Logical puzzles

[Magicwand]

i dont get this puzzle answer.
dont they already have that same info by looking at other's red spot?

[Redundant]

Magicwand

You can see that there is new information by looking at whether everyone knows the statement "If there were only one person with a red spot, would he know".

Without the visitor, they don't know this statement, but with the visitor's statement they do.

[HermanHiddema]

Magicwand

You can see that there is new information by looking at whether everyone knows the statement "If there were only one person with a red spot, would he know".

Without the visitor, they don't know this statement, but with the visitor's statement they do.

Another important aspect is that the tourist provides an element of timing. He synchronizes the islanders, so to speak. Suppose that the tourist had, instead of speaking, simply left a sign on the island saying "There is at least one person with a red spot on the island". And over the course of a few days or weeks, all islanders come upon this sign and read it. Then the puzzle wouldn't work anymore...

[Magicwand]

Magicwand

You can see that there is new information by looking at whether everyone knows the statement "If there were only one person with a red spot, would he know".

Without the visitor, they don't know this statement, but with the visitor's statement they do.

yes..only time the info is useful is when there is only 1 person.
but if more than 1 person then it is same info as tourist's info everyone are aware of the fact that there are at least one red spot.
if that is the case then there is only 1 possiblity for that island. there was only 1 red spot when tourist announce the info.

what am i missing?

[flOvermind]

what am i missing?

Let's assume there are 10 people with a red spot, and let's assume you are one of them.

You see 9 people with a red spot. That is, you know there are either 9 or 10 people with a red spot, since you don't know whether you have a red spot or not.

Now if I ask you "How many red spots are there?" you'll answer "9 or 10". Let's call that question A.

But what if I ask you "What would this other person with a red spot answer to question A?" (let's call that question B)? For all you know, this person might only see 8 red spots (you don't know that you have one yourself). And the person doesn't know either, so the answer has to be "either '8 or 9', or '9 or 10'". Of course, if you'd actually ask that person, the answer would be "9 or 10". But you can't know that, since you don't know you have a spot yourself.

Now what if I ask you "What would this other person with a red spot answer to question B?". And by the same reasoning, you'll have to answer "7,8,9 or 10". And again you can call that question C, and if I ask you what someone else would answer to question C, the answer will be "6,7,8,9 or 10".

You can play that game until you're down at an answer containing 0. But with comment of the tourist, everyone will know there is "at least 1" red spot, and everyone knows that everyone knows there is "at least 1" red spot, and everyone knows that everyone knows that, and so on... So before the tourist, while everyone might have known that there is at least 1 red spot, and also that everyone knew that there is at least one red spot, and so on, nobody could say whether (everyone knew that (everyone knew that (everyone knew that *repeat 10 times* (there is at least 1 red spot)))))))))). That's the additional information provided.

[tj86430]

Magicwand

You can see that there is new information by looking at whether everyone knows the statement "If there were only one person with a red spot, would he know".

Without the visitor, they don't know this statement, but with the visitor's statement they do.

yes..only time the info is useful is when there is only 1 person.
but if more than 1 person then it is same info as tourist's info everyone are aware of the fact that there are at least one red spot.
if that is the case then there is only 1 possiblity for that island. there was only 1 red spot when tourist announce the info.

what am i missing?

I think someone already explained this, but let me try:

- If there is only one person with the red spot, (s)he will leave after the announcement, since (s)he realises (s)he has red spot
- If there are exactly two persons with the red spot, each will see one person with the red spot. If that person was the only person with the red spot (s)he would leave after the announcement (see above). Because the other one didn't leave when it was due, they will both know they have red spots and leave the day after
- If there are exactly three persons with the red spot, each will see two persons with the red spot. If those persons were the only persons with the red spot they would leave after the announcement + 1 day (see above). Because the other two didn't leave when it was due, they will all know they have red spots and leave the day after
- etc

e: slow

[entropi]

Magicwand

You can see that there is new information by looking at whether everyone knows the statement "If there were only one person with a red spot, would he know".

Without the visitor, they don't know this statement, but with the visitor's statement they do.

yes..only time the info is useful is when there is only 1 person.
but if more than 1 person then it is same info as tourist's info everyone are aware of the fact that there are at least one red spot.
if that is the case then there is only 1 possiblity for that island. there was only 1 red spot when tourist announce the info.

what am i missing?

Magicwand, exactly that is the beauty of that puzzle.

If there are two red spots. Before the tourist says anything: Each one of them already knows that there is at least one red spot (all the non-red spots know that there are either two or three red spots, but that's not important).

But none of the red spots expect the other to leave, why? Let's call the red spots A and B.
A knows that B has a red spot.
But A also knows that B has no idea of his own spot. Therefore he does not expect B to leave next morning.

Once the visitor gives the information that there is at least one red spot, A learns the following:
If B is the only red spot, he will learn the color of his spot and leave the island next morning.

But the next morning B does not leave the island because he does not understand his color. This means that there must be at least one more red spot. Since A sees no red spot other than B, the only possibility is that A has a red spot himself.

So the information comes from the fact that A expects B to leave next morning but B does not leave.



Then imagine the case there is still one more red spot: A, B, C

A would think as follows:
I see only two red spots B and C. If I have a non-red spot, then B and C will leave in two days (for the reasons explained above).
But since B and C don't leave in two days, A understands that there must be at least one more red spot on the island, which cannot be anybody else than himself.
Thus, they all leave together after 3 days.
So goes on...

[Magicwand]

i think i understand..but it is profound!
much harder than go problem.

[entropi]

Another one:

Bandits fall upon a village with 100 inhabitants.

The bandit chief organizes a meeting with all of them and says:
"I have red and black hats. Tomorrow morning I will randomly put either a red or a black hat on each one of you. You won't be able to see your own hat but you will be able to see all the others.
Then I will ask each one of you one after the other the color of his head. Everybody will be able to hear his answer. He who knows his color will survive, otherwise he will be killed.

During the night, the logic master of the village comes up with an idea. They agree on something such that everyone survives with a probabilty of 99,5%. What is the idea?

[robinz]

Bandits and villagers:

It simply needs to be agreed that the first villager to be asked will say "red" if he sees, among the other 99, an odd number of red hats, and "black" if he seems an even number of red hats. (Or the other way round - the choice is arbitrary, although somewhat crucial for the unfortunate first villager!) Then this first villager has a simple 50-50 chance of giving he "correct" answer as regards his own hat colour, but the other 99 villagers can now each determine their own colour simply by counting those they can see (other than the first villager) and comparing it to the message given by the first one.

PS: isn't it nice of bad guys in these kind of puzzles to always actually pose a solveable puzzle, rather than just killing the lot of them

[entropi]

Bandits and villagers:

It simply needs to be agreed that the first villager to be asked will say "red" if he sees, among the other 99, an odd number of red hats, and "black" if he seems an even number of red hats. (Or the other way round - the choice is arbitrary, although somewhat crucial for the unfortunate first villager!) Then this first villager has a simple 50-50 chance of giving he "correct" answer as regards his own hat colour, but the other 99 villagers can now each determine their own colour simply by counting those they can see (other than the first villager) and comparing it to the message given by the first one.

PS: isn't it nice of bad guys in these kind of puzzles to always actually pose a solveable puzzle, rather than just killing the lot of them

correct

[HermanHiddema]

A rather more cruel bandit, one who laughs at the idea of giving his prisoners a 99.5% chance of all being let go, decides to put his prisoners through a rather more cruel puzzle

He tells his 100 prisoners that tomorrow, he will allow them a chance to win their freedom. He has placed in a room 100 boxes, marked on the outside with the numbers 1 through 100. In each box, he has put a piece of paper with the name of a single prisoner on it. Tomorrow, each prisoner will be given the chance to inspect the contents of at most 50 boxes. If he finds the piece of paper with his name on it in one of those 50 boxes, he has succeeded. If he does not find his name, he has failed. If any prisoner fails, all of them will die. If they all succeed, they will be set free.

The prisoners are allowed to devise a strategy tonight, but tomorrow they will be unable to communicate during the procedure. Prisoners cannot leave messages in any way. They cannot mark boxes, or leave them open, move or remove papers, fold papers, etc, etc. Each prisoner will find the room, the boxes and their contents in exactly the same state.

If each prisoner opens 50 boxes at random, he will have 1/2 = 50% chance of finding his name. Altogether, randomly opening boxes gives the prisoners a 1 in 2^100 chance of surviving. About 0.000000000000000000000000000008%. Long odds.

So now, it is up to you: Devise a strategy that maximizes the prisoners' survival chances...

[tj86430]

A rather more cruel bandit, one who laughs at the idea of giving his prisoners a 99.5% chance of all being let go, decides to put his prisoners through a rather more cruel puzzle

He tells his 100 prisoners that tomorrow, he will allow them a chance to win their freedom. He has placed in a room 100 boxes, marked on the outside with the numbers 1 through 100. In each box, he has put a piece of paper with the name of a single prisoner on it. Tomorrow, each prisoner will be given the chance to inspect the contents of at most 50 boxes. If he finds the piece of paper with his name on it in one of those 50 boxes, he has succeeded. If he does not find his name, he has failed. If any prisoner fails, all of them will die. If they all succeed, they will be set free.

The prisoners are allowed to devise a strategy tonight, but tomorrow they will be unable to communicate during the procedure. Prisoners cannot leave messages in any way. They cannot mark boxes, or leave them open, move or remove papers, fold papers, etc, etc. Each prisoner will find the room, the boxes and their contents in exactly the same state.

If each prisoner opens 50 boxes at random, he will have 1/2 = 50% chance of finding his name. Altogether, randomly opening boxes gives the prisoners a 1 in 2^100 chance of surviving. About 0.000000000000000000000000000008%. Long odds.

So now, it is up to you: Devise a strategy that maximizes the prisoners' survival chances...

Is the result of a single prisoner known to others immediately after (s)he has opened all his/her boxes? (i.e. if e.g. the first one fails, will the others immediately know that he has failed?)

[flOvermind]

Is the result of a single prisoner known to others immediately after (s)he has opened all his/her boxes? (i.e. if e.g. the first one fails, will the others immediately know that he has failed?)

That doesn't really matter, they can just assume everyone before them has succeeded.

Assume the other prisoners don't know the result of the previous prisoners. They can still safely assume that all of them succeeded. Because if they are wrong, their future actions can't affect the outcome anyway, so it doesn't matter what they do.

[tj86430]

Is the result of a single prisoner known to others immediately after (s)he has opened all his/her boxes? (i.e. if e.g. the first one fails, will the others immediately know that he has failed?)

That doesn't really matter, they can just assume everyone before them has succeeded.

Assume the other prisoners don't know the result of the previous prisoners. They can still safely assume that all of them succeeded. Because if they are wrong, their future actions can't affect the outcome anyway, so it doesn't matter what they do.

Ok, in that case my solution attempt:

The first 50 prisoners all open the same boxes (e.g. numbers 1-50, doesn't matter which as long as they are the same for all). If they all succeed, the rest will open the other 50 boxes