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1

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.


2

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 ...


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 ...


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 $x,x'$ ...


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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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