Interesting. Looks like it can accept multiple variations on each answer. I was wondering about that

I’m not actually going to do it, but I like the system design

Hmm.

Rainbow tables…

Yeah. It does seem counterintuitive, but it’s a result of the uncertainty that what they see is what others do. So they have to communicate a number, and the only way they can is leaving or not each night to count up to it.

I thought about it more and concluded that if the guru had said “I see only blue and brown eyed people” then everyone (but her) could leave the island using the same logic, regardless of how many of each color there was (greater than zero of course because otherwise she wouldn’t see that color). Same for any number of colors too as long as she lists them all and makes it clear that’s all of them and doesn’t include herself.

Same way it expands to two: When there are three blue eyes, then each of them guesses they might have brown or something and there could be only two blue on the island, in which case as described those two would have left on the second night.

But they didn’t… So there must be three total. Same with 4, the 3 you can see would have left on night 3 if each of them saw the other two not leave on night 2…

Leaving or not is the only communication, and what the guru really did was start a timer. It has to start at 1 even though everyone can see that there’s more than one simply due to the constraints of the riddle - if the guru were allowed to say ‘I see at least 50 blue eyed people’ then it would start at 50 because there’s no other fixed reference available. Everyone knows there’s either 99 or 100, but they don’t know which of those it is, so need a way to count to there. They also think everyone else can see anything from 98 to 101 depending, so it’s not as straightforward as thinking the count could start at 99.

What does it mean?

I think if I tried that, it would only be a day or two before I went digging into the source code.