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A particular island has 100 inhabitants. None of them know their own eye colour. They are all perfect logicians. Every night at midnight, a ferry stops on the island, and every islander who has figured out their own eye colour leaves. Every islander knows the eye colour of every other islander, and knows exactly who is still on the island, but they don't otherwise communicate. All the islanders know all the rules in this paragraph.

As it happens, every islander has blue eyes.

On the summer solstice, at noon, it becomes common knowledge among the islanders that at least one person on the island has blue eyes.

So far this sounds almost exactly like the classic "hardest logic puzzle in the world", and we might assume that nobody will leave the island until the 100th visit of the ferry, at which point all islanders leave.

But at noon exactly 50 days after the summer solstice, the island volcano erupts, killing 50 of the islanders.

Assuming no further unforeseen events, do the remaining islanders make it off the island, and if so, when?

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  • $\begingroup$ Isn't that the remaining islander without -s? On start of day 49, 52 islanders remain. One of them leaves before midnight, so 51 remain. If 50 die, one remains. Right? $\endgroup$ Aug 13, 2021 at 7:54
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    $\begingroup$ @GeorgeMenoutis what makes you think that any islanders leave before day 50? That doesn't happen in the solution to the original eruptionless version of this puzzle. $\endgroup$
    – fblundun
    Aug 13, 2021 at 7:58
  • $\begingroup$ Ahh, I'm so rusty. Scratch that. $\endgroup$ Aug 13, 2021 at 8:01
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    $\begingroup$ This is a novel and interesting variant of the puzzle! $\endgroup$
    – justhalf
    Aug 15, 2021 at 6:21

3 Answers 3

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My head hurts a little from thinking about this but I think this is what happens

Fifty days after the summer solstice, there have been $50$ ferry visits which means that, for any ordering of the islanders, $I_1, I_2, \ldots, I_{100}$,
$I_1$ knows that $I_2$ knows that $I_3$ knows that...$I_{49}$ knows that $I_{50}$ knows that there are at least $51$ islanders with blue eyes.
When the volcano erupts, killing $50$ islanders, we are now in the situation where for any ordering of the islanders, $I_1, I_2, \ldots, I_{50}$,
$I_1$ knows that $I_2$ knows that $I_3$ knows that...$I_{49}$ knows that $I_{50}$ knows that there is at least one blue-eyed islander and so we are essentially back in the situation where the oracle appears on the first day to $50$ blue-eyed islanders.
Hence, on the $50$th night after the eruption, all remaining islanders will leave (provided there are no further fatal eruptions).

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I agree with hexomino’s answer overall:

On the 50th night after the eruption, all survivors leave.

but I suggest a hopefully clearer explanation:

As in the original puzzle, the key point is: If one day it’s common knowledge that at least $k$ islanders have blue eyes (for $k≥1$), but nobody leaves that night, then the next day, it’s common knowledge that at least $k+1$ islanders must have blue eyes. (If there had been exactly $k$, then those with blue eyes would have seen only $(k-1)$, and so would have known they were blue-eyed themselves.)

So (until the eruption) on the $n$th day after the solstice, it’s common knowledge that at least $n+1$ people have blue eyes. In particular, on the 50th day after the solstice, it’s common knowledge that at least 51 people have blue eyes. So even after 50 are killed by the eruption, it’s common knowledge that at least one blue-eyed islander survives. Now just like before, the known number will tick up day by day, and by noon 49 days after the eruption, the 50 survivors will all know that they all have blue eyes. So at midnight that night — the 50th midnight since the eruption — everyone will leave.

Technical point: “X is common knowledge” is a useful piece of terminology for this puzzle — it means “everyone knows X, and everyone also knows that X is common knowledge”, so it implies “everyone knows that everyone knows X”, and so on.

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  • $\begingroup$ I think this is the most intuitive way of seeing the answer. $\endgroup$
    – fblundun
    Aug 13, 2021 at 21:46
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My answer:

The Islanders will never leave.

The shortest solution to the original puzzle is a proof-by-induction of: "If there are N blue-eyed Islanders, they will leave on day N". (We prove that if it's true for N, it's true for N+1, and also that it's true for the base case of N=1.)

The significance of the volcano is:

It breaks the induction for N=51. If there are 51 blue-eyed Islanders, then each will think, "I see 50 blue-eyed Islanders. If they don't leave on day 50, then I know that my eyes are blue and I can leave on day 51." But if the volcano kills 50 of the blue eyed Islanders at noon on day 50, then the survivor can't know whether or not they would have left at midnight, so he can't deduce his own eye color on day 51.

This breaks the induction for all future stages. If there were 52 blue-eyed Islanders, then they could no longer rely on the assumption, "If there were 51 blue-eyed Islanders, then the surviving one would have left on day 51," so they can't deduce their eye color on day 52.

EDIT: The above answer is wrong because

I miscounted days. If there had been 50 blue-eyed Islanders, they would have left on the midnight before the eruption, so the induction wouldn't be broken.

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    $\begingroup$ Even if the day count is right, I suppose they would just restart the induction from 1? $\endgroup$
    – justhalf
    Aug 14, 2021 at 5:06
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    $\begingroup$ I don't think the induction can restart. A simpler version is the case where there are two blue-eyed people on the island, and the oracle says "There is at least one person with blue eyes", and then one of the two islanders immediately drops dead of a heart attack. The second one has no way of deducing that there's still a blue-eyed person. $\endgroup$
    – Mazement
    Aug 14, 2021 at 23:26
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    $\begingroup$ Ah, you're right! Also this fits with Peter's answer, that if after eruption you still have at least one commonly-known blue-eyed person, the induction can continue, otherwise no. $\endgroup$
    – justhalf
    Aug 15, 2021 at 6:20

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