Show number of affected blocks using CFB DES decryption of altered ciphertext

Question

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.

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) — Xace, May 16 '18 at 10:21
@PaulUszak This seems to come from a course on crypto, see my answer for a link to a previous question. — Maarten Bodewes, May 16 '18 at 11:27

score 1 · Answer 1 · answered May 16 '18 at 11:32

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".

answered May 16 '18 at 11:32

Maarten Bodewes

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Ugh, did I write "statistical proof"? Help, I'm turning alpha! – Maarten Bodewes May 16 '18 at 11:32
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. – Maarten Bodewes May 16 '18 at 11:58

Show number of affected blocks using CFB DES decryption of altered ciphertext

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