Error during transmission with one plaintext block (Cipher block operation) I was given a graph like this, and the problem was stated as: if something happens to P2 then only P2 would not be retrieved on the decryption side, and all other P's can be retrieved normally. I do not understand how so? If something happens to P2 then its encrypted XOR will be wrong, then leads to a wrong C2. Also, P2 will get XOR with the encrypted P3 to create C3, won't that be wrong too, and the same with all other blocks?

Note: P is plaintext, C is ciphertext, D is decryption, and E is encryption.
First, Let us write the encryptions as an equation;

\begin{align}
c_1 &amp; = \text{IV} \oplus \operatorname{Enc}(k,p_1) \\
c_2 &amp; = P_1 \oplus \operatorname{Dec}(k,p_2)\\
\end{align}
and the last

\begin{align}
c_3 &amp; = p_{n-1} \oplus \operatorname{Dec}(k,p_3) \\
c_4 &amp; = P_n \oplus \operatorname{Dec}(k,p_4)
\end{align}

As one can see the chaining(?) is constructed(?) within two by two blocks. However, there are more serious problems than the original question.


The decryption of $p_2$; $$p_2 = \operatorname{Dec}(p_1 \oplus c_2)$$ If one look at the encryption of $p_2$ they will see that $\operatorname{Dec}$ is used. Therefore, unless the block cipher is a very sepcial cipher that $\operatorname{Enc}=\operatorname{Dec}$ than this is not possible to get $p_2$. Thas is the blockcipher chooses an involution for every key, see OEIS A000085. 
This is similar to the first; This time always $\operatorname{Dec}$ is used. So we don't expect even decryption.


Now back to corrected one, that is for encryption $\operatorname{Enc}$ is used and for decryption $\operatorname{dec}$. Then, as stated above, the blocks are operated two by two and this means that an error can only affect at most two blocks. $c_1$ affects two blocks and $c_2$ affects one block, etc. It is also not clear how the odd number of blocks are encrypted. One can guess that there is a special padding mechanism.