given 1000 bottles of juice, one of them contains poison and tastes
bitter. Spot the spoiled bottle in minimum sips?

• Take 500 of the bottles and put one drop from each of them into an empty bottle. Taste the juice in that bottle. If it's bitter, you know the poison is in one of them; if not, it's in one of the other 500. Now, take 250 of the bottles you've chosen and put one drop from each into another empty bottle and taste it. Now, you've narrowed it down to 250 bottles. Repeat, and narrow it down to 125... then 63... then 32... 16...8... 4... 2... and finally 1. And you only had to taste the juice 10 times, instead of possibly 999.
• 12 years agoHelpfull: Yes(96) No(10)
• Take 500 of the bottles and put one drop from each of them into an empty bottle. Taste the juice in that bottle. If it's bitter, you know the poison is in one of them; if not, it's in one of the other 500. Now, take 250 of the bottles you've chosen and put one drop from each into another empty bottle and taste it. Now, you've narrowed it down to 250 bottles. Repeat, and narrow it down to 125... then 63... then 32... 16...8... 4... 2... and finally 1. And you only had to taste the juice 10 times, instead of possibly 999.
• 12 years agoHelpfull: Yes(11) No(13)
• In any case, minimum 999 bottles are to be checked to be sure about spoiled bottle.
• 12 years agoHelpfull: Yes(5) No(48)
• the minimum no.of sips will be 10.
  bcoz if we are considering the each bottle which is numbered as in binary digits with 0's and1's then we can represent 1000bottles with 10 digits since 2pow10 is 1024..so consider from the small no..suppose we have 2bottles then we require only 1person either bottle 0 or 1 was tasted if there are 3bottles then we require 2members for checking that is if 0 means we are sipping that if 1 we are tasting that one and the no.of digits will represent the persons....
• 11 years agoHelpfull: Yes(5) No(3)
• 1 sip is the minimum try , if the selected bottle is poison.
• 11 years agoHelpfull: Yes(4) No(7)
• Take 500 of the bottles and put one drop from each of them into an empty bottle. Taste the juice in that bottle. If it's bitter, you know the poison is in one of them; if not, it's in one of the other 500. Now, take 250 of the bottles you've chosen and put one drop from each into another empty bottle and taste it. Now, you've narrowed it down to 250 bottles. Repeat, and narrow it down to 125... then 63... then 32... 16...8... 4... 2... and finally 1. And you only had to taste the juice 10 times, instead of possibly 999.
  
  We can do it in log(n) and that would be 10 sips. Of course it can be done in 1000 sips by checking each bottle but to do it in 10 sips you can take one drop from 500 bottles and mix them, if it is sour than the bottle is in those 500 or it is in different 500.Then out of those 500 you take 250 and do the same and rest is the binary search .
• 11 years agoHelpfull: Yes(3) No(1)
• Take 500 of the bottles and put one drop from each of them into an empty bottle. Taste the juice in that bottle. If it's bitter, you know the poison is in one of them; if not, it's in one of the other 500. Now, take 250 of the bottles you've chosen and put one drop from each into another empty bottle and taste it. Now, you've narrowed it down to 250 bottles. Repeat, and narrow it down to 125... then 63... then 32... 16...8... 4... 2... and finally 1. And you only had to taste the juice 10 times, instead of possibly 999.
• 10 years agoHelpfull: Yes(3) No(0)
• divide bottles into a=300,b=300,c=300,d=100 groups.
  taste 1 sip combination of a with that of b.
  if bitter taste found, then move into that group. otherwise taste any one of a or b with c to find bitter one. if found, move into c group. else move into d group.
  Now, analysing the worst case, it will take 2 sips atmost.
  Now we have 300 bottles remaining(not considering d group as it will not be worst case). divide into a=100,b=100,c=100. It will take just 1 sip to find out the bitter group. Total sips=3
  100 remaining. divide as a=30,b=30,c=30,d=10. Same case as 1st one. total sips=5
  30 remaining. divide as a=10,b=10,c=10. same as 2nd case. total sips=6
  10 remaining. divide as a=3,b=3,c=3,d=1. same as 1st case. total sips=8
  3 remaining. divide as a=1,b=1,c=1. sames as 2nd case. totals sips =9
  
  So, in worst case, mainimum sips required will be 9
• 10 years agoHelpfull: Yes(2) No(3)
• Hello, First Detective will call everybody(n-1) and gather all information and then every else(n-1) call him and get all the information.
  So total calls made:: 2(n-1)
  (Told My brainy Wife Kunju) :):)
• 10 years agoHelpfull: Yes(1) No(3)
• Friends Ur explanations are cool.. But why consider 500 bottles at the first time why not 501 or more than 500?? If after taking one drop from each of those 500 bottles anyone finds it bitter, then it will be bitter too if he choose 502 bottles.. Hope U understand my ques.. I understand clearly the logic which is like binary search... Although It would be easier to calculate minimum sips, still I'm in little bit doubt..
• 11 years agoHelpfull: Yes(0) No(3)
• minimum=1
  maximum=10;
• 10 years agoHelpfull: Yes(0) No(2)
• 1
  if we take the first sip and if its contains poison then we got it...hence MINIMUM is 1 and MAXIMUM is 999
• 10 years agoHelpfull: Yes(0) No(1)
• 1
  since the question says that you have to identify the poisoned bottle in minimum sips.therefore if he tastes first bottle and found it bitter means he has found the poisoned bottle in 1st sip itself.Hence the answer is 1.
• 10 years agoHelpfull: Yes(0) No(0)
• in 1 sip out of 2 specimen the poisonous one can be found out.
  so it will be n no. of sips where 1000
• 9 years agoHelpfull: Yes(0) No(0)
• i think we had to taste the juice with a minimum of 10 times since if consider this 1000 bottles each with binary digits 0 or 1 then we represent these 1000 bottles with 10 digits since 2 pow 10=1024.so ans might be 10
• 9 years agoHelpfull: Yes(0) No(0)