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The symbol of the circle with the + in it is one of many symbols for exclusive-or. XOR, EOR, EXOR, ⊻, ⊕, ↮, and ≢. Binary OR is true when either input is true; binary XOR is true when exactly one input is true. If both inputs are true, the XOR result is false. One property of this is that if either input bit flips, the output bit will also flip. That's sort ...


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It's exclusive or (or XOR), corresponds to $x \oplus y := x+y \pmod 2$ for single bits, i.e., scalars. It is sometimes used for binary vectors as well, whereby two bitvectors of length $n$ $$ \mathbf{x}=(x_1,\ldots,x_n)$$ and $$ \mathbf{y}=(y_1,\ldots,y_n)$$ result in $$ \mathbf{x}\oplus \mathbf{y}=(x_1\oplus y_1,\ldots,x_n \oplus y_n) $$ i.e., a bitwise ...


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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 $x,x'$ ...


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For example, let say that you have a message: "100 dollars should be moved." and you encrypt it with OTP. Then everybody can just take the first character "1" and change it to a 9 by XOR the 1 from the cipher-text and then XOR a "9" with the key you got. What you are describing is what happens if an attacker introduces changes in the ciphertext by a ...


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Is CBC mode in OTP more secure? No. If your one time pad satisfies the required properties (it's truly random, the attacker has no information about it, and it's only used once), then OTP already has perfect secrecy; playing around with how it works can't make things better. If your one time pad doesn't satisfy the required properties, then all bets ...


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XORing a key and message is called a one time pad. It is perfectly secure, providing confidentiality, when used correctly. That last part is the hard part, along with finding a situation in which you only need confidentiality.


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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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