Puzzler - Egg Drop

You are given two identical eggs and access to a 100-story building. You want to determine the highest floor from which an egg will not break when dropped. 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.

If an egg breaks when dropped from floor n, then it would also have broken from any floor above that. If an egg survives a fall, then it will survive any fall shorter than that.

The question is: What strategy should you adopt to minimize the number egg drops it takes to find the solution? (And what is the worst case for the number of drops it will take?)

Solution: Suppose we start by dropping the first egg from floor 50 and it breaks. The only way to determine the correct floor is to then start with floor 1, then 2, 3, ... until our second egg breaks. If the highest floor is 49 then we get 50 total egg drops -- far from optimal. Generalizing, if our first drop is from floor n and the egg breaks, then in the worst case we require n total egg drops. We know that the solution is between 1 and n-1 inclusive. On the other hand, if the egg does not break, we know the solution is between n+1 and 100. Within this range we can use the same technique but instead drop from n-1 floors above floor n. If the egg then breaks, then in the worst case we still total n egg drops. If it doesn't break we continue by taking our third drop to be n-2 floors above the second drop, and so on.. So assuming the first egg does not break, our successive drops are n, 2n-1, 3n-3, 4n-6, ..., n(n+1)/2. -- given by n, n+(n-1), n+(n-1)+(n-2), ..., n+...+1. When the egg finally does break on the kth drop, we drop the second egg incrementally from the floor of the (k-1)th drop up to the floor of the kth drop and in the worst case we get n total drops. Now we simply need to minimize n, which is done by taking the smallest integer n such that n(n+1)/2 > 99. Our solution is thus 14 total drops.

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Puzzler - Prisoners and Lightbulbs

There are 100 prisoners in solitary confinement. There's a central living room with one light bulb that is  initially off. No prisoner can see the light bulb from his or her own cell. Everyday, the warden picks a prisoner at random to visit the living room. While there, the prisoner can toggle the bulb if he or she wishes. The prisoner also has the option of asserting that all 100 prisoners have been to the living room by now. If this assertion is false, all 100 prisoners are shot. However, if it is indeed true, all prisoners are set free. So the assertion should only be made if the prisoner is 100% certain of its validity. The prisoners are allowed to get together one night in the courtyard, to discuss a plan. What plan should they agree on, so that eventually, someone will make a correct assertion?

Solution: Here is one sub-optimal solution: Suppose each of the prisoners are assigned a different number from 0-99. Each day is taken modulo 100 and if the prisoner is assigned the same number as the day he turns the light on (or keeps it on) and otherwise he turns the light off (or keeps it off). Whenever a prisoner is called to the room and sees that the light is on, he makes note that the previous days' prisoner has been to the room. Once any of the prisoners has tallied every other prisoner being in the room, he can declare so and they will be set free. For example, say prisoner 50 is called to the room on day 134=34 (mod 100) and sees that the light is on. He concludes that prisoner 33 was in the room the previous day and then proceeds to turn the light off.

There is a simple way to alter this solution to make it way better -- the expected wait goes from ~10,000 to ~1,500 days. Can you think of it?

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