Is it necessary keys to have equal propabillities for the system to have perfect secrecy?

tonythestark

2/10/24, 7:37 PM

Shannon's theorem for perfect secrecy states that $$\forall x \in M, y \in C:\quad P[x|y]=P[x] $$ I know we need $|M|\leq |C| \leq |K|$. If $|Μ|=|C|=|K|$ all keys should have equal probs.
If $|Μ|<|Κ|$ is it still necessary? I think not because : $$P[x|c_i]=P[x|c_j] \implies \sum_{k_i: E_{k_i}(m_i)=c_i}P[K=k_i]=\sum_{k_i:E_{k_i}(m_i)=c_i}P[k=k_j]$$ The difference is that when $|Μ|=|C|=|K|$ there is exactly one $k $ s.t: $Ε_k(m)=c$ while now there may be more .

Is my conclusion valid?

1 + 0

perfect-secrecy

Score:1

Crypto

poncho

2/10/24, 8:04 PM

Is my conclusion valid?

Yes, it is.

I'll give you a simple example of such a system: we have a 1 bit plaintext, 2 bits of key and 1 bit of ciphertext; the encryption method is:

$$C = K_0 \oplus K_1 \oplus P$$

(where $K_0, K_1$ are the two bits of the key); with the probabilities of the key bits are:

00 with probability 0.1
01 with probability 0.2
10 with probability 0.3
11 with probability 0.4

Simple calculation shows that this scheme achieves perfect secrecy, even though the key is nonuniformly distributed.

+ 0

Elon Musk

I sit in a Tesla and translated this thread with Ai:

EN: Is it necessary keys to have equal propabillities for the system to have perfect secrecy?

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