Question: Problem 3 . Let E : { 0 , 1 } n { 0 , 1 } n { 0 , 1 } n be

Problem 3. Let E:{0,1}n{0,1}n{0,1}n be a function family. For each Kin{0,1}n, let
CBCMAC[EK]:({0,1}n)+{0,1}n
be the standard CBC-MAC construction. Let F:({0,1}n{0,1}n)({0,1}n)+{0,1}n be
defined as follows: for all K1,K2in{0,1}n and Min({0,1}n)+, let
FK1,K2(M)=F(K1,K2,M)=EK2(CBCMAC[EK1](M)).
One can prove that if E is a good PRF then F is a good (VIL-)PRF.
A widely used standard employs this construction as its PRF. However, this standard also ex-
poses V=MSB(s,EK1(0n)) as an operational guarantee that the correct key K1 is begin used. ?1
This turns out to be a very bad choice.
Assume that s=n2. Given V, give a PRF-attack that takes something like 2n1 oracle queries to
distinguish F from a random function. (Hint: consider messages of the form 0n||(x||R) for fixed
xin{0,1}s and random Rin{0,1}n-s, and messages of the form (xo+V)||R' for random R'.)
Notice that this is the square root of the number of queries one would hope it would take.
?1 The function MSB(s,x) returns the most-significant s bits of string x.
Problem 3 . Let E : { 0 , 1 } n { 0 , 1 } n { 0 ,

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