Logic/maths puzzles

[Supergrunch]

I think an important part of this is that the only time they can leave is at a set point each night when a boat comes to get them. You need the time period of the daily boat to work it out.

I guess so. Or you can just treat each day/night as a discrete entity.

[Will]

Oh yeah completely, the key thing is that it's not available to them whenever they want and there is a clear cycle to when they can and can't leave. It's not that they can suddenly decide to leave and go right then, which seemed to be the issue Spambot was having with the problem.

 

Anyone trying to work it out should probably ignore reading into this - I'm confusing myself now.

[MoogleViper]

All the blue eyed people would leave on day 100?

[Supergrunch]

All the blue eyed people would leave on day 100?

Correct. But why?

[MoogleViper]

Correct. But why?

 

Well I assume he says the same thing everyday. Therefore they know that there are 100 blue eyed people. Each blue eyed person can only see 99 blue eyed people. Therefore they will be blue eyed.

 

I assume that's correct.

[Supergrunch]

With the eye problem it's only just the first person that makes the statement and then nothing else.

Well I assume he says the same thing everyday. Therefore they know that there are 100 blue eyed people. Each blue eyed person can only see 99 blue eyed people. Therefore they will be blue eyed.

No' date=' the green eyed person [i']only[/i] makes the statement on the first day, and never again. And as I stated in the first post with the problem, the people aren't aware of the proportions of eye colours on the island. So a given blue eyed person can see one person with green eyes, 99 with blue, and 100 with brown, but as far as they know, their own eyes could be any of these colours or something else entirely. However, the statement made by the green eyed person somehow changes this.

[Zell]

Here's my take on the first puzzle if people can't get their heads around it.

 

You can treat the collision of two ants as simply two ants walking through each other. With that in mind, the first ant, a1, will walk 2L before falling off at a speed v, so it will take 2L/v units of time to fall off. Clearly, it will be the last ant to fall off, and so the total time for all the ants to fall of is 2L/v.

 

 

Another puzzle.

 

Four prisoners are arrested, but there is no space in the jail. The jailer comes up with a puzzle where if they succeed they all go free but if they fail they are all executed. The jailer sits three men in a line, and the other man behind a screen.

 

 

He puts a hat on each prisoner, he explains to the prisoners there are two red and two blue hats. The prisoners can see the hats in front of them, but not on themselves or behind. The fourth man can not see the other prisoners and the other prisoners can not see him. They cannot turn their heads to look behind them and they cannot communicate with each other.

 

If any of the prisoners can figure out the colour of their hat and shout it out, then they all go free. If the guess if wrong, they are executed. Is there a way for the prisoners to guarantee escape?

[Dannyboy-the-Dane]

Gah, I can't figure out how him speaking up changes anything ...

 

Here's my take on the first puzzle if people can't get their heads around it.

 

You can treat the collision of two ants as simply two ants walking through each other. With that in mind, the first ant, a1, will walk 2L before falling off at a speed v, so it will take 2L/v units of time to fall off. Clearly, it will be the last ant to fall off, and so the total time for all the ants to fall of is 2L/v.

 

 

Another puzzle.

 

Four prisoners are arrested, but there is no space in the jail. The jailer comes up with a puzzle where if they succeed they all go free but if they fail they are all executed. The jailer sits three men in a line, and the other man behind a screen.

 

 

He puts a hat on each prisoner, he explains to the prisoners there are two red and two blue hats. The prisoners can see the hats in front of them, but not on themselves or behind. The fourth man can not see the other prisoners and the other prisoners can not see him. They cannot turn their heads to look behind them and they cannot communicate with each other.

 

If any of the prisoners can figure out the colour of their hat and shout it out, then they all go free. If the guess if wrong, they are executed. Is there a way for the prisoners to guarantee escape?

If A and B have the same colour, C can figure out what his own colour is. If A and B are not the same colour ... well, then I don't know if there's a way.

 

[Will]

Might get some of this wrong but I think its generally ok.

 

If A and B have the same hat on, C knows the colour of their hat and can free everyone.

If A and B have different hats on, C won't know their colour, but B will know that their hat is different to A and so can name their colour and free them all.

 

I think thats it actually. Am I right?

[Dannyboy-the-Dane]

Might get some of this wrong but I think its generally ok.

 

If A and B have the same hat on' date=' C knows the colour of their hat and can free everyone.

If A and B have different hats on, C won't know their colour, but B will know that their hat is different to A and so can name their colour and free them all.

 

I think thats it actually. Am I right?[/quote']

But they aren't allowed to communicate, so C couldn't tell A and B that their hats are different ...?

[Supergrunch]

For the hats problem:

 

If both B and A have red hats, C must have a blue hat and so will say so, vice versa if B and A both have blue hats. So these scenarios are straightfoward, now let's consider a mixed case (i.e. where B and A have hats of different colours):

 

The first thing to note is that if C says nothing, B and A will know the case is mixed. Thus if B sees that A has a blue hat, he will know his hat is the other colour and thus red, and will be able to say so, and vice versa if A has a red hat.

 

Thus the prisoners will always be able to escape: if B and A have hats of the same colour, C will know what colour hat he has and thus will say so, and if B and A have hats of different colours, they will be able to infer this fact from C's lack of talking, and then by looking at A's hat B will know his hat colour.

 

Edit: Yeah, same solution as will'.

[Will]

But if they were the same then C would know his colour. If C says nothing then B must know that their hat can't be the same as A, and so knows what colour hat they have on.

[Dannyboy-the-Dane]

Ah, right. They're also all logical individuals. Didn't take that into account.

[SPAMBOT4000]

Eyes problem...

Firstly, Will yeah that was the issue I was having with it to begin with, thanks for clearing it up.

 

This is how far I got with the actual puzzle after seeing the answer above. Well first, if I make it simpler and have 2 blues and 2 brown then... on the first night both the blue eyed people would see 1 blue and 2 brown and nobody would go. On the second day, they know that if there were only 1 person with blue eyes then they would have gone on the first night (because they wouldn't have seen any other blue eyes and come to the conclusion that they were the one with blue eyes) So on the second night both the blue eyed people would leave as they know there must be more than one person with blue eyes and as they can't see anyone else with blue eyes, it must be them.

 

I get that far... but I don't get how the same logic applies for 100 people. Why the 100th night? Surely they would all realise that there will be 98 (?) completely redundant days of waiting? Even someone with blue eyes viewing the others would see 99 blue eyes and would assume (supposing that they themselves didn't have blue eyes) that all those 99 would see 98 others with blue eyes. Wouldn't it be the 2nd or 3rd night instead? *Brain explodes*

 

 

And Zell...

B knows that if C saw two blue hats in front of him, he'd know that he had to be red and call out. So... if C hasn't called out a colour then B knows he must be the opposite colour of the hat in front of him

 

Edit: Completely ninja'd. Took too long writing out my thoughts about the eyes puzzle.

[Supergrunch]

Eyes problem...

Firstly, Will yeah that was the issue I was having with it to begin with, thanks for clearing it up.

 

This is how far I got with the actual puzzle after seeing the answer above. Well first, if I make it simpler and have 2 blues and 2 brown then... on the first night both the blue eyed people would see 1 blue and 2 brown and nobody would go. On the second day, they know that if there were only 1 person with blue eyes then they would have gone on the first night (because they wouldn't have seen any other blue eyes and come to the conclusion that they were the one with blue eyes) So on the second night both the blue eyed people would leave as they know there must be more than one person with blue eyes and as they can't see anyone else with blue eyes, it must be them.

 

I get that far... but I don't get how the same logic applies for 100 people. Why the 100th night? Surely they would all realise that there will be 98 (?) completely redundant days of waiting? Even someone with blue eyes viewing the others would see 99 blue eyes and would assume (supposing that they themselves didn't have blue eyes) that all those 99 would see 98 others with blue eyes. Wouldn't it be the 2nd or 3rd night instead? *Brain explodes*

Have a think about the 7 person case (3 brown, 3 blue, 1 green), and see where that gets you.

[Will]

Surely they would all realise that there will be 98 (?) completely redundant days of waiting? Even someone with blue eyes viewing the others would see 99 blue eyes and would assume (supposing that they themselves didn't have blue eyes) that all those 99 would see 98 others with blue eyes. Wouldn't it be the 2nd or 3rd night instead? *Brain explodes*

 

This is the problem that pops up every time this comes up. I'm sure there's some reason that the jump doesn't work but thinking about it now I can't see why for any situation with 3 or more of the colour they don't all leave on the 3rd night.

Edited January 27, 2010 by will'

[Dannyboy-the-Dane]

I'm not even sure I understand the premise correctly - at least not based on the proposed theories.

[Will]

This is the problem that pops up every time this comes up. I'm sure there's some reason that the jump doesn't work but thinking about it now I can't see why for any situation with 3 or more of the colour they don't all leave on the 3rd night.

 

Thinking about it some more I don't think the jump works because it assumes viewing the problem from the pov of a person with the selected colour. The jump doesn't work if you have the other colour eyes and it needs to work for every islander.

[Supergrunch]

This is the problem that pops up every time this comes up. I'm sure there's some reason that the jump doesn't work but thinking about it now I can't see why for any situation with 3 or more of the colour they don't all leave on the 3rd night.

(you got your quotation attribution wrong)

 

Anyway' date=' there is a decent reason for this. I'll explain the 7 person case for people (3 blue, 3 brown, 1 green):

Let's choose a random blue eyed person to focus on - it doesn't matter which, as they're all identical. They are able to see two people with blue eyes, but aren't sure of the colour of their own eyes. As a null hypothesis, let's say this person assumes their own eyes aren't blue, meaning the island only has two blue eyed people. To refute this hypothesis, the person has to wait - if it were true, we would have an anlogous situation to the 5 person case (2 blue, 2 brown, 1 green), as all that matters is the number of blue eyed people. So our our person imagines this situation, and imagines himself in the shoes of one of the putative 2 blue eyed people. In this imagined situation, the blue eyed person would wait one day to see if there was only one blue eyed person, but instead finds out that there are 2, and he must be one of them. Thus the blue eyed person at the start waits to see whether this happens on the second day, and only when it doesn't is his null hypothesis invalidated, meaning he must have blue eyes.

 

Blergh, that's hell to explain.

[SPAMBOT4000]

(you got your quotation attribution wrong)

 

Anyway, there is a decent reason for this. I'll explain the 7 person case for people (3 blue, 3 brown, 1 green):

Let's choose a random blue eyed person to focus on - it doesn't matter which, as they're all identical. They are able to see two people with blue eyes, but aren't sure of the colour of their own eyes. As a null hypothesis, let's say this person assumes their own eyes aren't blue, meaning the island only has two blue eyed people. To refute this hypothesis, the person has to wait - if it were true, we would have an anlogous situation to the 5 person case (2 blue, 2 brown, 1 green), as all that matters is the number of blue eyed people. So our our person imagines this situation, and imagines himself in the shoes of one of the putative 2 blue eyed people. In this imagined situation, the blue eyed person would wait one day to see if there was only one blue eyed person, but instead finds out that there are 2, and he must be one of them. Thus the blue eyed person at the start waits to see whether this happens on the second day, and only when it doesn't is his null hypothesis invalidated, meaning he must have blue eyes.

 

Blergh, that's hell to explain.

 

I get why that works but why does it work further than that?

Take the 100 blue eyed people.

Each of them sees 99 others with blue eyes. Now if they assume that they themselves do not have blue eyes and that they were now to imagine what another blue eyed person sees, the least amount of blue eyed people anyone else could be seeing is 98. And they all know this. So if argh *KABOOM* Lost it again.

[Supergrunch]

I get why that works but why does it work further than that?

Take the 100 blue eyed people.

Each of them sees 99 others with blue eyes. Now if they assume that they themselves do not have blue eyes and that they were now to imagine what another blue eyed person sees, the least amount of blue eyed people anyone else could be seeing is 98. And they all know this. So if argh *KABOOM* Lost it again.

Basically, each blue eyed person takes the 98 person case as a null hypothesis, then investigates this, realising that each of the putative 98 people would have to be imagining the 97 person case, each of whom would be imagining the 96 person case... And this line of reasoning goes all the way down the the one person case where it's obvious.

[SPAMBOT4000]

I sort of get how it would theoretically spiral down like that but I don't really understand, it just seems like there's something missing. If all of the blue eyes know that the minimum anyone could be seeing is 98 then I don't get how doing nothing for 98/97 days (when they all actually realise noone will go in this time) puts them at any advantage.

[Dannyboy-the-Dane]

But ... does that really work? Is their reasoning actually failproof?

[Supergrunch]

I sort of get how it would theoretically spiral down like that but I don't really understand, it just seems like there's something missing. If all of the blue eyes know that the minimum anyone could be seeing is 98 then I don't get how doing nothing for 98/97 days (when they all actually realise noone will go in this time) puts them at any advantage.

Because there's no other way to determine their own eye colour. If there are n blue eyed people, then they all have to wait n-1 days to rule out there only being n-1 blue eyed people. In the original case, a given blue eyed person realises that everyone is either seeing 99 other people (like him), or (if he doesn't have blue eyes), 98. There's no way of ruling out there being 98 people without thinking about the reasoning that would go on in that situation, and so everyone has to wait to ensure that the embedded hypotheticals are false.

But ... does that really work? Is their reasoning actually failproof?

Yep.

Edited January 27, 2010 by Supergrunch

[SPAMBOT4000]

Ahh thanks. Bit of a headfuck really, been pondering this most of the day! Soo when the green eyed guy says "I see someone with blue eyes" that is more of just a trigger to start counting the days rather than actually passing on the information of being able to see a blue eyed person?