Can you solve the egg drop riddle? - Yossi Elran

8,274,578 views ・ 2017-11-07

TED-Ed


Please double-click on the English subtitles below to play the video.

00:08
The city has just opened its one-of-a-kind Fabergé Egg Museum
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with a single egg displayed on each floor of a 100-story building.
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And the world's most notorious jewel thief already has her eyes on the prize.
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Because security is tight and the eggs are so large,
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she'll only get the chance to steal one
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by dropping it out the window into her waiting truck
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and repelling down before the police can arrive.
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All eggs are identical in weight and construction,
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but each floor's egg is more rare and valuable than the one below it.
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While the thief would naturally like to take the priceless egg at the top,
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she suspects it won't survive a 100-story drop.
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Being pragmatic, she decides to settle for the most expensive egg she can get.
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01:00
In the museum's gift shop, she finds two souvenir eggs,
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perfect replicas that are perfectly worthless.
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The plan is to test drop them
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to find the highest floor at which an egg will survive the fall
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without breaking.
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Of course, the experiment can only be repeated
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until both replica eggs are smashed.
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And throwing souvenirs out the window too many times
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is probably going to draw the guards' attention.
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What's the least number of tries it would take
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to guarantee that she find the right floor?
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Pause here if you want to figure it out for yourself!
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Answer in: 3
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Answer in: 2
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Answer in: 1
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If you're having trouble getting started on the solution,
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it might help to start with a simpler scenario.
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Imagine our thief only had one replica egg.
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She'd have a single option:
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To start by dropping it from the first floor
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and go up one by one until it breaks.
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Then she'd know that the floor below that
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is the one she needs to target for the real heist.
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But this could require as many as 100 tries.
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Having an additional replica egg gives the thief a better option.
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She can drop the first egg from different floors at larger intervals
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in order to narrow down the range where the critical floor can be found.
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And once the first breaks,
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she can use the second egg to explore that interval floor by floor.
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Large floor intervals don't work great.
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In the worst case scenario, they require many tests with the second egg.
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Smaller intervals work much better.
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For example, if she starts by dropping the first egg from every 10th floor,
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once it breaks, she'll only have to test the nine floors below.
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That means it'll take at most 19 tries to find the right floor.
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But can she do even better?
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After all, there's no reason every interval has to be the same size.
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Let's say there were only ten floors.
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The thief could test this whole building with just four total throws
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by dropping the first egg at floors four,
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seven,
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and nine.
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If it broke at floor four, it would take up to three throws of the second egg
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to find the exact floor.
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If it broke at seven,
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it would take up to two throws with the second egg.
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And if it broke at floor nine,
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it would take just one more throw of the second egg.
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Intuitively, what we're trying to do here is divide the building into sections
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where no matter which floor is correct,
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it takes up to the same number of throws to find it.
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We want each interval to be one floor smaller than the last.
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This equation can help us solve for the first floor we need to start with
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in the 100 floor building.
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There are several ways to solve this equation,
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including trial and error.
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If we plug in two for n, that equation would look like this.
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If we plug in three, we get this.
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So we can find the first n to pass 100
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by adding more terms until we get to our answer,
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which is 14.
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And so our thief starts on the 14th floor,
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moving up to the 27th,
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the 39th,
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and so on,
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for a maximum of 14 drops.
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Like the old saying goes, you can't pull a heist without breaking a few eggs.
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