%%% -*-latex-*-
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-%%% $Id: wrestlers.tex,v 1.5 2002/01/13 14:55:31 mdw Exp $
+%%% $Id: wrestlers.tex,v 1.6 2003/07/13 11:16:27 mdw Exp $
%%%
%%% Description of the Wrestlers Protocol
%%%
%%%----- Revision history ---------------------------------------------------
%%%
%%% $Log: wrestlers.tex,v $
+%%% Revision 1.6 2003/07/13 11:16:27 mdw
+%%% Much documentation progress.
+%%%
%%% Revision 1.5 2002/01/13 14:55:31 mdw
%%% More incomplete stuff.
%%%
%%% Initial versions of documentation.
%%%
-\newif\iffancystyle\fancystylefalse
+\newif\iffancystyle\fancystyletrue
\iffancystyle
\documentclass
[a4paper, article, 10pt, numbering, noherefloats, notitlepage]
{strayman}
\usepackage[palatino, helvetica, courier, maths=cmr]{mdwfonts}
- \usepackage[margin]{amsthm}
+ \usepackage[mdwmargin]{mdwthm}
+ \PassOptionsToPackage{dvips}{xy}
\else
- \documentclass[a4paper]{article}
- \usepackage{a4wide}
- \usepackage{amsthm}
+ \documentclass{llncs}
\fi
+\usepackage{mdwtab, mathenv, mdwlist, mdwmath, crypto}
\usepackage{amssymb, amstext}
-\usepackage{mdwtab, mathenv}
+\usepackage{tabularx}
+\usepackage{url}
+\usepackage[all]{xy}
\errorcontextlines=999
\showboxbreadth=999
\title{The Wrestlers Protocol: proof-of-receipt and secure key exchange}
\author{Mark Wooding \and Clive Jones}
-\bibliographystyle{alpha}
-
-\newtheorem{theorem}{Theorem}
-\renewcommand{\qedsymbol}{$\square$}
-
-\newcommand{\getsr}{\stackrel{\scriptscriptstyle R}{\gets}}
-\newcommand{\inr}{\in_{\scriptscriptstyle R}}
-\newcommand{\oracle}[1]{\mathcal{#1}}
-
-\newcommand{\expt}[2]{\mathbf{Exp}^{\text{\normalfont#1}}_{#2}}
-\newcommand{\adv}[2]{\mathbf{Adv}^{\text{\normalfont#1}}_{#2}}
-\newcommand{\ord}{\mathop{\operator@font ord}}
-\newcommand{\poly}[1]{\mathop{\operator@font poly}({#1})}
-\newcommand{\negl}[1]{\mathop{\operator@font negl}({#1})}
-\newcommand{\compind}{\stackrel{c}{\approx}}
+\bibliographystyle{mdwalpha}
\newcolumntype{G}{p{0pt}}
-
-\newcommand{\xor}{\oplus}
-\renewcommand{\epsilon}{\varepsilon}
-
-\newenvironment{program}
- {\begin{tabular}[C]{|G|}\hlx{hv[][\tabcolsep]}\begin{tabbing}}
- {\end{tabbing}\\\hlx{v[][\tabcolsep]h}\end{tabular}}
+\def\Nupto#1{\N_{<{#1}}}
+\let\Bin\Sigma
+\let\epsilon\varepsilon
+\let\emptystring\lambda
+\def\bitsto{\mathbin{..}}
+\turnradius{4pt}
+\def\fixme{\marginpar{FIXME}}
+\def\messages{%
+ \basedescript{%
+ \desclabelwidth{2.5cm}%
+ \desclabelstyle\pushlabel%
+ \let\makelabel\cookie%
+ }%
+}
+\let\endmessages\endbasedescript
\begin{document}
\maketitle
\begin{abstract}
- Fill this in later.
+ The Wrestlers Protocol\footnote{%
+ `The Wrestlers' is a pub in Cambridge which serves good beer and
+ excellent Thai food. It's where the authors made their first attempts at
+ a secure key-exchange protocol which doesn't use signatures.} %
+ is a key-exchange protocol with the interesting property that it leaves no
+ evidence which could be used to convince a third party that any of the
+ participants are involved. We describe the protocol and prove its security
+ in the random oracle model.
+
+ Almost incidentally, we provide a new security proof for the CBC encryption
+ mode. Our proof is much simpler than that of \cite{Bellare:2000:CST}, and
+ gives a slightly better security bound.
+
+ % I've not yet decided whose key-exchange model to use, but this ought to
+ % be mentioned.
\end{abstract}
\tableofcontents
\newpage
%%%--------------------------------------------------------------------------
\section{Introduction}
+\label{sec:intro}
% Some waffle here about the desirability of a key-exchange protocol that
% doesn't leave signatures lying around, followed by an extended report of
% the various results.
%%%--------------------------------------------------------------------------
-\section{A simple authentication protocol}
-% Present the basic Diffie-Hellman-based authenticator, and prove that an
-% authentication oracle is useless if the hash function has appropriate
-% properties.
-
-We begin by introducing a simple proof-of-identity protocol, based on the
-Diffie-Hellman key-exchange protocol \cite{Diffie:1976:NDC} and prove some of
-its useful properties.
-
-\subsection{Introduction}
-
-We start by selecting a security parameter $k$. We choose a cyclic group $G
-= \langle g \rangle$ with $q = |G|$ prime, in which the decision
-Diffie-Hellman problem is assumed to be hard \cite{Boneh:1998:DDP}; we
-require that $q$ is a $k$-bit number. Suitable groups include elliptic
-curves over finite fields and prime-order subgroups of prime fields.
-Throughout, we shall write the group operation as multiplication, and we
-shall assume that group elements are interchangeable with their binary
-representations.
-
-Alice can choose a private key $\alpha$ from the set $\{ 1, 2, \ldots, q - 1
-\}$ and publish her corresponding public key $a = g^\alpha$.
-
-A simplistic proof-of-identity protocol might then proceed as follows:
-\begin{enumerate}
-\item Bob chooses a random exponent $\beta$, also from $\{ 1, 2, \ldots, q -
- 1 \}$, and sends a \emph{challenge} $b = g^\beta$ to Alice.
-\item Alice computes her \emph{response} $b^\alpha$ and returns it to Bob.
-\item If Alice's response is equal to $a^\beta$ then Bob is satisfied.
-\end{enumerate}
-If Bob always plays by the rules then this protocol is evidently secure,
-since computing $g^{\alpha\beta}$ given $g^\alpha$ and $g^\beta$ is precisely
-the Diffie-Hellman problem.
-
-This protocol has a flaw, though: by using Alice as an oracle for the
-function $x \mapsto x^\alpha$, he could conceivably acquire information about
-Alice's private key which he could later use to impersonate her. As an
-example, Bob can submit $-g^\beta$ as a challenge: if the reply
-$(-g^\beta)^\alpha = g^{\alpha\beta}$, Bob can conclude that $\alpha$ is
-even.
-
-We fix the protocol by requiring that Bob prove that he already knows the
-answer to his challenge. We introduce a hash function $h$ whose properties
-we shall investigate later. The \emph{Wrestlers Authentication Protocol} is
-then:
-\begin{enumerate}
-\item Bob chooses a secret $\beta \getsr \{ 1, 2, \ldots, q - 1 \}$. He
- computes his \emph{challenge} $b \gets g^\beta$ and also a \emph{check
- value} $c \gets \beta \xor h(a^\beta)$. He sends the pair $(b, c)$.
-\item Alice receives $(b', c')$. She computes $r \gets b'^\alpha$ and
- $\gamma \gets c' \xor h(R)$. If $b' = g^\gamma$, she sends $r$ back as her
- \emph{response}; otherwise she sends the distinguished value $\bot$.
-\item Bob receives $r'$. If $r' = a^\beta$ then he accepts that he's talking
- to Alice.
-\end{enumerate}
-We show that the introduction of the check value has indeed patched up the
-weakness described above, and that the check value itself hasn't made an
-impersonator's job much easier.
-
-\subsection{Analysis in the random oracle model}
-
-Here, we prove that the Wrestlers Authentication Protocol is secure if $h$ is
-implemented by a public \emph{random oracle} \cite{Bellare:1993:ROP}.
-
-Fix a private key $\alpha \inr \{ 1, 2, \ldots, q - 1\}$ and the corresponding public
-key $a = g^\alpha$, and consider a probabilistic polynomial-time adversary
-$A$, equipped with two oracles. One is a random oracle $\oracle{H}\colon
-\{0, 1\}^* \to \{0, 1\}^k$, which implements a function $h$ randomly selected
-from the set of all functions with that signature; the other is an
-authentication oracle $\oracle{O}\colon G \times \{0, 1\}^k \to G \cup \{
-\bot \}$ defined by:
-\[
-\oracle{O}^(b, c) = \begin{cases}
- b^\alpha & if $b = g^{c \xor h(b^\alpha)}$ \\
- \bot & otherwise
-\end{cases}
-\]
-The authentication oracle will play the part of Alice in the following. We
-first show that access to this authentication oracle can't help an adversary
-learn how to impersonate Alice.
-
-\begin{theorem}
- The Wrestlers Authentication Protocol with random oracle is computational
- zero knowledge \cite{Brassard:1989:SZK}.
- \label{thm:wap-czk}
-\end{theorem}
-\begin{proof}
- \newcommand{\Hlist}{\textit{$\oracle{H}$-list}}
- \newcommand{\Hsim}{\textit{$\oracle{H}$-sim}}
- \newcommand{\Osim}{\textit{$\oracle{O}$-sim}}
- \renewcommand{\H}{\oracle{H}}
- \renewcommand{\O}{\oracle{O}}
- %
- Let $A$ be any probabilistic polynomial-time adversary with access to a
- random. We shall construct a probabilistic polynomial-time adversary $A'$
- without either oracle such that the probability distributions on the output
- of $A$ and $A'$ are equal. Specifically, $A'$ runs $A$ with simulated
- random and authentication oracles whose behaviour is computationally
- indistinguishable from the originals. The algorithm for $A'$ and the
- simulated oracles is shown in figure~\ref{fig:wap-czk-sim}.
-
- \begin{figure}
- \begin{program}
- \quad \= \kill
- Adversary $A'^{\H(\cdot)}$: \+ \\
- $\Hlist \gets \emptyset$; \\
- $r \gets A^{\Hsim(\cdot), \Osim(\cdot, \cdot)}$; \\
- \textbf{return} $r$; \- \\[\medskipamount]
- %
- Simulated random oracle $\Hsim(q)$: \+ \\
- \textbf{if} $\exists r : (q, r) \in \Hlist$ \textbf{then}
- \textbf{return} $r$; \\
- $r \getsr \H(q)$; \\
- $\Hlist \gets \Hlist \cup \{ (q, r) \}$; \\
- \textbf{return} $r$; \- \\[\medskipamount]
- %
- Simulated authentication oracle $\Osim(b, c)$: \+ \\
- \textbf{if} $\exists q, r : (q, r) \in \Hlist \wedge
- b = g^{c \xor r} \wedge
- q = a^{c \xor r}$ \textbf{then}
- \textbf{return} $a^{c \xor r}$; \\
- \textbf{else} \textbf{return} $\bot$;
- \end{program}
- %
- \caption{The algorithm and simulated oracles for the proof of
- theorem~\ref{thm:wap-czk}.}
- \label{fig:wap-czk-sim}
- \end{figure}
-
- Firstly we show that $A'$ runs in polynomial time. The only modifications
- to $\Hlist$ are in $\Hsim$, which adds at most one element for each oracle
- query. Since $\Hlist$ is initially empty and $A$ can only make $\poly{k}$
- queries to $\Hsim$, this implies that $|\Hlist|$ is always
- polynomially-bounded. Each query can be answered in polynomial time:
- indeed, the search implied by the test $\exists r : (q, r) \in \Hlist$ can
- be performed in $O(\log q)$ time by indexing $\Hlist$ on $q$ using a radix
- tree. Now, since $|\Hlist|$ is polynomially bounded, the test $\exists q,
- r : (q, r) \in \Hlist \wedge b = g^{c \xor r} \wedge q = a^{c \xor r}$ at
- the start of $\Osim$ runs in polynomial time even though it requires an
- exhaustive search of the history of $A$'s queries to its simulated random
- oracle.
-
- Next, we examine the behaviour of the simulated oracles. Since $\Hsim$ is
- implemented in terms of the public random oracle $\H$, and returns the same
- answers to queries, its probability distribution must be identical.
-
- We turn our attention to the simulation of the authentication oracle.
- Consider a query $(b, c)$ made by $A$ to its authentication oracle.
-
- Firstly, suppose that there is a $\beta$ such that $b = g^\beta$ and $c =
- \beta \xor r$ where $r$ is the response to a previous query made by $A$ to
- its random oracle on $a^\beta$. Then the true authentication oracle $\O$
- is satisfied because $b = g^\beta = g^{c \xor r} = g^{c \xor h(a^\beta)} =
- g^{c \xor h(b^\alpha)}$, and returns $b^\alpha$. Similarly, the simulated
- authentication oracle $\Osim$ will find the pair $(a^\beta, r) \in \Hlist$,
- recover $\beta = c \xor r$, confirm that $b = g^\beta$, and return
- $a^\beta$ as required. These events all occur with probability 1.
-
- Conversely, suppose that there is no such $\beta$. Then the simulated
- oracle $\Osim$ will reject the query, returning $\bot$, again with
- probability 1. However, a genuine authentication oracle $\O$ will succeed
- and return $b^\alpha$ with probability $2^{-k}$. To see this, note that $b
- = g^\gamma$ for some $\gamma \in \{ 0, 1, \ldots, q - 1 \}$, and the
- probability that $h(b^\alpha) = c \xor \gamma$ is precisely $2^{-k}$, since
- $c$ and $\gamma$ are $k$-bit strings. Since this probability is
- negligible, we have shown that the distributions of the simulated oracles
- $\Hsim$ and $\Osim$ are computationally indistinguishable from the genuine
- oracles $\H$ and $\O$. The theorem follows.
-\end{proof}
+\section{Preliminaries}
+\label{sec:prelim}
+% Here we provide definitions of the various kinds of things we use and make,
+% and describe some of the notation we use.
-Our next objective is to show that pretending to be Alice is hard without
-knowledge of her secret $\alpha$. We say that an adversary $A$
-\emph{impersonates} in the Wrestlers Authentication Protocol with probability
-$\epsilon$ if
-\[ \Pr[A^{\oracle{H}(\cdot)}(g^\alpha, g^\beta,
- \beta \xor h(g^{\alpha\beta}))
- = g^{\alpha\beta}]
- = \epsilon
-\]
-where the probability is taken over all choices of $\alpha, \beta \in \{ 1,
-2, \ldots, q - 1 \}$ and random oracles $\oracle{H}$.
-
-\begin{theorem}
- Assuming that the Diffie-Hellman problem is hard in $G$, no probabilistic
- polynomial-time adversary can impersonate in the Wrestlers Authentication
- Protocol with random oracle with better than negligible probability.
- \label{thm:wap-dhp}
-\end{theorem}
-\begin{proof}
- \newcommand{\Hlist}{\textit{$\oracle{H}$-list}}
- \newcommand{\Hqueries}{\textit{$\oracle{H}$-queries}}
- \newcommand{\Hsim}{\textit{$\oracle{H}$-sim}}
- \renewcommand{\H}{\oracle{H}}
- %
- We prove the theorem by contradiction. Given a polynomial-time adversary
- $A$ which impersonates Alice with non-negligible probability $\epsilon$, we
- construct an adversary which solves the Diffie-Hellman problem with
- probability no less than $\epsilon \poly{k}$, i.e.,
- \[ \Pr[A'(g^\alpha, g^\beta) = g^{\alpha\beta}] \ge \epsilon \poly{k}. \]
- Thus, if there is any polynomial-time $A$ which impersonates with better
- than negligible probability, then there is a polynomial-time $A'$ which
- solves Diffie-Hellman with essentially the same probability, which would
- violate our assumption of the hardness of Diffie-Hellman.
-
- The idea behind the proof is that the check value is only useful to $A$
- once it has discovered the correct response to the challenge, which it must
- have done by solving the Diffie-Hellman problem. Hence, our Diffie-Hellman
-
- \begin{figure}
- \begin{program}
- \quad \= \kill
- Adversary $A'(a, b)$: \+ \\
- $\Hlist \gets \emptyset$; \\
- $\Hqueries \gets \emptyset$; \\
- $r \getsr \{ 0, 1 \}^k$; \\
- $x \gets A^{\Hsim(\cdot)}(a, b, r)$; \\
- $c \getsr (\Hlist \cup \{ x \}) \cap G$; \\
- \textbf{return} $c$; \- \\[\medskipamount]
- %
- Simulated random oracle $\Hsim(q)$: \+ \\
- \textbf{if} $\exists r : (q, r) \in \Hlist$
- \textbf{then} \textbf{return} $r$; \\
- $r \gets \{ 0, 1 \}^k$; \\
- $\Hlist \gets \Hlist \cup \{ (q, r) \}$; \\
- $\Hqueries \gets \Hqueries \cup \{ q \}$; \\
- \textbf{return} $r$;
- \end{program}
- %
- \caption{The algorithm for the proof of theorem~\ref{thm:wap-dhp}}
- \label{fig:wap-dhp-rdc}
- \end{figure}
-
- The adversary $A'$ is shown in figure~\ref{fig:wap-dhp-rdc}. It generates
- a random check-value and runs the impersonator $A$.
-
- The simulated random oracle $\Hsim$ gathers together the results of all of
- the random oracle queries made by $A$. The final result returned by $A'$
- is randomly chosen from among all of the `plausible' random oracle queries
- and $A$'s output, i.e., those queries which are actually members of the
- group. The check value given to $A$ is most likely incorrect (with
- probability $1 - 2^{-k}$): the intuition is that $A$ can only notice if it
- actually computes the right answer and then checks explicitly.
-
- If $A$ does compute the correct answer $g^{\alpha\beta}$ and either returns
- it or queries $\Hsim$ on it, then $A'$ succeeds with probability $\epsilon
- / |(\Hlist \cup \{ x \}) \cap G| = \epsilon \poly{k}$, since $|\Hlist|$ is
- no greater than the number of random oracle queries made by $A$ (which must
- be polynomially bounded).\footnote{%
- This polynomial factor introduces a loss in the perceived security of the
- authentication protocol. It appears that this loss is only caused by the
- possibility of the adversary $A$ being deliberately awkward and checking
- the value of $c$ after having computed the answer. However, we can't see
- a way to improve the security bound on the scheme without imposing
- artificial requirements on $A$.} %
+\subsection{Bit strings}
+
+Most of our notation for bit strings is standard. The main thing to note is
+that everything is zero-indexed.
+
+\begin{itemize}
+\item We write $\Bin = \{0, 1\}$ for the set of binary digits. Then $\Bin^n$
+ is the set of $n$-bit strings, and $\Bin^*$ is the set of all bit strings.
+\item If $x$ is a bit string then $|x|$ is the length of $x$. If $x \in
+ \Bin^n$ then $|x| = n$.
+\item If $x, y \in \Bin^n$ are strings of bits of the same length then $x
+ \xor y \in \Bin^n$ is their bitwise XOR.
+\item If $x$ and $y$ are bit strings then $x \cat y$ is the result of
+ concatenating $y$ to $x$. If $z = x \cat y$ then we have $|z| = |x| +
+ |y|$.
+\item The empty string is denoted $\emptystring$. We have $|\emptystring| =
+ 0$, and $x = x \cat \emptystring = \emptystring \cat x$ for all strings $x
+ \in \Bin^*$.
+\item If $x$ is a bit string and $i$ is an integer satisfying $0 \le i < |x|$
+ then $x[i]$ is the $i$th bit of $x$. If $a$ and $b$ are integers
+ satisfying $0 \le a \le b \le |x|$ then $x[a \bitsto b]$ is the substring
+ of $x$ beginning with bit $a$ and ending just \emph{before} bit $b$. We
+ have $|x[i]| = 1$ and $|x[a \bitsto b]| = b - a$; if $y = x[a \bitsto b]$
+ then $y[i] = x[a + i]$.
+\item If $x$ is a bit string and $n$ is a natural number then $x^n$ is the
+ result of concatenating $x$ to itself $n$ times. We have $x^0 =
+ \emptystring$ and if $n > 0$ then $x^n = x^{n-1} \cat x = x \cat x^{n-1}$.
+\end{itemize}
+
+\subsection{Other notation}
+
+\begin{itemize}
+\item If $n$ is any natural number, then $\Nupto{n}$ is the set $\{\, i \in
+ \Z \mid 0 \le i < n \,\} = \{ 0, 1, \ldots, n \}$.
+\item The symbol $\bot$ (`bottom') is different from every bit string and
+ group element.
+\item We write $\Func{l}{L}$ as the set of all functions from $\Bin^l$ to
+ $\Bin^L$, and $\Perm{l}$ as the set of all permutations on $\Bin^l$.
+\end{itemize}
+
+\subsection{Algorithm descriptions}
+
+Most of the notation used in the algorithm descriptions should be obvious.
+We briefly note a few features which may be unfamiliar.
+\begin{itemize}
+\item The notation $a \gets x$ denotes the action of assigning the value $x$
+ to the variable $a$.
+\item The notation $a \getsr X$, where $X$ is a finite set, denotes the
+ action of assigning to $a$ a random value $x \in X$ according to the
+ uniform probability distribution on $X$; i.e., following $a \getsr X$,
+ $\Pr[a = x] = 1/|X|$ for any $x \in X$.
+\end{itemize}
+The notation is generally quite sloppy about types and scopes. In
+particular, there are implicit coercions between bit strings, integers and
+group elements. Any simple injective mapping will do for handling the
+conversions. We don't think these informalities cause much confusion, and
+they greatly simplify the presentation of the algorithms.
+
+\subsection{Random oracles}
+
+We shall analyse the Wrestlers Protocol in the random oracle model
+\cite{Bellare:1993:ROP}. That is, each participant including the adversary
+is given oracle access (only) to a uniformly-distributed random function
+$H\colon \Bin^* \to \Bin^\infty$ chosen at the beginning of the game: for any
+input string $x$, the oracle can produce, on demand, any prefix of an
+infinitely long random answer $y = H(x)$. Repeating a query yields a prefix
+of the same random result string; asking a new query yields a prefix of a new
+randomly-chosen string.
+
+We shan't need either to query the oracle on very long input strings nor
+shall we need outputs much longer than a representation of a group index.
+Indeed, since all the programs we shall be dealing with run in finite time,
+and can therefore make only a finite number of oracle queries, each with a
+finitely long result, we can safely think about the random oracle as a finite
+object.
+
+Finally, we shall treat the oracle as a function of multiple inputs and
+expect it to operate on some unambiguous encoding of all of the arguments in
+order.
+
+\subsection{Symmetric encryption}
+
+\begin{definition}[Symmetric encryption]
+ \label{def:sym-enc}
+ A \emph{symmetric encryption scheme} $\mathcal{E} = (E, D)$ is a pair of
+ algorithms:
+ \begin{itemize}
+ \item a randomized \emph{encryption algorithm} $E\colon \keys \mathcal{E}
+ \times \Bin^* \to \Bin^*$; and
+ \item a deterministic \emph{decryption algorithm} $E\colon \keys
+ \mathcal{E} \times \Bin^* \to \Bin^* \cup \{ \bot \}$
+ \end{itemize}
+ with the property that, for any $K \in \keys \mathcal{E}$, any plaintext
+ message $x$, and any ciphertext $y$ returned as a result of $E_K(x)$, we
+ have $D_K(y) = x$.
+\end{definition}
+
+\begin{definition}[Chosen plaintext security for symmetric encryption]
+ \label{def:sym-cpa}
+ Let $\mathcal{E} = (E, D)$ be a symmetric encryption scheme. Let $A$ be
+ any algorithm. Define
+ \begin{program}
+ Experiment $\Expt{lor-cpa-$b$}{\mathcal{E}}(A)$: \+ \\
+ $K \getsr \keys \mathcal{E}$; \\
+ $b' \getsr A^{E_K(\id{lr}(b, \cdot, \cdot))}$; \\
+ \RETURN $b'$;
+ \next
+ Function $\id{lr}(b, x_0, x_1)$: \+ \\
+ \RETURN $x_b$;
+ \end{program}
+ An adversary $A$ is forbidden from querying its encryption oracle
+ $E_K(\id{lr}(b, \cdot, \cdot))$ on a pair of strings with differing
+ lengths. We define the adversary's \emph{advantage} in this game by
+ \begin{equation}
+ \Adv{lor-cpa}{\mathcal{E}}(A) =
+ \Pr[\Expt{lor-cpa-$1$}{\mathcal{E}}(A) = 1] -
+ \Pr[\Expt{lor-cpa-$0$}{\mathcal{E}}(A) = 1]
+ \end{equation}
+ and the \emph{left-or-right insecurity of $\mathcal{E}$ under
+ chosen-plaintext attack} is given by
+ \begin{equation}
+ \InSec{lor-cpa}(\mathcal{E}; t, q_E, \mu_E) =
+ \max_A \Adv{lor-cpa}{\mathcal{E}}(A)
+ \end{equation}
+ where the maximum is taken over all adversaries $A$ running in time $t$ and
+ making at most $q_E$ encryption queries, totalling most $\mu_E$ bits of
+ plaintext.
+\end{definition}
+
+\subsection{The decision Diffie-Hellman problem}
+
+Let $G$ be some cyclic group. The standard \emph{Diffie-Hellman problem}
+\cite{Diffie:1976:NDC} is to compute $g^{\alpha\beta}$ given $g^\alpha$ and
+$g^\beta$. We need a slightly stronger assumption: that, given $g^\alpha$
+and $g^\beta$, it's hard to tell the difference between the correct
+Diffie-Hellman value $g^{\alpha\beta}$ and a randomly-chosen group element
+$g^\gamma$. This is the \emph{decision Diffie-Hellman problem}
+\cite{Boneh:1998:DDP}.
+
+\begin{definition}
+ \label{def:ddh}
+ Let $G$ be a cyclic group of order $q$, and let $g$ be a generator of $G$.
+ Let $A$ be any algorithm. Then $A$'s \emph{advantage in solving the
+ decision Diffie-Hellman problem in $G$} is
+ \begin{equation}
+ \begin{eqnalign}[rl]
+ \Adv{ddh}{G}(A) =
+ & \Pr[\alpha \getsr \Nupto{q}; \beta \getsr \Nupto{q} :
+ A(g^\alpha, g^\beta, g^{\alpha\beta}) = 1] - {} \\
+ & \Pr[\alpha \getsr \Nupto{q}; \beta \getsr \Nupto{q};
+ \gamma \getsr \Nupto{q} :
+ A(g^\alpha, g^\beta, g^\gamma) = 1].
+ \end{eqnalign}
+ \end{equation}
+ The \emph{insecurity function of the decision Diffie-Hellman problem in
+ $G$} is
+ \begin{equation}
+ \InSec{ddh}(G; t) = \max_A \Adv{ddh}{G}(A)
+ \end{equation}
+ where the maximum is taken over all algorithms $A$ which run in time $t$.
+\end{definition}
+
+%%%--------------------------------------------------------------------------
+
+\section{The protocol}
+\label{sec:protocol}
+
+The Wrestlers Protocol is parameterized. We need the following things:
+\begin{itemize}
+\item A cyclic group $G$ whose order~$q$ is prime. Let $g$ be a generator
+ of~$G$. We require that the (decision?\fixme) Diffie-Hellman problem be
+ hard in~$G$. The group operation is written multiplicatively.
+\item A symmetric encryption scheme $\mathcal{E} = (E, D)$. We require that
+ $\mathcal{E}$ be secure against adaptive chosen-plaintext attacks. Our
+ implementation uses Blowfish \cite{Schneier:1994:BEA} in CBC mode with
+ ciphertext stealing. See section~\ref{sec:cbc} for a description of
+ ciphertext stealing and an analysis of its security.
+\item A message authentication scheme $\mathcal{M} = (T, V)$. We require
+ that $\mathcal{M}$ be (strongly) existentially unforgeable under
+ chosen-message attacks. Our implementation uses RIPEMD-160
+ \cite{Dobbertin:1996:RSV} in the HMAC \cite{Bellare:1996:HC} construction.
+\item An instantiation for the random oracle. We use RIPEMD-160 again,
+ either on its own, if the output is long enough, or in the MGF-1
+ \cite{RFC2437} construction, if we need a larger output.\footnote{%
+ The use of the same hash function in the MAC as for instantiating the
+ random oracle is deliberate, with the aim of reducing the number of
+ primitives whose security we must assume. In an application of HMAC, the
+ message to be hashed is prefixed by a secret key padded out to the hash
+ function's block size. In a `random oracle' query, the message is
+ prefixed by a fixed identification string and not padded. Interference
+ between the two is then limited to the case where one of the HMAC keys
+ matches a random oracle prefix, which happens only with very tiny
+ probability.}%
+\end{itemize}
+
+An authenticated encryption scheme with associated data (AEAD)
+\cite{Rogaway:2002:AEAD, Rogaway:2001:OCB, Kohno:2003:CWC} could be used
+instead of a separate symmetric encryption scheme and MAC.
+
+\subsection{Symmetric encryption}
+
+The same symmetric encryption subprotocol is used both within the key
+exchange, to ensure secrecy and binding, and afterwards for message
+transfer. It provides a secure channel between two players, assuming that
+the key was chosen properly.
+
+A \id{keyset} contains the state required for communication between the two
+players. In particular it maintains:
+\begin{itemize}
+\item separate encryption and MAC keys in each direction (four keys in
+ total), chosen using the random oracle based on an input key assumed to be
+ unpredictable by the adversary and a pair of nonces chosen by the two
+ players; and
+\item incoming and outgoing sequence numbers, to detect and prevent replay
+ attacks.
+\end{itemize}
+
+The operations involved in the symmetric encryption protocol are shown in
+figure~\ref{fig:keyset}.
+
+The \id{keygen} procedure initializes a \id{keyset}, resetting the sequence
+numbers, and selecting keys for the encryption scheme and MAC using the
+random oracle. It uses the nonces $r_A$ and $r_B$ to ensure that with high
+probability the keys are different for the two directions: assuming that
+Alice chose her nonce $r_A$ at random, and that the keys and nonce are
+$\kappa$~bits long, the probability that the keys in the two directions are
+the same is at most $2^{\kappa - 2}$.
+
+The \id{encrypt} procedure constructs a ciphertext from a message $m$ and a
+\emph{message type} $\id{ty}$. It encrypts the message giving a ciphertext
+$y$, and computes a MAC tag $\tau$ for the triple $(\id{ty}, i, y)$, where
+$i$ is the next available outgoing sequence number. The ciphertext message
+to send is then $(i, y, \tau)$. The message type codes are used to
+separate ciphertexts used by the key-exchange protocol itself from those sent
+by the players later.
+
+The \id{decrypt} procedure recovers the plaintext from a ciphertext triple
+$(i, y, \tau)$, given its expected type code $\id{ty}$. It verifies that the
+tag $\tau$ is valid for the message $(\id{ty}, i, y)$, checks that the
+sequence number $i$ hasn't been seen before,\footnote{%
+ The sequence number checking shown in the figure is simple but obviously
+ secure. The actual implementation maintains a window of 32 previous
+ sequence numbers, to allow out-of-order reception of messages while still
+ preventing replay attacks. This doesn't affect our analysis.}%
+and then decrypts the ciphertext $y$.
+
+\begin{figure}
+ \begin{program}
+ Structure $\id{keyset}$: \+ \\
+ $\Xid{K}{enc-in}$; $\Xid{K}{enc-out}$; \\
+ $\Xid{K}{mac-in}$; $\Xid{K}{mac-out}$; \\
+ $\id{seq-in}$; $\id{seq-out}$; \- \\[\medskipamount]
+ Function $\id{gen-keys}(r_A, r_B, K)$: \+ \\
+ $k \gets \NEW \id{keyset}$; \\
+ $k.\Xid{K}{enc-in} \gets H(\cookie{encryption}, r_A, r_B, K)$; \\
+ $k.\Xid{K}{enc-out} \gets H(\cookie{encryption}, r_B, r_A, K)$; \\
+ $k.\Xid{K}{mac-in} \gets H(\cookie{integrity}, r_A, r_B, K)$; \\
+ $k.\Xid{K}{mac-out} \gets H(\cookie{integrity}, r_B, r_A, K)$; \\
+ $k.\id{seq-in} \gets 0$; \\
+ $k.\id{seq-out} \gets 0$; \\
+ \RETURN $k$;
+ \next
+ Function $\id{encrypt}(k, \id{ty}, m)$: \+ \\
+ $y \gets (E_{k.\Xid{K}{enc-out}}(m))$; \\
+ $i \gets k.\id{seq-out}$; \\
+ $\tau \gets T_{k.\Xid{K}{mac-out}}(\id{ty}, i, y)$; \\
+ $k.\id{seq-out} \gets i + 1$; \\
+ \RETURN $(i, y, \tau)$; \- \\[\medskipamount]
+ Function $\id{decrypt}(k, \id{ty}, c)$: \+ \\
+ $(i, y, \tau) \gets c$; \\
+ \IF $V_{k.\Xid{K}{mac-in}}((\id{ty}, i, y), \tau) = 0$ \THEN \\ \ind
+ \RETURN $\bot$; \- \\
+ \IF $i < k.\id{seq-in}$ \THEN \RETURN $\bot$; \\
+ $m \gets D_{k.\Xid{K}{enc-in}}(y)$; \\
+ $k.\id{seq-in} \gets i + 1$; \\
+ \RETURN $m$;
+ \end{program}
+
+ \caption{Symmetric-key encryption functions}
+ \label{fig:keyset}
+\end{figure}
+
+\subsection{The key-exchange}
+
+The key-exchange protocol is completely symmetrical. Either party may
+initiate, or both may attempt to converse at the same time. We shall
+describe the protocol from the point of view of Alice attempting to exchange
+a key with Bob.
+
+Alice's private key is a random index $\alpha \inr \Nupto{q}$. Her public
+key is $a = g^\alpha$. Bob's public key is $b \in G$. We'll subscript the
+variables Alice computes with an~$A$, and the values Bob has sent with a~$B$.
+Of course, if Bob is following the protocol correctly, he will have computed
+his $B$ values in a completely symmetrical way.
+
+There are six messages in the protocol, and we shall briefly discuss the
+purpose of each before embarking on the detailed descriptions. At the
+beginning of the protocol, Alice chooses a new random index $\rho_A$ and
+computes her \emph{challenge} $r_A = g^{\rho_A}$. Eventually, the shared
+secret key will be computed as $K = r_B^{\rho_A} = r_A^{\rho_B} =
+g^{\rho_A\rho_B}$, as for standard Diffie-Hellman key agreement.
+
+Throughout, we shall assume that messages are implicitly labelled with the
+sender's identity. If Alice is actually trying to talk to several other
+people she'll need to run multiple instances of the protocol, each with its
+own state, and she can use the sender label to decide which instance a
+message should be processed by. There's no need for the implicit labels to
+be attached securely.
+
+We'll summarize the messages and their part in the scheme of things before we
+start on the serious detail. For a summary of the names and symbols used in
+these descriptions, see table~\ref{tab:kx-names}. The actual message
+contents are summarized in table~\ref{tab:kx-messages}. A state-transition
+diagram of the protocol is shown in figure~\ref{fig:kx-states}. If reading
+pesudocode algorithms is your thing then you'll find message-processing
+procedures in figure~\ref{fig:kx-messages} with the necessary support procedures
+in figure~\ref{fig:kx-support}.
+
+\begin{table}
+ \begin{tabularx}{\textwidth}{Mr X}
+ G & A cyclic group known by all participants \\
+ q = |G| & The prime order of $G$ \\
+ g & A generator of $G$ \\
+ E_K(\cdot) & Encryption under key $K$, here used to denote
+ application of the $\id{encrypt}(K, \cdot)$
+ operation \\
+ \alpha \inr \Nupto{q} & Alice's private key \\
+ a = g^{\alpha} & Alice's public key \\
+ \rho_A \inr \Nupto{q} & Alice's secret Diffie-Hellman value \\
+ r_A = g^{\rho_A} & Alice's public \emph{challenge} \\
+ c_A = H(\cookie{cookie}, r_A)
+ & Alice's \emph{cookie} \\
+ v_A = \rho_A \xor H(\cookie{expected-reply}, r_A, r_B, b^{\rho_A})
+ & Alice's challenge \emph{check value} \\
+ r_B^\alpha = a^{\rho_B}
+ & Alice's reply \\
+ K = r_B^{\rho_A} = r_B^{\rho_A} = g^{\rho_A\rho_B}
+ & Alice and Bob's shared secret key \\
+ w_A = H(\cookie{switch-request}, c_A, c_B)
+ & Alice's \emph{switch request} value \\
+ u_A = H(\cookie{switch-confirm}, c_A, c_B)
+ & Alice's \emph{switch confirm} value \\
+ \end{tabularx}
+
+ \caption{Names used during key-exchange}
+ \label{tab:kx-names}
+\end{table}
+
+\begin{table}
+ \begin{tabular}[C]{Ml}
+ \cookie{kx-pre-challenge}, r_A \\
+ \cookie{kx-cookie}, r_A, c_B \\
+ \cookie{kx-challenge}, r_A, c_B, v_A \\
+ \cookie{kx-reply}, c_A, c_B, v_A, E_K(r_B^\alpha)) \\
+ \cookie{kx-switch}, c_A, c_B, E_K(r_B^\alpha, w_A)) \\
+ \cookie{kx-switch-ok}, E_K(u_A))
+ \end{tabular}
+
+ \caption{Message contents, as sent by Alice}
+ \label{tab:kx-messages}
+\end{table}
- It remains to show that if $A$ never queries its random oracle on
- $g^{\alpha\beta}$ then it cannot distinguish the random check value $r$
- from the correct one $\beta \xor \H(g^{\alpha\beta})$, and hence won't
- notice that $A'$ is lying to it. But this is obvious: $A$'s probability of
- guessing the random value that would have been returned had it actually
- queried $\Hsim$ on the input $g^{\alpha\beta}$ is equal to the probability
- of it guessing $\beta \xor r$ instead, since $\Hsim$ chooses its responses
- uniformly at random. This completes the proof.
-\end{proof}
+\begin{messages}
+\item[kx-pre-challenge] Contains a plain statement of Alice's challenge.
+ This is Alice's first message of a session.
+\item[kx-cookie] A bare acknowledgement of a received challenge: it restates
+ Alice's challenge, and contains a hash of Bob's challenge. This is an
+ engineering measure (rather than a cryptographic one) which prevents
+ trivial denial-of-service attacks from working.
+\item[kx-challenge] A full challenge, with a `check value' which proves the
+ challenge's honesty. Bob's correct reply to this challenge informs Alice
+ that she's received his challenge correctly.
+\item[kx-reply] A reply. This contains a `check value', like the
+ \cookie{kx-challenge} message above, and an encrypted reply which confirms
+ to Bob Alice's successful receipt of his challenge and lets Bob know he
+ received Alice's challenge correctly.
+\item[kx-switch] Acknowledges Alice's receipt of Bob's \cookie{kx-reply}
+ message, including Alice's own reply to Bob's challenge. Tells Bob that
+ she can start using the key they've agreed.
+\item[kx-switch-ok] Acknowlegement to Bob's \cookie{kx-switch} message.
+\end{messages}
+
+\begin{figure}
+ \small
+ \let\ns\normalsize
+ \let\c\cookie
+ \[ \begin{graph}
+ []!{0; <4.5cm, 0cm>: <0cm, 1.5cm>::}
+ *++[F:<4pt>]\txt{\ns Start \\ Choose $\rho_A$} ="start"
+ :[dd]
+ *++[F:<4pt>]\txt{
+ \ns State \c{challenge} \\
+ Send $(\c{pre-challenge}, r_A)$}
+ ="chal"
+ [] "chal" !{!L(0.5)} ="chal-cookie"
+ :@(d, d)[l]
+ *+\txt{Send $(\c{cookie}, r_A, c_B)$}
+ ="cookie"
+ |*+\txt{Receive \\ $(\c{pre-challenge}, r_B)$ \\ (no spare slot)}
+ :@(u, u)"chal-cookie"
+ "chal" :@/_0.8cm/ [ddddl]
+ *+\txt{Send \\ $(\c{challenge}, $\\$ r_A, c_B, v_A)$}
+ ="send-chal"
+ |<>(0.67) *+\txt\small{
+ Receive \\ $(\c{pre-challenge}, r_B)$ \\ (spare slot)}
+ "chal" :@/^0.8cm/ "send-chal" |<>(0.33)
+ *+\txt{Receive \\ $(\c{cookie}, r_B, c_A)$}
+ :[rr]
+ *+\txt{Send \\ $(\c{reply}, c_A, c_B, $\\$ v_A, E_K(r_B^\alpha))$}
+ ="send-reply"
+ |*+\txt{Receive \\ $(\c{challenge}, $\\$ r_B, c_A, v_B)$}
+ "chal" :"send-reply"
+ |*+\txt{Receive \\ $(\c{challenge}, $\\$ r_B, c_A, v_B)$}
+ "send-chal" :[ddd]
+ *++[F:<4pt>]\txt{
+ \ns State \c{commit} \\
+ Send \\ $(\c{switch}, c_A, c_B, $\\$ E_K(r_B^\alpha, w_A))$}
+ ="commit"
+ |*+\txt{Receive \\ $(\c{reply}, c_B, c_A, $\\$ v_B, E_K(b^{\rho_A}))$}
+ :[rr]
+ *+\txt{Send \\ $(\c{switch-ok}, E_K(u_A))$}
+ ="send-switch-ok"
+ |*+\txt{Receive \\ $(\c{switch}, c_B, c_A, $\\$ E_K(b^{\rho_A}, w_B))$}
+ "send-reply" :"commit"
+ |*+\txt{Receive \\ $(\c{reply}, c_B, c_A, $\\$ v_B, E_K(b^{\rho_A}))$}
+ "send-reply" :"send-switch-ok"
+ |*+\txt{Receive \\ $(\c{switch}, c_B, c_A, $\\$ E_K(b^{\rho_A}, w_B))$}
+ :[dddl]
+ *++[F:<4pt>]\txt{\ns Done}
+ ="done"
+ "commit" :"done"
+ |*+\txt{Receive \\ $(\c{switch-ok}, E_K(u_B))$}
+ "send-chal" [r] !{+<0cm, 0.75cm>}
+ *\txt\itshape{For each outstanding challenge}
+ ="for-each"
+ !{"send-chal"+DL-<8pt, 8pt> ="p0",
+ "for-each"+U+<8pt> ="p1",
+ "send-reply"+UR+<8pt, 8pt> ="p2",
+ "send-reply"+DR+<8pt, 8pt> ="p3",
+ "p0" !{"p1"-"p0"} !{"p2"-"p1"} !{"p3"-"p2"}
+ *\frm<8pt>{--}}
+ \end{graph} \]
+
+ \caption{State-transition diagram for key-exchange protocol}
+ \label{fig:kx-states}
+\end{figure}
+
+We now describe the protocol message by message, and Alice's actions when she
+receives each. Since the protocol is completely symmetrical, Bob should do
+the same, only swapping round $A$ and $B$ subscripts, the public keys $a$ and
+$b$, and using his private key $\beta$ instead of $\alpha$.
+
+\subsubsection{Starting the protocol}
+
+As described above, at the beginning of the protocol Alice chooses a random
+$\rho_A \inr \Nupto q$, and computes her \emph{challenge} $r_A = g^{\rho_A}$
+and her \emph{cookie} $c_A = H(\cookie{cookie}, r_A)$. She sends her
+announcement of her challenge as
+\begin{equation}
+ \label{eq:kx-pre-challenge}
+ \cookie{kx-pre-challenge}, r_A
+\end{equation}
+and enters the \cookie{challenge} state.
+
+\subsubsection{The \cookie{kx-pre-challenge} message}
+
+If Alice receieves a \cookie{kx-pre-challenge}, she ensures that she's in the
+\cookie{challenge} state: if not, she rejects the message.
+
+She must first calculate Bob's cookie $c_B = H(\cookie{cookie}, r_B)$. Then
+she has a choice: either she can send a full challenge, or she can send the
+cookie back.
+
+Suppose she decides to send a full challenge. She must compute a \emph{check
+value}
+\begin{equation}
+ \label{eq:v_A}
+ v_A = \rho_A \xor H(\cookie{expected-reply}, r_A, r_B, b^{\rho_A})
+\end{equation}
+and sends
+\begin{equation}
+ \label{eq:kx-challenge}
+ \cookie{kx-challenge}, r_A, c_B, v_A
+\end{equation}
+to Bob. Then she remembers Bob's challenge for later use, and awaits his
+reply.
+
+If she decides to send only a cookie, she just transmits
+\begin{equation}
+ \label{eq:kx-cookie}
+ \cookie{kx-cookie}, r_A, c_B
+\end{equation}
+to Bob and forgets all about it.
+
+Why's this useful? Well, if Alice sends off a full \cookie{kx-challenge}
+message, she must remember Bob's $r_B$ so she can check his reply and that
+involves using up a table slot. That means that someone can send Alice
+messages purporting to come from Bob which will chew up Alice's memory, and
+they don't even need to be able to read Alice's messages to Bob to do that.
+If this protocol were used over the open Internet, script kiddies from all
+over the world might be flooding Alice with bogus \cookie{kx-pre-challenge}
+messages and she'd never get around to talking to Bob.
+
+By sending a cookie intead, she avoids committing a table slot until Bob (or
+someone) sends either a cookie or a full challenge, thus proving, at least,
+that he can read her messages. This is the best we can do at this stage in
+the protocol. Against an adversary as powerful as the one we present in
+section~\fixme\ref{sec:formal} this measure provides no benefit (but we have
+to analyse it anyway); but it raises the bar too sufficiently high to
+eliminate a large class of `nuisance' attacks in the real world.
+
+Our definition of the Wrestlers Protocol doesn't stipulate when Alice should
+send a full challenge or just a cookie: we leave this up to individual
+implementations, because it makes no difference to the security of the
+protocol against powerful adversaries. But we recommend that Alice proceed
+`optimistically' at first, sending full challenges until her challenge table
+looks like it's running out, and then generating cookies only if it actually
+looks like she's under attack. This is what our pseudocode in
+figure~\ref{fig:kx-messages} does.
-We conclude here that the Wrestlers Authentication Protocol is a secure
-authentication protocol in the random oracle model: impersonation is hard if
-the Diffie-Hellman problem is hard, and proving one's identity doesn't leak
-secret key information.
+\subsubsection{The \cookie{kx-cookie} message}
-\subsection{Requirements on hash functions}
+When Alice receives a \cookie{kx-cookie} message, she must ensure that she's
+in the \cookie{challenge} state: if not, she rejects the message. She checks
+the cookie in the message against the value of $c_A$ she computed earlier.
+If all is well, Alice sends a \cookie{kx-challenge} message, as in
+equation~\ref{eq:kx-challenge} above.
-Having seen that the Wrestlers Authentication Protocol is secure in the
-random oracle model, we now ask which properties we require from the hash
-function. This at least demonstrates that the protocol isn't deeply flawed,
-and suggests an efficient implementation in terms of conventional hash
-functions.
+This time, she doesn't have a choice about using up a table slot to remember
+Bob's $r_B$. If her table size is fixed, she must choose a slot to recycle.
+We suggest simply recycling slots at random: this means there's no clever
+pattern of \cookie{kx-cookie} messages an attacker might be able to send to
+clog up all of Alice's slots.
-Here we investigate more carefully the properties required of the hash
-function, and provide a more quantitative analysis of the protocol.
+\subsubsection{The \cookie{kx-challenge} message}
-Looking at the proofs of the previous two sections, we see that the random
-oracles are mainly a device which allow our constructions to `grab hold' of
-the hashing operations performed by the adversary.
+
+
+\begin{figure}
+ \begin{program}
+ Procedure $\id{kx-initialize}$: \+ \\
+ $\rho_A \getsr [q]$; \\
+ $r_a \gets g^{\rho_A}$; \\
+ $\id{state} \gets \cookie{challenge}$; \\
+ $\Xid{n}{chal} \gets 0$; \\
+ $k \gets \bot$; \\
+ $\id{chal-commit} \gets \bot$; \\
+ $\id{send}(\cookie{kx-pre-challenge}, r_A)$; \- \\[\medskipamount]
+ Procedure $\id{kx-receive}(\id{type}, \id{data})$: \\ \ind
+ \IF $\id{type} = \cookie{kx-pre-challenge}$ \THEN \\ \ind
+ \id{msg-pre-challenge}(\id{data}); \- \\
+ \ELSE \IF $\id{type} = \cookie{kx-cookie}$ \THEN \\ \ind
+ \id{msg-cookie}(\id{data}); \- \\
+ \ELSE \IF $\id{type} = \cookie{kx-challenge}$ \THEN \\ \ind
+ \id{msg-challenge}(\id{data}); \- \\
+ \ELSE \IF $\id{type} = \cookie{kx-reply}$ \THEN \\ \ind
+ \id{msg-reply}(\id{data}); \- \\
+ \ELSE \IF $\id{type} = \cookie{kx-switch}$ \THEN \\ \ind
+ \id{msg-switch}(\id{data}); \- \\
+ \ELSE \IF $\id{type} = \cookie{kx-switch-ok}$ \THEN \\ \ind
+ \id{msg-switch-ok}(\id{data}); \-\- \\[\medskipamount]
+ Procedure $\id{msg-pre-challenge}(\id{data})$: \+ \\
+ \IF $\id{state} \ne \cookie{challenge}$ \THEN \RETURN; \\
+ $r \gets \id{data}$; \\
+ \IF $\Xid{n}{chal} \ge \Xid{n}{chal-thresh}$ \THEN \\ \ind
+ $\id{send}(\cookie{kx-cookie}, r_A, \id{cookie}(r_A)))$; \- \\
+ \ELSE \+ \\
+ $\id{new-chal}(r)$; \\
+ $\id{send}(\cookie{kx-challenge}, r_A,
+ \id{cookie}(r), \id{checkval}(r))$; \-\-\\[\medskipamount]
+ Procedure $\id{msg-cookie}(\id{data})$: \+ \\
+ \IF $\id{state} \ne \cookie{challenge}$ \THEN \RETURN; \\
+ $(r, c_A) \gets \id{data}$; \\
+ \IF $c_A \ne \id{cookie}(r_A)$ \THEN \RETURN; \\
+ $\id{new-chal}(r)$; \\
+ $\id{send}(\cookie{kx-challenge}, r_A,
+ \id{cookie}(r), \id{checkval}(r))$; \- \\[\medskipamount]
+ Procedure $\id{msg-challenge}(\id{data})$: \+ \\
+ \IF $\id{state} \ne \cookie{challenge}$ \THEN \RETURN; \\
+ $(r, c_A, v) \gets \id{data}$; \\
+ \IF $c_A \ne \id{cookie}(r_A)$ \THEN \RETURN; \\
+ $i \gets \id{check-reply}(\bot, r, v)$; \\
+ \IF $i = \bot$ \THEN \RETURN; \\
+ $k \gets \id{chal-tab}[i].k$; \\
+ $y \gets \id{encrypt}(k, \cookie{kx-reply}, r^\alpha)$; \\
+ $\id{send}(\cookie{kx-reply}, c_A, \id{cookie}(r),
+ \id{checkval}(r), y)$
+ \next
+ Procedure $\id{msg-reply}(\id{data})$: \+ \\
+ $(c, c_A, v, y) \gets \id{data}$; \\
+ \IF $c_A \ne \id{cookie}(r_A)$ \THEN \RETURN; \\
+ $i \gets \id{find-chal}(c)$; \\
+ \IF $i = \bot$ \THEN \RETURN; \\
+ \IF $\id{check-reply}(i, \id{chal-tab}[i].r, v) = \bot$ \THEN \\ \ind
+ \RETURN; \- \\
+ $k \gets \id{chal-tab}[i].k$; \\
+ $x \gets \id{decrypt}(k, \cookie{kx-reply}, y)$; \\
+ \IF $x = \bot$ \THEN \RETURN; \\
+ \IF $x \ne b^{\rho_A}$ \THEN \RETURN; \\
+ $\id{state} \gets \cookie{commit}$; \\
+ $\id{chal-commit} \gets \id{chal-tab}[i]$; \\
+ $w \gets H(\cookie{switch-request}, c_A, c)$; \\
+ $x \gets \id{chal-tab}[i].r^\alpha$; \\
+ $y \gets \id{encrypt}(k, (x, \cookie{kx-switch}, w))$; \\
+ $\id{send}(\cookie{kx-switch}, c_A, c, y)$; \-\\[\medskipamount]
+ Procedure $\id{msg-switch}(\id{data})$: \+ \\
+ $(c, c_A, y) \gets \id{data}$; \\
+ \IF $c_A \ne \cookie(r_A)$ \THEN \RETURN; \\
+ $i \gets \id{find-chal}(c)$; \\
+ \IF $i = \bot$ \THEN \RETURN; \\
+ $k \gets \id{chal-tab}[i].k$; \\
+ $x \gets \id{decrypt}(k, \cookie{kx-switch}, y)$; \\
+ \IF $x = \bot$ \THEN \RETURN; \\
+ $(x, w) \gets x$; \\
+ \IF $\id{state} = \cookie{challenge}$ \THEN \\ \ind
+ \IF $x \ne b^{\rho_A}$ \THEN \RETURN; \\
+ $\id{chal-commit} \gets \id{chal-tab}[i]$; \- \\
+ \ELSE \IF $c \ne \id{chal-commit}.c$ \THEN \RETURN; \\
+ \IF $w \ne H(\cookie{switch-request}, c, c_A)$ \THEN \RETURN; \\
+ $w \gets H(\cookie{switch-confirm}, c_A, c)$; \\
+ $y \gets \id{encrypt}(y, \cookie{kx-switch-ok}, w)$; \\
+ $\id{send}(\cookie{switch-ok}, y)$; \\
+ $\id{done}(k)$; \- \\[\medskipamount]
+ Procedure $\id{msg-switch-ok}(\id{data})$ \+ \\
+ \IF $\id{state} \ne \cookie{commit}$ \THEN \RETURN; \\
+ $y \gets \id{data}$; \\
+ $k \gets \id{chal-commit}.k$; \\
+ $w \gets \id{decrypt}(k, \cookie{kx-switch-ok}, y)$; \\
+ \IF $w = \bot$ \THEN \RETURN; \\
+ $c \gets \id{chal-commit}.c$; \\
+ $c_A \gets \id{cookie}(r_A)$; \\
+ \IF $w \ne H(\cookie{switch-confirm}, c, c_A)$ \THEN \RETURN; \\
+ $\id{done}(k)$;
+ \end{program}
+
+ \caption{The key-exchange protocol: message handling}
+ \label{fig:kx-messages}
+\end{figure}
+
+\begin{figure}
+ \begin{program}
+ Structure $\id{chal-slot}$: \+ \\
+ $r$; $c$; $\id{replied}$; $k$; \- \\[\medskipamount]
+ Function $\id{find-chal}(c)$: \+ \\
+ \FOR $i = 0$ \TO $\Xid{n}{chal}$ \DO \\ \ind
+ \IF $\id{chal-tab}[i].c = c$ \THEN \RETURN $i$; \- \\
+ \RETURN $\bot$; \- \\[\medskipamount]
+ Function $\id{cookie}(r)$: \+ \\
+ \RETURN $H(\cookie{cookie}, r)$; \- \\[\medskipamount]
+ Function $\id{check-reply}(i, r, v)$: \+ \\
+ \IF $i \ne \bot \land \id{chal-tab}[i].\id{replied} = 1$ \THEN \\ \ind
+ \RETURN $i$; \- \\
+ $\rho \gets v \xor H(\cookie{expected-reply}, r, r_A, r^\alpha)$; \\
+ \IF $g^\rho \ne r$ \THEN \RETURN $\bot$; \\
+ \IF $i = \bot$ \THEN $i \gets \id{new-chal}(r)$; \\
+ $\id{chal-tab}[i].k \gets \id{gen-keys}(r_A, r, r^{\rho_A})$; \\
+ $\id{chal-tab}[i].\id{replied} \gets 1$; \\
+ \RETURN $i$;
+ \next
+ Function $\id{checkval}(r)$: \\ \ind
+ \RETURN $\rho_A \xor H(\cookie{expected-reply},
+ r_A,r, b^{\rho_A})$; \- \\[\medskipamount]
+ Function $\id{new-chal}(r)$: \+ \\
+ $c \gets \id{cookie}(r)$; \\
+ $i \gets \id{find-chal}(c)$; \\
+ \IF $i \ne \bot$ \THEN \RETURN $i$; \\
+ \IF $\Xid{n}{chal} < \Xid{n}{chal-max}$ \THEN \\ \ind
+ $i \gets \Xid{n}{chal}$; \\
+ $\id{chal-tab}[i] \gets \NEW \id{chal-slot}$; \\
+ $\Xid{n}{chal} \gets \Xid{n}{chal} + 1$; \- \\
+ \ELSE \\ \ind
+ $i \getsr [\Xid{n}{chal-max}]$; \- \\
+ $\id{chal-tab}[i].r \gets r$; \\
+ $\id{chal-tab}[i].c \gets c$; \\
+ $\id{chal-tab}[i].\id{replied} \gets 0$; \\
+ $\id{chal-tab}[i].k \gets \bot$; \\
+ \RETURN $i$;
+ \end{program}
+
+ \caption{The key-exchange protocol: support functions}
+ \label{fig:kx-support}
+\end{figure}
%%%--------------------------------------------------------------------------
-\section{A key-exchange protocol}
-% Present the Wrestlers protocol in all its glory. Show, by means of the
-% previous proofs, that the Wrestlers protocol is simulatable in the
-% authenticated model using a much simpler protocol. Show that the simpler
-% protocol is SK-secure.
+\section{CBC mode encryption}
+\label{sec:cbc}
+
+Our implementation of the Wrestlers Protocol uses Blowfish
+\cite{Schneier:1994:BEA} in CBC mode. However, rather than pad plaintext
+messages to a block boundary, with the ciphertext expansion that entails, we
+use a technique called \emph{ciphertext stealing}
+\cite[section 9.3]{Schneier:1996:ACP}.
+
+\subsection{Standard CBC mode}
+
+Suppose $E$ is an $\ell$-bit pseudorandom permutation. Normal CBC mode works
+as follows. Given a message $X$, we divide it into blocks $x_0, x_1, \ldots,
+x_{n-1}$. Choose a random \emph{initialization vector} $I \inr \Bin^\ell$.
+Before passing each $x_i$ through $E$, we XOR it with the previous
+ciphertext, with $I$ standing in for the first block:
+\begin{equation}
+ y_0 = E_K(x_0 \xor I) \qquad
+ y_i = E_K(x_i \xor y_{i-1} \ \text{(for $1 \le i < n$)}.
+\end{equation}
+The ciphertext is then the concatenation of $I$ and the $y_i$. Decryption is
+simple:
+\begin{equation}
+ x_0 = E^{-1}_K(y_0) \xor I \qquad
+ x_i = E^{-1}_K(y_i) \xor y_{i-1} \ \text{(for $1 \le i < n$)}
+\end{equation}
+See figure~\ref{fig:cbc} for a diagram of CBC encryption.
+
+\begin{figure}
+ \[ \begin{graph}
+ []!{0; <0.85cm, 0cm>: <0cm, 0.5cm>::}
+ *+=(1, 0)+[F]{\mathstrut x_0}="x"
+ :[dd] *{\xor}="xor"
+ [ll] *+=(1, 0)+[F]{I} :"xor"
+ :[dd] *+[F]{E}="e" :[ddd] *+=(1, 0)+[F]{\mathstrut y_0}="i"
+ "e" [l] {K} :"e"
+ [rrruuuu] *+=(1, 0)+[F]{\mathstrut x_1}="x"
+ :[dd] *{\xor}="xor"
+ "e" [d] :`r [ru] `u "xor" "xor"
+ :[dd] *+[F]{E}="e" :[ddd]
+ *+=(1, 0)+[F]{\mathstrut y_1}="i"
+ "e" [l] {K} :"e"
+ [rrruuuu] *+=(1, 0)+[F--]{\mathstrut x_i}="x"
+ :@{-->}[dd] *{\xor}="xor"
+ "e" [d] :@{-->}`r [ru] `u "xor" "xor"
+ :@{-->}[dd] *+[F]{E}="e" :@{-->}[ddd]
+ *+=(1, 0)+[F--]{\mathstrut y_i}="i"
+ "e" [l] {K} :@{-->}"e"
+ [rrruuuu] *+=(1, 0)+[F]{\mathstrut x_{n-1}}="x"
+ :[dd] *{\xor}="xor"
+ "e" [d] :@{-->}`r [ru] `u "xor" "xor"
+ :[dd] *+[F]{E}="e" :[ddd]
+ *+=(1, 0)+[F]{\mathstrut y_{n-1}}="i"
+ "e" [l] {K} :"e"
+ \end{graph} \]
+
+ \caption{Encryption using CBC mode}
+ \label{fig:cbc}
+\end{figure}
+
+\begin{definition}[CBC mode]
+ \label{def:cbc}
+ Let $P\colon \keys P \times \Bin^\ell to \Bin^\ell$ be a pseudorandom
+ permutation. We define the symmetric encryption scheme
+ $\Xid{\mathcal{E}}{CBC}^P = (\Xid{E}{CBC}^P, \Xid{D}{CBC}^P)$ for messages
+ in $\Bin^{\ell\Z}$ by setting $\keys \Xid{\mathcal{E}}{CBC} = \keys P$ and
+ defining the encryption and decryption algorithms as follows:
+ \begin{program}
+ Algorithm $\Xid{E}{CBC}^P_K(x)$: \+ \\
+ $I \getsr \Bin^\ell$; \\
+ $y \gets I$; \\
+ \FOR $i = 0$ \TO $|x|/\ell$ \DO \\ \ind
+ $x_i \gets x[\ell i \bitsto \ell (i + 1)]$; \\
+ $y_i \gets P_K(x_i \xor I)$; \\
+ $I \gets y_i$; \\
+ $y \gets y \cat y_i$; \- \\
+ \RETURN $y$;
+ \next
+ Algorithm $\Xid{D}{CBC}^P_K(y)$: \+ \\
+ $I \gets y[0 \bitsto \ell]$; \\
+ $x \gets \emptystring$; \\
+ \FOR $1 = 0$ \TO $|y|/\ell$ \DO \\ \ind
+ $y_i \gets y[\ell i \bitsto \ell (i + 1)]$; \\
+ $x_i \gets P^{-1}_K(y_i) \xor I$; \\
+ $I \gets y_i$; \\
+ $x \gets x \cat x_i$; \- \\
+ \RETURN $x$;
+ \end{program}
+\end{definition}
+
+\begin{theorem}[Security of standard CBC mode]
+ \label{thm:cbc}
+ Let $P\colon \keys P \times \Bin^\ell \to \Bin^\ell$ be a pseudorandom
+ permutation. Then,
+ \begin{equation}
+ \InSec{lor-cpa}(\Xid{\mathcal{E}}{CBC}; t, q_E + \mu_E) \le
+ 2 \cdot \InSec{prp}(P; t + q t_P, q) +
+ \frac{q (q - 1)}{2^\ell - 2^{\ell/2}}
+ \end{equation}
+ where $q = \mu_E/\ell$ and $t_P$ is some small constant.
+\end{theorem}
+
+\begin{note}
+ Our security bound is slightly better than that of \cite[theorem
+ 17]{Bellare:2000:CST}. Their theorem statement contains a term $3 \cdot q
+ (q - 1) 2^{-\ell-1}$. Our result lowers the factor from 3 to just over 2.
+ Our proof is also much shorter and considerably more comprehensible.
+\end{note}
+
+The proof of this theorem is given in section~\ref{sec:cbc-proof}
+
+\subsection{Ciphertext stealing}
+
+Ciphertext stealing allows us to encrypt any message in $\Bin^*$ and make the
+ciphertext exactly $\ell$ bits longer than the plaintext. See
+figure~\ref{fig:cbc-steal} for a diagram.
+
+\begin{figure}
+ \[ \begin{graph}
+ []!{0; <0.85cm, 0cm>: <0cm, 0.5cm>::}
+ *+=(1, 0)+[F]{\mathstrut x_0}="x"
+ :[dd] *{\xor}="xor"
+ [ll] *+=(1, 0)+[F]{I} :"xor"
+ :[dd] *+[F]{E}="e" :[ddddd] *+=(1, 0)+[F]{\mathstrut y_0}="i"
+ "e" [l] {K} :"e"
+ [rrruuuu] *+=(1, 0)+[F]{\mathstrut x_1}="x"
+ :[dd] *{\xor}="xor"
+ "e" [d] :`r [ru] `u "xor" "xor"
+ :[dd] *+[F]{E}="e" :[ddddd]
+ *+=(1, 0)+[F]{\mathstrut y_1}="i"
+ "e" [l] {K} :"e"
+ [rrruuuu] *+=(1, 0)+[F--]{\mathstrut x_i}="x"
+ :@{-->}[dd] *{\xor}="xor"
+ "e" [d] :@{-->}`r [ru] `u "xor" "xor"
+ :@{-->}[dd] *+[F]{E}="e" :@{-->}[ddddd]
+ *+=(1, 0)+[F--]{\mathstrut y_i}="i"
+ "e" [l] {K} :@{-->}"e"
+ [rrruuuu] *+=(1, 0)+[F]{\mathstrut x_{n-2}}="x"
+ :[dd] *{\xor}="xor"
+ "e" [d] :@{-->}`r [ru] `u "xor" "xor"
+ :[dd] *+[F]{E}="e"
+ "e" [l] {K} :"e"
+ [rrruuuu] *+=(1, 0)+[F]{\mathstrut x_{n-1} \cat 0^{\ell-t}}="x"
+ :[dd] *{\xor}="xor"
+ "e" [d] :`r [ru] `u "xor" "xor"
+ "e" [dddddrrr] *+=(1, 0)+[F]{\mathstrut y_{n-1}[0 \bitsto t]}="i"
+ "e" [dd] ="x"
+ "i" [uu] ="y"
+ []!{"x"; "e" **{}, "x"+/4pt/ ="p",
+ "x"; "y" **{}, "x"+/4pt/ ="q",
+ "y"; "x" **{}, "y"+/4pt/ ="r",
+ "y"; "i" **{}, "y"+/4pt/ ="s",
+ "e";
+ "p" **\dir{-};
+ "q" **\crv{"x"};
+ "r" **\dir{-};
+ "s" **\crv{"y"};
+ "i" **\dir{-}?>*\dir{>}}
+ "xor" :[dd] *+[F]{E}="e"
+ "e" [l] {K} :"e"
+ "e" [dddddlll] *+=(1, 0)+[F]{\mathstrut y_{n-2}}="i"
+ "e" [dd] ="x"
+ "i" [uu] ="y"
+ []!{"x"; "e" **{}, "x"+/4pt/ ="p",
+ "x"; "y" **{}, "x"+/4pt/ ="q",
+ "y"; "x" **{}, "y"+/4pt/ ="r",
+ "y"; "i" **{}, "y"+/4pt/ ="s",
+ "x"; "y" **{} ?="c" ?(0.5)/-4pt/ ="cx" ?(0.5)/4pt/ ="cy",
+ "e";
+ "p" **\dir{-};
+ "q" **\crv{"x"};
+ "cx" **\dir{-};
+ "c" *[@]\cir<4pt>{d^u};
+ "cy";
+ "r" **\dir{-};
+ "s" **\crv{"y"};
+ "i" **\dir{-}?>*\dir{>}}
+ \end{graph} \]
+
+ \caption{Encryption using CBC mode with ciphertext stealing}
+ \label{fig:cbc-steal}
+\end{figure}
+
+\begin{definition}[CBC stealing]
+ \label{def:cbc-steal}
+ Let $P\colon \keys P \times \Bin^\ell \to \Bin^\ell$ be a pseudorandom
+ permutation. We define the symmetric encryption scheme
+ $\Xid{\mathcal{E}}{CBC-steal}^P = (\Xid{G}{CBC}^P, \Xid{E}{CBC-steal}^P,
+ \Xid{D}{CBC-steal}^P)$ for messages in $\Bin^{\ell\Z}$ by setting $\keys
+ \Xid{\mathcal{E}}{CBC-steal} = \keys P$ and defining the encryption and
+ decryption algorithms as follows:
+ \begin{program}
+ Algorithm $\Xid{E}{CBC-steal}^P_K(x)$: \+ \\
+ $I \getsr \Bin^\ell$; \\
+ $y \gets I$; \\
+ $t = |x| \bmod \ell$; \\
+ \IF $t \ne 0$ \THEN $x \gets x \cat 0^{\ell-t}$; \\
+ \FOR $i = 0$ \TO $|x|/\ell$ \DO \\ \ind
+ $x_i \gets x[\ell i \bitsto \ell (i + 1)]$; \\
+ $y_i \gets P_K(x_i \xor I)$; \\
+ $I \gets y_i$; \\
+ $y \gets y \cat y_i$; \- \\
+ \IF $t \ne 0$ \THEN \\ \ind
+ $b \gets |y| - 2\ell$; \\
+ $y \gets $\=$y[0 \bitsto b] \cat
+ y[b + \ell \bitsto |y|] \cat {}$ \\
+ \>$y[b \bitsto b + t]$; \- \\
+ \RETURN $y$;
+ \next
+ Algorithm $\Xid{D}{CBC-steal}^P_K(y)$: \+ \\
+ $I \gets y[0 \bitsto \ell]$; \\
+ $t = |y| \bmod \ell$; \\
+ \IF $t \ne 0$ \THEN \\ \ind
+ $b \gets |y| - t - \ell$; \\
+ $z \gets P^{-1}_K(y[b \bitsto b + \ell])$; \\
+ $y \gets $\=$y[0 \bitsto b] \cat
+ y[b + \ell \bitsto |y|] \cat {}$ \\
+ \>$z[t \bitsto \ell]$; \- \\
+ $x \gets \emptystring$; \\
+ \FOR $1 = 0$ \TO $|y|/\ell$ \DO \\ \ind
+ $y_i \gets y[\ell i \bitsto \ell (i + 1)]$; \\
+ $x_i \gets P^{-1}_K(y_i) \xor I$; \\
+ $I \gets y_i$; \\
+ $x \gets x \cat x_i$; \- \\
+ \IF $t \ne 0$ \THEN \\ \ind
+ $x \gets x \cat z[0 \bitsto t] \xor y[b \bitsto b + t]$; \- \\
+ \RETURN $x$;
+ \end{program}
+\end{definition}
+
+\begin{theorem}[Security of CBC with ciphertext stealing]
+ \label{thm:cbc-steal}
+ Let $P\colon \keys P \times \Bin^\ell \to \Bin^\ell$ be a pseudorandom
+ permutation. Then
+ \begin{equation}
+ \InSec{lor-cpa}(\Xid{\mathcal{E}}{CBC-steal}; t, q_E, \mu_E) \le
+ 2 \cdot \InSec{prp}(P; t + q t_P, q) +
+ \frac{q (q - 1)}{2^\ell - 2^{\ell/2}}
+ \end{equation}
+ where $q = \mu_E/\ell$ and $t_P$ is some small constant.
+\end{theorem}
+
+\begin{proof}
+ This is an easy reducibility argument. Let $A$ be an adversary attacking
+ $\Xid{\mathcal{E}}{CBC-steal}^P$. We construct an adversary which attacks
+ $\Xid{\mathcal{E}}{CBC}^P$:
+ \begin{program}
+ Adversary $A'^{E(\cdot)}$: \+ \\
+ $b \gets A^{\Xid{E}{steal}(\cdot)}$; \\
+ \RETURN $b$;
+ \- \\[\medskipamount]
+ Oracle $\Xid{E}{steal}(x_0, x_1)$: \+ \\
+ \IF $|x_0| \ne |x_1|$ \THEN \ABORT; \\
+ \RETURN $\id{steal}(|x_0|, E(\id{pad}(x_0), \id{pad}(x_1)))$;
+ \next
+ Function $\id{pad}(x)$: \+ \\
+ $t \gets |x| \bmod \ell$; \\
+ \RETURN $x \cat 0^{\ell-t}$;
+ \- \\[\medskipamount]
+ Function $\id{steal}(l, y)$: \+ \\
+ $t \gets l \bmod \ell$; \\
+ \IF $t \ne 0$ \THEN \\ \ind
+ $b \gets |y| - 2\ell$; \\
+ $y \gets $\=$y[0 \bitsto b] \cat
+ y[b + \ell \bitsto |y|] \cat y[b \bitsto b + t]$; \- \\
+ \RETURN $y$;
+ \end{program}
+ Comparing this to definition~\ref{def:cbc-steal} shows that $A'$ simlates
+ the LOR-CPA game for $\Xid{\mathcal{E}}{CBC-steal}$ perfectly. The theorem
+ follows.
+\end{proof}
+\subsection{Proof of theorem~\ref{thm:cbc}}
+\label{sec:cbc-proof}
-% messages
-%
-% pre-challenge: g^b
-% cookie: g^b, h(cookie, g^b')
-% challenge: g^b, h(g^b'), b + h(reply-check, g^b, g^b', g^ab')
-% reply: g^b, h(g^b'), b + h(reply-check, g^b, g^b', g^ab'), E_k(g^a'b)
-% switch: h(g^b), h(g^b'), E_k(g^a'b, h(switch-request, g^a, g^a'))
-% switch-ok: E_k(g^ab, h(switch-confirm, g^a, g^a'))begin{
+Consider an adversary $A$ attacking CBC encryption using an ideal random
+permutation $P(\cdot)$. Pick some point in the attack game when we're just
+about to encrypt the $n$th plaintext block. For each $i \in \Nupto{n}$,
+let $x_i$ be the $i$th block of plaintext we've processed; let $y_i$ be the
+corresponding ciphertext; and let $z_i = P^{-1}(y_i)$, i.e., $z_i = x_i \xor
+I$ for the first block of a message, and $z_i = x_i \xor y_{i-1}$ for the
+subsequent blocks.
+Say that `something went wrong' if any $z_i = z_j$ for $i \ne j$. This is
+indeed a disaster, because it means that $y_i = y_j$ , so he can detect it,
+and $x_i \xor y_{i-1} = x_j \xor y_{j-1}$, so he can compute an XOR
+difference between two plaintext blocks from the ciphertext and thus
+(possibly) reveal whether he's getting his left or right plaintexts
+encrypted. The alternative, `everything is fine', is much better. If all
+the $z_i$ are distinct, then because $y_i = P(z_i)$, the $y_i$ are all
+generated by $P(\cdot)$ on inputs it's never seen before, so they're all
+random subject to the requirement that they be distinct. If everything is
+fine, then, the adversary has no better way of deciding whether he has a left
+oracle or a right oracle than tossing a coin, and his advantage is therefore
+zero. Thus, we must bound the probability that something went wrong.
-We now describe the full Wrestlers Protocol, in a multi-party setting. Fix a
-cyclic group $G = \langle g \rangle$ of order $q = |G|$. Each player $P_i$
-chooses a private key $\alpha_i \in \{ 1, 2, \ldots, q - 1 \}$ and publishes
-the corresponding public key $a_i = g^{\alpha_i}$ to the other players.
+Assume that, at our point in the game so far, everything is fine. But we're
+just about to encrypt $x^* = x_n$. There are two cases:
+\begin{itemize}
+\item If $x_n$ is the first block in a new message, we've just invented a new
+ random IV $I \in \Bin^\ell$ which is unknown to $A$, and $z_n = x_n \xor
+ I$. Let $y^* = I$.
+\item If $x_n$ is \emph{not} the first block, then $z_n = x_n \xor y_{n-1}$,
+ but the adversary doesn't yet know $y_{n-1}$, except that because $P$ is a
+ permutation and all the $z_i$ are distinct, $y_{n-1} \ne y_i$ for any $0
+ \le i < n - 1$. Let $y^* = y_{n-1}$.
+\end{itemize}
+Either way, the adversary's choice of $x^*$ is independent of $y^*$. Let
+$z^* = x^* \xor y^*$. We want to know the probability that something goes
+wrong at this point, i.e., that $z^* = z_i$ for some $0 \le i < n$. Let's
+call this event $C_n$. Note first that, in the first case, there are
+$2^\ell$ possible values for $y^*$ and in the second there are $2^\ell - n +
+1$ possibilities for $y^*$. Then
+\begin{eqnarray}[rl]
+ \Pr[C_n]
+ & = \sum_{x \in \Bin^\ell} \Pr[C_n \mid x^* = x] \Pr[x^* = x] \\
+ & = \sum_{x \in \Bin^\ell}
+ \Pr[x^* = x] \sum_{0\le i<n} \Pr[y^* = z_i \xor x] \\
+ & \le \sum_{0\le i<n} \frac{1}{2^\ell - n}
+ \sum_{x \in \Bin^\ell} \Pr[x^* = x] \\
+ & = \frac{n}{2^\ell - n}
+\end{eqnarray}
+Having bounded the probability that something went wrong for any particular
+block, we can proceed to bound the probability of something going wrong in
+the course of the entire game. Let's suppose that $q = \mu_E/\ell \le
+2^{\ell/2}$; for if not, $q (q - 1) > 2^\ell$ and the theorem is trivially
+true, since no adversary can achieve advantage greater than one.
+Let's give the name $W_i$ to the probability that something went wrong after
+encrypting $i$ blocks. We therefore want to bound $W_q$ from above.
+Armed with the knowledge that $q \le 2^{\ell/2}$, we have
+\begin{eqnarray}[rl]
+ W_q &\le \sum_{0\le i<q} \Pr[C_i]
+ \le \sum_{0\le i<q} \frac{i}{2^\ell - i} \\
+ &\le \frac{1}{2^\ell - 2^{\ell/2}} \sum_{0\le i<q} i \\
+ &= \frac{q (q - 1)}{2 \cdot (2^\ell - 2^{\ell/2})}
+\end{eqnarray}
+Working through the definition of LOR-CPA security, we can see that $A$'s
+(and hence any adversary's) advantage against the ideal system is at most $2
+W_q$.
+By using an adversary attacking CBC encryption as a statistical test in an
+attempt to distinguish $P_K(\cdot)$ from a pseudorandom permutation, we see
+that
+\begin{equation}
+ \InSec{prp}(P; t + q t_P, q) \ge
+ \frac{1}{2} \cdot
+ \InSec{lor-cpa}(\Xid{\mathcal{E}}{CBC}; t, q_E, \mu_E) -
+ \frac{q (q - 1)}{2 \cdot (2^\ell - 2^{\ell/2})}
+\end{equation}
+where $t_P$ expresses the overhead of doing the XORs and other care and
+feeding of the CBC adversary; whence
+\begin{equation}
+ \InSec{lor-cpa}(\Xid{\mathcal{E}}{CBC}; t, q_E, \mu_E) \le
+ 2 \cdot \InSec{prp}(P; t, q) + \frac{q (q - 1)}{2^\ell - 2^{\ell/2}}
+\end{equation}
+as required.
+\qed
%%%----- That's all, folks --------------------------------------------------
-\bibliography{cryptography,mdw-crypto}
+\bibliography{mdw-crypto,cryptography,cryptography2000,rfc}
\end{document}
%%% Local Variables:
~mdw / tripe / commitdiff