[wolfgang]

You are one of 30 prisoners on death row with the execution date set for tomorrow. Your king is a ruthless man who likes to toy with his people's miseries. He comes to your cell today and tells you:

“I’m gonna give you prisoners a chance to go free tomorrow. You will all stand in a row (queue) before the executioner and we will put a dot on your backs, either a white or a black one. Of course you will not be able to see the color of your own dot; you will only be able to see the prisoner in front of you with his dot; you will not be allowed to look back or communicate together in any way (talking, touching.....).

Starting with the last person in the row, the one who can see only one who is in front of him, he will be asked a simple question: WHAT IS THE COLOR OF YOUR DOT?

He will be only allowed to answer “BLACK” or “WHITE”. If he says anything else you will ALL be executed immediately.

If he guesses the right color of the dot on his back he is set free, otherwise he is put to death. And we move on to the one in front of him and ask him the same question and so on…

Now since you all can communicate freely during the night,how can you find a way to guarantee the freedom of 29 prisoners tomorrow?

[Guest]

Planned intonation. For instance, if I emphasize "Color" in the phrase it means your dot is white. If I emphasize "Dot" in the phrase, then your dot is black!

[Guest]

@GraceAlone, they're only allowed to say "black" or "white"

They can sing or state the color. The first prisoner has 50/50 chance, but he can state the color of the dot on next prisoner. That prisoner then knows his color and if the same as the next can sing the color, or otherwise simply state the color. If sung, the next prisoner knows his is the same or if only stated then his is opposite.

Edited October 25, 2010 by wolfcry911

[Guest]

I would plan it using emphasis on the last letter of the word...if the last letter is emphasized, then the next person's color is white...if it is soft, then their color is black. the first person states the color of the one in front of them. the second person says the same color, but if the third person's color is white, they would emphasize the last letter, and if it was black, would use a soft last letter...for example, person 1 says black because person 2 has a black spot, person 3 has a white spot, so person 2 says blacK, with emphasis on the letter k. person 4 has a white spot just like person 3 who now knows that his spot is white due to the emphasis on the letter K from person 2...so person 3 says WhiTe...with emphasis on the T. continue this pattern down the line until everyone is safe except the possiblilty of the first person.

[wolfgang]

and if the answer is recorded and he has only to press the button(either white or black )!!!

moreover ..singing is a kind of hint...

[Guest]

Use parity

[wolfgang]

I would plan it using emphasis on the last letter of the word...if the last letter is emphasized, then the next person's color is white...if it is soft, then their color is black. the first person states the color of the one in front of them. the second person says the same color, but if the third person's color is white, they would emphasize the last letter, and if it was black, would use a soft last letter...for example, person 1 says black because person 2 has a black spot, person 3 has a white spot, so person 2 says blacK, with emphasis on the letter k. person 4 has a white spot just like person 3 who now knows that his spot is white due to the emphasis on the letter K from person 2...so person 3 says WhiTe...with emphasis on the T. continue this pattern down the line until everyone is safe except the possiblilty of the first person.

and what if the answer is recorded....and he has only to press one button(white or black)??

[wolfgang]

Use parity

??????

[Guest]

Use Parity.

The prisoners decide on a convention for the last one in line. If he sees an even number of black dots he says "black". If he sees an odd number of black dots he says "white". Let's suppose he says "black" (even number of black dots)

Then prisoner 29 counts the black dots in front of him. If he sees an even number he knows his own dot is white. If he sees an odd number he knows his own dot is black. In either case he knows his own color and can announce it.

Then prisoner 28 can do likewise by counting the black dots he can see and using the answers from 30 and 29 to deduce his own color.

And so on down the line.

This way prisoners 1 to 29 are guaranteed freedom and Prisoner 30 has a 50/50 chance.

??????

[Gmaster479]

Hawkeye is 100% correct. See topic from an almost exact replica of this problem entitled 'Hats on Death Row'

I'm surprised that no one recognized this puzzle was a repeat. Still a really good brain teaser

[araver]

Hawkeye is 100% correct. See topic from an almost exact replica of this problem entitled 'Hats on Death Row'

I'm surprised that no one recognized this puzzle was a repeat. Still a really good brain teaser

Nope, it's not a repeat. As the title says "A mdification to( Hats on death row)".

Prisoners can only see the NEXT prisoner in line, not ALL the prisoners.

[Gmaster479]

ahh I did misread it. My bad. The wording was so similar I didn't think of that

Hawkeye's solution is correct for the original problem then I believe. Again...my bad

Edited October 25, 2010 by Gmaster479

[araver]

and if the answer is recorded and he has only to press the button(either white or black )!!!

moreover ..singing is a kind of hint...

the prisoners should think of bribing whoever's in charge of the dots to manipulate the distribution

Because otherwise, it's impossible to ALWAYS rescue 29 assuming a random distribution of black and white dots.

The first prisoner always has a 50/50 chance of success whatever he says cause he does not see his dot (Schroedinger's cat)

His only information is about the next prisoner in line.

He has no knowledge of what other prisoners wear on their backs so his answer cannot encode any information about the rest of 28 dots.

The 2nd prisoner knows the dot of the 3rd prisoner in line and can possibly gain information from the previous volunteer in order to survive.

However he is restricted to say what he has on his back in order to live.

So he can indeed survive if he receives information from behind, but cannot transmit any information for others, except his dot color. Which in a random distribution has no value.

So the third prisoner does not have any new information except what he sees on the 4th prisoner back.

There's no way the first prisoner's answer affects his dot (since he did not see it) and there's no way the second prisoner's answer affects his dot since he has to say his own dot color to live. So we're back to guessing from an information-theory point of view.

... Regarding bribing Of course 30 Black or 30 Whites would clearly save all. If it's too obvious, then a bicolor sequence (BB...BBBBWW..WW) would save all but one, if they keep just stating what they see.

Edited October 25, 2010 by araver

[wolfgang]

In the original version,,the last one can see all the other 19....but here...no one can see but the one in front of him

and let me suppose that each prisoner has two buttons(white and black)and when he press one of them a recorder will say that color.

[Guest]

First one is up to fate, he'll hold 3 seconds before answering if he sees white in front of him otherwise he'll answer instantly...

[wolfgang]

the prisoners should think of bribing whoever's in charge of the dots to manipulate the distribution

Because otherwise, it's impossible to ALWAYS rescue 29 assuming a random distribution of black and white dots.

The first prisoner always has a 50/50 chance of success whatever he says cause he does not see his dot (Schroedinger's cat)

His only information is about the next prisoner in line.

He has no knowledge of what other prisoners wear on their backs so his answer cannot encode any information about the rest of 28 dots.

The 2nd prisoner knows the dot of the 3rd prisoner in line and can possibly gain information from the previous volunteer in order to survive.

However he is restricted to say what he has on his back in order to live.

So he can indeed survive if he receives information from behind, but cannot transmit any information for others, except his dot color. Which in a random distribution has no value.

So the third prisoner does not have any new information except what he sees on the 4th prisoner back.

There's no way the first prisoner's answer affects his dot (since he did not see it) and there's no way the second prisoner's answer affects his dot since he has to say his own dot color to live. So we're back to guessing from an information-theory point of view.

... Regarding bribing Of course 30 Black or 30 Whites would clearly save all. If it's too obvious, then a bicolor sequence (BB...BBBBWW..WW) would save all but one, if they keep just stating what they see.

the colors should be at random!

[Guest]

Aah! I misread the question. Thanks for the clarification, Wolfgang.

[wolfgang]

First one is up to fate, he'll hold 3 seconds before answering if he sees white in front of him otherwise he'll answer instantly...

GREAT!!

[Guest]

Hawkeye is 100% correct. See topic from an almost exact replica of this problem entitled 'Hats on Death Row'

I'm surprised that no one recognized this puzzle was a repeat. Still a really good brain teaser

This wont work as the puzzle said

you will only be able to see the prisoner in front of you with his dot;

Hence you can only see 1 persons dot

[wolfgang]

Nope, it's not a repeat. As the title says "A mdification to( Hats on death row)".

Prisoners can only see the NEXT prisoner in line, not ALL the prisoners.

thank you dear ....

[wolfgang]

Aah! I misread the question. Thanks for the clarification, Wolfgang.

uw dear :-)

[Guest]

GREAT!!

...

Do I win? XD

[wolfgang]

...

Do I win? XD

I like your way of thinking....yes...u win !!