ASF Puzzles & Conundrums Thread

[McLovin]

So the captain the question/answer asked by the captain/servant could be rephrased as "There is at least one blue eyed slave".

If there is one blue eyed slave, he will leave on the first day because he can only see two brown eyed slaves.

If there are two blue eyed slaves then they will not leave on the first day. From the point of view of a blue eyed slave he sees one brown eyed slave and one blue eyed slave but does not know his own eye colour. On the second day, because the other blue eyed slave has not left indicates to him that he must be blue eyed also; if the other blue eyed slave had seen two sets of brown eyes he would have left the day before.

If all three are blue eyed, then on the first and second day none of them would leave. On the third day, from the point of view of a blue eyed slave, he can see two other blue eyed slave. They have not left on the second day which must mean he also has blue eyes.

So they all leave on the third day knowing they have blue eyes.

 

[trainspotter]

What McLovin said ... when did Wednesday come into it?

 

[cynic]

What McLovin said ...

+1
McLovin did explain it better.

However, it seems pretty clear to me that you arrived at the solution first.

... when did Wednesday come into it?

I think it arrived just after the Tuesday following the preceding Monday!

 

[McLovin]

View attachment 62551

Of course! Why didn't I think of that! Just take your eyeballs out and look at them.

 

[bellenuit]

So the captain the question/answer asked by the captain/servant could be rephrased as "There is at least one blue eyed slave".

If there is one blue eyed slave, he will leave on the first day because he can only see two brown eyed slaves.

If there are two blue eyed slaves then they will not leave on the first day. From the point of view of a blue eyed slave he sees one brown eyed slave and one blue eyed slave but does not know his own eye colour. On the second day, because the other blue eyed slave has not left indicates to him that he must be blue eyed also; if the other blue eyed slave had seen two sets of brown eyes he would have left the day before.

If all three are blue eyed, then on the first and second day none of them would leave. On the third day, from the point of view of a blue eyed slave, he can see two other blue eyed slave. They have not left on the second day which must mean he also has blue eyes.

So they all leave on the third day knowing they have blue eyes.

Yes, McLovin and cynic, you are correct. This is perhaps the most concise explanation I have come across and thanks for saving me having to write up a solution.

This is a subset of very controversial puzzle and many do not agree with the answer. The original has more than just three islanders. However, if you don't accept the above answer for just 3 slaves, you will not accept it for more than 3. Equally, if you accept it for 3, you will see how it can logically be expanded to more than 3.

The original has a large number of slaves and they can be brown eyed and (at least one) blue eyed. The solution is basically the same, except you continue the logic beyond 3 to the total number of blue eyed slaves. The generalised result is that if there are x blue eyed people on the island, irrespective of the number of brown eyed people, then they will all leave on day x.

One person who disagrees with the result has compared the problem to this other one.

It's a Sunday. A very smart and logical prisoner is to be executed at midday on one of the 5 weekdays that follow and is told by the executioner that he (the prisoner) will not be able to determine beforehand what day this will be. The prisoner is very pleased on hearing this and says that means you can't execute me at all. When asked why is that, he replies: "You cannot execute me on Friday, since that is the last execution day, so I would know beforehand on Friday morning that Friday is going to be the day. If you cannot execute me on Friday, then Thursday is the last day available, but then I would know Thursday morning that Thursday is to be the day, so it can't be Thursday either." He continues using the same logic to rule out the possibility of being executed Wednesday, Tuesday or Monday.

To his surprise he is taken out Wednesday and executed.

So like the blue eyed problem, the solution appears logical and consistent with the problem parameters, but we feel there is a gotcha somewhere there.

 

[trainspotter]

TRUE OR FALSE

Are all Caucasian babies born with blue eyes?



I can't believe my logic did not make sense bellenuit?

 

[skc]

Yes, McLovin and cynic, you are correct. This is perhaps the most concise explanation I have come across and thanks for saving me having to write up a solution.

This is a subset of very controversial puzzle and many do not agree with the answer. The original has more than just three islanders. However, if you don't accept the above answer for just 3 slaves, you will not accept it for more than 3. Equally, if you accept it for 3, you will see how it can logically be expanded to more than 3.

The original has a large number of slaves and they can be brown eyed and (at least one) blue eyed. The solution is basically the same, except you continue the logic beyond 3 to the total number of blue eyed slaves. The generalised result is that if there are x blue eyed people on the island, irrespective of the number of brown eyed people, then they will all leave on day x.

One person who disagrees with the result has compared the problem to this other one.

It's a Sunday. A very smart and logical prisoner is to be executed at midday on one of the 5 weekdays that follow and is told by the executioner that he (the prisoner) will not be able to determine beforehand what day this will be. The prisoner is very pleased on hearing this and says that means you can't execute me at all. When asked why is that, he replies: "You cannot execute me on Friday, since that is the last execution day, so I would know beforehand on Friday morning that Friday is going to be the day. If you cannot execute me on Friday, then Thursday is the last day available, but then I would know Thursday morning that Thursday is to be the day, so it can't be Thursday either." He continues using the same logic to rule out the possibility of being executed Wednesday, Tuesday or Monday.

To his surprise he is taken out Wednesday and executed.

So like the blue eyed problem, the solution appears logical and consistent with the problem parameters, but we feel there is a gotcha somewhere there.

Thanks Bellenuit... your questions always managed to lose me a few hours of precious sleep time!

I am still trying to get my head around the answer. Yes it seems to make sense but here's what I can't figure out.

The only useful information from the overheard conversation is that there is at least one blue eye slave on the island, and that each slave knows that. However, this is not at all different to the situation BEFORE the conversation took place. Each slave sees two pairs of blue eyes... therefore each know themselves that there is at least one pair of blue eye. At the same time, each is certain that the other 2 slaves see at least one pair of blue eye. So... the captain's conversation has added no new information.

If I was to follow this logical... how can no new information suddenly trigger a sequence of inductions? I am not questioning the answer... but I wonder if someone can offer an explanation to this?

 

[cynic]

...
If I was to follow this logical... how can no new information suddenly trigger a sequence of inductions? I am not questioning the answer... but I wonder if someone can offer an explanation to this?

It does provide new information!

Although we know that each slave is able to recognise the presence of at least one blue eyed slave, the slaves don't know that their fellows have this same awareness until after all heard the question and answer regarding all slaves eye colour not being brown!

 

[bellenuit]

I can't believe my logic did not make sense bellenuit?

Tr,

You obviously were correct to say the third day, but I just couldn't see how your answer arrived at that conclusion.

There is the possibility that the slaves think that one of them has brown eyes but by the third day they would have deduced they all have blue eyes because one slave can see the other 2 have blue eyes. The second slave can see the other 2 have blue eyes. The third slave can see the other 2 have blue eyes so with CERTAINTY they can say that they ALL have blue eyes.

I'm sorry if I missed something in your answer that I should see.

Although McLovin set the answer out in a concise manner, the actual decision process (assuming the slaves are A, B and C) that arrives at the answer involves a nested set of hypotheses.

A hypothesises that he (A) has brown eyes and then puts himself in B's shoes assuming his hypothesis is correct. He then is thinking since B is evaluating all possibilities to deduce his (B's) eye colour, then B would, as part of his (B's) deductive process, also hypothesise that he (B) has brown eyes. So if both hypotheses are correct, C must know that his eye colour has to be blue as both A and B are brown and there is one blue eyed slave. So C would have left the island on Monday.

Since C didn't leave the island on Monday, B must know his hypothesis that he (B) has brown eyes is incorrect, so he would have left the island on Tuesday (he obviously had to wait until Monday was by to be sure C didn't leave).

But B leaving the island on Tuesday is predicated on A's hypothesis that he (A) has brown eyes. So when B didn't leave the island on Tuesday, A now knows that his hypothesis about the colour of his own eyes is incorrect, so A must have blue eyes. He must wait until Tuesday is by to be sure B has not left, so he leaves on Wednesday. Since each of the three slaves will have used the same deduction process they will each conclude on Wednesday that each have blue eyes and will thus all leave on Wednesday.

While I could see the above reasoning being implicit in Cynics and McLovin's responses, though not stated that way, I could not see that being implicit in your answer. Sorry if that wasn't the case.

 

[cynic]

It does provide new information!

Although we know that each slave is able to recognise the presence of at least one blue eyed slave, the slaves don't know that their fellows have this same awareness until after all heard the question and answer regarding all slaves eye colour not being brown!

Just to further elaborate, it's not just about knowing that there are blue eyed slaves present, it's about knowing that all slaves know this and that all slaves know that all slaves know!!

 

[trainspotter]

It does provide new information!

Although we know that each slave is able to recognise the presence of at least one blue eyed slave, the slaves don't know that their fellows have this same awareness until after all heard the question and answer regarding all slaves eye colour not being brown!

It's called "common knowledge". Prior to this they were not allowed to discuss their eye colour and had no way of knowing their own as their were no reflective materials to look into. Once the official asked the question and the captain answered in the affirmative then it became "common knowledge". Therefore they could deduce that if by the third day no one had left the island then with CERTAINTY they knew they all had blue eyes as they could see each others eye colour. I think that's how it goes?

Or so Wiki reckons:-

What's most interesting about this scenario is that, for k > 1, the outsider is only telling the island citizens what they already know: that there are blue-eyed people among them. However, before this fact is announced, the fact is not common knowledge.

For k = 2, it is merely "first-order" knowledge. Each blue-eyed person knows that there is someone with blue eyes, but each blue eyed person does not know that the other blue-eyed person has this same knowledge.

For k = 3, it is "second order" knowledge. After 2 days, each blue-eyed person knows that a second blue-eyed person knows that a third person has blue eyes, but no one knows that there is a third blue-eyed person with that knowledge, until the third day arrives.

In general: For k > 1, it is "(k − 1)th order" knowledge. After k − 1 days, each blue-eyed person knows that a second blue-eyed person knows that a third blue-eyed person knows that.... (repeat for a total of k − 1 levels) a kth person has blue eyes, but no one knows that there is a "kth" blue-eyed person with that knowledge, until the kth day arrives. The notion of common knowledge therefore has a palpable effect. Knowing that everyone knows does make a difference. When the outsider's public announcement (a fact already known to all) becomes common knowledge, the blue-eyed people on this island eventually deduce their status, and leave

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

 

[skc]

Just to further elaborate, it's not just about knowing that there are blue eyed slaves present, it's about knowing that all slaves know this and that all slaves know that all slaves know!!

But they do know beforehand. A sees B and C having blue eyes. A also knows that B sees C with blue eyes. A also knows that C sees B with blue eyes. Therefore, A knows that all slaves know there is at least one blue eye slave on the island. It is already common knowledge before the captain has spoken...

What am I missing still..

 

[bellenuit]

I am still trying to get my head around the answer. Yes it seems to make sense but here's what I can't figure out.

The only useful information from the overheard conversation is that there is at least one blue eye slave on the island, and that each slave knows that. However, this is not at all different to the situation BEFORE the conversation took place. Each slave sees two pairs of blue eyes... therefore each know themselves that there is at least one pair of blue eye. At the same time, each is certain that the other 2 slaves see at least one pair of blue eye. So... the captain's conversation has added no new information.

skc, it took me a while to get my head around that too, but then I realised it is providing new information. Some of the others have explained why it provides new information. This is my way of putting it. If you look at McLovin's answer:

If there is one blue eyed slave, he will leave on the first day because he can only see two brown eyed slaves.

If there are two blue eyed slaves then they will not leave on the first day. From the point of view of a blue eyed slave he sees one brown eyed slave and one blue eyed slave but does not know his own eye colour. On the second day, because the other blue eyed slave has not left indicates to him that he must be blue eyed also; if the other blue eyed slave had seen two sets of brown eyes he would have left the day before.

If all three are blue eyed, then on the first and second day none of them would leave. On the third day, from the point of view of a blue eyed slave, he can see two other blue eyed slave. They have not left on the second day which must mean he also has blue eyes.

Each of these paragraphs are predicated on the one before it being true. For the 3 blue eyed slaves scenario, they leave on the third day because if there were only two they would have left on the second day and they didn't. But for the 2 blue eyed slaves scenario, they leave on the second day because if there were only one he would have left on the first day, but he didn't. For the one blue eyed slave only, he needs that external information to arrive at a deduction that he has blue eyes. In other words, if there was just one blue eyed person on the island, without hearing what the official said, he would not be able to deduce what colour eyes he has. It is on hearing that piece of information from the official that he could determine his eye colour, which then allowed the two blue eyed scenario to be resolved which in turn allowed the 3 blue eyed scenario to be resolved.

I mentioned that this is a controversial problem and many don't accept the conclusion. IMO they are wrong. The conclusion is correct.

 

[cynic]

But they do know beforehand. A sees B and C having blue eyes. A also knows that B sees C with blue eyes. A also knows that C sees B with blue eyes. Therefore, A knows that all slaves know there is at least one blue eye slave on the island. It is already common knowledge before the captain has spoken...

What am I missing still..

Take it a step or two further.

A,B and C already know that each of their counterparts know that each of them can see at least one blue eyed slave.

What each of them doesn't know beforehand is that all slaves know that all slaves know that each will see at least one blue eyed slave!

 

[keithj]

But they do know beforehand. A sees B and C having blue eyes. A also knows that B sees C with blue eyes. A also knows that C sees B with blue eyes. Therefore, A knows that all slaves know there is at least one blue eye slave on the island. It is already common knowledge before the captain has spoken...

What am I missing still..

Try thinking about it with only 2 blue eyed slaves....
There's 4 possible scenarios -
Brown Brown
Blue Brown
Brown Blue
Blue Blue

They both can eliminate 1st scenario. And each can eliminate either scenario 2 or 3. So each has to consider 2 possible (but different) scenarios.
Then the captains makes a seemingly redundant comment that there is at least one blue eyed....
If the 2nd or 3rd scenario is correct then one slave will know the exact scenario and would leave (because they can see only a brown eyed slave & therefore they must be the blue eyed one).

As no-one leaves each slave knows the other has eliminated a scenario on their behalf.

Add a 3rd slave and an extra day.....

 

[skc]

Try thinking about it with only 2 blue eyed slaves....
There's 4 possible scenarios -
Brown Brown
Blue Brown
Brown Blue
Blue Blue

They both can eliminate 1st scenario. And each can eliminate either scenario 2 or 3. So each has to consider 2 possible (but different) scenarios.
Then the captains makes a seemingly redundant comment that there is at least one blue eyed....
If the 2nd or 3rd scenario is correct then one slave will know the exact scenario and would leave (because they can see only a brown eyed slave & therefore they must be the blue eyed one).

As no-one leaves each slave knows the other has eliminated a scenario on their behalf.

Add a 3rd slave and an extra day.....

OK... I think the above explained it.

What they used to know was that, they know other knows that there is at least one blue eye.

What they didn't used to know, was whether the blue-eye slave that other knows was the same slave.

Tricky.

 

[bellenuit]

This is a lot easier than the blue-eyed island puzzle.

The Special Room

A special room is on the 4th floor of a big house. The room has been locked for the last 50 years (a long story, that's what makes it special) and contains 3 light fittings at head height. There is one on each wall except the wall where the the door is. Each light fitting is controlled by a separate switch (one for each light), but due to an architectural oddity the switch panel is on the ground floor by the entrance to the house.

You have just bought the house and come to collect the keys from the agent late one afternoon. The agent explains about the special room. She checked it that morning and being the first to enter it in 50 years found it obviously very musty but otherwise OK. She explains about the light fittings and said she checked them out earlier that day and all the globes were working. On the way out she shows you the panel by the front door that has the three switches. Then she leaves. You check out the house and when you get to the special room, you realise that you didn't turn on the light switches. It is almost dark, but you can make out the 3 light fittings.

You head down to turn the lights on for the special room and you realise the agent forgot to tell you which switch controls which light. Not having your phone with you and being lazy and logical, you decide to work it out for your self.

The problem is: with all switches initially off and knowing all the light globes are working, what is the least number of trips you need to make from the switch panel to the special room on the 4th floor to determine which switch controls which fitting. Explain your answer.

You can assume that no special equipment is to be used, you are alone without help and that you cannot determine from outside the house or anywhere else in the house the status of a light in the special room other that going to the special room. There are no tricks to the question.

 

[Boggo]

You head down to turn the lights on for the special room and you realise the agent forgot to tell you which switch controls which light. Not having your phone with you and being lazy and logical, you decide to work it out for your self.

The problem is: with all switches initially off and knowing all the light globes are working, what is the least number of trips you need to make from the switch panel to the special room on the 4th floor to determine which switch controls which fitting. Explain your answer.

This is assuming that you are very fit you could do it in one trip.
You could turn on two of the switches and allow the lights to warm up, then switch one off and run up the stairs like buggery before the one you had on cooled down.
You should now have one on, one hot and the one that hasn't been switched on would still be cold ???????

 

[pixel]

This is assuming that you are very fit you could do it in one trip.
You could turn on two of the switches and allow the lights to warm up, then switch one off and run up the stairs like buggery before the one you had on cooled down.
You should now have one on, one hot and the one that hasn't been switched on would still be cold ???????

Smart thinking, 99.
Methinks you got it in one

 

[Boggo]

Smart thinking, 99.
Methinks you got it in one

Not sure yet, that 4th floor, big house bit could be a factor