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I was looking at the Wikipedia article about Block cipher mode of operation and I was wondering how can you from the encryption $C_i = E_K(P_i \oplus P_{i-1}\oplus C_{i-1})$ get the decryption $P_i = D_K(C_i) \oplus P_{i-1}\oplus C_{i-1}$ ?

Is there even a way to get decryption from encryption (and vice versa) or not?

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    $\begingroup$ Take the encryption equation, apply $D_K$ to both sides, solve for $P_i$. $\endgroup$ – Mikero Nov 9 '20 at 21:25
  • $\begingroup$ @Mikero Ok, but what is $D_K( E_K(P_i \oplus P_{i-1}\oplus C_{i-1}))$? I'm a beginner and I have no idea what should I get from this. $\endgroup$ – Jan Nov 9 '20 at 21:36
  • $\begingroup$ Oh, $D_K (E_K ( x))=x$, doesn't it? Ok, then I know how to solve this, thanks. $\endgroup$ – Jan Nov 9 '20 at 21:42
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How the PPCBC mode works

Propagating Cipher Block Chaining mode of operation (PCBC) works as with message indexes starts from 1;

  • For PCBC encryption we have;

$$C_i = E_K(P_i \oplus P_{i-1}\oplus C_{i-1})\text{ and } \color{red}{C_0 = P_0 \oplus IV}$$

  • For PCBC decryption ve have;

$$P_i = D_K(C_i) \oplus P_{i-1}\oplus C_{i-1}\text{ and } \color{red}{C_0 = P_0 \oplus IV}$$

given $C_1$ to decrypt $P_1$

$$P_1 = D_K(C_1) \oplus P_{0}\oplus C_{0} = P_1$$

given $C_2$ to decrypt $P_2$

$$P_2 = D_K(C_2) \oplus P_{1} \oplus C_{1}$$ Now we can solce since we know, $P_1$ and we have all $C_i$s.

and so on...

The red part was missing in your question.


The Correctness of the PCBC mode

An encryption scheme must satisfy the correctness requirement; for every key $k$ output by the key generation algorithm and for every message $m \in \mathcal{M}$ ($\mathcal{M}$ is the message space), the following must hold;

$$ D_k(E_k(m)) = m$$


To see that PCBC has the correctness; take

$$C_i = E_K(P_i \oplus P_{i-1}\oplus C_{i-1})$$ take decryption on the both sides

$$D_K(C_i)= \color{red}{D_K}(\color{red}{E_K}(P_i \oplus P_{i-1}\oplus C_{i-1}))$$ cancel Enc with Dec.

$$D_K(C_i)= P_i \oplus P_{i-1}\oplus C_{i-1}$$

Therefore:

$$P_i = D_K(C_i) \oplus P_{i-1}\oplus C_{i-1}$$

Since we know, all $C_{i}$s, and previously decrypted $P_{i-1}$.

If it is the first case $(i=1)$, then we already know $\color{red}{C_0 = P_0 \oplus IV}$

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