Question

Here’s something I’ve been thinking about: suppose you have a number, x, that can be infinitely large, and you have to find out what it is. All you know is if another number, y, is larger or smaller than x. What would be the fastest/best way to find x?

An evil adversary chooses a really large number somehow … say:

int x = 9^9^9^9^9^9^9^9^9^9^9^9^9^9^9

and provides isX, isBiggerThanX, and isSmallerThanx functions. Example code might look something like this:

int c = 2
int y = 2
while(true)
    if isX(y) return true
    if(isBiggerThanX(y)) fn()
    else y = y^c

where fn() is a function that, once a number y has been found (that’s bigger than x) does something to determine x (like divide the number in half and compare that, then repeat). The thing is, since x is arbitrarily large, it seems like a bad idea to me to use a constant to increase y.

This is just something that I’ve been wondering about for a while now, I’d like to hear what other people think

Answer 1

Use a binary search as in the usual “try to guess my number” game. But since there is no finite upper end point, we do a first phase to find a suitable one:

• Initially set the upper end point arbitrarily (e.g. 1000000, though 1 or 1^100 would also work — given the infinite space to work in, all finite values are equally disproportionate).
• Compare the mystery number X with the upper end point.
• If it’s not big enough, double it, and try again.
• Once the upper end point is bigger than the mystery number, proceed with a normal binary search.

The first phase is itself similar to a binary search. The difference is that instead of halving the search space with each step, it’s doubling it! The cost for each phase is O(log X). A small improvement would be to set the lower end point at each doubling step: we know X is at least as high as the previous upper end point, so we can reuse it as the lower end point. The size of the search space still doubles at each step, but in the end it will be half as large as would have been. The cost of the binary search will be reduced by only 1 step, so its overall complexity remains the same.

Some notes

A couple of notes in response to other comments:

It’s an interesting question, and computer science is not just about what can be done on physical machines. As long as the question can be defined properly, it’s worth asking and thinking about.

The range of numbers is infinite, but any possible mystery number is finite. So the above method will eventually find it. Eventually is defined such as that, for any possible finite input, the algorithm will terminate within a finite number of steps. However since the input is unbounded, the number of steps is also unbounded (it’s just that, in every particular case, it will “eventually” terminate.)