# Show that the equal difference property exits for a modified DES encryption system

My cryptology professor gave us this problem on a past homework assignment and I only managed to get one of the 3 parts correct. Needless to say, I want to know how to do the other two parts as well. The question is as follows:

Consider the following DES-like encryption system that operates on 16-bit strings. The system takes the input and divides it into two 8-bit string $L_0$ and $R_0$. One round of encryption starts with $L_{i-1}$ and $R_{i-1}$ and the output is $L_i=R_{i-1}$ and $R_i=L_{i-1}\oplus S(R_{i-1})\oplus K$, where $K$ is the key.

The function S rotates the bits of $R_{i-1}$ to the right. More precisely, if $R_{i-1}=b_0\|b_1\|b_2\|b_3\|b_4\|b_5\|b_6\|b_7$ then $S(R_{i-1})=b_7\|b_0\|b_1\|b_2\|b_3\|b_4\|b_5\|b_6$.

The two parts that I was not able to get are:

1) Explain briefly why if $A$ and $B$ are bit strings, then $S(A\oplus B)=S(A)\oplus S(B)$

2) If $M$ is the plaintext, let $E_K(M)$ denote the process of encrypting $M$ using one round of the above process. Show that $E_K$ has the equal difference property, namely that if $A\oplus B = C\oplus D$, then $E_K(A)\oplus E_K(B)=E_K(C)\oplus E_K(D)$.

Anyway, if someone could provide a complete explanation (answer included) on how to the 2 above questions, I would greatly appreciate it! Thanks in advance!

• On this website, basic $\TeX$ is easy as $E_K(A)\oplus E_K(B)=E_K(C)\oplus E_K(D)$. I often use this reference card, and a lot of it works. $\;$ Hint for 1: break down $A$ and $B$ as bit strings, apply definition of $S$, and watch the desired property unfold. $\;$ For 2: Break 16-bit strings into two 8-bit ones (not individual bits); perhaps, evaluate the Exclusive-OR of what's on both sides of the equality, and simplify until you get zero; use result in 1, associativity+commutativity of $⊕$, and that $\forall X, X⊕X=0$.
– fgrieu
Nov 1, 2014 at 16:27
• @fgrieu: thanks for the link to the reference card! I will definitely be using that in the future! Nov 2, 2014 at 20:09

For 1) you need to show that $S(X)$ is linear with respect to xor.

We can define $S(X)$ as:

$S(X) = (x_{7}, x_{0}, x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6})$

And define $A \oplus B$ as:

$A \oplus B = (a_0 \oplus b_0, a_1 \oplus b_1,a_2 \oplus b_2,a_3 \oplus b_3,a_4 \oplus b_4,a_5 \oplus b_5,a_6 \oplus b_6,a_7 \oplus b_7,)$

So we have:

$S(A \oplus B) = (a_7 \oplus b_7,a_0 \oplus b_0,a_1 \oplus b_1,a_2 \oplus b_2,a_3 \oplus b_3,a_4 \oplus b_4,a_5 \oplus b_5,a_6 \oplus b_6)$

$=(a_{7}, a_{0}, a_{1}, a_{2}, a_{3} a_{4}, a_{5}, a_{6}) \oplus (b_{7}, b_{0}, b_{1}, b_{2}, b_{3} b_{4}, b_{5}, b_{6})$

Which by definition is:

$=S(A) \oplus S(B)$

For number 2) we'll first show that if there is some difference $D = A \oplus B$ then $E_k(D) = E_k(A) \oplus E_k(B)$.

We can define $E_K(M)$ as:

$M_R || (M_L \oplus S(M_R))$

So given:

$E_k(A) \oplus E_k(B)$

$=\big[A_R || (A_L \oplus S(A_R))\big] \oplus \big[B_R||(B_L \oplus S(B_R))\big]$

$=(A_R \oplus B_R)||(A_L \oplus S(A_R) \oplus B_L \oplus S(B_R))$

And using the property proved in 1) you get:

$=(A_R \oplus B_R)||(A_L \oplus B_L \oplus S(A_R \oplus B_R))$

And because $D = A \oplus B$:

$=(D_R||(D_L \oplus S(D_R))$

$=E_k(D)$

Now we can show the equal difference property holds:

$A \oplus B = C \oplus D$

$E_k(A \oplus B) = E_k(C \oplus D)$

$E_k(A) \oplus E_k(B) = E_k(C) \oplus E_k(D)$

• awesome answer! Very clearly explained and easily understood! Nov 2, 2014 at 20:26