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I am taking cryptography course in coursera by Stanford University. and I have following question

I have a question why append 0 in G(k) i.e., G'(k) = G(k) || 0 is considered as not secure PRG as 0 in message is copied with out encryption. Whereas in semantically secure E'(k,m) = 0 || E(k,m) i.e., prepend 0 is considered as semantically secure. Here why appending 0 does not break semantically secure?

Kindly request to clarify.

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  • $\begingroup$ Have you found an efficient adversary that breaks G'? ​ ​ $\endgroup$ – user991 May 18 '17 at 6:05
  • $\begingroup$ I am not sure I understand the question. Are you asking why encrypted data is more secure than not encrypted data? The pretend/append of 0 in this case seems unimportant to me. $\endgroup$ – eckes May 18 '17 at 6:19
  • $\begingroup$ @eckes I believe it's a test to understand semantic security $\endgroup$ – gusto2 May 18 '17 at 8:30
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I consider the SU Cryptography course on Coursera really good, up to certain level is very comprehensive. You should discuss these topics on the Coursera forum, IMHO you will be guided more precisely.

You have two questions here and you should know the DEFINITIONS of the secure PRNG and semantically secure encryption. Security of the message differs between two with what needs to be secure. For PRNG it's the output, for encryption it's the plaintext message.

I have a question why append 0 in G(k) i.e., G'(k) = G(k) || 0 is considered as not secure PRG as 0 in message is copied with out encryption.

The expression ( G'(k) = G(k) || 0 ) effectively means you're appending zeros to the generated message. It means you can predict there will be 0 at the end of the (otherwise random) message. It is pretty unsecure PRNG.

Whereas in semantically secure E'(k,m) = 0 || E(k,m) i.e., prepend 0 is considered as semantically secure. Here why appending 0 does not break semantically secure?

Even you have prepend anything to the ciphertext, it still doesn't reveal anything about the plaintext message.

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