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This question already has an answer here:

  • Why is it that a one time pad is vulnerable is the key is shorter than the message?

  • I know that if a key is reused across multiple plaintexts, you can XOR two existing ciphertexts to get the pad, but I'm not entirely sure how this works in practice.

Any help/ explanation is appreciated. Thanks.

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marked as duplicate by Squeamish Ossifrage, forest, otus, Maarten Bodewes Mar 11 at 1:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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If the key bits are shorter then the message you need to reuse some part of the previous key bits. If this is known, and we can assume by Kerckhoff Principles, then the enemy can crib-drag on the same message. A more dedicated enemy will always try this case and also the reuse case; the same key is used with another message, too.

As an example;

m = xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
k = KkkkkkkkkkkkkkkkkkkkkkkkKkkkkkkkkkkkkkkkkk

where $x$ represents the message bits, $k$ represents the key bits and $K$ represents the start of the key bits. Divide the message into parts;

m1 = xxxxxxxxxxxxxxxxxxxxxxxx
k  = Kkkkkkkkkkkkkkkkkkkkkkkk
m2 = xxxxxxxxxxxxxxxxxx
k  = Kkkkkkkkkkkkkkkkkk

now, execute crib-dragging.

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