text:
\section{Introduction}
	The concept of a random binary sequence carries different meanings in different fields. In measure-theoretic probability theory, it is often defined as a sequence of independent Bernoulli random variables (with parameter $p=1/2$). With this definition, a random sequence considers in some sense every possible sequence at once, which allows for many different kinds of propositions. Three of these in particular focus on the running sum of the sequence: The strong law of large numbers (SLLN), the law of iterated logarithm (LIL), and the central limit theorem (CLT).

	In algorithmic information theory however, being random is a property that an individual binary sequence can possess. Many different definitions of randomness have been proposed and compared (for some examples, see Downey and Hirschfeldt \cite[Chapter 6-7]{Downey_book}). From the beginning, the SLLN and the LIL have played a crucial role as a filter for ``bad'' notions of randomness. For example, the first definition of sequence randomness was one by von Mises \cite{von Mises} (1919). Von Mises' idea was that a sequence was random if all ``reasonably selected'' infinite subsequences satisfy the law of large numbers, but he did not provide a formal definition of what a reasonably selected subsequence actually means. Although Church \cite{Church} would later provide such formalism using the then newly developed computability theory, Ville \cite{Ville} had already shown that for any reasonable formalization, there would always be so-called von Mises-random sequences with some rare property, namely that the proportion of $1$s tends to $1/2$ much faster than expected. In particular, these sequences would not satisfy the LIL. Because of this, von Mises' definition has since been considered too weak to capture ``true'' randomness.

	In probability theory, both the SLLN and the LIL are statements of the form ``the set of all sequences satisfying property $P$ has measure 1''. It is therefore possible to check if a given sequence has property $P$. The classic CLT is not of this form, as it is a statement about the convergence of some cumulative distribution functions (CDFs). However, there exists several alternative versions which are applicable to single sequences, the most fundamental of which is the so-called almost sure central limit theorem. In this work, we prove the ASCLT for all Martin-L\"of sequences, and also prove an alternative version of the CLT for individual sequences by generating an infinite number of samples from a single sequence and making statements about the collective behavior of these samples. Our paper starts with the latter in Section \ref{as_section} and then proves the ASCLT for MLR sequences in Section \ref{ASCLT_section}.

	\section{Preliminaries} \label{preliminaries}
	For an easy-to-understand introduction into recursive functions, Martin-L\"of randomness and related concepts, see for example Shen \cite{Shen}.

	~

	\textbf{Notation and measure theory}
	Unless stated otherwise, a digit or sequence of digits always refers to a \textit{binary} digit $\in \{0,1\}$ or binary sequence of digits.
	We use $x_i$ or $\omega_i$ to denote a binary digit (i.e., $\omega_i\in \{0,1\}$). Let $2^{\{n\}}$, $2^{<\N}$, and $2^{\omega}$ respectively be the set of all strings of length $n$, the set of all finite strings, and the set of all infinite sequences.
	Any sequence $\omega$ from one of these sets has an associated real number $0.\omega_1 \omega_2\omega_3\ldots\in [0,1]$ (with all tail bits set to zero from some point on for any finite sequence). This association works essentially in both directions, as for almost all numbers in $[0,1]$ this expansion is unique. Let $x=x_1x_2x_3,\ldots x_n$ be a finite string, then $|x|$ denotes the length of this string, in this case $|x|=n$. The cylinder of $x$, $\Omega_x \subset 2^{\N}$ is the set of all infinite sequences starting with $x$ (having prefix $x$). Recall that the Lebesgue measure $\mu$ on $[0,1]$ coincides with the fair coin-tossing measure, defined by the equality $\mu(\Omega_x)=2^{-|x|}$ for all strings $x$.

	~

	\textbf{Computability theory} A countable set $A$ (of strings, numbers, $\ldots)$ is called \textit{recursively enumerable} (r.e.) if there is an algorithm which enumerates the elements of $A$. This algorithm may run indefinitely, as long as any given element of $A$ is eventually enumerated.

	Informally, we call a function $f$ \textit{recursive} or \textit{computable} if there is an algorithm which computes $f$, meaning that if $a$ is some valid input for $f$, then this algorithm will on input $a$ give $f(a)$ as output. This intuition is sufficient for functions from $\N$ to $\N$, but becomes inadequate when the domain or range of $f$ is an uncountable space. In the literature (see for example \cite{Gacs, Galatolo}), a function $f:A\rightarrow B$ between computable metric spaces $A,B$ is computable the inverse image of ideal balls in $Y$ are recursively enumerable opens in $X$. However, since we limit ourselves to functions $f:2^{\N} \rightarrow\R$, we use the following equivalent definition based on Turing machines (see Soare \cite{Soare_book}): A function $f:2^{\N} \rightarrow\R$ if there is a Turing machine $A$ which takes an oracle $\omega\in 2^{\N}$ and $\epsilon \in \Q^+$ as input and outputs $r\in \Q$ such that $|f(\omega)-r |<\epsilon$.

	For a given sequence $(O_{n})$ of computable objects (numbers/functions), we can ask whether this sequence is \textit{uniformly computable} (in $n$), which intuitively means there is some algorithm $A$ such that $A(n)=O_n$. For example, a sequence of reals $(a_n)$ is uniformly computable in $n$ if there is an Turing machine which on input $n$ and rational $\epsilon>0$, outputs $r\in \Q$ such that $|r-a_n|<\epsilon$.

	Lastly, let $a_{1:\infty}$ be a sequence of numbers which converges to some limit $a$, then we say that $a_{1:\infty}$ \textit{converges effectively} to $a$ if there is some algorithm which on any rational input $\epsilon>0$, outputs an $N\in \mathbb{N}$ such that $|a_n-a|<\epsilon$ for any $n\geq N$. Similarly, a sequence $a_{1:\infty}$ \textit{diverges effectively} to $+\infty$ if there is an algorithm which outputs $N$ on input $M$ such that $a_n>M$ for all $n\geq N$.

	~

	\textbf{Algorithmic randomness} We will consider both Martin-L\"of randomness and the slightly weaker Schnorr randomness: A set $U=\Omega_{x_1}\cup \Omega_{x_2}\cup \ldots$ is \textit{effectively open} if the set of strings $\{x_1,x_2,\ldots\}$ is recursively enumerable.
	A sequence of effectively open sets $U_{1:\infty}$ is called uniformly r.e. if there is a recursive function $g$ such that $g(i)$ outputs an enumeration $\{x_1,x_2,\ldots\}$ where $U_i=\Omega_{x_1}\cup \Omega_{x_2}\cup \ldots$.
	A Martin-L\"of test is then a uniformly r.e. sequence $U_{1:\infty}$ such that $\lambda(U_i)\leq 2^{-i}$. A Schnorr test is a Martin-L\"of test with the added requirement that the measures $\mu(U_n)$ are uniformly computable in $n$. Finally, a sequence is Martin-L\"of random (MLR), resp. Schnorr random, if it is not contained in the intersection $\bigcap_i U_i$ of any Martin-L\"of, resp. Schnorr, test.
	\section{An alternative central limit theorem}\label{as_section}
	\subsection{Main theorem and proof}

	In this section, we explain and prove an alternative version of the CLT, where infinitely many samples are generated from a single infinite sequence which at infinity resemble samples from the normal distribution.

	Naturally, we will work with functions between the Cantor space and $\R$. Since the fair coin-tossing measure is a probability measure on $2^\N$, these functions can be seen as random variables. In order to avoid confusion when switching between the viewpoints of random variables and functions, we will refer to such function as Sequence-Based Variables, or SBVs for short:
	\begin{definition} A \textit{sequence-based variable} $X$ is a computable function $2^\omega\rightarrow \R$. A pair of computable functions $(I,f)$ is called a \textit{representation} of $X$ if $X=f\circ I$, where $I:2^\N\rightarrow 2^{\{N\}}: (\omega_1,\omega_2,\omega_3,\ldots)\mapsto (\omega_{i_1},\omega_{i_2},\ldots)$, ($N\in \mathbb{N}$ or $\{N\}=\N$) and $f:2^{\{N\}}\rightarrow \mathbb{R}$. The function $I$ in such a representation is referred to as a \textit{selection function} and set of the indices $\{i_1,i_2,\ldots\}$ as the \textit{selected indices}.
	\end{definition}
	A first example of an SBV $X$ is simply the sum of the first $n$ digits of a sequence:
	\[  \underbrace{\omega_1\qquad \omega_2\qquad \omega_3 \qquad \ldots \qquad \omega_n}_{X=\omega_1+\omega_2+\ldots+\omega_n}\qquad \omega_{n+1} \qquad \omega_{n+2} \qquad \ldots. \]
	Any SBV has of course many different representations (we can always add extra inputs to $f$ or change the order of the selected indices). However, one obvious representation for our example above is $(I,f)$ where $I$ simply selects the first $n$ digits in order and $f$ is the function of arity $n$ that adds its inputs.

	As mentioned before, SBVs inherit all concepts related to random variables by applying them to a sequence of i.i.d. Bernoulli RV (as an example, the SBV defined above inherits a binomial distribution $B(n,1/2)$). Hence, we can talk about independent SBVs, the expectation of an SBV, etc. We also have the following properties:
	\begin{itemize}[noitemsep]
		\item If two SBVs have representations such that sets of the selected indices are disjoint, then these SBVs are independent.
		\item If $X$ is an SBV with representation $(I,f)$ and $f$ has finite arity $n$, then
		\[\mathbb{E}\left[X\right]=\sum_{(\omega_{1},\omega_{2},\ldots,\omega_{n})\in 2^{\{n\}} } \frac{1}{2^n}f(\omega_{1},\omega_{2},\ldots,\omega_{n}), \]

		which demonstrates that changing the selection function does not change the expectation.
	\end{itemize}
	For an SBV $X$, we will use $X$ to denote both the true SBV (the function) as well as the output of applying $X$ to a specific sequence $\omega$. It will be clear from context whether $X$ must be seen as a function or an output of that function.

	~

	Now, we look at appropriate SBVs to generate our samples. Recall the CLT (see for example Feller \cite[p244]{Feller_book}): Let $(B_{n})$ be a sequence of iid (independent and identically distributed) random variables with mean $\mu$ and variance $\sigma^2$ and let $\Phi$ be the CDF of the standard normal distribution (mean $0$ and variance $1$). Letting $S_n=B_1+B_2+\ldots+B_n$, then for any $a\in \mathbb{R}$,
	\[ P\left( \frac{S_n-n\mu}{\sigma\sqrt{n}}<a \right)\rightarrow \Phi(a),  \]
	often written as $\frac{S_n-n\mu}{\sigma\sqrt{n}}\xrightarrow{D} N(0,1)$.

	In case the $B_i$ are Bernoulli random variables with parameter $p=1/2$, the statement becomes $\frac{2S_n-n}{\sqrt{n}}\xrightarrow{D} N(0,1)$. Based on this, we use the following set of SBVs to generate our samples\footnote{There are of course multiple alternatives to this particular set of SBVs, some of which are briefly discussed in Section \ref{variations_section}.}:
	\begin{align}\label{sampling_scheme}
		\underbrace{\omega_1}_{ X_1=\frac{2\omega_1-1}{\sqrt{1}} }\quad\underbrace{\omega_2\quad \omega_3}_{ X_2=\frac{2(\omega_2+\omega_3)-2}{\sqrt{2}} }\quad \underbrace{\omega_4\quad \omega_5\quad \omega_6}_{ X_3=\frac{2(\omega_4+\omega_5+\omega_6)-3}{\sqrt{3}} }\qquad \underbrace{ \omega_7 \quad \omega_8\quad \omega_9\quad \omega_{10}}_{X_4=\frac{2(\omega_7+\omega_8+\omega_9+\omega_{10})-4}{\sqrt{4}}}\qquad \underbrace{\omega_{11} \quad\ldots}_{X_5=\ldots },
	\end{align}
	always using $n$ digits in the definition of $X_n$. We will act as if these $X_i$ are our random samples from some distribution. Let $\widehat{D}_k$ be the empirical distribution of the first $k$ samples $X_1,\ldots, X_k$ (meaning $\widehat{D}_k$ is a RV whose CDF is the empirical distribution function of $\{X_1,\ldots,X_k \}$). We will show that if $\omega$ is MLR, then the empirical moments of these distributions converge to the corresponding moments of the normal distribution. Formally for any MLR sequence $\omega$, we will show that
	\begin{align}\label{moments_conv}
		\mathcal{E}\left[ \widehat{D}_k^m \right]=\frac{X_1^m+X_2^m+\ldots+X_k^m}{k} \rightarrow \nu_{m}, \qquad \text{as }k\rightarrow\infty, \text{ for all } m\in \mathbb{N}_0,
	\end{align}
	where $\nu_{m}$ denotes the $m$-th moment of the standard normal distribution (see Papoulis \cite[p148]{Papoulis_book}):
	\begin{align}\label{normalmoments}
		\nu_m=\begin{cases}
			0& \text{if } m \text{ is odd},\\
			\displaystyle
			(m-1)!! & \text{if } m \text{ is even\footnotemark}.
		\end{cases}
	\end{align}
	\footnotetext{$(m-1)!!$ denotes the double factorial: $(m-1)!!=(m-1)\cdot (m-3)\cdot (m-5)\cdot \ldots \cdot 3\cdot 1$.}

	We use the notation $\mathcal{E}$ instead of $\mathbb{E}$ to stress that $\mathcal{E}\left[ \widehat{D}_k^m \right]$ is an SBV (its value depends on $\omega$), not a number.

	As shown later, if (\ref{moments_conv}) holds for a given sequence, then the corresponding empirical distribution functions $\widehat{F}_k$ defined as
	\begin{align}\label{emp_dist_func}
		\widehat{F}_k(t)= \frac{\pmb{1}_{\left\{X_1\leq t \right\}} +\pmb{1}_{\left\{X_2\leq t \right\}}+\ldots +\pmb{1}_{\left\{X_k\leq t \right\}}}{k},\qquad t\in \mathbb{R}
	\end{align}
	will converge uniformly to the CDF of the standard normal distribution. Note that for any $k$ and $t$, $\widehat{F}_k(t)$ is an SBV. These results are summarized in Theorem \ref{main_result}.

	~

	To prove the convergence of the moments (\ref{moments_conv}), we use an adapted version of the SLLN due to Kolmogorov.

	\begin{theorem}[Kolmogorov's SLLN for MLR sequences]\label{kolmogorov}
		Let $(X_n)$ be a sequence of independent sequence-based variables, uniformly in $n$. Let $\mu_n$ and $\sigma_n^2$ respectively be the average and variance of $X_n$\footnote{These exist and are uniformly computable: Since $2^{\N}$ is compact and $X_n$ are computable, it is an easy to show that $\int X_n(\omega)\, d\omega$ and $\int X_n^2(\omega)\, d\omega$ are uniformly computable in $n$.}
		Assume the $X_n$ have finite averages $\mu_n$ and variances $\sigma_n^2$. Also assume the following holds
		\begin{enumerate}
			\item The sequences $(\mu_n)$ converges effectively to some $\mu \in\R$.
			\item $\sum_k k^{-2}\sigma_k^2$ is a finite, computable number.
		\end{enumerate}
		Then for all Martin-L\"of random sequences
		\begin{align}\label{converge}
			\bar{X}_n \rightarrow \mu \qquad \text{as }n\rightarrow \infty,
		\end{align}
		where $\bar{X}_n=\frac{1}{n}\sum_{i\leq n}X_i$.
	\end{theorem}
	The proof of this theorem can be found at the end of this section.
	The proof of the original statement by Kolmogorov can be found in Sen \cite[p67]{Sen}, where of course conditions \textit{1} and \textit{2} were not present. These conditions are necessary to make an effective version of the proof in \cite{Sen}, but are perhaps not required for the theorem to hold. This paper does not make any attempts to answer whether or not they are indeed required.

	Another minor difference: Usually, the condition $\mu_n\rightarrow \mu$ is not present and the theorem simply states that
	\begin{align}\label{converge_original}
		\bar{X}_n-\bar{\mu}_n \rightarrow 0 \qquad \text{a.s. as }n\rightarrow \infty,
	\end{align}
	where $\bar{\mu}_n=\frac{1}{n}\sum_{i\leq n}\mu_i$. It is an easy exercise to show that (\ref{converge}) and (\ref{converge_original}) are equivalent when $\mu_n \rightarrow \mu$ and that the convergence of the $(\bar{\mu}_n)$ is effective when the $(\mu_n)$ converge effectively.

	~

	The conditions  are necessary to make an effective version of the proof by Shen. It might be possible to weaken these conditions.

	~

	We will, for any $m\in \mathbb{N}_0$, apply Theorem \ref{kolmogorov} to the sequence $X_1^m, X_2^m,\ldots$ where the $X_i$ are those defined in (\ref{sampling_scheme}) to show that (\ref{moments_conv}) holds. However, before we can apply the theorem, we need some knowledge on the expectation and variance of the $X_i^m$.

	Note that the SBVs in (\ref{sampling_scheme}) can also be defined by first transforming the original sequence $\omega$ into a Rademacher sequence $r$ by the equation $r_i=2\omega_i-1$ (replacing any $0$s with $-1$s) and then defining
	\begin{align}\label{alternative}
		\underbrace{r_1}_{ X_1=\frac{r_1}{\sqrt{1}} }\qquad\underbrace{r_2\qquad r_3}_{ X_2=\frac{r_2+r_3}{\sqrt{2}} }\qquad \underbrace{r_4\qquad r_5\qquad r_6}_{ X_3=\frac{r_4+r_5+r_6}{\sqrt{3}} }\qquad \underbrace{ r_7 \qquad r_8\qquad r_9\qquad r_{10}}_{X_4=\frac{r_7+r_8+r_9+r_{10}}{\sqrt{4}}}\qquad \underbrace{r_{11} \qquad\ldots}_{X_5=\ldots }.
	\end{align}

	Hence, we use the following lemma, which will allow us to use Theorem \ref{kolmogorov}.

	\begin{lemma}\label{rad_lemma}
		Let $r_i$ be iid Rademacher distributed and define $S_n=r_1+r_2+\ldots +r_n$. Then for any fixed $m\geq 1$,
		\begin{align*}
			\mathbb{E}[S_n^m]=\begin{cases}
				0& \text{if } m \text{ is odd},\\
				\displaystyle
				(m-1)!!\cdot n^{m/2}+\mathcal{O}\left(n^{\frac{m}{2}-1}\right) & \text{if } m \text{ is even}.
			\end{cases}
		\end{align*}
	\end{lemma}

	The proof of this lemma is given at the end of this section, as we first demonstrate how it applies to our situation. Letting $X_i$ as always denote the SBVs in (\ref{sampling_scheme}) (and alternatively (\ref{alternative})),
	\begin{align}\label{expectationSBV}
		\mathbb{E}\left[X_n^m\right]=\mathbb{E}\left[\left(\frac{r_1+\ldots+r_n}{\sqrt{n}}\right)^m\right]=\begin{cases}
			0& \text{if } m \text{ is odd},\\
			\displaystyle
			(m-1)!!+\mathcal{O}\left(\frac{1}{n}\right) & \text{if } m \text{ is even}.
		\end{cases}
	\end{align}

	Hence, we can see that these expectations converge to the moments of the standard normal distribution. This fact alone can be proven much more easily, but (\ref{expectationSBV}) also gives a computable upper bound on the error, which which implies that the convergences are effective. Using this expression, it can also be seen that the variance $\text{Var}\,X_n^m$ converges to $(2m-1)!!$ if $m$ is odd and to $(2m-1)!!-[(m-1)!!]^2$ if $m$ is even. In particular, the variances remain bounded. This implies that $\sum_k k^{-2}\text{Var}\,X_k^m <\infty$ and that this sum is computable, since we have an upper bound for the size of the tail. Thus, we may apply Theorem \ref{kolmogorov} to the sequence $(X_{n}^m)$, which finally implies that (\ref{moments_conv}) holds for all MLR sequences.

	~

	Next, consider one of the sequences $\omega$ for which (\ref{moments_conv}) holds. We show that for such a sequence the empirical distribution function converges uniformly to $\Phi$. This result follows directly from the following theorem.

	\begin{theorem}\label{method_of_moments}
		The following statements hold
		\begin{enumerate}[noitemsep]
			\item The standard normal distribution is completely determined by its moments, meaning that if some r.v. $X$ has the same moments as the standard normal distribution, then $X\sim N(0,1)$.
			\item  Suppose that the distribution of $X$ is determined by its
			moments, that the $X_n$ have moments of all orders, and that $\lim_n \mathbb{E}[X_n^m] =
			\mathbb{E}[X^m]$ for $m = 1, 2, \ldots$. Then $X_n \xrightarrow{D} X$.
			\item Let $X_n,X$ be random variables such that $X_n \xrightarrow{D} X$ and let $F_n,F$ respectively denote their CDF. If $F$ is continuous, then
			\[ \sup_{x}|F_n(x)-F(x)|\rightarrow 0 \qquad \text{as }n\rightarrow \infty. \]
		\end{enumerate}
	\end{theorem}

	Indeed, for such a fixed sequence, the empirical distributions $\widehat{D}_n$ form a sequence of random variables whose moments converge to those of the normal distribution. Hence, we can apply Theorem \ref{method_of_moments} to find that $\widehat{F}_n\rightarrow\Phi$ uniformly for all MLR sequences $\omega$, where
	$\widehat{F}_n(t)$ is understood as the output of the SBV $\widehat{F}_n(t)$ on input $\omega$.
	The proof of Statements \textit{1} and \textit{2} can be found in Billingsley \cite[Ex. 30.1 \& Thm 30.2]{Billingsley_book}. Statement \textit{3} is proven in Chow \cite[p260]{Chow_book}. An alternative proof of the pointwise convergence of $\widehat{F}_n(t)$ can be found by applying Theorem \ref{kolmogorov} to the sequence $\pmb{1}_{\left\{X_k\leq t \right\}}$. We summarize these results in the following theorem.

	\begin{theorem}\label{main_result}
		Let $\omega$ be a Martin-L\"of random sequence and define the sequence $(X_n)$ as in \eqref{sampling_scheme}.
		Let $\widehat{D}_k$ be the random variable assigning probability $1/k$ to any outcome $X_i,i=1,\ldots,k$ and let $\widehat{F}_k$ be the corresponding CDF. Then $\widehat{D}_k$ approximates the normal distribution in the following ways:
		\begin{enumerate}[noitemsep]
			\item  Let $\nu_m$ denote the $m$-th moment of the standard normal distribution. Then for every $m\in \N_0$:
			\[ \mathbb{E}\left[\widehat{D}_k^m\right]=\frac{X_1^m+X_2^m+\ldots+X_k^m}{k}\rightarrow \nu_m \qquad \text{ as }k\rightarrow \infty. \]
			In other words, all moments of $\widehat{D}_k$ converge to the corresponding moments of the standard normal distribution.
			\item The CDFs $\widehat{F}_k$ converge uniformly to the CDF $\Phi$ of the standard normal distribution, i.e.
			\begin{align*}
				\sup_{t}|\widehat{F}_k(t)-\Phi(t)|\rightarrow 0 \qquad \text{as }k\rightarrow \infty.
			\end{align*}
		\end{enumerate}
	\end{theorem}

		An alternative way to prove Theorem 3.5 is to consider the sequence $(X_k)$ as the image of a map $f:2^\N \rightarrow \R^\N:\omega\mapsto (X_k)$. If we then endow the space of real sequences $\R^\N$ with the push-forward measure $\mu_f$ defined as $\mu_f(A)=\mu(f^{-1}(A))$ (with $\mu$ the Lebesgue measure on $2^\N$), we can apply \cite[Theorem 3.9]{Hoyrup2} which says that if $x$ is ML-random, so is $f(x)$. Theorem \ref{main_result} then reduces to proving that a ML-random sequence $(X_k)$ in this $\mathbb{R}^\N$ has the property that $\mathbb{E}\left[\widehat{D}_k^m\right]$ as defined above converges to $\nu_m$ for all $m\in \mathbb{N}$. Because of the nature of the measure $\mu_f$, this would likely require proving a statement similar to Theorem \ref{kolmogorov}.

	~

	Theorem \ref{kolmogorov} and, therefore, Theorem \ref{main_result} also hold for Schnorr random sequences: When looking at the proof of Theorem \ref{kolmogorov} (see below), the measures of the $K_{M}^{\epsilon}$ are computable since
	\[ K_{M}^{\epsilon}= \left\{ \omega \mid \sup\limits_{ M\leq k\leq N  }\, |\bar{X}_k-\mu|>\epsilon \right\}\cup\left\{ \omega \mid \sup\limits_{ k\geq N  }\, |\bar{X}_k-\mu|>\epsilon \right\} \]
	for any $N\geq M$, which splits $K_{M}^{\epsilon}$ into a set described by a finite amount of strings, and a set whose measure effectively tends to $0$ as $N\rightarrow \infty$. The measures of the $U_n=\cup_{l=1}^\infty K_{M(n,l)}^{2^{-l}}$ are then also uniformly computable since the measure of the tail $\cup_{l=L}^\infty K_{M(n,l)}^{2^{-l}}$ is bounded by $2^{-n-L+1}$.

	~

	We end this section with the proofs of Theorem \ref{kolmogorov} and Lemma \ref{rad_lemma}.
	\begin{proof}[Proof of Theorem \ref{kolmogorov}, based on Sen \cite{Sen}]
		Let $T_k=\sum_{n\leq k}(X_n-\mu_n)$ and $T_0=0$. We start from the following inequality (see Sen \cite[Eq 2.3.65]{Sen}):
		\begin{align}\label{derivedineq}
			P\left(\max\limits_{M\leq k\leq N}\,\frac{1}{k} |T_k|>t\right)\leq \frac{1}{t^2}\left[ \frac{1}{M^2}\left( \sum_{k=1}^{M}\sigma_k^2 \right)+\sum_{k=M+1}^{N} k^{-2}\sigma_k^2\right] .
		\end{align}
		Let for any $k\geq 1$ $D_k=\sum_{n\geq k}n^{-2}\sigma_n^2$. Note that by assumption, $D_1=\sum_{n}n^{-2}\sigma_n^2<\infty$ and hence, $(D_k)$ is a bounded decreasing sequence with $\lim_kD_k=0$ (since $D_1$ is computable, this convergence is effective). Also,
		\begin{align*}
			\frac{1}{M^2}\sum_{k=1}^{M}\sigma_k^2&=\frac{1}{M^2}\sum_{k=1}^{M}k^2[D_k-D_{k+1}]\leq \frac{1}{M^2}\sum_{k=1}^{M}(2k-1)D_k.
		\end{align*}
		Since $M^{-2}\sum_{k=1}^{M}(2k-1)\rightarrow 1$ and $D_k\rightarrow 0$, it is easy to show that the rhs (right-hand side) converges to $0$ (and does so effectively since $D_k$ also converges effectively to $0$). Hence, the lhs converges effectively to $0$. Finally, we can show that $\bar{X}_k\rightarrow \mu$ for all MLR sequences. Fix $\epsilon>0$ and note that by the triangle inequality, $|\bar{X}_k-\mu|>\epsilon$ implies that either $|\bar{X}_k-\bar{\mu}_k|> \epsilon/2$ or $|\bar{\mu}_k-\mu|> \epsilon/2$. As mentioned before, it is easy to show that $\bar{\mu}_k\rightarrow\mu$ and that this convergence is effective if and only if the convergence of $(\mu_n)$ is effective. Since the latter is part of our assumptions, we can for any $\epsilon$, find $M_0$ such that $|\bar{\mu}_k-\mu|< \epsilon/2$ for all $k\geq M_0$. For such a $k$, we have
		\[ P\left( |\bar{X}_k-\mu|> \epsilon \right)\leq P\left( |\bar{X}_k-\bar{\mu}_k|> \frac{\epsilon}{2} \right)=P\left(\frac{1}{k} |T_k|> \frac{\epsilon}{2}  \right).  \]
		Using the previous argument simultaneously for all $k$ between $M\geq M_0$ and some $N$, we find that
		\begin{align*}
			P\left( \max\limits_{M\leq k\leq N}\,|\bar{X}_k-\mu|> \epsilon \right) \leq  P\left(\max\limits_{M\leq k\leq N}\,\frac{1}{k} |T_k|>\frac{\epsilon}{2}\right)\leq \frac{4}{\epsilon^2}\left[ \frac{1}{M^2}\left( \sum_{k=1}^{M}\sigma_k^2 \right)+\sum_{k=M+1}^{N} k^{-2}\sigma_k^2\right].
		\end{align*}

		Letting $N$ tend to infinity and keeping $M$ fixed gives
		\begin{align}\label{K_eps}
			P\left(\sup\limits_{k\geq M}\, |\bar{X}_k-\mu|>\epsilon\right)\leq \frac{4}{\epsilon^2}\left[ \frac{1}{M^2}\left( \sum_{k=1}^{M}\sigma_k^2 \right)+D_{M+1}\right].
		\end{align}
		For any $\epsilon>0$ the rhs effectively tends to $0$ as $M\rightarrow \infty$ and hence, the r.e. sets $K_M^{\epsilon}=\{ \omega \in 2^{\N}\mid \sup\limits_{k\geq M}\, |\bar{X}_k-\mu|>\epsilon \}$
		allow us to define the Martin-L\"of test: let $M(n,l)$ be a computable function such that
		\begin{align*}
			\mu\left(K_{M(n,l)}^{2^{-l}}\right)=P\left(\sup\limits_{k\geq M(n,l)}\, |\bar{X}_k-\mu|>2^{-l}\right)<2^{-(n+l)}
		\end{align*}
		and define $U_n=\cup_{l=1}^\infty K_{M(n,l)}^{2^{-l}}$. The $U_n$ are uniformly r.e. since the $K_{M(n,l)}^{2^{-l}}$ are uniformly r.e. in $n,l$ and by construction, $\mu(U_n)<2^{-n}$. Hence, the $U_n$ define a Martin-L\"of test whose intersection contains all sequences for which $\bar{X}_n \nrightarrow \mu$.
	\end{proof}

	\begin{proof}[Proof of Lemma \ref{rad_lemma}]
		The case where $m$ is odd is trivial: Note that the distribution of $S_n$ (and therefore $S_n^m$) is completely symmetric around $0$. Hence $\mathbb{E}[S_n^m]=0$.

		~

		Now for the case where $m$ is even: Using the multinomial theorem (Spiegel \cite[p3]{Spiegel_book}), we find
		\begin{align*}
			S_n^m=&(r_1+r_2+\ldots+r_n)^m\\
			=& \sum_{k_1+k_2+\ldots+k_n=m} \binom{m}{k_1,k_2,\ldots,k_n}r_1^{k_1}r_2^{k_2}\ldots r_n^{k_n},
		\end{align*}
		where \[\binom{m}{k_1,k_2,\ldots,k_n}=\frac{m!}{k_1!\cdot k_2!\cdot \ldots \cdot k_n!}. \]
		Taking expectation, we find
		\begin{align*}
			\mathbb{E}[S_n^m]=\sum_{k_1+k_2+\ldots+k_n=m} \binom{m}{k_1,k_2,\ldots,k_n}\mathbb{E}[r_1^{k_1}]\mathbb{E}[r_2^{k_2}]\ldots \mathbb{E}[r_n^{k_n}].
		\end{align*}
		Since $\mathbb{E}[r_i^{k_i}]$ is $0$ if $k_i$ is odd and $1$ if $k_i$ is even, this simplifies to
		\begin{align*}
			\mathbb{E}[S_n^m]=\sum_{\substack{k_1+k_2+\ldots+k_n=m\\\text{all }k_i\text{ even}}} \binom{m}{k_1,k_2,\ldots,k_n}
		\end{align*}
		Simplifying the summation such that we only sum over distinct partitions $(k_1,k_2,\ldots,k_l)$ of $m$ yields
		\begin{align*}
			\mathbb{E}[S_n^m]=&\sum_{\substack{p=(k_1,k_2,\ldots,k_l)\\\text{partition of }m, \\\text{all }k_i\text{ even}
			}} \binom{m}{k_1,k_2,\ldots,k_l}K_{p}^{(n)},
		\end{align*}
		with
		\begin{align*}
			K_p^{(n)}=K_{(k_1,k_2,\ldots,k_l)}^{(n)}=\begin{aligned}
				&\text{ The number of ways we can write }k'_1+k'_2+\ldots+k'_n=m\\
				& \text{ where the non-zero terms are exactly the }k_i.
			\end{aligned}
		\end{align*}
		While an explicit formula for $K_p^{(n)}$ is quite cumbersome, we only give the following bounds, which suffice for the coming discussion:
		\[  \binom{n}{l}\leq K_{(k_1,k_2,\ldots,k_l)}^{(n)} \leq \frac{n!}{(n-l)!}<n^l. \]
		The lower (resp. upper) bound is achieved by assuming that all the $k_i$ are the same (resp. different). In particular, $K_{p}^{(n)}=\mathcal{O}(n^l)$ as $n\rightarrow \infty$.

		Note that $(k_1,k_2,\ldots,k_l)$ is a partition of $m$ with even coefficients if and only if  $\left(\frac{k_1}{2},\frac{k_2}{2},\ldots,\frac{k_l}{2}\right)$ is a partition of $\frac{m}{2}$. Hence
		\begin{align*}
			\mathbb{E}[S_n^m]=&\sum_{\substack{p=(k_1,k_2,\ldots,k_l)\\\text{partition of }\frac{m}{2}}} \binom{m}{2k_1,2k_2,\ldots,2k_l}K_{2p}^{(n)}=\sum_{l=1}^{m/2} \sum_{\substack{p=(k_1,k_2,\ldots,k_l)\\\text{partition of }\frac{m}{2} \\\text{of length }l}} \binom{m}{2k_1,2k_2,\ldots,2k_l}K_{2p}^{(n)},
		\end{align*}

		Note that there is only 1 partition of length $l=m/2$, namely $(1,\ldots,1)$. In that case, $K_{2p}^{(n)}=\binom{n}{\frac{m}{2}}$. Hence,
		\begin{align*}
			\mathbb{E}[S_n^m]=& \sum_{l=1}^{m/2} \sum_{\substack{p=(k_1,k_2,\ldots,k_l)\\\text{partition of }\frac{m}{2} \\\text{of length }l}} \binom{m}{2k_1,2k_2,\ldots,2k_l}K_{2p}^{(n)}\\
			=& \binom{n}{\frac{m}{2}}\frac{m!}{2^{m/2}} + \sum_{l=1}^{m/2-1} \sum_{\substack{p=(k_1,k_2,\ldots,k_l)\\\text{partition of }\frac{m}{2} \\\text{of length }l}} \binom{m}{2k_1,2k_2,\ldots,2k_l}K_{2p}^{(n)}\\
			=& \frac{m!}{\left(\frac{m}{2}\right)!2^{m/2} }n^{m/2}+\mathcal{O}\left(n^{\frac{m}{2}-1}\right)\\
			=& (m-1)!!\cdot n^{m/2}+\mathcal{O}\left(n^{\frac{m}{2}-1}\right).
		\end{align*}
	\end{proof}

	\subsection{Alternative versions}\label{variations_section}

	Recall that our way of defining the sampling scheme (\ref{sampling_scheme}) was mostly arbitrary. As the only requirement was that the samples resembled $\frac{2S_n-n}{\sqrt{n}}$ for some large $n$, nothing forced us to require that $X_k$ uses exactly $k$ digits. A slightly more general approach is to let $X_k$ use $n(k)$ digits for some computable function $n:\mathbb{N} \mapsto \mathbb{N}$. For example, if $n(k)=2k$, the SBVs become
	\begin{align*}
		\underbrace{\omega_1\quad \omega_2}_{ X_1=\frac{2(\omega_1+\omega_2)-2}{\sqrt{2}} }\qquad\underbrace{ \omega_3 \quad \omega_4\quad \omega_5\quad \omega_6}_{ X_2=\frac{2(\omega_3+\omega_4+ \omega_5+\omega_6)-4}{\sqrt{4}} }\qquad\underbrace{\omega_7\quad \omega_8 \quad \omega_9\quad \omega_{10}\quad \omega_{11} \quad \omega_{12}}_{ X_3=\frac{2(\omega_7+ \omega_8 + \omega_9+ \omega_{10}+ \omega_{11} + \omega_{12})-6}{\sqrt{6}} }\qquad \underbrace{\omega_{13}\quad \ldots}_{ X_4=\ldots }.
	\end{align*}

	In this section, we consider two cases: $n(k)\rightarrow \infty$ effectively and $n(k)=N$ for some fixed $N\in \mathbb{N}$.

	~

	\underline{$n(k)\rightarrow  \infty$ effectively}: As a generalization of (\ref{expectationSBV}), we have
	\begin{align*}
		\mathbb{E}\left[X_k^m\right]=\begin{cases}
			0& \text{if } m \text{ is odd},\\
			\displaystyle
			(m-1)!!+\mathcal{O}\left(\frac{1}{n(k)}\right) & \text{if } m \text{ is even}.
		\end{cases}
	\end{align*}
	Hence, the averages (and variances) still converge to the same values. As nothing truly changes, Theorem \ref{main_result} still holds for these SBVs.

	~

	$\underline{n(k)=N}:$ As an example, if $N=4$ the SBVs become
	\begin{align*}
		\underbrace{\omega_1\quad \omega_2\quad \omega_3\quad \omega_4}_{ X_1=\frac{2(\omega_1+\omega_2+\omega_3+ \omega_4)-4}{\sqrt{4}} }\qquad\underbrace{ \omega_5\quad \omega_6\quad \omega_7\quad \omega_8}_{ X_2=\frac{2(\omega_5+ \omega_6+ \omega_7+ \omega_8)-4}{\sqrt{4}} }\qquad\underbrace{\omega_9\quad \omega_{10}\quad \omega_{11} \quad \omega_{12}}_{ X_3=\frac{2(\omega_9+ \omega_{10}+ \omega_{11} + \omega_{12})-4}{\sqrt{4}} }\qquad \underbrace{\omega_{13}\quad \ldots}_{ X_4=\ldots }.
	\end{align*}
	Of course, the corresponding moments $\mathbb{E}\left[\widehat{D}_k^m \right]$ and distribution function $\widehat{F}_k$ no longer converge to those of the normal distribution, but instead to those of a normalized binomial distribution. However, as $N$ tends to infinity, these moments and distribution function themselves tend to those of the normal distribution function. Hence, we have for all MLR sequences:
	\[ \lim\limits_{N\rightarrow \infty}\lim\limits_{k\rightarrow \infty} \frac{X_1^m+X_2^m+\ldots+X_k^m}{k} \rightarrow \nu_{m},\qquad \lim\limits_{N\rightarrow \infty}\lim\limits_{k\rightarrow \infty} \widehat{F}_k(t)=\Phi(t) \quad t\in \R. \]

	Note that the only condition that is required for this version of the CLT, is that the original sequence $\omega$ is normal. Schnorr random sequences are of course all normal, see Shen \cite[Thm 167 \& 168]{Shen}.

	~

		Sampling scheme (\ref{sampling_scheme}) can also be generalized by introducing an in-between function $f$ in the following way:
		\begin{align}\label{sampling_scheme2}
			\underbrace{\omega_1 \ldots \omega_k}_{ X_1=\frac{f(\omega_{1:k})}{\sqrt{1}} }\quad\underbrace{\omega_{k+1}\ldots \omega_{2k}\quad \omega_{2k+1}\ldots \omega_{3k}}_{ X_2=\frac{f(\omega_{(k+1):2k})+f(\omega_{(2k+1):3k})}{\sqrt{2}} }\quad \underbrace{\omega_{3k+1}\ldots \omega_{4k}\quad \omega_{4k+1}\ldots \omega_{5k}  \quad \omega_{5k+1}\ldots \omega_{6k}}_{ X_3=\frac{f(\omega_{(3k+1):4k})+f(\omega_{(4k+1):5k})+f(\omega_{(5k+1):6k})}{\sqrt{3}} }
		\end{align}
		with the added condition that when $f$ is applied to standard Bernoulli random variables $\mathbb{E}[f]=0$ and $\mathbb{E}[f^2]=1$. The natural question is then to ask what conditions can be placed on $f$ such that Theorem \ref{main_result} still holds. Another question is whether the theorem still holds if the definition of $X_k$ uses slightly overlapping digits of $\omega$. We leave these questions for future work.

	\section{The almost sure central limit theorem}\label{ASCLT_section}

	In this section, the $X_k$ no longer denote the SBVs in \eqref{sampling_scheme}, but some other SBVs such that $E[X_k]=0$ and $E[X_k^2]=1$ (in particular, $X_k=\omega_k$).

	We will show that the classic ASCLT (see Brosamler \cite{Brosamler})
	\begin{align}\label{ASCLT}
		\lim_{n\rightarrow\infty }\frac{1}{\log n}\sum_{k=1}^n\frac{1}{k} I\left\{\frac{S_k}{\sqrt{k}}\leq x \right\}\rightarrow \Phi(x)\quad \text{ for all }x\in \R
	\end{align}
	holds for all Martin-L\"of (and even Schnorr sequences). Our proof is a based on the one presented by Jonsson \cite{Jonsson}. Since we only show this basic form, we can substantially shorten that proof (apart from adding some computability restrictions).

	~

	We start this section with three preliminary lemmas. Although slightly restated in the context of MLR sequences, the proofs of these lemmas are identical to those found in \cite{Jonsson} (respectively Theorem 2.4, Theorem 2.25 \& Lemma 3.2).

	\begin{lemma}\label{portmanteau}
		Let $(d_k)$ be a sequence of positive uniformly computable real numbers with $D_n=\sum_{k\leq n}d_k$. Also, let $(X_k)$ be a sequence of SBVs and let $G$ be a probability measure on $\R$ with $C_G$ its continuity points, then the following three conditions are equivalent:
		\begin{enumerate}
			\item \label{cond:first}  $\frac{1}{D_n} \sum_{k\leq n}d_k f(X_k)\rightarrow \int f\,dG$ for all MLR sequences and all bounded and continuous $f$.
			\item \label{cond:second} $\frac{1}{D_n} \sum_{k\leq n}d_k f(X_k)\rightarrow \int f\,dG$ for all MLR sequences and all bounded Lipschitz-functions $f$.
			\item \label{cond:third} $\frac{1}{D_n} \sum_{k\leq n}d_k I\{X_k\leq x\}\rightarrow G(x)$ for all MLR sequences and all $x\in C_G$.
		\end{enumerate}
	\end{lemma}

	\begin{lemma}\label{notes2.25}
		Let $(p_n)$ and $(q_n)$ be two positive sequences and let $P_n=\sum_{k\leq n} p_n$, $Q_n=\sum_{k\leq n} q_n$. Then if $p_n/q_n\rightarrow 1$, we have
		\[  \frac{1}{P_n}\sum_{k=1}^n p_ks_k\rightarrow s\qquad \Leftrightarrow \qquad  \frac{1}{Q_n}\sum_{k=1}^n q_ks_k\rightarrow s \]
		for any bounded sequence $(s_n)$ and $s\in \R$.
	\end{lemma}

	\begin{lemma}\label{notes3.2}
		Let $(d_k)$ and $(D_n)$ be as in Lemma \ref{portmanteau} and assume that
		\[D_n\rightarrow \infty \quad \text{ and }\quad D_{n+1}/D_n\rightarrow 1  \]
		effectively as $n\rightarrow \infty$, then for each computable $a>1$ there exists a computable subsequence $n_k$ such that
		\[ D_{n_k}\geq a^k  \quad \text{and}\quad D_{n_k}/a^k \rightarrow 1. \]
	\end{lemma}
	Now, we start with the actual proof. By Lemma \ref{notes2.25}, it is sufficient to show that
	\[ \lim_{n\rightarrow\infty }\frac{1}{\log (n+1)}\sum_{k=1}^n\log\left(1+\frac{1}{k}\right) I\left\{\frac{S_k}{\sqrt{k}}\leq x \right\}\rightarrow \Phi(x)\quad \text{ for all }x\in \R \]
	holds for all MLR numbers. Letting $d_k=log\left(1+\frac{1}{k}\right), D_n=\sum_{k\leq n}d_k$ and $Y_k=S_k/\sqrt{k}$, the LHS can be rewritten as $\lim_{n\rightarrow\infty }\frac{1}{D_n}\sum_{k=1}^nd_k I\left\{Y_k\leq x \right\}$. By Lemma \ref{portmanteau}, this is equivalent to
	\begin{align}\label{toshowTn}
		\frac{1}{D_n}\sum_{k\leq n}d_k f\left(\frac{S_k}{\sqrt{k}} \right)\rightarrow \int f\,d \Phi
	\end{align}
	for all MLR sequences and all bounded, Lipschitz continuous $f$. It follows from $Y_k \xrightarrow{D} N(0,1)$ and Portmanteau's lemma that
	\[ Ef(Y_k)\rightarrow \int f\,d\Phi \quad \text{as } k\rightarrow\infty \]
	for such functions $f$. It is easy to verify that this implies
	\[ \frac{1}{D_n}\sum_{k=1}^{n}d_k Ef(Y_k)\rightarrow \int f\,d\Phi \quad \text{as } k\rightarrow\infty. \]
	It therefore remains to prove that
	\begin{align}\label{defTn}
		T_n=\frac{1}{D_n}\sum_{k=1}^{n}d_k\left(f(Y_k)-Ef(Y_k) \right)\rightarrow 0 \quad \text{as } n\rightarrow\infty
	\end{align}
	for all MLR numbers. Let
	\[ \xi_k=f(Y_k)-Ef(Y_k), \]
	so that $T_n=\frac{1}{D_n}\sum_{k\leq n}d_k\xi_k$. The main idea of the proof is to show the following lemma.
	\begin{lemma}\label{expTn}
		Let $T_n$ be as defined in \eqref{defTn}, then the following holds.
		\begin{align}\label{Tnbound}
			E[T_n^2]\leq  C \frac{\log \log n}{\log n}.
		\end{align}
		for large $n$ and some $C>0$.
	\end{lemma}
	This lemma is a reformulated version of Proposition 4.3 in \cite{Jonsson} and the proof given at the end of this section is based on the one given in \cite{Jonsson}. First, we show how this result can be used to prove \eqref{ASCLT}.

	Let $a>1$ and apply Lemma \ref{notes3.2} with this $a$ and $d_k=\log \frac{k+1}{k}$. Since $D_{n_k}/a^k\geq 1$,
	the inequality above for this subsequence then becomes
	\begin{align}
		ET_{n_k}^2\leq C \frac{k \log a}{a^k}.
	\end{align}
	Hence we have for any $M\in \N$,
	\begin{align*}
		E\sum_{k\geq M}T_{n_k}^2=\sum_{k\geq K} ET_{n_k}^2\leq C \sum_{k\geq M} \frac{k \log a}{a^k}=L_{a,M}<\infty,
	\end{align*}
	with $(a, M)\mapsto L_{a,M}$ computable and $L_{a,M}\rightarrow 0$ as $M\rightarrow \infty$. By Markov's inequality, we have
	\[ \mu\left\{\omega \mid \sup_{k\geq M} T_{n_k}^2>\epsilon \right\}\leq  \mu\left\{\omega \mid \sum_{k\geq M} T_{n_k}^2>\epsilon \right\} = P\left(\sum_{k\geq M} T_{n_k}^2>\epsilon \right)\leq  \frac{E\sum_{k\geq K} T_{n_k}^2}{\epsilon}=\frac{L_{a,M}}{\epsilon}. \]
	Similarly to the proof of Theorem \ref{main_result}, define $K_M^{\epsilon}$ as $\{ \omega \in 2^{\N}\mid \sup\limits_{k\geq M}T_{n_k}^2 >\epsilon \}$ and a function $M(n,l)$ such that $\mu(K^{2^{-l}}_{M(n,l)})<2^{-n-l}$. As explained before, the sets $U_n=\cup_{l=1}^\infty K_{M(n,l)}^{2^{-l}}$ define a Martin-L\"of test (and even a Schnorr test) whose intersection contains all sequences for which $T_{n_k}\nrightarrow 0$. Thus, $T_n\rightarrow 0$ for all Schnorr random sequences.

	Now we show that convergence for the subsequence $T_{n_k}$ implies convergence for the whole sequence $T_n$.
	Consider an arbitrary $n$ and assume that $n_k < n \leq n_{k+1}$. Since $D_{n_k}\sim a^k$, we have
	\[D_{n_{k+1}}/D_{n_k}\rightarrow a. \]
	By assumption there exists an $M$ such that $-M < \xi_k$ (assume $M>0$), for all $k$. Define, for
	$n\geq 1$,
	\[ T_n'=\frac{1}{D_n}\sum_{k\leq n}d_k(\xi_k+M)=T_n+M. \]
	We have that $ T_{n_k}'\rightarrow M $
	for all Schnorr random sequences. Moreover, by positivity
	\[ D_{n_k}T_{n_k}'\leq D_{n}T_{n}'\leq D_{n_{k+1}}T_{n_{k+1}}' \]
	which gives
	\[ \frac{D_{n_k}}{D_{n_{k+1}}}T_{n_k}'\leq  T_n' \leq  \frac{D_{n_{k+1}}}{D_{n_k}}T_{n_{k+1}}'. \]
	By letting $a\rightarrow 1$, we can see that $T_n'\rightarrow M$ and hence $T_n\rightarrow 0$.

	~

	This finally proves that for all Schnorr random numbers, $T_n\rightarrow 0$ which in turn shows that \eqref{toshowTn} holds, finishing the proof.

	~

	We end this section with the proof of Lemma \ref{expTn}.

	\begin{proof}[Proof of Lemma \ref{expTn}]
		Note that for any numbers $a_k\geq 0$
		\begin{align}\label{easy_ineq2}
			\sum_{k\leq l}a_ka_l \leq 	\left(\sum_k a_k\right)^2\leq 2\sum_{k\leq l}a_ka_l .
		\end{align}
		Applying the second inequality to $T_n$ gives
		\[E[T_n^2]\leq \frac{2}{D_n^2}\sum_{1\leq k\leq l\leq  n}d_kd_lE(\xi_k\xi_l)\leq  \frac{2}{D_n^2}\sum_{1\leq k\leq l\leq  n}d_kd_l|E(\xi_k\xi_l)|.  \]
		For convenience, we define for all $k,l$ with $k\leq l$
		\[ f_{k,l}=f\left( \frac{S_l-S_k}{\sqrt{l}}\right) \]
		Next, let $K$ be the Lipschitz constant of $f$ and assume $|f|\leq K$. Since $f$ is bounded, so is $\xi_k$; say $|\xi_k|\leq C$. Using this, the Lipschitz continuity of $f$ and the fact that $f_{k,l}$ is independent of $\xi_k$, we obtain
		\begin{align*}
			\left|E(\xi_k\xi_l) \right|&=\left|E(\xi_k(\xi_l-f_{k,l})) \right|\\
			&= \left|E \left[ \xi_k  \left( f\left( \frac{S_l}{\sqrt{l}}\right)-E\left[f\left(\frac{S_l}{\sqrt{l}}\right)\right]-f_{k,l} \right) \right]  \right|\\
			&= \left|E\left[\xi_k  \left( f\left( \frac{S_l}{\sqrt{l}}\right)-f_{k,l} \right)  \right]\right|\\
			&\leq C E\left|f\left( \frac{S_l}{\sqrt{l}}\right)-f_{k,l}\right|\\
			&\leq C E\left[   \frac{ K \left|S_k \right|}{\sqrt{l}}  \right]\\
			&=  CK \sqrt{\frac{k}{l}} E\left[  \frac{  \left|S_k \right|}{\sqrt{k}}  \right].
		\end{align*}
		Now, for all $k$ we show that $E\left(\frac{\left|S_k\right|}{\sqrt{k}}\right)\leq 2$.
		Indeed, let $D=\{ \omega \mid \frac{\left|S_k(\omega)\right|}{\sqrt{k}}<1 \}$, then
		\begin{align*}
			E\left(\frac{\left|S_k\right|}{\sqrt{k}}\right)=E_D\left(\frac{\left|S_k\right|}{\sqrt{k}}\right)+E_{D^C}\left(\frac{\left|S_k\right|}{\sqrt{k}}\right)
			\leq 1+\frac{1}{k}E\left[ S_k \right]^2
			= 1+\frac{1}{k}E\left[\sum_{i\leq k} X_i^2  \right]\leq 2.
		\end{align*}
		With that, we finally obtain
		\[ \left|E(\xi_k\xi_l) \right|\leq C'(k/l)^{1/2}. \]

		By the inequality above and using the definition of $D_n$ and $d_n$,
		\[E[T_n^2]\leq  \frac{2}{D_n^2}\sum_{1\leq k\leq l\leq  n}d_kd_l|E(\xi_k\xi_l)|\leq \frac{C}{(\log(n+1))^2}\sum_{1\leq k\leq l\leq  n}\log\frac{k+1}{k}\log\frac{l+1}{l}\left(\frac{k}{l}\right)^{1/2}.  \]
		To obtain a bound for the sum on the RHS, we define for a fixed $n$ and any $k$ the following sets:
		\begin{align*}
			A_k&=\{l\mid k\leq l\leq n\text{ and }l\leq k\log n \}\\
			B_k&=\{l\mid k\leq l\leq n\text{ and }l> k\log n \}.
		\end{align*}
		Note that for large $k$ ($k>n / \log n$), $A_k=\{ k,k+1,\ldots,n \}$ and $B_k=\emptyset$.
		\begin{align*}
			E[T_n^2]&\leq  \frac{C}{(\log(n+1))^2}\sum_{1\leq k\leq l\leq  n}\log\frac{k+1}{k}\log\frac{l+1}{l}\left(\frac{k}{l}\right)^{1/2}\\
			&=\frac{C}{(\log(n+1))^2}\left[\sum_{k=1}^n\sum_{l\in A_k }\log\frac{k+1}{k}\log\frac{l+1}{l}\left(\frac{k}{l}\right)^{1/2}+\sum_{k=1}^n\sum_{l\in B_k }\log\frac{k+1}{k}\log\frac{l+1}{l}\left(\frac{k}{l}\right)^{1/2}\right]
		\end{align*}
		Intuitively, we show that if $l\in A_k$ then $\log \frac{l+1}{l}$ is small, and if $l\in B_k$ then $k/l$ is small.

		Starting with $A_k$, note that $A_k=\{k,k+1,\ldots, n_k \}$ for some $n_k\leq n$. Hence
		\[ \sum_{l\in A_k}\log \frac{l+1}{l}=\log \frac{n_k+1}{k}=\log \frac{n_k+1}{n_k}+\log \frac{n_k}{k}\leq \log 2 +\log\log n, \]
		where we used the definition of $A_k$. For large $n$, the above is clearly bounded by $2\log\log n$. Hence
		\begin{align*}
			\sum_{k=1}^n\sum_{l\in A_k }\log\frac{k+1}{k}\log\frac{l+1}{l}\left(\frac{k}{l}\right)^{1/2}&\leq \sum_{k=1}^n\sum_{l\in A_k }\log\frac{k+1}{k}\log\frac{l+1}{l}\\
			&\leq \sum_{k=1}^n\log\frac{k+1}{k}\sum_{l\in A_k }\log\frac{l+1}{l}\leq 2\log (n+1 )\log \log n.
		\end{align*}
		For the indices $l\in B_k$, $k/l<1/ \log n$. Combining this with \eqref{easy_ineq2} where $d_k=\log \frac{k+1}{k}$ gives
		\begin{align*}
			\sum_{k=1}^n\sum_{l\in B_k }\log\frac{k+1}{k}\log\frac{l+1}{l}\left(\frac{k}{l}\right)^{1/2}&\leq \frac{1}{(\log n)^{1/2}}\sum_{k=1}^n\sum_{l\in B_k }\log\frac{k+1}{k}\log\frac{l+1}{l}\\
			&\leq \frac{1}{(\log n)^{1/2}}\sum_{1\leq k\leq l\leq n} \log\frac{k+1}{k}\log\frac{l+1}{l}\\
			&\leq \frac{(\log n+1)^2}{(\log n)^{1/2}}.
		\end{align*}
		Combining the two expressions, we finally obtain for some $C>0$
		\begin{align*}
			E[T_n^2]&=\frac{C}{(\log(n+1))^2}\left[\sum_{k=1}^n\sum_{l\in A_k }\log\frac{k+1}{k}\log\frac{l+1}{l}\left(\frac{k}{l}\right)^{1/2}+\sum_{k=1}^n\sum_{l\in B_k }\log\frac{k+1}{k}\log\frac{l+1}{l}\left(\frac{k}{l}\right)^{1/2}\right]\\
			&\leq C\left[\frac{\log \log n}{\log n+1} + \frac{1}{(\log n)^{1/2}} \right]\leq C' \frac{\log \log n}{\log n},
		\end{align*}
		which completes the proof.
	\end{proof}

	~

	\textbf{Acknowledgement} We would like to thank Edward De Brouwer for his contributions in the early parts of the research leading up to this paper.

	~

	This research received funding from:
	\begin{itemize}[noitemsep]
		\item Research Council KU Leuven: C14/18/092 SymBioSys3; CELSA-HIDUCTION CELSA/17/032
		\item Flemish Government:
		\begin{itemize}[noitemsep]
			\item IWT: Exaptation, PhD grants
			\item FWO 06260 (Iterative and multi-level methods for Bayesian multirelational factorization with features); Elixir I002819N
			\item This research received funding from the Flemish Government (AI Research Program).
			\item VLAIO PM: Augmanting Therapeutic Effectiveness through Novel Analytics
		\end{itemize}
		\item EU: "MELLODDY" This project has received funding from the Innovative Medicines Initiative 2 Joint Undertaking under grant agreement No 831472. This Joint Undertaking receives support from the European Union's Horizon 2020 research and innovation programme and EFPIA.
	\end{itemize}

redpajama_set_name: RedPajamaArXiv


text:
@interface PodsDummy_Pods_DYSCategories_Example : NSObject
@end
@implementation PodsDummy_Pods_DYSCategories_Example
@end

redpajama_set_name: RedPajamaGithub


The exact moment may have been lost in the midst of time, but the decision or suggestion to get Horace Andy in to collaborate with Massive Attack pretty early on was one of those ideas of total genius, like getting Peter Falk in to play himself in Wim Wenders' Wings of Desire.
Horace Andy has such an eerie, unearthly perfect voice. He is almost a ghost-like presence on the great Massive Attack records, providing echoes of old Studio One tunes, like 'Skylarking', 'Spying Glass', 'Money Money', and 'You Are My Angel', drifting in and out of the rhythmic mix like a benign spectral presence.
One of the absolute joys of my life was finding in the late '80s a copy of Horace Andy's 'Oh Lord Why Lord' on a 7" which I got for 10p in the days when the basements of Record And Tape Exchange were filled with abandoned singles. It was a great time to pick up lost gems like the Disco Dub Band's mutation of 'For The Love Of Money', the Peppers' 'Pepper Box', and the Soul Runners' 'Charley' for next to nothing.
My copy of 'Oh Lord Why Lord' had the label scratched off, and just the credits typed on – literally typed on somehow. And what a beautiful recording it is! What I didn't realise at the time was that it was a Parliament song, from their first LP, with a co-credit by Ruth Copeland, our girl from the North Country, and based on a sacred religious tune oft appropriated in pop.
Partly as a result of the compilations that followed on from his appearances with Massive Attack, now Horace Andy's performance of 'Oh Lord Why Lord' can be heard again. I have it on a Studio One selection called Mr Bassie. The title track reminds us that in its Augustus Pablo version, 'Mr Bassie' was one of the earliest Rough Trade singles, and in the mid-'80s Horace Andy himself would record an LP for the label.
This would have followed on from a series of records Horace Andy made for the legendary New York label Wackies. Alongside classic recordings by the likes of Wayne Jarrett, Lovejoys, Junior Delahaye, Sugar Minott, and complementary adventurous dub excursions, Wackies put out some of the best ever Horace Andy sets, and certainly Dancehall Style and Exclusively in salvaged forms sound superb. And interestingly one of the things that once irritated me about reggae, which is the endless revisiting of old refrains and rhythms, is one of the more appealing aspects of these Wackies sets, again suggesting this idea of Horace Andy as a ghostly voice echoing not just through Massive Attack's catalogue.
The Berlin-based Basic Channel outlet has been carrying out a superb salvage operation of Wackies releases, and all the CDs are immaculately packaged in stark digipacks. You would expect nothing else from the Basic Channel organisation. The label may have been avowedly committed to preserving vinyl, but its presentation of CD releases has been brilliantly unique. It's something the Ghost Box operation understands too.
The CD compilation of early Basic Channel minimalist techno infestation came defiantly clad in a plain cardboard cover, with a dearth of information. As the label evolved into Chain Reaction, the CD releases initially came uniformly in a metal tin, plainly projecting the label name, barely mentioning the artists' name, leaving the often amazing electronic music inside to speak for itself. And these tins allegedly destroyed the music inside, but they looked great.
These early Chain Reaction CDs included ones by Vainqueur and Various Artists. The latter's Decay Product set made explicit connections to the dub traditions at the heart of the Basic Channel ethos. In a famous Kodwo Eshun article for The Wire, Various Artists' Thorsten Profrock says: "Dub was a kind of resonance to my soul because everything I like within techno music, all these effects, these dub people had it. You see a line between dub and Basic Channel". The true realisation of this approach would come on the almost 20-minute suite that is his 'No. 8' – a uniquely absorbing road movie of a track – and its successor 'No. 9' which were recorded for the London-based Fat Cat label.
Another, slightly later, classic Chain Reaction set was Mutila by Vladislav Delay, a Finnish innovator and jazz percussionist who no doubt reluctantly became a leading light in the clicks and cuts, microhouse or glitch, electronic scene. The fascinating thing about this music was its seemingly self-contained life completely outside the mainstream, but occasionally you would hear an unexpected r'n'b or dancehall track and find strange connections that seem to have come from nowhere.
Basic Channel's guiding spirits, the purposefully anonymous and therefore exceedingly enigmatic Mark Ernestus and Moritz Von Oswald (Maurizio), would retreat further inside the dub organism as Rhythm & Sound, explicitly connecting the techno twitches and throbs with the dub decays and delays, the exaggerations and extractions. It is no coincidence surely that the outlet for this, Burial Mix, would be a massive influence on some of the participants in the bass continuum for years to come. Just as the space and echo of dub transplanted to an industrial Berlin would conjure up images of abandoned warehouses and desolate factory spaces, rather like the music of Joy Division and Martin Hannett captured a decaying Manchester now disappeared, so new variants like dubstep captured the conflict between inner city disease and desire.
Just as Basic Channel/Chain Reaction helped invent the future, so Maurizio and others still active in the German electronic underground have links back to the new wave of sounds that punk inspired. Maurizio, for example, played in the legendary Palais Schaumberg with Thomas Fehlmann, albeit after the immortal David Cunningham-produced debut and after the departure of Holger Hiller. Hiller himself would be also be a recurring figure in odd places where electronic adventuresomeness was taking place, and occasional solo sets are highly recommended (the presence of Billy MacKenzie on 'Oben Im Eck' is surely enough to encourage investigation?), and the Hiller name occurs in the credits for the likes of Renegade Soundwave and Chicks On Speed.
Almost running parallel to Palais Schaumberg were Liaisons Dangereuses and Malaria! And from these outfits Beate Bartel and Gudrun Gut would continue to make the case for electronic exploration long after they were in a very early incarnation of Einsturzende Neubauten, and after they worked together as Matador where they very deliberately set out to empower women to experiment with electronica. Gudrun still runs the excellent Monika label, home to some of the best pop music around, with Barbara Morgenstern and Cobra Killer among others.
And Horace Andy is still somewhere singing: "You live in the city. You mind your own business. What you see you don't see. But some people they always see. They never mind their own business".


The Play series… entertaining instructional programs illustrating the methods and techniques made famously recognizable by the world's greatest guitarists, this release concentrates on the awesome Stevie Ray Vaughan. 'Master of the Stratocaster' Stephen "Stevie" Ray Vaughan was the leading musician in the blues rock sound. Before his tragic early death at 35, he was associated with numerous artists such as Johnny Copeland, The Fabulous Thunderbirds, and David Bowie. He won multiple Grammys and W.C. Handy awards and was posthumously inducted into the Blues Hall of Fame.


Over the past year, more and more companies have been turning to "HR as PR": Making changes to compensation, benefits, or employer-employee relations not only to gain an edge in the competition for talent, but also to win the hearts and minds of consumers. The goal of this strategy is to say: "We care about our employees, so you should like us as a company and buy more from us." We have seen this strategy play out in terms of companies raising their minimum wage, expanding parental leave or other benefits, and aggressively pursuing diversity and inclusion initiatives. In the US, we've also seen a dramatic increase in companies' willingness to engage in the political debate, particularly since the start of the Trump administration.
What is happening right now in tech is different: Rather than advocating for internal policies, employees are putting pressure on their companies to become vocal opponents of the Trump administration—by having CEOs make public statements, by turning down certain government contracts, by signing on to legal briefs contesting Trump's policies. Of the 127 companies that signed onto the amicus brief filed in support of Washington state's legal challenge to the immigration executive order, the majority are tech companies.
Other employee petitions call for Oracle to oppose US president Donald Trump's second travel ban, and to let men who work at US regional supermarket Publix grow beards. Employee petitions are now so popular there's a website, coworker.org, devoted to hosting them. In some cases, the campaigns work: Starbuck's relaxed its rules about visible tattoos and unnatural hair color for baristas after thousands signed petitions asking for a change. Sometimes, they fail disastrously. Interns at one (unnamed) company described in a blog about being fired en masse after signing a petition asking for a more relaxed dress code.
Given this shift, companies that come under pressure from employees to take public stances on politically charged issues need to make two choices: what they will say and what they will do. It may seem like the easiest win in this situation is to have the CEO speak out and take a public position, but not make significant changes to how the company actually operates, but this approach can backfire by raising employees' expectations and then failing to deliver. That's the worst of all possible outcomes.
Rather, the choice companies face is much tougher: You can say and do nothing, which will slowly erode any competitive advantage you have in the labor market and for customers, or you can determine what is critical for your employees and customers, and both speak and act to support it.


If you like our Deli's fresh made-in-store pizza, you're already familiar with this great crust. It has a nicely chewy texture from whole wheat flour and a pleasing mild flavor. Easy to work with, it is a great choice for beginners to homemade pizza.


Prettify your household by employing a suggestions associated with Amazing Contemporary Flower Pots Creative Flowers Pendant Lighting Balcony Plant Flower Pot Aisle Stair Living Room Pendant Light YL picture. There is completely advantageous determination which you can take up out of Amazing Contemporary Flower Pots Creative Flowers Pendant Lighting Balcony Plant Flower Pot Aisle Stair Living Room Pendant Light YL image. Certain key elements to help you upgrade a residence can be had simply out of Amazing Contemporary Flower Pots Creative Flowers Pendant Lighting Balcony Plant Flower Pot Aisle Stair Living Room Pendant Light YL graphic. You can improve your opinions as a result of mastering Amazing Contemporary Flower Pots Creative Flowers Pendant Lighting Balcony Plant Flower Pot Aisle Stair Living Room Pendant Light YL picture carefully. We highly recommend that you investigate this Amazing Contemporary Flower Pots Creative Flowers Pendant Lighting Balcony Plant Flower Pot Aisle Stair Living Room Pendant Light YL picture as you require a lot of tricks to remodel your household. By having a residence prefer Amazing Contemporary Flower Pots Creative Flowers Pendant Lighting Balcony Plant Flower Pot Aisle Stair Living Room Pendant Light YL pic displays, surely, you will be able to rest really gently at your residence.
Amazing Contemporary Flower Pots Creative Flowers Pendant Lighting Balcony Plant Flower Pot Aisle Stair Living Room Pendant Light YL photo can make your home exudes a warmth designed to get your guest visitors feel at ease. If you would like to have more options, you can actually look into Contemporary Flower Pots pic gallery which provides quite a few wonderful pattern creative ideas. By using high quality photos included, Contemporary Flower Pots pic stock are extremely correct being used for the reason that your private reference. Of this Amazing Contemporary Flower Pots Creative Flowers Pendant Lighting Balcony Plant Flower Pot Aisle Stair Living Room Pendant Light YL photo, this actual resolution is 763 x 635 although the actual size is exactly 56 KB. If you need to collect this Amazing Contemporary Flower Pots Creative Flowers Pendant Lighting Balcony Plant Flower Pot Aisle Stair Living Room Pendant Light YL picture, it is possible to push here. And if you would like upload this approach Amazing Contemporary Flower Pots Creative Flowers Pendant Lighting Balcony Plant Flower Pot Aisle Stair Living Room Pendant Light YL graphic for the web site, remember to add some original source link. Because of Amazing Contemporary Flower Pots Creative Flowers Pendant Lighting Balcony Plant Flower Pot Aisle Stair Living Room Pendant Light YL photograph illustrates a wonderful style and design, a lot of people like it so much. You can understand coming from 1 viewers who have got witout a doubt experienced Amazing Contemporary Flower Pots Creative Flowers Pendant Lighting Balcony Plant Flower Pot Aisle Stair Living Room Pendant Light YL picture ever since the website admin published the application at September 17, 2017 at 1:00 am. Hopefully you will be kept entertained simply by design that will Amazing Contemporary Flower Pots Creative Flowers Pendant Lighting Balcony Plant Flower Pot Aisle Stair Living Room Pendant Light YL image indicates.


Dun Fell Hush © Copyright Peter McDermott and licensed for reuse under this Creative Commons Licence.
From OED transitive verb "to hush" originally from noun "hush": the sound of swift but smooth flowing water.
To send or let forth (water) with a rush; spec. in Lead mining, to send a rush of water over a sloping surface, in order to uncover ore, and separate it from the earth and stones in which it is embedded, or for similar purposes.
The sites of hushing have become known as "hushes".


text:
Q: Use the sheet number to check the active sheet Before some code runs, I want to check it is the correct sheet.
I can use the sheet name, but my concern is that if someone changes the sheet name, the code won't run. The sheet index also seems to change if the sheet is moved.
Therefore I want to use something that doesn't change.
I believe the sheet number and sheet ID never change.
So I was hoping to use one of them, but I can't see a way of doing that.
What I want in non-coding language is:
If active sheet number = 4 then run the code or
If active sheet ID = 0123456789 then run the code.

Thanks to JPV's link, I have an answer and learned a few things.

*

*You can't use the sheet number to check if the correct sheet is active.

*You can use the getSheetId(), but the return you get is not useable.

*To make it useable, you need to add getSheetId().toString().

*To do the if statement, I needed two equal signs.

if (SpreadsheetApp.getActiveSheet().getSheetId().toString() == 0123456789) {do this when true}

*I've been trying to solve this for days, so I should have come here and asked for help earlier.

The main thing I didn't know was the "toString()" part. It always seems so easy and obvious once I know.
Thank you, JPV.

A: Thanks to JPV's link, I have an answer and learned a few things.

*

*You can't use the sheet number to check if the correct sheet is active.

*You can use the getSheetId(), but the return you get is not useable.

*To make it useable, you need to add getSheetId().toString().

*To do the if statement, I needed two equal signs.

if (SpreadsheetApp.getActiveSheet().getSheetId().toString() == 0123456789) {do this when true}

*I've been trying to solve this for days, so I should have come here and asked for help earlier.

The main thing I didn't know was the "toString()" part. It always seems so easy and obvious once I know.
Thank you, JPV.

redpajama_set_name: RedPajamaStackExchange


text:
Neobaryssinus phalarus is a species of beetle in the family Cerambycidae. It was described by Monné and Martins in 1976.

References

Acanthocinini
Beetles described in 1976

redpajama_set_name: RedPajamaWikipedia


jersey ace china hot on sale now.
Buy cheap jersey ace china online!
Discount shop jersey ace china.
jersey ace china 2018 outlet store.
jersey ace china,Wholesale Cheap Jerseys From China, More Buy More Save.


Mahatma Gandhi - "You may never know what results come of your actions, but if you do nothing, there will be no results."
The EKansh team was happy to see that just the interaction in a fair like this can alter the mindsets and expectations of potential employers as many of them relaxed their criteria and broadened the scope for candidates over the two days.It also raised confidence and awareness levels of PwD and their families. We did not compromise our view that a Fair should be just that -fair and open to all. We did not limit entries based on any criterion. We learnt yet again that NGOs and organizations working for PwD must scale up the training courses. PwD must begin to aspire and work for better qualifications and opportunities. While there are hiccups and upsets, they are negligibile compared to the long term victories that far outnumber them.
2. Ansec HR Services Pvt. Ltd.
A big THANKS to friends and families of the EKansh Team Members!


text:
India slowly slipping into the dark; lifestyle the main cause says experts
IG Fresh Produce to invest Rs 100 crore in Arunachal Pradesh for Kiwi cultivation
Livelihood of millions of women will be hit by proposed amendments to tobacco products act
India-China soldiers engaged in brief face-off in Tawang sector, Arunachal Pradesh
Innovation and investment facility launched in Arunachal to boost startup ecosystem
Subhayan Dutta
India is slowly but quite steadily moving to a future where it will lose itself to the throes of the soon-to-be epidemic named cancer. According to a data released by national cancer registry, India will see an estimated 1.45 million new cancer cases in 2016 alone, killing 736,000 of the patients as barely 12.5% of those suffering from the condition get diagnosed early.
The report estimates that one in eight men and one in nine women in India will be developing some form of cancer in their lifetime. It comes at a time when medical costs are rising sharply.
There has been a steep rise of 350,000 first-time cancer cases since 2011 when the records were last updated. The crisis only seems to worsen. By 2020, it is expected to go up to 1.73 million cases in a year, making it top the list of non-communicable illnesses plaguing India in the coming years.
The two most common cancers found in men are mouth and lung. Women mostly suffer from breast and cervix cancers. The highest incidence among men has been reported from Mizoram's Aizawl district while Papumpare district of Arunachal Pradesh has the highest rate of cancer among women.
Experts, thankfully, say that 60% of the cancers are preventable. 40% of this is due to tobacco use and remaining 20% of infections. They advise getting vaccinated and maintaining a healthy lifestyle.
According to the National Cancer Institute (NCI) which is part of the US Department of Health and Human Services (USDHHS), every 13th new cancer patient in the world is an Indian and breast, cervical and oral cancers top the list in the country.
The scenario for women is also quite gloomy.
Ranga Rao Rangaraju, director (oncology services) at Max Super Specialty Hospital told IANS, "Out of 12.5 lakh (1.25 million) new cancer patients each year in India, over seven lakh are women."
"Nearly 3.5 lakh women face deaths owing to cancer each year and the figure is likely to go up to 4.5 lakh by 2025," Dr Rangaraju added.
The emergence of modern lifestyle hugely impacts this cause. Imbalance in the circadian clock due to sleep loss, lack of exercise, wrong eating habits, work-related stress, smoking and alcohol abuse are some of the key reasons behind high mortality rates among women with cancer.
Women would mainly suffer for their increased focus on professional life, ending up in late marriages and childbearing. Some of them opt for hormone replacement therapies which further increases the risk of cancer. A recent study also reveals that India has seen a 7.5 percent rise in the number of cancer patients primarily due to poor lifestyle choices.
Related Topics:000 first-time cancer, 1.45 million new cancer cases in 2016, 350, Aizwal, Arunachal Pradesh, hormone replacement therapies, infections, Lifestyle, lung and breast cancer, Mizoram, national cancer registry, tobacco, US Department of Health and Human Services
Infographic: Rise of Social media
Indian passenger cars fail crash test terribly, raising concerns

redpajama_set_name: RedPajamaCommonCrawl


text:
import React from 'react'
import Box from "@mui/material/Box";
import Title from "../components/Title";
import Grid from "@mui/material/Grid";
import FormControl from "@mui/material/FormControl";
import FormLabel from "@mui/material/FormLabel";
import RadioGroup from "@mui/material/RadioGroup";
import FormControlLabel from "@mui/material/FormControlLabel";
import Radio from "@mui/material/Radio";
import TextField from "../components/TextField";
import {FontAwesomeIcon} from "@fortawesome/react-fontawesome";
import BrowseOpenDialog from "../components/BrowseOpenDialog";
import {get, isNil, set} from "lodash";
import MediaSelectField from "../components/MediaSelectField";
import Stack from "@mui/material/Stack";
import RedirectButton from "../components/RedirectButton";
import Button from "../components/Button";
import ConfirmDialog from "../components/ConfirmDialog";
import {Api} from "../utils/Api";

const initialState = {
    confirmOpen: false,
    sourceMedia: null,
    sourcePath: null,
    sourceType: 'ImageFile'
}

export default function Verify() {
    const [state, setState] = React.useState({ ...initialState })
    const [destinationPath, setDestinationPath] = React.useState(null)

    const api = new Api()

    const {
        confirmOpen,
        sourceMedia,
        sourcePath,
        sourceType
    } = state

    const verifyDisabled = isNil(sourcePath) || isNil(destinationPath)

    const handleChange = ({name, value}) => {
        if (name === 'sourceType') {
            state.sourcePath = null
            state.sourceMedia = null
        }
        set(state, name, value)
        setState({...state})
    }

    const handleMediaChange = (media) => {
        state.sourceMedia = media
        state.sourcePath = media.path
        setState({...state})
    }

    const handleVerify = async () => {
        const response = await fetch('api/verify', {
            method: 'POST',
            headers: {
                'Accept': 'application/json',
                'Content-Type': 'application/json',
            },
            body: JSON.stringify({
                title: `Verify ${(sourceType === 'ImageFile' ? 'image file' : 'disk')} '${isNil(sourceMedia) ? sourcePath : sourceMedia.model}' and image file '${destinationPath}'`,
                sourceType,
                sourcePath,
                destinationPath
            })
        });
        if (!response.ok) {
            console.error('Failed to verify')
        }
    }

    const handleConfirm = async (confirmed) => {
        setState({
            ...state,
            confirmOpen: false
        })
        if (!confirmed) {
            return
        }
        await handleVerify()
    }

    const handleCancel = () => {
        setState({ ...initialState })
    }

    const handleUpdate = async () => {
        await api.list()
    }

    return (
        <Box>
            <ConfirmDialog
                id="confirm-verify"
                open={confirmOpen}
                title="Verify"
                description={`Do you want to verify '${isNil(sourceMedia) ? sourcePath : sourceMedia.model}' and '${destinationPath}'?`}
                onClose={async (confirmed) => await handleConfirm(confirmed)}
            />
            <Title
                text="Verify"
                description="Verify image files or physical disk and image file comparing them byte by byte."
            />
            <Grid container spacing="2" direction="row" alignItems="center" sx={{mt: 2}}>
                <Grid item xs={12} lg={6}>
                    <FormControl>
                        <FormLabel id="source-type-label">Source</FormLabel>
                        <RadioGroup
                            row
                            aria-labelledby="source-type-label"
                            name="source-type"
                            value={sourceType || ''}
                            onChange={(event) => handleChange({
                                name: 'sourceType',
                                value: event.target.value
                            })}
                        >
                            <FormControlLabel value="ImageFile" control={<Radio />} label="Image file" />
                            <FormControlLabel value="PhysicalDisk" control={<Radio />} label="Physical disk" />
                        </RadioGroup>
                    </FormControl>
                </Grid>
            </Grid>
            <Grid container spacing="2" direction="row" alignItems="center" sx={{mt: 2}}>
                <Grid item xs={12} lg={6}>
                    {sourceType === 'ImageFile' && (
                        <TextField
                            id="source-image-path"
                            label={

                                    <FontAwesomeIcon icon="file" style={{marginRight: '5px'}} /> Source image file

                            }
                            value={sourcePath || ''}
                            endAdornment={
                                <BrowseOpenDialog
                                    id="browse-source-image-path"
                                    title="Select source image file"
                                    onChange={(path) => handleChange({
                                        name: 'sourcePath',
                                        value: path
                                    })}
                                />
                            }
                            onChange={(event) => handleChange({
                                name: 'sourcePath',
                                value: get(event, 'target.value'
                                )
                            })}
                        />
                    )}
                    {sourceType === 'PhysicalDisk' && (
                        <MediaSelectField
                            label={

                                    <FontAwesomeIcon icon="hdd" style={{marginRight: '5px'}} /> Source physical disk

                            }
                            id="source-media-path"
                            path={sourcePath || ''}
                            onChange={(media) => handleMediaChange(media)}
                        />
                    )}
                </Grid>
            </Grid>
            <Grid container spacing="2" direction="row" alignItems="center" sx={{mt: 2}}>
                <Grid item xs={12} lg={6}>
                    <TextField
                        id="destination-path"
                        label={

                                <FontAwesomeIcon icon="file" style={{marginRight: '5px'}} /> Destination image file

                        }
                        value={destinationPath || ''}
                        endAdornment={
                            <BrowseOpenDialog
                                id="browse-destination-path"
                                title="Select destination image file"
                                onChange={(path) => setDestinationPath(path)}
                            />
                        }
                        onChange={(event) => setDestinationPath(get(event, 'target.value'))}
                    />
                </Grid>
            </Grid>
            <Grid container spacing="2" direction="row" alignItems="center" sx={{mt: 2}}>
                <Grid item xs={12} lg={6}>
                    <Box display="flex" justifyContent="flex-end">
                        <Stack direction="row" spacing={2} sx={{mt: 2}}>
                            <RedirectButton
                                path="/"
                                icon="ban"
                                onClick={async () => handleCancel()}
                            >
                                Cancel
                            </RedirectButton>
                            <Button
                                icon="sync-alt"
                                onClick={async () => handleUpdate()}
                            >
                                Update
                            </Button>
                            <Button
                                disabled={verifyDisabled}
                                icon="check"
                                onClick={async () => handleChange({
                                    name: 'confirmOpen',
                                    value: true
                                })}
                            >
                                Start verify
                            </Button>
                        </Stack>
                    </Box>
                </Grid>
            </Grid>
        </Box>
    )
}


text:
Fashion & BeautyLifestyleThe Latest
On the Scene: Ksubi x Hidji Debuts Second Capsule Collection at NYFW
by Bianca Brutus September 9, 2021
by Bianca Brutus September 9, 2021 0 comment
The second collection comes two years after "Seven Deadly Sins."
On September 7, Ksubi x Hidji unveiled their latest collaboration at the old Opening Ceremony store in Soho. The installation comes two years after their last collaboration. The second collaborative collection, "Family and Friends." consists of 27 pieces of men's and women's clothing with the trademark Ksubi print. The inspiration for the collection was "the school yeard and childhood innocence."
Founded in the '90s, Ksubi is a premier Australian brand known for its colorful streetwear and distinct prints. With stores across North America, Europe, and Australia, the brand has grown tremendously over the last few years to be a staple name in the design realm.
The pop-up is additionally in collaboration with Travis Rogers of DNTWATCHTV who designed and produced an art installation for the space. The pop-up also showcases the talent of 10 international emerging artists. The Ksubi x Hidji limited-edition collection will remain on display at the pop-up until September 11.
HidjiKsubiNYFW
Bianca Brutus
Cooking at Home with Peacock's 'Top Chef Family Style'
Exclusive: Samuel Mancini Talks about New Song 'Infatuation' [Interview]
80purppp Performs Latest Single "Uni" in New York...
On The Scene: Drake Live At The Apollo...
MISTER PARADISE ANNOUNCES EXTENSIVE NEW MENU WITH SOPHISTICATED,...
Actor Mo McRae To Release Directorial Debut 'A...
Exclusive: HEESU Talks Creating Dance Music That Honors...
'The Last of Us' Season 1, Episode 2...


One thing I've always believed is that we are never done learning. Even the CEO of the biggest company in the world doesn't know it all. No one does. Still, it today's chaotic world, it's hard to find the time to read a book or take a course that could help us grow, especially when you're a business owner. Podcasts offer the perfect answer to this problem.
Being Boss is a podcast for creative entrepreneurs, side hustlers, and freelancers, with hosts Emily Thompson and Kathleen Shannon. Being Boss covers an array of topics from how to manage your finances as a business owner to health and wellness. With a fun, friendly and honest tone, the podcast is easy to relate to, no matter what industry you are in.
Test out a few podcasts to see which ones resonate with you most and incorporate them into your day.
Kristen Harold's global marketing career has spanned across technology, music and media. In approaching any aspect of marketing, Kristen believes in strategically applying data insights without forgetting the people behind it to tell the most effective story. Currently the Owner of KMH Marketing, Kristen focuses on building successful marketing strategies, audience growth and engagement for small businesses. Outside of work, Kristen enjoys painting, traveling, exploring New York and spending time with her dog Katie.


Medication / Prescription Errors | Brooks, LeBoeuf, Bennett, Foster & Gwartney, P.A.
Medication is a common part of many elderly people's lives and in nursing home facilities, routine prescription regimens are critical to maintaining resident well-being. However, when these regimens are not treated with the vigilance and respect they deserve, nursing home residents can be harmed due to medication or prescription errors. When negligence contributes to these errors, families may seek financial compensation for these inexcusable mistakes.
If your elderly or disabled loved one has been harmed by a prescription or medication error, our team at Brooks, LeBoeuf, Bennett, Foster & Gwartney, P.A. is ready to assist your family. Our compassionate, proven Tallahassee nursing home neglect attorneys are industry recognized for the quality of their work and know what it takes to hold these facilities and their administrators accountable for their wrongdoing.
There's Too Much at Stake. Call us at (850) 222-2000 today to speak to our award-winning counselors.
Implicating a beneficial medication regimen depends on precision and timeliness. Many nursing home facilities have efficient, safe systems for distributing and administering medication, but it only takes one error to put a resident's health at risk.
If you believe that one of the above factors played a role in your loved one's medication error, it may be time to seek legal counsel. At Brooks, LeBoeuf, Bennett, Foster & Gwartney, P.A., we're ready to examine your loved one's medical record, assess all relevant evidence and testimony, and develop a thorough and compelling claim for relief on your behalf.
Want to learn more about what our firm can do for your family at this critical time? Contact us today to request a free consultation.


Away from the harbour itself there is much to recommend Howth. Historians will love the ruined abbey, nearby Baily Lighthouse and 15th Century castle. You can take a bracing stroll along the piers, sightsee aboard an open top tram, watch seals and dolphins in the waters along the shore and take in breathtaking views from cliff top walks. Of course Howth's working fishing port means that fish and seafood lovers are absolutely spoilt when it comes to dining out and the pub scene is second only to Dublin itself, if a little more relaxing.
The Harbourguides Crew would like to say a special thanks to Aidan and Joan Macmanus from the excellent King Sitric for providing us with photos of beautiful Howth.
Office situated Northern End of Auction Hall.
You could try calling Howth YC on +353 1 832 2141 or email them on reception@hyc.ie Hope you have some success.
Would there be a lost and found for the harbour ladies restroom? Long shot I know but I left something in there and would love to retrieve it.
I believe the German yacht Ruby Tuesday is arriving soon in Howth Marina.
WE met that boat last summer in Iceland.
Would you please be so kind as to tell them that the crew of the French sailing boat Brecqhou which was alongside them in Bolungarvik would be glad to get in contact with them?
They can contact us at brecqhou@gmail.com or phone 00 33 6 81 06 18 38.
Hi myself and my fiancee are getting married on September 1st in howth and are looking to get the gate open to the pier so we can drive down and get some pictures 30mins at most, just wondering how we'd best go about this or is it at all possible?
Hello Harbour Masters and lovers of the seaside!
I am part of a Sustainability campaign in collaboration with other designers and scientists. Our website and campaign is 'Seastainability' and together we research, investigate, campaign and promote sustainability.
We regularly arrange beach cleans and individually do our best to tackle environmental and pollution issues, however serious intervention is needed to resolve the massive amount of plastic debris and broken fishing nets which is nesting into our harbours and beaches.
Much of this debris is stuck in the harbour corners or settled on boulder rock and it is not safe or accessible for members of the public to clear it. Professional clearing is needed regularly, we need more bins in these rubbish spots and regulations need to be put in place to stop broken nets being discarded into the sea.
The negative effects are damaging wildlife, our food supply and tourism!
King Citric best fish i have had in a long tíme. 10 out of 10.
Had a trip out to Marine Parts Direct in Swords, fantastic place.
Deer park still keeping up the štandard of the good old Traditional hotel.
When I was employed by Rock Fall Co I was in charge for a period in deepening the Howth harbour by drilling and blasting probably in late 1960's or early 1970's. Could you please let me know the exact year of that operation?
I am writing my memoir and this information will be valuable. I shall appreciate your assistance in this regard.
Whilst entering Howth marina, in our Sadler 26, 4' 10" draught, at low tide in June this year (2011) we ran aground when only just off the main pontoons. When I say ran aground, I mean hit a submerged rock and then had to be towed off by the club launch. The following day another vessel was left high and dry for the duration of the tide in roughly the same spot.


text:
articles | 03 July 2013
Cyprus President Anastasiades meets ECB President today
The practical difficulties the banking sector is facing regarding the implementation of a loan agreement Cyprus has signed with its international lenders will be the main issue of discussion at today's meeting, in Frankfurt, between the President of the Republic of Cyprus Nicos Anastasiades and the President of the European Central Bank (ECB) Mario Draghi.
President Anastasiades will be accompanied at the meeting by Finance Minister Harris Georgiades, and the Governor of the Cyprus Central Bank Panicos Demetriades. In addition, during the meeting the two sides will also discuss the future of Bank of Cyprus.
Speaking last week in Brussels, after the European Council's conclusion, the President of Cyprus said that the meeting would be in search of solutions. Moreover, he said that talks will not center on changing the terms of the bailout agreement, but rather on the ECB's contribution to problems that have emerged, after the European Council's decision.
Last March Cyprus agreed with international lenders on 10 billon euro aid package, under which it closed down the Popular Bank, and imposed painful losses on deposits of more than 100,000 euros held at the euro zone state`s biggest lender, Bank of Cyprus. Both institutions were heavily exposed to the Greek debt.


text:
Put More Strings in Your Writing Bow
Posted on May 4, 2014 by Joe Moore
@jamesscottbell
Today's post is brought to you by a new string in my writing bow: the latest story in the saga of 1950s Los Angeles boxer, Irish Jimmy Gallagher. And it is FREEtoday for Kindle.
So what's the meaning of this bow business?
I'm a fan of the Parker novels by Donald Westlake (writing as Richard Stark). I've seen all the film versions, like Point Blank with Lee Marvin, The Outfit with Robert Duvall, and Paybackwith Mel Gibson.
Payback, a 1999 release, is particularly good. But I recently became aware that the director, Brian Helgeland, had the film taken away from him. His version did not test well, so a new third act was written under the eye of Gibson, who was one of the producers.
A few years ago, Helgeland was given permission to release his director's cut. I recently watched it. It is darker and perhaps truer to the feel of the novels. I do think, however, Mel and Paramount were correct. The 1999 version is more satisfying.
But I digress. The director's cut DVD has an interview with the late, great Westlake on the genesis of Richard Stark and the Parker novels.
Westlake was putting out one hardcover book a year under his own name. Wanting to make a living as a writer, he decided he needed "another string in his bow." He decided to try the paperback original market, which was mostly for a male audience in those days.
He wanted the books to be lean and dark. "Without adverbs," he said. "Stark."
That's how he came up with the last name for his pseudonym.
He chose Richard because he liked the iconic noir actor Richard Widmark.
That's how Richard Stark was born.
Then he needed a name for his character. He chose Parker. With a wry smile he said he wished he'd chosen another name, because he spent so much time trying to come up with other ways to say, "Parker parked the car."
In any event, his agent showed the first book, The Hunter, to Gold Medal, the leading PBO publisher of the day. Rejected. So they tried Pocket Books. An editor with the wonderful name of Bucklin Moonliked it.
The original manuscript ended with Parker in jail. He did not, in other words, get away with it (it being the killing of some bad guys in order to get money owed him from a heist). Moon asked Westlake if he would consider changing the ending and making it a series, and could he turn out three books a year?
Westlake jumped at the chance.
What happened over the next several years is that Richard Stark started selling better than Donald Westlake, which irked Westlake the author . . .but pleased Westlake the guy who wanted to make a living.
And so Parker became one of the great characters of hard noir.
When self-publishing took off in 2008, I said it felt like the mass market boom of the 50s, where many literary authors made extra money. Like Evan Hunter writing as Ed McBain. Or Gore Vidal writing as Edgar Box.
And that gives us a lesson: We can, like Westlake, have more strings in our bows. Self-publishing offers that opportunity. But unlike Westlake and writers of that era, we don't need to use a pseudonym. Indie publishing distinguishes brands by way of cover design, book description and categorization. Writers can therefore gain fans for material unlike other things they're doing. Some cross-pollination of fans is not only possible, but probable. Reader have found me by way of my vigilante nun series and gone on to sample my historicals. Imagine that.
[NOTE: When I did my zombie legal thrillers, I was a traditional-only author, so I chose a pseudonym, K. Bennett. I've grown to like the sullen, mysterious K. He may write some more.]
We can freely write in multiple forms and genres, short and long, and the tide will lift all the boats. Back when I started getting paid for writing, there was only one stream available for the professional scribe. Now there are three: traditional, indie and a river made up of both.
Which is good news for writers of every stripe, especially those who want to stretch and grow and make some actual money, too.
So how many strings are in your bow?
This entry was posted in Donald Westlake, Richard Stark, self-publishing by Joe Moore. Bookmark the permalink.
About Joe Moore
#1 Amazon and international bestselling author. Co-president emeritus, International Thriller Writers.
View all posts by Joe Moore →
7 thoughts on "Put More Strings in Your Writing Bow"
Steve Hooley on May 4, 2014 at 12:50 pm said:
I enjoyed SUCKER PUNCHED tremendously. I read it the first night I learned of it. The price was good, too. Thanks. I was pulled in from the first flurry of fists, and didn't come up for air until the surprise and resonant ending. I hope there will be more.
I'm looking forward to seeing the comments today on your "string theory." I've been exploring the use of short story as another string, particularly as a vehicle to bridge from one novel to the next – use of back story with a secondary character as the protagonist, publicity for the next novel, dual use of research, etc.
I hope some of your readers will comment on use of the short story from the perspective of the novelist.
James Scott Bell on May 4, 2014 at 2:13 pm said:
Thanks for the kind words about the story, Steve. As for the use of short stories, you probably know that many "A List" authors are using short form digitals to fill in "gaps" or do backstory, etc. A good move UNLESS it appears to be nothing but a marketing ploy for an upcoming novel. Then readers will pummel the author. Every offering needs to stand on its own vie-a-vis quality.
So, yes, explore. That's the nice thing about having digital. We can try things.
Elizabeth Poole on May 4, 2014 at 1:53 pm said:
At the moment, none. I'm unpublished on all fronts.
But that's about to change. This year I'm planning on self publishing a series of novellas. It's going to be in episodic format, much like how TV is structured, with one long arc to be completed at the end of the season.
I am very excited this short form is now viable. I've always had ideas that fit into series, but I've often felt concerned that I would set it all up only to get the book rejected or the series cancelled.
Shorter forms are good for improving your craft, too. It's not the same thing to write a novel, but it's less time to write a novella, edit it profusely, sent it to an editor, and edit it some more. I think it has the possibility on boosting your writing skills exponentially.
I've been reading a ton of wonderful books on self publishing, and I'm very excited to be able to be in control of my writing career like this.
I plan to continue work on my novel length fiction and submit it to agents, but in the mean time I'll have another string in my bow, already out there and building my audience.
Thanks for the wonderful post!
And thanks again for Self Publishing Attack! It's been one of my go-to books on the subject.
Elizabeth, this serial idea is gaining popularity in digital. I do think the challenge will be the wrap up, just as it has been for TV shows whose last episodes have been less than satisfying.
Glad to hear you're pursuing traditional full-length fiction but, at the same time, looking at self-publishing shorter work. So many possibilities for writers now.
Yes, I think wrapping everything up in a satisfying manner is one of the hardest parts. Also planning enough out that I have a clear direction, but not so much that I'm trapped in a storyline.
At the moment I'm using Write from the Middle, as well as the doorway method you talk about in Plot and Structure to give me some plot points to hit but keep things open in the middle.
I've always loved the idea of serial fiction, so I'm glad I can take advantage of it. 😀
tom combs on May 4, 2014 at 5:54 pm said:
Thank for "Sucker Punched" – looking forward to reading it soon.
I am indie releasing my debut medical thriller "Nerve Damage" within the month. It has been professionally edited (Jodie Renner) and is the product of years of writing and study (including more than 30 courses at the Loft Literary Center, your books and "Story Masters" conference held in Mpls).
I sent one query to an agent on an author's recommendation but have otherwise planned on indie release from the start.
I'm unsure whether this is my first string or perhaps the launch of my first arrow? I hope it finds appreciative readers!
I've followed TKZ for several years. Thank you for providing guidance and writing wisdom. I much appreciate it!
Welcome to the stream, Tom. If a book has the goods it will find readers, and vice versa. It may be a trickle at first, but keep feeding it with new, quality material. That's really the only "secret."
Have a little fun, too. My Gallagher stories are fun for me to write, so I hope they're fun for the readers as well.


text:
angular.module('BlocksApp').controller('DAOController', function($stateParams, $rootScope, $scope, $http, $location) {
    $scope.$on('$viewContentLoaded', function() {
        // initialize core components
        App.initAjax();
    });
    var activeTab = $location.url().split('#');
    if (activeTab.length > 1)
      $scope.activeTab = activeTab[1];

    $scope.dao = {"balance": 0, "extra_balance": 0};

    //fetch dao stuff
    $http({
      method: 'POST',
      url: '/daorelay',
      data: {"action": "info"}
    }).success(function(data) {
      $scope.dao = data;
    });

    $scope.form = {};
    $scope.errors = {};
    $scope.showTokens = false;
    $scope.getBalance = function(a) {
        var addr = a.toLowerCase();

        $scope.form.addrInput="";
        $scope.errors = {};

        $scope.form.tokens.$setPristine();
        $scope.form.tokens.$setUntouched();
        if (isAddress(addr)) {
          $http({
            method: 'POST',
            url: '/daorelay',
            data: {"action": "balanceOf", "addr": addr}
          }).success(function(data) {
            console.log(data)
            $scope.showTokens = true;
            $scope.dao.tokens = data.tokens;
          });
        } else
            $scope.errors.address = "Invalid Address";

    }

})
.directive('createdTokens', function($http) {
  return {
    restrict: 'E',
    templateUrl: '/views/created-tokens.html',
    scope: false,
    link: function(scope, elem, attrs){
      // fetch created tokens
      scope.getCreatedTokens = function(last) {
        var data = {"action": "createdTokens"};
        if (last)
          data.last_id = last;

        $http({
          method: 'POST',
          url: '/daorelay',
          data: data
        }).success(function(data) {
          scope.created_tokens = data;
        });
      }

      scope.getCreatedTokens();
    }

  }
})
.directive('transferTokens', function($http) {
  return {
    restrict: 'E',
    templateUrl: '/views/transfer-tokens.html',
    scope: false,
    link: function(scope, elem, attrs){
      // fetch created tokens
      scope.getTransferTokens = function(last) {
        var data = {"action": "transferTokens"};
        if (last)
          data.last_id = last;

        $http({
          method: 'POST',
          url: '/daorelay',
          data: data
        }).success(function(data) {
          scope.transfer_tokens = data;
        });
      }

      scope.getTransferTokens();
    }

  }
})