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Small-Scale Question Sunday for August 11, 2024

Do you have a dumb question that you're kind of embarrassed to ask in the main thread? Is there something you're just not sure about?

This is your opportunity to ask questions. No question too simple or too silly.

Culture war topics are accepted, and proposals for a better intro post are appreciated.

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So I'm a few days late, but I'm struggling with the riddle. I can't get it to work out in my head.

I know the answer is that all n blue-eyed people leave on the nth ferry. I understand why that works for low values of n, and how it builds on itself, as each blue-eyed person expects the (n-1) scenario to play out, and then when it doesn't they realize they have blue eyes.

But at some point, everyone can see multiple blue-eyed people. And everyone can see that everyone else can see multiple blue-eyed people. At that point I can't see how the Guru has provided anyone with any new information or any new common knowledge. Nor has the empty outbound ferry on Day 1. So I don't understand how all the blue-eyed people know to leave on Day 100.

The note provides the inductive base case.

(blue, brown, note falls from the sky saying someone has blue eyes)

(1, 0, False): No information on their eyes. They never leave.

(1, 0, True): No one else could possibly have blue eyes. They leave on day 1.

(1, 1, False): Same as (1, 0, False). No one leaves.

(1, n, False): Same as (1, n-1, False). No on leaves.

(2, 0, True): On day 1, each reasons that if they are brown in (1, 1, True), the other person will leave. The other person doesn't leave. They each leave on day 2.

(n, 0, True): On day n-1, each reasons that if they are brown in (n-1, 0, True), the other n-1 people will leave on day n-1. This doesn't happen. All n people leave on day n.

(2, 0, False): On day 1, each reasons that whether they are blue or brown in (1, 1, False), the other person will never leave. The other person does not leave. This gives no information. No one ever leaves.

Etc