This is a really good article but you aren't going to like it if you're on the pro-SJW side.

I think it's a very good rationalization of the rational side of anti-SJW reasoning particularly as applied to Free Software.

Dude is literally being slandered and called a Nazi and SJW people in the comments here are defending the slander and attacking the poster ad hominem while making straw man arguments. It is amusing but also makes me lose hope that rehabilitation of SJW types is possible.

I personally think we need to divide communities into pro-SJ and anti-SJ spaces. SJ is self destructive, least of which because they tend to lose lawsuits when courts get involved. Defending literal slander is yet another example of that...

Suppose you wanted to find a root of f(x) = x^3 - a mod N. If there is a root smaller than N^(1/3), this reduces to simply computing the integer cube root of a, since there is no wrap-around.

But imagine if the problem is slightly changed to f(x) = (x + padding)^3 - a mod N, for some large but known padding. Now you can't simply compute an integer cube root, and the above doesn't work anymore. But you can look for a product of f(x) by some other polynomial g(x): f(x)g(x) necessarily contains the roots of f(x), but may have smaller coefficients such that evaluating f(x)g(x) at the root you're looking for does not wrap around N.

Now look into what this multiplication looks like: f(x)g(x) = c_0 f(x) + c_1 x f(x) + c_2 x^2 f(x) + ..., where c_i are the unknown coefficients of g(x). This is a sum of integer multiples of known coefficient vectors and, to find the set of coefficients c_i such that this sum is a small as possible is exactly the same as finding a short vector in the lattice (f(x), xf(x), x^2f(x), ...), where the vectors are the coefficients of each (shifted) polynomial. You also have to weight the coefficients according to their exponent, to account for the fact that the coefficient of, say, x^3 needs to be smaller than x to avoid wrap-around.

Once lattice reduction gives you back an f(x)g(x) with small enough coefficients you compute its integer roots, which is doable very quickly, and check which of them work for f(x). That's the gist of it. The full-fledged method also includes powers of f(x) modulo powers of N instead of simply multiplying it by g(x), and that was the trick Coppersmith introduced, but if you understand the simple version the full version is straightforward.