%%% -*-latex-*-
%%%
-%%% $Id: wrestlers.tex,v 1.1 2001/02/16 21:43:33 mdw Exp $
+%%% $Id: wrestlers.tex,v 1.5 2002/01/13 14:55:31 mdw Exp $
%%%
%%% Description of the Wrestlers Protocol
%%%
%%%----- Revision history ---------------------------------------------------
%%%
%%% $Log: wrestlers.tex,v $
+%%% Revision 1.5 2002/01/13 14:55:31 mdw
+%%% More incomplete stuff.
+%%%
+%%% Revision 1.4 2001/06/29 19:36:05 mdw
+%%% Some progress made on laptop.
+%%%
+%%% Revision 1.3 2001/06/22 19:41:31 mdw
+%%% Restart with different structure and rather more formal objectives.
+%%%
+%%% Revision 1.2 2001/02/22 09:09:05 mdw
+%%% Partially through reworking.
+%%%
%%% Revision 1.1 2001/02/16 21:43:33 mdw
%%% Initial versions of documentation.
%%%
-\documentclass{article}
-\usepackage{amssymb}
-\title{The Wrestlers Protocol}
-\author{Mark Wooding \\ Clive Jones}
-\def\from{\leftarrow}
+\newif\iffancystyle\fancystylefalse
+
+\iffancystyle
+ \documentclass
+ [a4paper, article, 10pt, numbering, noherefloats, notitlepage]
+ {strayman}
+ \usepackage[palatino, helvetica, courier, maths=cmr]{mdwfonts}
+ \usepackage[margin]{amsthm}
+\else
+ \documentclass[a4paper]{article}
+ \usepackage{a4wide}
+ \usepackage{amsthm}
+\fi
+
+\usepackage{amssymb, amstext}
+\usepackage{mdwtab, mathenv}
+
+\errorcontextlines=999
+\showboxbreadth=999
+\showboxdepth=999
+\makeatletter
+
+\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}}
+
+\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}}
\begin{document}
\maketitle
-
\begin{abstract}
- We present a simple key-exchange protocol with of mutual authentication and
- perfect forward-secrecy, which doesn't leave any long-lasting evidence of
- participation in the exchange. The protocol's security depends on the
- intractability of the Diffie-Hellman problem (in some cyclic group), and on
- the strength of a hash function.
+ Fill this in later.
\end{abstract}
+\tableofcontents
+\newpage
-\section{Introduction}
+%%%--------------------------------------------------------------------------
-Current key-agreement protocols are all very well for securing generally
-`honest' traffic (e.g., transmission of credit-card details to a merchant),
-but they're less satisfactory if you actually have something to hide.
-
-In the UK, new key exchange protocols have been particularly motivated by the
-new Regulation of Investigatory Powers Act, which allows `authorized persons'
-to intercept communications and demand long-term encryption keys.
-
-Let's suppose that Alice and Bob are shady characters, and that their
-communications are of great interest to the draconian r\'egime in which they
-live. (They might be international arms smugglers, for example, because they
-export cryptographic toolkits.)
-
-Alice could just invent a session key and transmit it to Bob, encrypted under
-his public key, each time she wanted to talk to him. However, once the
-secret police turn up at Bob's house and demand his private key, the game is
-over and all of the communications can be recovered.
-
-Alice and Bob would clearly be better off using a system which offers forward
-secrecy, for example, Diffie-Hellman. However, in order to prevent active
-attacks, the messages in the Diffie-Hellman exchange must be authenticated.
-The way this usually works is that Alice and Bob pick a group $G$ of order
-$q$ generated by $g$. When Alice and Bob want to communicate, they choose
-exponents $\alpha$ and $\beta$ respectively ($1 < \alpha, \beta < q$), Alice
-sends $S_A(g^\alpha)$ to Bob, Bob sends $S_B(g^\beta)$ to Alice, and, after
-verifying each other's signatures, they each compute a shared key $K =
-g^{\alpha \beta}$. They dispose of their secrets $\alpha$ and $\beta$
-forthwith, and destroy $K$ when the conversation finishes. Now the secret
-police can demand all they like: they still can't decrypt old sessions, and
-Alice and Bob, however badly tortured, can't help them. The secret police
-might not even be all owed to demand their long-term signing keys: for
-example, the RIPA grants special protection to authentication-only keys.
-
-This is fine, except that sending $S_A(g^\alpha)$ is a wonderful way of
-shouting `Alice was here' to all of the spooks tapping Bob's network
-connection. The Wrestlers Protocol\footnote{%
- Named after the excellent pub in Cambridge where most of the design was
- done.}
-fixes these problems. It provides perfect forward secrecy, just like
-Diffie-Hellman, without leaving signatures around for the spooks.
+\section{Introduction}
+% 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.
-As a building-block, we construct a simple authentication protocol based on
-Diffie-Hellman key exchange. As before, let's use a group $G$ of order $q$
-(for some prime $q$), generated by a group element $g$.
+\subsection{Introduction}
-A Diffie-Hellman key exchange allows two parties to compute the same value,
-with different knowledge. We'll use this to make an authentication protocol.
+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 chooses a private key $1 < a < q$. Her public key is $A = g^a$. She
-can prove her knowledge of $a$ to Bob like this:
+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 makes up a random $1 < \beta < q$. He sends a challenge $C =
- g^\beta$ to Alice.
-\item Alice computes the response $R = C^a = g^{a \beta}$. This would
- be the shared key if we were doing proper Diffie-Hellman, but we aren't.
- Instead, she just sends $R$ back to Bob.
-\item Bob checks that $R = A^\beta$. If it is, he accepts that the person
- he's talking to has Alice's private key, and hence is presumably Alice.
+\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}
-
-This protocol has nice properties. It's not terribly difficult to implement,
-given the usual tools like modular exponentiation or elliptic curve
-point-addition.
-
-An eavesdropper -- let's call her Eve, for tradition's sake -- doesn't learn
-anything terribly interesting from watching the exchange. She sees $C$ going
-one way and then $C^a$ coming back. If she finds this illuminating, she can
-program her computer to generate random numbers $\gamma$ and show her pairs
-$C = g^\gamma$ and $R = A^\gamma = C^a$. So Eve learns nothing useful she
-couldn't have worked out for herself. In fact, she doesn't even learn that
-Alice is involved in the conversation! Bob can fake up an authentication
-with Alice by secretly agreeing which value of $\beta$ he's going to use with
-an accomplice.
-
-Bob's in a better position than Eve. If he computes his challenges honestly
-then he doesn't learn much except that he's talking to Alice, because as
-we've seen, she only tells him $R$, which he knew already. However, if Bob
-carefully chooses a challenge $C$ without knowing its discrete log $\beta$,
-then Alice's response might tell him useful information about her private
-key that he couldn't have worked out just by sitting at home computing
-discrete logs.
-
-We can fix this little problem easily enough if we make Bob transmit a hash
-of his expected answer. Let $H \colon \mathbb{Z} \to \{\,0, 1\,\}^n$ be a
-hash function. The property we require from $H$ is that Bob can't compute
-$H(g^{a \beta})$ given only $C = g^\beta$ and $A = g^a$ with more than
-negligible probability; a random function would fit the bill fine. This
-does, of course, also assume that the Diffie-Hellman problem is difficult.
-The new protocol looks very much like the old one:
+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 random $1 < \beta < q$. He computes $C = g^\beta$ and $R
- = A^\beta$, and sends $C, H(R)$ to Alice.
-\item Alice computes $R' = C^a$ and checks that it matches the hash which Bob
- sent. If it doesn't, he's trying to cheat, and she should refuse to
- answer. Otherwise, she sends her response $R'$ back to Bob.
-\item Bob checks that Alice's reply matches the one he computed back in step
- 1. If it does, he knows that he's talking to Alice.
+\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}
+
+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 $A$.} %
+
+ 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}
+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.
-\section{The Wrestlers Protocol}
+\subsection{Requirements on hash functions}
-We observe a useful side-effect of the authentication protocol just
-described: Bob should be convinved that Alice received his challenge $C$
-correctly. The idea of the Wrestlers Protocol is to use this to construct a
-full Diffie-Hellman key exchange with mutual authentication. We maintain the
-useful properties of the previous protocol.
+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.
-Before they can use the protocol, Alice and Bob must agree on a group $G$ as
-before. Alice chooses a private key $1 < a < q$, and publishes her public
-key $A = g^a$; Bob similarly chooses a private key $1 < b < q$ and publishes
-his public key $B = g^b$.
+Here we investigate more carefully the properties required of the hash
+function, and provide a more quantitative analysis of the protocol.
-Here's the actual protocol in summary:
-\begin{enumerate}
-\item $A \to B$:\quad $g^\alpha$
-\item $A \from B$:\quad $g^\beta$, $H(g^\alpha, g^\beta, g^{a \beta})$
-\item $A \to B$:\quad $g^{a \beta}$, $H(g^\beta, g^\alpha, g^{\alpha b})$
-\item $A \from B$:\quad $g^{\alpha b}$
-\end{enumerate}
+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.
-And now in detail:
-\begin{enumerate}
+%%%--------------------------------------------------------------------------
+
+\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.
+
+
+% 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{
+
+
+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.
-\item Alice invents a temporary secret $1 < \alpha < q$. She computes her
- challenge $C_A = g^\alpha$, and sends it to Bob.
-
-\item Bob receives the $C_A$, and stores it away. He invents a temporary
- secret $1 < \beta < q$ of his own, and computes both his challenge $C_B =
- g^\beta$ and the expected response $R_B = A^\beta = g^{a \beta}$. He
- hashes both challenges (hers first) and the expected response $R_B$, and
- sends his challenge and the hash back to Alice.
-
-\item Alice reads Bob's challenge. She computes her response $R_B' = C_B^a =
- g^{a \beta}$ and ensures that the Bob's hash is correct. If it isn't, she
- stops talking to Bob. If the hash matches, she sends back her response,
- together with a hash of Bob's challenge, her original challenge from step
- 1, and her expected response $R_A = B^\alpha = g^{\alpha b}$.
-\item Bob reads Alice's response. If it's wrong then he stops talking.
- Otherwise he computes his response to Alice's challenge $R_A' = C_A^b =
- g^{\alpha b}$ and checks Alice's hash. If the hash is wrong, he also stops
- talking. Otherwise he sends the response back to Alice.
-\end{enumerate}
-Finally, Alice checks Bob's response, stopping the conversation if it's
-wrong. Then both sides compute their shared key $K = C_A^\beta = C_B^\alpha
-= g^{\alpha \beta}$, and discard their temporary secrets.
-
-The protocol is essentially symmetrical: each side sends and receives both a
-challenge and hash pair, and a response, but it doesn't look that simple
-because the hashes include both sides' challenges. Looking at it from one
-side at a time will make things clearer, so let's just take Alice's point of
-view.
-
-Alice constructs her challenge in step 1, and sends it off. She receives a
-challenge and hash in step 2. When she computes the response to the
-challenge, she verifies the hash she received. If it matches, she knows that
-\begin{itemize}
-\item whoever she's talking to hasn't attempted to cheat her by sending a
- challenge for which he doesn't know the answer; and
-\item he has successfully received her challenge.
-\end{itemize}
-Because she's now received a challenge, she can work out her hash. She sends
-off her response to the challenge, together with the hash, and awaits the
-response.
-
-In step 4, the response arrives. If it's correct, she knows that it's from
-Bob, and that he (Bob) received her challenge OK. Tying everything else
-together is the tricky bit.
-
-If we assume that Bob is playing by the rules, the fact that he's sent his
-response means that he verified it against Alice's hash and decided that
-\begin{itemize}
-\item Alice wasn't trying to cheat him and find out about his private key;
- and
-\item Alice correctly received his challenge.
-\end{itemize}
-Because Bob wouldn't have replied if these weren't true, Alice can therefore
-believe that she has received Bob's challenge correctly.
-
-To summarize: Alice has managed to get a challenge to Bob, and he responded;
-Alice has also received Bob's challenge correctly.
-
-What if Bob isn't honest? The only hole in the protocol which can be
-exploited by Bob is that he can send a response \emph{even though} it doesn't
-match Alice's hash. This means that the protocol will continue even if Alice
-is attempting to cheat Bob and find information about his private key: this
-is a penalty Bob has to pay for not following the rules. The protocol still
-aborts if an adversary interferes with the challenges: if Alice isn't given
-Bob's challenge accurately, her response will be wrong, and Bob can abort the
-exchange; similarly, if Bob isn't given Alice's challenge, she will detect
-this and abort the exchange.
-
-
-\section{Conclusions and further work}
-
-We have presented a new key exchange protocol based upon a novel use of
-Diffie-Hellman key exchange as a means of authentication.
-
-The arguments given in the previous section sound fairly convincing, but they
-don't provide a formal proof of the security of the Wrestlers Protocol. The
-authors are unaware of a logic system for verifying protocols which correctly
-capture the properties of hash functions.
%%%----- That's all, folks --------------------------------------------------
+\bibliography{cryptography,mdw-crypto}
\end{document}
%%% Local Variables:
%%% mode: latex
%%% TeX-master: "wrestlers"
-%%% End:
+%%% End:
~mdw / tripe / blobdiff