The Tower and the Towns Puzzle

[pdurrant]

Originally Posted by GeoffC

Spoiler:

Why ?

He sees 8 - he doesn't know whether it is 10 or 13
He sees 11, therefore it must be 13. They are both out.

However on the next day, he still only sees 8 - and still he doesn't know whether it is 10 or 13.

and the same for each of the remaining days until old age creeps in or one of them takes the plunge and guesses 50:50 at being right.....

I think the thing you've missed is that

Spoiler:

(given the first chap can see 2 towns)

if the second chap sees eight, and so can't get them both freed on the first night, this gives information to the first chap, who on the second day knows that the second chap doesn't see 11. It's the first chap who now knows that there are 10 towns in total and gets them free.

The information passed from prisoner to prisoner each night is "I don't know enough to say how many towns there are", passedin the form of the prisoners still being prisoners the next day.

And each day, the information that this conveys is different, reducing the number of possibilities for the number of towns that the other person can see and can think that the other prisoner can see.

HTH.

[pdurrant]

Originally Posted by =X=

Okay I'm simple minded but the only way I can see them 'knowing' the answer without guessing is if one side sees all 13 towns. Since only one side has to give the right answer the person that sees 0. Would not provide an answer

Clue:

Spoiler:

How about if one of them could see 12 towns? Would he have to guess about how many towns there were?

[GeoffC]

There's a lot of 'givens' ....

[omk3]

Originally Posted by GeoffC

Spoiler:

Why ?

He sees 8 - he doesn't know whether it is 10 or 13
He sees 11, therefore it must be 13. They are both out.

However on the next day, he still only sees 8 - and still he doesn't know whether it is 10 or 13.

and the same for each of the remaining days until old age creeps in or one of them takes the plunge and guesses 50:50 at being right.....

Spoiler:

Okay, I started from the simplest case, but let's take this one:

A sees 8. So A knows that B sees either 2 or 5.
A knows that if B sees 2, he will know that A sees either 8 or 11.

Working on the assumption that B sees 2:
A knows that B will wait one day to see if A sees in fact 11, because if he does, they will be free on day 1.
As they are not, A knows that B now knows that A does not see 11.
Therefore, B, if he sees 2, will know on day 2 that A must see 8 instead of 11, so he will tell the guard the towns are 10 and they will be freed.
If day 2 passes without event, A knows that B does not see 2 towns after all.
Assumption disproved.

As B could only see either 2 or 5 towns, and 2 is out, he must see 5. So A now asks to be freed, and tells the guard the towns are 13.

No guess work, just a waiting game.

[obs20]

Spoiler:

The first night they are captive so on the second night

If wise guy A sees 13 then wise guy B sees 0
they go home that second night

Same if wise guy A sees 12 then wise guy B sees 1
they go home that night because wise guy A knows there cannot be 10

Same if Wise guy A sees 11 and wise guy b sees 2
If wise guy A sees 10 then wise guy B sees 3 or 0
because wise guy B knows that if Wise guy A
saw 13 they'd gone home the second night. They go home the third night

If wise guy A sees 9 then wise guy B sees 4 or 1
because wise guy B knows that if Wise guy A
saw 12 they'd gone home. On the third night Wise Guy A knows that wise guy B didn't answer so he now knows that wise guy B saw 4 the go home the fourth night

If wise guy A sees 5 then wise guy B sees 8 or 5
because wise guy B knows that if Wise guy B
saw 8 they'd gone home the forth night. Do he must have seen 5 and they go home the fifth night

[pdurrant]

Originally Posted by obs20

Spoiler:

The first night they are captive so on the second night

If wise guy A sees 13 then wise guy B sees 0
they go home that second night

Same if wise guy A sees 12 then wise guy B sees 1
they go home that night because wise guy A knows there cannot be 10

Same if Wise guy A sees 11 and wise guy b sees 2
If wise guy A sees 10 then wise guy B sees 3 or 0
because wise guy B knows that if Wise guy A
saw 13 they'd gone home the second night. They go home the third night

If wise guy A sees 9 then wise guy B sees 4 or 1
because wise guy B knows that if Wise guy A
saw 12 they'd gone home. On the third night Wise Guy A knows that wise guy B didn't answer so he now knows that wise guy B saw 4 the go home the fourth night

If wise guy A sees 5 then wise guy B sees 8 or 5
because wise guy B knows that if Wise guy B
saw 8 they'd gone home the forth night. Do he must have seen 5 and they go home the fifth night

Very much along the right lines, but you have the count of nights wrong, and a few typos. Looking at the question, I can see that I didn't state explicitly that the fifth evening was the fifth evening after their arrival at the castle. I'll see if I can fix it.

But I'm sure you'll be able to come up with the answer I intended, now that I've clarified the count of the nights.

[=X=]

Originally Posted by GeoffC

You are as simple minded as I am ....

Ha ha, let just hope it's not you and I are not put to the test together.

=X=

[=X=]

Originally Posted by pdurrant

Clue:

Spoiler:

How about if one of them could see 12 towns? Would he have to guess about how many towns there were?

I kind of cheated and read omk3 first post. Once I read his logic, even though he didn't answer the problem I, knew I was wrong and also knew the answer, but at that point it was not a challenge since he gave me the direction I needed to solve the teaser.

Spoiler:

It looks like a dampened harmonic oscillator. Where the final answer is reached on day 5. If the day it is solved lands on an odd day the answer is 13, but if it lands on an even day the answer is 10.

It's been years since I've done this kind of math and though I do see the pattern I can't figure it out until I break open a math book.

So if the problem was solved on
Day 1=13 (11,12,13 are eliminated)
Day 2=10 (0,1,2 are eliminated)
Day 3=13 (8,9,10 are eliminated)
Day 4=10 (3,4,5 are eliminated)
Day 5=13 (only 6 and 7 are left)

By day 5 the only two numbers left is 6 and 7.

[obs20]

Originally Posted by pdurrant

Very much along the right lines, but you have the count of nights wrong, and a few typos. Looking at the question, I can see that I didn't state explicitly that the fifth evening was the fifth evening after their arrival at the castle. I'll see if I can fix it.

But I'm sure you'll be able to come up with the answer I intended, now that I've clarified the count of the nights.

Okay

Spoiler:

Then if we go one more night it's 6 and 7. Alot like the last prison puzzle.

[dreams]

I love how this one has everyone jumping in. pdurrant, these puzzles are great!

I may have to wait for some time to try this one out, but it looks to be fun.

Congrats to all those who were able to figure this one out so quick.

[pdurrant]

Originally Posted by obs20

Okay

Spoiler:

Then if we go one more night it's 6 and 7. Alot like the last prison puzzle.

Yes, exactly right. And yes, it is a lot like the prison puzzle.

[GeoffC]

You know, I'm still not convinced ....

Spoiler:

A sees 8. So A knows that B sees either 2 or 5.
A knows that if B sees 2, he will know that A sees either 8 or 11.


Each night provides the same scenario .....
How does that give them the answer of 10 or 13 - it still leaves them each night not knowing ..... there is no new information between night one and any of the succeeding nights .... !?

[pdurrant]

Originally Posted by GeoffC

You know, I'm still not convinced ....

Spoiler:

A sees 8. So A knows that B sees either 2 or 5.
A knows that if B sees 2, he will know that A sees either 8 or 11.


Each night provides the same scenario .....
How does that give them the answer of 10 or 13 - it still leaves them each night not knowing ..... there is no new information between night one and any of the succeeding nights .... !?

OK, I'll try to explain using the example you've started, looking only at what wise man A can know.

Spoiler:

BTW, the new information is that the other person could not work out the answer.

1st day
A sees 8 towns. A knows that B sees 2 or 5 towns, but does not know which.
A knows that

if B sees 2 towns, B will know that A sees 8 or 11 towns
if B sees 5 towns, B will know that A sees 5 or 8 towns.

A knows that A doesn't know how many towns there are, and that B doesn't know how many towns there are. However, B does not know that A does not know how many towns there are. If B sees 2 towns, B will know that A sees 8 or 11 towns, and if A sees 11 towns, A would know there were 13 towns.

A knows that A doesn't know how many towns there are. But there's a possibility that B knows that there's a possibility that A knows.

1st evening: Neither knows, they stay imprisoned.

2nd day
A now knows that B knows that A does not see 11 towns (because if A saw 11 towns, they would both now be free). A knows that if B sees 2 towns, B will now know that A sees 8 towns. But A also knows that it's still a possibility that B sees 5 towns.

A knows that A still doesn't know how many towns there are. But A also knows that there's a possibility that B now knows how many towns there are — if B sees 2 towns, he knows that A sees 8.

2nd evening: A doesn't know. As it happens, B doesn't see 2 towns, he sees 5, so he still doesn't know whether A sees 8 or 5 towns. They stay imprisoned.

3rd day:
A now knows that B does not see 2 towns, as if B saw 2 towns, B would have known on the second day that A saw 8 towns not 11 towns, and they'd both now be free.

So A now knows that B sees 5 towns.

3rd evening: A knows that B sees 5 towns, so tells the Gaoler that there are 13 towns, and A & B are freed.


HTH

[LazyScot]

Originally Posted by GeoffC

You know, I'm still not convinced ....

Spoiler:

A sees 8. So A knows that B sees either 2 or 5.
A knows that if B sees 2, he will know that A sees either 8 or 11.


Each night provides the same scenario .....
How does that give them the answer of 10 or 13 - it still leaves them each night not knowing ..... there is no new information between night one and any of the succeeding nights .... !?

A different version of the explanation pdurrant gave above...

Spoiler:

The new information on the 2nd night is that they are still imprisoned. This sounds like it isn't information, but it is, just encoded in a strange way (double dutch, perhaps....). The reason that it is information is that it tells you that the other wise man couldn't work out whether the answer 10 or 13.

This means you, on the second night, now know that the other wise man cannot see certain numbers of towns. This is the information that is interchanged on the second night.

[GeoffC]

... walks away shrugging shoulders, stage right, shaking head .... muttering ..........