Lecture-5-17-10

# Lecture-5-17-10 - H(K| C = H(K H(M H(C Recall two basic...

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Achieving Perfect Secrecy We have Perfect Secrecy when the ciphertext gives no information about the plaintext A random cryptographic system Thus the number of keys must be at least as large as the number of ciphertexts for a Fxed key --> different plaintexts must go to different ciphertexts C yields no information about M means that M and C are independent Random Variables Thus the number of ciphertexts must be at least as large as the number of plaintexts next

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The Ciphertext is independent of the plaintext thus the Ciphertext gives no information about the plaintext Q.E.D. next
How to construct perfect secrecy systems In the simplest case Solution: Ciphertexts Show that this wheel gives a perfect secrecy system Sample quiz problem next next

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For the known plaintext attack we have H(K| C,M ) = H(K) - H(C|M ) For the ciphertext only attack we have

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Unformatted text preview: H(K| C) = H(K) + H(M) - H(C) Recall two basic equations next next next Recall the experiment we carried out in class ? The sequence that gave the number of guesses for each letter may be viewed as an encoding of the sentence next The Unicity distance The least amount of ciphertext the opponent needs to know to uniquely recover the plaintext = 0 ? next The unicity distance equation then becomes Computation of the Unicity Distance for monoalphabetic ciphers Thus the encryption of such a message will use only 14 letters of the permuted alphabet and we get In conclusion: for a monoalphabetic cipher 23 characters of ciphertext should be sufFcient to uniquely determine the plaintext next SAMPLE PROBLEMS next Sample quiz problems 1 2 next next next...
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## This note was uploaded on 09/22/2010 for the course MATH MATH187 taught by Professor Math187 during the Spring '10 term at UCSD.

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Lecture-5-17-10 - H(K| C = H(K H(M H(C Recall two basic...

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