Throwing cats out of windows

Question

Imagine you\'re in a tall building with a cat. The cat can survive a fall out of a low story window, but will die if thrown from a high floor. How can you figure out the lon

Answer 1

I first read this problem in Steven Skiena's Algorithm Design Manual (exercise 8.15). It followed a chapter on dynamic programming, but you don't need to know dynamic programming to prove precise bounds on the strategy. First the problem statement, then the solution below.

Eggs break when dropped from great enough height. Given an n-story building, there must be a floor f such that eggs dropped from floor f break, but eggs dropped from floor f-1 survive. (If the egg breaks from any floor, we'll say f = 1. If the egg survives from any floor, we'll say f = n+1).

You seek to find the critical floor f. The only operation you can perform is to drop an egg off some floor and see what happens. You start out with k eggs, and seek to drop eggs as few times as possible. Broken eggs cannot be reused (intact eggs can). Let E(k,n) be the minimum number of egg droppings that will always suffice.

1. Show that E(1,n) = n.
2. Show that E(k,n) = Θ(n**(1/k)).
3. Find a recurrence for E(k,n). What is the running time of the dynamic program to find E(k,n)?

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Only 1 egg

Dropping the egg from each floor starting at the first will find the critical floor in (at-worst) n operations.

There is no faster algorithm. At any time in any algorithm, let g the highest floor from which the egg has been seen not to break. The algorithm must test floor g+1 before any higher floor h > g+1, else if the egg were to break from floor h, it could not distinguish between f=g+1 and f=h.

2 eggs

First, let's consider the k=2 eggs case, when n = r**2 is a perfect square. Here's a strategy that takes O(sqrt(n)) time. Start by dropping the first egg in increments of r floors. When the first egg breaks, say at floor ar, we know the critical floor f must be (a-1)r < f <= ar. We then drop the second egg from each floor starting at (a-1)r. When the second egg breaks, we have found the critical floor. We dropped each egg at most r time, so this algorithm takes at-worst 2r operations, which is Θ(sqrt(n)).

When n isn't a perfect square, take r = ceil(sqrt(n)) ∈ Θ(sqrt(n)). The algorithm remains Θ(sqrt(n)).

Proof that any algorithm takes at least sqrt(n) time. Suppose there is a faster algorithm. Consider the sequence of floors from which it drops the first egg (so long as it doesn't break). Since it drops fewer than sqrt(n), there must be an interval of at least n/sqrt(n) which is sqrt(n). When f is in this interval, the algorithm will have to investigate it with the second egg, and that must be done floor-by-floor recalling the 1-egg case. CONTRADICTION.

k eggs

The algorithm presented for 2 eggs can be easily extended to k eggs. Drop each egg with constant intervals, which should be taken as the powers of the kth root of n. For example, for n=1000 and k=3, search intervals of 100 floors with the first egg, 10 with the second egg, and 1 with the last egg.

Similarly, we can prove that no algorithm is faster Θ(n**(1/k)) by inducting from the k=2 proof.

Exact solution

We deduce the recurrence by optimising where to drop the first egg (floor g), presuming we know optimal solutions for smaller parameters. If the egg breaks, we have the g-1 floors below to explore with k-1 eggs. If the egg survives, we have n-g floors above to explore with k eggs. The devil chooses the worst for us. Thus for k>1 the recurrence

E(k,n) = min(max(E(k,n-g), E(k-1,g))) minimised over g in 1..n