How to Escape a Monster Using Calculus
Nick Berry of DataGenetics poses a puzzle of logic, strategy, and math. Since Halloween is coming up, it involves a monster. I think he's kind of cute, but that's not the point at all. He will eat you if he catches you!
Imagine you are sitting in a rowing boat in the middle of circular lake. Standing on the edge of the lake is a hungry monster. The monster wants to eat you. You want to escape.
You know that, if you can reach land, you can outrun the monster, but first of all, you need to row to shore. Here’s the issue: whilst you can outrun the monster on foot, he can run faster than you can row the boat. Much faster. The monster can run four times faster than you can row. He’s also a clever monster and will always do that smartest thing to attempt to catch you; moving around the shoreline, as needed, to maximize his advantage. If he touches you, you’re dead!
Can you escape?
Knowing how circles work and knowing that pi is less than four, rowing straight for the shore opposite the monster will not help you. For the purpose of this game, you cannot do something odd like jump out of the boat and swim, or call 911 for a helicopter. Yes, there is a way to do it. I was tempted to take his word for it, thinking I'd never understand the math involved, but as Nick explains the solution, it makes sense even if you didn't study calculus. Try to figure it out on your own first, then read the winning strategy at Datagenetics.
But wait, there's more! What if the monster could run even faster? Nick wondered exactly how much faster a monster would have to run than you could row to make it impossible for you to escape no matter how much calculus you used.
The fastest monster we can escape is determined when the monster arrives at the shore at exactly the same time we do. Let's define the ratio of our two speeds by the letter K. (The monster runs K-times faster than we can row). What is the maximum value for K?
Well, that involves more math, but the answer is there.
Check out previous wizardry from Datagenetics.
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