Generalised Two-Egg Puzzle

Question

Here is the Problem Description :

Suppose that we wish to know which stories in a N-story building are safe to drop eggs from, and which will cause the eggs to break on landing. We make a few assumptions: An egg that survives a fall can be used again.

• A broken egg must be discarded.
• The effect of a fall is the same for all eggs.
• If an egg breaks when dropped, then it would break if dropped from a higher window.
• If an egg survives a fall then it would survive a shorter fall.
• It is not ruled out that the first-floor windows break eggs, nor is it ruled out that the Nth-floor windows do not cause an egg to break.

Given an N story building and a supply of d eggs, find the strategy which minimizes (in the worst case) the number of experimental drops required to determine the breakfloor.

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I have seen and solved this problem for 2 eggs where answer comes out to be 14 for N=100. I tried to understand the generalized solution from wiki using DP but couldn't Understand what are they trying to do. Please tell how they arrived at the DP and how it is working ?

EDIT :

The Recurrence given in this Article for the The highest floor that can be tested with d drops and e eggs is as follows :

f[d,e] = f[d-1,e] + f[d-1,e-1] + 1

The recurrence is fine but i not able to understand how it is derived ?

The explanation is not clear to me....i just want someone to explain this recurrence to me in more clear words.

Answer 1

(1) Consider the case that the first drop breaks the egg. Then you can determine the breakfloor if and only if it is at most f[d-1, e-1]. Therefore you can't start higher than f[d-1, e-1] + 1 (and shouldn't start lower, of course).

(2) If your first drop doesn't breaks the egg, you are in the case of f[d-1, e], just starting at the floor of your first drop + 1, instead of floor 1.

So, the best you can do is to start dropping eggs at floor f[d-1, e-1] + 1 (because of (1)), and you can get up to f[d-1, e] floors higher than that (because of (2)). That's

f[d, e] = f[d-1, e-1] + 1 + f[d-1, e]

Answer 2

From Wiki Egg Dropping puzzle we know that the the state transfer equation is:

W(n,k) = 1 + min{ max(W(n − 1, x − 1), W(n,k − x)) } , x = 1, 2, ..., k

W(n,1)=1, W(1,k)=k

n = number of test eggs available

k = number of (consecutive) floors yet to be tested

Below is my understanding.

We have k floors, n eggs, assume we use an egg to test in x floor. there are only two possible results:

1. it breaks, so the problem recursively come to: x-1 floors, n-1 eggs, which reflects to W(n-1,x-1)
2. it doesn't break, so the problem recursively come to: k-x floors, n eggs, which reflects to W(n,k-x)

Since the problem requires the worst case, we have to choose the bigger one to ensure the worst case works, that's why we add an max between W(n-1,x-1) and W(n,k-x).

Besides, as we just assumed testing in x floor, x can be from 1 to k, in this situation, we definitely need to choose the minimum to ensure the min experimental drops to find out N, that's why we add an min between {max(W(n − 1, x − 1), W(n,k − x)): x = 1, 2, ..., k}

Finally, as we have used 1 drop in x floor, so the equation must add 1, which reflects to the first part of the equation.

Hope that solves your puzzle :-)