I have encrypted a file using -des -cfb and now I want to view how many blocks it contains once I decrypt it.

I have done this on the Linux terminal. I know that 9 blocks will be effected if there is an error in the ciphertext.

I need to show proof that this is true. But I'm unsure how to do this in a Linux terminal.

  • $\begingroup$ Where does the 9 come from? $\endgroup$
    – Paul Uszak
    May 16, 2018 at 10:16
  • $\begingroup$ If there is an error in the ciphertext transmission c1 9 blocks will be corrupted during the decyption . How can I show that it is 9. (64/8 = 8, +1 = 9) $\endgroup$
    – Xace
    May 16, 2018 at 10:21
  • $\begingroup$ @PaulUszak This seems to come from a course on crypto, see my answer for a link to a previous question. $\endgroup$
    – Maarten Bodewes
    May 16, 2018 at 11:27

1 Answer 1


Block size = 1 byte for CFB thus 9 bytes are affected. Why this is true has been answered before. Even for CFB8 it takes 8 shifts to shift the wrong ciphertext value out of the buffer.

To proof this is true just change a single byte of a binary output file using a hex editor (hexedit) and then decrypt. Count the number of differing number of bytes (diff). If you get 8 bytes then the last byte refused to budge - it may regenerate the same value by chance. Redo from start, multiple times if you want "statistical proof".

  • $\begingroup$ Ugh, did I write "statistical proof"? Help, I'm turning alpha! $\endgroup$
    – Maarten Bodewes
    May 16, 2018 at 11:32
  • $\begingroup$ Note that this is an assignment and that this is not a Unix Q/A site, I have left out the explicit commands that you're required to perform. $\endgroup$
    – Maarten Bodewes
    May 16, 2018 at 11:58

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