Many time pad attack (XOR)

Question

The question is quite similar to Many time pad attack and I was trying to rely on the top answer, but still am a bit confused, so any explanation and help will be much appreciated. Assume I have a bunch of ciphertexts with the same key. I need to decipher one of them. At the moment I understand that I need to XOR the desired one with each of the rest, but I am confused whether I should do it one by one or XOR it will all at once? Where do I go from there checking whether there's been a space or no, surely all of them will have spaces in different places? How do I recover the symbols then? Thank you so much.

Answer 1

A character is usually encoded as an ASCII. This means that it uses up one byte. That's a number from $0 - 255$. It can be represented as a hexadecimal $\text{0x00} - \text{0xFF}$. All your operations must be done character by character. From now on by "message", "key" and "cipher" i mean a single $0-255$ number. $$ message1 \oplus key = cipher1 \\ message2 \oplus key = cipher2 $$ when we xor two ciphered letters $$ cipher1 \oplus cipher2 = \\ message1 \oplus key \oplus message2 \oplus key = \\ message1 \oplus message2 $$ So when we xor one character of the cipher with corresponding character of a second cipher we remove the key from the equation, and receive a character of first message xored with character of second message. Now we do some deduction. We look for numbers looking like $\text{0x42}$ I did some calculations previously and I know that we can get this number within a-z(space)A-Z only by xoring "b" (lowercase b) and " " (space). Rest is pretty much a lot of guessing.

To get that byte of the key all you need is to xor byte of message with byte of cipher. $$ key = message \oplus cipher $$ Example: We have two ciphertexts from which we take the third character: $$48A6\color{red}{48}3C\\7692\color{red}{0A}A2$$ We xor them together: $$\text{0x48} \oplus \text{0x0A} = \text{0x42}$$ I made myself a table of all possible letters xored with space, and I see that $\text{0x42}$ is in that table. This means that it's possible that one of the letters is "b" ($\text{0x62}$) and second is " " ($\text{0x20}$) Let's assume that " " is encrypted in first ciphertext and get the key: $$\text{0x48} \oplus \text{0x20} = \text{0x68}$$

If we assume that " " is encrypted in second ciphertext we get another possibility of the key: $$\text{0x0A} \oplus \text{0x20} = \text{0x2A}$$

So the key can be: $$\_\_\_\_\color{red}{68}\_\_$$ or $$\_\_\_\_\color{red}{2A}\_\_$$

We just decreased the number of possibilities for this byte of the key from 256 to 2.