AcesUp46
Friday, November 24th, 2006, 2:41 PM
QUOTE (LongLiveYorke @ Friday, November 24th, 2006, 2:28 PM)

Would you like to explain your answer?
I worked backwards. Let n be the number of glasses of wine.
Trivial. When n=2, you just need 1 taster.
When n=3, you need just 2 tasters.
When n=4, you still just need 2 tasters. You can leave out 1 glass (say Glass 4) cos if that Glass is poisonous, then if no one dies, that glass is the one. So 3 glasses left. So you let Taster A taste Glass 1 and Glass 2. While Taster B tasters Glass 3 and Glass 2. If ONLY Taster A dies, then Glass 1 is the culprit. If ONLY Taster B dies, then Glass 3 is the culprit. If BOTH die, then Glass2 is culprit. If NONE die, then Glass 4 it is.
Note that 2 = 2^1.
2^1 < 3 < 2^2 ...Hence 2 tasters is sufficient for n=3 & 4
At this point, I figure that there are plateus at which x number of tasters are sufficient, beyond which, you move to the next level of tasters. ie 2 tasters may be sufficient for n<= 4. And then say 5 tasters is sufficient for n<= 10 for example.
Anyway, I realized that effectively, you're working with powers of 2. And note that 2^9 < 1000 < 2^10
Thus, 10 tasters, since you're need to operate at the 10th power of 2.
EDIT: Don't look at this problem in a linear way. Instead formulate a matrix where each individual taster tastes his unique selection of wine. No 2 tasters are tasting the same portfolio of wine. Hence, by looking at which tasters died, you can pinpoint which bottle of wine is poisonous since your matrix is set up in such a way that all the dead tasters share ONLY 1 glass of wine. Hope my answer is clear enough....