Solutions_HW4_547 (1)

Then an adversary might be able to recover p but he

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Unformatted text preview: adversary (the encryption is similar to one-time pad). Distinguishing m0 and m1 , where m0 ￿= m1 , under Π would imply learning something about the pad p used. Since this is impossible, Π is CPA secure. On the other hand, suppose that Π2 is the CPA-secure and Π1 it not. Then an adversary might be able to recover p, but he won’t be able to recover m ⊕ p, thus he will still not be able to recover any information concerning the message m. Distinguishing m0 and m1 , where m0 ￿= m1 , under Π would imply being able to distinguish m0 ⊕ p and m1 ⊕ p since p might be known. But Π2 is CPA secure, hence Π must be CPA secure. Exercise 4.4 (a) Let m1 ￿m2 be any message with m1 , m2 ∈ {0, 1}n . Then, the tag on m1 ￿m2 is identical to the tag on m2 ￿m1 . Thus, an adversary A can ask for a tag on m1 ￿m2 and output the message m2 ￿m1 together with the tag received. Fall 2010 Comp 547: Cryptography and Data Security 3 (b) As with the previous item, the tag ￿r, t￿ on m1 ￿m2 is acceptable also for m2 ￿m1 . (c) There is an attack on this scheme that does not request any tags. Let m1 ∈ {0, 1}n/2 be arbitrary, and set r := ￿1￿￿m1 . Then ￿r, 0n ￿ is a valid tag on m1...
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This document was uploaded on 02/24/2014.

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