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

I've got it! What you need to do is to use a parallel search with bisection and continued fractions.

Bisection will give you a limit toward a specific real number, as represented as a power of two, and continued fractions will take the real number and find the nearest rational number.

How you run them in parallel is as follows.

At each step, you have l and u being the lower and upper bounds of bisection. The idea is, you have a choice between halving the range of bisection, and adding an additional term as a continued fraction representation. When both l and u have the same next term as a continued fraction, then you take the next step in the continued fraction search, and make a query using the continued fraction. Otherwise, you halve the range using bisection.

Since both methods increase the denominator by at least a constant factor (bisection goes by factors of 2, continued fractions go by at least a factor of phi = (1+sqrt(5))/2), this means your search should be O(log(q)). (There may be repeated continued fraction calculations, so it may end up as O(log(q)^2).)

Our continued fraction search needs to round to the nearest integer, not use floor (this is clearer below).

The above is kind of handwavy. Let's use a concrete example of r = 1/31:

1. l = 0, u = 1, query = 1/2. 0 is not expressible as a continued fraction, so we use binary search until l != 0.

2. l = 0, u = 1/2, query = 1/4.

3. l = 0, u = 1/4, query = 1/8.

4. l = 0, u = 1/8, query = 1/16.

5. l = 0, u = 1/16, query = 1/32.

6. l = 1/32, u = 1/16. Now 1/l = 32, 1/u = 16, these have different cfrac reps, so keep bisecting., query = 3/64.

7. l = 1/32, u = 3/64, query = 5/128 = 1/25.6

8. l = 1/32, u = 5/128, query = 9/256 = 1/28.4444....

9. l = 1/32, u = 9/256, query = 17/512 = 1/30.1176... (round to 1/30)

10. l = 1/32, u = 17/512, query = 33/1024 = 1/31.0303... (round to 1/31)

11. l = 33/1024, u = 17/512, query = 67/2048 = 1/30.5672... (round to 1/31)

12. l = 33/1024, u = 67/2048. At this point both l and u have the same continued fraction term 31, so now we use a continued fraction guess. query = 1/31.

SUCCESS!

For another example let's use 16/113 (= 355/113 - 3 where 355/113 is pretty close to pi).

[to be continued, I have to go somewhere]

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On further reflection, continued fractions are the way to go, never mind bisection except to determine the next term. More when I get back.