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One thing I know about the Vernam cipher is that the key must not be repeated over the plain-text, for example:

Plain-text (hex):   48 65 6C 6C 6F 57 6F 72 6C 64 // HelloWorld
One-time-pad (hex): 47 72 61 70 65 47 72 61 70 65 // GrapeGrape
Cipher-text (hex):  15 23 13 28 10 16 29 19 28 01 // ??????????

Notice that in the one-time-pad, the sequence 47 72 61 70 65 is repeated.

How does this constitute a vulnerability, and how can the one-time-pad be discovered through its repetition?

Side-note: I'd be especially interested to know how it would be possible to determine a single repeated byte pad, for example:

Plain-text (hex):   48 65 6C 6C 6F 57 6F 72 6C 64 // HelloWorld
One-time-pad (hex): 56 56 56 56 56 56 56 56 56 56 // VVVVVVVVVV
Cipher-text (hex):  1E-33-3A-3A-39-01-39-24-3A-32 // ??????????
Matthew Layton
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    This appears to be a duplicate of this question - is there some aspect of your question that is not answered there? – Ella Rose Jan 21 '19 at 22:01
  • @kelalaka what if you don't know what you're looking for? i.e. there's no "crib", so-to-speak. For example, it's just arbitrary data with no meaning, but still encrypted? – Matthew Layton Jan 22 '19 at 15:47
  • Did you notice that crib-dragging uses a dictionary of the language of the original message? The words are used to verification of the guess, right? – kelalaka Jan 22 '19 at 15:53

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