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sasc:laboratoare:04 [2017/03/16 14:26]
marios.choudary
sasc:laboratoare:04 [2017/03/16 16:21] (current)
marios.choudary
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   * c) Let $K_3 = \{0, 1\}^{n+1}$. Construct a new PRF $F_3 : K_3 \times X \to Y$ with the following property: the PRF $F_3$ is secure, however if the adversary learns the last bit of the key then the PRF is no longer secure. This shows that leaking even a single bit of the secret key can completely destroy the PRF security property.   * c) Let $K_3 = \{0, 1\}^{n+1}$. Construct a new PRF $F_3 : K_3 \times X \to Y$ with the following property: the PRF $F_3$ is secure, however if the adversary learns the last bit of the key then the PRF is no longer secure. This shows that leaking even a single bit of the secret key can completely destroy the PRF security property.
 <note tip> <note tip>
-Hint: Let $k_3 = k \| b$ where $k \in \{0,​1\}^{n}$ and $b \in \{0,1\}$. Set $F_3(k_3,x)$ to be the same as $F (k, x)$ for all $x \neq 0^{n}$. Define $F_3\left(k_3,​ 0^{n}\right)$ so that $F_3$ is a secure PRF, but becomes easily distinguishable from a random function if the last bit of the secret key $k_3$ is known to the adversary. Prove that your $F_3$ is a secure PRF by arguing the contra-positive,​ as in part (b).+Hint: Let $k_3 = k \| b$ where $k \in \{0,​1\}^{n}$ and $b \in \{0,1\}$. Set $F_3(k_3,​x)$ so that $F_3$ is a secure PRF, but becomes easily distinguishable from a random function if the last bit of the secret key $k_3$ is known to the adversary. Prove that your $F_3$ is a secure PRF by arguing the contra-positive,​ as in part (b).
 </​note>​ </​note>​
-  * d) Construct a new PRF $F_4 : K_3 × X \to Y$ that remains secure if the attacker learns any single bit of the key. Your function $F_4$ may only call $F$ once. Briefly explain why your PRF remains secure if any single bit of the key is leaked. 
  
 ==== Exercise 3 (2p) ==== ==== Exercise 3 (2p) ====
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 We defined security of encryption scheme E against an eavesdropper in two ways: We defined security of encryption scheme E against an eavesdropper in two ways:
  
-  * $Pr[A(c_b)=b] \le negl(n)$ +  * $Pr[A(c_b)=b] \le \frac{1}{2} + negl(n)$ 
-  * $|Pr[A(c_1)=1] - Pr[A(c_0)=1]| \le \frac{1}{2} + negl(n)$+  * $|Pr[A(c_1)=1] - Pr[A(c_0)=1]| \le negl(n)$
  
 where $A(c_i) = j$ means that when the adversary receives the encryption of message $i$ he returns the bit $b'​=j$. where $A(c_i) = j$ means that when the adversary receives the encryption of message $i$ he returns the bit $b'​=j$.
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