Proof that OTP is a perfect cipher

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Discussion Overview

The discussion revolves around understanding the proof that the One-Time Pad (OTP) is a perfect cipher. Participants are examining specific aspects of the proof, particularly the relationship between the ciphertext, plaintext, and key as expressed through the XOR operation. The focus is on the theoretical underpinnings of the OTP and its properties as a cipher.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant seeks clarification on the proof of the OTP being a perfect cipher, specifically regarding the equations C = y and K = (x XOR y).
  • Another participant suggests rephrasing the question in terms of arithmetic operations on single bits to facilitate understanding.
  • A participant provides the definition of a perfect cipher and attempts to simplify the probability expressions involved, indicating difficulty in calculating certain probabilities related to the ciphertext and plaintext.
  • One participant clarifies the relationship between the message bit, ciphertext bit, and key bit, questioning why C = y is equivalent to K = x XOR y.

Areas of Agreement / Disagreement

Participants are engaged in a collaborative exploration of the proof, with some expressing confusion and seeking clarification. There is no consensus yet on the specific mathematical aspects being discussed, and multiple interpretations of the relationships between the variables are present.

Contextual Notes

Participants have not resolved the mathematical steps involved in the proof, particularly regarding the probabilities associated with the events described. The discussion is limited by the assumptions made about the definitions of the variables and the XOR operation.

Bipolarity
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I'm trying to understand the proof that the OTP is a perfect cipher.
I tried to understand the proof shown in http://web.mit.edu/6.857/OldStuff/Fall97/lectures/lecture2.pdf but I don't understand the part C = y and K = (x XOR y).

Could someone assist me in understanding the proof?

Thanks!

BiP
 
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It looks like that's a statement about the behavior of bits and the exclusive-or operation. Write your question in the form of a question about arithmetic operations on single bits.
 
The definition of a perfect cipher:

[tex]P(m*=M | E_{k}(M) = c) = P(m*=M)[/tex]

where [itex]m*[/itex] is the interceptor's guess at the message and [itex]M[/itex] is the original plaintext message. [itex]k[/itex] is the key used in the encryption, [itex]c[/itex] is the ciphertext, and [itex]E[/itex] is the encryption algorithm, in this case the XOR operator.

Simplifying, we get

[tex]P(m*=M | M \oplus k = c) = P(m*=M)[/tex]

[tex]\frac{P(m*=M \bigcap M \oplus k = c)}{P(M \oplus k = c)} = P(m*=M)[/tex]

I have trouble figuring out the probability of [itex]m*=M \bigcap M \oplus k = c[/itex] and [itex]M \oplus k = c[/itex].
 
As I understand the question.

[itex]M[/itex] = the message bit
[itex]C[/itex] = the corresponding bit of the cypher text
[itex]K[/itex] = the corresponding bit of the key

Question: If [itex]M = x[/itex], why is the event [itex]C = y[/itex] equivalent to the event [itex]K = x \oplus y[/itex]?

The cypher bit is formed by [itex]C = M \oplus K[/itex]

MKC
0 0 0
0 1 1
1 0 1
1 1 0

For example, If [itex]M = 1[/itex] then [itex]C = 1[/itex] exactly when [itex]K = 1 \oplus 1 = 0[/itex]
 
Thank you!

BiP
 

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