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The property of Indistinguishability in the presence of an eavesdropper as described in the following experiment:

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(tract from: Introduction to Modern Cryptography)

It must be interpreted as a scenario of Ciphertext-only attack or Known-plaintext attack? Thanks

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  • $\begingroup$ The adversary outputs the two messages in step one, so it must be a chosen-plaintext. $\endgroup$
    – mikeazo
    Aug 9 '17 at 19:36
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    $\begingroup$ @mikeazo No. A CPA is when the adversary obtains the encryption of many plaintexts of its choice. Here it does not even obtain one since it doesn't know which plaintext was encrypted. $\endgroup$
    – fkraiem
    Aug 9 '17 at 22:57
  • $\begingroup$ @fkraiem, yes, you are right $\endgroup$
    – mikeazo
    Aug 10 '17 at 2:14
  • $\begingroup$ Take a look at the CPA indistinguishability experiment in Section 3.4.2 (Introduction to Modern Cryptography Second Edition). In a CPA-scenario, the game works the same, but the adversary can additionally query the encryption function at any point in time. $\endgroup$
    – Patrick K
    Aug 10 '17 at 8:07
  • $\begingroup$ The definition og KPA is: "Known-plaintext attack: Here, the adversary learns one or more pairs of plaintexts/ciphertexts encrypted under the same key. The aim of the adversary is then to determine the plaintext that was encrypted to give some other ciphertext (for which it does not know the corresponding plaintext)." So "the adversary learns one or more pairs of plaintexts/ciphertexts encrypted under the same key" -> IS TRUE "for which it does not know the corresponding plaintext" -> IS TRUE Maybe it's just a KPA. What do you think? $\endgroup$
    – akanai
    Aug 10 '17 at 12:19
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It's CTO, you are trying to figure out the plaintext given two ciphertexts. M1 and M0 are just bits of either 1 or 0.

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  • $\begingroup$ You may want to note that $m_0$ and $m_1$ may edit be more than 1 bit long and contain arbitrary adversary-chosen values. $\endgroup$
    – SEJPM
    Oct 22 '18 at 11:31

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