# Conditional entropy between ciphertext, plaintext and the key

If we have a (possibly imperfect) cryptosystem that generates ciphertext $$C$$ from plaintext $$P$$ and key $$K$$, we have:

$$H(C, P, K) = H(C | P, K) + H(P, K)$$

where $$H$$ is entropy.

My question is why following is true:

$$H(C | P, K) = 0$$

It seems because each key and plaintext uniquely define a ciphertext but I want to prove this mathematically (theories in entropy).

• Your question might be more useful to other readers if you add an explanation for the symbols used. Mar 8 at 19:54
• Edited to explain symbols. Two partial answers below could be merged. Mar 21 at 14:39

$$H(C|P,K) = - \sum_{c \ \in \ C \\ m \ \in \ P \\ k \ \in \ K} p(c, m, k) \log\left(\frac{p(c,m,k)}{p(m,k)}\right)$$
(maybe this is clearer on the Wikipedia, sorry for my rubbish formatting skills) then notice that, if $$c$$ is totally conditioned on $$(m, k)$$ - because, like you said, each ciphertext corresponds to a unique plaintext-key pair with a bijective mapping - then it must be that $$p(c,m,k) = p(m,k)$$. Consequently, the fraction inside the logarithm evaluates to $$1$$ always, and so the logarithm evaluates to $$0$$, always. Then each term in the sum is $$0$$ and the over all conditional entropy is $$0$$, as you expect for a distribution where $$(m,k)$$ totally defines $$c$$.
As mentioned above, for probabilistic encryption, where a nonce or IV is used, this wouldn't be the case, because there would be more inputs to define $$c$$, so the entropy of those inputs would 'filter through' to $$c$$ leading to non-zero conditional entropy.