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Guest Message by DevFuse
 

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Hats on a death row!! One of my favorites puzzles!


Best Answer bonanova, 08 January 2008 - 06:25 PM

Spoiler for Here's my idea
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#21 PDR

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Posted 02 January 2008 - 12:24 AM

oops = thought it through and realized my logic didn't work. I think I'm on the right track - but didn't quite get it.
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#22 unreality

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Posted 02 January 2008 - 02:25 AM

all you know is if there are more or less, so it only works if it's a tie in front of you, that's the only way you could deduce your own color. And then that would tell nothing to the people in front of you, so 10 would live for sure and 10 would have a 50/50. Yours doesnt work

the best one I've found so far is the one I posted a while back where 15 people are saved no matter what, but I'm thinking for the 19-saved solution
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#23 roolstar

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Posted 02 January 2008 - 10:21 AM

Try this:
The first speaker counts the number of red vs black hats on the remaining 19. Whichever there are more of, is the color he says his is. Since there are 19 remaining, there must be more of one color or the other.
2nd speaker also counts the number of red vs black on the remaining 18. Based on the first person's pronouncement, he knows which color there should be more of on the 18 plus his own. He compares what he sees and knows which color his own hat must be.
3rd-20th speaker should be keeping track of the the changing number left based on each previous speakers correct pronouncement and can compare that to the number they see in front of themself. They will be able to correctly deduce their own color.



Very nice try PDR! This was my first intuition when I was first presented with this puzzle. And you are on the right track in your thinking process.

I'm feeling that someone else will get this one soon...
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#24 spoxjox

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Posted 02 January 2008 - 07:50 PM

Okay, with all the clues and attempts, I think I've found a solution. If it's wrong, it's my fault, but if it's right, credit goes mostly to others who have already tried (and to those who gave hints).
Spoiler for solution

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#25 spoxjox

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Posted 02 January 2008 - 11:13 PM

ADDENDUM

You are the first to guess, and thus the "sacrificial lamb". The guard, who is truthful, offers to answer any single yes-or-no question that you put to him. This is your chance to escape execution! But you decide it's better to sacrifice yourself, if necessary, than to leave your buddies to execution. What single yes-or-no question can you ask that will insure your own survival as well as that of your comrades? (Assume the guard knows all the conditions of the execution.)
Spoiler for solution

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#26 roolstar

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Posted 03 January 2008 - 11:13 AM

You are the first to guess, and thus the "sacrificial lamb". The guard, who is truthful, offers to answer any single yes-or-no question that you put to him. This is your chance to escape execution! But you decide it's better to sacrifice yourself, if necessary, than to leave your buddies to execution. What single yes-or-no question can you ask that will insure your own survival as well as that of your comrades? (Assume the guard knows all the conditions of the execution.)



Great job there spoxjox!

But as you said, YOU ARE the first to guess (post) an answer for this puzzle (on the forum) so your solution will be dissected and scrutinized…

But let me first clarify: for this puzzle your solution is correct, but I invite you to find an answer that will be suitable for ANY number of prisoners (more than 1).

Because your technique does not work for 19 prisoners; in scenarios where we have 7 Black hats and 11 Red ones for example in front of the "sacrificial lamb" in the back.

Now don’t get me wrong, your solution works perfectly for the initial puzzle (taking into consideration that 0 hats is an even number of hats of course), but is there a universal way for any number of prisoners?

PS: It’s incredibly close to your technique!
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#27 spoxjox

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Posted 03 January 2008 - 05:03 PM

But let me first clarify: for this puzzle your solution is correct, but I invite you to find an answer that will be suitable for ANY number of prisoners (more than 1).

Because your technique does not work for 19 prisoners; in scenarios where we have 7 Black hats and 11 Red ones for example in front of the "sacrificial lamb" in the back.

Now don’t get me wrong, your solution works perfectly for the initial puzzle (taking into consideration that 0 hats is an even number of hats of course), but is there a universal way for any number of prisoners?

PS: It’s incredibly close to your technique!


How about this:
Spoiler for solution

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#28 Jason

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Posted 07 January 2008 - 12:48 AM

Here's my solution, not sure if it's allowed or not...

Spoiler for Here's my solution.. probably something wrong with it...


How's this one?
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#29 PDR

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Posted 07 January 2008 - 01:10 AM

How's this one?

Great idea, but perhaps a bit complicated to ensure each person is counting seconds correctly. I like it, so here's a twist:
Spoiler for A twist on Jason...

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#30 roolstar

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Posted 07 January 2008 - 11:19 AM

Here's my solution, not sure if it's allowed or not...


Good thinking you guys, and there are even more ideas to solve it this way: there is the tone of voice (high or low pitch), the volume (screaming or normal), timing (suggested)....

But we wanted to stay restricted to a system where a simple and neutral "BLACK" & "WHITE" will provide a bullet proof technique to save ALL BUT ONE of the prisoners (one of them will have a 50-50 chance of surviving).

And here's my final challenge:

FIND A WAY THAT WILL SOLVE THE PROBLEM FOR ANY NUMBER OF PRISONERS (20 or 19).

PS: Bonanova already found it... And you are very close: just twist the most successful technique for a bit.
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