mathematics - Exponential time in a Blue Eyes variant

I created a variant of the Blue Eyes problem where the logic took 4901 days (over 13 years) to totally play out. I thought this was nice, but I was hoping for an even longer period of time.

Of course, I could trivially I could do the following: Suppose $n$ is the number of inhabitants, and the maximum size number on their heads is $n$, and they each have forehead number either $\lfloor n/2 \rfloor$ or $\lfloor n/2 \rfloor + 1$. Then the amount of time it takes to discover their numbers is roughly proportional to $n^2/2$. Thus, by increasing $n$ I can increase the amount of time quadratically. But I was wondering if it was possible to do better yet.

Is there a way to design a Blue-Eyes-like logic puzzle where the amount of time for the islanders to figure out the answer is exponential in the size of the problem?

A couple notes

• I don't know if this is possible -- my attempts so far haven't done very well. Even rough ideas or explanations as to why it might not be possible would be appreciated.


• Anything about the problem can be changed I'd say: for example, perhaps not everyone can see everyone else. However, the core mechanic of ever increasing common knowledge should remain. Ideally, it would also be a simple to state puzzle, and perhaps with some other twists and turns for any potential puzzle solver.


• Suppose the islanders are immortal -- just in case the solution to your puzzle outpaces a person's expected lifetime.

Answer

This might not qualify as a blue-eyes puzzle because it does not use common knowledge, but it involves chains of deductions based on nothing happening for a particular amount of time:

$n$ villagers wear either black or white hats. They sit in a line, so that each villager can see all the hats in front of them, but not the hats behind them. If a villager ever deduces their own hat color, they get up and leave the line at noon on the next day (without anyone in front noticing).

One day, a light is installed where all villagers can see it. The light remains on as long as at least one villager in the line has a black hat.

If the villagers all have black hats, nothing happens for the first $2^{n-1}-1$ days. On the $2^{n-1}$th day, the first villager leaves. $2^{n-2}$ days later, the second villager leaves. This continues until the last villager leaves on the $2^n-1$th day.