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

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?

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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.

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