New answers tagged collision-resistance
0
If $P = (x,y)$ has order $q$, then $$(q-1)P = -P = (x,-y).$$
When $q=2$ (equivalently, $y=0$), these two points coincide: $P=-P$.
1
Yes. Fix the $x$ co-ordinate and let $c=x^3-17$. The equation $y^2\equiv c\pmod p$ has at most two solutions (it will have zero if $c$ is a quadratic non-residue, two if $c$ is a quadratic residue and one if $c\equiv 0\pmod p$). If it has two solutions $y_1$, $y_2$ they will be additive inverses: $y_1\equiv -y_2\pmod p$. In the standard formulation of the ...
3
Can we claim that if $n < \lceil{q/2}\rceil$, then there do not exist $y_2 \neq y_1$ such that $(x,y_2)$ is a valid curve point?
No, such a claim would be false. If $(x, y_1)$ is a valid point, that is, if $y_1^2 = x^3 - 17$, then $(x, q-y_1)$ is also a valid point. Hence, unless $y_1 = 0$, there will always be a second point with the same $x$ ...
1
No by our definition that we're using it is indeed not (strongly) collison-free. Because with the Algorithm $A$ we've already constructed a very fast way to compute such $x_1,x_2$ with $H(x_1)=H(x_2)$ and virtually by definition this contradicts the condition collision-free.
2
I am using AES-CBC as a hash function which is encrypting a block of length n. The blocks, m = (m1, m2, ..., mn). The IV is one block long and the encryption key is length 128, 192 or 256 bits.
Questions:
What is the key? Is it fixed in advance, or is it something secret? BTW: if it's 'something secret', you don't have a standard hash function (where the ...
Top 50 recent answers are included