How can I find the original message of two encrypted (Vernam Cipher) text without the key [duplicate]

Question

Can I get the steps or the solution if possible?

Answer 1

You know that the two ciphertext you are given are formed as $$C_1 = S_1 \oplus R\\C_2 = S_2 \oplus R$$ with the same $R$ in both cases. So you can get rid of $R$ by calculating $$C^\ast = C_1 \oplus C_2 = S_1 \oplus S_2$$ Then you know that there are only two words, $111$ and $0000$. Thus, in $C^\ast$ $1$ can only occur if one sentence contains $A$ while the other contains $B$. From there, since the two words have different lengths, you should be able to determine candidate sentences.

Answer 2

I think I've gotten the solution After reading the following posts:

• is it possible to get key in One time pad if key is reused?

• How does one attack a two-time pad (i.e. one time pad with key reuse)?

• Taking advantage of one-time pad key reuse?

and the answer of @Emme above/below, I came out with this solution:

Let M1 and M2 be the original messages.

S1 ⊕ S2 = M1 ⊕ M2 since R is the same for S1 and S2

We calculate S1 ⊕ S2

S1 ⊕ S2 = 0 0 0 1 1 1 0 1 1 1 1 1 1 0 0 0 0 0 1 1 1

Since the language has just two words, A and B, I could guess the value of M1 by taking a random combination of A and B.

let M1 = AABBBA. Converting M1 gives M1 = 111 111 0000 0000 0000 111

let, S1 ⊕ S2 = S3. Then, M1 ⊕ S3 = M2

We calculate M1 ⊕ S3

M1 ⊕ S3 = 111 0000 111 111 0000 0000

Therefore, M2 = M1 ⊕ S3 = 111 0000 111 111 0000 0000

Decoding M2 gives M2 = ABAABB

Therefore, M1 = AABBBA and M2 = ABAABB

Proof

Remeber we said S1 ⊕ S2 = M1 ⊕ M2 since R is the same for S1 and S2
If we calculate M1 ⊕ M2, we will have:
111 111 0000 0000 0000 111 ⊕ 111 0000 111 111 0000 0000 = 0 0 0 1 1 1 0 1 1 1 1 1 1 0 0 0 0 0 1 1 1
Which is the same as S1 ⊕ S2 as seen above