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

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.

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  • $\begingroup$ The image is misleading. It is still taking b bits. You can get it from $b$ msb. $\endgroup$ – kelalaka Jun 18 at 16:54
  • $\begingroup$ @kelalaka what does "$b$ msb" mean exactly? $\endgroup$ – yolo expectz Jun 18 at 16:56
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    $\begingroup$ $b$ Most Significant Bits. $\endgroup$ – kelalaka Jun 18 at 16:57
  • $\begingroup$ @kelalaka and do these bits become a prefix or a suffix to the $y$ bits? $\endgroup$ – yolo expectz Jun 18 at 17:03
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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.

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