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You are the ruler of a medieval empire and you are about to have a celebration tomorrow. The celebration is the most important party you have ever hosted. You've got 1000 bottles of wine you were planning to open for the celebration, but you find out that one of them is poisoned.
The poison exhibits no symptoms until death. Death occurs within ten to twenty hours after consuming even the minutest amount of poison.
You have over a thousand slaves at your disposal and just under 24 hours to determine which single bottle is poisoned.
You have a handful of prisoners about to be executed, and it would mar your celebration to have anyone else killed.
What is the smallest number of prisoners you must have to drink from the bottles to be absolutely sure to find the poisoned bottle within 24 hours? -
Senior Member
Array Dammit. Lemme guess, even just one missing slave from the "over a thousand" slaves would mar the celebration? "Yes! Rampaging bears are the answer to all of our cultural missteps!"
"Exactly. Paris Hilton? Bear attack. Emo? Bear attack. Reality television? Bear attack. Ann Coulter? Two bear attacks and a swarm of angry locusts." -Faye and Dora, Questionable Content 2003-2008 -
Wow, I'm still third top poster... # Posts Per Day: 15.18 -
 Originally Posted by Neinteen Ten.
heh. your turn -
Senior Member
Array Wait... how would you get all the bottles tested in time? The poison works anywhere from 10 to 20 hours... "Yes! Rampaging bears are the answer to all of our cultural missteps!"
"Exactly. Paris Hilton? Bear attack. Emo? Bear attack. Reality television? Bear attack. Ann Coulter? Two bear attacks and a swarm of angry locusts." -Faye and Dora, Questionable Content 2003-2008 -
 Originally Posted by Parry and Sunder Wait... how would you get all the bottles tested in time? The poison works anywhere from 10 to 20 hours... each prisoner will take, as quickly as possible, about 500 drinks from different bottles, in a specific sequence such that you can tell the bottle based on which prisoners die. multiple prisoners will be dying. -
 Originally Posted by noodle each prisoner will take, as quickly as possible, about 500 drinks from different bottles, in a specific sequence such that you can tell the bottle based on which prisoners die. multiple prisoners will be dying. To further elaborate, think binary code. Having 10 prisoners actually lets you test 2^10 bottles. So some of them don't get to drink from 500 bottles.
I always wondered though how one would get the tasters to swig from 500 bottles of wine and not throw up first. -
 Originally Posted by pokey To further elaborate, think binary code. Having 10 prisoners actually lets you test 2^10 bottles. So some of them don't get to drink from 500 bottles.
I always wondered though how one would get the tasters to swig from 500 bottles of wine and not throw up first. and some get to drink from a full 512 bottles. i was trying to explain without having to get into a bitmask explanation.
also, i imagine that you could get away with not drinking the wine, but holding it in your mouth to absorb the poison -
Senior Member
Array In the end, I had to look up the answer to really understand what you guys were talking about. I get it now though. It really is binary sequence.
So for bottle one, only prisoner one would drink it; for bottle two, the second prisoner would drink it; for bottle three, both prisoners one and two would drink, and so on and so forth, up until the final combination of prisoners four and six-through-ten. "Yes! Rampaging bears are the answer to all of our cultural missteps!"
"Exactly. Paris Hilton? Bear attack. Emo? Bear attack. Reality television? Bear attack. Ann Coulter? Two bear attacks and a swarm of angry locusts." -Faye and Dora, Questionable Content 2003-2008 -
 Originally Posted by Parry and Sunder In the end, I had to look up the answer to really understand what you guys were talking about. I get it now though. It really is binary sequence.
So for bottle one, only prisoner one would drink it; for bottle two, the second prisoner would drink it; for bottle three, both prisoners one and two would drink, and so on and so forth, up until the final combination of prisoners four and six-through-ten. yes, although i'd like to note that this isn't the only solution, its just the best, easiest to visualize and already-been-solved-and-easy-to-apply-here solution. -
It works fine until the guests find out your filthy prisoners have all been passing the wine bottles around beforehand. -
There are 3 deities, True, False, and Random.
True always speaks truthfully, False always speaks in lies, and Random can answer both ways (randomly... duh).
Determine the identities of the three deities by asking three Yes/No questions (each question must be put to exactly one god).
The deities know your language but will answer all questions in their language, in which the words for yes and no are "A" and "B", not necessarily in that order. You do not know which word means which.
Whether Random speaks true or false should be thought of like a flip of a coin. Not predetermined and completely up to chance, possible to have 100 truths (heads) and 1 false (tails), but very unlikely.
Good luck. Wow, I'm still third top poster... # Posts Per Day: 15.18 -
 Originally Posted by Neinteen There are 3 deities, True, False, and Random.
True always speaks truthfully, False always speaks in lies, and Random can answer both ways (randomly... duh).
Determine the identities of the three deities by asking three Yes/No questions (each question must be put to exactly one god).
The deities know your language but will answer all questions in their language, in which the words for yes and no are "A" and "B", not necessarily in that order. You do not know which word means which.
Whether Random speaks true or false should be thought of like a flip of a coin. Not predetermined and completely up to chance, possible to have 100 truths (heads) and 1 false (tails), but very unlikely.
Good luck. go to #1 and ask them "if #2 is random, would you say 'A'"?
if the answer is 'A', then #3 is not random.
if the answer is 'B', then #2 is not random.
go to the not random one and ask them "if you were true, would you say 'A'"?
if the answer is 'A', he's true.
if the answer is 'B', he's false.
ask the same deity "if #1 was random, would you say 'A'"?
if they're truth and the answer is 'A', #1 is random.
if they're truth and the answer is 'B', #2 is random.
if they're false and the answer is 'A', #2 is random.
if they're false and the answer is 'B', #1 is random.
the one not tagged yet can be figured out by whichever attribute is left over.
hooray for counterfactuals -
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