Puzzles to puzzle you

[Ashutosh]

Here's a hint for blue eyed puzzle: Consider the same problem with instead of 900/100 brown/blue eyed people, 999/1 brown/blue eyed people. Too easy? Now try 998/2. What about 997/4? Can you see something now?

[jitendragarg]

Already thought of it that way. For 999/1 its fairly simple.
Now, for 998/2, there are 2 persons, A and B. Person A will see colour of B's eyes, and he will think that foreigner was talking about B, and vice-versa, meaning no suicides. Similar logic can be extended for 900/100 eye colour ratio. I know my logic may be faulty, and I am missing something very basic here, but till now, I am still happy with my solution.

[Ashutosh]

For the case 998/2, here's what happens:

Say A and B have blue eyes. Before the vistor's statement, this is what A might have been thinking:

"This dude B has very pretty blue eyes. I wonder if I have such pretty eyes."

B's thoughts are completely symmetric.

Now what can change after visitor's statement? Here are A's thought just after the statement was made:

"Do I have blue eyes? Let's see. Say I don't. Then, I know that B is the only guy with blue eyes. Clearly he now knows that since that idiot visitor made it known that there is someone with blue eyes. So he should kill himself tomorrow. That would be so awesome to watch ."

B's thoughts are entirely symmetric. So both A and B wait for the ritualistic suicide. But the fateful moment comes and no one commits suicide. Here's what A's thinking now:

"! Why isn't B dead? Surely because I do have blue eyes and he was waiting thinking that I might die today. Crap! Now I know that I have blue eyes. I'm shooting myself."

B thinks the same and they both immediately commit suicide.

Can you extend the argument to the case 900/100?

[jitendragarg]

Lets do this. Extend your argument, to three person. Say, A and B are blue eyed, and C is not. After the visitor's statement, what will C thinks?

"A and B both have blue eyes, so it will be awesome to watch them die." At the same moment A and b are both thinking other will commit suicide.

Next day, C finds that none of them commited the suicide yet. So, what will he think? Exactly the same as A and B, that he might be also blue eyed. So, by that logic, C should also commit suicide, since no one will tell him that he is not blue eyed.

So, till the time, there is either A or B's suicide, every C person thinks he might be blue eyed and kill himself. Only when A or B will kill himself, then only they will know that the blue eyed person mentioned by visitor is dead now. Till then every brown eyed person will commit suicide.

Also, in that case, as soon as A is dead, B will know that blue eyed person is dead, so, he will think he is not the blue eyed person, unless mentioned by someone else.

Result, B will still live, and since he is still alive every other C will keep commiting suicide. So, the whole group will kill themselves, before B realises that he is the last standing blue eyed person.

So, by that logic, every person will be dead, and not just the blue eyed person.


Now, you can similarly expand my logic to 900/100 people. There will be every time a blue eyed person who thinks he is brown eyed, since all other blue eyed people are dead. So, by the logic you suggested, every person will die, which was not mentioned in any of the situations in the question. So, technically, the solution should be "everyone is dead", and not "all blue eyed people are dead".

[Ashutosh]

A correction. In the case 998/2, both blued eyed men commit suicide 2 days after the visitor's speech (and not 1 day) since the rule says you die a day after you know your eye colour. In the case 900/100, it is exactly 100 days after which all blue eyed people commit suicide all together. To prove this properly we may use induction. For instance in the case 997/3, say blued eyed men are A, B and C. Then A will argue thus:

"Say I didn't have blue eyes. Then A and B will use the 998/2 argument and I should see them die in 2 days from now."

B and C have similar thoughts. But 2 days later no one dies so A, B and C simultaneously deduce that they've blue eyes. Hence on the third day they kill themselves. You can see now how induction works.

The reason why the brown eyed people don't kill themselves is simple. This argument will not work if the visitor hadn't made his statement. The brown eyed men still don't share the "common knowledge" that there are brown eyed people on the island. To see the meaning of common knowledge more clearly, suppose there were only 2 brown eyed people say A and B on the island. Denote by S, the statement "There is at least one brown eyed person on the island". Then while A knows S and B knows S, A doesn't know that B knows S and B diesn't know that A knows S. This is the crux of matter. What the visitor did was that he made S common knowledge.

[Ashutosh]

I forgot to mention that the fact that everyone on the island has either blue or brown eyes isn't known to the natives. Of course, after 99 days all blue eyed people learn this fact but being true men they do not disclose this to the brown eyed men.

[jitendragarg]

I really don't get it. How does it denies my logic? Consider the scenario, that no one knows that there are actually brown eyed or blue eyed people, as you mentioned in your logic. In that case, a person X, no matter what his eye colour is, will still know that there are both brown eyed and blue eyed people, since they can actually see other. And since none of them is blind, person X can easily deduce, that person Y has seen other people's eye colour too. Saying that is not the case, is a remote possibility of having a particular tribe where no one actually ever know any other person of the tribe. And for all possible deductions, normal scenarios are considered, and not some remote exceptions.

So, the answer you gave me, gets invalid at the exact moment when you say, and i quote,

The brown eyed men still don't share the "common knowledge" that there are brown eyed people on the island. To see the meaning of common knowledge more clearly, suppose there were only 2 brown eyed people say A and B on the island. Denote by S, the statement "There is at least one brown eyed person on the island". Then while A knows S and B knows S, A doesn't know that B knows S and B diesn't know that A knows S. This is the crux of matter. What the visitor did was that he made S common knowledge.

Consider the same scenario for three people. Since A, B, C, all knows statement S. And A knows that B and C know each other's eye colour too. So, its clear that A can deduce the simple fact that B and C know the statement S too. Your argument is valid in case of two people. But for any no. of people more than 2, the statement is invalid. So, visitor did nothing exactly. He didn't made anything a common knowledge, as it already was. Only thing he did was to make them talk about the eye colour.

[Ashutosh]

I didn't define common knowledge precisely. Instead of typing it in my words, I'll give you the wikipedia link on common knowledge and a version of the island puzzle:

http://en.m.wikipedia.org/wiki/Common_knowledge_(logic)

[jitendragarg]

Ah! Leave it. That wiki article is missing a very simple logic, but I guess we have discussed it way too far, so lets continue with other puzzles.

[Ashutosh]

Here's another puzzle:

A person X chooses a number between 1 and 16 - so he has 16 choices. You have to guess the number he chose. You can ask him 7 yes/no questions.

Such a question may be:

"Is your number bigger than or equal to 9?"

X is an honest person and won't lie more than once. He may be too honest and could even answer all 7 questions correctly. You don't know beforehand. At the end of these 7 questions, you have to tell X what number he chose.

So the question is: Find a strategy to do this.

If this seemed too easy, show that 6 questions aren't enough to do the job.