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+%%% -*-latex-*-
+%%%
+%%% $Id: wrestlers.tex,v 1.1 2001/02/16 21:43:33 mdw Exp $
+%%%
+%%% Description of the Wrestlers Protocol
+%%%
+%%% (c) 2001 Mark Wooding
+%%%
+
+%%%----- Revision history ---------------------------------------------------
+%%%
+%%% $Log: wrestlers.tex,v $
+%%% 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}
+
+\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.
+\end{abstract}
+
+\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{A simple authentication protocol}
+
+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 $g$.
+
+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.
+
+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:
+\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.
+\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:
+\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.
+\end{enumerate}
+
+
+\section{The Wrestlers Protocol}
+
+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.
+
+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'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}
+
+And now in detail:
+\begin{enumerate}
+
+\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 --------------------------------------------------
+
+\end{document}
+
+%%% Local Variables: 
+%%% mode: latex
+%%% TeX-master: "wrestlers"
+%%% End: 
