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• Poison Wine
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  I came up with a different answer, mainly because the 5 week limit allowed me to use time to encode additional information (as it takes exactly one month for the poison to work):
  All bottles are numbered 000 -> 999
  Prisoner 1 = Drinks all bottles ending in 0 on day 1, 1 on day 2, 2 on day 3. Shorthand this to P0 = 0,1,2
  P2 = 3,4,5
  P3 = 6,7,8
  P4 = 0,1,2 in the tens digit
  P5 = 3,4,5 tens
  P6 = 6,7,8 tens
  P7 = 0,1,2 in hundreds digit
  P8 = 3,4,5 hundreds
  P9 = 6,7,8 hundreds
  P10 = 9 in ones, 9 in tens, 9 in hundreds
  
  4 weeks later, over the course of 3 days, 3 people die. The day + person gives each digit.

  Posted by oberman77 on November 14, 2007 at 18:10:55 Pacific Time

• Black or White?
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  What's interesting about this puzzle is that it does not require the second and third guessers to see the ball-color of the previous guessers.
  
  All of the necessary information can be passed forward verbally, as in "Point of Gnome Return" from issue 11. The first guesser sees both other ball-colors, the second guesser need only see the third ball-color, and the third guesser might as well be blind.

  Posted by sodium11 on November 15, 2007 at 07:14:16 Pacific Time

• Poison Wine
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  My solution was similar in spirit to Oberman77's, but not as elegant, as it requires a fourth day of testing:
  
  Number the bottles 000-999, number the prisoners 0-9. Day one, they solve for the hundreds digit of the poisoned bottle, day two the tens digit, and day three the ones digit.
  
  The problem occurs when the first two digits are different, and the last is equal to either of the first two. The prisoner whose death would have determined the last digit is already dead. Thus, a fourth day of testing is needed, when each prisoner drinks from the bottles matching his number+1 in the ones digit.
  
  Oberman77's solution demonstrates that the problem can be solved with only nine prisoners: If P10 is eliminated, any 9's in the serial number of the poisoned bottle can be deduced from the survival of the other prisoners.
  
  It is clear from Oberman77's solution, as well as the official solution, that the scenario allows for the collection of more data than is necessary to solve the puzzle.

  Posted by sodium11 on November 15, 2007 at 07:36:23 Pacific Time

• Poison wine
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  I would offer a slightly different way of explaining the solution. Imagine that you divided the entire wine collection into a 2 x 2 x 2 x 2 x 2 array (i.e., a 5-dimensional array). (You would have some extra spaces because 2^5 is 1024, but no matter.) You then divide the 10 prisoners into 5 pairs, which member of the pair taking a sip from one half of the array; each prisoner's half would be unique and based their assignment to the array. After 1 month, five prisoners will die, which will provide the king the specific coordinates (x,y,z,a,b) in the array that correspond to the tainted wine. To figure it out, you just have to image a 5-dimensional cube!

  Posted by walaszek on November 16, 2007 at 04:51:34 Pacific Time

• Poison Wine
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  Now, now, no need to overwork the problem.
  
  The poison is so strong that *any amount* is fatal. It is also *immediately* fatal after exactly 1 month (4 weeks).
  
  You've then got a week of testing during which a *single drop* from any bottle will prove immediately fatal.
  
  You can plow through 100 drops per prisoner in a couple hours, cover the entire batch, *and* if one of them falls over dead the moment the poison hits their mouth, you can stop testing.

  Posted by deadguy on November 17, 2007 at 15:30:39 Pacific Time

• human multiplexer
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  This is a simple binary number trick. dead=1 alive =0. Label the bottles 0 through 999, and attach a binary value to each prisoner.example prisoner one will drink from bottles 1,3,5..etc prisoner two from bottles 2,3 6,7.. etc all the way up to prisoner 10 who drinks from bottle 512 on.
  the result will be a death pattern that can be expressed as the binary number of the poisoned bottle.

  Posted by m-troniks on November 18, 2007 at 10:34:51 Pacific Time

• Poison Wine-Even Odds
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  The binary method, although the simplest method, leaves certain prisoners with a greater chance of death (Prisoner 1: 50%, Prisoner 2: 25%...). There is a different approach: combinations (still using a binary bookkeeping system). There are 252 different combinations of 5 prisoners that can be made. Similarly: 210 combos for groups of 4 or 6 prisoners, 120 combos for groups of 3 or 7, and 45 combos for groups of 2 or 8. This gives you 252+ 2*210+ 2*120+ 2*45 = 1002 different possibilities. Just enough that you can be called the merciful king.

  Posted by strasberg on November 21, 2007 at 08:25:35 Pacific Time

• False Positives
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  To arbitrarily complicate the matter, what about the chances of one or more prisoners dying naturally and providing a corrupted result?
  
  Since we can determine the poisoned bottle with only one day of testing, there are still 3 days left to do error checking (assuming a granularity of one day for death accuracy).
  
  How many false positive deaths in the first round of testing can happen and still allow a 100% certain identification of the poisoned bottle?

  Posted by thatguyoverthere on December 21, 2007 at 20:27:19 Pacific Time

• Black or White?
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  I don't know about this puzzle, because it seemed vaguely worded to me. For instance, it never explicitly stated that the sages could look at the stones their colleagues had.
  
  Here is something strange I found assumed in the solution. In this game, there really is only three possible results for stones not chosen. They are:
  
  1.) 2 black.
  2.) 2 white.
  3.) 1 black, 1 white.
  
  Now, for the first option, once the sages see that the remaining stones are both black, they can easily deduce the color of their stones.
  
  The second option logically points to the fact that 2 black stones were picked, along with 1 white stone. Rather than pass, why wouldn't the first sage look at the odds of him being wrong? It's 2 to 1, a 50% chance that he would be right or wrong! With those odds, on a test with no apparent penalties for guessing, it would seem criminal to NOT guess.
  
  The same goes for the third possibility: the sages collectively have 1 black stone and 2 white stones. Again, 2 to 1 odds favoring white.
  
  So I don't see why the first sage would pass at all when he stands good odds of guessing a right answer.

  Posted by Corus on December 29, 2007 at 12:46:21 Pacific Time