How would you do a binary search on a continuous space without discretizing?

Replace A[x] with f(x) and done.

A modified binary search called ternary search is well known used to find when derivative = 0, numerically.

https://en.m.wikipedia.org/wiki/Ternary_search

What would you search for, the zero/root in the derivative? Imagine the derivative is a function like y=x^2-eps with some small eps, say 0.000001. For binary search like that to work, you need to start with two values x1 and x2 such that sign(y1) != sign(y2). But with the derivative above those values are hard to find with random initialization.

More generally, for binary search to find a value x such that f(x)=0, we need to have f(any value less than x)<0 and f(any value greater than x)>0. We can't guarantee these properties for general derivatives.

> What would you search for

I imagine you could search until the incremental change between your current and last result is below some threshold epsilon, as a kind of stopping criterion.

IEEE 754 is by default limited by 64 bits, so yes.

This is probably what you're thinking of:

https://en.m.wikipedia.org/wiki/Bisection_method

Because the search space is tremendously huge.

Try doing binary search on a space of nontrivial dimensionality (ie, not 1, 2, or 3) and you will have your answer.

Binary search only works on sorted data.

Because of the dimensionality of the problem. Bracketing is easy to do in 1D space, not so in N-dimensional space, with large N.

We do, it’s called global optimization. It works for small problems with ~ 1000 variables or so. Check BARON.

Thanks!

Binary search only works in 1D parameter space