# Tag Info

7

$g^x \cdot g^y \;\;\; = \;\;\; (\hspace{.02 in}g\cdot g\cdot g\cdot \ldots$ [$\hspace{.02 in}x$ of them] $\ldots \cdot g\cdot g\cdot g) \: \cdot \: (\hspace{.02 in}g\cdot g\cdot g\cdot \ldots$ [$\hspace{.03 in}y$ of them] $\ldots \cdot g\cdot g\cdot g)$ $= \;\;\; g\cdot g\cdot g\cdot \ldots$ [$\hspace{.02 in}x\hspace{-0.05 in}+\hspace{-0.05 in}y$ of them] ...

4

$\pi$ is the transcendental number 3.1415926... It's there in the formula to show this specific number was not chosen with a specific cryptographical backdoor in mind; it seems unlikely that anyone was able to select the value of $\pi$ (unless Carl Sagan was correct, of course :-)

3

In DH if you want to compute $g^a$ from $K$ you have to know $b$ (which the legitimate receiver of $g^a$ clearly knowns, so this does not really make sence). This party can compute the inverse of $b$, namely $b^{-1}$, and then compute $g^a=K^{b^{-1}}$. Note that this is not the same as $(K^b)^{-1}=(g^{ab})^{-1}$ (as I will discuss below). But that is not ...

3

That depends on the groups you are working in. Using a $\Sigma$-protocol If you have a group $G$ of prime order $q$ where the DDH is hard and you have a DH tuple $(g,g^u,g^v,g^w)$ with $w\equiv uv \pmod q$, then if your prover knows one of these values, say $u$, then we can write the DH tuple as $(g,g^u,h,h^u)$ and he is able to convince a verifier that ...

1

All this work is done modulo the prime $p$ (which is 11 in your toy example); in this field, addition, subtraction and multiplication is done in the usual way (except you do a modulo $p$ at the end); however division is defined differently. We define $x = 6/4\ (\bmod 11)$ to be that value such that $x \times 4 = 6\ (\bmod 11)$. Now, we see that \$7 \times ...

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