I know the definition of CDH:
The Computational Diffie-Hellman problem: Given $y_1 = g^{x_1}$ and $y_2 = g^{x_2}$ (but not $x_1$ and $x_2$), find $y = g^{x_1·x_2}$.
- What happens if I knew one of the $x_1$, would it still be hard?
- Considering the CDH, is the problem still hard for $g^{x_1x_2}/g^{x_1}$ or $g^{x_1x_2}*g^{x_1}$ or even just $zg^{x_1x_2}$ for some integer $z$. (For cases when I don't know $x_1$ and $x_2$ and the other case when I do know one like $x_1$?)
I think for the first one it would reduce to only needing to solve $y_2 = g^{x_2}$ which is a hard discrete $\log$ problem. And we know that if CDH is hard, DL is hard [2].
If $x_1$ were known so would $y = y_2^{x_1}$ be. So both should be unknown (which is true if DL is hard). And if it's hard to compute $g^{x_1\cdot x_2}$ from $y_1$ and $y_2$ alone the same holds for $y =\frac{g^{x_1x_2}}{g^{x_1}}$, because $y_1 \cdot y = g^{x_1 x_2}$ and $y_1$ is given, and multiplication is feasible. The same holds for $y = g^{x_1 x_2}\cdot g^{x_1}$, as we can divide by $y_1 = g^{x_1}$. Division by an integer is also easy (supposing $z$ is indeed in the multiplicative group in which we work).