The 2 Eggs Problem: Cracking the Code of Optimal Strategy

Imagine you’re standing in front of a 100-story building. In your pocket are two identical eggs. These eggs have a peculiar property: they will break if dropped from a certain floor or higher, but remain intact if dropped from any floor below that.

The challenge?
Find the highest floor from which you can drop an egg without it breaking.
The catch?
You only have two eggs, and you want to minimize the worst-case number of drops.

This classic 2 Eggs Problem is a brilliant puzzle from the world of algorithms and optimization. It’s all about strategy, efficiency, and minimizing risk. Let’s crack it open.

The Rules of the Game

• If an egg breaks when dropped from a floor, it would break from any higher floor.
• If an egg doesn’t break, it wouldn’t break from any lower floor.
• Both eggs are identical.
• The goal is to find the critical floor (the tipping point) with the minimum number of drops in the worst case.

The Brute Force Approach

You could start at the 1st floor and move up one by one, dropping an egg at each floor until it breaks.
But that could take up to 100 drops in the worst case.
Surely, we can do better!

The Smart Approach: Minimizing the Worst Case

Here’s where the strategy gets clever.

1. You balance the risk by choosing floors wisely.
2. With the first egg, you take bigger leaps to narrow down the range.
3. Once the first egg breaks, you use the second egg to do a linear search in the remaining floors.

But how do you choose which floors to test first?

The Math Behind It

You need a strategy that minimizes the worst-case number of drops.
This can be done by making your first drop from floor x, your second from x + (x - 1), your third from x + (x - 1) + (x - 2), and so on.

This ensures that after each drop, you cover fewer floors, accounting for fewer remaining attempts.

You need the sum of floors tested to at least cover all 100 floors:

x + (x - 1) + (x - 2) + … + 1

This is the same as:

(x * (x + 1)) / 2 >= 100

Now, solve for x.

Solving the inequality:

x * (x + 1) >= 200

Try x = 14:

14 * (14 + 1) = 14 * 15 = 210

So, x = 14 works.

This means the first drop should be from the 14th floor, then you move up by 13, then 12, and so on.

Step-by-Step Example

Here’s what the sequence of floors looks like:

You stop once you reach or exceed 100.

How it works:

• Drop the first egg at floor 14.

  • If it breaks, you now search floors 1-13 with the second egg (up to 14 drops in total).
  • If it doesn’t break, move to floor 27.

• If it breaks at 27, search 15-26 with the second egg.

• Continue following the increments until you find the critical floor.

Why It Works

By decreasing the interval by 1 each time, you balance the number of drops:

• If the egg breaks early, you have more drops left to search the lower floors.
• If it breaks later, you’ve eliminated many floors already.

The worst-case number of drops is 14.
This is much better than the 100 drops you’d have with a brute-force approach.

Generalizing the Problem

The 2 Eggs Problem is a simplified case of the Egg Dropping Puzzle, which can have:

• More floors
• More eggs

With more eggs, you can optimize the strategy even further using dynamic programming or mathematical recursion.

For k eggs and n floors, there’s a general solution that minimizes the number of drops by balancing these factors.

The Takeaway

The 2 Eggs Problem teaches us:

• How to balance risk and reward when information is incomplete.
• How mathematical reasoning can minimize risk and optimize outcomes.
• How algorithms help us make better decisions, even outside of computer science.

It’s a fun and insightful way to appreciate how mathematics meets strategy in problem-solving.