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 $\epsilon \le \epsilon' + (q_V + 1)
+ + q_V + 1$, $t = t' + O(q)$, 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.
$A$ 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
\[ \InSec{suf-cma}(F; t, q_T, q_V) \le
\InSec{prf}(F; t + O(q), q) + (q_V + 1)2^{-L}. \]%
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
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 $A'^{T(\cdot), V(\cdot, \cdot)}$ \+ \\
$K \getsr \{0, 1\}^k$; \\
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
- 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 \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 $K$
~mdw / doc / ips / blobdiff