The “guess the number” game for arbitrary rational numbers?

Question

I once got the following as an interview question:

I\'m thinking of a positive integer n. Come up with an algorithm that can guess it in O(lg n) quer

Answer 1

Okay, I think I figured out an O(lg² q) algorithm for this problem that is based on Jason S's most excellent insight about using continued fractions. I thought I'd flesh the algorithm out all the way right here so that we have a complete solution, along with a runtime analysis.

The intuition behind the algorithm is that any rational number p/q within the range can be written as

a₀ + 1 / (a₁ + 1 / (a₂ + 1 / (a₃ + 1 / ...))

For appropriate choices of a_i. This is called a continued fraction. More importantly, though these a_i can be derived by running the Euclidean algorithm on the numerator and denominator. For example, suppose we want to represent 11/14 this way. We begin by noting that 14 goes into eleven zero times, so a crude approximation of 11/14 would be

0 = 0

Now, suppose that we take the reciprocal of this fraction to get 14/11 = 1 ³/₁₁. So if we write

0 + (1 / 1) = 1

We get a slightly better approximation to 11/14. Now that we're left with 3 / 11, we can take the reciprocal again to get 11/3 = 3 ²/₃, so we can consider

0 + (1 / (1 + 1/3)) = 3/4

Which is another good approximation to 11/14. Now, we have 2/3, so consider the reciprocal, which is 3/2 = 1 ¹/₂. If we then write

0 + (1 / (1 + 1/(3 + 1/1))) = 5/6

We get another good approximation to 11/14. Finally, we're left with 1/2, whose reciprocal is 2/1. If we finally write out

0 + (1 / (1 + 1/(3 + 1/(1 + 1/2)))) = (1 / (1 + 1/(3 + 1/(3/2)))) = (1 / (1 + 1/(3 + 2/3)))) = (1 / (1 + 1/(11/3)))) = (1 / (1 + 3/11)) = 1 / (14/11) = 11/14

which is exactly the fraction we wanted. Moreover, look at the sequence of coefficients we ended up using. If you run the extended Euclidean algorithm on 11 and 14, you get that

11 = 0 x 14 + 11 --> a0 = 0 14 = 1 x 11 + 3 --> a1 = 1 11 = 3 x 3 + 2 --> a2 = 3 3 = 2 x 1 + 1 --> a3 = 2

It turns out that (using more math than I currently know how to do!) that this isn't a coincidence and that the coefficients in the continued fraction of p/q are always formed by using the extended Euclidean algorithm. This is great, because it tells us two things:

1. There can be at most O(lg (p + q)) coefficients, because the Euclidean algorithm always terminates in this many steps, and
2. Each coefficient is at most max{p, q}.

Given these two facts, we can come up with an algorithm to recover any rational number p/q, not just those between 0 and 1, by applying the general algorithm for guessing arbitrary integers n one at a time to recover all of the coefficients in the continued fraction for p/q. For now, though, we'll just worry about numbers in the range (0, 1], since the logic for handling arbitrary rational numbers can be done easily given this as a subroutine.

As a first step, let's suppose that we want to find the best value of a₁ so that 1 / a₁ is as close as possible to p/q and a₁ is an integer. To do this, we can just run our algorithm for guessing arbitrary integers, taking the reciprocal each time. After doing this, one of two things will have happened. First, we might by sheer coincidence discover that p/q = 1/k for some integer k, in which case we're done. If not, we'll find that p/q is sandwiched between 1/(a₁ - 1) and 1/a₀ for some a₁. When we do this, then we start working on the continued fraction one level deeper by finding the a₂ such that p/q is between 1/(a₁ + 1/a₂) and 1/(a₁ + 1/(a₂ + 1)). If we magically find p/q, that's great! Otherwise, we then go one level down further in the continued fraction. Eventually, we'll find the number this way, and it can't take too long. Each binary search to find a coefficient takes at most O(lg(p + q)) time, and there are at most O(lg(p + q)) levels to the search, so we need only O(lg²(p + q)) arithmetic operations and probes to recover p/q.

One detail I want to point out is that we need to keep track of whether we're on an odd level or an even level when doing the search because when we sandwich p/q between two continued fractions, we need to know whether the coefficient we were looking for was the upper or the lower fraction. I'll state without proof that for a_i with i odd you want to use the upper of the two numbers, and with a_i even you use the lower of the two numbers.

I am almost 100% confident that this algorithm works. I'm going to try to write up a more formal proof of this in which I fill in all of the gaps in this reasoning, and when I do I'll post a link here.

Thanks to everyone for contributing the insights necessary to get this solution working, especially Jason S for suggesting a binary search over continued fractions.

Answer 2

You can sort rational numbers in a given interval by for example the pair (denominator, numerator). Then to play the game you can

1. Find the interval [0, N] using the doubling-step approach
2. Given an interval [a, b] shoot for the rational with smallest denominator in the interval that is the closest to the center of the interval

this is however probably still O(log(num/den) + den) (not sure and it's too early in the morning here to make me think clearly ;-) )