text:
\section{Introduction}
DeFi platforms constructed over blockchains such as Ethereum have seen a recent boom of activity and interest. Their growing ecosystem allows for increasingly complex financial interactions executed in a fully decentralized manner. The main building blocks used to construct these platforms are the smart contracts that define the rules of interaction in code.

Smart contracts enable a wide range of applications, such as auctions, voting systems, and second layer protocols (e.g., payment channels) that operate above the blockchain layer. They typically provide rules that allow them to act as an automated adjudicator in case conflicts between participants arise.

For many applications, interactions with smart contracts are time dependent and are even subject to deadlines, meaning that in some cases, transactions added after a specific moment will effectively be rejected. For example in the case of auctions, a bid must be received before the end of the auction otherwise it is not valid.
Another example appears in the context of payment channels~\cite{gudgeon2020sok} where participants have a limited interval of time to dispute the division of funds if they disagree with their peers.

A major weakness of such deadlines is that in cases where the blockchain is congested, users that submit transactions will not have them included in blocks in time. In fact, several attacks and failures can be attributed directly to this weakness (we provide some examples below). One mitigation often employed by participants is to offer higher fees for transactions with deadlines which means users are usually overpaying. Another is to extend the deadlines which greatly delays processing and settlement within the context of the relevant smart contract. In many cases, transaction fees and deadlines are decided upon in advance, before the exact conditions that will prevail when the transaction is actually transmitted are known, which causes participants to take wider safety margins and increases costs further.
Due to the well-known scalability issues of blockchains~\cite{bano2019sok}, we expect congested periods to become increasingly more common, which will directly impact the design of time-sensitive smart contracts.

\vspace{2pt} \noindent{\bf Our Contributions.}
In this work we present a mechanism aimed at solving these issues. We propose to set short deadlines that are automatically extended if congestion occurs. We lay the theoretical foundations of congestion monitoring in blockchains and formalize the notion of challenge-response protocols in this context.
We then propose two different protocols to detect congestion over multiple blocks: the `$L$ Consecutive Blocks' protocol defines uncongestion by the existence of $L$ consecutive uncongested blocks; its generalization, the `Sliding Window (K-out-of-N)' protocol, defines uncongestion by the existence of N consecutive uncongested blocks with K uncongested blocks among them.
We show that the Sliding Window protocol is more resilient to attacks than the $L$ Consecutive Blocks protocol when attacked by miners. Furthermore, we propose a new opcode for Ethereum that will provide the required functionality; we also provide an implementation (not requiring new opcodes) in Solidity, using opcodes introduced by Ethereum Improvement Proposal 1559 (EIP 1559)~\cite{buterin2019eip}.

\paragraph{Examples of Congestion Attacks and Related Failures.}
A recent well-known example of congestion-related failure took place on \emph{Crypto Black Thursday} (March 12th, 2020) when the price of Ethereum dropped by more than 50\% in less than 24 hours~\cite{cryptoblackthursday2}. This led to a panic-sale of coins and increased congestion. At the peak, during a 2-3 hour window, the Ethereum blockchain's fees climbed to \$1.65 on average, more than $10$ times their cost on previous days.

The drop in ETH price triggered many MakerDAO auctions to liquidate collateral (typically collateral on short positions must be sold if prices fluctuate too much). The tokens to be sold were purchased at almost no cost due to the inability of many bidders to send transactions and participate.
This has, allegedly, been leveraged by one user to gain \$8.3 million worth of ETH~\cite{cryptoblackthursday1}.

Several studies~\cite{censorship_attacks,harris2020flood} deal with different types of attacks designed to prevent a party from responding on time to a challenge~\cite{censorship_attacks}. Harris and Zohar~\cite{harris2020flood} present an attack where the attacker forces many victims at once to flood the blockchain with claims for their funds. The ensuing congestion allows the attacker to steal the funds that cannot be claimed before the deadline.
Our protocol will prevent these issues by extending the deadlines until the congestion passes.

\section{Related Work}\label{sec:relatedwork}
Congestion is a real-world problem faced by the most prominent cryptocurrencies.
In addition to the popular examples of Crypto Black Thursday or Cryptokitties, widely discussed online~\cite{cryptoblackthursday1,cryptoblackthursday2,consensys_2018}, Sokolov~\cite{sokolov2021ransomware} examined periods of congestion caused by ransomware.

One way to deal with congestion is to improve the scalability of the underlying consensus protocol~\cite{sompolinsky2018phantom,sompolinsky2015secure,sompolinsky2016spectre,croman2016scaling,eyal2016bitcoin} or to introduce higher-level layers that help to scale. Solutions ranging from sharding~\cite{wang2019sok}, off-chain payment channels~\cite{gudgeon2020sok} or layer-zero optimization~\cite{tanana2019avalanche}
(i.e., network-level optimization) have been considered. All these solutions improve the number of transactions per second that the network can process, but congestion may still occur even at higher rates.

Other methods that help to ensure that time-sensitive transactions are processed are rather ad-hoc. For example, the \emph{replace by fee}~\cite{replace-by-fee} and \emph{child pays for parent}~\cite{cpfp} mechanisms allow users to add or change the fees of their transactions.
Bitcoin's fee mechanism---equivalent to a first-price auction---is sub-optimal and often results in users paying more than what is necessary. EIP 1559 was made to change this mechanism in Ethereum~\cite{roughgarden2020eip,buterin2019eip}.
EIP 1559 implements a \emph{base fee} that is burned. This base fee can be seen as an indication of the level of congestion in recent blocks, and we utilize this in our implementation.

Another line of research that could potentially prevent transaction fees from spiking considers order-fairness consensus protocols~\cite{kelkar2020order,kursawe2020wendy}. The idea is to ensure that transactions are ordered in the blockchain in the same order they arrived in. This also helps to avoid problems such as front-running~\cite{daian2019flash}.

\section{Preliminaries and Definitions} \label{sec:FormalDef}
\subsection{Challenge-Response Protocols} \label{background:challenge-response}
A challenge-response protocol is an implementation of a pattern in which some party must respond to a challenge within a fixed time limit. This pattern consists of a challenge that takes effect at time \textbf{$T_c$} and a response deadline \textbf{$T_{rd}$} which is the latest time by which response to the challenge will be accepted. We call the time period between \textbf{$T_c$} and \textbf{$T_{rd}$} the challenge window.
Responding to the challenge during the challenge window yields different results compared to responding \emph{after} the deadline.
The protocol we propose inspects the challenge window period and extends it (by extending \textbf{$T_{rd}$}) as long as the blockchain stays congested.

\subsection{Blockchain Congestion}\label{sec:definitions}
Our protocol has two components. First it relies on a mechanism to define what it means for a block to be congested. We then use this definition to define an uncongested \emph{period}. Intuitively, a period will be (un)congested if some threshold of blocks is (un)congested.
We start by defining block congestion before moving on to presenting different period congestion definitions and choosing one that meets our requirements.

\paragraph{Blocks and Transactions.} A block $\mathbf{B} = \{\mathsf{tx}_1,\cdots,\mathsf{tx}_n\}$ is as a set of transactions
(we ignore the order of transactions in the block as well as other data---such as nonce---as they are irrelevant to our problem).
Transactions pending to be included in a block are kept locally by each node in their \emph{mempool} until they are included in the chain.
Each transaction $\mathsf{tx}$ has a size \size{tx}, and a fee density \fee{tx}. The fee paid by the transaction is therefore $\size{tx}\cdot\fee{tx}$.
We define the total weight of transactions in a block $\mathbf{B}$ with a fee density above $\theta$ as  $\weight{\theta}{\mathbf{B}} \coloneqq \sum_{tx\in B:~\fee{tx}\geq \theta} \size{tx}$.

Blocks can contain transactions with total size bounded by \ensuremath{\mathcal{B}}, i.e., $\weight{\text{0}}{\mathbf{B}} \le \ensuremath{\mathcal{B}}$.
For simplicity,
we treat every block as \emph{full}, i.e., for any $block$ $\weight{\text{0}}{\mathbf{B}} = \ensuremath{\mathcal{B}}$ (if necessary,
we fill them artificially with transactions with a fee of 0).

The total amount of fees collected from a block by the miner is $\utility{\mathbf{B}} \coloneqq \sum_{tx\in B} \size{tx}\cdot\fee{tx}$.
If the size of the mempool is bigger than the maximum block size $\weight{\text{0}}{\mathbf{B}}$, we assume that honest miners choose the transactions in a way to maximize the fees they get.

\paragraph{Period.}\label{def:period}
A period $Pe=(b_1,b_2,...,b_n)$ in the blockchain is a non-empty sequence of \textbf{consecutive} blocks. We denote the length (number of blocks) of the period by $|Pe|=n$, and write for $i\in\{1,...,n\}$: $Pe[i] = b_i\in Pe$.
For a period $P_2$, we say that period $P_1$ is included in $P_2$ and note $P_1\subseteq P_2$ if every block in $P_1$ is included in $P_2$.

In this work, we want to capture the notion of congestion: a phenomenon where there's a spike in the number of transactions waiting in the mempool. Since the mempool is not part of the blockchain, we instead rely on the data in the blocks in order to define congestion. We propose the following definition for \emph{block congestion}.

\begin{definition}[$(\theta,\gamma)$-congestion] \label{block_congestion_def}
	We say that a single block $\mathbf{B}$ is $(\theta,\gamma)$-conges-ted if
	$\weight{\theta}{\mathbf{B}} \geq \gamma \cdot \ensuremath{\mathcal{B}}$
	and denote $\IsBlockCongested{\theta,\gamma}{\mathbf{B}}=1$, where $\mathcal{C}_{\theta,\gamma}$ is the corresponding indicator function.
\end{definition}

Per this definition, all transactions above fee density $\theta$ are examined and required to make up at least a $\gamma$-fraction of the block in terms of size.
Intuitively, for $\gamma=1$ the definition captures that if a block is $(\theta,1)$-congested, a transaction needs to have a fee density that is at least $\theta$ in order to have a better chance of being included.
In other words, we use the price of entering a transaction to the blockchain as a reliable signal for congestion.

For a block $\mathbf{B}$ and a fee density $\theta \geq 0$, we define the \emph{$\theta$-weight threshold} $\gamma_{\mathbf{B}}(\theta)$ as the maximum fraction of the block weight under which
the block is $(\theta,\gamma)$-congested. From definition~\ref{block_congestion_def} it is clear that $\gamma_{\mathbf{B}}(\theta) = \frac{\weight{\theta}{\mathbf{B}}}{\ensuremath{\mathcal{B}}}$.
Similarly, for a block $\mathbf{B}$ and a fraction $\gamma \geq 0$, we define the \emph{$\gamma$-fee density threshold} $\theta_{\mathbf{B}}(\gamma)$ as the maximum fee density under which the block is $(\theta,\gamma)$-congested ($\theta_{\mathbf{B}}(\gamma) \coloneqq \max\{\theta \mid \IsBlockCongested{\theta,\gamma}{\mathbf{B}}=1\}$).

\paragraph{Block Manipulation.} One of the key measures we are interested in is when is an adversary able to manipulate blocks' congestion signals. When a miner mines a block, they can choose to include transactions from their mempool or to add dummy transactions that move money between their accounts and pay a fee (to themselves), making the fees appear different than they ought to be.
However, miners cannot manipulate blocks at arbitrary heights, and doing so would incur a cost.
The miner's chance of mining a new block depends on its relative computational power. Therefore, as is standard, we denote the computational power of an adversary by $\alpha$. Each block has a probability $\alpha$ to be mined by the adversary, and $1-\alpha$ to be mined by the other miners.
Furthermore, giving up mempool transactions means missing out their fees and hence induces a loss that we compute in the next two propositions.

\begin{proposition}\label{prop:one}

	An adversary manipulating a block $\mathbf{B}$ to make it $(\theta_1,\gamma_1)$-conges-ted when it is not, will lose a potential profit of at-least  $\ensuremath{\mathcal{B}}\cdot\int_{1-(\gamma_1-\gamma_{\mathbf{B}}(\theta_1))}^{1} \theta_{\mathbf{B}}(\gamma) \,d\gamma$.
\end{proposition}

\newcommand{\proofpropositionone}{
	\begin{figure}[ht]
		\centering
		\includegraphics[scale=0.48]{plots/tight1.PNG}
		\caption{Cumulative function of the $\theta$-weight threshold of a block $\mathbf{B}$ before and after miner manipulation}
		\label{fig:tight1}
	\end{figure}

	\begin{proof}
		Given a block $\mathbf{B}$ which is \textbf{not} $(\theta_1,\gamma_1)$-congested ($\IsBlockCongested{\mathnormal{\theta_1},\mathnormal{\gamma_1}}{\mathbf{B}}=0$), it is possible to manipulate it into a block $\overline{\mathbf{B}}$ which is $(\theta_1,\gamma_1)$-congested ($\IsBlockCongested{\mathnormal{\theta_1},\mathnormal{\gamma_1}}{\overline{\mathbf{B}}}=1$) by replacing
		some of its transactions with dummy transactions that have fee density $\ge \theta_1$.
		In order to maximize its revenue, the adversary will remove the transactions with the lowest fee density.
		The minimum portion of transactions that the adversary needs to remove is $\gamma_1-\gamma_{\mathbf{B}}(\theta_1)$ (by the definition of
		$\gamma_{\mathbf{B}}$), and by doing so the miner misses the rewards associated with removing these legitimate transactions.

		Using the notations from above, we compute a lower bound on the miner's loss, which can be expressed by
		$\utility{\overline{\mathbf{B}}}\leq \utility{\mathbf{B}} - \ensuremath{\mathcal{B}}\cdot~\int_{1-(\gamma_1-\gamma_{\mathbf{B}}(\theta_1))}^{1} \theta_{\mathbf{B}}(\gamma) \,d\gamma$ (see Figure~\ref{fig:tight1}).
		Note this is only a lower bound since a miner can remove transactions only in their entirety and not parts of them.
\end{proof}}
\ifsubmission
\else
\proofpropositionone
\fi

\begin{proposition}\label{prop:two}
	An adversary manipulating a block $\mathbf{B}$ to reverse its signal from $(\theta_1,\gamma_1)$-congested to not congested will lose a potential profit of at-least  $\ensuremath{\mathcal{B}}\cdot\int_{\gamma_1}^{\gamma_{\mathbf{B}}(\theta_1)}(\theta_{\mathbf{B}}(\gamma)-\theta_1) \,d\gamma$.
\end{proposition}

\newcommand{\proofpropositiontwo}{
	\begin{figure}[ht]
		\centering
		\includegraphics[scale=0.48]{plots/tight2.PNG}
		\caption{Cumulative function of the $\theta$-weight threshold of a block $\mathbf{B}$ before and after miner manipulation}
		\label{fig:tight2}
	\end{figure}

	\begin{proof}
		The cost of reverting a block signal to uncongested depends on the state of the mempool. It is likely that transactions with fee density slightly lower than $\theta_1$ will be available in the miner's mempool allowing him to substitute the original transactions with fee density $\geq \theta_1$ for these and lose the fee difference.
		This allows us to give a bottom bound on the loss which can be expressed by
		$\utility{\overline{\mathbf{B}}}\leq \utility{\mathbf{B}} - \ensuremath{\mathcal{B}}\cdot\int_{\gamma_1}^{\gamma_{\mathbf{B}}(\theta_1)}(\theta_{\mathbf{B}}(\gamma)-\theta_1) \,d\gamma$ (see Figure~\ref{fig:tight2}).
\end{proof}}

\ifsubmission{The proofs for both propositions can be found in Appendix~\ref{app:proofs1}.}
\else
\proofpropositiontwo
\fi

Before moving on to define period congestion, we note that there exist other ways in which block congestion could be defined. For example, in Section~\ref{sec:Implementation}, we take the EIP 1559 \emph{base fee} as a measure of congestion and use it to implement our suggested protocol. We include several other examples that are less efficient in Appendix~\ref{app:block_congestion_def_examples}.

\paragraph{Congestion Vector of a Period.} To determine whether a period $Pe$ is uncongested we will refer to the congestion vector $Pe^c \coloneqq(\IsBlockCongested{}{\mathnormal{Pe[i]}})_{i=1}^{n} \in \{0,1\}^n$ which consists of the congestion signal of its blocks.
Intuitively, if most of the blocks in the period are congested then the period is congested and vice-versa. However, we must also account for the fact that an adversary may be able to change the congestion signal of some of the blocks, as already discussed.
We will consider different protocols to define period uncongestion, a situation in which the period is considered not congested. An uncongestion period protocol is a function that we denote by
$\IsPeriodUnCongested:\{0,1\}^*\rightarrow \{0,1\}$.
This function takes as input a binary series representing the congestion signal of the blocks in the examined time period.
It will return 0 if the period is congested and 1 otherwise.
This function can furthermore be extended to also provide auxiliary information such as a proof $\pi$ in the case where the period is uncongested.
For the efficiency of the protocol, we will strive for a definition that can provide a compact and easy-to-verify proof.
Throughout the rest of the paper,
we use $B(n,p)$ to denote the binomial distribution with parameters $n$ and $p$.
\begin{definition}[Period Manipulation]
	For a period $Pe$ and an adversary with a fraction $\alpha$ of the total computational power, we associate a manipulated period $\hat{Pe}$ defined as follows.
	For $i\in\{1,\dots,|Pe|\}$ the adversary can replace $Pe[i]$ with probability $\alpha$, with a block that has a congestion signal of their choice.
	We denote by $\overline{m}=m_{|Pe|}(\alpha) \sim B(|Pe|,\alpha)$ the vector that indicates which of the blocks in the given period the adversary controls, meaning the adversary can replace the $Pe[i]$ block's congestion signal iff $\overline{m}[i]=1$.
	We then define the adversary's manipulation set $S_{\overline{m},Pe} \coloneqq \{\hat{Pe^c} \in \{0,1\}^{|Pe|}\mid \forall 1\leq i \leq |Pe|: \overline{m}[i]=0 \Rightarrow \hat{Pe^c}[i] = Pe^c[i]\}$. Intuitively, $S_{\overline{m},Pe}$ corresponds to the set of possible congestion vectors that the adversary could create by changing the signal of the blocks that it controls.
\end{definition}

In a real world setting, even if there is a long period of uncongestion, it could be the case that one or more of the blocks are fuller than the others due to some randomness in the transactions' arrival time (e.g., there was a temporary high transaction volume).
To account for this randomness,
we make a simplifying assumption that blocks are congested independently with probability $p$ and say that the blockchain is \emph{$p-$congested}. We note that, in reality, congestion is often changing and is usually correlated when considering several consecutive blocks. We leave more complex models of congestion for future work. In our case, the congestion vector of a period $Pe$ chosen at random has a binomial distribution: $Pe^c \sim B(n,p)$.
When studying attacks where the adversary tries to convert a congested period to an uncongested one, we will assume that $p$ is close to one (i.e., most of the blocks are congested),
whereas when studying the opposite case, we will consider $p$ to be close to zero.

Our protocol consists in extending the deadline of challenge-response in the event of a congestion period.
However, to avoid an edge case where the deadline is extended indefinitely, we define $\hat{M}$, an upper bound on the total length of the extended period.

\begin{definition}[$\hat{M}$-maximum Extension] \label{def:M}
	Given a challenge-response protocol in a $p-$congested blockchain
	where the challenge starts at block height $h$, we say that $\hat{M}$ is the maximum extension of the challenge if the deadline cannot be extended further than height $h + \hat{M}$.
\end{definition}

\subsection{Desirable Properties of Protocols}\label{desirableProperties}

In this section, we define some properties that we aim for our protocol to achieve.

In the rest of the paper we use the notation $D \xleftarrow{} s$ to denote that $s$ was selected randomly from the distribution $D$.
We start by describing the two types of attack that we will consider---a congestion attack and an uncongestion attack---before defining the
\emph{robustness} of the protocol, which captures the security of the protocol against either attack.

\begin{definition}[Congestion/Uncongestion Attack on $Pe$]\label{def:attacks}
	Given a period $Pe$, chosen at random in a $p$-congested blockchain,
	we say that the adversary wins a congestion, resp. uncongestion, attack on $Pe$
	if it can manipulate $Pe$ into an congested, resp. uncongested, period.
\end{definition}

\begin{definition}[$(\alpha, p, q, n)$-congestion Robustness]
	\label{def:con_rob}
	We say an uncongestion period protocol $\IsPeriodUnCongested:\{0,1\}^*\rightarrow \{0,1\}$ is \textbf{$(\alpha, p, q,n)$-congestion robust} if, given an adversary with a relative computational power $\alpha$,
	his probability of winning a congestion attack, i.e., of successfully manipulating a period $Pe$ of $n$ blocks into a congested period $\hat{Pe}$, is less than $q$.
	$$ B(n,p)\xleftarrow{} Pe~:~
	P_r(\exists \hat{Pe} \in S_{\overline{m}, Pe}~s.t.~\IsPeriodUnCongested(\hat{Pe})=0) \leq q.$$
\end{definition}

\begin{definition}[$(\alpha, p, q, n)$-uncongestion Robustness]
	\label{def:uncon_rob}
	We say an uncongestion period protocol $\IsPeriodUnCongested:\{0,1\}^*\rightarrow \{0,1\}$ is \textbf{$(\alpha, p, q,n)$-uncongestion robust} if, given an adversary with a relative computational power $\alpha$,
	his probability of winning an uncongestion attack, i.e., of successfully manipulating a period $Pe$ of $n$ blocks into an uncongested period $\hat{Pe}$, is less than $q$.
	$$ B(n,p)\xleftarrow{} Pe~:~
	P_r(\exists \hat{Pe} \in S_{\overline{m}, Pe}~s.t.~\IsPeriodUnCongested(\hat{Pe})=1) \leq q.$$

\end{definition}

\begin{definition}[Monotonicity]
	A congestion protocol is \textbf{monotone} if
	for every two periods $Pe_1$ and $Pe_2$, if $Pe_1\subseteq Pe_2$ and $Pe_1$ is considered uncongested, then so is $Pe_2$, i.e.,
	$\forall Pe_1\subseteq Pe_2: \IsPeriodUnCongested(Pe_1^c)=1 \rightarrow \IsPeriodUnCongested(Pe_2^c)=1$.
\end{definition}
A monotone protocol is easier to verify
as the prover only needs to select a portion of blocks from the time period $Pe$ in order to prove uncongestion.
Furthermore, a monotone protocol requires only sporadic access to the blockchain. A prover can go offline and prove uncongestion when they come back online by choosing any uncongested period from the time they were offline.
In the case of a non-monotonic protocol, if the prover is offline during an uncongested period, they cannot prove the uncongestion of the longer period, after they came back online, they missed the uncongested period.

\paragraph{Efficiency Properties.}
We define two properties that capture the efficiency of the protocol.
\begin{itemize}
	\item \textbf{Concise proof size}  The evidence needed to prove uncongestion of a period should be as concise as possible.
	\item \textbf{Concise refresh information} The extra information needed to be kept when checking the congestion signal of a period that
	has already been extended due to congestion should be as concise as possible. Ideally, when we extend a period from $Pe_1$ to $Pe_2$ in order to check $Pe_2$ for congestion, we should not have to recheck every block in $Pe_1$ but, rather, aggregate this information.
\end{itemize}

In the next section we will discuss different period congestion protocols with the goal of finding one that will be proof efficient and robust against an attacker with reasonable hash rate with high probability.

\section{Uncongested Period Protocols}\ \label{sec:UncongestedPeriodProtocols}
In this section, we examine different protocols that fit the definition of congestion of a period $Pe$.
We start by presenting ``naive'' protocols and discuss why they are not good enough, i.e., why they lack the desirable properties defined in Section~\ref{desirableProperties}.

\subsection{Strawman Protocols}

\begin{definition}[Cumulative M]
	Period $Pe$ is uncongested if there exists M blocks which are uncongested:
	$\UCP{CM}{Pe^c}=1 \leftrightarrow \left(\sum_{b\in Pe} (1-\IsBlockCongested{}{\mathnormal{b}})\geq M\right)$.
\end{definition}
This protocol is monotonic but is not sufficiently robust to adversarial attacks:
if we wait long enough, the probability of the adversary controlling M blocks becomes overwhelming (even if $\alpha$ is small). We solve this in the next strawman by considering the percentage of blocks instead of a fixed number.

\begin{definition}[Percentage]
	A period $Pe$ is uncongested if $x\%$ of its blocks are not congested:
	$\UCP{PC}{Pe^c}=1 \leftrightarrow \left(\sum_{b\in Pe} (1-\IsBlockCongested{}{\mathnormal{b}})\geq \frac{x}{100}\cdot|Pe|\right) $.
\end{definition}
This protocol is much more robust but has the drawback of not being monotonic. For example, if all blocks are uncongested during the first part of the period and congestion begins in the second part, then the beginning of the period is uncongested while the whole period may not be.

We now suggest the following monotonic rule:
\begin{definition}[$L$ Consecutive Blocks]
	A period $Pe$ is uncongested if there exists at least L
	consecutive uncongested blocks included in it:\\
	$\UCP{L}{Pe^c}=1 \leftrightarrow \left(\exists~ 1 \leq i \leq |Pe|-L+1~s.t.~\forall~0\leq j \leq L-1:Pe^c[i+j]=0\right) $.
\end{definition}
\newcommand{\proofconscuivemonotone}{
	\begin{newProposition}{3}
		The $L$ Consecutive Blocks protocol is monotonic.
	\end{newProposition}

	\begin{proof}
		Given a period $Pe_1$,
		uncongested according to the $L$ Consecutive Blocks protocol, which is included in period $Pe_2$:
		\begin{align*}
			&\UCP{L}{Pe_1^c}=1\Rightarrow \\
			& \exists~ 1 \leq i_1 \leq |Pe_1|-L+1~s.t.~\forall~0\leq j \leq L-1:Pe_1^c[i_1+j]=0 \\
			& Pe_1\subseteq Pe_2 \Rightarrow \\
			& \exists~ 1 \leq k \leq |Pe_2|-|Pe_1|+1~s.t.~\forall~1\leq d \leq |Pe_1|:Pe_1^c[d]=Pe_2^c[k+d-1]\\
			&\Rightarrow for~ i_2=k+i_1-1,~\forall~0\leq j \leq L-1:Pe_2^c[i_2+j]=0 \\
			& \Rightarrow \UCP{L}{Pe_2^c}=1
		\end{align*}
\end{proof}}

\ifsubmission{
	We show that this protocol is monotonic and inspect its efficiency in Appendix~\ref{app:proofs2}.
	We now evaluate its robustness. }
\else{We first show that this protocol is monotonic, before evaluate its efficiency.
	\begin{proposition}
		The $L$ Consecutive Blocks protocol is monotonic.
	\end{proposition}
	\proofconscuivemonotone}
\fi

\newcommand{\LconsecEffInfo}{
	We now evaluate its efficiency.
	\paragraph{Proof size.}
	In order to provide evidence for the uncongestion of period $Pe$ of size $n$, it is enough to point to the location of the first block in a series of $L$ consecutive uncongested blocks, given formally by $\pi=\min\{i\in\{1,...,n-L+1\} \mid \forall~0\leq j \leq L-1:Pe^c[i+j]=0\}$.

	\paragraph{Refresh Information.}
	Given a congested period $Pe$, and $\hat{Pe}$ that extends it,
	in order to determine the congestion level of the extended period $\UCP{L}{\hat{Pe^c}}$, it is enough to check only $\hat{Pe} \setminus\left( Pe[:-(L-1)]\right)$, the period beginning $L$-1 blocks before $Pe$ ends.
}
\ifsubmission{
}\else{\LconsecEffInfo
}
\fi

\subsubsection*{Evaluation of the Robustness of the $L$ Consecutive Blocks protocol.}

We examine situations where the adversary attempts to manipulate the congestion signal for a given period. We separate this into two attacks: uncongestion and congestion attacks (as in Definition~\ref{def:attacks}).
We strive to achieve a high defense rate against both attacks, meaning finding a value $L$ that will give a low probability for an adversary to succeed in each of the attacks separately.

\paragraph{Evaluation of the Uncongestion Attack.}
In order to compute the probability of an attacker to successfully manipulate $Pe$ into an uncongested period, we define the following matrix $T_{(L+1)\times(L+1)}$:
\begin{equation}
	\forall~0 \leq i,j\leq L:~~ T_{i,j} =
	\begin{cases*}
		(1-\alpha)\cdot p & if $j = 0 \wedge i\neq L$ \\
		\alpha + (1-\alpha)\cdot (1-p) & if $j = i+1$ \\
		1 & if $i = j = L$ \\
		0        & otherwise
	\end{cases*}
\end{equation}

and denote by $e_i$ the $i^{th}$ unit vector of dimension $L$+1 (i.e., $e_i$ has a 1 in the $i^{th}$ coordinate and 0's elsewhere).
\begin{theorem}
	The probability of an attacker with a relative computational power $\alpha$ to successfully manipulate $Pe$ into an uncongested period in a p-congested network
	equals    $e_1 \cdot T^n \cdot e_{L+1}^t$.
\end{theorem}
\begin{proof}
	We note that, at each block, the attacker has a probability $\alpha$ to mine the next block, which allows them to decide its congestion level. In this context, this means setting the block to be uncongested. In addition, the congestion signal of a block not mined by the attacker depends on the prevailing congestion state which is expressed by $p$. The probability of an honest block being congested, resp. uncongested, is hence equal to $(1-\alpha)\cdot p$, resp.  $\alpha+(1-\alpha)\cdot(1-p)$. We define the following Markov chain that describes a random walk on $Pe$'s blocks and whose states represent the number of consecutive blocks that are uncongested at a point in time.

	\begin{figure}[ht]
		\centering
		\includegraphics[scale=0.6, trim={200cm 0.8cm 200cm .5cm}]{plots/markov1.pdf}
	\end{figure} \label{fig:markov1}

	The initial state is $0$ since it corresponds to the 0 consecutive uncongested blocks at the beginning of the walk.
	With each step, we move from state $i$ to state $i+1$, for $i<L$, if the block is uncongested, and return to state $0$ if it is not.
	If we reach state $L$, we stay there since it means the adversary has reached the goal of $L$ consecutive uncongested blocks in $Pe$ and can manipulate it to an uncongested period.

	T is the corresponding transition matrix; hence the probability of reaching state $L$ in $|Pe|=n$ steps is expressed by $e_1 \cdot T^n \cdot e_{L+1}^t$.
\end{proof}

\paragraph{Evaluation of the Congestion Attack.}
For the attack in the opposite direction we
define $\hat{T}_{(L+1)\times(L+1)}$ as follows:
\begin{equation}
	\forall~0 \leq i,j\leq L:~~ \hat{T}_{i,j} =
	\begin{cases*}
		\alpha + (1-\alpha) \cdot p & if $j = 0 \wedge i\neq L$ \\
		(1-\alpha) \cdot (1-p) & if $j = i+1$ \\
		1 & if $i = j = L$ \\
		0        & otherwise
	\end{cases*}
\end{equation}

\begin{theorem}
	The probability of an attacker with a relative computational power $\alpha$ to successfully manipulate $Pe$ into a congested period, in a p-congested network
	equals    $1-e_1 \cdot \hat{T^n} \cdot e_{L+1}^t$.
\end{theorem}
\newcommand{\prooflconsecutivecongestion}{
	\begin{proof}
		This time, if the attacker succeeds in mining a block, they will make it congested. Therefore the probability for a block to be uncongested is $(1-\alpha)\cdot(1-p)$.
		As before, we define the following Markov chain whose states represent the number of consecutive blocks that are uncongested at a point in time in $Pe$:

		\begin{figure}[ht]
			\centering
			\includegraphics[scale=0.6, trim={200cm 0.8cm 200cm .5cm}]{plots/markov2.pdf}
		\end{figure} \label{fig:markov2}

		$\hat{T}$ is the corresponding transition matrix.
		Therefore, the probability for the adversary to succeed in the congestion attack is equivalent to the probability that the rest of the miners will not reach the $L$ state in $n$ steps, which is expressed by: $1-e_1 \cdot \hat{T^n} \cdot e_{L+1}^t$.
\end{proof}}
\prooflconsecutivecongestion

Now that we have the attacks' success rates, we examine the robustness of the protocol against both attacks for different values of $L$.

Although attacks are potentially expensive for the adversary (who needs to change the contents of its block and, hence, loses transaction fees), we still desire a low success probability for the attack even for strong attackers.
We assume in the following evaluations that the attacker controls $33\%$ of the computational power.

Given that congestion may cause period extension, we need a value for $L$ that gives protection also against attacks over longer periods. We examine the behavior of the protocol for periods as long as $\hat{M}$ using different values for $L$.

The value $p$ should represent realistic network conditions. For our analysis we pick $p=0.85$ when studying the uncongestion attack, to simulate more congested settings, or $p=0.15$ when studying the congestion attack, to simulate relatively uncongested settings.
Other values can be plugged in, if needed, for other conditions. We start by examining the robustness of the protocol for a period of 1 day.

Figures~\ref{fig:consec-eth}-\ref{fig:consec-bitcoin} present
the probability of success in both attacks for two different period lengths: 6450 blocks in Figure~\ref{fig:consec-eth} and 144 blocks in Figure~\ref{fig:consec-bitcoin}. These periods correspond, roughly, to a single day in Ethereum and a single day in Bitcoin. The red curves correspond to the congestion attack and the blue curves to the uncongestion attack.
We compute these probabilities for different values of $L$.

\begin{figure}[ht]
	\begin{subfigure}[t]{.5\linewidth}
		\includegraphics[scale=0.38]{plots/attack_success_graph_L_consecutive_eth.pdf}
		\caption{$|Pe|=6450\sim 1$ day in Ethereum}
		\label{fig:consec-eth}
	\end{subfigure}
	\label{fig:consec}
	\begin{subfigure}[t]{.5\linewidth}
		\includegraphics[scale=0.38] {plots/attack_success_graph_L_consecutive_bitcoin.pdf}
		\caption{$|Pe|=144\sim 1$ day in Bitcoin}
		\label{fig:consec-bitcoin}
	\end{subfigure}
	\caption{Attack success rate as a function of $L$, for $\alpha=0.33$}
\end{figure}

The results in both figures show there is no value $L$ that gives a probability of success less than $1\%$ for both attacks. Formally, it shows that the $L$ Consecutive Blocks protocol cannot be simultaneously $(0.33, 0.15, 0.01, 1~day)$-congestion~robust and
$(0.33, 0.85, 0.01, 1~day)$-uncongestion~robust.
Therefore,
we find the $L$ Consecutive Blocks protocol not sufficiently secure. Intuitively, this is because more robust estimates of congestion are typically obtained over longer observation windows. The $L$ Consecutive Blocks protocol obtains longer observations if $L$ is increased, but then the requirement for consecutive blocks to be uncongested is too strict and is not robust. As a result of this insight, we propose a new protocol that generalizes the $L$ Consecutive Blocks protocol and allows for longer observation windows with a relaxed condition for uncongestion.

\subsection{\textbf{Sliding Window (K-out-of-N) Protocol}}\label{sliding}

\begin{definition}[K-out-of-N Sliding Window] A period $Pe$ is uncongested if there exists a period $\hat{Pe}$ of length $N$ included in it in which at least $K$ blocks are uncongested.

	$$\UCP{SW}{Pe^c}=1 \leftrightarrow\left( ~\exists~\hat{Pe}\subseteq Pe: |\hat{Pe}|=N \wedge \left(\sum_{b\in \hat{Pe}} (1-\IsBlockCongested{}{\mathnormal{b}})\geq K\right)\right)$$
\end{definition}

We note that the $L$ Consecutive Blocks protocol is a special case in which $L=N=K$.
\newcommand{\slidingwindowmonotonicity}{
	\begin{newProposition}{4}
		The Sliding Window protocol is monotonic.
	\end{newProposition}
	\begin{proof}
		Given an uncongested period $Pe_1$, according to the Sliding Window protocol, which is included in period $Pe_2$:
		\begin{align*}
			& \UCP{SW}{Pe_1^c}=1\Rightarrow \left(\exists~\hat{Pe}\subseteq Pe_1: |\hat{Pe}|=N \wedge \left(\sum_{b\in \hat{Pe}} \IsBlockCongested{}{\mathnormal{b}}\geq K\right)\right)\\
			& Pe_1\subseteq Pe_2 \Rightarrow  \left(\hat{Pe}\subseteq Pe_2\right) \wedge \left(\sum_{b\in \hat{Pe}} \IsBlockCongested{}{\mathnormal{b}}\geq K\right)\\
			&\Rightarrow \UCP{SW}{Pe_2^c}=1
		\end{align*}
\end{proof}}
\slidingwindowmonotonicity

We now evaluate its efficiency.
\paragraph{Proof Size.}
In order to provide evidence for the uncongestion of a period $Pe$ of size $n$, it is enough to point to a window in which uncongestion occurs. Formally, to present $\pi=i \in \{1,...,n-K+1\}$ s.t. $\sum_{l=i}^{i+N}(1-\IsBlockCongested{}{\mathnormal{Pe[l]}})\geq K$.

\paragraph{Refresh Information.}
Given a congested period $Pe$, and $\hat{Pe}$ that extends it,
in order to determine the congestion level of the extended period $\UCP{SW}{\hat{Pe^c}}$, it is enough to check only windows that overlap blocks in $\hat{Pe} \setminus Pe$.

\subsubsection*{\textbf{Evaluation of the Sliding Window Protocol's Robustness.}}
We consider the two attacks in Definition~\ref{def:attacks}.
We first note that the two attacks may differ in their consequences. While the congestion attack can cause a delay in the response deadline (i.e., a deadline will be extended even if it is not really needed), the uncongestion attack might lead participants to miss the chance to respond on time,
as the deadline will not be extended even if the network is congested. The damage in each case depends on the particular use case.
For example, in the case of payment channels, not responding in time is more severe and may lead to financial losses.
We strive to achieve a high level of security against both types of attack, i.e., to find values for parameters ($N, K$) that will yield a low probability of success for both.

We begin by presenting upper bounds on the probabilities of success in each of the attacks.
\begin{theorem}
	\label{thm:bound_un}
	The probability of an attacker with a relative computational power $\alpha$ to successfully manipulate $Pe$ into an uncongested period, in a p-congested network, is bounded above by $(n-N+1)\cdot \sum_{j=K}^{N}\binom{N}{j}\cdot q^{j}\cdot(1-q)^{N-j}$, for $q=\alpha+(1-p)\cdot(1-\alpha)$.
\end{theorem}

\newcommand{\proofslidingwindowuncongestion}{
	\begin{proof}
		The probability for a block to be uncongested during this attack is $q=\alpha+(1-p)\cdot(1-\alpha)$.
		In a period of size $n$, there are $n-N+1$ different sliding windows. We denote by $A_i$ the event in which there are $K$ out of $N$ uncongested blocks in the $i^{th}$ sliding window.
		Therefore, the probability of a single sliding window being uncongested is $P(A_i)=\sum_{j=K}^{N}\binom{N}{j}\cdot q^{j}\cdot(1-q)^{N-j}$.
		To succeed in the uncongestion attack, at least one of the sliding windows has to be uncongested, which is expressed by $P(\cup_{i=1}^{n-N+1}A_i)$.
		We use the union bound to bound this probability and get: \\ $P(\cup_{i=1}^{n-N+1}A_i) \leq \sum_{i=1}^{n-N+1}P(A_i) = (n-N+1)\cdot \sum_{j=K}^{N}\binom{N}{j}\cdot q^{j}\cdot(1-q)^{N-j}$
\end{proof}}
\proofslidingwindowuncongestion
\begin{theorem}
	\label{thm:bound_co}
	The probability of an attacker with a relative computational power $\alpha$ to successfully manipulate $Pe$ into a congested period, in a p-congested network is bounded above by $(\sum_{j=0}^{K-1}\binom{N}{j}\cdot q^{j}\cdot(1-q)^{N-j})^{\lfloor \frac{n}{N} \rfloor}$, for $q=(1-p)\cdot(1-\alpha)$.
\end{theorem}
\newcommand{\slidingwindowcongestion}{
	\begin{proof}
		The probability for a block to be uncongested is $q=(1-p)\cdot(1-\alpha)$.
		We denote by $B_i$ the event in which there are less than $K$ uncongested blocks in the $i^{th}$ sliding window.
		The probability of a single sliding window being congested is $P(B_i)=\sum_{j=0}^{K-1}\binom{N}{j}\cdot q^{j}\cdot(1-q)^{N-j}$.
		To succeed in the congestion attack, all sliding windows in the period must be congested, which is expressed by $P(\cap_{i=1}^{n-N+1}B_i)$.
		We bound this probability by $P(\cap_{i=1}^{\lfloor \frac{n}{N} \rfloor}B_{N \cdot (i-1) + 1})$, i.e., we consider a subset of events $B_i$ that are independent from each other (removing overlapping windows). We compute the intersection of the pairwise independent events and get: $P(\cap_{i=1}^{n-N+1}B_i) \leq P(\cap_{i=1}^{\lfloor \frac{n}{N} \rfloor}B_{N \cdot (i-1) + 1}) = \prod_{i=1}^{\lfloor \frac{n}{N} \rfloor}P(B_{N \cdot (i-1) + 1}) = (\sum_{j=0}^{K-1}\binom{N}{j}\cdot q^{j}\cdot(1-q)^{N-j})^{\lfloor \frac{n}{N} \rfloor}$.
\end{proof}}
\slidingwindowcongestion

We would like to compute the robustness of the protocol for 1 day to 1 hour sliding windows.
We examine the situation where a period $Pe$ of size $n$ is chosen at random and the blockchain is $p-congested$ for values of $p=0.85$ (relatively congested) and $p=0.15$ (relatively uncongested) against an attacker with computational power $\alpha\leq0.33$.
In the evaluation, we allow periods to be extended up to two weeks, a reasonable time for congestion to pass. We set the $\hat{M}$-maximum extension (see Definition~\ref{def:M}) accordingly (90300 blocks in Ethereum and 2016 blocks in Bitcoin).

We first evaluate the attack over Ethereum, computing the above bounds for different sliding window sizes.
We begin with a sliding window of 1 day ($N=6450$), setting $K=\frac{N}{2} = 3225$.
Figure~\ref{fig:sliding2w_upper_eth} presents the two upper bounds for the different possible period lengths $N\leq n \leq \hat{M}$. For the protocol to be considered secure, we need low values in both curves for the different period lengths (since periods might be extended).
As can be seen, the probabilities in the graph are extremely low, showing the protocol to be very secure. We emphasize that the blue curve is not horizontal, as shown in the graph; all of its values are smaller than $10^{-323}$. Note that these are only upper bounds; the actual probabilities are even lower.

\begin{figure}[ht]
	\centering
	\includegraphics[scale=0.43]{plots/sliding_window_upper_bound_eth.pdf}
	\caption{Upper bounds on the attacks' success rates as a function of the period length, for $\hat{M}=90300, N = 6450, K=3225, \alpha=0.33$}
	\label{fig:sliding2w_upper_eth}
\end{figure}

We evaluate the attack for smaller sliding windows. The following table
summarizes our results:

\begin{center}
	\setlength{\tabcolsep}{10pt}
	\resizebox{9,2cm}{!}{%
		\begin{tabular}{ c  c  c  c }
			\hline
			\textbf{N} & \textbf{K} & \textbf{Uncongestion} & \textbf{Congestion}\\
			\hline
			6450 (1 day)   & 3225        & $<10^{-323}$   & $1.44\times 10^{-29}$
			\\
			3225 (12 hours)   & 1612    & $1.26\times 10^{-10}$   & $8.06\times 10^{-16}$
			\\
			1612 (6 hours)   & 815       & $7.14\times 10^{-5}$   & $1.08\times 10^{-7}$
			\\
			806 (3 hours)   & 421       & $8.87\times 10^{-3}$   & $3.16\times 10^{-3}$
			\\ \hline
		\end{tabular}
	}
\end{center}

The wider the sliding window is, the greater the protection.
For smaller sliding windows, such as 1 hour ($N=269$), we can achieve a $99\%$ defense rate against each attack if we lower the attackers' computation power to $\alpha \leq 0.27$ (instead of 0.33).
We provide examples of $N, K$ values and the level of protection they provide (an upper bound), but these are configurable and subject to the user's discretion. One can choose to increase the level of protection from one attack at the expense of the other, or to set a larger initial period length ($>N$) to increase the protection.

Next, we want to know what happens with smaller periods such as in Bitcoin, which has longer block intervals. To do so, we set $\hat{M}=2016$ and begin with a sliding window of 1 day ($N=144$).

We use a simulation to draw 100,000 samples $Pe^c \sim B(\hat M,p)$ of congestion vectors and to compute the success rates of both attacks among the samples (Figures~\ref{fig:sliding2w}a,c,d). We use error plots to plot the standard error of the data; however, the errors are very small and therefore are almost invisible in the graphs.

\begin{figure}[ht]
	\begin{subfigure}[t]{.5\linewidth}
		\centering
		\includegraphics[scale=0.39] {plots/sliding_window_graph1_0.33_0.85_0.15_100000_2016_144.pdf}
		\caption{Attack success rate as a function of K, for $n = 2016, N = 144,\alpha=0.33$}
		\label{fig:sliding2w-a}
	\end{subfigure}
	\hspace{1em}%
	\begin{subfigure}[t]{.5\linewidth}
		\centering
		\includegraphics[scale=0.38]{plots/sliding_window_upper_bound_bitcoin.pdf}
		\caption{Upper bounds on the attacks' success rates as a function of the period length, for $N = 144, K=89, \alpha=0.33$}
		\label{fig:sliding2w-b}
	\end{subfigure}\\[1ex]
	\begin{subfigure}[t]{.5\linewidth}
		\centering
		\includegraphics[scale=0.37]{plots/sliding_window_graph1_alphalst_0.85_0.15_100000_2016_144_89.pdf}
		\caption{Congestion attack success rate as a function of the period length, for $N = 144, K=89$}
		\label{fig:sliding2w-c}
	\end{subfigure}
	\hspace{1em}%
	\begin{subfigure}[t]{.5\linewidth}
		\centering
		\includegraphics[scale=0.37] {plots/sliding_window_graph_alphalst_multi_n.pdf}
		\caption{Congestion attack success rate as a function of the period length, for $\alpha=0.2$}
		\label{fig:sliding2w-d}
	\end{subfigure}
	\caption{Evaluation of the attacks' success rates for $\hat{M}=2016$ (2 weeks in Bitcoin)}
	\label{fig:sliding2w}
\end{figure}

Figure~\ref{fig:sliding2w-a} presents the probability of success in each of the attacks for different $K$ values. As the graph shows, choosing $K=89$ gives protection against both attacks. We compute the upper bounds (from Theorems~\ref{thm:bound_un}-\ref{thm:bound_co}) for this value of $K$ in Figure~\ref{fig:sliding2w-b}.
The presented bounds as they appear in the graph are loose compared to the simulation results and afford a low level of defense, especially against the congestion attack. These bounds give us useful, but non-tight, upper bounds on the results for periods that are of longer length, for which the probability is extremely small. To get more precise results, we use more simulations to compute the congestion attacks' success rate for different period lengths and present the results in Figure~\ref{fig:sliding2w-c}. Each curve corresponds to a different value of $\alpha$, the computational power of the attacker. The defense rate against congestion attacks is extremely low for short periods. For $\alpha=0.33$, we reach a $>99\%$ defense rate only for periods of $\sim~11$ days or more. Lowering the computational power of the adversary naturally improves these results. For example, considering an attacker with computational power $\alpha=0.2$ results in an above $0.9995$ defense rate for period lengths starting from 2 days.

Finally, in Figure~\ref{fig:sliding2w-d} we consider an attacker with a computational power $\alpha=0.2$ and show the congestion attack success rate for different choices of $N, K$ correspondig to sliding windows of lengths 24/12/6/3 hours. We do not present the uncongestion attack results which had above $99\%$ defense rate for any $N\leq n \leq \hat{M}=2016$.

We conclude that the longer the periods are, the higher and more effective the protection against attacks is. In Ethereum, we obtained very high defense rates even when choosing short sliding window sizes and against strong attackers. In Bitcoin, on the other hand, we need to compromise on the window size and on the attackers' power to achieve higher defense.

We defined uncongested period protocols and suggested a concrete one, the Sliding Window protocol, which meets our requirements (as defined in Section~\ref{desirableProperties}). In the next section, we will describe how to use an uncongested period protocol to adjust the challenge-response protocol to deal with congested periods.

\subsection{Application to Challenge-Response Protocols}
A challenge-response protocol consists of a challenge that takes effect at time \textbf{$T_c$} and a response deadline \textbf{$T_{rd}$} (see section~\ref{background:challenge-response}).
We link $T_c$ and $T_{rd}$ to their corresponding block height and denote by $b(\textbf{$T$})$ the block at height $T$.

The parties involved in the challenge decide in advance on an uncongestion period protocol $\IsPeriodUnCongested{}$ to use. We recall that $\IsPeriodUnCongested:\{0,1\}^*\rightarrow \{0,1\}$ accepts a congestion vector (a binary series representing the congestion signal of blocks in a period) and returns 1 if the period is congested and 0 otherwise.
To apply the uncongestion period protocol, the parties adjust \textbf{$T_{rd}$} to a short deadline that gives them a reasonable time to respond to the challenge assuming an optimal case with no congestion.

The response deadline \textbf{$T_{rd}$} is applied only in the event that the challenge window $Pe=(b(\textbf{$T_c$}),b(\textbf{$T_c+1$}),...,b(\textbf{$T_{rd}$}))$ is uncongested.
In the case where the challenge window is congested, we repeatedly extend \textbf{$T_{rd}$}, 1 block at a time, as long as it remains congested.
To avoid an edge case where the deadline is extended indefinitely, we define \textbf{$\hat{T}_{rd} = T_c + \hat{M}$},
an upper bound on the deadline (see Definition~\ref{def:M}).
The challenge-response protocol adjustment is summarized in the algorithm below.

\begin{algorithm}[ht]
	\DontPrintSemicolon
	$T_c\gets init$\;
	$T_{rd}\gets init$\;
	$Pe=(b(\textbf{$T_c$}),b(\textbf{$T_c+1$}),...,b(\textbf{$T_{rd}$}))$\;
	$Pe^c\gets congestion\_vector(Pe)$\;
	\While{$\IsPeriodUnCongested(Pe^c)=0$ and $T_{rd}<\textbf{$\hat{T}_{rd}$}$}{
		$T_{rd}\gets T_{rd}+1$\;
		$Pe=(b(\textbf{$T_c$}),b(\textbf{$T_c+1$}),...,b(\textbf{$T_{rd}$}))$\;
	}
\end{algorithm}

We emphasize that the extension of the deadline is not necessarily carried out at the exact moment of the deadline (since smart contract actions need to be triggered by a transaction to the contract). Instead, a transaction that is submitted afterwards is determined to be either before or after the deadline given any possible extensions that are due. The uncongestion period protocol is specified in advance in the smart contract, and the deadline calculation is triggered either by a late response to the challenge or by the challenger that claims that a response did not arrive in time.

\section{Implementation}
\label{sec:Implementation}
We provide an implementation of the Sliding Window protocol as an Ethereum smart contract using the EIP 1559 \emph{base fee} to determine block congestion.
EIP 1559 implements a \emph{base fee} that is adjusted up and down by the protocol according to how congested the network is.
The EVM supports fetching the \emph{base fee} of the highest (current) block.
We suggest extending this to fetch the \emph{base fee} of any block, and to add an opcode that checks whether a block is congested (without such opcodes, it is not possible to fully implement the mechanisms put forward in this paper).
This opcode will receive as inputs a block and a maximum base fee (chosen by a user) and will return whether the maximum base fee exceeds the block's base fee.

In the implementation, we set the sliding window size equal to the initial deadline of the examined period (before being granted any extension).

The full \href{https://github.com/stonecoldpat/slidingwindow}{github}\footnote{\href{https://github.com/stonecoldpat/slidingwindow}{https://github.com/stonecoldpat/slidingwindow}} repository includes the smart contracts, the new opcode, and the tests. In addition, we include the Solidity code of the contracts in Appendix~\ref{app:code}.

\section{Conclusion}
\label{sec:Conclusion}
In this paper, we tackled a problem that arises when challenge-response protocols face congested periods.
When the network experiences congestion,
users will often miss the response deadline, which can lead to serious issues including financial loss. We formalized the problem and proposed a new protocol called the Sliding Window as a solution. Our protocol defines a reliable way to detect congested periods by looking only at the data available on-chain.
We then used this to extend the challenge-response deadline when congestion occurs.
We studied the security of the protocol for different parameters.
Our results showed that it is possible to decrease the
time settlement (deadline) of challenge-response protocols significantly, while expanding the security of the protocol to deal with cases of congestion.

For future work, it would be interesting to evaluate and optimize this protocol and its security analysis for more realistic congestion settings---in particular, settings in which congestion is correlated between consecutive blocks---and to provide more experimental analysis of these settings. Is is also of interest to explore whether Ethereum's proposed base fee can be used as a sufficiently robust congestion signal.

\section{Acknowledgments}
	Ayelet Lotem and Aviv Zohar are partially supported by grants from the Israel Science Foundation (grants 1504/17 \& 1443/21) and by a grant from the HUJI Cyber Security Research Center in conjunction with the Israel National Cyber Bureau.

\bibliographystyle{splncs04}

redpajama_set_name: RedPajamaArXiv


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The oppositionist was detained immediately after returning to Russia
Politicians from the countries of the European Union and the United States, as well as representatives of human rights organizations, demanded the release of opposition leader Alexei Navalny from custody, Forbes reports. On January 17, he returned to Russia after a five-month stay in Germany, where he received treatment after being poisoned with a nerve agent from the Novichok group. The oppositionist was detained at passport control at Sheremetyevo airport.
The foreign ministers of Latvia, Lithuania and Estonia in a joint statement called Navalny's detention unacceptable and called on the EU to impose sanctions against Russia if he is not released. European Council President Charles Michel also condemned Navalny's detention. "I call on the Russian authorities to release him immediately."– he wrote on Twitter.
The head of the US Department of State Mike Pompeo also spoke out in defense of the opposition politician, Interfax reports. According to him, the detention is an attempt "To silence Navalny and other oppositionists"… "The Russian people deserve a government that supports a diversity of ideas, an independent judiciary, and human rights to freedom of speech and assembly."– wrote Pompeo. The international human rights organization Amnesty International recognized Navalny as a prisoner of conscience.
According to the Federal Penitentiary Service, Alexei Navalny has been on the wanted list since December 29, 2020 for multiple violations of the probationary period, during which he was in the case of embezzlement of Yves Rocher funds. He remains in custody, and the further measure of restraint will be determined by the court.
Tags alexei, countries, demand, navalny, Newbie, release, Russia, The EU, USA
COVID-19: WHO regrets that ten countries concentrate 95% of vaccinations
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Copenhagen | Europe must show more vaccine solidarity at a time when 95% of doses have been used in ten countries, asked the regional management of the World Health Organization (WHO) on Thursday.
• Read also: All developments in the COVID-19 pandemic
• Read also: Quebec ready to wait up to 90 days for the 2nd dose
It is necessary that "all countries capable of contributing, giving and supporting equitable access and deployment of vaccines do so," said WHO's European director, Hans Kluge, during a press briefing. online, highlighting the "huge" efforts made by the organization and its partners to ensure that every country can get vaccines.
According to him, 95% of the COVID-19 vaccines administered worldwide have been in ten countries he did not name.
According to the data analysis site Our world in data, these are the United States, China, United Kingdom, Israel, United Arab Emirates, Italy, Russia , Germany, Spain and Canada.
• Read also: WHO meeting on variants, their experts investigate in China
In just over a month, some 28 million doses of COVID-19 vaccine have been injected in about 46 countries, WHO director of emergency health matters Michael Ryan said on Wednesday.
Despite the start of vaccination campaigns, the speed of transmission observed in some countries, in particular because of new, more contagious variants, in particular those called English and South African, is worrying.
"The situation is alarming", reaffirmed Mr. Kluge.
Twenty-five countries in the European zone, including Russia, have recorded cases linked to this new strain "VOC 202012/01".
"With increased transmissibility and similar disease severity, the variant is causing concern: without increased control to slow its spread, there will be an increased impact on healthcare facilities already under stress and under pressure," he said. -he explains.
Among the 53 countries in the region, 31 have started their immunization campaign and countries benefiting from the Covax mechanism, which distributes COVID vaccines to underprivileged countries, are expected in early February.
For Mr. Kluge, it is important to remain optimistic.
"2021 will be another year of the coronavirus, but it will be a more predictable year, the situation will be easier to control", he said
Tags concentrate, countries, Covid, regrets, vaccinations
28 million doses of vaccines injected in 46 countries, according to WHO
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In 36 days, some 28 million doses of COVID-19 vaccines have been injected in about 46 countries around the world, World Health Organization director of health emergencies Michael Ryan said on Wednesday.
Despite the start of vaccination campaigns, he was concerned about the speed of transmission observed in some countries, particularly because of new, more contagious variants.
• Read also: Residents and staff of CHSLDs vaccinated next week
• Read also: 10 million people have received an injection in the United States
• Read also: All the developments of the pandemic
"We are entering the second year (of the pandemic) and it could even get even harder, when you see the rates of transmission," he said during one of the regular question and answer sessions that the WHO organizes for the general public.
As for the vaccination campaigns, he regretted that they are mainly done by rich countries. "I think we are at 28 million doses of vaccine administered so far. Five different vaccines or platforms were used, "said Dr. Ryan.
"Approximately 46 countries are currently vaccinating, but there is only one low income," said Michael Ryan, while 38 of these 46 are rich countries.
"There are populations who want and need vaccines and who are not going to receive them unless, and until, we share better," he said, adding "everyone must do more".
WHO and the Vaccine Alliance (Gavi) set up the Covax mechanism to distribute COVID vaccines to underprivileged countries, but the system suffers from a beggar-thy-neighbor tendency of rich countries and a lack of funding.
WHO's goal is to deliver doses for up to 20% of the population of Covax participating countries by the end of the year. The UN agency hopes to send the first vaccines in late January or February.
The pandemic, which broke out in China at the end of 2019, killed 1,964,557 people, according to a report established by AFP on Wednesday from official sources.
Tags according, countries, doses, injected, million, Vaccines
Entertainment UK
Madonna 'visits 5 countries in 3 weeks' for winter break amid Covid-19 pandemic
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Madonna has reportedly been in five countries in just three weeks for her winter holiday amid the coronavirus pandemic.
The singer seemingly hasn't let the worldwide pandemic slow her down as she jetted all around the globe.
It's claimed the songstress, 62, jetted from Los Angeles to London with her boyfriend Ahlamalik Williams, 26, with four of her children – David Banda, Mercy James, twins Estere and Stella – on Christmas Eve.
In London, they picked up Madonna's UK-based son Rocco, 20, whom she shares with Guy Ritchie.
The famous family then made the trip to Malawi via a stop over in Egypt on December 28.
(Image: madonna/Instagram)
They met President Lazarus Chakwera and the locals as well as taking a trip to the hospital Madonna founded and Jacaranda School for Orphans near Blantyre during their week in Malawi.
Four of her six children were adopted from Malawi.
The group touched down in Kenya on Wednesday where they have spent time with the Pokot Tribe in the Baringo Valley.
A source close said Madonna and her family have been taking regular Covid-19 tests, according to The Sun.
Madonna's family were understood to have jetted around via private jet.
Also her personal photographer Ricardo Gomes was on hand to capture her magnificent breaks in Malawi and Kenya.
Madonna has been sharing a glorious slew of sun-soaked snaps from her holiday on Instagram with her 15.6million fans.
Travellers arriving in Egypt, Malawi and Kenya must present a negative Covid-19 test taken up to 72 hours before.
"High-value business travellers" didn't have to quarantine in the UK "subject to specific criteria being met" in December.
The Mirror reached out to Madonna's representative for further comment.
Tags break, countries, Covid, madonna, visits, weeks, winter
UAE-Qatar ties: Countries need to rebuild trust to solve issues, Gargash says
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UAE draft law on human rights being framed, Gargash says
Anwar Gargash
Image Credit: AFP
Dubai: The UAE has ended its differences with Qatar but still there is a need to rebuild trust, Dr. Anwar Mohammed Gargash, UAE Minister of State for Foreign Affairs said on Sunday.
"There are pending issues with Qatar that still need cooperation to be solved, however, the reopening of airspace to Qatar is a positive step to rebuild trust," Gargash said.
He said the joint Gulf action has returned to its normal path.
On UAE's relations with Turkey, Gargash said the UAE, the largest trade partner of Turkey in the Arab region, seeks to have normal ties with Ankara.
"Turkey's support for the Muslim Brotherhood reflects negatively on its relations with Arab countries," he said, adding Turkey needs to reconsider its relations with the Muslim Brotherhood.
"There is complete coordination between the UAE, Saudi Arabia and Egypt with regards to the regional issues, and we support diplomatic solution to the Iranian dossier," Gargash said.
He referred to threats to Saudi Arabia's security and said: "Saudi Arabia's security lies within the UAE's security, and we cannot ignore missile attacks against the Kingdom."
Gargash reiterated the UAE's trust in the Saudi role in solving inter-differences.
"The UAE enhances its position as an attractive hub for development and adopts an approach for interfaith tolerance," he said.
Tags countries, gargash, issues, qatar, rebuild, solve, trust
The countries that would lose 50 percent of their population in the coming years due to an unexpected global demographic decline | The State
No Comments on The countries that would lose 50 percent of their population in the coming years due to an unexpected global demographic decline | The State
The most recent statistic from The World Bank points out that today almost 7.6 billion people share this world. In general, the population growth trend has been on the rise for many decades; for example, in 1950 there were some 2,600 million people in the world according to the United Nations. But scientists have observed that that trend will reverse in the coming years.
In fact, some countries would lose 50 percent of their population in the coming years due to an unexpected global demographic decline.
Research by the Institute for Health Metrics and Evaluation (IHME) at the University of Washington advances that In 40 years, the world population will reach a maximum of 9.7 billion people, but it will gradually decline to reach 8.8 billion by the year 2100..
The implications of this decline that will challenge historical population trends are many, from difficulties in keeping economies active until geopolitical and environmental changes. For example, by 2035, in just 14 years, experts anticipate that China will become the largest economy in the world, but after 2050 its population growth will decline to 51 percent less.
Research published in The Lancet predicts that some countries would lose up to half of their current population in the next 80 years, particularly Spain, Italy, Portugal, Japan and South Korea, which would imply important challenges in their economies, such as a reduction in the workforce and an increase in their pension system.
"Our findings suggest that continuing trends in female educational attainment and access to contraception will accelerate declines in fertility and slow population growth", The authors write.
Why is it not so convenient to reduce the population?
Although a decline in world population could have "Positive implications for the environment, climate change and food production", the effects on the economy can be "negative for the workforce, economic growth and social support systems in the parts of the world with the greatest fertility declines ".
The IHME experts assure that the population reduction implies "economic, fiscal and geopolitical risks" that some nations are already anticipating. For example, "geopolitical power is also linked to military might and, for now at least, armies require individuals to serve in them"They write.
Parallel, the researchers suggest that countries with lower fertility rates have four options: "create an enabling environment for women to have children and pursue their careers; restrict access to reproductive health services; increase participation in the labor force, especially in older ages; and promote immigration ".
That is why if nations like Canada, Australia and the United States manage to maintain "their long-term working-age population through migration, they would do well".
"Policy options for Adapting to continued low fertility, while maintaining and improving female reproductive health, will be crucial for years to come"Conclude the authors of the research funded by the Bill and Melinda Gates Foundation.
Tags coming, countries, Curious news, curious notes, Curious Science News, percent, Population, population growth, their, would, years
Pope Francis urges countries to strive for peace as he leads just 200 masked faithful
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Pope Francis urges countries to strive for peace as he leads just 200 masked faithful at an early evening Covid-restricted 'midnight' mass in St Peter's Basilica
Pope celebrated the mass in St Peter's Basilica, Vatican City, on Christmas Eve
Mostly employees of the state attended papal ceremony whilst wearing masks
Mass was held at 7.30pm instead of midnight to meet Italy's curfew rules
By Nicole Conner For Mailonline
Published: 23:41 GMT, 24 December 2020 | Updated: 02:01 GMT, 25 December 2020
Pope Francis has led a smaller midnight mass than in previous years as he urged peace to masked believers.
The spiritual leader of 1.3billion Catholics across the globe celebrated Christmas Eve mass in St Peter's Basilica.
Fewer than 200 people, wearing face coverings, attended the mass, and they were mostly employees of the tiny state of Vatican City.
Christmas during the pandemic has started early as Pope Francis (pictured) leads midnight mass to masked followers
Fewer than 200 people, wearing face coverings, attended the mass, and they were mostly employees of the tiny state of Vatican City (pictured)
The mass, traditionally held at midnight, had been moved forward by two hours to 7.30pm to meet Italy's curfew rules.
During the message, he expressed his desire to visit crisis-hit Lebanon and urged political leaders in South Sudan to continue working for peace.
Before the pandemic hit, several thousand believers and tourists had obtained precious tickets to attend the papal mass.
But on Christmas Eve, St Peter's Square, which is usually filled with people, was deserted.
Tough new coronavirus restrictions were imposed on Thursday over the Christmas and New Year period across Italy, the country hardest hit by the virus in Europe.
The country had nearly 71,000 deaths and more than two million cases since the beginning of the pandemic.
The mass, traditionally held at midnight, had been moved forward by two hours to 7.30pm to meet Italy's curfew rules
Pope Francis, who just celebrated his 84th birthday, will address his eighth Christmas message 'Urbi et orbi' ('to the city and the world') on Friday by video
Pope Francis, who just celebrated his 84th birthday, will address his eighth Christmas message 'Urbi et orbi' ('to the city and the world') on Friday by video.
He will do it from the apostolic palace to prevent a crowd from gathering in St Peter's Square.
Christmas celebrations across the world have been changed due to the pandemic this year.
Germany has been forced to cancel its famous Christmas markets, while in Kuwait, churches were closed until January 10.
In Belgium, residents are largely limited to welcoming a single visitor.
While in the Philippines some are choosing to spend the holidays alone because of the risk of catching the virus on public transport
The Christmas Eve mass at the Church of the Nativity in Bethlehem will be closed to the public this year and broadcast online.
Only clergy and select individuals allowed inside the basilica, which was sterilised earlier Thursday ahead of the service.
Tags #christmas, countries, dailymail, francis, leads, masked, News, peace, Pope Francis, strive, urges
Headlines UK Liverpool London
Singapore and The Philippines are the latest countries to ban UK flights
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Two more countries have banned incoming UK flights ahead of Christmas Eve after the identification of a new Covid strain in Britain.
Singapore will ban entry to UK travellers from Wednesday night. Health officials in the Philippines say they will also suspend flights from the UK from Christmas Eve, until December 31.
The rules for UK travel to Singapore and The Philippines
Singapore will ban entry to UK travellers from 23.59pm Wednesday night. The ban will include anyone who has been in the UK in the previous 14 days and transit through the country.
Singapore Citizens and Permanent Residents are still able to enter Singapore, but must undergo COVID-19 tests upon arrival in Singapore and at the end of their mandatory 14 day quarantine period.
The Philippines has temporarily suspended flights from the UK to the Philippines starting at 12:01am on 24 December until 31 December 2020.
Passengers not arriving directly from the UK who have been in the UK within 14 days prior to arrival to the Philippines, including those just transiting through the Philippines, are prohibited from entering the country during this period.
It brings the total number of countries restricting entry to UK travellers to 55.
The latest bans could also have a wider impact on British travellers, with Singapore a popular stop-off for those flying from the UK to Australia.
Singapore health officials say transit from the UK through the Southeast Asian city-state will not be allowed under the new travel ban.
The ministry of health website said: 'All long-term pass holders and short-term visitors with recent travel history to the UK within the last 14 days will not be allowed entry into Singapore, or transit through Singapore.
'This will also apply to all those who had obtained prior approval for entry into Singapore.'
Transit through the Philippines for UK travellers is also banned.
It comes after Greece and Cyprus both said they will require passengers from the UK to have three separate Covid tests to gain entry to stem the spread of a mutant strain of the virus – while all other EU member states have slammed their doors on Britain.
Both nations require evidence of a negative PCR (polymerase chain reaction) test prior to arrival, followed by a rapid lateral test after landing and then a third after a period of self-isolation.
PCR tests need to be processed in a lab, while lateral flow tests can deliver results in 15 minutes but are less accurate.
This is particularly the case when people administer the test themselves, as they often do not push the swab deep enough to get enough of a sample.
Singapore (pictured: Passengers at Jewel Changi airport in Singapore) will ban entry to UK travellers from Wednesday night. Health officials in the Philippines have also announced plans to suspend flights from the UK from Christmas Eve.
Greece and Cyprus are the only EU members not to have imposed blanket bans on UK travellers.
Today the European Commission advised all other 25 states to lift their embargo on trains, planes and lorries from Britain to avoid further disruption.
Tough new measures adopted by Greece and Cyprus
Travellers from Britain will have to meet the following requirements to enter both countries.
Evidence of a negative PCR test before arrival;
Rapid lateral flow test on arrival followed by 10 days of self-isolation;
A second PCR test afterwards.
PCR on arrival, followed by seven days in self-isolation;
However, as the Commission's statement is only advisory, individual countries still have the option of continuing with their own blanket bans.
European Commissioner for Justice Didier Reynders said: 'Given the current uncertainties and in light of the precautionary principle, member states should take co-ordinated action to discourage non-essential travel between the UK and the EU.
'At the same time, blanket travel bans should not prevent thousands of EU and UK citizens from returning to their homes.
'While precautions are needed to contain the spread of the new coronavirus variant, with today's recommendation, we therefore ensure that the restrictions are co-ordinated and provide for the necessary exemptions for citizens and residents returning home and other essential travellers.'
EU Transport Commissioner Adina Valean added: 'Within the EU, it is crucial that transport workers are exempted from any restrictive measures, as quarantine and testing.'
Emanuel Macron was the first EU leader to ban travellers from the UK on Sunday, putting in place a 48-hour embargo on all travel. This included freight lorries at Dover, causing disruption to cross-Channel trade – including fresh food supplies.
Other EU countries quickly halted UK flights, as data suggested the new variant could be 70% more infectious. Dozens of countries around the world followed Europe's lead – but the US is allowing British visitors with a negative Covid test.
People queue to enter the departures area at Heathrow Airport in chaotic scenes yesterday
Which countries have banned flights from the UK?
All flights from UK banned –
Austria, Belgium, Bulgaria, Croatia, Czech Republic, Denmark, Estonia, Finland, France, Germany, Ireland, Italy, Latvia, Lithuania, Luxembourg, Malta, Netherlands, Poland, Romania, Slovakia, Slovenia, Sweden.
UK travellers allowed with a negative test –
Cyprus, Greece – three negative tests required plus self-isolation.
Still allowing own nationals to enter –
Hungary, Portugal, Spain.
Norway, Switzerland, Turkey,
Argentina, Canada, Colombia, Chile, Dominican Republic, Ecuador, El Salvador, Grenada, Hong Kong, India, Israel, Iran, Jamaica, Kuwait, Morocco, Oman, Pakistan, Peru, Russia, Saudi Arabia, Sudan, Philippines and Singapore.
Liberia, US.
The EU Commission announced the outcome of its talks early this afternoon, after countries including France and Germany had repeatedly called for a united approach across the whole bloc.
Countries closed their doors to Britain after Mr Johnson announced on Saturday that the new variant was up to 70 per cent more transmissible than the original strain as he put London and parts of the South East and East of England into a two-week Christmas lockdown, with nearly 18 million people in a new Tier 4.
Dozens of nations outside Europe have also blocked flights from Britain, including Canada, Russia, India, Saudi Arabia and Argentina.
Health Secretary Matt Hancock admitted the new variant coronavirus was 'out of control' and said the new restrictions may have to remain in place for months.
Concerns about the rapid spread of the disease were underlined with the publication of the latest official figures showing there had been a further 35,928 lab-confirmed cases of coronavirus in the UK as of 9am on Sunday.
Millions of people have been forced to tear up their festive plans, with Mr Johnson effectively cancelling Christmas for those in Tier 4.
The Europe-wide flight ban was evident at Heathrow's Terminal Two where just two flights to the Continent had been listed on the departure board.
With more than 40 countries banning UK visitors over fears of the fast-spreading mutant strain of Covid 19 a morning and afternoon flight to the Portuguese capital Lisbon was the only European destination still open to travellers.
But even those flights were closed to UK tourists with only Portuguese nationals allowed on to board.
Passengers also had to present paperwork to show they had a negative Covid 19 test before they could check their bags.
The scenes of chaos with hundreds of people desperately trying to leave Heathrow before European countries closed their borders had given way to business as usual for staff.
Large queues of people trying to travel to the EU yesterday as travel bans came into force
Passengers at London's Heathrow Airport attempted to make the last flight to Dublin last night before the Covid-19 travel ban
PCR vs lateral flow Covid tests: Chaos as UK and France clash over type of testing used for truckers
The French government is demanding that any travellers from the UK, including truckers, take PCR tests before arriving in the country, which can take up to three days to return a result.
WHAT IS THE DIFFERENCE BETWEEN A PCR TEST AND A LATERAL FLOW?
A PCR test can cost upwards of £180 per person, with the swab needing to be processed in a lab.
The UK, on the other hand, favours faster tests which are not lab based and give a result within 15 minutes.
These rapid coronavirus tests, known as lateral flow tests, are ones that can be done on the spot using portable equipment.
They are faster and cheaper than lab-based PCR tests, which the government uses to diagnose people, but are less accurate.
LATERAL FLOW TESTS ARE RAPID – BUT CAN SACRIFICE ACCURACY
In a lateral flow test a swab is used to get a sample from the person's nose or throat and it is then processed in a small machine that tries to detect the coronavirus by mixing the sample with something the virus would react with.
If there is a reaction in the mixture it suggests that the person is carrying coronavirus. If not, they get a negative result. This process can be completed in as little as 15 minutes.
You take your own swab though a professional on site processes it through the machine.
However, as the swabs are often taken by people themselves, the accuracy of the test could be hampered as they may not push the swab deep enough to get enough of a sample.
Results from trials have varied wildly and show the tests perform better when the swabs are done by trained medics and worse when people do them themselves.
PCR TESTS CAN TAKE SEVERAL DAYS TO GET RESULTS – BUT ARE MORE ACCURATE
These lateral flow tests differ from the gold standard PCR test – known scientifically as polymerase chain reaction testing.
PCR tests also use a swab but this is then processed using high-tech laboratory equipment to analyse the genetic sequence of the sample to see if any of it matches the genes of coronavirus.
This is a much more long-winded and expensive process, involving multiple types of trained staff, and the analysis process can take hours, with the whole process from swab to someone receiving their result taking days.
It is significantly more accurate, however. In ideal conditions the tests are almost 100 per cent accurate at spotting the virus, although this may be more like 70 per cent in the real world.
This compares to a much lower sensitivity in lateral flow tests, with a trial of one type used in Liverpool suggesting they miss around 50 per cent of the people who would test positive with PCR.
SO, WHAT IS THE BENEFIT OF LATERAL FLOW TESTING?
Extreme accuracy may be a drawback for PCR now that so many people have been infected, however, with the tests able to detect shreds of the virus in people who recovered weeks ago and are no longer infectious, which may lead them to have to self-isolate unnecessarily.
Lateral flow tests are more likely to miss people who are carrying the virus but, experts say, do have value as a way of weeding out people carrying large amounts of the virus and therefore most likely to be spreading the disease.
All airlines had contacted passengers to inform them their flights were cancelled and not to turn up at the airport.
The only queues were at the Emirates Airline check in desk with more than 300 people waiting patiently in line to fly to Dubai.
Many families were in the long line as airline staff checked passports and paperwork to prove they had undergone a nose and throat PCR swab test for Covid 19 and returned a negative result.
Mother-of-two Alison Chambers was heading for a Christmas break with her two sons and husband.
She said: 'I don't mind how long I have to queue up, I'm just grateful that we can still get away. Fortunately, Dubai is still accepting tourists provided they have had a negative test. It is such a relief.'
Another passenger added:' Given what is happening in this country I feel very luck to be leaving. '
The departure boards at Terminal Two told the story of the Europe wide flight ban with the words 'cancelled' or check with airline' alongside destinations such as Nice, Paris and Zurich.
Virgin Atlantic, United Airlines and Delta were still operating transatlantic flights to the US.
Only US citizens and holders of a green card, or permanent residents, have been allowed to travel since the start of the Pandemic.
Despite the Governor of New York calling for all flights from the UK to be banned from landing at JFK airport Virgin operated their morning flight as usual.
Staff at the check in said passengers were not asked to present a negative Covid test, but a member of staff said:' That is for now. It is changing all the time.'
US citizens returning the New York have to quarantine for 14 days on arrival.
Passengers on a Virgin flights to Los Angeles, Delta to Atlanta and a United flight to Washington DC were not required to have undergone a test before boarding.
Other airlines operating out of Terminal Two, such as Middle Eastern Airlines, required negative tests before passengers were allowed to check in and pass-through security.
At Terminal 5, the home of British Airways, all passengers on the two flights to New York's JFK Airport had to have undergone a test to show they were not carrying the virus.
Those who had not previously had a test were able to book a short notice PCR swab test for £99.00. Results were delivered within 80 minutes.
A saliva test was also available as well as an antigen test to show if a passenger had previously had the virus and developed antibodies.
Passengers on other BA flights to the US, including Miami, Florida and Dallas in Texas were not required to have had any test.
New York Governor Andrew Cuomo had demanded that flights from the UK be halted – but had his request rejected by the federal government.
British Airways and Delta voluntarily agreed that passengers on their flights would be tested prior to departure.
Virgin will insist on a test from Christmas Eve and the rule will apply to code share airline Delta.
The on-site test for BA passengers was being operated by a private firm called Collinson.
The company were also operating a test centre on the ground floor at Terminal Two where a small queue had formed by mid-morning.
Many of those waiting for an appointment were taking part in the recently launched 'test to release' scheme which allowed visitors to quarantine for only five days instead of the 14 days imposed by the Government.
A British Airways spokesman said:' We have agreed to pre-departure testing for passengers on flights from the UK to New York from December 22nd in line with the request. We continue to work closely with local health authorities around the world.'
The French government has pledged to 'resume movement' as soon as possible, with the Port of Dover saying inbound lorries are now coming into the UK
After a pointless press conference on Monday evening, Boris Johnson (pictured left) is said to be drawing up plans to send extra testing capacity to the Port of Dover in a bid to end chaos brought about by France's travel ban. Emmanuel Macron (pictured right) is expected to announce his plan to end the travel ban later
Boris Johnson 'is convinced Macron imposed ban to pile on pressure over Brexit'
Downing Street was 'incandescent' when France announced its UK travel ban on Sunday night and is convinced Emmanuel Macron is trying to use the crisis to force Britain to cave in during Brexit trade talks, it was claimed today.
Number 10 yesterday publicly dismissed suggestions that there may be a Brexit element to Mr Macron's decision to ban all traffic and travellers from Britain because of the new mutant strain of coronavirus.
But privately officials and ministers are said to believe Mr Macron is at least in part motivated by a desire to pile the pressure on the UK as trade negotiations go down to the wire.
A transport industry source told The Times that officials in Number 10 were 'incandescent' when the ban was announced and that they were caught off guard because they thought it would only apply to passengers and not to freight as well.
Meanwhile, Kent has been turned into a giant car park today with up to 1,500 lorries now filling the motorway, side streets and laybys in Dover – as No10 sources accuse Paris of using the travel ban to strong-arm Brexit talks and trapped French drivers slam Emmanuel Macron for 'abandoning them in a foreign country'.
France announced a travel ban on arrivals from the UK on Sunday night after the identification of a new Covid-19 strain in the south east of England. More than 1,500 lorries are now backed up in Kent, unable to make the crossing, leaving drivers spending a second night sleeping in their cabs.
Photos show 873 lorries parked at the disused Manston Airfield – which was previously reserved for a No Deal Brexit.
The 48 hour closure is due to end at 11pm UK time tonight but it has triggered a raft of similar moves with more than 40 countries banning flights from the UK.
The controversial ban has now come under fire with Boris Johnson said to be convinced French President Macron is using the crisis to force Britain to cave in during Brexit trade talks.
Though Downing Street publicly insists the port crisis is nothing to do with Brexit, officials and ministers are privately convinced Macron is in part motivated by a desire to pile the pressure on the UK as trade negotiations go down to the wire.
The French ban caught the UK government completely by surprise, it has been claimed, something Transport Minister Grant Shapps alluded to yesterday when he described the move as 'slightly surprising'.
Tags CORONAVIRUS, countries, dailymail, European Union, Flights, France, Greece, latest, News, philippines, Singapore, Spain
Abu Dhabi update: Passengers from these countries don't need to quarantine
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Image Credit: Supplied
Abu Dhabi: Passengers from select countries will not be required to undergo quarantine, according to a statement from Abu Dhabi Media Office.
As of December 22, 2020, citizens from 'Green Countries' that do no not have to quarantine upon arrival to Abu Dhabi include:
"Travellers from green countries will be allowed into the emirate when they receive a negative test result, while travellers from other countries must self-isolate for 10 days. Country classifications will be reviewed every two weeks," said the media office.
All other passengers will need to undertake a PCR test on the sixth day if they are staying six or more consecutive days, as well as on day 12 if they are staying for 12 or more consecutive days.
Travel Procedures
Before making any travel arrangements to Abu Dhabi, travellers are urged to carry out the following procedures:
Contact your airline for important travel information and familiarise yourself with the guidelines and regulations set by relevant authorities.
Ensure you meet entry visa requirements to visit the UAE.
If you are travelling with Etihad Airways, you'll be covered by Etihad's COVID-19 global wellness insurance cover. Cover is included with every Etihad Airways ticket for travel up to March 31, 2021.
Ahead of departure
Travellers to Abu Dhabi will be required to present a valid negative COVID-19 PCR test received within 96 hours prior to their scheduled departure.
Airport arrival
Upon entering the emirate, travellers will be given a PCR test at the airport and must wait for the negative result to be received. International travellers from countries on the green list will no longer be required to self-isolate upon receiving a negative result. Travellers from other countries will be required to self-isolate for 10 days only.
All travellers will need to take a PCR test on day six for those staying for six days or more, and on day 12 for those staying for 12 days or more.
Departure from Abu Dhabi
Ensure you are familiar with the travel and health requirements of the destination you are headed to.
Tags countries, dhabi, passengers, quarantine, these, update
More than 30 countries have limited flights with the UK due to a new strain of coronavirus
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Hi Christopher Robin!
It has already been found in patients in Denmark, Australia, the Netherlands and Italy.
On December 20, the UK government tightened coronavirus restrictions in London and the southeast and east of England due to a new strain of coronavirus. According to medical experts, it spreads 70% faster than usual. Sky News reports that the new strain has already been identified in cases in Denmark, Australia, the Netherlands, South Africa and Italy. Countries have begun to restrict air traffic with the UK, writes the BBC.
The list of states that have closed borders for Great Britain is as follows:
– France suspended any communication with the UK, including through the Eurotunnel, for 48 hours from midnight on December 20;
– Germany stopped flights with the UK until December 31;
– Italy closed flights with the UK until January 6, 2021;
– Ireland suspended flights for at least 48 hours starting December 20;
– Canada banned flights from the UK for 72 hours starting December 21;
– Belgium suspended air and rail services from the UK for at least 24 hours starting December 20;
– Netherlands banned all flights from the UK until January 1, 2021;
– Turkey restricted air traffic with Great Britain, Denmark and the Netherlands until further notice;
– Switzerland suspended flights with the UK and South Africa until further notice;
– Bulgaria closed flights with the UK until January 31, 2021;
– government India decided to ban flights with the UK from 21 December to 31 December;
– Denmark, Norway and Croatia banned flights from the UK for 48 hours starting December 21;
– Austria restricted air traffic with Great Britain until further notice;
– Lithuania Latvia and Estonia closed flights with the UK until January 31, 2021;
– Saudi Arabia suspends air traffic with all countries for a week, starting from December 21;
– Israel denied entry to foreign nationals arriving from the UK, Denmark and South Africa until further notice;
– Spain announced that it would restrict flights if the EU does not come up with a common decision on the UK;
– Portugal will accept only those arrivals from the UK who have local citizenship;
– Salvador will not accept travelers. who have been in the UK for the past month;
– Czech will isolate anyone who has been in the UK for at least 24 hours in transit;
– Colombia, Morocco, Chile, Finland, Hong Kong and Argentina also banned flights from the UK;
– Sweden will announce restrictive measures shortly.
Tags AMERICA, Asia, CORONAVIRUS, countries, covid-19, Europe, Flights, Great Britain, Limited, pandemic, strain, The EU, travels

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