Extensions of Coppersmith algorithm

The Coppersmith algorithm can be made even more general. There are two main extensions, first to an unknown modulus, then to multivariate polynomials.

Unknown modulus

This extension of Coppersmith allows one to find small roots modulo an unknown some unknown factor of a number. More specifically, suppose that we have some unknown factor$b$of$N$such that $b<N^\beta$and some monic polynomial$f$of degree$d$such that $f(x_0)=0\pmod b$for some $x_0<N^{\frac{\beta^2}d}$. Then we can find$x_0$in time polynomial in $\log N,d$.

One key reason why this is possible is because we dont need to explicitly know the modulus to determine if a small root exists, i.e. $\left\lVert f(Bx)\right\rVert_2\leq\frac b{\sqrt{d+1}}\leq\frac{N^{\beta}}{\sqrt{d+1}}$is sufficient for a root less than$B$to exist. The algorithm here is extremely similar to the Coppersmith algorithm, except we add more polynomials into the lattice. The polynomials that we will use are

$$\begin{align*} g_{i,j}(x)&=N^{h-j}f(x)^jx^i&0\leq i<d,0\leq j<h\\ g_{i,h}(x)&=f(x)^hx^i&0\leq i<t \end{align*}$$

The lattice$L$generated by these polynomials would have

$$\dim(L)=dh+t=n\quad\text{vol}(L)=N^{\frac{dh(h+1)}{2}}B^{\frac{(n-1)n}2}$$

As we require the shortest vector to have length at most $\frac{N^{\beta h}}{\sqrt{n}}$, repeating the computations from the previous section, we obtain

$$B<\sqrt{\frac{4\delta-1}4}\left(\frac1n\right)^{\frac1{n-1}}N^{\frac{2\beta h}{n-1}-\frac{dh(h+1)}{n(n-1)}}$$

It turns out that the maxima of $\frac{2\beta h}{n-1}-\frac{dh(h+1)}{n(n-1)}$is $\frac{\beta^2}d$as $n,h\to\infty$. One way to achieve this is by setting$n=\frac d\beta h$and we obtain

$$B<\sqrt{\frac{4\delta-1}4}\left(\frac1n\right)^{\frac1{n-1}}N^{\frac{(h-1)\beta^2}{dh-\beta}}$$

and this indeed achieves the $O\left(N^{\frac{\beta^2}d}\right)$bound. Similar to the Coppersmith algorithm, one chooses a sufficiently big $h,n$such that the remainding bits can be brute forced in constant time while the algorithm still remains in polynomial time.

Exercises

1) We show that the maximum of$\frac{2\beta h}{n-1}-\frac{dh(h+1)}{n(n-1)}$is indeed $\frac{\beta^2}d$. We can assume that $d\geq2,h\geq1,0<\beta\leq1,n\geq2$. Since $\frac h{n-1}\left(2\beta-d\frac{h+1}{n}\right)$and $\frac h{n-1}<\frac{h+1}n$, the maximum occurs when $h,n\to\infty$and $\frac h{n-1},\frac{h+1}n\to x$, hence we have reduced this to maximizing $2\beta x-dx^2$which achieves its maximum of $\frac{\beta^2}d$ at $x=\frac\beta d$.