- #1

- 775

- 1

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Bipolarity
- Start date

- #1

- 775

- 1

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

- #2

Stephen Tashi

Science Advisor

- 7,642

- 1,496

- #3

- 775

- 1

[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].

- #4

Stephen Tashi

Science Advisor

- 7,642

- 1,496

[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]

- #5

- 775

- 1

Thank you!!!!

BiP

BiP

Share: