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Consider a very simple symmetric block encryption algorithm, in which 32-bits blocks of plaintext are encrypted using a 64-bit key.

Encryption is defined as $C = (P\oplus K_L) \boxplus K_R$

where $C$ = ciphertext; $K$ = secret key;

$K_L$ = leftmost 32 bits of K;

$K_R$ = rightmost 32 bits of K;

$\oplus$ = bitwise exclusive or;

$\boxplus$ is addition mod $2^{32}$enter image description here

Show the decryption equation. That is, show the equation for P as a function of $C$, $K_L$ and $K_R$

below is my attempt, I am kinda lost.

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  • $\begingroup$ What are you confused about? You show an equation at the bottom; are you dissatisfied by it? The only thing "wrong" I can immediately see is notational: using $\otimes$ for modular addition and $\ominus$ for modular subtraction, which would appear a tad inconsistent. $\endgroup$
    – poncho
    Feb 10, 2015 at 11:36
  • $\begingroup$ Looks good to me. $\endgroup$
    – Ken
    Feb 19, 2015 at 23:24

1 Answer 1

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The ciphertext is $C=((p \oplus K_0) + K_1) \bmod 2^K)$, so whatever the result of the Xor's operations, the ciphertext will also be 32 bits.

Since the size of plaintext is 32 bits which is similar to the size of $K_0,K_1$. So, the module operation is just an extra operation because $X \bmod 2^K$ is always $X$ for $X < 2^K$.

So, $C=(p \oplus K_0) \oplus K_1$.

To get the plaintext, perform the XOR operations in reverse order to the Ciphertext. Plaintext, $P=(C - R_1) \oplus R_0$.

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  • $\begingroup$ Welcome to Cryptography.SE. FYI, we have $\LaTeX$/MathJax is installed on our site. Check the edits. $\endgroup$
    – kelalaka
    Mar 27, 2022 at 15:33

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