# Egg Dropping Puzzle – Dynamic Programming

The following is a description of the instance of this famous puzzle involving n=2 eggs and a building with H=36 floors.

Suppose that we wish to know which storeys in a 36-storey building are safe to drop eggs from, and which will cause the eggs to break on landing (using U.S. English terminology, in which the first floor is at ground level). 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 36th-floor windows do not cause an egg to break.
If only one egg is available and we wish to be sure of obtaining the right result, the experiment can be carried out in only one way. Drop the egg from the first-floor window; if it survives, drop it from the second-floor window. Continue upward until it breaks. In the worst case, this method may require 36 droppings. Suppose 2 eggs are available. What is the least number of egg-droppings that is guaranteed to work in all cases?
To derive a dynamic programming functional equation for this puzzle, let the state of the dynamic programming model be a pair s = (n,k), where

n = number of test eggs available, n = 0, 1, 2, 3, …, N − 1.
k = number of (consecutive) floors yet to be tested, k = 0, 1, 2, …, H − 1.
For instance, s = (2,6) indicates that two test eggs are available and 6 (consecutive) floors are yet to be tested. The initial state of the process is s = (N,H) where N denotes the number of test eggs available at the commencement of the experiment. The process terminates either when there are no more test eggs (n = 0) or when k = 0, whichever occurs first. If termination occurs at state s = (0,k) and k > 0, then the test failed.

Now, let

W(n,k) = minimum number of trials required to identify the value of the critical floor under the worst-case scenario given that the process is in state s = (n,k).
Then it can be shown that[17]

W(n,k) = 1 + min{max(W(n − 1, x − 1), W(n,k − x)): x = 1, 2, …, k }
with W(n,1) = 1 for all n > 0 and W(1,k) = k for all k. It is easy to solve this equation iteratively by systematically increasing the values of n and k.

An interactive online facility is available for experimentation with this model as well as with other versions of this puzzle (e.g. when the objective is to minimize the expected value of the number of trials.)[17]

Reference: Wikipedia