# Why does a perfect secrecy can be achieved when decryption correctness is not totally required?

By Shanon theorem, a perfect secrecy encryption scheme must use a key space of equal size as the message space.

But when the correctness requirement is weakened such that $Pr[Dec_k(Enc_k(m))=m]=1/2$ we know that the key space may be smaller than the message space.

Generally, what is the lower bound of the key space when the scheme correctness requirement is $Pr[Dec_k(Enc_k(m))=m] >= 2^{-t}$? (and why?)

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You can omit part of the information from the output and guess them. Those guesses will be incorrect sometimes. This boils down to using lossy compression before encrypting. –  CodesInChaos Jan 31 at 15:45
Can you please give a more concrete answer? (with a lower bound..) –  Bush Jan 31 at 16:52