\begin{slide}
\topic{informal security notion}
- \head{Strong MACs, 1: informal security notion}
+ \resetseq
+ \head{Strong MACs, \seq: informal security notion}
Our standard notion of security for MACs is \emph{strong unforgeability
against chosen message attack}, or SUF-CMA, for short
\begin{slide}
\topic{strong MACs}
- \head{Strong MACs, 2: the experiment}
+ \head{Strong MACs, \seq: the experiment}
We perform the following experiment with the adversary.
\begin{program}
\end{slide}
\begin{slide}
- \head{Strong MACs, 3: wrapping up the notation}
+ \head{Strong MACs, \seq: wrapping up the notation}
The \emph{success probability} of an adversary $A$ against the MAC
$\mathcal{M}$ in the sense of SUF-CMA is
\end{itemize}
\end{slide}
-%% IFF systems
-\begin{exercise}
- %% A nice open-ended question. Need to sort out IFF details (e.g.,
- %% aircraft numbering or whatever.
- An \emph{identify friend-or-foe} (IFF) system works as follows. Each
- `friendly' aircraft knows a common secret key. When a new aircraft is
- detected, a \emph{challenge} $x$ is sent. If a correct \emph{response} is
- returned, the new aircraft is presumed friendly; otherwise it is attacked.
- Discuss the security requirements of IFF systems and construct a formal
- model. Decide which primitive is best suited to the job and relate the
- security notions. What sorts of resource constraints can be imposed on
- adversaries attacking an IFF system?
- \answer%
- No prizes for guessing that a MAC is what's wanted, this being the MAC
- section.
-\end{exercise}
-
\xcalways\subsection{Basic results}\x
\begin{slide}
\topic{PRFs are MACs}
- \head{PRFs are MACs, 1}
+ \resetseq
+ \head{PRFs are MACs, \seq}
- If $F_K\colon \{0, 1\}^* \to \{0, 1\}^L$ is a $(t, q, \epsilon)$-secure
- PRF, then it's also a $(t', q_T, q_V, \epsilon')$-secure MAC, with $q = q_T
- + q_V + 1$, $t = t' + O(q)$, and $\epsilon \le \epsilon' + (q_V + 1)
- 2^{-L}$. The constant hidden by the $O(\cdot)$ is small and depends on the
- model of computation.
+ If $F_K\colon \{0, 1\}^* \to \{0, 1\}^L$ is a PRF then it's also a MAC.
+ Quantitatively:
+ \[ \InSec{suf-cma}(F; t, q_T, q_V) \le
+ \InSec{prf}(F; t + O(q), q_T + q_V) +
+ \frac(q_V + 1}{2^L}. \]
+ The constant hidden by the $O(\cdot)$ is small and depends on the model of
+ computation.
Suppose $A$ can break $F$ used as a MAC in time $t$ and with $q_T$ and
$q_V$ queries to its tagging and verification oracles respectively.
\end{slide}
\begin{slide}
- \head{PRFs are MACs, 2: the distinguisher}
+ \head{PRFs are MACs, \seq: the distinguisher}
\begin{program}
Distinguisher $D^{F(\cdot)}$: \+ \\
\end{slide}
\begin{slide}
- \head{PRFs are MACs, 3: wrapping up}
+ \head{PRFs are MACs, \seq: wrapping up}
The distinguisher simulates the tagging and verification oracles for the
MAC forger, using its supplied oracle. If the forger succeeds, then the
$A$ does when it's given a random function. But we know that the
probability of it successfully guessing the MAC for a message for which it
didn't query $T$ can be at most $(q_V + 1) 2^{-L}$. So
- \[ \Adv{prf}{F}(D) \le \Succ{suf-cma}{F}(A) - (q_V + 1) 2^{-L}. \]
+ \[ \Adv{prf}{F}(D) \ge \Succ{suf-cma}{F}(A) - (q_V + 1) 2^{-L}. \]
Let $q = q_T + q_V + 1$; then counting, rearranging, maximizing yields
\[ \InSec{suf-cma}(F; t, q_T, q_V) \le
\InSec{prf}(F; t + O(q), q) + (q_V + 1)2^{-L}. \]%
\end{slide}
+\begin{slide}
+ \head{PRFs are MACs, \seq: MACs aren't PRFs}
+
+ The converse of our result is not true. Suppose $\mathcal{M} = (T, V)$ is
+ a deterministic MAC. Choose some integer $n$. Then define $\mathcal{M}' =
+ (T', V')$, as follows:
+ \[ T'_K(x) = 0^n \cat T_K(x); \qquad
+ V'_K(x, \tau) = \begin{cases}
+ 1 & if $T'_K(x) = \tau$ \\
+ 0 & otherwise
+ \end{cases}.
+ \]
+ $T'$ is obviously not a PRF: an adversary checking for the string of $n$
+ zero bits on the output will succeed with advantage $1 - 2^{-qn}$.
+
+ However, $\mathcal{M}'$ is a secure MAC. Suppose $A'$ attacks
+ $\mathcal{M}'$.
+ \begin{program}
+ Adversary $A^{T(\cdot), V(\cdot)}$: \+ \\
+ $(m, \tau') \gets A'^{\id{tag}(\cdot), \id{verify}(\cdot)}$; \\
+ \PARSE $\tau'$ \AS $n\colon z, \tau$; \\
+ \RETURN $(m, \tau)$;
+ \next
+ Oracle $\id{tag}(m)$: \+ \\
+ \RETURN $0^n \cat T(m)$; \- \\[\smallskipamount]
+ Oracle $\id{verify}(m, \tau')$: \+ \\
+ \PARSE $\tau'$ \AS $n\colon z, \tau$; \\
+ \IF $z \ne 0^n$ \THEN \RETURN $0$; \\
+ \ELSE \RETURN $V(m, \tau)$;
+ \end{program}
+\end{slide}
+
\begin{exercise}
\begin{parenum}
\item Suppose that $F\colon \{0, 1\}^k \times \{0, 1\}^* \to \{0, 1\}^L$ is
a $(t, q, \epsilon)$-secure PRF. Let $T^{(\ell)}_K(x)$ be the leftmost
- $\ell$~bits of $F_K(x)$. Demonstrate the security of $T^{(\ell)}(\cdot)$
- as a MAC.
+ $\ell$~bits of $F_K(x)$ for $\ell \le L$. Demonstrate the security of
+ $T^{(\ell)}(\cdot)$ as a MAC.
\item Discuss the merits of truncating MAC tags in practical situations.
\end{parenum}
\answer%
to the tagging oracle.
\item Once a verification query succeeds, all subsequent verification
queries also succeed and the adversary returns a correct forgery (e.g.,
- by simply repeating the succeessful query).
+ by simply repeating the successful query).
\end{itemize}
It's clear that any adversary can be transformed into one which has these
properties and succeeds with probability at least as high.
return it; otherwise we guess randomly from the remaining $2^L - q$
possibilities. Now
\begin{eqnarray*}[rl]
- e_q &= e_{q-1} + (1 - e_{q-1}) \frac{1}{2^L - q} \\
+ e_q &= e_{q-1} + \frac{1 - e_{q-1}}{2^L - q} \\
&= \frac{q}{2^L} + \frac{2^L - q}{2^L} \cdot \frac{1}{2^L - q} \\
&= \frac{q + 1}{2^L}
\end{eqnarray*}
\xcalways\subsection{The HMAC construction}\x
\begin{slide}
- \head{The HMAC construction \cite{Bellare:1996:KHF}, 1: motivation}
-
+ \resetseq
+ \head{The HMAC construction \cite{Bellare:1996:KHF}, \seq: motivation}
+
It ought to be possible to construct a decent MAC using a hash function.
Many attempts have failed, however. For example, these constructions are
- weak if used with Merkle-Damg\aa{}rd iterated hashes:
+ weak if used with standard one-pass Merkle-Damg\aa{}rd iterated hashes.
\begin{itemize}
- \item Secret prefix: $T_K(m) = H(K \cat m)$.
- \item Secret suffix: $T_K(m) = H(m \cat K)$.
+ \item Secret prefix: $T_K(m) = H(K \cat m)$. Given $H(K \cat m)$, it's
+ easy to compute $H(K \cat m \cat p \cat m')$ for a padding string $p$ and
+ arbitrary suffix $m'$.
+ \item Secret suffix: $T_K(m) = H(m \cat K)$. Finding a collision $H(m) =
+ H(m')$ yields $H(m \cat K) = H(m' \cat K)$. We saw earlier that
+ adversaries which know collisions \emph{exist} even if we don't know how
+ to describe them.
\end{itemize}
It would be nice to have a construction whose security was provably related
\end{slide}
\begin{slide}
- \head{The HMAC construction, 2: definition of NMAC}
+ \head{The HMAC construction, \seq: definition of NMAC}
Let $H\colon \{0, 1\}^* \to \{0, 1\}^k$ be an iterated hash, constructed
from the compression function $F\colon \{0, 1\}^k \times \{0, 1\}^L \to
\end{slide}
\begin{slide}
- \head{The HMAC construction, 3: security of NMAC}
+ \head{The HMAC construction, \seq: security of NMAC}
Consider a function $F\colon \{0, 1\}^k \times \{0, 1\}^* \to \{0, 1\}^k$.
We say that $F$ is \emph{$(t, q, \epsilon)$-weakly collision resistant} if,
for any adversary $A$ constrained to run in time $t$ and permitted $q$
oracle queries,
\[ \Pr[K \getsr \{0, 1\}^k;
- (x, y) \gets A^{F_K(\cdot)}] \le \epsilon :
- x \ne y \land F_K(x) = F_K(y) \]%
+ (x, y) \gets A^{F_K(\cdot)} :
+ x \ne y \land F_K(x) = F_K(y)] \le \epsilon \]%
If $H_K$ is a $(t, q_T, q_V, \epsilon)$-secure MAC on $k$-bit messages, and
moreover $(t, q_T + q_V, \epsilon')$-weakly collision resistant, then
\end{slide}
\begin{slide}
- \head{The HMAC construction, 4: NMAC security proof}
+ \head{The HMAC construction, \seq: NMAC security proof}
Let $A$ be an adversary which forges a $\Xid{T}{NMAC}^H$ tag in time $t$,
using $q_T$ tagging queries and $q_V$ verification queries with probability
- $\epsilon$. We construct an adversary $A'$ which forces a $H$ tag for a
- $k$-bit in essentially the same time.
+ $\epsilon$. We construct an adversary $A'$ which forges an $H$-tag for a
+ $k$-bit message in essentially the same time.
\begin{program}
Adversary $A'^{T(\cdot), V(\cdot, \cdot)}$ \+ \\
$K \getsr \{0, 1\}^k$; \\
\end{slide}
\begin{slide}
- \head{The HMAC construction, 5: from NMAC to HMAC}
+ \head{The HMAC construction, \seq: from NMAC to HMAC}
Implementing NMAC involves using strange initialization vectors and
generally messing about with your hash function. HMAC is an attempt to
\end{slide}
\begin{slide}
- \head{The HMAC construction, 6: comparison with NMAC}
+ \head{The HMAC construction, \seq: comparison with NMAC}
Comparing the two constructions, we see that
\[ \Xid{T}{HMAC}^H_K =
$i \in \{0, 1\}$) using the $H_K$ construction. Verification in each
case consists of computing the tag and comparing to the one offered.
\begin{eqlines*}
- T^0_K(x) = H_K(x); \qquad T^1_K(x) = H_K(x \cat K)
+ T^0_K(x) = H_K(x); \qquad T^1_K(x) = H_K(x \cat K); \\
V^i_K(x, \tau) = \begin{cases}
1 & if $\tau = T^i_K(x)$ \\
0 & otherwise
- \end{cases}
+ \end{cases}.
\end{eqlines*}
Decide whether each of these constructions is secure. A full proof is
rather hard: an informal justification would be good.
function $F$. For the sake of simplicity, we allow the adversary $A$
to query on \emph{padded} messages, rather than the raw unpadded
messages. We count the number $q'$ of individual message blocks.
-
- As the game with $A$ progresses, we can construct a directed
- \emph{graph} of the query results so far. We start with a node
- labelled $I$. When processing an $H$-query, each time we compute $t'
- = F(t \cat x_i)$, we add a node $t'$, and an edge $x_i$ from $t$ to
- $t'$. The `bad' event occurs whenever we add an edge to a previously
- existing node. We must show, firstly, that the
- adversary cannot distinguish $H$ from a random function unless the bad
- event occurs; and, secondly, that the bad event doesn't occur very
- often.
+
+ As the game with $A$ progresses, we can construct a directed \emph{graph}
+ of the query results so far. We start with a node labelled $I$. When
+ processing an $H$-query, each time we compute $t' = F(t \cat x_i)$, we add
+ a node $t'$, and an edge $x_i$ from $t$ to $t'$. The `bad' event occurs
+ whenever we add an edge to a previously existing node. We must show,
+ firstly, that the adversary cannot distinguish $H$ from a random function
+ unless the bad event occurs; and, secondly, that the bad event doesn't
+ occur very often.
The latter is easier: our standard collision bound shows that the bad
- event occurs during the game with probability at most $q'(q' - 1)/2$.
-
+ event occurs during the game with probability at most $q'(q' - 1)/2^{t+1}$.
+
The former is trickier. This needs a lot more work to make it really
rigorous, but we show the idea. Assume that the bad event has not
- occured. Consider a query $x_0, x_1, \ldots, x_{n-1}$. If it's the
- same as an earlier query, then $A$ learns nothing (because it could
- have remembered the answer from last time). If it's \emph{not} a
- prefix of some previous query, then we must add a new edge to our
- graph; then either the bad event occurs or we create a new node for
- the result, and because $F$ is a random function, the answer is
- uniformly random. Finally, we consider the case where the query is a
- prefix of some earlier query, or queries. But these were computed at
- random at the time.
+ occurred. Consider a query $x_0, x_1, \ldots, x_{n-1}$. If it's the same
+ as an earlier query, then $A$ learns nothing (because it could have
+ remembered the answer from last time). If it's a \emph{prefix} of some
+ earlier query, then the answer is the value of some internal node which
+ hasn't been revealed before; however, the value of that internal node was
+ chosen uniformly at random (we claim). Finally, if the query is not a
+ prefix of any previous query, then we add a new edge to our graph. If the
+ bad event doesn't occur, we must add a new node for the result, and the
+ value at that node will be uniformly random, because $F$ is a random
+ function being evaluated at a new point -- this is the only time we add new
+ nodes to the graph, justifying the claim made earlier.
At the end of all of this, we see that
\[ \InSec{prf}(T^0; t, q) \le
- \InSec{prf}(F; t, q') + \frac{q'(q' - 1)}{2} \]%
+ \InSec{prf}(F; t, q') + \frac{q'(q' - 1)}{2^{t+1}} \]%
and hence
\[ \InSec{suf-cma}(\mathcal{M}^0; t, q) \le
- \InSec{prf}(F; t, q') + \frac{q'(q' - 1)}{2} + \frac{1}{2^t}. \]%
+ \InSec{prf}(F; t, q') + \frac{2 q'(q' - 1) + 1}{2^{t+1}}. \]%
Now we turn our attention to $T^1$. It's clear that we can't simulate
$T^1$ very easily using an oracle for $F$, since we don't know $K$
(and indeed there might not be a key $K$). The intuitive reason why
- $T^1$ is insecure is that $F$ might have leak useful information if
- its input matches its key. This doesn't affect the strength of $F$ as
- a PRF because you have to know the key before you can exploit this
- leakage; but $T^1$ already knows the key, and this can be exploited to
- break the MAC.
+ $T^1$ is insecure is that $F$ might leak useful information if its input
+ matches its key. This doesn't affect the strength of $F$ as a PRF
+ because you have to know the key before you can exploit this leakage; but
+ $T^1$ already knows the key, and this can be exploited to break the MAC.
To show that this is insecure formally, let $F'$ be defined as
follows:
as $\G1$, except that if $A$ makes any query of the form $p \cat K
\cat q$ with $|p| = t$ and $|q| = \ell - k$ then the game halts
immediately, and let $F_2$ be the event that this occurs. By
- Lemma~\ref{lem:shoup} (page~\pageref{lem:shoup}), then, $|{\Pr[S_2]} -
+ Lemma~\ref{lem:shoup} (slide~\pageref{lem:shoup}), then, $|{\Pr[S_2]} -
\Pr[S_1]| \le \Pr[F_2]$. Let game~$\G3$ be the same as $\G2$ except
that we give $A$ an oracle for $F_K$ rather than $F'_K$. Since $F$
and $F'$ differ only on queries of the form $p \cat K \cat q$, we have
as $p \cat K' \cat q$ with $|p| = t$, $|K'| = k$ and $|q| = \ell - k$;
then, if $F_{K'}(0) = F(0) \land F_{K'}(1) = F(1) \land \cdots \land
F_{K'}(n - 1) = F(n - 1)$, $B$ immediately returns $1$, claiming that
- its oracle $F$ is the function $F_{K'}$; if this never occirs, $B$
+ its oracle $F$ is the function $F_{K'}$; if this never occurs, $B$
returns $0$. Clearly, if $B$ is given an instance $F_K$ of $F$ then
it succeeds with probability $\Pr[F_2]$; however, if $F$ is a random
function then $B$ returns $1$ with probability at most $q 2^{-nk}$.
queries, and takes time $t + O(n q)$. Wrapping everything up, we get
\[ \InSec{prf}(F'; t, q) \le
2\cdot\InSec{prf}(F; t + O(q n), q + n) + \frac{q}{2^{nk}}. \]%
- This completes the proof of generic insecurty for $\mathcal{M}^1$.
+ This completes the proof of generic insecurity for $\mathcal{M}^1$.
\end{exercise}
\xcalways\subsection{Universal hashing}\x
\begin{slide}
\topic{almost-universal hash functions}
- \head{Universal hashing, 1: definition}
+ \resetseq
+ \head{Universal hashing, \seq: definition}
Consider a family of hash functions $H\colon \keys H \times \dom H \to
\ran H$. We define
If $\InSec{uh}(H) \le \epsilon$ then we say that $H$ is
\emph{$\epsilon$-almost universal}. Note that the concept of
almost-universality is not quantified by running times.
+
+ Such families definitely exist: we don't need to make intractability
+ assumptions. We'll see some examples later.
If $H$ is $1/|{\ran H}|$-almost universal, then we say that $H$ is
\emph{universal}. Sometimes it's said that this is the best possible
\begin{slide}
\topic{dynamic view}
- \head{Universal hashing, 2: a dynamic view}
+ \head{Universal hashing, \seq: a dynamic view}
Suppose that $H$ is $\epsilon$-almost universal. Consider this experiment:
\begin{program}
\end{slide}
\begin{slide}
- \head{Universal hashing, 3: the dynamic view (cont.)}
+ \head{Universal hashing, \seq: the dynamic view (cont.)}
Now we treat $\rho$ as a random variable, selected from some distribution
$P$ on the set $\{0, 1\}^*$. We see that
\begin{slide}
\topic{composition}
- \head{Universal hashing, 4: composition}
+ \head{Universal hashing, \seq: composition}
Suppose that $G$ is $\epsilon$-almost universal, and $G'$ is
$\epsilon'$-almost universal, and $\dom G = \ran G'$. We define the
\end{slide}
\begin{slide}
- \topic{the collision game}
- \head{Universal hashing, 5: the collision game}
-
- Suppose that, instead of merely a pair $(x, y)$, our adversary was allowed
- to return a \emph{set} $Y$ of $q$ elements, and measure the probability
- that $H_K(x) = H_K(y)$ for some $x \ne y$ with $x, y \in Y$, and for $K
- \inr \keys H$.
-
- Let $\InSec{uh-set}(H; q)$ be maximum probability acheivable for sets $Y$
- with $|Y| \le q$. Then
- \[ \InSec{uh-set}(H; q) \le \frac{q(q - 1)}{2} \cdot \InSec{uh}(H) .\]
+ \topic{domain and range extension of a PRF}
+ \head{Universal hashing, \seq: PRF domain extension}
+
+ Suppose that $F\colon \keys F \times \{0, 1\}^L \to \{0, 1\}^l$ is a PRF.
+ We'd like to have a PRF for a bigger domain $\{0, 1\}^n$.
+
+ If $H\colon \{0, 1\}^k \times \{0, 1\}^n \to \{0, 1\}^L$ is an
+ almost-universal hash function. Then we can define $F'$ by
+ \[ F'_{K, h}(x) = F_K(H_h(x)). \]
+ Now we can prove that
+ \[ \InSec{prf}(F'; t, q) \le
+ \InSec{prf}(F; t, q) +
+ \frac{q(q - 1)}{2} \InSec{ah}(H). \]
+ Immediately this tells us that $F'$ is a MAC for $n$-bit messages.
\end{slide}
\begin{proof}
- This is rather tedious. We use the dynamic view. Suppose $A$ returns $(x,
- Y)$ with $|Y| = q$, and succeeds with probability $\epsilon$. Consider
+ Suppose $A$ attacks $F'$. Consider the distinguisher $B$ which attacks
+ $F$:
\begin{program}
- Adversary $A'$: \+ \\
- $(x, Y) \gets A$; \\
- $y \getsr Y$; \\
- \RETURN $(x, Y \setminus \{y\})$;
+ Distinguisher $B^{F(\cdot)}$: \+ \\
+ $h \getsr \{0, 1\}^k$; \\
+ \RETURN $A^{F(H_h(\cdot))}()$;
\end{program}
- The worst that can happen is that $A'$ accidentally removes the one
- colliding element from $Y$. This occurs with probability $2/q$. So
- \[ \Succ{uh-set}{H}(A') \ge \frac{q - 2}{q} \Succ{uh-set}{H}(A). \]
- Rearranging and maximzing gives
- \[ \InSec{uh-set}(H; q) \le
- \frac{q}{q - 2} \cdot \InSec{uh-set}(H; q - 1). \]
- Note that $\InSec{uh-set}(H; 2) = \InSec{uh}(H)$ is our initial notion. A
- simple inductive argument completes the proof.
+ Suppose $F$ is an oracle for $F_K(\cdot)$. Then $A$ plays its standard
+ attack game and all is well. Conversely, suppose $F$ is a random
+ function. Then the only way $A$ can distinguish its oracle $F(H_h(\cdot))$
+ from a random function is if it issues two queries $x_i \ne x_j$ such that
+ $H_h(x_i) = H_h(x_j)$. But for each pair $(i, j)$ this happens with
+ probability at most $\InSec{ah}(H)$. The stated bound follows.
\end{proof}
-\begin{slide}
- \topic{a MAC}
- \head{Universal hashing, 6: a MAC}
-
- Suppse that $H\colon \{0, 1\}^k \times \{0, 1\}^* \to \{0, 1\}^l$ is an
- almost universal hash fucntion, and $F\colon \{0, 1\}^{k'} \times \{0,
- 1\}^l \to \{0, 1\}^L$ is a PRF\@. Define a MAC $\Xid{\mathcal{M}}{UH}^{H,
- F} = (\Xid{T}{UH}^{H, F}, \Xid{V}{UH}^{H, F})$ where:
- \begin{eqnarray*}[rl]
- \Xid{T}{UH}^{H, F}_{K, K'}(m) &= F_{K'}(H_K(m)) \\
- \Xid{V}{UH}^{H, F}_{K, K'}(m, \tau) &= \begin{cases}
- 1 & if $\tau = F_{K'}(H_K(m))$ \\
- 0 & otherwise
- \end{cases}.
- \end{eqnarray*}
- We have
- \begin{eqnarray*}[Ll]
- \InSec{suf-cma}(\Xid{\mathcal{M}}{UH}^{H, F}; t, q_T, q_V) \\
- & \le
- (q_V + 1) \biggl(\InSec{prf}(F; t, q_T + 1) + \frac{1}{2^L} +
- \frac{q_T(q_T - 1)}{2} \cdot \InSec{uh}(H)\biggr).
- \end{eqnarray*}
-\end{slide}
-
-\begin{proof}
- We shall prove the result for $q_V = 0$ and $q_T = q$, and appeal to the
- earlier result on verification oracles.
-
- Suppose $A$ attacks the scheme $\Xid{\mathcal{M}}{UH}^{H, F}$ in time $t$,
- issuing $q$ tagging queries. Consider a distinguisher $D$, constructed
- from a forger $A$:
- \begin{program}
- Distinguisher $D^{F(\cdot)}$: \+ \\
- $K \getsr \{0, 1\}^k$; \\
- $\Xid{T}{list} \gets \emptyset$; \\
- $(m, \tau) \gets A^{\id{tag}(K, \cdot)}$; \\
- \IF $m \notin \Xid{T}{list} \land \tau = F(H_K(m))$
- \THEN \RETURN $1$; \\
- \ELSE \RETURN $0$; \- \\[\smallskipamount]
- Oracle $\id{tag}(K, m)$: \+ \\
- $\Xid{T}{list} \gets \Xid{T}{list} \cup \{m\}$; \\
- \RETURN $F(H_K(m))$; \- \\[\smallskipamount]
- \end{program}
- Note that $A$ isn't provided with a verification oracle: that's because we
- didn't allow it any verification queries.
-
- We can see immediately that
- \[ \Pr[K \getsr \{0, 1\}^{k'} : D^{F_K(\cdot)} = 1] =
- \Succ{suf-cma}{\Xid{\mathcal{M}}{UH}^{H, F}}(A). \]%
-
- We must now find an upper bound for $\Pr[F \getsr \Func{l}{L} :
- D^{F(\cdot)}]$. Suppose that the adversary returns the pair $(m^*,
- \tau^*)$, and that its tagging oracle queries and their answers are $(m_i,
- \tau_i)$ for $0 \le i < q$. Consider the event $C$ that $H_K(m) =
- H_K(m')$ for some $m \ne m'$, with $m, m' \in \{m^*\} \cup \{\,m_i \mid 0
- \le i < q\,\}$.
-
- If $C$ doesn't occur, then $F$ has not been queried before at $H_K(m)$, but
- there's a $2^{-L}$ probability that the adversary guesses right anyway. If
- $C$ does occur, then we just assume that the adversary wins, even though it
- might not have guessed the right tag.
-
- By our result on the collision game, $\Pr[C] \le q \cdot \InSec{uh}(H)$.
- Then
- \[ \Succ{prf}{F}(D) \ge
- \Succ{suf-cma}{\Xid{\mathcal{M}}{UH}^{H, F}}(A) -
- \frac{1}{2^L} - \frac{q(q - 1)}{2} \cdot \InSec{uh}(H). \]%
- The result follows.
-\end{proof}
+\begin{exercise}
+ Show how to extend the \emph{range} of a PRF. Specifically, suppose you're
+ given a PRF $F\colon \keys{F} \times \{0, 1\}^L \to \{0, 1\}^t$, but want
+ an $l$-bit output. Show how to achieve this.
+\answer
+ There are two obvious techniques.
+ \begin{enumerate}
+ \item Suppose we have a PRG $G\colon \{0, 1\}^t \to \{0, 1\}^l$. Then we
+ can easily use $G \compose F$; it's easy to see how this is secure.
+ Indeed, we can use $F$ itself as a PRG, by defining $G(x) = F_x(0) \cat
+ F_x(1) \cat \cdots$.
+ \item Let $m = \lceil \log_2 (l/t) \rceil$. Use the construction from the
+ slide to build a PRF $F'$ with domain $\{0, 1\}^{L+m}$. Then we define
+ $F''_K(x) = F'_K(x \cat 0) \cat F'_K(x \cat 1) \cat \cdots \cat F'_K(x
+ \cat m - 1)$.
+ \end{enumerate}
+\end{exercise}
\begin{slide}
\topic{almost XOR-universality}
- \head{Almost XOR-universality, 1: definition}
+ \resetseq
+ \head{Almost XOR-universality, \seq: definition}
Consider a family of hash functions $H\colon \keys H \times \dom H \to
\{0, 1\}^L$. Define
technique as for almost universal functions.
If $H$ is $2^{-L}$-almost XOR universal then we say that $H$ is
- \emph{XOR-universal}. This is the best acheivable.
+ \emph{XOR-universal}. This is the best achievable.
\end{slide}
\begin{proof}
\begin{slide}
\topic{composition}
- \head{Almost XOR-universality, 2: composition}
+ \head{Almost XOR-universality, \seq: composition}
We extend our result about composition of almost-universal functions.
Suppose that $G$ is $\epsilon$-almost XOR universal, and $G'$ is
\end{slide}
\begin{slide}
+ \head{Almost XOR-universality, \seq: examples of AXU hash functions}
+
+ Let $\F = \gf{2^l}$ be a finite field. Given a message $m = (m_0, \ldots,
+ m_{n-1} \in \F^n$, we can hash it in several ways.
+ \begin{itemize}
+ \item Inner-product: choose $k = (k_0, \ldots, k_{n-1}) \in \F^n$.
+ \[ \Xid{H}{ip}_{k}(m) = m \cdot k = \sum_{0\le i<n} k_i m_i. \]
+ This hash is XOR-universal: $\InSec{axu}(\Xid{H}{ip}) = 1/2^l$.
+ \item Polynomial evaluation: choose $x \in \F$.
+ \[ \Xid{H}{pe}_{x}(m) = \sum_{0\le i<n} m_i x^{i+1}. \]
+ Here we have only $\InSec{axu}(\Xid{H}{pe}) = n/2^l$.
+ \end{itemize}
+\end{slide}
+
+\begin{proof}
+ \begin{itemize}
+ \item Suppose we have messages $m \ne m'$ and $\delta \in \F$. Then $m
+ \cdot k + m' \cdot k = \delta$. We must have some $m_j \ne m'_j$. It
+ follows that
+ \[ k_j = \biggl( \delta + \sum_{\begin{script}
+ 0\le i<n \\
+ i \ne j
+ \end{script}
+ k_i (m_i + m'_i)
+ \biggr) \biggm/ (m'_i - m_j) \]
+ But we choose $k$ uniformly at random, so the probability that this holds
+ is $2^{-l}$.
+ \item Again, we have distinct messages and some $\delta \in \F$. We have a
+ polynomial equation
+ \[ \delta + \sum_{0\le i <n} (m_i - m'_i) x^{i+1} = 0 \]
+ Since there is some $m_j \ne m'_j$, this has degree at most $n$, so
+ there are at most $n$ distinct roots $x \in \F$. But we choose $x$
+ uniformly at random, so $x$ is one of these roots with probability at
+ most $n/2^l$.
+ \end{itemize}
+\end{proof}
+
+\begin{exercise}
+ Suppose $H$ is an $\epsilon$-AXU hash function with $l$-bit output. Let
+ $G_h(x)$ be the first $t$ bits of $H_h(x)$. Show that $G$ is
+ $2^{l-t}\epsilon$-AXU.
+\answer
+ Let $m \ne m'$ be distinct messages and let $\delta \in \{0, 1\}^t$ be some
+ XOR difference. Then
+ \begin{eqnarray*}
+ \Pr[h \gets \keys{G} : G_h(m) \xor G_h(m') = \delta]
+ &= \sum_{\delta' \in \{0, 1\}^{l-t}}
+ \Pr[h \gets \keys{H} : H_h(m) \xor H_h(m') = \delta \cat \delta']
+ &\le \sum_{\delta' \in \{0, 1\}^{l-t}} \InSec{axu}(H)
+ &= 2^{l-t} \InSec{axu}(H).
+ \end{eqnarray*}
+\end{exercise}
+
+\begin{slide}
\topic{a better MAC}
- \head{Almost XOR-universality, 3: a better MAC}
+ \head{Almost XOR-universality, \seq: a better MAC}
The security result for the UH-based MAC contains a factor $q_T$, which
it'd be nice to remove. Our new scheme uses an AXU hash $H\colon \keys H
\next
Algorithm $\Xid{V}{XUH}^{H, F}_{K, K'}(m, \tau)$: \+ \\
$(s, \sigma) \gets \tau$; \\
- \IF $\sigma = H_K(m) \xor F_{K'}(i)$ \THEN \RETURN $1$; \\
+ \IF $\sigma = H_K(m) \xor F_{K'}(s)$ \THEN \RETURN $1$; \\
\ELSE \RETURN $0$;
\end{program}
Note that verification is stateless.
\end{slide}
\begin{slide}
- \head{Almost XOR-universality, 4: security of AXU-based MACs}
+ \head{Almost XOR-universality, \seq: security of AXU-based MACs}
For the stateful scheme presented earlier, provided $q_T \le 2^l$, we have
\begin{eqnarray*}[Ll]
\InSec{suf-cma}(\Xid{\mathcal{M}}{XUH}^{H, F}; t, q_T, q_V) \\
- & \le (q_V + 1)(\InSec{prf}(F; t, q_T + 1) + \InSec{xuh}(H) + 2^{-L}).
+ & \le (q_V + 1)(\InSec{prf}(F; t, q_T + 1) + \InSec{xuh}(H)).
\end{eqnarray*}
\end{slide}
\begin{slide}
- \head{Almost XOR-universality, 5: randomized AXU-based MAC}
+ \head{Almost XOR-universality, \seq: randomized AXU-based MAC}
We can avoid statefulness by using randomization. This new scheme is
$\Xid{\mathcal{M}}{XUH$\$$}^{H, F} = (\Xid{T}{XUH$\$$}^{H, F},
\next
Algorithm $\Xid{V}{XUH$\$$}^{H, F}_{K, K'}(m, \tau)$: \+ \\
$(s, \sigma) \gets \tau$; \\
- \IF $\sigma = H_K(m) \xor F_{K'}(i)$ \THEN \RETURN $1$; \\
+ \IF $\sigma = H_K(m) \xor F_{K'}(s)$ \THEN \RETURN $1$; \\
\ELSE \RETURN $0$;
\end{program}
\begin{eqnarray*}[Ll]
\InSec{suf-cma}(\Xid{\mathcal{M}}{XUH$\$$}^{H, F}; t, q_T, q_V) \\
& \le (q_V + 1)
- \Bigl(\InSec{prf}(F; t, q_T + 1) + \InSec{xuh}(H) + 2^{-L} +
+ \Bigl(\InSec{prf}(F; t, q_T + 1) + \InSec{xuh}(H) +
\frac{q_T(q_T - 1)}{2^{l+1}}\Bigr).
\end{eqnarray*}
\end{slide}
$F$ is random, and let $N$ be the event that the nonce $s$ returned by $A$
is not equal to any nonce $s_i$ returned by the tagging oracle. Suppose
$N$ occurs: then the random function $F$ has never been queried before at
- $F$, and $\Pr[F(s) = \sigma \xor H_K(m)]$ is precisely $2^{-L}$.
+ $F$, and $\Pr[S \mid N] = \Pr[F(s) = \sigma \xor H_K(m)]$ is precisely
+ $2^{-L}$.
So suppose instead that $N$ doesn't occur. Then, since the $s_i$ are
distinct, there is a unique $i$ such that $s = s_i$. For $A$ to win, we
\[ H_K(m_i) \xor H_K(m) = \sigma \xor \sigma_i. \]
Since the $s_i$ are distinct and $F$ is a random function, the $\sigma_i$
are independent uniformly-distributed random strings from $\{0, 1\}^L$.
- Hence the collision-finder $C$ succeeds with probability $\Pr[S \land
- \lnot N] \le \InSec{xuh}(H)$.
+ Hence the collision-finder $C$ succeeds with probability $\Pr[S \mid
+ \bar{N}] \le \InSec{xuh}(H)$.
Wrapping up, we have
\begin{eqnarray*}[rl]
\Adv{prf}{F}(D)
& \ge \Succ{suf-cma}{\Xid{\mathcal{M}}{XUH}^{H, F}}(A) -
- (\Pr[S \mid N] \Pr[N] + \Pr[S \mid \lnot N] \Pr[\lnot N]) \\
+ (\Pr[S \mid N] \Pr[N] + \Pr[S \mid \bar{N}] \Pr[\bar{N}]) \\
& \ge \Succ{suf-cma}{\Xid{\mathcal{M}}{XUH}^{H, F}}(A) -
- (2^{-L} + \InSec{xuh}(H)).
+ (2^{-L} \Pr[N] + \InSec{xuh}(H) \Pr[\bar{N}]) \\
+ & \ge \Succ{suf-cma}{\Xid{\mathcal{M}}{XUH}^{H, F}}(A) -
+ \InSec{xuh}(H).
\end{eqnarray*}
Maximizing and rearranging yields the required result.
\end{proof}
-\begin{remark*}
- Note that our bound has a $2^{-L}$ term in it that's missing from
- \cite{Goldwasser:1999:LNC}. We believe that their proof is wrong in its
- handling of the XOR-collision probability.
-\end{remark*}
-
\endinput
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