RW, love it! The Great British tradition of rounds does not however extend to requiring everyone to have to turn up at the same time.

Once again though, how do you set up the 'triple click' thing?

Bigtone53 wrote:Once again though, how do you set up the 'triple click' thing?

To find out about any BBCode tricks used in any post, press the "quote" button in the upper corner of the post and you'll see exactly how that post was written, complete with code tags. You may then of course use the "back" button to avoid replying to the thread.

RW

- RW
- 2010 Supporter
**Posts:**1000**Joined:**16 March 2006

Here is one of my favorites from when I was a kid:

10 prisoners are to be executed tomorrow morning. The king decides to give them a chance to save themselves. He tells them that they will be lined up single file such that every prisoner can see the prisoners in front of him, but none of the prisoners behind him. Each will have a hat placed on their head, colored either white or black, but they will not know what is on their own head, nor how many total black and white hats there are. Starting from the back of the line, the executioner will ask the prisoners in turn to announce to all what color they believe their hat is. If they guess correctly, they live. Otherwise, they die.

Then prisoners start mumbling amongst themselves, trying to concoct a plan. Just as the king is about to leave he turns back around and tells them, "oh, no one will speak until executioner asks you for your answer and then you are to say only 'black' or 'white'. If anyone breaks this rule, you will all be killed. If you design a system that passes information based on the cadence of your voice, or the pitch, or long pauses or any other hanky-panky, you will all be killed."

The prisoners commiserate and decide the king is serious. They will play it straight. How many prisoners can be saved no matter what the distribution of white and black hats?

10 prisoners are to be executed tomorrow morning. The king decides to give them a chance to save themselves. He tells them that they will be lined up single file such that every prisoner can see the prisoners in front of him, but none of the prisoners behind him. Each will have a hat placed on their head, colored either white or black, but they will not know what is on their own head, nor how many total black and white hats there are. Starting from the back of the line, the executioner will ask the prisoners in turn to announce to all what color they believe their hat is. If they guess correctly, they live. Otherwise, they die.

Then prisoners start mumbling amongst themselves, trying to concoct a plan. Just as the king is about to leave he turns back around and tells them, "oh, no one will speak until executioner asks you for your answer and then you are to say only 'black' or 'white'. If anyone breaks this rule, you will all be killed. If you design a system that passes information based on the cadence of your voice, or the pitch, or long pauses or any other hanky-panky, you will all be killed."

The prisoners commiserate and decide the king is serious. They will play it straight. How many prisoners can be saved no matter what the distribution of white and black hats?

- re'born
**Posts:**551**Joined:**31 May 2007

Reasoning below

They agree the code Black = odd and white = even, and agree to count black hats (hopefully before the king stops them talking!)

When it is time to speak, the poor guy on the end says Black if he sees an odd number of black hats and White if he sees an even number of black hats. He has a 50:50 chance of being right. The next guy though, knowing whether there are an odd or even number of black hats amongst the nine guys left and able to see everyone elses, know for sure what he has. The next guy also knows whether there was an odd or even number of black hats amongst the 9 of them originally and now knows what the guy behind him had so he can be certain about his hat. And so to the front.

They agree the code Black = odd and white = even, and agree to count black hats (hopefully before the king stops them talking!)

When it is time to speak, the poor guy on the end says Black if he sees an odd number of black hats and White if he sees an even number of black hats. He has a 50:50 chance of being right. The next guy though, knowing whether there are an odd or even number of black hats amongst the nine guys left and able to see everyone elses, know for sure what he has. The next guy also knows whether there was an odd or even number of black hats amongst the 9 of them originally and now knows what the guy behind him had so he can be certain about his hat. And so to the front.

- Bigtone53
**Posts:**413**Joined:**19 September 2005

Here are the codes for the "triple click" trick:

The effect:

For your 10-digit puzzle and re'born's prisoner puzzle, just have to say it has been posted and solved in Djape's forum many months ago, in the following 2 threads:

http://www.djape.net/sudoku/forum/viewtopic.php?p=695#p695

http://www.djape.net/sudoku/forum/viewtopic.php?t=95

But the fibobia one is marvellous and well done re'born for solving it elegantly and RW for providing a humourous alternative answer.

However, if you allow one to get pedantic there is something wrong with this riddle:

Apparently all 4 of A, B, C, D each has made only 1 statement there, so nothing can be drawn from these statements.

Perhaps you should have written:

A said "I've had 4. E's had 3. B's had 5. C's had 1. D's had 2."...

- Code: Select all
`[quote="Triple click below to see what I"][color=white]Insert your text here![/color][/quote]`

The effect:

Triple click below to see what I wrote:Insert your text here!

For your 10-digit puzzle and re'born's prisoner puzzle, just have to say it has been posted and solved in Djape's forum many months ago, in the following 2 threads:

http://www.djape.net/sudoku/forum/viewtopic.php?p=695#p695

http://www.djape.net/sudoku/forum/viewtopic.php?t=95

But the fibobia one is marvellous and well done re'born for solving it elegantly and RW for providing a humourous alternative answer.

However, if you allow one to get pedantic there is something wrong with this riddle:

Bigtone53 wrote:It is well known that certain people suffer from fibobia after drinking beer. If you suffer from this unfortunate affliction, then of any 5 statements you may make, the number of false ones is the same as the number of pints you have had. A,B C D and E all suffer from fibobia and I asked them one night how many pints of beer they had had. These were the answers

A said " I've had 4, E 3, B 5,C 1 and D 2"

B said " I've had 4, C 3, A 5, E 1 and D 2"

C said " I've had 3, E 2, B 5, A 1 and D 4"

D said " I've had 4, E 3, B 5, C 2 and A 1"...

Apparently all 4 of A, B, C, D each has made only 1 statement there, so nothing can be drawn from these statements.

Perhaps you should have written:

A said "I've had 4. E's had 3. B's had 5. C's had 1. D's had 2."...

Last edited by udosuk on Fri Oct 05, 2007 6:47 am, edited 1 time in total.

- udosuk
**Posts:**2698**Joined:**17 July 2005

Triple click below to see what re'born wrote:The key observation for me was to note that the sum of digits must be 10. This immediately implies that for n>=5, then the number in the nth position (I label the positions as the 0th through 9th positions) is either 0 or 1. From here I noted that if 1 occurs in the nth-position, then this will imply a sum of digits of at least 1 + n + n*r, where r is the position in which n lays. If we can deduce that r<>1, then we get a sum of digits at least 4 + n + n*r.

Re'born, although the last sentence is probably true, I think you should have explained a bit how you came to this conclusion. Let's say if I'm a teacher marking you on this proof I wouldn't give you full marks for this "logical gap".

- udosuk
**Posts:**2698**Joined:**17 July 2005

udosuk wrote:Triple click below to see what re'born wrote:The key observation for me was to note that the sum of digits must be 10. This immediately implies that for n>=5, then the number in the nth position (I label the positions as the 0th through 9th positions) is either 0 or 1. From here I noted that if 1 occurs in the nth-position, then this will imply a sum of digits of at least 1 + n + n*r, where r is the position in which n lays. If we can deduce that r<>1, then we get a sum of digits at least 4 + n + n*r.

Re'born, although the last sentence is probably true, I think you should have explained a bit how you came to this conclusion. Let's say if I'm a teacher marking you on this proof I wouldn't give you full marks for this "logical gap".

Hee Hee. Fair enough, Herr Professor.

Triple click below to see what re'born wrote:If r<>1, then r=0 and there is a 1 in the nth position, an n in the 0th position and at least a 2 in the 1st position, hence at least a 1 in 2nd position. So we have 1 + n + 2 + 1 = 4 + n (the +n*r is missing since r = 0).

- re'born
**Posts:**551**Joined:**31 May 2007

re'born wrote:Hee Hee. Fair enough, Herr Professor.Triple click below to see what re'born wrote:If r<>1, then r=0 and there is a 1 in the nth position, an n in the 0th position and at least a 2 in the 1st position, hence at least a 1 in 2nd position. So we have 1 + n + 2 + 1 = 4 + n (the +n*r is missing since r = 0).

Nah, still can't pay you full marks for this: "If r<>1, then r=0".

- udosuk
**Posts:**2698**Joined:**17 July 2005