Vol. 12: Puzzle This
Poisoned wine and the three wise sages.
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Showing messages 1 through 18 of 18.

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 15:46:21 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 23:27:19 Pacific Time

False Positives
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In the days of yore, as determined by the "so last millenia" styles the King and his knight are sporting, they didn't necessarily have binary numbers on the mind, nor matrix algebra. He probably would have simply had prisoner 1 drink from every other bottle, prisoner two drink from every 4th bottle, prisoner 3 from every 8th bottle, etc. Same as binary, but no geeks required.
And only 999 bottles need to be tested. The bottle with the spider web can be drunk by the King immediately, to enjoy the whole macabre affair.
And don't forget the King is evil, and looks English judging by the garb. He would probably arrest 1000 Scots or Irish to test the bottles, and ignore tricky mathematics altogether.
On the other hand, he could simply find the one bottle with a little wine missing. You know for sure if the assassin was Irish like me, he would have had a gulp before poisoning it. Aha!Posted by DrillPressRelease on February 16, 2008 at 06:30:04 Pacific Time

Poison WineEven 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 11:25:35 Pacific Time

Poison WineEven Odds
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If you wanted to minimize the number dead I believe you can get away with killing only 1 prisoner.
Since the poison works in exactly 1 month, lets take the long value or a month as 31 days. The party is in 5 weeks (35 days). This leaves a window of 4 days.
Number the bottles 11000 and give the prisoner a drop from each bottle every 5 mins. 5000mins ~= 3.5 days
Now 31 days after the first drop was administered if the prisoner dies you know it was the first bottle, 5 mins later, the second and so on.
Of course this won't work at all unless you know precisely how long a month is. However if you can be guaranteed that a month is exactly either 28,29,30 or 31 days you can shorten the interval between drops administered to one minute and start monitoring 28, 29, 30 and 31 days later.Posted by smoil on December 03, 2007 at 17:28:26 Pacific Time

Poison WineEven Odds
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Sorry, probability angle is wrong. This actually will result in more deaths. No longer merciful.Posted by strasberg on November 21, 2007 at 11:29:17 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 mtroniks on November 18, 2007 at 13:34:51 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 18:30:39 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 5dimensional 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 5dimensional cube!Posted by walaszek on November 16, 2007 at 07:51:34 Pacific Time

Poison wine
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If you want to minimize the possible casualties slightly, you could use a 3D matrix and solve each axis over the course of 3 days. That reduces the highest death count to 3.Posted by suburbanist on December 20, 2007 at 14:21:54 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 000999, number the prisoners 09. 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 10:36:23 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 ballcolor 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 ballcolors, the second guesser need only see the third ballcolor, and the third guesser might as well be blind.Posted by sodium11 on November 15, 2007 at 10:14:16 Pacific Time

Black or White?
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What happens when all sages have picked a white marble?Posted by notsobright on November 29, 2007 at 10:45:20 Pacific Time

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 21:10:55 Pacific Time

Poison Wine
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Here's my idea, feel free to tell me if it's crazy.
As the drinker of the poison will die exactly 5 weeks after consuming it, if that is precise to the minute, then time can be used as the variable that determines the poisonous bottle.
Have each prisoner drink from 100 of the bottles, at a rate of one tasting per bottle per 30 minutes, labeling each bottle by when it was sampled, and by which jailbird. In just over 2 days, this will cover all 1000 bottles (if said prisoners don't die from exhaustion or being sauced first). A month or so later, the time at which a particular prisoner dies should equate to a specific bottle.
Being as the longest month is 31 days, and 5 weeks is 35 days, you could even pace your prisoners consumption, though what would be fun about that?Posted by chiaseth@yahoo.com on November 15, 2007 at 21:57:38 Pacific Time

Poison Wine
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I think the idea was that they would die 1 month to the day, but not necessarily to the minute or second. Otherwise you would just need one prisoner (or 3 to be reasonably sure). Just have one prisoner drink from a different bottle each minute, then time his death to the minute and it tells you which bottle was poisioned. Have 3 prisoners do this just in case one dies of natural causes. Thus if one dies at one time and two dies another time, you can use the time of death of the two and be fairly certain, leaving only the possibility that two prisoners would die from natural causes at the EXACT same time  the risk you have to take if you want to drink potentially tainted wine.Posted by DeuceSevin on January 07, 2008 at 16:55:41 Pacific Time

Poison Wine
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Assuming a month to be four weeks, only eight prisoners would be needed.
Using the same binary coding as described in the original solution,but spaced over six days(the 7th is the day of the party, too late!).The prisoners would only need to test 167 sips of wine a day over six days. This would only require 8 binary bits. The date of death would be in six parts.Posted by carterchas on November 28, 2007 at 04:53:03 Pacific Time

Poison Wine
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Assuming a month to be four weeks, only eight prisoners would be needed.
Using the same binary coding as described in the original solution,but spaced over six days(the 7th is the day of the party, too late!).The prisoners would only need to test 167 sips of wine a day over six days. This would only require 8 binary bits. The date of death would be in six parts.Posted by carterchas on November 28, 2007 at 04:52:49 Pacific Time
Showing messages 1 through 18 of 18. 
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