So based on my understanding using the same key for encrypting two completely random plain texts when using an otp cipher will still retain perfectly secret. This is based on the answers on this question.

The answers to those questions also say that using that data couldn't then be used to encrypt more messages. My understanding is that if the messages have been transmitted securely using 2-time pad cipher. This should be equal to handing them off at a physical location.

What I am wondering is why not be able to send new one time pads? If the message are perfectly secure an enemy wouldn't be able to determine what the secrets are? How would you brute force crack the messages since they hold perfect secrecy?

• lets say I use pad1 to encrypt two new pads pad2 and pad3. then using pad2 encrypt m1. I now longer use pad2 for any purpose I then use pad3 for two new pads, pad4, 5. pad4 for encrypting m2. lets say I continue this pattern for indefinetly. what I am trying to find is where somebody could break this pattern. Commented Jan 9, 2016 at 1:14

Quite simply, it's exactly the same as a two-time pad with one extra step.

So, as you know the problem with re-using a one time pad is that it leaks information.

If we encrypt m1 and m2 with pad1, that gives us:

m1 `xor` pad1 = c1 and m2 `xor` pad1 = c2

So if we capture c1 and c2, we can then:

c1 `xor` c2 to get m1 `xor` m2

If however, we re-use pad1 to transmit pad2, and then use pad2 to encrypt m2, we get the following:

• m1 `xor` pad1 = c1
• pad2 `xor` pad1 = c2
• m2 `xor` pad2 = c3

So now, as long as I capture c1, c2, and c3, I can do the following:

c1 `xor` c2 = m1 `xor` pad2

Which effectively gives me the ciphertext I would have if I had used pad2 on m1 directly. So now I effectively have the ciphertexts for:

m1 `xor` pad2 and m2 `xor` pad2

Which when `xor'ed` then gives me:

m1 `xor` m2

Voila.

• lets say I use pad1 to encrypt two new pads pad2 and pad3. then using pad2 encrypt m1. I now longer use pad2 for any purpose I then use pad3 for two new pads, pad4, 5. lets say I continue this pattern for indefinetly Commented Jan 9, 2016 at 1:07
• @5hammer It's the same scenario, with n additional steps. Eventually, if you send two non-random messages, you can always get back to the xor of those messages. So, while you can make it harder, you've destroyed the security of the pad, and might as well use an easier, more secure mechanism. Commented Jan 9, 2016 at 14:46