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41761fdc 1\xcalways\section{Message authentication codes}\x
2
3\xcalways\subsection{Definitions and security notions}\x
4
5\begin{slide}
7
8 A MAC is a pair of algorithms $\mathcal{M} = (T, V)$:
9 \begin{itemize}
10 \item The \emph{tagging} algorithm $T\colon \{0, 1\}^k \times \{0, 1\}^* 11 \to \{0, 1\}^L$ is a probabilistic algorithm which, given a key and a
12 string, returns a \emph{tag}. We write $\tau \in T_K(m)$.
13 \item The \emph{verification} algorithm $V\colon \{0, 1\}^k \times \{0, 14 1\}^* \times \{0, 1\}^L \to \{0, 1\}$ is a deterministic algorithm which,
15 given a key, a message and a tag, returns $1$ if the tag is valid, or $0$
16 otherwise; i.e., we require that $V_K(m, \tau) = 1 \iff \tau \in T_K(m)$.
17 \end{itemize}
18 The basic idea is that it's hard for an adversary to \emph{forge} a tag for
19 a message it's not seen before.
20\end{slide}
21
22\begin{slide}
23 \topic{informal security notion}
53aa10b5 24 \resetseq
25 \head{Strong MACs, \seq: informal security notion}
41761fdc 26
27 Our standard notion of security for MACs is \emph{strong unforgeability
28 against chosen message attack}, or SUF-CMA, for short
29 \cite{Abdalla:2001:DHIES, Bellare:2000:AER}. Let $A$ be an adversary which
30 is attempting to attack the MAC $\mathcal{M} = (T, V)$.
31
32 We play a game with the adversary. We invent a key $K \inr \{0, 1\}^k$.
33 The adversary \emph{wins} if, after requesting tags for some messages of
34 its choice, and checking some guesses, it can return a pair $(m, \tau)$
35 such that:
36 \begin{itemize}
37 \item the tag is correct, i.e., $V_K(m, \tau) = 1$; and
38 \item the tag is not one returned by the adversary's tagging oracle for
39 that message.
40 \end{itemize}
41\end{slide}
42
43\begin{slide}
44 \topic{strong MACs}
53aa10b5 45 \head{Strong MACs, \seq: the experiment}
41761fdc 46
47 We perform the following experiment with the adversary.
48 \begin{program}
49 Experiment $\Expt{suf-cma}{\mathcal{M}}(A)$: \+ \\
50 $K \getsr \{0, 1\}^k$; \\
51 $\Xid{T}{list} \gets \emptyset$; \\
52 $(m, \tau) \gets A^{\id{tag}(\cdot), V_K(\cdot, \cdot)}$; \\
53 \IF $V_K(m, \tau) \land (m, \tau) \notin \Xid{T}{list}$
54 \THEN \RETURN $1$; \\
55 \ELSE \RETURN $0$; \- \$\smallskipamount] 56 Oracle \id{tag}(m): \+ \\ 57 \tau \gets T_K(m); \\ 58 \Xid{T}{list} \gets \Xid{T}{list} \cup \{(m, \tau)\}; \\ 59 \RETURN \tau; 60 \end{program} 61\end{slide} 62 63\begin{slide} 53aa10b5 64 \head{Strong MACs, \seq: wrapping up the notation} 41761fdc 65 66 The \emph{success probability} of an adversary A against the MAC 67 \mathcal{M} in the sense of SUF-CMA is 68 \[ \Succ{suf-cma}{\mathcal{M}}(A) = 69 \Pr[\Expt{suf-cma}{\mathcal{M}}(A) = 1].$%
70 The \emph{insecurity} of a MAC $\mathcal{M}$ in the SUF-CMA sense is then
71 $\InSec{suf-cma}(\mathcal{M}; t, q_T, q_V) = 72 \max_A \Succ{suf-cma}{\mathcal{M}}(A)$%
73 where the maximum is taken over all adversaries $A$ running in time $t$ and
74 making $q_T$ tagging and $q_V$ verification queries.
75
76 If $\InSec{suf-cma}(\mathcal{M}; t, q_T, q_V) \le \epsilon$ then we say
77 that $\mathcal{M}$ is a \emph{$(t, q_T, q_V, \epsilon)$-secure MAC in the
78 SUF-CMA sense}.
79\end{slide}
80
81\begin{slide}
82 \topic{other notions}
83 \head{Other security notions for MACs}
84
85 There are a number of weaker security notions in use:
86 \begin{itemize}
87 \item The definition of a \emph{weak MAC} restricts the adversary from
88 returning any message with which it queried its tagging oracle. The
89 strong MAC definition considers this OK, as long as the tag is different
90 from any returned by the oracle for that message.
91 \item Some definitions of MACs don't equip the adversary with a
92 verification oracle. Our definition considers these to be $(t, q_T, 0, 93 \epsilon)$-secure.
94 \item You can have a MAC with a bounded domain $\{0, 1\}^L$ rather than
95 $\{0, 1\}^*$ as shown previously.
96 \item Further quantification is possible, e.g., counting the total number
97 of bytes queried, or the maximum size of a tagging query.
98 \end{itemize}
99\end{slide}
100
41761fdc 101\xcalways\subsection{Basic results}\x
102
103\begin{slide}
104 \topic{PRFs are MACs}
53aa10b5 105 \resetseq
41761fdc 107
108 If $F_K\colon \{0, 1\}^* \to \{0, 1\}^L$ is a $(t, q, \epsilon)$-secure
109 PRF, then it's also a $(t', q_T, q_V, \epsilon')$-secure MAC, with $q = q_T 110 + q_V + 1$, $t = t' + O(q)$, and $\epsilon \le \epsilon' + (q_V + 1) 111 2^{-L}$. The constant hidden by the $O(\cdot)$ is small and depends on the
112 model of computation.
113
114 Suppose $A$ can break $F$ used as a MAC in time $t$ and with $q_T$ and
115 $q_V$ queries to its tagging and verification oracles respectively.
116
117 If we can construct an adversary which distinguishes $F_K$ from a random
118 function using $A$ as an essential component, then we can prove the
119 result.
120\end{slide}
121
122\begin{slide}
53aa10b5 123 \head{PRFs are MACs, \seq: the distinguisher}
41761fdc 124
125 \begin{program}
126 Distinguisher $D^{F(\cdot)}$: \+ \\
127 $\Xid{T}{list} \gets \emptyset$; \\
128 $(m, \tau) \gets A^{T_F(\cdot), V_F(\cdot, \cdot)}$; \\
129 \IF $m \notin \Xid{T}{list} \land \tau = F(m)$
130 \THEN \RETURN $1$; \\
131 \ELSE \RETURN $0$; \- \$\smallskipamount] 132 Oracle T_F(m): \+ \\ 133 \Xid{T}{list} \gets \Xid{T}{list} \cup \{m\}; \\ 134 \RETURN F(m); \- \\[\smallskipamount] 135 Oracle V_F(m, \tau): \+ \\ 136 \IF \tau = F(m) \THEN \RETURN 1; \\ 137 \ELSE \RETURN 0; 138 \end{program} 139\end{slide} 140 141\begin{slide} 53aa10b5 142 \head{PRFs are MACs, \seq: wrapping up} 41761fdc 143 144 The distinguisher simulates the tagging and verification oracles for the 145 MAC forger, using its supplied oracle. If the forger succeeds, then the 146 distinguisher returns 1; otherwise it returns zero. 147 148 The probability that the distinguisher returns 1 given an instance F_K of 149 the PRF is precisely \Succ{suf-cma}{F}(A). 150 151 The probability that it returns 0 given a random function depends on what 152 A does when it's given a random function. But we know that the 153 probability of it successfully guessing the MAC for a message for which it 154 didn't query T can be at most (q_V + 1) 2^{-L}. So 155 \[ \Adv{prf}{F}(D) \le \Succ{suf-cma}{F}(A) - (q_V + 1) 2^{-L}.$
156 Let $q = q_T + q_V + 1$; then counting, rearranging, maximizing yields
157 $\InSec{suf-cma}(F; t, q_T, q_V) \le 158 \InSec{prf}(F; t + O(q), q) + (q_V + 1)2^{-L}.$%
159\end{slide}
160
53aa10b5 161\begin{slide}
162 \head{PRFs are MACs, \seq: MACs aren't PRFs}
163
164 The converse of our result is not true. Suppose $\mathcal{M} = (T, V)$ is
165 a deterministic MAC. Choose some integer $n$. Then define $\mathcal{M}' = 166 (T', V')$, as follows:
167 $T'_K(x) = 0^n \cat T_K(x); \qquad 168 V'_K(x, \tau) = \begin{cases} 169 1 & if T'_K(x) = \tau \\ 170 0 & otherwise 171 \end{cases}. 172$
173 $T'$ is obviously not a PRF: an adversary checking for the string of $n$
174 zero bits on the output will succeed with advantage $1 - 2^{-qn}$.
175
176 However, $\mathcal{M}'$ is a secure MAC. Suppose $A'$ attacks
177 $\mathcal{M}'$.
178 \begin{program}
179 Adversary $A^{T(\cdot), V(\cdot)}$: \+ \\
180 $(m, \tau') \gets A'^{\id{tag}(\cdot), \id{verify}(\cdot)}$; \\
181 \PARSE $\tau'$ \AS $n\colon z, \tau$; \\
182 \RETURN $(m, \tau)$;
183 \next
184 Oracle $\id{tag}(m)$: \+ \\
185 \RETURN $0^n \cat T(m)$; \- \$\smallskipamount] 186 Oracle \id{verify}(m, \tau'): \+ \\ 187 \PARSE \tau' \AS n\colon z, \tau; \\ 188 \IF z \ne 0^n \THEN \RETURN 0; \\ 189 \ELSE \RETURN V(m, \tau); 190 \end{program} 191\end{slide} 192 41761fdc 193\begin{exercise} 194 \begin{parenum} 195 \item Suppose that F\colon \{0, 1\}^k \times \{0, 1\}^* \to \{0, 1\}^L is 196 a (t, q, \epsilon)-secure PRF. Let T^{(\ell)}_K(x) be the leftmost 53aa10b5 197 \ell~bits of F_K(x) for \ell \le L. Demonstrate the security of 198 T^{(\ell)}(\cdot) as a MAC. 41761fdc 199 \item Discuss the merits of truncating MAC tags in practical situations. 200 \end{parenum} 201 \answer% 202 \begin{parenum} 203 \item The follows exactly the same pattern as the PRFs are MACs' proof in 204 the slides: T^{(\ell)} is a (t, q_T, q_V, \epsilon + (q_V + 205 1)2^{-\ell})-secure MAC, where q_T + q_V = q. 206 \item Obviously, truncating tags saves bandwidth. There is a trade-off 207 between tag size and security, as the 2^{-\ell} term shows. Note that 208 this term represents the probability of the adversary guessing the 209 correct tag when it's actually attacking a random function, and 210 therefore, when this occurs, the adversary has one by accident'. 211 Including sequence numbers in packets ensures that replay of accidental 212 forgery (or honest messages) will be detected. Hence, for some 213 applications, setting \ell = 32 or even lower is of no particular harm. 214 Perhaps more significantly, if the PRF isn't actually as good as it ought 215 to be, and (say) leaks key material very slowly, then truncating its 216 output can actually improve security. 217 \end{parenum} 218\end{exercise} 219 220\begin{exercise} 221 A traditional MAC construction is the \emph{CBC-MAC}: it works like this. 222 Suppose F\colon \{0, 1\}^k \times \{0, 1\}^l \to \{0, 1\}^l is a PRF. 223 Split a message~x into l-bit blocks x_0, x_1, \ldots, x_{n-1} 224 (applying some sort of padding if you need to). Then we define the CBC-MAC 225 as F^{(n)}_K(x), where 226 \[ F^{(1)}_K(x) = F_K(x); 227 \qquad F^{(i+i)}(x) = F_K(x_i \xor F^{(i)}_K(x)).$%
228 In \cite{Bellare:1994:SCB}, Mihir Bellare, Joe Kilian and Phil Rogaway
229 introduced the world to the concrete-security approach and, almost
230 incidentally, proved that the CBC-MAC is a PRF (and therefore a MAC) for
231 any \emph{fixed sized} input.
232
233 Show that the CBC-MAC is easily broken if you're allowed to choose messages
234 of different lengths.
236 Request tags $\tau$ for the message $x = x_0, x_1, \ldots, x_{n-1}$ and
237 $\tau'$ for $x' = x'_0 \xor \tau, x'_1, \ldots, x'_{n'-1}$. Let $y = x_0, 238 x_1, \ldots, x_{n-1}, x'_0 , x'_1, \ldots, x'_{n'-1}$. Note that
239 $F^{(n)}_K(y) = \tau$, and $F^{(n+1)}_K(y) = F_K(x'_0 \xor F^{(n)}_K(x)) = 240 F^{(1)}_K(x')$. Hence, $F^{(n+n')}_K(y) = \tau'$, and we have a correct
241 forgery.
242\end{exercise}
243
244\begin{slide}
245 \topic{verification oracles}
247
248 We can leave verification oracles out of our analysis. This simplifies
249 analysis, but produces slightly less satisfactory quantitative results.
250
251 Suppose that $\mathcal{M} = (T, V)$ is a $(t, q_T, 0, \epsilon)$-secure
252 MAC. Then, for any $q_V$,
253 $\InSec{suf-cma}(\mathcal{M}; t, q_T, q_V) \le 254 (q_V + 1)\InSec{suf-cma}(\mathcal{M}; t, q_T, 0).$%
255 This bound is \emph{tight}: it's not possible for a general result like
256 this to do better.
257\end{slide}
258
259\begin{proof}
260 Consider an adversary $A$ which uses $q_V$ verification queries. We assume
261 the following properties of $A$'s behaviour:
262 \begin{itemize}
263 \item No verification query contains a message and a tag for that message
264 received from the tagging oracle.
265 \item If a verification query succeeds, the message is not given in a query
266 to the tagging oracle.
267 \item Once a verification query succeeds, all subsequent verification
268 queries also succeed and the adversary returns a correct forgery (e.g.,
53aa10b5 269 by simply repeating the successful query).
41761fdc 270 \end{itemize}
271 It's clear that any adversary can be transformed into one which has these
272 properties and succeeds with probability at least as high.
273
274 Let $V$ be the event that at least one verification query succeeds, and let
275 $S$ be the event that $A$ succeeds. Then
276 \begin{eqnarray*}[rl]
277 \Succ{suf-cma}{\mathcal{M}}(A)
278 &= \Pr[S \mid V] \Pr[V] + \Pr[S \mid \lnot V] \Pr[\lnot V] \\
279 &= \Pr[V] + \Pr[S \mid \lnot V] \Pr[\lnot V].
280 \end{eqnarray*}
281 Now consider these two adversaries:
282 \begin{program}
283 Adversary $A'^{T(\cdot), V(\cdot, \cdot)}$: \+ \\
284 $i \gets 0$; \\
285 $(m, \tau) \gets A^{T(\cdot), \Xid{V}{sim}(\cdot, \cdot)}$; \\
286 \RETURN $(m^*, \tau^*)$; \- \$\smallskipamount] 287 Oracle \Xid{V}{sim}(m, \tau): \+ \\ 288 i \gets i + 1; \\ 289 \IF i < q_V \THEN \RETURN V(m, \tau); \\ 290 (m^*, \tau^*) \gets (m, \tau); \\ 291 \RETURN 1; 292 \next 293 Adversary Z^{T(\cdot), V(\cdot, \cdot)}: \+ \\ 294 \\ 295 (m, \tau) \gets A^{T(\cdot), \Xid{V}{zero}(\cdot, \cdot)}; \\ 296 \RETURN (m, \tau); \- \\[\smallskipamount] 297 Oracle \Xid{V}{zero}(m, \tau): \+ \\ 298 \RETURN 0; 299 \end{program} 300 The adversary A' uses q_V - 1 verification queries. It ignores the 301 output of A, returning instead A's q_V-th verification query. Thus, 302 by our assumptions on the behaviour of A, we have that A' succeeds 303 precisely whenever one of A's verification queries succeeds. Thus: 304 \[ \Pr[V] = \Succ{suf-cma}{\mathcal{M}}(A') 305 \le \InSec{suf-cma}(\mathcal{M}; t, q_T, q_V - 1).$%
306 Similarly, the adversary $Z$ succeeds with precisely the same probability
307 as $A$, given that all of its verification queries failed; i.e.,
308 $\Pr[S \mid \lnot V] \Pr[\lnot V] = \Succ{suf-cma}{\mathcal{M}}(Z) 309 \le \InSec{suf-cma}(\mathcal{M}; t, q_T, 0).$%
310 Because $A$ was chosen arbitrarily, we can maximize:
311 \begin{eqnarray*}[rl]
312 \InSec{suf-cma}(\mathcal{M}; t, q_T, q_V)
313 & \le \InSec{suf-cma}(\mathcal{M}; t, q_T, q_V - 1) +
314 \InSec{suf-cma}(\mathcal{M}; t, q_T, 0) \\
315 & \le (q_V + 1)\InSec{suf-cma}(\mathcal{M}; t, q_T, 0)
316 \end{eqnarray*}
317 as required.
318
319 To show that the bound is tight, consider a random function $F$ used as a
320 MAC, with $L$-bit output. Then we claim that $\InSec{suf-cma}(F; t, q_T, 321 q_V) = (q_V + 1)/2^L$. To save typing, let $e_q = \InSec{suf-cma}(F; t, 322 q_T, q)$. We prove the claim by induction on $q$. Firstly, note that if
323 $q = 0$ then necessarily $e_q = 1/2^L$. Now suppose that $e_{q-1} = 324 q/2^L$. We're now allowed an extra query, so rather than returning the
325 result, we feed it to the verification oracle. If it answers yes' then we
326 return it; otherwise we guess randomly from the remaining $2^L - q$
327 possibilities. Now
328 \begin{eqnarray*}[rl]
53aa10b5 329 e_q &= e_{q-1} + \frac{1 - e_{q-1}}{2^L - q} \\
41761fdc 330 &= \frac{q}{2^L} + \frac{2^L - q}{2^L} \cdot \frac{1}{2^L - q} \\
331 &= \frac{q + 1}{2^L}
332 \end{eqnarray*}
333 as claimed.
334\end{proof}
335
336\xcalways\subsection{The HMAC construction}\x
337
338\begin{slide}
53aa10b5 339 \resetseq
340 \head{The HMAC construction \cite{Bellare:1996:KHF}, \seq: motivation}
341
41761fdc 342 It ought to be possible to construct a decent MAC using a hash function.
343 Many attempts have failed, however. For example, these constructions are
53aa10b5 344 weak if used with standard one-pass Merkle-Damg\aa{}rd iterated hashes.
41761fdc 345 \begin{itemize}
53aa10b5 346 \item Secret prefix: $T_K(m) = H(K \cat m)$. Given $H(K \cat m)$, it's
347 easy to compute $H(K \cat m \cat p \cat m')$ for a padding string $p$ and
348 arbitrary suffix $m'$.
349 \item Secret suffix: $T_K(m) = H(m \cat K)$. Finding a collision $H(m) = 350 H(m')$ yields $H(m \cat K) = H(m' \cat K)$. We saw earlier that
351 adversaries which know collisions \emph{exist} even if we don't know how
352 to describe them.
41761fdc 353 \end{itemize}
354
355 It would be nice to have a construction whose security was provably related
356 to some plausible property of the underlying hash function.
357\end{slide}
358
359\begin{slide}
53aa10b5 360 \head{The HMAC construction, \seq: definition of NMAC}
41761fdc 361
362 Let $H\colon \{0, 1\}^* \to \{0, 1\}^k$ be an iterated hash, constructed
363 from the compression function $F\colon \{0, 1\}^k \times \{0, 1\}^L \to 364 \{0, 1\}^k$. We define a keyed version of $H$. Let $K \in \{0, 1\}^k$;
365 then we compute $H_K(x)$ as follows:
366 \begin{enumerate}
367 \item Pad and split $x$ into the $L$-bit blocks $x_0$, $x_1$, \ldots,
368 $x_{n-1}$ as before.
369 \item Set $I_0 = K$. Let $I_{i+1} = F(I_i \cat x_i)$ for $0 \le i < n$.
370 \item The result $H_K(x) = I_n$.
371 \end{enumerate}
372 The NMAC (nested-MAC) construction requires two independent $k$-bit keys
373 $K_0$ and $K_1$. The construction itself is simply:
374 $\Xid{T}{NMAC}^H_{K_0, K_1}(x) = H_{K_0}(H_{K_1}(x)).$
375 NMAC is deterministic, so verification consists of computing the tag and
376 comparing.
377\end{slide}
378
379\begin{slide}
53aa10b5 380 \head{The HMAC construction, \seq: security of NMAC}
41761fdc 381
382 Consider a function $F\colon \{0, 1\}^k \times \{0, 1\}^* \to \{0, 1\}^k$.
383 We say that $F$ is \emph{$(t, q, \epsilon)$-weakly collision resistant} if,
384 for any adversary $A$ constrained to run in time $t$ and permitted $q$
385 oracle queries,
386 $\Pr[K \getsr \{0, 1\}^k; 387 (x, y) \gets A^{F_K(\cdot)}] \le \epsilon : 388 x \ne y \land F_K(x) = F_K(y)$%
389
390 If $H_K$ is a $(t, q_T, q_V, \epsilon)$-secure MAC on $k$-bit messages, and
391 moreover $(t, q_T + q_V, \epsilon')$-weakly collision resistant, then
392 $\Xid{T}{NMAC}^H$ is a $(t, q_T, q_V, \epsilon + \epsilon')$-secure MAC.
393\end{slide}
394
395\begin{slide}
53aa10b5 396 \head{The HMAC construction, \seq: NMAC security proof}
41761fdc 397
398 Let $A$ be an adversary which forges a $\Xid{T}{NMAC}^H$ tag in time $t$,
399 using $q_T$ tagging queries and $q_V$ verification queries with probability
400 $\epsilon$. We construct an adversary $A'$ which forces a $H$ tag for a
401 $k$-bit in essentially the same time.
402 \begin{program}
403 Adversary $A'^{T(\cdot), V(\cdot, \cdot)}$ \+ \\
404 $K \getsr \{0, 1\}^k$; \\
405 $(m, \tau) \gets A^{T(H_K(\cdot)), V(H_K(\cdot), \cdot)}$; \\
406 \RETURN $(H_K(m), \tau)$;
407 \end{program}
408 $A'$ might fail even though $A$ succeeded only if the message it returns,
409 $H_K(m)$, collides with one of its tagging queries. But since $H_K$ is
410 $(t, q_T + q_V, \epsilon')$-weakly collision resistant, this happens with
411 at most probability $\epsilon'$. Hence, $A'$ succeeds with probability at
412 least $\epsilon - \epsilon'$. Rearrangement yields the required result.
413\end{slide}
414
415\begin{slide}
53aa10b5 416 \head{The HMAC construction, \seq: from NMAC to HMAC}
41761fdc 417
418 Implementing NMAC involves using strange initialization vectors and
419 generally messing about with your hash function. HMAC is an attempt to
420 capture the provable security properties using a plain ol' hash function.
421
422 Suppose $H$ is an iterated hash function with a $k$-bit output and $L$-bit
423 blocks (with $L \ge k$). We set $\id{ipad}$ to be the byte $\hex{36}$
424 repeated $L/8$ times, and $\id{opad}$ to be the byte $\hex{5C}$ repeated
425 $L/8$ times. Select a key $K$ of $L$ bits: if your key is shorter, pad it
426 by appending zero bits; if it's longer, hash it with $H$ and then pad.
427
428 The HMAC tagging function is then defined as
429 $\Xid{T}{HMAC}^H_K(m) = 430 H(K \xor \id{opad} \cat H(K \xor \id{ipad} \cat m)).$%
431\end{slide}
432
433\begin{slide}
53aa10b5 434 \head{The HMAC construction, \seq: comparison with NMAC}
41761fdc 435
436 Comparing the two constructions, we see that
437 $\Xid{T}{HMAC}^H_K = 438 \Xid{T}{NMAC}^{H'}_{F(I \cat K \xor \id{opad}), 439 F(I \cat K \xor \id{ipad})}.$%
440 Here, $I$ is $H$'s initialization vector, $F$ is the compression function;
441 $H'$ denotes a keyed hash function that is like' $H$ but performs padding
442 as if there were an extra initial block of message data for each message.
443
444 The proof of NMAC assumes that the two keys are random and independent.
445 This isn't the case in HMAC, but a little handwaving about pseudorandomness
446 of the compression function makes the problem go away.
447\end{slide}
448
449\begin{exercise}
450 Suppose that $F\colon \{0, 1\}^k \times \{0, 1\}^{t+\ell} \to \{0, 451 1\}^t$ is a PRF. Let $x \in \{0, 1\}^*$ be a message. We define the
452 function $H_K(x)$ as follows:
453 \begin{itemize}
454 \item Pad $x$ to a multiple of $\ell$ bits using some injective
455 mapping. Break the image of $x$ under this mapping into $\ell$-bit
456 blocks $x_0, x_1, \ldots, x_{n-1}$.
457 \item For $0 \le i \le n$, define $H^{(i)}_K(x)$ by
458 $H^{(0)}_K(x) = I; \qquad 459 H^{(i+1)}_K(x) = F_K(H^{(i)}(x) \cat x_i)$%
460 where $I$ is some fixed $t$-bit string (e.g., $I = 0^t$).
461 \item Then set $H_K(x) = H^{(n)}_K(x)$.
462 \end{itemize}
463 We define two (deterministic) MACs $\mathcal{M}^i = (T^i, V^i)$ (for
464 $i \in \{0, 1\}$) using the $H_K$ construction. Verification in each
465 case consists of computing the tag and comparing to the one offered.
466 \begin{eqlines*}
53aa10b5 467 T^0_K(x) = H_K(x); \qquad T^1_K(x) = H_K(x \cat K); \\
41761fdc 468 V^i_K(x, \tau) = \begin{cases}
469 1 & if $\tau = T^i_K(x)$ \\
470 0 & otherwise
53aa10b5 471 \end{cases}.
41761fdc 472 \end{eqlines*}
473 Decide whether each of these constructions is secure. A full proof is
474 rather hard: an informal justification would be good.
476 $\mathcal{M}^0$ is secure; $\mathcal{M}^1$ isn't, under the sole
477 assumption that $F$ is a PRF.
478
479 To see that $\mathcal{M}^0$ is secure, it suffices to show that $T^0$
480 is a PRF. This is actually quite involved. Given an adversary $A$
481 attacking $T^1$ as a PRF, we construct an adversary $B$ attacking $F$,
482 which simply computes $H$ as required, using the oracle supplied. To
483 complete the proof, we need to show a bound on the
484 information-theoretic security of $H$ when instantiated using a random
485 function $F$. For the sake of simplicity, we allow the adversary $A$
487 messages. We count the number $q'$ of individual message blocks.
488
489 As the game with $A$ progresses, we can construct a directed
490 \emph{graph} of the query results so far. We start with a node
491 labelled $I$. When processing an $H$-query, each time we compute $t' 492 = F(t \cat x_i)$, we add a node $t'$, and an edge $x_i$ from $t$ to
493 $t'$. The `bad' event occurs whenever we add an edge to a previously
494 existing node. We must show, firstly, that the
495 adversary cannot distinguish $H$ from a random function unless the bad
496 event occurs; and, secondly, that the bad event doesn't occur very
497 often.
498
499 The latter is easier: our standard collision bound shows that the bad
500 event occurs during the game with probability at most $q'(q' - 1)/2$.
501
502 The former is trickier. This needs a lot more work to make it really
503 rigorous, but we show the idea. Assume that the bad event has not
53aa10b5 504 occurred. Consider a query $x_0, x_1, \ldots, x_{n-1}$. If it's the
41761fdc 505 same as an earlier query, then $A$ learns nothing (because it could
506 have remembered the answer from last time). If it's \emph{not} a
507 prefix of some previous query, then we must add a new edge to our
508 graph; then either the bad event occurs or we create a new node for
509 the result, and because $F$ is a random function, the answer is
510 uniformly random. Finally, we consider the case where the query is a
511 prefix of some earlier query, or queries. But these were computed at
512 random at the time.
513
514 At the end of all of this, we see that
515 $\InSec{prf}(T^0; t, q) \le 516 \InSec{prf}(F; t, q') + \frac{q'(q' - 1)}{2}$%
517 and hence
518 $\InSec{suf-cma}(\mathcal{M}^0; t, q) \le 519 \InSec{prf}(F; t, q') + \frac{q'(q' - 1)}{2} + \frac{1}{2^t}.$%
520
521 Now we turn our attention to $T^1$. It's clear that we can't simulate
522 $T^1$ very easily using an oracle for $F$, since we don't know $K$
523 (and indeed there might not be a key $K$). The intuitive reason why
524 $T^1$ is insecure is that $F$ might have leak useful information if
525 its input matches its key. This doesn't affect the strength of $F$ as
526 a PRF because you have to know the key before you can exploit this
527 leakage; but $T^1$ already knows the key, and this can be exploited to
528 break the MAC.
529
530 To show that this is insecure formally, let $F'$ be defined as
531 follows:
532 $F'_K(x) = \begin{cases} 533 K & if x = p \cat K \cat q where 534 |p| = t and |q| = \ell - k \\ 535 F_K(x) & otherwise 536 \end{cases}.$%
537 We choose a simple injective padding scheme: if $x$ is a message then
538 we form $x' = x \cat 1 \cat 0^n$, where $0 \le n < \ell$ and $|x'|$ is
539 a multiple of $\ell$. If $T^1$ is instantiated with this PRF then it
540 is insecure as a MAC: submitting a tagging query for the empty string
541 $\emptystring$ reveals the key $K$, which can be used to construct a
542 forgery.
543
544 To complete the proof, we must show that $F'$ is a PRF. Let $A$ be an
545 adversary attacking $F'$. We consider a series of games; for each
546 $\G{i}$, let $S_i$ be the event that $A$ returns $1$ in that game.
547 Game~$\G0$ is the standard attack game with $A$ given an oracle for a
548 random function; game~$\G1$ is the same, except that $A$ is given an
549 oracle for $F'_K$ for some $K \inr \{0, 1\}^k$. Then
550 $\Adv{prf}{F'}(A) = \Pr[S_1] - \Pr[S_0]$. Let game~$\G2$ be the same
551 as $\G1$, except that if $A$ makes any query of the form $p \cat K 552 \cat q$ with $|p| = t$ and $|q| = \ell - k$ then the game halts
553 immediately, and let $F_2$ be the event that this occurs. By
53aa10b5 554 Lemma~\ref{lem:shoup} (slide~\pageref{lem:shoup}), then, $|{\Pr[S_2]} - 41761fdc 555 \Pr[S_1]| \le \Pr[F_2]$. Let game~$\G3$ be the same as $\G2$ except
556 that we give $A$ an oracle for $F_K$ rather than $F'_K$. Since $F$
557 and $F'$ differ only on queries of the form $p \cat K \cat q$, we have
558 $\Pr[S_3] = \Pr[S_2]$. But $\Pr[S_3] - \Pr[S_0] = \Adv{prf}{F}(A) \le 559 \InSec{prf}(F; t, q)$. Hence, $\Adv{prf}{F'}(A) \le \InSec{prf}{F}(A) 560 - \Pr[F_2]$.
561
562 Finally, we bound the probability of $F_2$. Fix an integer $n$.
563 Consider an adversary $B$ attacking $F$ which runs as follows. It
564 initially requests $F(0), F(1), \ldots, F(n - 1)$ from its oracle. It
565 then runs $A$, except that, for each oracle query $x$, it parses $x$
566 as $p \cat K' \cat q$ with $|p| = t$, $|K'| = k$ and $|q| = \ell - k$;
567 then, if $F_{K'}(0) = F(0) \land F_{K'}(1) = F(1) \land \cdots \land 568 F_{K'}(n - 1) = F(n - 1)$, $B$ immediately returns $1$, claiming that
53aa10b5 569 its oracle $F$ is the function $F_{K'}$; if this never occurs, $B$
41761fdc 570 returns $0$. Clearly, if $B$ is given an instance $F_K$ of $F$ then
571 it succeeds with probability $\Pr[F_2]$; however, if $F$ is a random
572 function then $B$ returns $1$ with probability at most $q 2^{-nk}$.
573 Hence, $\Adv{prf}{F}(B) \le \Pr[F_2] - q 2^{-nk}$. $B$ issues $q + n$
574 queries, and takes time $t + O(n q)$. Wrapping everything up, we get
575 $\InSec{prf}(F'; t, q) \le 576 2\cdot\InSec{prf}(F; t + O(q n), q + n) + \frac{q}{2^{nk}}.$%
53aa10b5 577 This completes the proof of generic insecurity for $\mathcal{M}^1$.
41761fdc 578\end{exercise}
579
580\xcalways\subsection{Universal hashing}\x
581
582\begin{slide}
583 \topic{almost-universal hash functions}
53aa10b5 584 \resetseq
41761fdc 586
587 Consider a family of hash functions $H\colon \keys H \times \dom H \to 588 \ran H$. We define
589 $\InSec{uh}(H) = 590 \max_{x \ne y} \Pr[K \getsr \keys H : H_K(x) = H_K(y)].$%
591 If $\InSec{uh}(H) \le \epsilon$ then we say that $H$ is
592 \emph{$\epsilon$-almost universal}. Note that the concept of
593 almost-universality is not quantified by running times.
594
595 If $H$ is $1/|{\ran H}|$-almost universal, then we say that $H$ is
596 \emph{universal}. Sometimes it's said that this is the best possible
597 insecurity: this isn't true.
598\end{slide}
599
600\begin{proof}[Counterexample]
601 Here's a function $H\colon \{0, 1, 2\} \times \{0, 1, 2, 3\} \to \{0, 1\}$
602 which is $\frac{1}{3}$-almost universal, though $|{\ran H}| = 2$:
603 \begin{quote} \item
604 \begin{tabular}[C]{c|cccc}
605 & 0 & 1 & 2 & 3 \\ \hlx{vhv}
606 0 & 0 & 1 & 0 & 1 \\
607 1 & 0 & 0 & 1 & 1 \\
608 2 & 0 & 1 & 1 & 0
609 \end{tabular}
610 \end{quote}
611\end{proof}
612
613\begin{slide}
614 \topic{dynamic view}
53aa10b5 615 \head{Universal hashing, \seq: a dynamic view}
41761fdc 616
617 Suppose that $H$ is $\epsilon$-almost universal. Consider this experiment:
618 \begin{program}
619 Experiment $\Expt{uh}{H}(A)$: \+ \\
620 $(x, y) \gets A$; \\
621 $K \getsr \keys H$; \\
622 \IF $x \ne y \land H_K(x) = H_K(y)$ \THEN \RETURN $1$; \\
623 \ELSE \RETURN $0$;
624 \end{program}
625 The adversary may make random decisions before outputting its selection
626 $x$, $y$. We show that $\Pr[\Expt{uh}{H}(A) = 1] \le \InSec{uh}(H) = 627 \epsilon$.
628
629 Let $\rho \in \{0, 1\}^*$ be $A$'s coin tosses: $A$ chooses $x$ and $y$ as
630 functions of $\rho$. For some \emph{fixed} $\rho$,
631 $\Pr[K \getsr \keys H : H_K(x(\rho)) = H_K(y(\rho))] \le 632 \InSec{uh}(H).$%
633\end{slide}
634
635\begin{slide}
53aa10b5 636 \head{Universal hashing, \seq: the dynamic view (cont.)}
41761fdc 637
638 Now we treat $\rho$ as a random variable, selected from some distribution
639 $P$ on the set $\{0, 1\}^*$. We see that
640 \begin{eqnarray*}[Ll]
641 \Pr[\rho \getsr P; K \getsr \keys H : H_K(x(\rho)) = H_K(y(\rho))] \\
642 &= \sum_{\rho \in \{0, 1\}^*}
643 P(\rho) \cdot \Pr[K \getsr \keys H : H_K(x(\rho)) = H_K(y(\rho))] \\
644 &\le \sum_{\rho \in \{0, 1\}^*} P(\rho) \cdot \InSec{uh}(H)
645 = \InSec{uh}(H).
646 \end{eqnarray*}
647 Thus, no adversary can succeed in producing a collision in an
648 $\epsilon$-almost universal hash with probability better than $\epsilon$.
649 But obviously the adversary can ignore its coin tosses and simply return
650 the best colliding pair. Hence the two notions are completely equivalent.
651\end{slide}
652
653\begin{slide}
654 \topic{composition}
53aa10b5 655 \head{Universal hashing, \seq: composition}
41761fdc 656
657 Suppose that $G$ is $\epsilon$-almost universal, and $G'$ is
658 $\epsilon'$-almost universal, and $\dom G = \ran G'$. We define the
659 composition $G \compose G'$ to be the family $H\colon (\keys G \times 660 \keys G') \times \dom G' \to \ran G$ by $H_{k, k'}(m) = 661 G_k(G'_{k'}(m))$.
662
663 Then $H$ is $(\epsilon + \epsilon')$-almost universal. To see this, fix $x 664 \ne y$, and choose $K = (k, k') \inr \keys G \times \keys G'$. Let $x' = 665 G'_{k'}(x)$ and $y' = G'_{k'}(y)$. Following our previous result, we see:
666 \begin{eqnarray*}[rl]
667 \Pr[H_K(x) = H_K(y)]
668 &= \Pr[G_k(G'_{k'}(x)) = G_k(G'_{k'}(y))] \\
669 &= \Pr[G_k(x') = G_k(y')] \\
670 &= \Pr[G_k(x') = G_k(y') \mid x' \ne y'] \Pr[G'_{k'}(x) \ne G'_{k'}(y)]\\
671 &\le \epsilon + \epsilon'.
672 \end{eqnarray*}
673\end{slide}
674
675\begin{slide}
676 \topic{the collision game}
53aa10b5 677 \head{Universal hashing, \seq: the collision game}
41761fdc 678
679 Suppose that, instead of merely a pair $(x, y)$, our adversary was allowed
680 to return a \emph{set} $Y$ of $q$ elements, and measure the probability
681 that $H_K(x) = H_K(y)$ for some $x \ne y$ with $x, y \in Y$, and for $K 682 \inr \keys H$.
683
53aa10b5 684 Let $\InSec{uh-set}(H; q)$ be maximum probability achievable for sets $Y$
41761fdc 685 with $|Y| \le q$. Then
686 $\InSec{uh-set}(H; q) \le \frac{q(q - 1)}{2} \cdot \InSec{uh}(H) .$
687\end{slide}
688
689\begin{proof}
690 This is rather tedious. We use the dynamic view. Suppose $A$ returns $(x, 691 Y)$ with $|Y| = q$, and succeeds with probability $\epsilon$. Consider
692 \begin{program}
693 Adversary $A'$: \+ \\
694 $(x, Y) \gets A$; \\
695 $y \getsr Y$; \\
696 \RETURN $(x, Y \setminus \{y\})$;
697 \end{program}
698 The worst that can happen is that $A'$ accidentally removes the one
699 colliding element from $Y$. This occurs with probability $2/q$. So
700 $\Succ{uh-set}{H}(A') \ge \frac{q - 2}{q} \Succ{uh-set}{H}(A).$
53aa10b5 701 Rearranging and maximizing gives
41761fdc 702 $\InSec{uh-set}(H; q) \le 703 \frac{q}{q - 2} \cdot \InSec{uh-set}(H; q - 1).$
704 Note that $\InSec{uh-set}(H; 2) = \InSec{uh}(H)$ is our initial notion. A
705 simple inductive argument completes the proof.
706\end{proof}
707
708\begin{slide}
709 \topic{a MAC}
53aa10b5 710 \head{Universal hashing, \seq: a MAC}
41761fdc 711
53aa10b5 712 Suppose that $H\colon \{0, 1\}^k \times \{0, 1\}^* \to \{0, 1\}^l$ is an
713 almost universal hash function, and $F\colon \{0, 1\}^{k'} \times \{0, 41761fdc 714 1\}^l \to \{0, 1\}^L$ is a PRF\@. Define a MAC $\Xid{\mathcal{M}}{UH}^{H, 715 F} = (\Xid{T}{UH}^{H, F}, \Xid{V}{UH}^{H, F})$ where:
716 \begin{eqnarray*}[rl]
717 \Xid{T}{UH}^{H, F}_{K, K'}(m) &= F_{K'}(H_K(m)) \\
718 \Xid{V}{UH}^{H, F}_{K, K'}(m, \tau) &= \begin{cases}
719 1 & if $\tau = F_{K'}(H_K(m))$ \\
720 0 & otherwise
721 \end{cases}.
722 \end{eqnarray*}
723 We have
724 \begin{eqnarray*}[Ll]
725 \InSec{suf-cma}(\Xid{\mathcal{M}}{UH}^{H, F}; t, q_T, q_V) \\
726 & \le
727 (q_V + 1) \biggl(\InSec{prf}(F; t, q_T + 1) + \frac{1}{2^L} +
728 \frac{q_T(q_T - 1)}{2} \cdot \InSec{uh}(H)\biggr).
729 \end{eqnarray*}
730\end{slide}
731
732\begin{proof}
733 We shall prove the result for $q_V = 0$ and $q_T = q$, and appeal to the
734 earlier result on verification oracles.
735
736 Suppose $A$ attacks the scheme $\Xid{\mathcal{M}}{UH}^{H, F}$ in time $t$,
737 issuing $q$ tagging queries. Consider a distinguisher $D$, constructed
738 from a forger $A$:
739 \begin{program}
740 Distinguisher $D^{F(\cdot)}$: \+ \\
741 $K \getsr \{0, 1\}^k$; \\
742 $\Xid{T}{list} \gets \emptyset$; \\
743 $(m, \tau) \gets A^{\id{tag}(K, \cdot)}$; \\
744 \IF $m \notin \Xid{T}{list} \land \tau = F(H_K(m))$
745 \THEN \RETURN $1$; \\
746 \ELSE \RETURN $0$; \- \$\smallskipamount] 747 Oracle \id{tag}(K, m): \+ \\ 748 \Xid{T}{list} \gets \Xid{T}{list} \cup \{m\}; \\ 749 \RETURN F(H_K(m)); \- \\[\smallskipamount] 750 \end{program} 751 Note that A isn't provided with a verification oracle: that's because we 752 didn't allow it any verification queries. 753 754 We can see immediately that 755 \[ \Pr[K \getsr \{0, 1\}^{k'} : D^{F_K(\cdot)} = 1] = 756 \Succ{suf-cma}{\Xid{\mathcal{M}}{UH}^{H, F}}(A).$%
757
758 We must now find an upper bound for $\Pr[F \getsr \Func{l}{L} : 759 D^{F(\cdot)}]$. Suppose that the adversary returns the pair $(m^*, 760 \tau^*)$, and that its tagging oracle queries and their answers are $(m_i, 761 \tau_i)$ for $0 \le i < q$. Consider the event $C$ that $H_K(m) = 762 H_K(m')$ for some $m \ne m'$, with $m, m' \in \{m^*\} \cup \{\,m_i \mid 0 763 \le i < q\,\}$.
764
765 If $C$ doesn't occur, then $F$ has not been queried before at $H_K(m)$, but
766 there's a $2^{-L}$ probability that the adversary guesses right anyway. If
767 $C$ does occur, then we just assume that the adversary wins, even though it
768 might not have guessed the right tag.
769
770 By our result on the collision game, $\Pr[C] \le q \cdot \InSec{uh}(H)$.
771 Then
772 $\Succ{prf}{F}(D) \ge 773 \Succ{suf-cma}{\Xid{\mathcal{M}}{UH}^{H, F}}(A) - 774 \frac{1}{2^L} - \frac{q(q - 1)}{2} \cdot \InSec{uh}(H).$%
775 The result follows.
776\end{proof}
777
778\begin{slide}
779 \topic{almost XOR-universality}
53aa10b5 780 \resetseq
41761fdc 782
783 Consider a family of hash functions $H\colon \keys H \times \dom H \to 784 \{0, 1\}^L$. Define
785 $\InSec{xuh}(H) = 786 \max_{x \ne y, \delta} 787 \Pr[K \getsr \keys H : H_K(x) \xor H_K(y) = \delta].$%
788 If $\InSec{xuh}(H) < \epsilon$ then we say that $H$ is
789 \emph{$\epsilon$-almost XOR-universal}, or \emph{AXU}. Setting $\delta = 790 0$ shows that
791 \begin{eqnarray*}[rl]
792 \InSec{xuh}(H)
793 & \ge \max_{x \ne y} \Pr[K \getsr \keys H : H_K(x) \xor H_K(y) = 0] \\
794 & = \InSec{uh}(H).
795 \end{eqnarray*}
796
797 We can take a dynamic view of almost XOR-universality using the same
798 technique as for almost universal functions.
799
800 If $H$ is $2^{-L}$-almost XOR universal then we say that $H$ is
53aa10b5 801 \emph{XOR-universal}. This is the best achievable.
41761fdc 802\end{slide}
803
804\begin{proof}
805 Fix some pair $x \ne y$. Choose $\delta \inr \{0, 1\}^L$. Then for any
806 \emph{fixed} $K \in \keys H$, $H_K(x)$ and $H_K(y)$ are fixed, so $H_K(x) 807 \xor H_K(y)$ is fixed, and $\Pr[H_K(x) \xor H_K(y) = \delta] = 2^{-L}$.
808 Using the same trick as for proving the dynamic view:
809 \begin{eqnarray*}[Ll]
810 \Pr[\delta \getsr \{0, 1\}^L; K \getsr \keys H :
811 H_K(x) \xor H_K(y) = \delta] \\
812 & = \sum_{K \in \keys H} \frac{1}{|{\keys H}|}
813 \Pr[\delta \getsr \{0, 1\}^L; K \getsr \keys H :
814 H_K(x) \xor H_K(y) = \delta] \\
815 & = \sum_{K \in \keys H} \frac{1}{|{\keys H}|} 2^{-L} = 2^{-L}.
816 \end{eqnarray*}
817 Since $H$ is arbitrary, this proves the lower bound on the almost
818 XOR-universality.
819\end{proof}
820
821\begin{slide}
822 \topic{composition}
53aa10b5 823 \head{Almost XOR-universality, \seq: composition}
41761fdc 824
825 We extend our result about composition of almost-universal functions.
826 Suppose that $G$ is $\epsilon$-almost XOR universal, and $G'$ is
827 $\epsilon'$-almost universal (it doesn't have to be almost XOR-universal),
828 and $\dom G = \ran G'$.
829
830 Then the composition $H = G \compose G'$ is $(\epsilon + \epsilon')$-almost
831 XOR-universal. The proof is simple, and very similar to the
832 almost-universal case.
833\end{slide}
834
835\begin{slide}
836 \topic{a better MAC}
53aa10b5 837 \head{Almost XOR-universality, \seq: a better MAC}
41761fdc 838
839 The security result for the UH-based MAC contains a factor $q_T$, which
840 it'd be nice to remove. Our new scheme uses an AXU hash $H\colon \keys H 841 \times \{0, 1\}^* \to \{0, 1\}^l$ and a PRF $F\colon \keys F \times \{0, 842 1\}^l \to \{0, 1\}^L$.
843
844 We first present a stateful version $\Xid{\mathcal{M}}{XUH}^{H, F}$.
845 Choose $(K, K') \inr \keys H \times \keys F$, and initialize a counter $i 846 \gets 0$. The tagging and verification algorithms are then:
847 \begin{program}
848 Algorithm $\Xid{T}{XUH}^{H, F}_{K, K'}(m)$: \+ \\
849 $\tau \gets (i, H_K(m) \xor F_{K'}(i))$; \\
850 $i \gets i + 1$; \\
851 \RETURN $\tau$;
852 \next
853 Algorithm $\Xid{V}{XUH}^{H, F}_{K, K'}(m, \tau)$: \+ \\
854 $(s, \sigma) \gets \tau$; \\
855 \IF $\sigma = H_K(m) \xor F_{K'}(i)$ \THEN \RETURN $1$; \\
856 \ELSE \RETURN $0$;
857 \end{program}
858 Note that verification is stateless.
859\end{slide}
860
861\begin{slide}
53aa10b5 862 \head{Almost XOR-universality, \seq: security of AXU-based MACs}
41761fdc 863
864 For the stateful scheme presented earlier, provided $q_T \le 2^l$, we have
865 \begin{eqnarray*}[Ll]
866 \InSec{suf-cma}(\Xid{\mathcal{M}}{XUH}^{H, F}; t, q_T, q_V) \\
867 & \le (q_V + 1)(\InSec{prf}(F; t, q_T + 1) + \InSec{xuh}(H) + 2^{-L}).
868 \end{eqnarray*}
869\end{slide}
870
871\begin{slide}
53aa10b5 872 \head{Almost XOR-universality, \seq: randomized AXU-based MAC}
41761fdc 873
874 We can avoid statefulness by using randomization. This new scheme is
875 $\Xid{\mathcal{M}}{XUH$\$$}^{H, F} = (\Xid{T}{XUH\$$}^{H, F},
876 \Xid{V}{XUH$\$$}^{H, F}): 877 \begin{program} 878 Algorithm \Xid{T}{XUH\$$}^{H, F}_{K, K'}(m)$: \+ \\
879 $s \getsr \{0, 1\}^l$; \\
880 $\tau \gets (s, H_K(m) \xor F_{K'}(s))$; \\
881 \RETURN $\tau$;
882 \next
883 Algorithm $\Xid{V}{XUH$\$$}^{H, F}_{K, K'}(m, \tau): \+ \\ 884 (s, \sigma) \gets \tau; \\ 885 \IF \sigma = H_K(m) \xor F_{K'}(i) \THEN \RETURN 1; \\ 886 \ELSE \RETURN 0; 887 \end{program} 888 \begin{eqnarray*}[Ll] 889 \InSec{suf-cma}(\Xid{\mathcal{M}}{XUH\$$}^{H, F}; t, q_T, q_V) \\
890 & \le (q_V + 1)
891 \Bigl(\InSec{prf}(F; t, q_T + 1) + \InSec{xuh}(H) + 2^{-L} +
892 \frac{q_T(q_T - 1)}{2^{l+1}}\Bigr).
893 \end{eqnarray*}
894\end{slide}
895
896\begin{proof}
897 We prove the result with $q_V = 0$ and $q_T = q$, and appeal to the result
898 on verification oracles. Let $m_i$ be the message specified in the $i$-th
899 tagging query ($0 \le i < q$), and let $(s_i, \sigma_i) = (s_i, H_K(m) \xor 900 F_{K'}(s_i))$ be the tag returned. We call the $s_i$ the \emph{nonce}.
901
902 We prove the result for the stateless scheme. The bound $q \le 2^l$
903 ensures that the nonces are all distinct (we have $s_i = i$). The security
904 bound for the randomized version merely has as an extra term upper bound
905 for the probability of a nonce collision.
906
907 Let $A$ be an adversary attacking the MAC in time $t$, and using $q$
908 tagging queries. Then we present the following pair of adversaries:
909 \begin{program}
910 Distinguisher $D^{F(\cdot)}$: \+ \\
911 $i \gets 0$; \\
912 $\Xid{T}{list} \gets \emptyset$; \\
913 $K \getsr \keys H$; \\
914 $(m, \tau) \gets A^{\id{tag}}$; \\
915 $(s, \sigma) \gets \tau$; \\
916 \IF $(m, \tau) \notin \Xid{T}{list} \land 917 \sigma = H_K(m) \xor F(s)$
918 \THEN \RETURN $1$; \\
919 \ELSE \RETURN $0$; \- \$\smallskipamount] 920 Oracle \id{tag}(m): \+ \\ 921 \tau \gets (i, H_K(m) \xor F(i)); \\ 922 \Xid{T}{list} \gets \Xid{T}{list} \cup \{(m, \tau)\}; \\ 923 i \gets i + 1; 924 \RETURN \tau; 925 \next 926 Collision-finder C: \+ \\ 927 i \gets 0; \\ 928 (m, \tau) \gets A^{\id{tag}}; \\ 929 (s, \sigma) \gets \tau; \\ 930 \IF s \ge i \lor m = m_s \THEN \ABORT; \\ 931 \RETURN (m, m_s, \sigma \xor \sigma_s); \- \\[\smallskipamount] 932 Oracle \id{tag}(m): \+ \\ 933 m_i \gets m; \\ 934 \sigma_i \getsr \{0, 1\}^L; \\ 935 \tau \gets (i, \sigma_i); \\ 936 i \gets i + 1; \\ 937 \RETURN \tau; 938 \end{program} 939 940 We need to find a lower bound on the advantage of D. If F is chosen 941 from the PRF then D returns 1 precisely when A finds a valid 942 forgery. We now examine the setting in which F is a random function. 943 944 Let S be the event that A succeeds in returning a valid forgery when 945 F is random, and let N be the event that the nonce s returned by A 946 is not equal to any nonce s_i returned by the tagging oracle. Suppose 947 N occurs: then the random function F has never been queried before at 948 F, and \Pr[F(s) = \sigma \xor H_K(m)] is precisely 2^{-L}. 949 950 So suppose instead that N doesn't occur. Then, since the s_i are 951 distinct, there is a unique i such that s = s_i. For A to win, we 952 must have m \ne m_i (for if m = m_i then the only valid tag is 953 H_K(m_i) \xor F(s_i) = \sigma_i, which was already returned by the 954 tagging oracle). If A's forgery is successful then 955 \[ \sigma = H_K(m) \xor F(s) 956 \qquad \text{and} \qquad 957 \sigma_i = H_K(m_i) \xor F(s_i)$%
958 but $s = s_i$, whence
959 $H_K(m_i) \xor H_K(m) = \sigma \xor \sigma_i.$
960 Since the $s_i$ are distinct and $F$ is a random function, the $\sigma_i$
961 are independent uniformly-distributed random strings from $\{0, 1\}^L$.
962 Hence the collision-finder $C$ succeeds with probability $\Pr[S \land 963 \lnot N] \le \InSec{xuh}(H)$.
964
965 Wrapping up, we have
966 \begin{eqnarray*}[rl]
968 & \ge \Succ{suf-cma}{\Xid{\mathcal{M}}{XUH}^{H, F}}(A) -
969 (\Pr[S \mid N] \Pr[N] + \Pr[S \mid \lnot N] \Pr[\lnot N]) \\
970 & \ge \Succ{suf-cma}{\Xid{\mathcal{M}}{XUH}^{H, F}}(A) -
971 (2^{-L} + \InSec{xuh}(H)).
972 \end{eqnarray*}
973 Maximizing and rearranging yields the required result.
974\end{proof}
975
976\begin{remark*}
977 Note that our bound has a $2^{-L}$ term in it that's missing from
978 \cite{Goldwasser:1999:LNC}. We believe that their proof is wrong in its
979 handling of the XOR-collision probability.
980\end{remark*}
981
982\endinput
983
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