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5

As fgrieu notes in the comments, the protocol might not even work reliably in the absence of adversaries: if the tag fails to receive the reader's reply of "True", the keys will get out of sync. (If that happens, the tag will just retry the next exchange with the old key, so this could be fixed by having the reader remember one more "subkey". But the ...


5

This is not a mathematical proof. A notable place it fails to be a proof is here: Pay attention to which cipher text I use, look up to match the message with the cipher-number below. $$ cipher1⊕cipher3=character1⊕character3⊕IV1⊕IV2$$ (Note that the cipher BOTH use the SAME KEY, but they remain secure because of the two different IV) This line is ...


3

There is of course but because of carry bits there will be data-dependent "nonlinear" terms. If I do it for 2 bits you can get the idea. It gets unwieldy but you can easily write code to do it for longer bitlengths. The list below is an XOR table expressed as integers: $$ \begin{array}{c|ccc} \oplus & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & ...


2

Well, if you construct what you described you basically create a function $f: \{0,1\}^{2m} \rightarrow \{0,1\}^m$. As you correctly pointed out these two strings give the same when xor'ed. So the messages $10101111$ and $00000101$ will result in the same xor and hence will get mapped to the same hash, resulting in a second preimage as you found two $x,x'$ ...


1

There's a lot of ways to attack this. The first thing to notice is that if you know the value of plaintext at index $i$, you can then deduce the value of all the plaintext bytes at index $i+8k$; for all integers $i$ (!). That's because the relation between plaintext and ciphertext bytes is $P_{i} = C_{i} \oplus P_{i-8}$; if you know $P_{i-8}$ (and, of ...



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