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Consider an encryption scheme ( KeyGen , Enc, Dec ) with message space M , key space K , and ciphertext space C . For

Consider an encryption scheme (KeyGen, Enc, Dec) with message space M, key space K, and ciphertext
space C. For any minM, we define a random variable C(m) over C corresponding to the encryption of
m. That is,C(m) is the random variable corresponding to c in the following encryption process:
klarrKeyGen()
clarrEnc(k,m)
Then, the notion of perfect indistinguishability we saw in the first lecture may be written as follows.
Definition 2.2(Perfect Indistinguishability (Probabilities)). The encryption scheme (KeyGen, Enc, Dec)
satisfies perfect indistinguishability if, for any m0,m1inM and any cinC, we have:
Pr[C(m0)=c]=Pr[C(m1)=c]
By the observations about advantage earlier, the following is an equivalent definition of perfect
indistinguishability.
Definition 2.3(Perfect Indistinguishability (Advantage)). The encryption scheme (KeyGen, Enc, Dec)
satisfies perfect indistinguishability if, for any m0,m1inM and any algorithm A, we have:
advAC(m0),C(m1)=0
The benefit of writing the definition in terms of the allowed distinguishing advantage is that it gives
us a simple parameter that we can change to weaken the definition. Consider, for instance, the following.
Definition 2.4(-Indistinguishability). For in[0,1], the encryption scheme (KeyGen, Enc, Dec) satisfies
-indistinguishability if, for any m0,m1inM and any algorithm A, we have:
advAC(m0),C(m1)
For very small values of , this is still a meaningful notion of security. One would hope that this
weakening of perfect indistinguishability might allow one to get around Shannon's theorem and use
smaller keys while still being relatively secure, but one would be wrong.
Problem 2.3(5 points). Prove that in any encryption scheme that satisfies correctness and -indistin-
guishability, we necessarily have |K|(1-)|M|.
Hint. Try drawing out the mappings like we did while proving Shannon's theorem. Think about how
many elements the supports of C(m0) and C(m1) can have in common. How small can you make this
intersection?
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