w 1935 uczestniczył w akcji bojkotu wyborów parlamentarnych. Na wiecach propagował ideę utworzenia ludowego frontu antyfaszystowskiego. Był też współorganizatorem imprez kulturalnych.

II wojna światowa
We wrześniu 1939 Kwarta, jako inwalida wojenny z czasów I wojny światowej, kiedy to doznał postrzałowego złamania prawego ramienia, nie brał udziału w kampanii wrześniowej, przygotował natomiast szpital dla rannych w Jędrzejowie.

W czasie okupacji niemieckiej był obserwowany i szykanowany. Zabroniono mu opuszczania miasta i odwiedzania chorych w porze nocnej. Jesienią 1939 w mieszkaniu doktora przeprowadzono rewizję, a następnie aresztowano go i przetrzymywano w magistrackim areszcie jako zakładnika. Do działalności konspiracyjnej włączył się wiosną 1940. Był współorganizatorem powstałej na terenie Jędrzejowa lewicowej organizacji "Młot i Sierp". Należał też do Straży Chłopskiej "Chłostry", przemianowanej później na Bataliony Chłopskie.

W nocy z 14 na 15 czerwca 1942 został aresztowany przez Gestapo, a następnie przewieziony do więzienia na ulicy Zamkowej w Kielcach. We wrześniu 1942 został umieszczony w niemieckim obozie koncentracyjnym Auschwitz-Birkenau, otrzymując numer 62526. Podczas pobytu w obozie zachorował na gruźlicę. Mimo umieszczenia go w obozowym szpitalu, tzw. Bloku 20, oraz zastosowania leczenia, choroba stale postępowała, przeżył jednak, dzięki fachowej opiece dr. Władysława Tondosa. Pracujący z nim Józef Mężyk, student medycyny, napisał: Wśród ciągłych intryg, szantaży, fantastycznego szpiegostwa, groźby, infekcji, oraliśmy ten beznadziejny ugór obydwaj, wychodząc obronną ręką z wszelkich haczyków, pułapek itd. Niezależnie od choroby pełnił posługę lekarską, pomagając obozowemu ordynatorowi. Prowadził także agitację socjalistyczną. Po wyzwoleniu obozu przez armię radziecką 27 stycznia 1945, został przewieziony wraz z grupą współwięźniów do szpitala na Białym Prądniku w Krakowie, skąd na początku lutego został zabrany do Jędrzejowa przez Antoniego Patrzałka.

Działalność po wojnie
14 marca 1945 został mianowany dyrektorem szpitala, natomiast 20 marca został przewodniczącym Powiatowej Rady Narodowej w Jędrzejowie. 23 maja 1945 zmarł na gruźlicę w szpitalu w Krakowie. Został pochowany w Jędrzejowie na cmentarzu parafialnym Świętej Trójcy.

28 czerwca 1945 Rada Narodowa przemianowała dawną ulicę Prostą w Jędrzejowie na ul. dr Ignacego Kwarty. W 2017 roku Rada Miejska w Jędrzejowie przywróciła nazwę "ulica Prosta".

W ocenie Instytutu Pamięci Narodowej nazwa ulicy imienia Ignacego Kwarty wypełnia normę ustawy o zakazie propagowania komunizmu.

Przypisy

Bibliografia
 S. Kłodziński, Dr Ignacy Kwarta, "Przegląd Lekarski", 35 (1) z 1978 r, s. 195-200.
 W.K. Urbański, Dr Ignacy Kwarta (l898-1945) – "Judym jędrzejowski".
 "Studia Kieleckie", nr 2 (10), Kielce 1976, s. 187-191.
 J. Roszko, S. Bratkowski, Ostatki staropolskie, Warszawa 1966, s. 78-81.
 A. Paszta, Ignacy Kwarta (1898-1945), w: Słownik biograficzny działaczy polskiego ruchu robotniczego, tom III: K (pod redakcją Feliksa Tycha), Muzeum Niepodległości w Warszawie, Warszawa 1992, s. 594.
 J. B. Gliński, Słownik biograficzny lekarzy i farmaceutów ofiar drugiej wojny światowej, Wydawnictwo Medyczne "Urban & Partner", Wrocław 1997, s. 215.

Więźniowie KL Auschwitz
Urodzeni w 1898
Zmarli w 1945
Ludzie związani z Jędrzejowem
Polscy lekarze
Ludzie związani z Warszawą
Ludzie związani ze Świeciem
Ludzie związani ze Lwowem
Więźniowie polityczni w II Rzeczypospolitej
Oficerowie batalionu KOP "Ludwikowo"

redpajama_set_name: RedPajamaWikipedia


In 2011, Samsung became the world's largest manufacturer of smartphones, when it overtook both Apple and Nokia, according to US-based research company HIS iSuppli. However, the success of Apple's iPhone 4S enabled Apple to outsell Samsung in the year's final quarter.
Samsung shipped 95 million smartphones in 2011, an increase of 278 percent from 25 million in 2010. Apple's shipments almost doubled from 47 million to 93 million units.
iSuppli warns that its table (below) is incomplete: it does not include companies that have not yet reported results, such as RIM, HTC and LG. However, the top two places are not in doubt.
Wayne Lam, IHS's senior analyst for wireless communications, said in a statement: "Samsung advanced in 2011 because of its strategy of offering a complete line of smartphone products, spanning a variety of price points, features and operating systems. This enabled Samsung to move past perennial market leader Nokia and to slightly exceed Apple's total for the year."
Apple overtook Samsung in the fourth quarter of 2011 by shipping 37 million smartphones, an increase of 117 percent. Samsung's smartphone shipments grew by only 28 percent to 36 million units, according to iSuppli.
Research company Strategy Analytics had the same ranking but listed Samsung's smartphone shipments at 36.5 million units.
Apple is suing Samsung in numerous countries to try to halt its advance, which is based mainly on the success of smartphones running Google's Android operating system.
Nokia, the former market leader, increased its smartphone shipments from 17 million to 20 million units in the fourth quarter, when it launched its Lumia range running Microsoft's Windows Phone operating system. Over the full year, however, Nokia's smartphone shipments slumped by 23 percent from 100 million to 77 million units, according to iSuppli.
Bloomberg estimated that Nokia sold 1.3 million Lumia phones: the average of estimates from 22 market analysts. Nokia declined to provide numbers, but said it had "sold well over 1 million Lumia devices to date". However, Nokia did say that it had sold 93.9 million mobile phones in the fourth quarter of 2011, a slight reduction on last year's 95 million.
Nokia said it was selling about a dozen Symbian Series 40 phones every second, and has now sold 1.5 billion of them. It estimated that about 675 million S40 phones were still in active use. Late last year, Nokia launched the Asha range of S40 feature phones, which can send messages, upload photos to Facebook and Twitter, and play Angry Birds.
Nokia's strategy is to move users from Symbian S40 phones to smartphones running Windows Phone 7.5. However, this is threatened by the growing availability of Android smartphones at very low prices.


Nominate a Young Conservationist! Zoo News : Nominate a Young Conservationist! Zoo News : Nominate a Young Conservationist!
Do you know an outstanding youth who is an advocate for wildlife and conservation? Nominate him or her for the Young Conservationist Award.
As part of the celebration of Zootopia, Zoo New England's annual fundraising gala, Zoo staff is excited to announce that applications are now being accepted for the Young Conservationist Award. The nominated youth should be between the ages of eight and 18, have demonstrated a positive commitment to Zoo New England and its conservation mission, and have engaged in some outstanding action to support this. The award, which includes a $2,500 scholarship, will be presented at Zootopia on May 12, 2018 to further the recipient's project, conservation work and education. The winner and two adults will receive complimentary tickets to Zootopia to participate in the awards ceremony. This year's presentation marks the sixth award presented to an exceptional youth committed to conservation.
Nominations for the Young Conservationist Award are being accepted through Friday, March 16, 2018 at 5:00 p.m. Email questions to YoungConservationistAward@zoonewengland.com.
Zootopia is a fundraising gala event that will "bring the Zoo" to the Royal Sonesta Boston for an unforgettable evening. The event, which will be held May 12 from 6:00 – 11:00 p.m., begins with a cocktail reception and auction followed by a spectacular dinner and dancing. Attendees are encouraged to wear animal-inspired attire.
Proceeds from Zootopia will directly benefit Zoo New England's Franklin Park Zoo and Stone Zoo and will help to further the Zoos' mission to inspire and educate the next generation of conservation stewards and leaders.
To purchase tickets or to learn about sponsorship opportunities, visit our event page.


What I did to my system, and im not sure if this affects sleep or not, is to disable BT from main menu as I dont need it. Others than that I have also disabled some of the live tiles on metro screen (finance, sports) and disabled auto update.
With these options my system draws almost zero power when sleeping. Even thou wifi works as often there are updates waiting for me when I resume it from sleep.
I'm having the same issues; sleep mode w/no use for a day or so, open it up to use, nothing, nada. Won't even turn on DEAD.
So after 20 hours my tablet lost 3.5%. One thing to note thou, I am not using it with a dock, so when its sleeping its detached from dock, maybe those that have am issue with drain, use it docked?
Don't know about you, but I know that the HDD dock I have is definitely draining power when sleeping because the HDD is not turned off properly by the device.
I get very close to those same drain results. In my case...while charged fully....after turning off the WIFI (and still connected to the keyboard dock).
Exactly. I have had a couple occasions where the machine completely drained in one day while "sleeping". It is because the screen comes on when you close the lid. I suspect it is hitting against the track pad or one or more keys and waking up the screen when closed. Since I almost always use mine with the dock attached, I just make sure I completely shut down if I am planning to leave it for any length of time.
Update...just went to turn the machine on, completely dead. After doing a COMPLETE SHUTDOWN yesterday. This has happened to me 2-3 times. Yikes.
just a question: is this a problem of the tablet NOT SLEEPING or WAKING FROM SLEEP?
I have a feeling my tablet doesn't go to sleep even though i press the sleep button. (if i connect a usb drive the light stays on, i go to my wifi router and it says my device is active) is that normal?


Solar Fields is the Ambient production alias for Swedish DJ, producer and performer Magnus Birgersson. The Solar Fields sound spread across much of the Ambient spectrum and even touches upon Psy-Trance. Birgersson has released nine Solar Fields studio albums.


Origin Fencing was established in 2003, utilising over 50 years of Australian fencing experience and expertise. The company is Queensland owned and operated, using local and Australian suppliers and contractors. Origin Fencing has established itself throughout Queensland as a large fencing contractor specialised in infrastructure, industrial, security, schools and public utilities.


This is probably my 5th time at Ikoi but I love this place because they have the best sashimi buffet in Singapore! You can find Japanese buffets ranging from $20-$30 out there but these buffets might not have sashimi or their sashimi is not of too great a quality or they only have the more common sashimi like salmon and tuna.
Over at Ikoi, you get SWORDFISH(my favourite), salmon, tuna, yellowtail, snapper, cuttlefish and octopus. Free flow of swordfish sashimi. That is more than enough to make your $35++ worth it.
Located at Miramar hotel, Ikoi is extremely popular. You can't get a seat at Ikoi if you don't reserve in advance. For weekends, its recommended that you reserve a few days in advance.
Complimentary teapot soup which was extremely bland..
As usual, the alcoholic had to get some alcohol.
It tasted just like absolut vodka.
They have an extremely wide variety and its definitely worth the money to dine here. We did not try most of the items as the sushi and the rice/udon/soba would have been extremely filling. Haha. Had to save our tum tums for sashimi!
The fresh sashimi, the efficient service, the wide variety and the fact that you can stay here for 4 hours more than justify the $35++. Definitely one of the best buffets in Singapore!
Many people choose great dining restaurants for a special occasion, so the food must not disappoint- in either selection or quality. You don't need to feature a huge menu, but it should be interesting, offering unique items that patrons wouldn't find at any other restaurant.
look over it all at the minute but I have book-marked it and also added in your RSS feeds, so when I have time I will be back to read a lot more, Please do keep up the superb work.
I love swordfish too! If i died and had to choose sashimi i would choose swordfish rather than salmon.
Ikoi Japanese restaurant is very tasty. You can have a taste of it.


text:
Tag: Lana Del Ray
INspired by…Music Videos
Creative inspiration comes from myriad sources. Taking our lead from a recent article in Dezeen, we're having an alternative look at inspiration on screen – exploring what music videos act as a creative spur to the Deirdre Dyson Kings Road team.
Company Manager Edison
Lana del Ray – 'Young and Beautiful'
I love the resonance of Lana Del Ray's voice and her unconventional beauty often mixed with an orchestral backdrop. It's a style of music that makes you feel dead and alive at the same time, that is both distinctive and unique.
I find her video for the song "Young and Beautiful" so inspiring.
It is highly influenced by the 'Toccata and Fuge in D Minor' scene from Disney's 'Fantasia'. The overlap of colours, the silhouettes, shadows and the play on dark and light both in the foreground and background. It reminds me how important lighting is to any interior and how it can either make or break a space and ambience. Lighting is so often overlooked in interior design and remains such an important element in creating atmosphere and drama. I love the understated luxury elements of her make up, hair and jewellery and the grandiose styling, amplifying height, scale and space. It all works so beautifully in combination with her voice.
Deirdre's LIGHT BETWEEN design evokes the style and mood of Lana's video perfectly.
Trade Sales Manager Chris
Tina Turner – 'Look Me In The Heart'
Hard to believe this video is now 30 years old as it doesn't seem that dated to me.
For years we have all been inspired by greys in our interiors which I love.
But the visual palette of this video by Tina Turner oozes colours in wonderful tones from saffron, aquas and Tina herself a beautiful magenta pink silk trouser suit.
Watching this I now notice flashes of pink and oranges which remind me of Deirdre Dyson's STEPS carpet design.
STEPS was inspired by a holiday Deirdre had in Mexico and the colours here, as well as the simple and spacious set of this video, makes me feel it is also reminiscent of a hot climate.
Watch the video, listen to this sweet song and consider how strong colours in the correct tones can create statements in our interiors without shouting.
Deirdre has over 5000 colours to choose from so introducing a carpet is an excellent way to bring colours you like into your own interiors schemes.
Social Media & Marketing Manager Sonia
Jack White – 'Over and Over and Over'
This video is really clever – it's set in one room but contains a multitude of vignettes in a tiny space.
Of course this is a set but I've always been intrigued by the way houses in cities like London are carved up into interesting living spaces, bursting with character and how cleverly people craft these corners into individual homes.
The turn of this stairs hints to more and forms a crucial link in the story as well as making you to want to explore this spare little space as it transforms over the course of the song. The 'illustrated' version of the room is inspired.
I really love colour pops and think the way the colour blue threads through the spot is inspired, from the metallic indigo on the first shot of the guitar pick ups through to the the paint dripping from the stars to the minimalist canvas flooding with Yves Klein blue, a colour favoured by Deirdre and used in a recent version of UP AND OVER.
The gold costume of the character sitting on the sofa at the beginning remind me of the colourways in Deirdre's original version of UP & OVER shot with bold, gold panels.
I just think the aesthetic is fabulous, dark to light, surreal and common place all in a little studio or flat anyone could own.
Design Technician – Nichola
OK Go – 'Obsession'
I couldn't stop watching the video, 'Obsession'.
The background of the video incorporates over 550 printers and numerous pieces of paper that were recycled later.
The patterns and spectrum of coloured paper used were stunning.
Right from the beginning I was drawn in by the brilliant use of white paper which then changed to a symphony of so many different colour combinations. It was surprising to see the contrast of subtle colours at the end of the video.
The speed and rhythm of the moving paper with the fabulous choreography of the band members works so well with the music.
I also liked how the band members were dressed in white which was such a contrast to the vivid backgrounds. Dancing whilst suspended in the air made the video even more unique.
All in all I found this to be very inspirational on many design levels – I could watch it again and again.
We thought Deirdre's CHEQUERED carpet, graded in wool with silk was reminiscent of the aesthetic created by the graded grid of paper in Nichola's chosen spot.
Deirdre Dyson Exclusive Carpets and Rugs CHEQUERED

redpajama_set_name: RedPajamaCommonCrawl


text:
I Just Watched: The Big Short – Film
Sunday, 30th July, 2017
I hope you are doing well. So the latest film I've watched is The Big Short. Last time I wrote about Bad Moms which I watched on Amazon Video. This is one of the blu rays which I got a while back, and was a bit of a "blind buy". I had no idea about this film, but I really like films about money and financial stuff, such as Wall Street and Money Never Sleeps, as well as Too Big To Fail and others, films about the financial crisis of 2007-08. I find them really interesting, although I know they won't be everyone's cup of tea.
This film has a lot of A-listers in which gives it a bit more of a Hollywood feel to it, however the actors have done a great job, because they play the parts so well, you don't think of them as the characters they've previously played or are synonymous for; e.g. It stars Christian Bale, Steve Carrell Ryan Gosling and Brad Pitt, with Steve Carrell playing quite a serious character and role, something I'm not used to seeing him playing, everything I've seen him do has been comedy.
So the film is all about the financial crisis of 2007-08, so I think to enjoy it you have to be interested in knowing all about it, it has a few story lines from the characters who each have their own interests and perspectives, but they don't overlap and are just connected with the events of the crisis. There's some cut away parts to the film when there's technical financial jargon talked about to explain things so people can understand which are quite funny; e.g. Margot Robbie in a bubble bath with a glass of champagne explaining Sub prime loans.
If you like films like Wall Street and more factual-based drama films then you may like this one, and if you don't then you probably won't enjoy this.
Bad Moms & Amazon Video
Common Goal For Footballers To Donate 1% Wages By Juan Mata
Lifestyle » I Just Watched: The Big Short – Film


Tractor Supply increased estimates today, and Cramer came out saying that if TSCO said things were getting better, then, by gosh things are getting better!
You can have mine. After all, I was all over this stock last October when nobody cared or nobody loved it, or no-one wanted to visit a store.
I'll give it to Cramer! It worked with BUCY!
Gold's spiking and a beneficiary of that is Bucyrus. But, I've been touted this name last July at 28, said to sell it at 66, when Cramer touted it, and then buy it back at 53 on Feb 1 when it corrected, so I'm letting that number go also.
Maybe it will run higher, but since Cramer gave me an exit on TSCO I'll take another name off. Booyah!
And enough already with WYNN. You got this name last March before it tripled, (and LVS at 1.77). Granted, at these depressed levels of earnings WYNN should sell at a big multiple but if earnings double to $2, is it then worth more than 40X that number? LVS is opening in Singapore at the end of the month, and usually, you get a sell the news type of reaction. And WYNN, has to have its plan in place for Foxwoods by April 26. All, I know is that the shorts are getting killed in this name, and when you hear them crying, you usually, get some sort of reprieve. When Sheldon talked up Vegas, Steve completely downplayed it. So hit the bid then, if you really believe it. Maybe Steve will sell some shares, to pay for some expansion, and then if he changes his mind, come December, pay another big dividend, and work the shorts some more!
And I know I've been a big bull on AAPL. Heck, I touted this number last March under 90. I ridiculed the WSJ last July when they said AAPL was overpriced at152. And I know from a fundamental standpoint, there is nothing wrong with the story, and the iPhone, and the iPad, and the Mac are still must have products, but every pimp on CNBC is now touting AAPL at 240, but where were they 200% ago, when they were really giving this number away?
All these stocks are now favorites of the market, all loved, all tipped and all touted. But none of these touts would show their face on CNBS when these stocks were cheap! The charts look great, the fundamentals are getting better, and they still don't have any distribution yet. But institutions can't sell their position in one day. It takes time. So if anyone wants to lighten up, they'll have to bleed stock into the buyers.
Today, with a straight face, Goldman Sachs came out with their annual report and said they didn't trade against their clients.
Today, with a straight face, Greenspan came out on Capitol Hill and said that they never favored any banks.
So today, with a straight face, I sold some stocks.
I just couldn't take the bullsh*t.
are you just cashing out of these names? are you still holding MGM, APWR, and some others?
The stocks were extended on the charts, and maybe they'll pull back.
And if they come in, I can use options to play them if I want.
Sometimes you sell things, and you don't have any really good reasons why you sold them.
I treat the casinos as a real estate play, and use a cap rate on the operating income net of depreciation. They still look expensive, but I am holding LVS thru Q1 earnings.
Anecdotally, I stayed at the Venitian during Sweet 16 weekend and it was full. I decided to get there one day early and booked a Bella suite for Wednesday at the last minute...they charged me $800/night for Wednesday because they were that full. The front desk adjusted my rate back down to $340 when I checked in.


Council's Governance Review Sub-Committee – which is now responsible for accountability and transparency – meets to plan for after the October election.
They will vote upon the 2015 meeting schedule and changes to the membership of Standing Committees for the next term of Council.
The meeting begins at 9:30 a.m. at City Hall.
As it falls under the oversight of the Ontario Ombudsman, this public meeting is open to the public.


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162; prince albert record friday night at the langley
6H ago by RotoWire Staff | RotoWire Blackhawks' Alex DeBrincat: Two more assists Wednesday.
The mandate imposed a tax penalty on consumers who went uninsured and was a key part of the health-care law.
13, replacing Mike Parker, who has been CEO since 2006 and chairman, president & CEO since 2016.
Prior to joining the Oilers training staff, Lowe was the Head Athletic Therapist of the Edmonton Eskimos of the Canadian Football League and was a member of Grey Cup championship teams in both 1982 and 1987.
12 – Trent Thornton fell to 0, allowing five runs on eight hits and two walks in 3 innings, as the Toronto Blue Jays lost to the Tampa Bay Rays 11 on Friday.
I'm super grateful to see other members of the communities stepping up and helping in their own backyard, Wilson said.
He can beat you with inside moves.
COLLEGE: Played four seasons at Alabama-Birmingham, appearing in 48 career games…Converted a school-record 59 field goals on 77 attempts and 159 extra points for 322 points over his college career…Punted five times for 196 yards …As a sophomore, ranked second in Conference-USA to Cairo Santos in field goal accuracy at 82 pct.
He missed 103 games this season because of a left hamstring injury.
The healthier American consumers are, the healthier the economy: Discretionary spending accounts for two-thirds of US GDP https://www.storiedelmonferrato.com It was an unbelievable time for me, Byrom tells .
They had obviously spoken about it beforehand and Chris would jump in when he needed to but he was just very good with the lads.
The skinny: 58; lost in FinalsThe first year of the Miami Heatles seemed destined to cruise to the title after destroying Derrick Rose's Chicago Bulls and the rest of their Eastern Conference Playoff opponents, four games to one in each series.
BACKGROUND: A five-star offensive line recruit out of high school, Collins was wooed by a number of national programs, but there was very little question where he would play his college ball, committing to hometown LSU prior to his junior season.
122 outside linebacker in the nation and the No.
17, Paul Pastelok, AccuWeather's lead long-range meteorologist, said.
Calvin Ridley makes too much sense for the Baltimore Ravens.
Seriously, where are the drawbacks?
2, while Australia jumped eight spots to No.
Thank you to the fans who make representing those 3 letters mean something.
To a large extent, yes.
Rushed for a team-high 68 yards on 24 carries and completed eight-of-13 passes for 60 yards in the Gators' win over Penn State in the Outback Bowl.
For the month, Washington led the East in assists and ranked third in field goal percentage .
Treadwell, one of the Rebels' top recruits in a loaded 2013 class, said Ole Miss is still learning when to try for big plays and when to be conservative when the defense warrants it.
She won't rest until she gets her hands on the 21-year-old villain who is described as pure evil.
Also named to the Atlanta Journal-Constitution Super 11 and Class AAAAA All-State team…
Bobrovsky's win was his 20th of the season…
If you would like to search for all players born on a certain day, for example all players born on December 25th in any year, choose the month and day with the drop down boxes and then choose the 'Month and Day Search' option.
Neville feels any new addition will have to fill some of the void in the dressing room left by the departure of club captain Vincent Kompany in the summer – and that City are only too aware of that requirement.
Highs on Wednesday will be in the 20s and low 30s depending on your location.
One of Czarnik's tallies came on the power play, and he added an even-strength goal as well on an assist from Tobias Rieder.
Injury Update .
Posted a team-high seven receptions, gaining 84 yards with two receiving touchdowns in a win vs.
We talked a little ball, but I don't really know the magnitude of how different it's going to be.
1 – Acquired C Sam Carrick and LW Spencer Abbott from Chicago in exchange for LW Kenton Helgesen and a seventh-round selection in the 2019 NHL Draft.
Her pictures prove that she is the star in making.
Rendon hit a sacrifice fly to center fielder Bader, Turner scored.
8 Notre Dame on Nov.
Reuse as many products, food containers and household items as you can, and recycle as much of your waste as is possible.
On monday jan 27 america membersThree dots icon down arrow icon email icon


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PHOTO: Cmmellow on Pixabay
Three steps to science diplomacy success
How South Korea can build a stronger diplomacy of influence
Olga Krasnyak
International relations, Science and technology | Asia, East Asia
For a country so reliant on soft power, why is South Korea neglecting science diplomacy as a tool of statecraft? Olga Krasnyak suggests three ways for Seoul to have diplomacy down to a science.
Science diplomacy offers governments significant potential to grow international influence and improve diplomatic relations. This potential has already been recognised by countries such as the US and the UK.
But South Korea – a world expert in the use of soft power – is yet to recognise the power of science diplomacy. If incorporated into Seoul's foreign policy agenda, science diplomacy would advance South Korea's image as an attractive, cooperative, and influential country.
With rising tensions on the Korean Peninsula and the return of great power politics between the US and China, strengthening South Korea's influence regionally and internationally is an urgent task.
On the North Korea nuclear crisis, South Korea is squeezed between Beijing and Washington, the former leveraging economic interdependence for its own security interests and the latter, South Korea's key strategic ally, threatening to launch a military operation against the North.
Beyond the Peninsula, the US-China rivalry places additional constraints on South Korea. Seoul is party to neither China's Belt and Road Initiative nor the recently revived Quadrilateral Security Dialogue between the US, Japan, India and Australia. This exclusion puts South Korea in a situation where proactive policy and skilful diplomacy are especially needed for it to maintain international influence.
More on this: Why science gets shut out of policy
Encouraging the development of science diplomacy to foster a stronger academic-practitioner community could strengthen South Korea's voice and promote its image. Science diplomacy could also be an extension of President Moon Jae-in's "innovative growth" agenda.
So, how could it be achieved? There are three things South Korea needs to do to encourage the growth of science diplomacy and reap the benefits.
First, Seoul needs to give science diplomacy a higher profile.
South Korea has recently demonstrated how sports and cultural diplomacy could help with North Korea during the Winter Olympics. Science diplomacy has the potential to be used as a universal tool to smooth the cultural gap between the two Koreas, thus securing the foreign policy objectives for both.
South Korea has already built an image as a strong, capable, and reliable international partner through intergovernmental organisations such as the UN and the World Health Organization and through international agreements such as the Antarctic Treaty. Future international treaties will require South Korean diplomacy to become more savvy and skilful.
By creating a high profile for science diplomacy and understanding both diplomacy for science (facilitating international science cooperation) and science in diplomacy (informing foreign policy objectives with scientific advice), South Korea can promote its future foreign policy concerns and increase its influence.
Second, Seoul should encourage the wider inclusion of non-state actors and develop public diplomacy as a whole.
The first ever meeting of South Korea's committee on public diplomacy took place in August 2017, at which a five-year 'master plan' was created. The plan includes 410 billion won (US $383 million) for public diplomacy projects in six fields, including policy, knowledge, and culture, with the biggest focus on culture.
Science diplomacy, however, was not officially recognised in this diplomacy master plan.
More on this: South Korea is failing North Korean studies
The plan also measures the success of public diplomacy through citizen participation. This means that Korean citizens are the target group, which overlooks opportunities for foreign communities to spread South Korean influence abroad.
For science diplomacy, foreign non-state actors are essential. These foreign communities might be represented by academic networks in other countries as well as foreign academics living and working in Korea.
The inclusiveness of non-state actors for science diplomacy can build on South Korea's nation-branding from already well-known, different kinds of soft power – such as K-pop or electronics – towards a more scientifically-relevant, academic image. Focusing on non-state actors for science diplomacy offers key advantages in the long-run and contributes to a higher profile.
The third ingredient for successful science diplomacy is establishing a network of 'science diplomacy centres' under the supervision of the Ministry of Foreign Affairs.
Initially proposed in a report by the Asia Institute, these centres of excellence would operate within the worldwide network of South Korean embassies and consulates. The role of embassies and consulates has shifted from pursuing only intergovernmental contacts to connecting with a wide range of non-state actors such as companies, various institutions, and individual professionals. This expanding network is a potential tool to promote Korea's science diplomacy.
The centres might also affirm a democratic, knowledge-based model with a balance between economic growth, political development, social justice, and human security. Additionally, they could coordinate joint scientific projects and diplomatic negotiations and monitor research opportunities for long-term sustainability.
By building goodwill and influence, these centres could become strategically autonomous based on science policy. And South Korea doesn't need to look too far to find an example of how to do this – 'innovation hubs' can already be seen in the United Kingdom's network of embassies and high commissions.
Taken together, these three steps might increase efficiency of traditional diplomacy, enhance scientific cooperation, and ultimately, allow South Korea to shape its future as an important geopolitical player and demonstrate that its soft power skills go well beyond K-pop and electronics.
This piece is based on the author's paper for the Korea Economic Institute of America (KEI) 'Science diplomacy: An underestimated toolkit of South Korea's foreign policy' and her associated public talk at KEI 'Knowledge is power: How science diplomacy can advance South Korea's foreign policy agenda'. Watch a video of the talk below.
Dr Olga Krasnyak is currently an Assistant Professor at the Department of International Relations of the National Research University Higher School of Economics, Moscow.
APEC's role in the Asia Pacific's rebuild Olga Krasnyak
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text:
package hr.bpervan.mt.model;

import hr.bpervan.mt.utils.StringUtils;

import java.util.DoubleSummaryStatistics;
import java.util.HashMap;
import java.util.Map;

/**
 * Created by Branimir on 17.6.2015..
 */
public class ItemBuilder {
    private int itemId;
    private String itemName;
    private int location;
    private Map<String, Double> map;

    private ItemBuilder(){
        this.itemId = -1;
        this.itemName = null;
        this.location = 0;
        this.map = null;
    }

    public static ItemBuilder getInstance(){
        return new ItemBuilder();
    }

    public ItemBuilder setItemId(int itemId){
        this.itemId = itemId;
        return this;
    }

    public ItemBuilder setItemName(String itemName){
        this.itemName = itemName;
        return this;
    }

    public ItemBuilder setLocation(int location){
        this.location = location;
        return this;
    }

    public ItemBuilder setCharacteristicsMap(Map<String, Double> map){
        this.map = map;
        return this;
    }

    public Item build(){
        return new Item(itemId, itemName, location, map);
    }
}

redpajama_set_name: RedPajamaGithub


Well, Boltorr Rd survived the move and now sits awaiting attention in the basement at Bloke Acres.
It will be a few more weeks before I can get to grips with the layout again, but I have a long todo list in my head.
At the top is getting the track laid and the DCC control sorted out. There is only one cross-over in the whole layout and this will be DCC controlled as well.
There is only room for a short fiddleyard at either end of the layout, but since traffic will be only a couple of two-coach or six wagon trains that should suffice. The limited length means I shall either have to use a sector table or a traverser.
Anyway, I've included a panorama view of the layout which, I hope, will illustrate my attempt to show a small section of railway situated in a three dimensional world. I've really gone all out to avoid the table-top look. The most important aspect was to continue the terrain below the track level.
The foreground is pretty much set. The background profile will be finished when the dam is placed and integrated into the scenery.
Eventually there will be a scale effect of over 100ft/30m from the Boltorr Brook to the highest points. I hope it will be effective.
Well, the last few months have been exciting. But now, Mrs Bloke and I are now installed in our new – well nearly a century old – house in a cool little town in New Jersey.
So Americans will go, "How quaint". And Brits will yawn.
Yes, the garage is still full of boxes, as is the dining room. But we're getting there. Working towards … as the education jargon used to say.
Our last move took two or three months to settle down. And even then, this time we moved, there were a dozen boxes which hadn't been opened since the last move.
That is a barnstar up there. Akin to a lucky horseshoe I would think. It'll be staying. We'll need all the luck we can get come November. But that's another story.
Anyhoooooo… it's now a ten minute – brisk – walk into town. Even the theatre is only about a fifteen minute walk away. Uber and GrubHub work here so we're doing our bit for the deconstruction of Capitalism, and the implementation of a Brave New World, or exploitation, if you like. (I'm going for the "more is more" theory of punctuation in that sentence). And yes, the full-stop outside the brackets is English, not American English, whatever that is.
Now the Important Things, apart from the love and devotion of a wonderful life partner, Mrs Bloke. The things that make my heart go Zing! Hopefully for a while longer.
Bicycles: Somehow we have to accommodate about seven+ or so alternative means of transport here. But it will be done! This is a cool little town, and I'm going to ride my bikes around it. And on occasions I'll be forced to report. See here.
Model railway: You know it's not a toy train? Not a train set? This is the world Jones ThatBloke made… It's been on hiatus for a few months so not much – actually nothing – has happened. But somehow it survived the move and the wibbly-wobbly way into our new – well, oldish, like I said before – basement/cellar/vault/panic room without any damage or other issue. Thanks to our movers, Simonik, who did a great job.
I have to rethink the ends of the layout, Boltorr Road, since this isn't much more than a diorama. The trains will still have to start from somewhere, negotiate my own little world and then move on. At this moment, I'm thinking a traverser or sector table. Maybe even a Nelevator?
Hifi: Not certain what's happening with that yet, but whatever, it'll be good – of course.
So, plenty to do. But lots of time to do it.
I'm really looking forward to making this place Mrs Bloke's and mine. Did I mention the Fibre Room?
Nice album of old stations around Bristol, UK, from Bristol247.
Just going through a redesign, so please bear with me.
It's amazing what you can do with WordPress.
You get on with life, the universe and everything and before you know it a couple of months have gone by and you haven't looked at, let alone written in your favourite blog.
But, this is notice that the Bloke is back on track.
The next few weeks are going to be quite intense, but that will become obvious as Linda and I work through them.
So hang on and keep checking back here.
In the meantime, here's yet another sunrise over Raritan Bay.
I really liked the effect I achieved on the Bolt Tor tunnel portal.
This was done by rolling air-curing clay very thinly, breaking it up into scale sized "stones" and individually gluing them onto the structure of the walls.
However, this was pretty labour intensive and I wasn't enthralled by the thought of sticking thousands of individual stones over possibly hundreds of square inches.
So I've gone for another method to represent the masonry on the reservoir barrier. Rolling the clay really thin – around a millimetre – and glueing the sheets onto the structure, then inscribing the stones while the clay is still plastic.
For my first test I glued the clay directly onto a piece of foam board. This resulted in the board warping as it dried. It seems the board I'm using is faced with a card veneer which absorbs water and deforms as it dries out.
I solved this by adding some extra bracing to the rear of any large flat area and sealing the card with an acrylic spray. This seems to work really well. The PVA glue I'm using still fixes the clay to the board.
I'm learning this as I go along, so I've had a couple of issues, the most serious being shrinkage of the clay layer causing a couple of cracks large enough to be embarrassing on a representation of a dam. This, as I found out, can be avoided by feathering the joints between adjacent pieces of clay.
I'll need to go back over a couple bits of the model to eliminate the cracks when it's all completely dry.
The "stones" were scribed by hand and a pointed piece of wood while the clay was still plastic. This leaves a fair amount of whiskery material sticking out of the wall, but this will be sanded back. So far it looks pretty good.
I found the most effective way of marking the random stone was to treat most blocks as irregular pentagons, filling in with rectangular and triangular blocks as appropriate. At the present rate, the rendering is going to take about three to four hours in total.
I know this is a two-year project.
But every now and again I just crave a tantalising glimpse of what will be in my universe.
Work is going on on the course of the Boltorr Brook from the reservoir barrier, under the rail bridge and onwards into the Tamar valley.
The major construction is that of the Boltorr dam. Once that is set into the landscape the rest of the path of the river will be set.
Simultaneous work is also happening for the abutments of the rail bridge. Actually, the space between the bridge and the dam doesn't look so great now so I need to see how it can all be fitted in without looking unfeasibly crowded.
I couldn't resist unpacking my Class 153, First Great Western DMU and posing it on the bridge.
The dam looms up behind it. I'm in the process of fitting engineering brick vaults up under the overflow arches, so the T-pins are doing their job.
I've been experimenting with facing the foam board of the structure with air-curing clay. I'm glad I've done a couple of test with this, because it seems essential to keep the thickness of the clay at no more than 1mm otherwise it's tending to distort the foam board. This particular board seems to be faced with cardboard, which is okay, but make it prone to warping in the presence of moisture.
However, my last test, braced board, lacquered to resist moisture, 1mm DAS clay and a base coat of "granite" looking matt acrylic is looking pretty good, so once the glue has cured, any excesses and drips cleaned and trimmed off I think I'll be moving on to the next step.
… The heart doesn't grieve over.
Construction is going well on the Boltorr dam.
Foam-board, PlastiKard, PVA and hot-glue are the essential parts of my builder's arsenal.
As I've mentioned before, the actual structure is quite large, although I've scaled it back as far as I can so it still looks mighty, but in scale with its surroundings. A real dam, even a small one would probably take up five or six foot of space to faithfully represent as well as a couple of feet of depth as well.
Although I've been as accurate as I can with the marking out, on occasions I've adjusted it to "look right" rather than worry too much about getting my laser level out. Btw; I actually don't have a laser level.
Looking up at the Boltorr Reservoir dam from "eye-level". Glue beads to be trimmed and gaps will – hopefully – be filled at the masonry stage.
But now the glue is almost set I'm starting to look at it from the perspective of the layout to see where I can leave gaps, and on the other hand, stuff I can't get away with.
Since the dam will be set right to the back of the scenery I'm not too concerned about the back of the dam. I am not an OCD completist.
However, checking the structure from the viewpoint of the layout I will need to represent "under the arches" and any flow control devices a dam like this would have.
I think the dam needs a run-off to keep the Boltorr Brook from drying up, so I just have to figure how to represent that.
I'm going to give the structure a couple of days for the glue to set and cure. First job will be trimming the beads of glue which are visible, then experimenting with getting a thin lamination of DAS clay to adhere to the foam-board to form the basis of rendering the granite masonry.
The contours of the layout east of – behind – the track are really going to be determined by the setting of the dam walls in the Boltorr Brook valley.
So construction is now quite a high priority.
This is going to be quite a large edifice, about 600mm long and 300mm high. In scale about 150ft/45m long and 75ft/23m high.
I considered using the same method to represent stonework as I had on the Bolt Tor tunnel portal, but, as effective as it is, it would be a considerable amount of work to render the entire reservoir structure in that manner.
Besides, the portal method produced a very rough stone texture, appropriate enough given it was originally built by French prisoners of war, whereas the dam retention walls are going to be rather more dressed and regular in a Victorian mason sort of way.
So, I'm building up the form the stonework from foam-board and plastic card. I'm ten going to roll DAS Clay as thinly as I can, hopefully under 2mm thick and laminate it onto the foam-board. When it's dry I'm going to carve the stone courses into the clay and, hopefully, achieve a realistic 3D effect.