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I couldn't understand the avalanche effect in DES. Could someone explain how avalanche effect happens in DES

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First of all, the avalanche effect is a desirable effect: it means that a very small change in the input will lead to a very big change in the output.

A security algorithm that doesn't provide this avalanche effect can lead to an easy statistical analysis: if the change of one bit from the input leads to the change of only one bit of the output, then it's easy to guess / try to find the input: just change one bit of it and verify if it was the right one.

How it works in DES: if you take a look at the DES, you'll see that it has 16 rounds and 2 permutations. The real thing happens in the rounds: for each round, the input bits are used to look up for bits that will be used as the output (the S-box). take a look at this presentation, for example. With the shifts that are made at each round, it means that the "index bits" (the ones used to do the table look up) come from the "data bits" from the previous round.

Changing a single bit in the input will cause more bits to be changed in the following rounds. Slide 6.41 of the above presentation shows how many bits will change at each round.

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  • $\begingroup$ the presentation link is broken $\endgroup$ Dec 13, 2017 at 15:43
  • $\begingroup$ The first paragraph is misleading. A big change is a vague term. This answer needs an update with proper changes. What we expect if one flips a bit in the input then every bit in the output has a 50% chance of flipping. For example, see here $\endgroup$
    – kelalaka
    Apr 20, 2020 at 17:07

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