Unbalanced Feistel network function

Question

I am trying to understand how an unbalanced Feistel network works. I came across this image:

It says that the function takes $b$ bits to $y$ bits. Which works on the first step, however clearly doesn't on the next one. What am I not getting? If you always apply the function to the side with $b$ bits would that break the cypher? I am guessing it does.

Answer 1

Let formalize it

• Round function $R:\{0,1\}^b \to \{0,1\}^y$ (This is bad naming, $F$ was better here as in DES)

• Input to each round is $b+y$ bit register/array $I$.

• Output of each round $O = (R(\texttt{MSB}(b,I)) \oplus \texttt{LSB}(y,I)) \mathbin\| \texttt{MSB}(b,I)$

Therefore $O$ is again $b+y$ bits register/array as an input to next round.

---

• $\texttt{MSB}(b,I)$ the Most Significant $b$ Bits of register $I$.
• $\texttt{LSB}(y,I)$ the Least Significant $y$ Bits of register $I$.

---

Example

Let $I=\texttt{[0,1,1,1,0,1,0,0,1,0,0,1,1,0,0,1]}$ be 16-bit register then

• $\texttt{MSB}(3,I) = \texttt{[0,1,1]}$, and
• $\texttt{LSB}(12,I) = \texttt{[1,0,1,0,0,1,0,0,1,1,0,0,1]}$

Note that here we used binary representation for $I$, not the array repsentation.