Two egg dropping puzzle variation: unknown/infinite floors

问题

Preface

This problem was inspired by a similar question last week on SO, that got deleted before it was clear what the real question was. I think this variation makes a nice problem that I wanted to share.

Two Egg Problem

A detailed definition and solution can be found here, but I will add a quick summary:

Definition

You are given two eggs, and access to a k-storey building. Both eggs are identical. The aim is to find out the highest floor f* from which an egg will not break when dropped out of a window from that floor. If an egg is dropped and does not break, it is undamaged and can be dropped again. However, once an egg is broken, that’s it for that egg. What is the fastes (least amount of drops) way to find f*?

Solution

The idea is to drop the first egg from floors sqrt(k), 2*sqrt(k), 3*sqrt(k)... k. If the egg breaks at floor i*sqrt(k) use the second egg to test the remaining floors between (i-1)*sqrt(k) and i*sqrt(k)-1. Overall this will result in at most 2*sqrt(k) drops so the complexity will be O(sqrt(k)).

Just for completeness: there is a method with less drops in the worst-case (details can be found here), which however has the same complexity of O(sqrt(k))

The Question: Two Egg Problem with infinite/unknown floors

Now imagine that you have no information about the number of floors k or k is infinite. Is it possible to find f* more efficient than just testing each floor in O(f*)?

In other words: Is there an efficient method to drop the two eggs whose runtime complexity is independent from k but only depends on the answer f*?

回答1:

There is a simple method that has O(sqrt(f*)) complexity. Make your nth step to be n floors up, that is, check floors 1, 3 (1 + 2), 6 (1 + 2 + 3), etc. This way at the nth step you will be on n*(n+1)/2 floor, and you will reach f* in n = O(sqrt(f*)) steps.

Then for the second egg you will need to go n single steps over your last step in stage 1, which will add another O(sqrt(f*)).

If O(sqrt(k)) was optimal for known k, this method has to be optimal in terms of complexity, too.

回答2:

The sub-optimal solution for the infinite problem for two eggs is to use the sequence 1, 2^2, 3^3,... ,i^2,... and start the second edge where the last value the first egg remains. So if the first edge remains at n^2, then the next one will do at most 2*n + 1 - 1 (minus 1 from the first) test to give the total of 3 * n in the worst case with n = sqrt(m).

来源：https://stackoverflow.com/questions/49851682/two-egg-dropping-puzzle-variation-unknown-infinite-floors