There exists an island on which all inhabitants of the island are perfection logicians. If a conclusion can be logically deduced, all will do it immediately.
No-one can communicate in any way with any other inhabitant of the island (no talking, to sign language, etc.)
No one on the island can see their own eyes or knows their own eye color. However, all inhabitants of the island can see and knows the eye color of all other inhabitants of the island.
All inhabitants of the island have one of three eye colors:
Some number (N) have green eyes
Some number (M) have brown eyes
One inhabitant, the Guru, has red eyes
M and N are both >= 100
Every night, just before midnight, a ferry comes to the island, and all island inhabitants that have figured out their own eye color board the ferry to leave the island.
Up until today however, no one has left the island. Today (Day 0) at noon, the Guru stands up in front of all the inhabitants of the island and is allowed to speak one phrase:
"I can see someone with green eyes"
Who leaves the island, and on what day(s)?
*edit* This puzzle is not my own. I will credit the source after the answer is revealed.
- Joined
- 29 Dec 08
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Originally posted by forkedknightI know this one, and I think you need to specify that the guru says this on day 0 and every day following (and of course, as a guru, is truthful).
There exists an island on which all inhabitants of the island are perfection logicians. If a conclusion can be logically deduced, all will do it immediately.
No-one can communicate in any way with any other inhabitant of the island (no talking, to sign language, etc.)
No one on the island can see their own eyes or knows their own eye color. Ho ...[text shortened]... ?
*edit* This puzzle is not my own. I will credit the source after the answer is revealed.
Originally posted by AThousandYoungThe Guru doesn't know his own eye color, and no-one knows how many (total) of each color of eyes there are, nor that there are 3 (or only 3) colors of eyes.
Assuming the Guru knows someone has red eyes, he will be the one to leave, because he knows nobody else has red eyes.
i.e., if I am Green-eyed inhabitant #14, I can see N-1 green-eyed people, M brown-eyed people, and 1 red-eyed person. I have no idea if my own eyes are green, brown, red, or possibly yellow.
(This all applies before day 0.)
Originally posted by forkedknightIf there were only one green-eyed person then that person would not be able to see any green-eyed people and would therefore deduce his own eye colour and leave on day 1. (The first midnight ferry)
There exists an island on which all inhabitants of the island are perfection logicians. If a conclusion can be logically deduced, all will do it immediately.
No-one can communicate in any way with any other inhabitant of the island (no talking, to sign language, etc.)
No one on the island can see their own eyes or knows their own eye color. Ho ...[text shortened]... ?
*edit* This puzzle is not my own. I will credit the source after the answer is revealed.
if there were 2 green-eyed people then they could each see only one green-eyed person and when that person did not leave (as above logic) would deduce that they had green-eyes and both leave on day 2.
Etcetera etcetera.
Therefore ALL the green-eyed people leave on day N.
The remaining people would know that they have either brown or red eyes but cannot leave without further information.