41761fdc |
1 | \xcalways\section{Public-key encryption}\x |
2 | |
3 | \xcalways\subsection{Syntax}\x |
4 | |
5 | \begin{slide} |
6 | \head{Syntax of public-key encryption schemes} |
7 | |
8 | A public-key encryption scheme $\mathcal{E} = (G, E, D)$ is a triple of |
9 | algorithms: |
10 | \begin{itemize} |
11 | \item A probabilistic \emph{key-generation} algorithm $G(k)$ which returns |
12 | a matching public/private key pair $(P, K)$. |
13 | \item A probabilistic \emph{encryption} algorithm $E(P, m)$ which returns a |
14 | ciphertext $c$. We write $E_P(m)$ for $E(P, m)$. |
15 | \item A deterministic \emph{decryption} algorithm $D(K, c)$. If $(P, K) |
16 | \in G(k)$ and $c \in E(P, m)$ then $m = D(K, c)$. We write $D_K(c)$ for |
17 | $D(K, c)$. |
18 | \end{itemize} |
19 | On those occasions it matters, we write $\mathcal{P} = \dom E_P$ as the |
20 | \emph{plaintext space}, and $\mathcal{C} = \dom D_K$ as the |
21 | \emph{ciphertext space}. Hence, $E_P\colon \mathcal{P} \to \mathcal{C}$ |
22 | and $D_K\colon \mathcal{C} \to \mathcal{P} \cup \{ \bot \}$ (allowing an |
23 | `invalid ciphertext' return). |
24 | \end{slide} |
25 | |
26 | \xcalways\subsection{Semantic security and indistinguishability}\x |
27 | |
28 | \begin{slide} |
29 | \head{Notation for oracles} |
30 | |
31 | We'll use the following decryption oracles throughout, to abbreviate the |
32 | properties of the various attacks: |
33 | \begin{tabular}[C]{l Mc Mc } |
34 | \hlx*{hv} |
35 | Attack & D_0(c) & D_1(c) \\ \hlx{vhv} |
36 | CPA & \bot & \bot \\ |
37 | CCA1 & D_K(c) & \bot \\ |
38 | CCA2 & D_K(c) & D_K(c) \\ \hlx*{vh} |
39 | \end{tabular} |
40 | In all cases, we forbid the adversary from querying a decryption oracle on |
41 | a challenge ciphertext, i.e., one returned to it by the experiment. |
42 | \end{slide} |
43 | |
44 | \begin{slide} |
45 | \topic{semantic security} |
46 | \head{Semantic security} |
47 | |
48 | Semantic security for $\mathcal{E} = (G, E, D)$, against an adversary |
49 | $A$ and attack $\id{atk} \in \{\text{cpa}, \text{cca1}, \text{cca2}\}$ |
50 | is measured using the following game \cite{Bellare:2000:CST}: |
51 | \begin{program} |
52 | Experiment $\Expt{sem-\id{atk}-$b$}{\mathcal{E}}(A)$: \+ \\ |
53 | $(P, K) \gets G$; \\ |
54 | $(\mathcal{M}, s) \gets A^{D_0(\cdot)}(\cookie{select}, P)$; \\ |
55 | $x_0 \getsr \mathcal{M}$; $x_1 \getsr \mathcal{M}$; \\ |
56 | $y \gets E_P(x_1)$; \\ |
57 | $(f, \alpha) \gets A^{D_1(\cdot)}(\cookie{predict}, y, s)$; \\ |
58 | \IF $f(x_b) = \alpha$ \THEN \RETURN $1$; \\ |
59 | \ELSE \RETURN $0$; |
60 | \end{program} |
61 | Here, $\mathcal{M}\colon \mathcal{P} \to [0, 1]$ is a \emph{distribution} |
62 | over the plaintext space, and $f\colon \mathcal{P} \to \ran f$ is a |
63 | \emph{function} on plaintexts, with $\alpha \in \ran f$. |
64 | \end{slide} |
65 | |
66 | \begin{slide} |
67 | \head{Semantic security (cont.)} |
68 | |
69 | We include the time required to sample the distribution $\mathcal{M}$ and |
70 | to compute the function $f$ in the adversary's running time. |
71 | |
72 | We require that $\mathcal{M}$ is \emph{valid}: i.e., that all messages in |
73 | $\supp \mathcal{M}$ have the same length. |
74 | \end{slide} |
75 | |
76 | \begin{slide} |
77 | \topic{indistinguishability} |
78 | \head{Indistinguishability} |
79 | |
80 | Indistinguishability for $\mathcal{E} = (G, E, D)$, against an adversary |
81 | $A$ and attack $\id{atk} \in \{\text{cpa}, \text{cca1}, \text{cca2}\}$ |
82 | is measured using the following game: |
83 | \begin{program} |
84 | Experiment $\Expt{ind-\id{atk}-$b$}{\mathcal{E}}(A)$: \+ \\ |
85 | $(P, K) \gets G$; \\ |
86 | $(x_0, x_1, s) \gets A^{D_0(\cdot)}(\cookie{find}, P)$; \\ |
87 | \IF $|x_0| \ne |x_1|$ \THEN \RETURN $0$; \\ |
88 | $y \gets E_P(x_b)$; \\ |
89 | $b' \gets A^{D_1(\cdot)}(\cookie{guess}, y, s)$; \\ |
90 | \RETURN $b'$; |
91 | \end{program} |
92 | In the first stage, the adversary has to choose two plaintexts. One is |
93 | then chosen by the experimenter, encrypted, and the corresponding |
94 | ciphertext given to the adversary. The adversary must decide which |
95 | plaintext was encrypted. |
96 | \end{slide} |
97 | |
98 | \begin{slide} |
99 | \topic{advantage and insecurity} |
100 | \head{Advantage and insecurity} |
101 | |
102 | For a public-key encryption scheme $\mathcal{E}$, under attack $\id{atk} |
103 | \in \{\text{cpa}, \text{cca1}, \text{cca2}\}$ by an adversary $A$, we |
104 | define $A$'s advantage by: |
105 | \begin{eqnarray*}[rl] |
106 | \Adv{ind-\id{atk}}{\mathcal{E}}(A) &= |
107 | \Pr[\Expt{ind-\id{atk}-$1$}{\mathcal{E}}(A) = 1] - |
108 | \Pr[\Expt{ind-\id{atk}-$0$}{\mathcal{E}}(A) = 1]; |
109 | \\ |
110 | \Adv{sem-\id{atk}}{\mathcal{E}}(A) &= |
111 | \Pr[\Expt{sem-\id{atk}-$1$}{\mathcal{E}}(A) = 1] - |
112 | \Pr[\Expt{sem-\id{atk}-$0$}{\mathcal{E}}(A) = 1]. |
113 | \end{eqnarray*} |
114 | We define insecurities for $\id{goal} \in \{\text{ind}, \text{sem}\}$ under |
115 | chosen plaintext attacks, and chosen ciphertext attacks $\id{cca} \in |
116 | \{\text{cca1}, \text{cca2}\}$ by: |
117 | \begin{eqnarray*} |
118 | \InSec{\id{goal}-cpa}(\mathcal{E}; t) &= |
119 | \max_A \Adv{\id{goal}-cpa}{\mathcal{E}}(A); |
120 | \\ |
121 | \InSec{\id{goal}-\id{cca}}(\mathcal{E}; t, q_D) &= |
122 | \max_A \Adv{\id{goal}-\id{cca}}{\mathcal{E}}(A). |
123 | \end{eqnarray*} |
124 | where the maxima are taken over adversaries $A$ which run in time $t$ and |
125 | issue $q_E$ encryption and $q_D$ decryption queries. |
126 | \end{slide} |
127 | |
128 | \begin{slide} |
129 | \topic{good news} |
130 | \head{Some good news} |
131 | |
132 | Now there's some good news: \emph{semantic security and (find-then-guess) |
133 | indistinguishability are almost equivalent}. |
134 | |
135 | More precisely, if we fix a collection of resource constraints $R$, we have |
136 | \[ \InSec{sem-\id{atk}}(\mathcal{E}; R) \le |
137 | \InSec{ind-\id{atk}}(\mathcal{E}; R) \le |
138 | 2 \cdot \InSec{sem-\id{atk}}(\mathcal{E}; R). \]% |
139 | It's useful to realise that a relatively strong notion like semantic |
140 | security is actually equivalent to a simpler notion like |
141 | indistinguishability. The latter is certainly easier to work with in |
142 | proofs of security. |
143 | \end{slide} |
144 | |
145 | \begin{proof} |
146 | \label{pf:pub-ind-eq-sem} |
147 | Proving that $\text{IND-\id{atk}} \implies \text{SEM-\id{atk}}$ is |
148 | pretty easy, so we do that first. Suppose that $A'$ attacks $\mathcal{E}$ |
149 | in the semantic security sense. Consider the find-then-guess adversary |
150 | $A$: |
151 | \begin{program} |
152 | Adversary $A^{D(\cdot)}(\cookie{find}, P)$: \+ \\ |
153 | $(\mathcal{M}, s') \gets A'^{D(\cdot)}(\cookie{select}, P)$; \\ |
154 | $x_0 \getsr \mathcal{M}$; $x_1 \getsr \mathcal{M}$; \\ |
155 | \RETURN $(x_0, x_1, (x_1, s'))$; |
156 | \next |
157 | Adversary $A^{D(\cdot)}(\cookie{guess}, y, s)$: \+ \\ |
158 | $(x_1, s') \gets s$; \\ |
159 | $(f, \alpha) \gets A'^{D(\cdot)}(\cookie{predict}, y, s')$; \\ |
160 | \IF $f(x_1) = \alpha$ \THEN \RETURN $1$; \\ |
161 | \ELSE \RETURN $0$; |
162 | \end{program} |
163 | Here, $A$ is simulating the semantic security experiment itself, drawing |
164 | plaintexts from $A'$'s distribution $\mathcal{M}$ and evaluating its chosen |
165 | function. |
166 | |
167 | We study the result of the experiment |
168 | $\Expt{ind-\id{atk}-$b$}{\mathcal{E}}(A)$. Firstly, suppose $b = 1$. |
169 | Then $y \in E_P(x_1)$, and $A$ succeeds with the same probability that $A'$ |
170 | wins in $\Expt{sem-\id{atk}-$1$}{\mathcal{E}}(A')$. Secondly, suppose $b |
171 | = 0$. Then $y \in E_P(x_0)$, but we still compare $A'$'s answer against |
172 | $x_1$, as in $\Expt{sem-\id{atk}-$0$}{\mathcal{E}}(A')$, up to |
173 | relabelling of $x_0$ and $x_1$. Hence, |
174 | \[ \Adv{ind-\id{atk}}{\mathcal{E}}(A) = |
175 | \Adv{sem-\id{atk}}{\mathcal{E}}(A'). \]% |
176 | |
177 | Now we show that $\text{SEM-\id{atk}} \implies \text{IND-\id{atk}}$. |
178 | Suppose that $A$ attacks $\mathcal{E}$ in the indistinguishability sense. |
179 | Then consider the adversary $A'$ attacking its semantic security: |
180 | \begin{program} |
181 | Adversary $A'^{D(\cdot)}(\cookie{select}, P)$: \+ \\ |
182 | $(x'_0, x'_1, s) \gets A^{D(\cdot)}(\cookie{find}, P)$; \\ |
183 | $\mathcal{M} \gets |
184 | \{ x'_0 \mapsto \frac{1}{2}, x'_1 \mapsto \frac{1}{2} \}$; \\ |
185 | \RETURN $(x'_0, x'_1, s')$; |
186 | \next |
187 | Adversary $A'^{D(\cdot)}(\cookie{predict}, y, s')$: \+ \\ |
188 | $(x'_0, x'_1, s) \gets s'$; \\ |
189 | $b \gets A^{D(\cdot)}(\cookie{guess}, y, s)$; \\ |
190 | \RETURN $(\lambda x.x, x'_b)$; |
191 | \end{program} |
192 | Here $A'$ is simply running the indistinguishability experiment. In the |
193 | $\cookie{select}$ stage, it runs $A$'s $\cookie{find}$ stage, and returns |
194 | the uniform distribution over $A$'s two chosen plaintexts. In the |
195 | $\cookie{predict}$ stage, $A'$ collects $A$'s $\cookie{guess}$ and returns |
196 | the identity function $\lambda x.x$ and the guessed plaintext. |
197 | |
198 | In the case of experiment $\Expt{sem-\id{atk}-$1$}{\mathcal{E}}(A')$, the |
199 | rules are fair, and we win with probability |
200 | \[ p = \frac{1}{2} + \frac{\Adv{ind-\id{atk}}{\mathcal{E}}(A)}{2}. \] |
201 | In the case of $\Expt{sem-\id{atk}-$0$}{\mathcal{E}}(A')$, however, we |
202 | \emph{lose} in the event that $x_0 = x'_b$. This happens if \emph{either} |
203 | $x_0 = x_1$ and we guess correctly (probability $p/2$), \emph{or} if $x_0 |
204 | \ne x_1$ and we guess incorrectly (probability $(1 - p)/2$). Hence, we see |
205 | that have |
206 | \[ \Adv{sem-\id{atk}}{\mathcal{E}}(A') = |
207 | \frac{1}{2} \Adv{ind-\id{atk}}{\mathcal{E}}(A) \]% |
208 | completing the proof. |
209 | \end{proof} |
210 | |
211 | \xcalways\subsection{Non-malleability}\x |
212 | |
213 | \begin{slide} |
214 | \topic{definition} |
215 | \head{Non-malleability} |
216 | |
217 | The intuition behind non-malleability is that an adversary can't easily |
218 | take some ciphertext and turn it into some other ciphertext such that the |
219 | plaintexts have a simple relationship to each other. |
220 | |
221 | Here's a relatively standard definition of non-malleability, from |
222 | \cite{Bellare:1998:RAN, Bellare:1999:NEE}: |
223 | \begin{program} |
224 | Experiment $\Expt{cnm-\id{atk}-$b$}{\mathcal{E}}(A)$: \+ \\ |
225 | $(P, K) \gets G$; \\ |
226 | $(\mathcal{M}, s) \gets A^{D_0(\cdot)}(\cookie{find}, P)$; |
227 | $x_0 \getsr \mathcal{M}$; $x_1 \getsr \mathcal{M}$; |
228 | $y \gets E_P(x_1)$; \\ |
229 | $(R, \vect{y}) \gets A^{D_1(\cdot)}(\cookie{guess}, y, s)$; |
230 | $\vect{x} \gets D_K(\vect{y})$; \\ |
231 | \IF $y \notin \vect{y} \land |
232 | R(x_b, \vect{x})$ \THEN \RETURN $1$; |
233 | \ELSE \RETURN $0$; |
234 | \end{program} |
235 | In the above, $\mathcal{M}$ is a valid distribution on plaintexts, and $R$ |
236 | is an $n$-ary relation on plaintexts. The adversary's advantage is |
237 | \[ \Adv{cnm-\id{atk}}{\mathcal{E}}(A) = |
238 | \Pr[\Expt{cnm-\id{atk}-$1$}{\mathcal{E}}(A) = 1] - |
239 | \Pr[\Expt{cnm-\id{atk}-$0$}{\mathcal{E}}(A) = 1]. \]% |
240 | \end{slide} |
241 | |
242 | \begin{slide} |
243 | \topic{more good news} |
244 | \head{Non-malleability (more good news)} |
245 | |
246 | The previous definition involved all sorts of nasties like distributions |
247 | and relations. Fortunately, there's an approximately equivalent notion, |
248 | based on indistinguishability with a single \emph{parallel} |
249 | chosen-ciphertext query: |
250 | \begin{program} |
251 | Experiment $\Expt{nm-\id{atk}-$b$}{\mathcal{E}}(A)$: \+ \\ |
252 | $(P, K) \gets G$; \\ |
253 | $(x_0, x_1, s) \gets A^{D_0(\cdot)}(\cookie{find}, P)$; |
254 | $y \gets E_P(x_b)$; \\ |
255 | $(\vect{y}, t) \gets A^{D_1(\cdot)}(\cookie{choose}, y, s)$; |
256 | \IF $y \in \vect{y}$ \THEN \RETURN $0$; \\ |
257 | $\vect{x} \gets D_K(\vect{y})$; |
258 | $b \gets A^{D_1(\cdot)}(\cookie{guess}, \vect{x}, t)$; |
259 | \RETURN $b$; |
260 | \end{program} |
261 | We define advantage by |
262 | \[ \Adv{nm-\id{atk}}{\mathcal{E}}(A) = |
263 | \Pr[\Expt{nm-\id{atk}-$1$}{\mathcal{E}}(A) = 1] - |
264 | \Pr[\Expt{nm-\id{atk}-$0$}{\mathcal{E}}(A) = 1]. \]% |
265 | \end{slide} |
266 | |
267 | \begin{proof} |
268 | Firstly, $\text{NM} \implies \text{CNM}$. Suppose $A'$ attacks |
269 | $\mathcal{E}$ in the CNM sense. Then we construct $A$ in the obvious way: |
270 | \begin{program} |
271 | Adversary $A^{D(\cdot)}(\cookie{find}, P)$: \+ \\ |
272 | $(\mathcal{M}, s') \gets A'^{D(\cdot)}(\cookie{find}, P)$; \\ |
273 | $x_0 \getsr \mathcal{M}$; $x_1 \getsr \mathcal{M}$; \\ |
274 | \RETURN $(x_0, x_1, (x_1, s'))$; |
275 | \next |
276 | Adversary $A^{D(\cdot)}(\cookie{choose}, y, s)$: \+ \\ |
277 | $(x_1, s') \gets s$; \\ |
278 | $(R, \vect{y}) \gets A'^{D(\cdot)}(\cookie{guess}, y, s')$; \\ |
279 | \RETURN $(\vect{y}, (x_1, R))$; |
280 | \next |
281 | Adversary $A^{D(\cdot)}(\cookie{guess}, \vect{x}, s)$: \+ \\ |
282 | $(x_1, R) \gets s$; \\ |
283 | \IF $R(x_1, \vect{x})$ \THEN \RETURN $1$; \\ |
284 | \ELSE \RETURN $0$; |
285 | \end{program} |
286 | Studying the behaviour of $A$, we see that it succeeds precisely when $A'$ |
287 | succeeds. Hence |
288 | \[ \Adv{nm-\id{atk}}{\mathcal{E}}(A) = |
289 | \Adv{cnm-\id{atk}}{\mathcal{E}}(A'). \]% |
290 | |
291 | Secondly, suppose that $A$ attacks $\mathcal{E}$ in the NM sense. Then we |
292 | construct $A'$ in a somewhat tricky manner: $A'$ aims to run the final |
293 | stage of $A$ from its relation $R$. |
294 | |
295 | We can suppose, without loss of generality, that $A$ doesn't require its |
296 | chosen ciphertext oracle $D$ during its final $\cookie{guess}$ phase. If |
297 | it is allowed an adaptive chosen ciphertext oracle, it can make its |
298 | parallel query through $D$ (admittedly at the cost of $|\vect{y}|$ |
299 | additional decryption queries). Hence, we don't need to provide |
300 | $A(\cookie{guess})$ with a decryption oracle. |
301 | |
302 | The relation isn't allowed to be probabilistic. Hence, we choose the coins |
303 | $\rho \in \{0, 1\}^n$ that $A(\cookie{guess})$ requires in advance. Since |
304 | $A$'s running time is bounded, $n$ must also be bounded. We encode the |
305 | values $(x'_0, x'_1, t, \rho)$ into the description of $R$ output by |
306 | $A'(\cookie{guess})$. |
307 | |
308 | \begin{program} |
309 | Adversary $A'^{D(\cdot)}(\cookie{find}, P)$: \+ \\ |
310 | $(x'_0, x'_1, s) \gets A^{D(\cdot)}(\cookie{find}, P)$; \\ |
311 | $\mathcal{M} \gets |
312 | \{ x'_0 \mapsto \frac{1}{2}, x'_1 \mapsto \frac{1}{2} \}$; \\ |
313 | \RETURN $(\mathcal{M}, (x'_0, x'_1, P, s))$; |
314 | \next |
315 | Adversary $A'^{D(\cdot)}(\cookie{guess}, y, s')$: \+ \\ |
316 | $(x'_0, x'_1, P, s) \gets s'$; \\ |
317 | $(\vect{y}, t) \gets A^{D(\cdot)}(\cookie{choose}, y, s)$; \\ |
318 | $\rho \getsr \{0, 1\}^n$; \\ |
319 | \RETURN $(R, \vect{y})$; \- \\[\smallskipamount] |
320 | Relation $R(x, \vect{x})$: \+ \\ |
321 | $b' \gets A(\cookie{guess}, \vect{x}, t)$ (with coins $\rho$); \\ |
322 | \IF $x = x'_{b'}$ \THEN \RETURN $1$; \\ |
323 | \ELSE \RETURN $0$; |
324 | \end{program} |
325 | |
326 | The analysis of $A'$'s success probability is very similar to the proof |
327 | that $\text{SEM} \implies \text{IND}$ above. When the CNM experiment's |
328 | hidden bit $b = 1$, $A'$ succeeds whenever $A$ correctly guesses which |
329 | plaintext was encrypted, which occurs with probability |
330 | \[ p = \frac{1}{2} + \frac{\Adv{nm-\id{atk}}{\mathcal{E}}(A)}{2}. \] |
331 | When $b = 0$, $A'$ fails when $A$ guesses correctly and $x_0 = x_1$ |
332 | (probability $p/2$), or when $A$ guesses wrongly and $x_0 \ne x_1$ |
333 | (probability $(1 - p)/2$). Hence, finally, |
334 | \[ \Adv{cnm-\id{atk}}{\mathcal{E}}(A) = |
335 | \frac{\Adv{nm-\id{atk}}{\mathcal{E}}(A)}{2}. \]% |
336 | |
337 | For any arbitrary resource bounds $R$, we have |
338 | \[ \InSec{cnm-\id{atk}}(\mathcal{E}; R) \le |
339 | \InSec{nm-\id{atk}}(\mathcal{E}; R) \le |
340 | 2 \cdot \InSec{cnm-\id{atk}}(\mathcal{E}; R'), \]% |
341 | where $R'$ is related to $R$, but may need to allow additional decryption |
342 | queries, to cope with NM-CCA2 adversaries which use decryption queries in |
343 | their final $\cookie{guess}$ stages. |
344 | |
345 | We conclude that the two notions are almost equivalent, as claimed. |
346 | \end{proof} |
347 | |
348 | \xcalways\subsection{Relations between security notions}\x |
349 | |
350 | \begin{slide} |
351 | \head{Relations between security notions \cite{Bellare:1998:RAN}} |
352 | |
353 | %% 3 3 |
354 | %% NM-CPA <--------- NM-CCA1 <--------- NM-CCA2 |
355 | %% | \__ | \__ ^ ^ |
356 | %% | \__ 6 | \__ 7 | | |
357 | %% 1| \/_ |1 \/_ 4| |1 |
358 | %% | 5/ \__ | / \_ | | |
359 | %% v __\ v __\ v | |
360 | %% IND-CPA <-------- IND-CCA1 <------- IND-CCA2 |
361 | %% 2 2 |
362 | |
363 | \[ \xymatrix @=2cm { |
364 | \txt{NM-CPA} \ar[d]_1 |
365 | \POS !<1ex, 0pt> \ar[dr]!<1ex, 0pt>|-@{/}^6 & |
366 | \txt{NM-CCA1} \ar[d]^1 \ar[l]_3 \ar[dr]|-@{/}^7 & |
367 | \txt{NM-CCA2} \ar[l]_3 \ar@<0.5ex>[d]^1 \\ |
368 | \txt{IND-CPA} & |
369 | \txt{IND-CCA1} \ar[l]^2 |
370 | \POS !<-1ex, 0pt> \ar[ul]!<-1ex, 0pt>|-@{/}^5 & |
371 | \txt{IND-CCA2} \ar[l]^2 \ar@<0.5ex>[u]^4 \\ |
372 | } \] |
373 | |
374 | \begin{list}{}{ |
375 | \settowidth{\labelwidth}{\textbf{Key}} |
376 | \leftmargin\labelwidth\advance\leftmargin\labelsep |
377 | \itemindent0pt\let\makelabel\textbf} |
378 | \item[Key] \begin{itemize} |
379 | \item An arrow $\xy\ar*+{A};<1.5cm, 0cm>*+{B}\endxy$ indicates an |
380 | \emph{implication}: if a scheme is secure in notion $A$ then it is also |
381 | secure in notion $B$. |
382 | \item A crossed arrow $\xy\ar|-@{/}*+{A};<1.5cm, 0cm>*+{B}\endxy$ |
383 | indicates a \emph{separation}: there exists a scheme secure in notion |
384 | $A$ but not in $B$. |
385 | \item The numbers refer to sections of the proof provided in the notes. |
386 | \end{itemize} |
387 | \end{list} |
388 | \end{slide} |
389 | |
390 | \begin{proof} |
391 | Most of these are fairly simple. We use the indistinguishability-based |
392 | characterization of non-malleability that we showed earlier, because it |
393 | makes life much easier. We assume that schemes meeting each of the |
394 | definitions exist, else the theorems are all vacuously true. |
395 | |
396 | \begin{enumerate} |
397 | |
398 | \item We show $\text{NM-\id{atk}} \implies \text{IND-\id{atk}}$ for |
399 | $\id{atk} \in \{\text{CPA}, \text{CCA1}, \text{CCA2}\}$. Let $[]$ denote |
400 | the empty vector. Suppose that $A$ attacks $\mathcal{E}$ in the |
401 | $\text{IND-\id{atk}}$ sense. |
402 | \begin{program} |
403 | Adversary $A'^{D(\cdot)}(\cookie{find}, P)$: \+ \\ |
404 | $(x_0, x_1, s) \gets A^{D(\cdot)}(\cookie{find}, P)$; \\ |
405 | \RETURN $(x_0, x_1, s)$; |
406 | \newline |
407 | Adversary $A'^{D(\cdot)}(\cookie{choose}, y, s)$: \+ \\ |
408 | \RETURN $([], (y, s))$; |
409 | \next |
410 | Adversary $A'^{D(\cdot)}(\cookie{guess}, \vect{x}, s')$: \+ \\ |
411 | $(y, s) \gets s'$; \\ |
412 | $b' \gets A^{D(\cdot)}(y, s)$; \\ |
413 | \RETURN $b'$; |
414 | \end{program} |
415 | Evidently $\Adv{nm-\id{atk}}{\mathcal{E}}(A') = |
416 | \Adv{ind-\id{atk}}{\mathcal{E}}(A)$, and hence |
417 | \[ \InSec{ind-\id{atk}}(\mathcal{E}; t, q_D) \le |
418 | \InSec{nm-\id{atk}}(\mathcal{E}; t, q_D), \]% |
419 | proving the implication. |
420 | |
421 | \item We show that $\text{IND-$\id{atk}'$} \implies \text{IND-\id{atk}}$ |
422 | for $(\id{atk}', \id{atk}) \in \{(\text{CCA1}, \text{CPA}), (\text{CCA2}, |
423 | \text{CCA1})\}$. Suppose that $A$ attacks $\mathcal{E}$ in the |
424 | $\text{IND-\id{atk}}$ sense. |
425 | \begin{program} |
426 | Adversary $A'^{D(\cdot)}(\cookie{find}, P)$: \+ \\ |
427 | $(x_0, x_1, s) \gets A^{D_0(\cdot)}(\cookie{find}, P)$; \\ |
428 | \RETURN $(x_0, x_1, s)$; |
429 | \next |
430 | Adversary $A'^{D(\cdot)}(\cookie{guess}, y, s)$: \+ \\ |
431 | $b' \gets A^{D_1(\cdot)}(\cookie{guess}, y, s)$; \\ |
432 | \RETURN $b'$; |
433 | \end{program} |
434 | The oracles $D_0$ and $D_1$ are defined according to \id{atk}, as |
435 | shown in table~\ref{tab:se-rel-oracle}. |
436 | \begin{table} |
437 | \begin{tabular}[C]{l Mc Mc} \hlx*{hv} |
438 | \id{atk} & D_0(x) & D_1(x) \\ \hlx{vhv} |
439 | CPA & \bot & \bot \\ |
440 | CCA1 & D(x) & \bot \\ \hlx*{vh} |
441 | \end{tabular} |
442 | \caption{Relations between notions of security for public key |
443 | encryption: decryption oracles} |
444 | \label{tab:se-rel-oracle} |
445 | \end{table} |
446 | Evidently $\Adv{ind-$\id{atk}'$}{\mathcal{E}}(A') = |
447 | \Adv{ind-\id{atk}}{\mathcal{E}}(A)$, and hence |
448 | \[ \InSec{ind-\id{atk}}(\mathcal{E}; t, q_D) \le |
449 | \InSec{ind-$\id{atk}'$}(\mathcal{E}; t, q_D), \]% |
450 | proving the implication. |
451 | |
452 | \item We show that $\text{NM-$\id{atk}'$} \implies \text{NM-\id{atk}}$ |
453 | for $(\id{atk}', \id{atk}) \in \{(\text{CCA1}, \text{CPA}), (\text{CCA2}, |
454 | \text{CCA1})\}$. Suppose that $A$ attacks $\mathcal{E}$ in the |
455 | $\text{NM-\id{atk}}$ sense. |
456 | \begin{program} |
457 | Adversary $A'^{D(\cdot)}(\cookie{find}, P)$: \+ \\ |
458 | $(x_0, x_1, s) \gets A^{D_0(\cdot)}(\cookie{find}, P)$; \\ |
459 | \RETURN $(x_0, x_1, s)$; |
460 | \newline |
461 | Adversary $A'^{D(\cdot)}(\cookie{choose}, y, s)$: \+ \\ |
462 | $(\vect{y}, t) \gets A^{D_1(\cdot)}(\cookie{choose}, y, s)$; \\ |
463 | \RETURN $(\vect{y}, t)$; |
464 | \next |
465 | Adversary $A'^{D(\cdot)}(\cookie{guess}, \vect{x}, t)$: \+ \\ |
466 | $b' \gets A^{D_1(\cdot)}(\cookie{guess}, \vect{x}, t)$; \\ |
467 | \RETURN $b'$; |
468 | \end{program} |
469 | The oracles $D_0$ and $D_1$ are defined according to $\id{atk}'$, as |
470 | shown in table~\ref{tab:se-rel-oracle}. Evidently |
471 | $\Adv{nm-$\id{atk}'$}{\mathcal{E}}(A') = |
472 | \Adv{nm-\id{atk}}{\mathcal{E}}(A)$, and hence |
473 | \[ \InSec{nm-\id{atk}}(\mathcal{E}; t, q_D) \le |
474 | \InSec{nm-$\id{atk}'$}(\mathcal{E}; t, q_D), \]% |
475 | proving the implication. |
476 | |
477 | \item We show that $\text{IND-CCA2} \implies \text{NM-CCA2}$. Suppose that |
478 | $A$ is an adversary attacking $\mathcal{E}$ in the NM-CCA2 sense. |
479 | \begin{program} |
480 | Adversary $A'^{D(\cdot)}(\cookie{find}, P)$: \+ \\ |
481 | $(x_0, x_1, s) \gets A^{D_0(\cdot)}(\cookie{find}, P)$; \\ |
482 | \RETURN $(x_0, x_1, s)$; |
483 | \next |
484 | Adversary $A'^{D(\cdot)}(\cookie{guess}, y, s)$: \+ \\ |
485 | $(\vect{y}, t) \gets A^{D(\cdot)}(\cookie{choose}, y, s)$; \\ |
486 | $\vect{x} = D(\vect{y})$; \\ |
487 | $b' \gets A^{D(\cdot)}(\cookie{guess}, \vect{x}, t)$; \\ |
488 | \RETURN $b'$; |
489 | \end{program} |
490 | Evidently $\Adv{ind-cca2}{\mathcal{E}}(A') = |
491 | \Adv{nm-cca2}{\mathcal{E}}(A)$, and hence |
492 | \[ \InSec{nm-cca2}(\mathcal{E}; t, q_D) \le |
493 | \InSec{ind-cca2}(\mathcal{E}; t, q_D), \]% |
494 | proving the implication. |
495 | |
496 | \item We show that $\text{IND-CCA1} \not\implies \text{NM-CPA}$. Suppose |
497 | that $\mathcal{E} = (G, E, D)$ is secure in the IND-CCA1 sense. Consider |
498 | the encryption scheme $\mathcal{E}' = (G', E', D')$: |
499 | \begin{program} |
500 | Algorithm $G'(k)$: \+ \\ |
501 | $(P, K) \gets G$; \\ |
502 | \RETURN $(P, K)$; |
503 | \next |
504 | Algorithm $E'_P(x)$: \+ \\ |
505 | $y \gets E_P(x)$; \\ |
506 | \RETURN $(0, y)$; |
507 | \next |
508 | Algorithm $D'_K(y')$: \+ \\ |
509 | $(b, y) \gets y'$; \\ |
510 | $x \gets D_K(y)$; \\ |
511 | \RETURN $x$; |
512 | \end{program} |
513 | This is a classic example of a malleable encryption scheme. |
514 | |
515 | Firstly, we show that $\mathcal{E}'$ is still IND-CCA1. Suppose that |
516 | $A'$ is an adversary attacking $\mathcal{E}'$ in the sense of IND-CCA1. |
517 | Consider $A$, attacking $\mathcal{E}$: |
518 | \begin{program} |
519 | Adversary $A^{D(\cdot)}(\cookie{find}, P)$: \+ \\ |
520 | $(x_0, x_1, s) \gets A'^{\Xid{D'}{sim}(\cdot)}(\cookie{find}, P)$; \\ |
521 | \RETURN $(x_0, x_1, s)$; \- \\[\smallskipamount] |
522 | Oracle $\Xid{D'}{sim}(y')$: \+ \\ |
523 | $(b, y) \gets y'$; \\ |
524 | $x \gets D(y)$; \\ |
525 | \RETURN $x$; |
526 | \next |
527 | Adversary $A^\bot(\cookie{guess}, y, s)$: \+ \\ |
528 | $b' \gets A'^\bot(\cookie{guess}, y, s)$; \\ |
529 | \RETURN $b'$; |
530 | \end{program} |
531 | Clearly, $A$ simulates the expected environment for $A'$ perfectly; hence |
532 | \[ \Adv{ind-cca1}{\mathcal{E}'}(A') = \Adv{ind-cca1}{\mathcal{E}}(A). \] |
533 | |
534 | Now, we show that $\mathcal{E}'$ is not NM-CPA. Consider the adversary |
535 | $C'$: |
536 | \begin{program} |
537 | Adversary $C'^{D(\cdot)}(\cookie{find}, P')$: \+ \\ |
538 | \RETURN $(0, 1, \bot)$; |
539 | \newline |
540 | Adversary $C'^{D(\cdot)}(\cookie{choose}, y', s)$: \+ \\ |
541 | $(b, y) \gets y'$; \\ |
542 | $\vect{y} \gets [(1, y)]$; \\ |
543 | \RETURN $(\vect{y}, \bot)$; |
544 | \next |
545 | Adversary $C'^{D(\cdot)}(\cookie{guess}, \vect{x}, t)$: \+ \\ |
546 | \RETURN $\vect{x}[0]$; |
547 | \end{program} |
548 | Since $(0, y) \ne (1, y)$, the experiment provides the plaintext |
549 | corresponding to the challenge ciphertext $y'$. |
550 | $\Adv{nm-cpa}{\mathcal{E}'}(C') = 1$. Hence, $\mathcal{E}'$ is insecure |
551 | in the NM-CPA sense. |
552 | |
553 | \item We show that $\text{NM-CPA} \not\implies \text{IND-CCA1}$. Suppose |
554 | that $\mathcal{E} = (G, E, D)$ is secure in the NM-CPA sense. Fix a |
555 | security parameter $k$. Consider the encryption scheme $\mathcal{E}' = |
556 | (G', E', D')$: |
557 | \begin{program} |
558 | Algorithm $G'(k)$: \+ \\ |
559 | $(P, K) \gets G(k)$; \\ |
560 | $u \getsr \{0, 1\}^k$; $v \getsr \{0, 1\}^k$; \\ |
561 | \RETURN $((P, u), (K, u, v))$; |
562 | \next |
563 | Algorithm $E'_{P'}(x)$: \+ \\ |
564 | $(P, u) \gets P'$; \\ |
565 | $y \gets E_P(x)$; \\ |
566 | \RETURN $(0, y)$; |
567 | \next |
568 | Algorithm $D'_{K'}(y')$: \+ \\ |
569 | $(b, y) \gets y'$; \\ |
570 | $(K, u, v) \gets K'$; \\ |
571 | \IF $b = 0$ \THEN $x \gets D_K(y)$; \\ |
572 | \ELSE \IF $y = u$ \THEN $x \gets v$; \\ |
573 | \ELSE \IF $y = v$ \THEN $x \gets K$; \\ |
574 | \ELSE $x \gets \bot$; \\ |
575 | \RETURN $x$; |
576 | \end{program} |
577 | The idea is that the decryption oracle can be used to leak the key by |
578 | following the little $u \to v \to K$ chain, but the parallel |
579 | non-malleability oracle can't do this. |
580 | |
581 | We first show that $\mathcal{E}'$ is still NM-CPA. Suppose that $A'$ is |
582 | an adversary attacking $\mathcal{E}'$ in the NM-CPA sense. Consider |
583 | this adversary $A$: |
584 | \begin{program} |
585 | Adversary $A^\bot(\cookie{find}, P)$: \+ \\ |
586 | $u \getsr \{0, 1\}^k$; $v \getsr \{0, 1\}^k$; \\ |
587 | $(x_0, x_1, s') \gets A'^\bot(\cookie{find}, (P, u))$; \\ |
588 | \RETURN $(x_0, x_1, (s', u, v))$; |
589 | \newline |
590 | Adversary $A^\bot(\cookie{choose}, y, s)$: \+ \\ |
591 | $(s', u, v) \gets s$; \\ |
592 | $(\vect{y}', t') \gets A'^\bot(\cookie{choose}, y, s')$; \\ |
593 | \FOR $j = 0$ \TO $|\vect{y}'| - 1$ \DO |
594 | $\vect{x}'[j] \gets \bot$; \\ |
595 | \FOR $j = 0$ \TO $|\vect{y}'| - 1$ \DO \\ \quad \= \+ \kill |
596 | $(b', y') \gets \vect{y}'[j]$; \\ |
597 | \IF $b' = 0$ \THEN $\vect{y}[j] \gets y'$; \\ |
598 | \ELSE \IF $y' = u$ \THEN \= $\vect{x}'[j] \gets v$; \\ |
599 | \> $\vect{y}[j] \gets \bot$; \\ |
600 | \ELSE \IF $y' = v$ \THEN \ABORT; \\ |
601 | \ELSE \= $\vect{x}'[j] \gets \bot$; \\ |
602 | \> $\vect{y}[j] \gets \bot$; \\ |
603 | \RETURN $(\vect{y}, (\vect{x}', t'))$; |
604 | \next |
605 | Adversary $A^\bot(\cookie{guess}, \vect{x}, t)$: \+ \\ |
606 | $(\vect{x}', t') \gets t$; \\ |
607 | \FOR $j = 0$ \TO $|\vect{x}| - 1$ \DO \\ |
608 | \quad \IF $\vect{x}'[j] = \bot$ \THEN |
609 | $\vect{x}'[j] \gets \vect{x}[j]$; \\ |
610 | $b' \gets A'^\bot(\cookie{guess}, \vect{x}', t')$; \\ |
611 | \RETURN $b'$; |
612 | \end{program} |
613 | Clearly, $A$ behaves indistinguishably from the NM-CPA experiment |
614 | expected by $A'$, unless $A$ aborts because $A'$ guesses $v$ during its |
615 | $\cookie{choose}$ phase. But $v$ is independent of $A'$'s view at that |
616 | moment, so the probability this occurs is $2^{-k}$. Hence |
617 | \[ \Adv{nm-cpa}{\mathcal{E}'} \le |
618 | \Adv{nm-cpa}{\mathcal{E}} + \frac{1}{2^k}. \]% |
619 | |
620 | Now to show that $\mathcal{E}'$ is insecure in the IND-CCA1 sense. |
621 | Consider this adversary: |
622 | \begin{program} |
623 | Adversary $C'^{D(\cdot)}(\cookie{find}, P')$: \+ \\ |
624 | $(P, u) \gets P'$; \\ |
625 | $v \gets D(u)$; $K \gets D(v)$; \\ |
626 | \RETURN $(0, 1, K)$; |
627 | \next |
628 | Adversary $C'^\bot(\cookie{guess}, y, K)$: \+ \\ |
629 | $b' \gets D_K(y)$; \\ |
630 | \RETURN $b'$; |
631 | \end{program} |
632 | The adversary $C'$ makes 2 decryption queries, and |
633 | $\Adv{ind-cca1}{\mathcal{E}'}(C') = 1$. Hence, $\mathcal{E}'$ is |
634 | insecure in the IND-CCA1 sense. |
635 | |
636 | \item We show that $\text{NM-CCA1} \not\implies \text{IND-CCA2}$. Suppose |
637 | that $\mathcal{E} = (G, E, D)$ is secure in the NM-CCA1 sense. Let |
638 | $\mathcal{M} = (T, V)$ be a secure MAC. Consider the encryption scheme |
639 | $\mathcal{E}' = (G', E', D')$: |
640 | \begin{program} |
641 | Algorithm $G'(k)$: \+ \\ |
642 | $(P, K) \gets G(k)$; \\ |
643 | $i \getsr \keys \mathcal{M}$; \\ |
644 | \RETURN $(P, (K, i))$; |
645 | \next |
646 | Algorithm $E'_P(x)$: \+ \\ |
647 | $y \gets E_P(x)$; \\ |
648 | \RETURN $(0, y, \bot)$; |
649 | \next |
650 | Algorithm $D'_{K'}(y')$: \+ \\ |
651 | $(b, y, \tau) \gets y'$; \\ |
652 | $(K, i) \gets K'$; \\ |
653 | \IF $b = 0$ \THEN $x \gets D_K(y)$; \\ |
654 | \ELSE \IF $\tau = \bot$ \THEN $x \gets T_i(y)$; \\ |
655 | \ELSE \IF $V_i(y, \tau) = 1$ \THEN $x \gets D_K(y)$; \\ |
656 | \ELSE $x \gets \bot$; \\ |
657 | \RETURN $x$; |
658 | \end{program} |
659 | Answering decryption queries only when presented with a correct tag |
660 | ensures that the NM-CCA1 adversary can't obtain anything useful with its |
661 | queries (hence the scheme remains NM-CCA1-secure), but the IND-CCA2 |
662 | adversary can use its adaptive queries to discover the required tag. |
663 | |
664 | Firstly, we show that $\mathcal{E}'$ is still NM-CCA1-secure. Suppose |
665 | $A'$ attacks $\mathcal{E}'$ in the sense of NM-CCA1. Consider the two |
666 | adversaries shown in \ref{fig:nm-cca1-sep-ind-cca2}: $A$ attacks the |
667 | original $\mathcal{E}$; $A''$ attacks the MAC $\mathcal{M}$. |
668 | \begin{figure} |
669 | \begin{program} |
670 | Adversary $A^{D(\cdot)}(\cookie{find}, P)$: \+ \\ |
671 | $i \getsr \keys F$; \\ |
672 | $(x_0, x_1, s') \gets A'^{\Xid{D}{sim}(\cdot)}(\cookie{find}, P)$; \\ |
673 | \RETURN $(x_0, x_1, (s', i))$; \- \\[\smallskipamount] |
674 | Adversary $A^\bot(\cookie{choose}, y, s)$: \+ \\ |
675 | $(s', i) \gets s$; \\ |
676 | $(\vect{y}', t') \gets A'^\bot(\cookie{choose}, y, s')$; \\ |
677 | \FOR $j = 0$ \TO $|\vect{y}'| - 1$ \DO |
678 | $\vect{x}'[j] \gets \bot$; \\ |
679 | \FOR $j = 0$ \TO $|\vect{y}'| - 1$ \DO \\ \quad \= \+ \kill |
680 | $(b', y', \tau') \gets \vect{y}'[j]$; \\ |
681 | \IF $b' = 0$ \THEN $\vect{y}[j] \gets y'$; \\ |
682 | \ELSE \IF $z' = \bot$ \THEN \= $\vect{x}'[j] \gets T_i(y')$; \\ |
683 | \> $\vect{y}[j] \gets \bot$; \\ |
684 | \ELSE \IF $V_i(y', \tau') \ne 1$ |
685 | \THEN $\vect{y}[j] \gets \bot$; \\ |
686 | \ELSE \IF $y' = y$ \THEN \ABORT; \\ |
687 | \ELSE $\vect{y}[j] \gets y'$; \- \\ |
688 | \RETURN $(\vect{y}, (\vect{x}', t'))$; \- \\[\smallskipamount] |
689 | Adversary $A^\bot(\cookie{guess}, \vect{x}, t)$: \+ \\ |
690 | $(\vect{x}', t') \gets t$; \\ |
691 | \FOR $j = 0$ \TO $|\vect{x}| - 1$ \DO \\ |
692 | \quad \IF $\vect{x}'[j] = \bot$ \THEN |
693 | $\vect{x}'[j] \gets \vect{x}[j]$; \\ |
694 | $b' \gets A'^\bot(\cookie{guess}, \vect{x}', t')$; \\ |
695 | \RETURN $b'$; |
696 | \next |
697 | Adversary $A''^{T(\cdot), V(\cdot)}$: \+ \\ |
698 | $(P, K) \gets G$; |
699 | $b \getsr \{0, 1\}$; \\ |
700 | $(x_0, x_1, s) \gets A'^{\Xid{D}{sim}(\cdot)}(\cookie{find}, P)$; \\ |
701 | $y \gets (0, E_P(x_b), \bot)$; \\ |
702 | $(\vect{y}, t) \gets A'^\bot(\cookie{choose}, y, s)$; \\ |
703 | \FOR $j = 0$ \TO $|\vect{y}| - 1$ \DO \\ \quad \= \+ \kill |
704 | $(b', y', \tau') \gets \vect{y}'[j]$; \\ |
705 | \IF $b' = 1 \land V(y', \tau') = 1$ |
706 | \THEN \RETURN $(y', \tau')$; \- \\ |
707 | \ABORT; \- \\[\smallskipamount] |
708 | Oracle $\Xid{D}{sim}(y')$: \+ \\ |
709 | $(b, y, \tau) \gets y'$; \\ |
710 | \IF $b = 0$ \THEN $x \gets D_K(y)$; \\ |
711 | \ELSE \IF $\tau = \bot$ \THEN $x \gets T(y)$; \\ |
712 | \ELSE \IF $V(y, \tau) = 1$ \THEN $x \gets D_K(y)$; \\ |
713 | \ELSE $x \gets \bot$; \\ |
714 | \RETURN $x$; |
715 | \end{program} |
716 | \caption{From the proof that $\text{NM-CCA1} \not\implies |
717 | \text{IND-CCA2}$ -- adversaries $A$ and $A''$, attacking $\mathcal{E}$ |
718 | in the NM-CCA1 sense and the MAC $\mathcal{M}$ respectively} |
719 | \label{fig:nm-cca1-sep-ind-cca2} |
720 | \end{figure} |
721 | |
722 | Suppose, without loss of generality, that if the challenge ciphertext $y$ |
723 | returned to $A'$ matches one of $A'$'s earlier decryption queries then |
724 | $A'$ wins without making its parallel chosen ciphertext query. |
725 | |
726 | Let $B$ be the event that $A$ aborts; let $S$ be the event that $A$ wins, |
727 | and let $S'$ be the event that $A'$ wins. Since unless it aborts, $A$ |
728 | implements the NM-CCA1 game for $\mathcal{E}'$ perfectly, we must have |
729 | \[ \Pr[A] = \Pr[A'] - \Pr[B]. \] |
730 | But $B$ is precisely the event in which $A''$ wins its game. Hence |
731 | \[ \Adv{nm-cca1}{\mathcal{E}'}(A') \le |
732 | \Adv{nm-cca1}{\mathcal{E}}(A) + \Adv{suf-cma}{\mathcal{E}}(A'') \]% |
733 | proving the claim that $\mathcal{E}'$ remains NM-CCA1 secure. |
734 | |
735 | Now to show that $\mathcal{E}'$ is not IND-CCA2 secure. |
736 | \begin{program} |
737 | Adversary $C'^{D(\cdot)}(\cookie{find}, P')$: \+ \\ |
738 | \RETURN $(0, 1, \bot)$; |
739 | \next |
740 | Adversary $C'^{D(\cdot)}(\cookie{guess}, y, s)$: \+ \\ |
741 | $\tau \gets D(1, y, \bot)$; \\ |
742 | $b' \gets D(1, y, \tau)$; \\ |
743 | \RETURN $b'$; |
744 | \end{program} |
745 | The adversary $C'$ makes 2 decryption queries, and |
746 | $\Adv{ind-cca2}{\mathcal{E}'}(C') = 1$. Hence, $\mathcal{E}'$ is |
747 | insecure in the IND-CCA2 sense. |
748 | |
749 | \end{enumerate} |
750 | All present and correct. |
751 | \end{proof} |
752 | |
753 | \begin{exercise} |
754 | In \cite{Goldwasser:1984:PE}, Shafi Goldwasser and Silvio Micali first |
755 | introduced the concept of semantic security as a definition of |
756 | \emph{computationally} secure encryption, and presented the first |
757 | provably-secure probabilistic encryption scheme. For this scheme, the |
758 | private key is a pair of $k$-bit primes $(p, q)$; the public key is their |
759 | product $n = pq$ and a \emph{pseudosquare} $z$. Given this starting point, |
760 | define a public-key encryption scheme, and relate its security in the |
761 | IND-CPA sense to the difficulty of the Quadratic Residuosity Problem. Show |
762 | that the encryption scheme is malleable. |
763 | |
764 | Hints: |
765 | \begin{parenum} |
766 | \item Encrypt the message one bit at a time. |
767 | \item Choosing a random element of $(\Z/n\Z)^*$ and squaring gives you a |
768 | random element of $Q_n$. |
769 | \item You will need to define a formal game for the Quadratic Residuosity |
770 | Problem. |
771 | \end{parenum} |
772 | \answer% |
773 | Encryption works as $\Xid{E}{GM}_{(n, z)}(x)$: \{ \FOREACH $1\colon x_i$ |
774 | \FROM $x$ \DO \{ $a_i \getsr (\Z/n\Z)^*$; $\vect{y}[i] \gets a_i^2 |
775 | z^{x_i}$~\} \RETURN $\vect{y}$;~\}. Decryption works as $\Xid{D}{GM}_{(p, |
776 | q)}(\vect{y})$: \{ $x \gets \emptystring$; \FOR $i = 0$ \TO $|\vect{y}| - |
777 | 1$ \DO \{ \IF $\jacobi{\vect{y}[i]}{p} = 1 \land \jacobi{\vect{y}[i]}{q} = |
778 | 1$ \THEN $x \gets x \cat 0$; \ELSE $x \gets x \cat 1$;~\} \RETURN $x$;~\}. |
779 | |
780 | To prove that this is meets IND-CPA, let $A$ be an adversary attacking the |
781 | GM scheme. Now, we present an algorithm which, given an odd composite |
782 | integer $n$ and an element $z$ with $\jacobi{x}{n} = 1$, decides whether $x |
783 | \in Q_n$: Algorithm $B(n, z)$: $(x_0, x_1, s) \gets A(\cookie{find}, (n, |
784 | z))$; $b \getsr \{0, 1\}$; $y \gets \Xid{E}{GM}_{(n, z)}(x_b)$; $b' \gets |
785 | A(\cookie{guess}, y, s)$; \IF $b = b'$ \THEN \RETURN $0$; \ELSE \RETURN |
786 | $1$;~\}. If $z$ is a nonresidue, then $B$ returns $0$ whenever $A$ |
787 | successfully guesses the right bit; if $z \in Q_n$ then the ciphertext $y$ |
788 | returned by $B$ is a string of random quadratic residues independent of the |
789 | challenge plaintext, and $A$ can have no advantage. Hence |
790 | $\InSec{ind-cpa}(\Xid{\mathcal{E}}{GM-$k$}; t) \le 2 \cdot \InSec{qrp}(k; |
791 | t)$. |
792 | |
793 | To prove malleability, simply note that multiplying an element |
794 | $\vect{y}[i]$ by $z$ toggles the corresponding plaintext bit. |
795 | \end{exercise} |
796 | |
797 | \xcalways\subsection{The ElGamal scheme}\x |
798 | |
799 | \begin{slide} |
800 | \topic{description} |
801 | \head{The ElGamal public-key encryption scheme} |
802 | |
803 | ElGamal's encryption scheme is based on Diffie-Hellman. Let $G = \langle g |
804 | \rangle$ be a cyclic group of order $q$. Plaintexts and ciphertexts in the |
805 | scheme are elements of $G$. |
806 | |
807 | The scheme $\Xid{\mathcal{E}}{ElGamal}^G = (\Xid{G}{ElGamal}^G, |
808 | \Xid{E}{ElGamal}^G, \Xid{D}{ElGamal}^G)$ is defined by: |
809 | \begin{program} |
810 | $\Xid{G}{ElGamal}^G$: \+ \\ |
811 | $\alpha \getsr \{1, 2, \ldots, q - 1\}$; \\ |
812 | \RETURN $(g^\alpha, \alpha)$; |
813 | \next |
814 | $\Xid{E}{ElGamal}^G_a(x)$: \+ \\ |
815 | $\beta \getsr \{1, 2, \ldots, q - 1\}$; \\ |
816 | \RETURN $(g^\beta, x a^\beta)$; |
817 | \next |
818 | $\Xid{D}{ElGamal}^G_\alpha(y)$: \+ \\ |
819 | $(b, c) \gets y$; \\ |
820 | $x \gets b^{-\alpha} c$; \\ |
821 | \RETURN $x$; |
822 | \end{program} |
823 | This scheme is secure in the IND-CPA sense if the Decisional Diffie-Hellman |
824 | problem is hard in $G$. |
825 | \end{slide} |
826 | |
827 | \begin{slide} |
828 | \topic{security proof} |
829 | \head{Security proof for ElGamal} |
830 | |
831 | Suppose $A$ is an adversary attacking the ElGamal scheme in the IND-CPA |
832 | sense. We construct from it an algorithm which solves the DDH problem |
833 | (i.e., given a triple $g^\alpha, g^\beta, c$, decides whether $c = |
834 | g^{\alpha\beta}$): |
835 | \begin{program} |
836 | Algorithm $D(a, b, c)$: \+ \\ |
837 | $(x_0, x_1, s) \gets A(\cookie{find}, a)$; \\ |
838 | $z \getsr \{0, 1\}$; \\ |
839 | $y \gets (b, x_z c)$; \\ |
840 | $z' \gets A(\cookie{guess}, y, s)$; \\ |
841 | \IF $z = z'$ \THEN \RETURN $1$; \\ |
842 | \ELSE \RETURN $0$; |
843 | \end{program} |
844 | \end{slide} |
845 | |
846 | \begin{slide} |
847 | \head{Security proof for ElGamal (cont.)} |
848 | |
849 | Let $\alpha$ and $\beta$ be the discrete logs of $a$ and $b$. |
850 | |
851 | If $c = g^{\alpha\beta}$, then $D$'s success probability is equal to $A$'s |
852 | probability of guessing the hidden bit correctly, which is |
853 | \[ \frac{\Adv{ind-cpa}{\Xid{\mathcal{E}}{ElGamal}^G}(A)}{2} + \frac{1}{2}. \]% |
854 | If $c$ is random, then $x_z c$ is uniformly distributed in $G$, and |
855 | independent of $b$, so $A$ answers correctly with probability exactly |
856 | $\frac{1}{2}$. |
857 | |
858 | Hence, $\Adv{ddh}{G}(D) = |
859 | \Adv{ind-cpa}{\Xid{\mathcal{E}}{ElGamal}^G}(A)/2$, and |
860 | \[ \InSec{ind-cpa}(\Xid{\mathcal{E}}{ElGamal}^G; t) \le |
861 | 2 \cdot \InSec{ddh}(G; t). \]% |
862 | \end{slide} |
863 | |
864 | \begin{slide} |
865 | \topic{other notes} |
866 | \head{Notes about ElGamal} |
867 | |
868 | \begin{itemize} |
869 | |
870 | \item We needed the Decisional Diffie-Hellman assumption to prove the |
871 | security. As noted before, this is a very strong assumption. Still, a |
872 | proof based on DDH is a lot better than nothing. |
873 | |
874 | We really do need the Decisional Diffie-Hellman assumption. An adversary |
875 | with a DDH algorithm can submit $x_0 \inr G$ and $x_1 = 1$; it receives a |
876 | ciphertext $(b, y)$, and returns $1$ if $(a, b, y)$ looks like a |
877 | Diffie-Hellman triple, or $0$ if it looks random. |
878 | |
879 | \item The plaintexts must be elements of the cyclic group $G$. For |
880 | example, if $G$ is a subgroup of $\F_p^*$, it's \emph{not} safe to allow |
881 | elements outside the subgroup as plaintexts: an adversary can compare |
882 | orders of ciphertext elements to break the semantic security of the |
883 | scheme. |
884 | |
885 | \item ElGamal is malleable. We can decrypt a challenge ciphertext $y = |
886 | (g^\beta, a^\beta x)$ by choosing a random $\gamma$ and requesting a |
887 | decryption of $y' = (g^{\beta\gamma}, a^{\beta\gamma} x^\gamma)$. |
888 | |
889 | \end{itemize} |
890 | \end{slide} |
891 | |
892 | \xcalways\subsection{Encrypting using trapdoor one-way functions}\x |
893 | |
894 | \begin{slide} |
895 | \head{Trapdoor one-way functions} |
896 | |
897 | Trapdoor one-way functions RSA ($x \mapsto x^e \bmod n$) and Rabin ($x |
898 | \mapsto x^2 \bmod n$) functions are well-studied. Can we make secure |
899 | schemes from them? |
900 | |
901 | We can't use them directly, however. For a start, the functions are |
902 | deterministic, so `encrypting' messages just by doing the modular |
903 | exponentiation on the plaintext will leak equality of plaintexts. |
904 | |
905 | How can we fix these schemes? |
906 | |
907 | We'll focus on RSA, because decryption is simpler; Rabin works in a very |
908 | similar way. |
909 | \end{slide} |
910 | |
911 | \begin{slide} |
912 | \topic{simple embedding schemes} |
913 | \head{Simple embedding schemes} |
914 | |
915 | If we restrict our attention to plaintext messages which are `about' the |
916 | input size of our one-way function, we'd like to be able to use ciphertexts |
917 | which are no bigger than the output size of the function. |
918 | |
919 | Perhaps if we encode the message in some way before passing it through the |
920 | function, we can improve security. Ideally, we'd like security against |
921 | chosen-ciphertext attacks. |
922 | |
923 | An encryption scheme which processes messages like this is called a |
924 | \emph{simple embedding scheme}. |
925 | \end{slide} |
926 | |
927 | \begin{slide} |
928 | \topic{\PKCS1 padding} |
929 | \head{\PKCS1 encryption padding for RSA \cite{RSA:1993:PRE}} |
930 | |
931 | Let $n = p q$ be an RSA modulus, with $2^{8(k-1)} < n < 2^{8k)}$ -- i.e., |
932 | $n$ is a $k$-byte number. Let $m$ be the message to be signed. |
933 | |
934 | We compute $\hat{m} = 0^8 \cat T \cat r \cat 0^8 \cat x$ for |
935 | some hash function $m$, where: |
936 | \begin{itemize} |
937 | \item $|\hat{m}| = 8k$, i.e., $\hat{m}$ is a $k$-byte string; |
938 | \item $0^8$ is the string of 8 zero-bits; |
939 | \item $T = 00000002$ is an 8-bit \emph{type} code; and |
940 | \item $r$ is a string of random \emph{non-zero} bytes (\PKCS1 requires at |
941 | least 8 bytes) |
942 | \end{itemize} |
943 | This bit string is then converted into an integer, using a big-endian |
944 | representation. Note that $\hat{m} < n$, since its top byte is zero. |
945 | \end{slide} |
946 | |
947 | \begin{slide} |
948 | \head{\PKCS1 encryption padding for RSA (cont.)} |
949 | |
53aa10b5 |
950 | Diagrammatically, \PKCS1 encryption padding looks like this: |
41761fdc |
951 | \begin{tabular}[C]{r|c|c|c|c|} \hlx{c{2-5}v} |
952 | \hex{00} & \hex{01} & |
953 | \ldots\ random nonzero bytes \ldots & |
954 | \hex{00} & |
955 | $m$ |
956 | \\ \hlx{vc{2-5}} |
957 | \end{tabular} |
958 | |
959 | Unfortunately, this scheme isn't capable of resisting chosen-ciphertext |
960 | attacks: Bleichenbacher \cite{Bleichenbacher:1998:CCA} shows how to decrypt |
961 | a ciphertext $y$ given an oracle (say, an SSL server) which returns whether |
962 | a given ciphertext is valid (i.e., its padding is correct). |
963 | \end{slide} |
964 | |
965 | \begin{slide} |
966 | \topic{Optimal Asymmetric Encryption Padding (OAEP)} |
967 | \head{Optimal Asymmetric Encryption Padding (OAEP) \cite{Bellare:1995:OAE}} |
968 | |
969 | OAEP is a simple embedding scheme for use with an arbitrary trapdoor |
970 | one-way function. It requires two hash functions, $H$ and $H'$, which we |
971 | model as random oracles. |
972 | |
973 | We assume that our one-way function $f_P$ operates on $n$-bit strings. Fix |
974 | a security parameter $k$. We require two random oracles $G\colon \{0, |
975 | 1\}^k \to \{0, 1\}^{n-k}$ and $H\colon \{0, 1\}^{n-k} \to \{0, 1\}^k$. |
976 | Given a message $x \in \{0, 1\}^{n-2k}$, we encrypt it as follows: |
977 | \begin{program} |
978 | Algorithm $\Xid{E}{OAEP}^{\mathcal{T}, G(\cdot), H(\cdot)}_P(x)$: \+ \\ |
979 | $r \getsr \{0, 1\}^k$; \\ |
980 | $s \gets (x \cat 0^k) \xor G(r)$; $t \gets r \xor H(s)$; \\ |
981 | $w \gets s \cat t$; \\ |
982 | \RETURN $f_P(w)$; |
983 | \next |
984 | Algorithm $\Xid{D}{OAEP}^{\mathcal{T}, G(\cdot), H(\cdot)}_K(y)$: \+ \\ |
985 | $w \gets T_K(y)$; \\ |
986 | \PARSE $w$ \AS $s, k\colon t$; \\ |
987 | $r \gets t \xor H(s)$; $x' \gets s \xor G(r)$; \\ |
988 | \PARSE $x'$ \AS $x, k\colon c$; \\ |
989 | \IF $c = 0^k$ \THEN \RETURN $x$; \\ |
990 | \ELSE \RETURN $\bot$; |
991 | \end{program} |
992 | \end{slide} |
993 | |
994 | \begin{slide} |
995 | \head{Optimal Asymmetric Encryption Padding (OAEP) (cont.)} |
996 | |
997 | %% x <- {0, 1}^{n-2k} r <-R {0, 1}^k |
998 | %% | | |
999 | %% | | |
1000 | %% v | |
1001 | %% [||]<--- 0^k | |
1002 | %% | | |
1003 | %% | | |
1004 | %% v | |
1005 | %% (+)<-----------[G]--------------| |
1006 | %% | | |
1007 | %% | | |
1008 | %% | v |
1009 | %% |-------------[H]------------>(+) |
1010 | %% | | |
1011 | %% | | |
1012 | %% v v |
1013 | %% < s = (x || 0^k) (+) G(r) > < t = r (+) H(s) > |
1014 | |
1015 | \vfil |
1016 | \[ \begin{graph} |
1017 | []!{0; <1cm, 0cm>:} |
1018 | {x \in \{0, 1\}^n}="x" [rrrr] {r \inr \{0, 1\}^k}="r" |
1019 | "r" :[ddd] *{\xor}="h-xor" |
1020 | :[d] *++=(2, 0)[F:thicker]{\strut t = r \xor H(s)} |
1021 | "x" :[d] *{\ocat}="cat" |
1022 | [r] *+[r]{\mathstrut 0^k} :"cat" |
1023 | :[d] *{\xor}="g-xor" |
1024 | :[dd] *++=(4, 0)[F:thicker]{\strut s = (x \cat 0^k) \xor G(r)} |
1025 | [u] :[rr] *+[F]{H'} :"h-xor" |
1026 | "r" [dd] :[ll] *+[F]{H} :"g-xor" |
1027 | \end{graph} \] |
1028 | \vfil |
1029 | \end{slide} |
1030 | |
1031 | \begin{slide} |
53aa10b5 |
1032 | \head{Security of OAEP, \seq: chosen plaintext attacks} |
41761fdc |
1033 | |
1034 | We write the encryption scheme formed by using the trapdoor one-way |
1035 | function $\mathcal{T}$ with OAEP embedding as |
1036 | $\Xid{\mathcal{E}}{OAEP}^{\mathcal{T}}$. |
1037 | |
1038 | The security of OAEP in the IND-CPA sense is given by: |
1039 | \[ \InSec{ind-cpa}(\Xid{\mathcal{E}}{OAEP}^{\mathcal{T}}; t, q_G, q_H) |
1040 | \le 2 \left (\frac{q_G}{2^k} + |
1041 | \InSec{owf}(\mathcal{T}; t + O(q_G q_H)) \right). \]% |
1042 | \end{slide} |
1043 | |
1044 | We omit the security proof of OAEP, because it's very similar to the proof |
1045 | for OAEP+ which is presented in full later. |
1046 | |
1047 | \begin{slide} |
1048 | \topic{possible malleability of OAEP} |
1049 | \head{OAEP is not (generically) non-malleable \cite{Shoup:2001:OAEPR}} |
1050 | |
1051 | If a one-way function might be \emph{XOR-malleable} -- i.e., given $f_P(x)$ |
1052 | and $\delta$, it's possible to construct $f_P(x \xor \delta)$ without |
1053 | knowledge of the trapdoor -- then the OAEP encryption scheme is malleable. |
1054 | |
1055 | \begin{graph} |
1056 | []!{0; <1cm, 0cm>:} |
1057 | *!UR{\parbox{0.5\linewidth}{ |
1058 | \raggedright |
1059 | We need to suppose, in addition, that the function leaks the leftmost |
1060 | $n-k$ bits of its input; e.g., $f(s \cat t) = s \cat f'(t)$. Then |
1061 | exploiting the differential (pictured right) only requires a pair of |
1062 | queries to the $H$ oracle, not $G$. Using its parallel |
1063 | chosen-ciphertext query, the adversary can decrypt its target |
1064 | plaintext. |
1065 | }} |
1066 | !{+(1.5, 0)} |
1067 | {x}="x" [rr] {r}="r" |
1068 | "r" :[ddd] *{\xor}="h-xor" ^{0} |
1069 | :[d] {t}^{H(s) \xor H(s \xor (\delta \cat 0^k))} |
1070 | "x" :[d] *{\ocat}="cat" _{\delta} |
1071 | [r] *+[r]{\mathstrut 0^k} :"cat" |
1072 | :[d] *{\xor}="g-xor" _{\delta \cat 0^k} |
1073 | :[dd] {s}_{\delta \cat 0^k} |
1074 | [u] :[r] *+[F]{H'} :"h-xor" |
1075 | "r" [dd] :[l] *+[F]{H} :"g-xor" |
1076 | \end{graph} |
1077 | |
1078 | Nevertheless, OAEP \emph{does} provide security against chosen-ciphertext |
1079 | attacks when used with the RSA and Rabin functions. |
1080 | \end{slide} |
1081 | |
1082 | \xcalways\subsection{Plaintext awareness and OAEP+}\x |
1083 | |
1084 | \begin{slide} |
1085 | \head{Plaintext awareness} |
1086 | |
1087 | The intuitive idea behind plaintext awareness is that it's hard to |
1088 | construct a new ciphertext for which you can't easily guess the plaintext |
1089 | (or guess that the ciphertext is invalid). Obviously, such an idea would |
1090 | imply security against chosen ciphertext attack -- since the adversary |
1091 | effectively knows the plaintext anyway, the decryption oracle is useless. |
1092 | |
1093 | The formalization introduces a \emph{plaintext extractor} -- an algorithm |
1094 | which, given a ciphertext and the random oracle queries of the program |
1095 | which created it, returns the corresponding plaintext. |
1096 | |
1097 | Note, then, that plaintext awareness (as currently defined) only works in |
1098 | the random oracle model. |
1099 | \end{slide} |
1100 | |
1101 | \begin{slide} |
1102 | \topic{definition of plaintext awareness} |
1103 | \head{Definition of plaintext awareness \cite{Bellare:1998:RAN}} |
1104 | |
1105 | Let $\mathcal{E} = (G, E, D)$ be a public-key encryption scheme. Consider |
1106 | an adversary $A$ which takes a public key as input and returns a ciphertext |
1107 | $y$. We run the adversary, providing it with an \emph{encryption} oracle, |
1108 | and record |
1109 | \begin{itemize} |
1110 | \item the set $Q_H$, which contains a pair $(q, h)$ for each random oracle |
1111 | query $q$ made by the adversary, together with its reply $h$; and |
1112 | \item the set $C$, which contains the responses (only: not the queries) |
1113 | from $A$'s encryption oracle. |
1114 | \end{itemize} |
1115 | We write $(y, Q_H, C) \gets \id{transcript}(A; z)$ to denote running $A$ on |
1116 | the arguments $z$ and collecting this information. |
1117 | |
1118 | Note that $Q_H$ doesn't include the oracle queries required by the |
1119 | encryption oracle. |
1120 | \end{slide} |
1121 | |
1122 | \begin{slide} |
1123 | \head{Definition of plaintext awareness (cont.)} |
1124 | |
1125 | An \emph{$\epsilon$-plaintext extractor} for $\mathcal{E}$ is an algorithm |
1126 | $X$ for which |
1127 | \begin{eqnarray*}[rl] |
1128 | \min_A \Pr[ |
1129 | & |
1130 | (P, K) \gets G; |
1131 | (y, Q_H, C) \gets \id{transcript}(A^{E(\cdot), H(\cdot)}; P); \\ |
1132 | & x \gets X(y, P, Q_H, C) : x = D_K(y)] \ge 1 - \epsilon . |
1133 | \end{eqnarray*} |
1134 | |
1135 | The scheme $\mathcal{E}$ is \emph{$\epsilon$-plaintext aware} (PA) if there |
1136 | exists an $\epsilon$-plaintext extractor for $\mathcal{E}$ \emph{and} |
1137 | $\mathcal{E}$ is IND-CPA. |
1138 | |
1139 | (The requirement that the scheme meet IND-CPA prevents trivial schemes such |
1140 | as the identity function from being considered plaintext aware.) |
1141 | \end{slide} |
1142 | |
1143 | \begin{slide} |
1144 | \topic{chosen-ciphertext security} |
1145 | \head{Plaintext awareness and chosen-ciphertext security |
1146 | \cite{Bellare:1998:RAN}} |
1147 | |
1148 | If a public key encryption scheme $\mathcal{E}$ is PA, then it is also |
1149 | IND-CCA2 (and hence NM-CCA2). This is proven using the plaintext extractor |
1150 | to simulate the decryption oracle. The encryption oracle in the PA |
1151 | definition is used to feed the challenge ciphertext to the plaintext |
1152 | extractor. |
1153 | |
53aa10b5 |
1154 | Quantitatively, if $\mathcal{E}$ is $\epsilon$-plaintext aware, and the |
41761fdc |
1155 | plaintext extractor runs in time $t_X(q_H)$, then |
1156 | \[ \InSec{ind-cca2}(\mathcal{E}; t, q_D, q_H) \le |
1157 | \InSec{ind-cpa}(\mathcal{E}; t + q_D t_X(q_H), q_H) + q_D \epsilon. \]% |
1158 | |
1159 | The converse is not true: IND-CCA2 does not imply plaintext awareness. |
1160 | \end{slide} |
1161 | |
1162 | \begin{proof} |
1163 | Firstly, we prove that $\text{PA} \implies \text{IND-CCA2}$. Let |
1164 | $\mathcal{E} = (G, E, D)$ be an $\epsilon$-plaintext aware public-key |
1165 | encryption scheme. Suppose $A'$ is an adversary attacking $\mathcal{E}$ in |
1166 | the IND-CCA2 sense. Let $X$ be the $\epsilon$-plaintext extractor for |
1167 | $\mathcal{E}$. We construct an adversary attacking $\mathcal{E}$ in the |
1168 | IND-CPA sense. |
1169 | \begin{program} |
1170 | Adversary $A^{H(\cdot)}(\cookie{find}, P)$: \+ \\ |
1171 | $\Xid{H}{map} \gets \emptyset$; $y^* \gets \bot$; \\ |
1172 | $(x_0, x_1, s') \gets A'^{\Xid{D}{sim}(\cdot), \Xid{H}{sim}(\cdot)} |
1173 | (\cookie{find}, P)$; \\ |
1174 | \RETURN $(x_0, x_1, (\Xid{H}{map}, P, s')$; \- \\[\smallskipamount] |
1175 | Oracle $\Xid{H}{sim}(x)$: \+ \\ |
1176 | $h \gets H(x)$; \\ |
1177 | $\Xid{H}{map} \gets \Xid{H}{map} \cup \{ x \mapsto h \}$; \\ |
1178 | \RETURN $h$; |
1179 | \next |
1180 | Adversary $A^{H(\cdot)}(\cookie{guess}, y^*, s)$: \+ \\ |
1181 | $(\Xid{H}{map}, P, s') \gets s$; \\ |
1182 | $b \gets A'^{\Xid{D}{sim}(\cdot), \Xid{H}{sim}(\cdot)} |
1183 | (\cookie{guess}, y^*, s')$; \\ |
1184 | \RETURN $b$; \- \\[\smallskipamount] |
1185 | Oracle $\Xid{D}{sim}(y)$: \+ \\ |
1186 | $x \gets X(y, P, \Xid{H}{map}, \{ y^* \})$; \\ |
1187 | \RETURN $x$; |
1188 | \end{program} |
1189 | Let $q_D$ be the number of decryption queries made by the adversary. We |
1190 | can see that, if the plaintext extractor does its job correctly, $A$ |
1191 | simulates the decryption oracle for $A'$ correctly. Hence |
1192 | \[ \Adv{ind-cpa}{\mathcal{E}}(A) \ge |
1193 | \Adv{ind-cca2}{\mathcal{E}}(A') - q_D \epsilon. \]% |
1194 | Notice how the encryption oracle from the plaintext awareness definition is |
1195 | used in the proof. |
1196 | |
1197 | We can show that $\text{IND-CCA2} \not\implies \text{PA}$ easily enough by |
1198 | amending an existing IND-CCA2 scheme $\mathcal{E}$ so that it has a |
1199 | `magical' ciphertext, attached to the public key. Here is our modified |
1200 | scheme $\mathcal{E}' = (G', E', D')$: |
1201 | \begin{program} |
1202 | Algorithm $G'^{H(\cdot)}(k)$: \+ \\ |
1203 | $(P, K) \gets G^{H(\cdot)}$; \\ |
1204 | $p \getsr \dom E^{H(\cdot)}_P$; \\ |
1205 | $c \getsr \ran E^{H(\cdot)}_P$; \\ |
1206 | \RETURN $((P, c), (K, c, p))$; |
1207 | \next |
1208 | Algorithm $E'^{H(\cdot)}_{P, c}(x)$: \+ \\ |
1209 | $y \gets E^{H(\cdot)}_P(x)$; \\ |
1210 | \RETURN $y$; |
1211 | \next |
1212 | Algorithm $D'^{H(\cdot)}_{K, c, p}(y)$: \+ \\ |
1213 | \IF $y = c$ \THEN $x \gets p$; \\ |
1214 | \ELSE $x \gets D^{H(\cdot)}_K(y)$; \\ |
1215 | \RETURN $x$; |
1216 | \end{program} |
1217 | Firstly, we show that it still meets IND-CCA2. Let $A'$ be an adversary |
53aa10b5 |
1218 | attacking $\mathcal{E}'$ in the IND-CCA2 sense. Then $A$, below, achieves |
41761fdc |
1219 | the same advantage against $\mathcal{E}$. |
1220 | \begin{program} |
1221 | Adversary $A^{D(\cdot), H(\cdot)}(\cookie{find}, P)$: \+ \\ |
1222 | $p \getsr \dom E^{H(\cdot)}_P$; $c \getsr \ran E^{H(\cdot)}_P$; \\ |
1223 | $(x_0, x_1, s') \gets A'^{\Xid{D}{sim}(\cdot), H(\cdot)} |
1224 | (\cookie{find}, (P, c))$; \\ |
1225 | \RETURN $(x_0, x_1, (p, c, s')$; |
1226 | \next |
1227 | Adversary $A^{D(\cdot), H(\cdot)}(\cookie{guess}, y, s)$: \+ \\ |
1228 | $(p, c, s') \gets s$; \\ |
1229 | $b \gets A'^{\Xid{D}{sim}(\cdot), H(\cdot)}(\cookie{guess}, y, s')$; \\ |
1230 | \RETURN $y$; |
1231 | \newline |
1232 | Oracle $\Xid{D}{sim}(y)$: \+ \\ |
1233 | \IF $y = c$ \THEN $x \gets p$; \\ |
1234 | \ELSE $x \gets D(y)$; \\ |
1235 | \RETURN $x$; |
1236 | \end{program} |
1237 | |
1238 | Now to show that $\mathcal{E}'$ is not plaintext-aware. Suppose that it |
1239 | is, and there is a plaintext extractor $X$. We show that $\mathcal{E}$ is |
1240 | not IND-CPA (contradicting the earlier assumption that $\mathcal{E}$ was |
1241 | IND-CCA2). |
1242 | \begin{program} |
1243 | Adversary $A''^{H(\cdot)}(\cookie{find}, P)$: \+ \\ |
1244 | $x_0 \getsr \dom E_P$; $x_1 \getsr \dom E_P$; \\ |
1245 | \RETURN $(x_0, x_1, (P, x_1))$; |
1246 | \next |
1247 | Adversary $A''^{H(\cdot)}(\cookie{guess}, y, s)$: \+ \\ |
1248 | $(P, x_1) \gets s$; \\ |
1249 | $x \gets X(y, (P, y), \emptyset, \emptyset)$; \\ |
1250 | \IF $x = x_1$ \THEN \RETURN $1$; |
1251 | \ELSE \RETURN $0$; |
1252 | \end{program} |
1253 | Since $x_0$ and $x_1$ are uniform in $\dom E_P$, the transcript $(y, (P, |
1254 | y), \emptyset, \emptyset)$ passed to the extractor is distributed exactly |
1255 | as for the $\mathcal{E}'$ adversary which, given the public key $(P, c)$ |
1256 | returns $c$; hence it succeeds with its usual probability $1 - \epsilon$. |
1257 | Then, $A''$'s advantage is |
1258 | \[ \Adv{ind-cpa}{\mathcal{E}}(A'') = |
1259 | 1 - 2\epsilon - \frac{1}{|{\dom E_P}|}. \]% |
1260 | This completes the proof. |
1261 | \end{proof} |
1262 | |
1263 | \begin{slide} |
1264 | \topic{old definition} |
1265 | \head{The old definition of plaintext awareness} |
1266 | |
1267 | An earlier definition of plaintext awareness omitted the encryption oracle |
1268 | provided to the adversary. The designers of OAEP proved that it met this |
1269 | definition of plaintext awareness and asserted, without proof, that this |
1270 | implied security against adaptive chosen-ciphertext attacks. |
1271 | |
1272 | The original definition of plaintext-awareness is sufficient to guarantee |
1273 | IND-CCA1 security, but it doesn't provide any sort of non-malleability. |
1274 | \end{slide} |
1275 | |
1276 | \begin{slide} |
1277 | \topic{OAEP+} |
53aa10b5 |
1278 | \resetseq |
1279 | \head{Plaintext-aware encryption: OAEP+ \cite{Shoup:2001:OAEPR}, \seq} |
41761fdc |
1280 | |
53aa10b5 |
1281 | OAEP+ is a simple embedding scheme, very similar to OAEP, which achieves |
41761fdc |
1282 | plaintext-awareness. |
1283 | |
1284 | We assume that our one-way function $f_P$ operates on $n$-bit strings. Fix |
1285 | a security parameter $k$. We require three random oracles $G\colon \{0, |
1286 | 1\}^k \to \{0, 1\}^{n-2k}$, $H\colon \{0, 1\}^{n-k} \to \{0, 1\}^k$, and |
1287 | $H'\colon \{0, 1\}^{n-k} \to \{0, 1\}^k$. |
1288 | \begin{program} |
1289 | Algorithm $\Xid{E}{OAEP+} |
1290 | ^{\mathcal{T}, G(\cdot), H(\cdot), H'(\cdot)}_P(x)$: \+ \\ |
1291 | $r \getsr \{0, 1\}^k$; \\ |
1292 | $c \gets H'(x \cat r)$; \\ |
1293 | $s \gets (x \xor G(r)) \cat c$; \\ |
1294 | $t \gets r \xor H(s)$; \\ |
1295 | $w \gets s \cat t$; \\ |
1296 | \RETURN $f_P(w)$; |
1297 | \next |
1298 | Algorithm $\Xid{D}{OAEP+} |
1299 | ^{\mathcal{T}, G(\cdot), H(\cdot), H'(\cdot)}_K(y)$: \+ \\ |
1300 | $w \gets T_K(y)$; \\ |
1301 | \PARSE $w$ \AS $s, k\colon t$; \\ |
1302 | $r \gets t \xor H(s)$; \\ |
1303 | \PARSE $s$ \AS $s', k\colon c$; \\ |
1304 | $x \gets s' \xor G(r)$; \\ |
1305 | \IF $c = H'(x \cat r)$ \THEN \RETURN $x$; \\ |
1306 | \ELSE \RETURN $\bot$; |
1307 | \end{program} |
1308 | \end{slide} |
1309 | |
1310 | \begin{slide} |
53aa10b5 |
1311 | \head{Plaintext-aware encryption: OAEP+, \seq} |
41761fdc |
1312 | |
1313 | %% x <- {0, 1}^{n-2k} r <-R {0, 1}^k |
1314 | %% | | |
1315 | %% | | |
1316 | %% | | |
1317 | %% |-------->[H']<----------------| |
1318 | %% | | | |
1319 | %% | | | |
1320 | %% v | | |
1321 | %% (+)<-------------[G]<-----------| |
1322 | %% | | | |
1323 | %% | | | |
1324 | %% v | | |
1325 | %% [||]<-------' | |
1326 | %% | | |
1327 | %% | | |
1328 | %% | v |
1329 | %% |-------------->[H]---------->(+) |
1330 | %% | | |
1331 | %% | | |
1332 | %% v v |
1333 | %% < s = (x (+) G(r)) || H'(x || r) > < t = r (+) H(s) > |
1334 | |
1335 | \vfil |
1336 | \[ \begin{graph} |
1337 | []!{0; <1cm, 0cm>:} |
1338 | {x \in \{0, 1\}^n}="x" [rrrr] {r \inr \{0, 1\}^k}="r" |
1339 | "x" [d] :[r] *+[F]{H'}="h'" "r" [d] :"h'" |
1340 | "r" :[dddd] *{\xor}="h-xor" |
1341 | :[d] *++=(2, 0)[F:thicker]{\strut t = r \xor H(s)} |
1342 | "x" :[dd] *{\xor}="g-xor" |
1343 | :[d] *{\ocat}="cat" |
1344 | :[dd] *++=(4, 0)[F:thicker] |
1345 | {\strut s = (x \xor G(r)) \cat H'(x \cat r)} |
1346 | [u] :[rr] *+[F]{H} :"h-xor" |
1347 | "r" [dd] :[ll] *+[F]{G} :"g-xor" |
1348 | "h'" :'[d]*=<8pt>\cir<4pt>{r_l} `d"cat" "cat" |
1349 | \end{graph} \] |
1350 | \vfil |
1351 | \end{slide} |
1352 | |
1353 | \begin{slide} |
53aa10b5 |
1354 | \head{Plaintext-aware encryption: OAEP+, \seq} |
41761fdc |
1355 | |
1356 | Let $\mathcal{T}$ be a trapdoor one-way function. Then the insecurity of |
1357 | the public-key encryption scheme $\Xid{\mathcal{E}}{OAEP+}^{\mathcal{T}}$ |
1358 | in the IND-CPA sense is given by |
1359 | \[ \InSec{ind-cpa}(\Xid{\mathcal{E}}{OAEP+}^{\mathcal{T}}; |
1360 | t, q_G, q_H, q_{H'}) |
1361 | \le |
1362 | 2 \left (\frac{q_G}{2^k} + |
1363 | \InSec{owf}(\mathcal{T}; t + O(q_G q_H)) \right). \]% |
1364 | Also, $\Xid{\mathcal{E}}{OAEP+}^{\mathcal{T}}$ is $(q_G + 1) |
1365 | 2^{-k}$-plaintext aware, the plaintext extractor running time being |
1366 | $O(q_{H'} t_f)$, where $t_f$ is the time required to execute the one-way |
1367 | function $f$. |
1368 | |
1369 | Tying up the various results, then, we can derive that |
1370 | \begin{eqnarray*}[Ll] |
1371 | \InSec{ind-cca2}(\Xid{\mathcal{E}}{OAEP+}^{\mathcal{T}}; |
1372 | t, q_D, q_G, q_H, q_{H'}) \\ |
1373 | & \le |
1374 | \frac{(2 + q_D) (q_G + 1) - 2}{2^k} + |
1375 | 2 \cdot \InSec{owf}(\mathcal{T}; t + O(q_G q_H + q_D q_{H'} t_f)). |
1376 | \end{eqnarray*} |
1377 | \end{slide} |
1378 | |
41761fdc |
1379 | \begin{proof}[Proof of the main OAEP+ result] |
1380 | |
1381 | We assume throughout that, before querying its $H'$ oracle at $x \cat r$, |
1382 | the adversary has queried $G$ at $r$. |
1383 | |
1384 | First, we show that OAEP+ meets the IND-CPA notion. Suppose that $A$ |
1385 | attacks the OAEP+ scheme in the IND-CPA sense. We shall play a series of |
1386 | games with the adversary, and vary the rules as we go along. The adversary |
1387 | remains the same throughout. |
1388 | |
1389 | At any point in a game, let $Q_G$ be the list of queries the adversary has |
1390 | made to the oracle $G$, let $Q_H$ be the queries made to $H$, and let |
1391 | $Q_{H'}$ be the queries made to $H'$. |
1392 | |
1393 | \paragraph{Game $\G0$} |
1394 | This is effectively the original attack game: the adversary chooses two |
53aa10b5 |
1395 | plaintexts: one is chosen uniformly at random, the adversary is given the |
41761fdc |
1396 | ciphertext and asked to identify which plaintext was chosen. Label the |
1397 | selected plaintext $x^*$, and the target ciphertext $y^*$. Let $c^*$, |
1398 | $s^*$, $t^*$ and $w^*$ be the intermediate results of the OAEP+ padding, as |
1399 | described above. Let $S_0$ be the event that the adversary wins the game. |
53aa10b5 |
1400 | Our objective is to bound the probability $\Pr[S_0] = |
41761fdc |
1401 | (\Adv{ind-cpa}{\Xid{\mathcal{E}}{OAEP+}^{\mathcal{T}}}(A) + 1)/2$. |
1402 | |
1403 | \paragraph{Game $\G1$} |
1404 | At the beginning of the game, we select at random: |
1405 | \begin{itemize} |
1406 | \item $s^* \inr \{0, 1\}^{n-k}$; |
1407 | \item $t^* \inr \{0, 1\}^k$; and |
1408 | \item $r^* \inr \{0, 1\}^k$. |
1409 | \end{itemize} |
1410 | We compute $w^* = s^* \cat t^*$ and $y^* = f_P(w^*)$. Note, then, that |
1411 | \emph{$y^*$ is fixed at the start of the game}. |
1412 | |
1413 | We also `rig' the random oracle $H$ so that $H(s^*) = r^* \xor t^*$. |
1414 | This isn't much of a fix, because this value is evidently uniformly |
1415 | distributed and independent of everything except $r^* \xor t^*$. |
1416 | |
1417 | We could rig $G$ and $H'$ too, once we knew $x^*$, so that $s^* = (x^* \xor |
1418 | G(r)) \cat H'(x^* \cat r)$. But we don't. Instead, consider the event |
1419 | $F_1$ that the adversary queries $G$ at $r^*$. Since we have opted for the |
1420 | bare-faced lie rather than oracle subterfuge, the target ciphertext $y^*$ |
1421 | and the oracles are entirely independent of the target plaintext $x^*$, |
1422 | whatever that might be. Hence, $A$'s probability of guessing which |
1423 | plaintext was chosen, $\Pr[S_1] = \frac{1}{2}$. |
1424 | |
1425 | When attempting to bound $\Pr[F_1]$, there are two cases to consider: |
1426 | \begin{enumerate} |
1427 | |
1428 | \item The adversary has not previously queried $H$ at $s^*$. In this |
1429 | case, the value of $r^*$ is independent of the adversary's view -- it has |
1430 | no information at all about $r^*$. This can occur, therefore, with |
1431 | probability at most $q_G 2^{-k}$. |
1432 | |
1433 | \item The adversary has previously queried $H$ at $s^*$. Then we can |
1434 | construct an inverter. Note that $f_P$ is a permutation (by assumption); |
1435 | hence, selecting a $y^*$ at random implies the values of $w^*$, and hence |
1436 | $s^*$ and $t^*$, even though they can't be computed easily. |
1437 | \begin{program} |
1438 | Inverter $I(P, y)$: \+ \\ |
1439 | $\Xid{G}{map} \gets \emptyset$; |
1440 | $\Xid{H}{map} \gets \emptyset$; |
1441 | $\Xid{H'}{map} \gets \emptyset$; \\ |
1442 | $z \gets \bot$; \\ |
1443 | $(x_0, x_1, s) \gets A^{G(\cdot), H(\cdot), H'(\cdot)} |
1444 | (\cookie{find}, P)$; \\ |
1445 | $b \gets A^{G(\cdot), H(\cdot), H'(\cdot)}(\cookie{guess}, y, s)$; \\ |
1446 | \RETURN $z$; \- \\[\smallskipamount] |
1447 | Oracle $H(s)$: \+ \\ |
1448 | \IF $s \in \dom \Xid{H}{map}$ \\ |
1449 | \THEN \RETURN $\Xid{H}{map}(s)$; \\ |
1450 | $h \getsr \{0, 1\}^k$; \\ |
1451 | $\Xid{H}{map} \gets \Xid{H}{map} \cup \{s \mapsto h\}$; \\ |
1452 | \RETURN $h$; |
1453 | \next |
1454 | Oracle $H'(s)$: \+ \\ |
1455 | \IF $s \in \dom \Xid{H'}{map}$ \\ |
1456 | \THEN \RETURN $\Xid{H'}{map}(s)$; \\ |
1457 | $h' \getsr \{0, 1\}^k$; \\ |
1458 | $\Xid{H'}{map} \gets \Xid{H'}{map} \cup \{s \mapsto h'\}$; \\ |
1459 | \RETURN $h$; \- \\[\smallskipamount] |
1460 | Oracle $G(r)$: \+ \\ |
1461 | \IF $r \in \dom \Xid{G}{map}$ \THEN \\ |
1462 | \RETURN $\Xid{G}{map}(r)$; \\ |
1463 | \FOR $s \in \dom \Xid{H}{map}$ \DO \\ \quad \= \+ \kill |
1464 | $t \gets r \xor \Xid{H}{map}(s);$ \\ |
1465 | $w \gets s \cat t$; \\ |
1466 | \IF $f_P(w) = y$ \THEN $z \gets w$; \- \\ |
1467 | $g \getsr \{0, 1\}^{n-2k}$; \\ |
1468 | $\Xid{G}{map} \gets \Xid{G}{map} \cup \{r \mapsto g\}$; \\ |
1469 | \RETURN $g$; |
1470 | \end{program} |
1471 | Clearly, $I$ succeeds whenever $A$ queries both $H$ at $s^*$ and $G$ at |
1472 | $r^*$. $I$'s running time is about $O(q_G q_H)$ because of all of the |
1473 | searching in the $G$ oracle. |
1474 | |
1475 | \end{enumerate} |
1476 | Thus, |
1477 | \[ \Pr[F_1] \le |
1478 | \frac{q_G}{2^k} + \InSec{owf}(\mathcal{T}; t + O(q_G q_H)). \]% |
1479 | Now observe that, if $F_1$ doesn't occur, Games $\G0$ and~$\G1$ are |
1480 | indistinguishable. So |
1481 | \[ \Pr[S_0 \land \lnot F_1] = \Pr[S_1 \land \lnot F_1], \] |
53aa10b5 |
1482 | so by Lemma~\ref{lem:shoup} (slide~\pageref{lem:shoup}), |
41761fdc |
1483 | \[ |{\Pr[S_0]} - \Pr[S_1]| \le \Pr[F_1]. \] |
1484 | Rearranging yields: |
1485 | \[ \Adv{ind-cpa}{\Xid{\mathcal{E}}{OAEP+}^{\mathcal{T}}}(A) |
1486 | \le 2 \left (\frac{q_G}{2^k} + |
1487 | \InSec{owf}(\mathcal{T}; t + O(q_G q_H)) \right). \]% |
1488 | But $A$ was arbitrary. The result follows. |
1489 | |
1490 | To complete the plaintext-awareness proof, we must present a plaintext |
1491 | extractor. |
1492 | \begin{program} |
1493 | Plaintext extractor $X(y, P, Q_G, Q_H, Q_{H'}, C)$: \+ \\ |
1494 | \FOR $z \in \dom Q_{H'}$ \DO \\ \quad \= \+ \kill |
1495 | \PARSE $z$ \AS $x, k\colon r$; \\ |
1496 | $c \gets Q_{H'}(z)$; \\ |
1497 | $s \gets (x \xor Q_G(r)) \cat c$; \\ |
1498 | \IF $s \in \dom Q_H$ \THEN \\ \quad \= \+ \kill |
1499 | $t \gets r \xor Q_H(s)$; \\ |
1500 | $w \gets s \cat t$; \\ |
1501 | $y' \gets f_P(w)$; \\ |
1502 | \IF $y = y'$ \THEN \RETURN $x$; \-\- \\ |
1503 | \RETURN $\bot$; |
1504 | \end{program} |
1505 | To obtain a lower bound on the success probability of $X$, consider an |
1506 | adversary $A$ which queries its various oracles and returns a ciphertext |
1507 | $y$. We aim to show that, if $A$ did not query all of its oracles as |
1508 | required for $X$ to work then its probability of producing a valid |
1509 | ciphertext (i.e., one for which the decryption algorithm does not return |
1510 | $\bot$) is small. |
1511 | |
1512 | We consider the decryption algorithm applied to $y$, and analyse the |
1513 | probability that $y$ is a valid ciphertext even though $A$ did not make all |
1514 | of the appropriate oracle queries. |
1515 | |
1516 | We shall take some of our notation from the decryption algorithm, which |
1517 | proceeds as follows: |
1518 | \begin{program} |
1519 | Algorithm $\Xid{D}{OAEP+}^ |
1520 | {\mathcal{T}, G(\cdot), H(\cdot), H'(\cdot)}_K(y)$: \+ \\ |
1521 | $w \gets T_K(y)$; \\ |
1522 | \PARSE $w$ \AS $s, k\colon t$; \\ |
1523 | $r \gets t \xor H(s)$; \\ |
1524 | \PARSE $s$ \AS $s' \cat k\colon c$; \\ |
1525 | $x \gets s' \xor G(r)$; \\ |
1526 | \IF $c = H'(x \cat r)$ \THEN \RETURN $x$; \\ |
1527 | \ELSE \RETURN $\bot$; |
1528 | \end{program} |
1529 | |
1530 | Firstly, suppose that $A$ did not query $H'$ at $x \cat r$. Then obviously |
1531 | there is only a $2^{-k}$ probability that $c$ is correct and the ciphertext |
1532 | will be accepted. By our assumption about adversary behaviour, if $A$ |
1533 | didn't query $G$ at $r$ then it also didn't query $H'$ at $x \cat r$ for |
1534 | any $x$. Hence, a decryption algorithm which rejected immediately if it |
1535 | discovered that $G$ hadn't been queried at $r$, or that $H'$ hadn't been |
1536 | queried at $x \cat r$, would be incorrect with probability at most |
1537 | $2^{-k}$. |
1538 | |
1539 | Secondly, suppose that $A$ did not query $H$ at $s$. Then $c$ is still |
1540 | fixed, but $r$ is now random and independent of $A$'s view. The |
1541 | probability that $G$ has been queried at $r$ is then at most $q_G 2^{-k}$. |
1542 | So a decryption algorithm which, in addition to the changes above, also |
1543 | rejected if it discovered that $H$ had not been queried at $s$ would be |
1544 | incorrect with probability at most $q_G 2^{-k}$. |
1545 | |
1546 | But our plaintext extractor $X$ is just such a decryption algorithm. |
1547 | Hence, $X$'s failure probability is |
1548 | \[ \epsilon = \frac{q_G + 1}{2^k}. \] |
1549 | \end{proof} |
1550 | |
1551 | \endinput |
1552 | |
1553 | %%% Local Variables: |
1554 | %%% mode: latex |
1555 | %%% TeX-master: "ips" |
1556 | %%% End: |