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80 views

One-time pad, zero key and Shannon

I'm supposed to prove that OTP without the zero key $k=0^n$ is not perfectly secret anymore. I understand that it's not because an attacker learns something by looking at the plaintext and ciphertext. ...
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1answer
153 views

One time pad: why is it useless in practice?

The symmetric cryptosystem One time pad (OTP) seems to be very beautiful since it is perfectly secret according to Shannon. Many books, however, point out that the main drawback is that one must ...
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2answers
117 views

Perfect Forward Secrecy in TLS

I read that TLS does PFS using Diffie Hellman. However, DH can be used even without certificates - so how is DHE-RSA better than plain DHE? Is DHE a insecure algorithm, that DHE-RSA is needed?
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How we can said a crypto system have perfect secrecy?

For example: I have 3 plaintexts ($a$, $b$, $c$) and 4 keys ($K_1$, $K_2$, $K_3$, $K_4$), making a table map to the cipher text, key as row and plaintext as column $\begin{matrix} \ \ \ \ \ \ \ \ \ \ ...
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1answer
102 views

Can a monoalphabetic substitution cipher attain perfect secrecy?

Can a monoalphabetic substitution cipher attain perfect secrecy? Definition of perfect secrecy: $${\rm Pr}[\,{\rm Enc}_k(m_1) = c\,] = {\rm Pr}[\,{\rm Enc}_k(m_2) = c\,]$$
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1answer
104 views

Hill cipher is not perfectly secure

I am on cryptography course and there is a homework question to show that Hill cipher doesn't have perfect security. So assume we have an cryptosystem $(P,C,K)$, where $P = C = \mathbb Z_{26}^N$ and ...
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83 views

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 $Pr[Dec_k(Enc_k(m))=m]=1/2$ ...
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0answers
86 views

Which is better ECDHE with TLS 1.0

I have a webserver which support only TLS 1.0 Which is the better cipher in this group for the best security? ...
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1answer
124 views

Perfect secrecy over Stirling numbers

For all $c_0\leftarrow m_0 \oplus k$ there exists a $k'$ such that $c_1 \leftarrow m_1 \oplus k'$, where $m_0 \neq m_1$ and $c_0 = c_1$. Assuming a truly random $k$, the first assignment is a ...