[Guest]

You have been presented with 9 bottles of antique, and rare wines. Out of these 9, one bottle is poisoned and you don't know which (of-course). You also have 2 mice which can be used for wine testing. If a mice tastes the poisoned wine , it'll die instantly.

Can you determine, with 2 mice and two test, which bottle is poisoned?

Addition: If you solve it, can you find the poisoned bottle from 240 bottles with 5 mice..but two test only

[soop]

Ok, I've almost got it...

so you split the wine into groups of 3 x 3. Then you mix a small amount of wine from two of the groups, and give a taste of one to one mouse, and the other to the other mouse.

If they both live, gat them to each taste one of the remaining wines. if they both live it's the last one, otherwise the dead mouse was drinking the poisoned wine.

However, if one of the mice dies while drinking the mixed wine, then you have one mouse and three suspect bottles.

You could mix two of the remaining wines I suppose.

Wait, I think I know... You can add one extra bottle to both mixes. If both mice die, it was the one you put in both mixtures. But that still doesn't help me with my problem. can you PM me the answer?

[Guest]

When you say two test only do you mean that each mice can only taste one lot of wine?

[Guest]

Or do you mean two tastes per mouse?

[Guest]

Think I've got it

Mix three of the wines and give the first mouse a taste. If he dies, give a taste of one of the three to the second one, if he dies, job done. If he doesn't die you know its one of the other two, so give him a taste of one other. He either dies (job done) or doesn't, in which case its the other one. If the first mice doesn't die, then give that mouse a taste of another three wines mixed together. Repeat above if he dies. If he doesn't die, repeat the above with the three that you haven't given to either mouse yet.

[Guest]

edit:forgot the spoiler

Edited August 5, 2010 by kristmark1

[Guest]

Test: Mouse1 tests Bottle 1 & 2. If it dies you use Mouse2 to find which of the two is poisoned.Mouse2 tests Bottle 3 & 4. If it dies Mouse1 is still alive and you use it to find which of the two is poisoned. Test 2: If both survive you go to test 2. Mouse1 tests Bottle 5 & 6. If it dies you use Mouse2 to find which of the two is poisoned-just like in test 1 test 2. Mouse2 tests Bottle 7 & 8. If it dies you (again) use the living one to find the poisoned bottle. If both mice are still pretty much alive it's obviously bottle 9.

Edit: cleonew's one is better but that one still works

Edited August 5, 2010 by kristmark1

[Guest]

Call the bottles 1,2,3,4,5,6,7,8 and 9 and the Mice A and B.

Give A a mix of 123 and give B a mix of 145.

If they both die, 1 is poisoned.

If only A dies, then it is either 2 or 3. Have B test one of them.

If only B dies, then it is either 4 or 5. Have A test one of them.

If both live then:

Give A a mix of 67 and B a mix of 68.

If they both die, 6 is poisoned.

If only A dies, 7 is poisoned.

If only B dies, 8 is poisoned.

If they both live, 9 is poisoned.

[Guest]

Addition: If you solve it, can you find the poisoned bottle from 240 bottles with 5 mice..but two test only

I could find the poison with 243 bottles of wine under those conditions.

There are 2 possible outcomes when one mouse drinks in a single test (life or death) so one mouse in one test can distinguish between 2 cases (or 2 bottles of wine) in one test.

There are 4 possible outcomes when two mice drink in a single test (Life/life, death/death, life/death, death/life) so two mice can distinguish between 4 cases (or 4 bottles of wine) in one test.

3 mice can distinguish between 8 cases in one test.

4 mice can distinguish between 16 cases in one test.

5 mice can distinguish between 32 cases in one test.

Now we have two tests. Optimally we'll use all five mice in the first test and cover all 32 ways that things could turn out.

There's only one way for all 5 mice to die leaving us with zero mice for test 2. This case covers 1 bottle.

There are 5 ways for 4 mice to die leaving us with 1 mouse for test 2. This case covers 5*2=10 bottles.

There are 10 ways for 3 mice to die leaving us with 2 mice for test 2. This case covers 10*4=40 bottles.

There are 10 ways for 2 mice to die leaving us with 3 mice for test 2. This case covers 10*8=80 bottles.

There are 5 ways for only one mouse to die leaving us with 4 mice for test 2. This case covers 5*16=80 bottles.

There's only one way for no mice to die leaving us with all 5 mice for test 2. This case covers 1*32=32 bottles.

Adding this all up, an optimal strategy (Which I am confident exists) can distinguish one poison bottle out of 243 using 5 mice and 2 tests.

I don't have time right this minute though to type out the exact strategy since it will likely take quite a bit of space.

[Guest]

First test give mouse A bottles 123 and mouse B bottles 345, if both die then 3 is the poisoned one, if one die test one of the two other bottles on the other mouse and see, if both live give mouse A bottles 6 7 and mouse B bottles 7 8 and you'll know which one was poisoned

Split the bottles to 9 groups (some 26 and some 27) and take two mice and do the same exact test as for part 1 on them, now you have 27 bottles that you know one of them has the poison, split them into 9 smaller groups and do the same test on mice 3 and 4 and you'll be left with one mouse (minimum) and 2 or 3 bottles, test it on two bottles and you'll know which is the poisoned one...

You can also do it on one mouse each time by splitting into 3 groups and testing group A then group B (if it lives through both then it's group C)...

For any number of bottles you will always need: ceil(log₃(n)) mice

(ceil means ceiling, as in ceil(1.05)=2)

Edited August 5, 2010 by Anza Power

[Guest]

Ok, since 5 mice and 240 bottles is just so big, I'll try to demonstrate the method with 3 mice and 27 bottles.

So for the first test we want to mix the bottles such that we use every combination of living and dead mice to full effect.

Since the case where all 3 mice die doesn't give us any mice to use for test 2, we want that case to incriminate only one bottle:

Mouse A drinks from: 1

Mouse B drinks from: 1

Mouse C drinks from: 1

There are 3 ways that two mice can die, leaving one mouse for test 2. Since one mouse can distinguish between 2 bottles, it's ok to have each result incriminating 2 bottles:

Mouse A drinks from: 1,2,3,4,5,

Mouse B drinks from: 1,2,3,    6,7

Mouse C drinks from: 1,    4,5,6,7

There are 3 ways that only one mouse can die, leaving 2 mice for test 2. Since 2 mice can distinguish between 4 bottles, it's ok to have each result incriminating 4 bottles:

Mouse A drinks from: 1,2,3,4,5,   ,8,9,10,11

Mouse B drinks from: 1,2,3,    6,7,          12,13,14,15

Mouse C drinks from: 1,    4,5,6,7,                     ,16,17,18,19

There is only one way which all mice can survive to move on to test 2. Since 3 mice can distinguish between 8 bottles, it's ok to have this result incriminating 8 bottles by having no mice taste 20-27.

If all mice die, the poison has to be in bottle 1.

If A and B both die, the poison was either in 2 or 3. Have C test one of them. All other cases where 2 mice die follow the same pattern.

If only A dies, the poison was either in 8,9,10 or 11. Have B try 8+9 and C try 8+10. The result will reveal the poison. All other cases where only one mouse dies are similar.

If all the mice survive the poison was either in 20,21,22,23,24,25,26 or 27. Have A try 20+21+22+24, B try 20+21+23+25 and C try 20+22+23+26. The result will reveal where the poison is.

Now expand this method to 5 mice and 243 bottles and remove the last 3 bottles from the final solution and voila!

[Guest]

For Part 1-

Both Tuckleton and Anza Power got it right but Tuckleton got there first

For Part 2-

Anza Power got it right..

divide the bottles into set of 9 and proceed in the same way as if you had just 9 bottles..

good work guys

Edited August 6, 2010 by sak_lko

[Guest]

For Part 2-

Anza Power got it right..

divide the bottles into set of 9 and proceed in the same way as if you had just 9 bottles..

Wait, that method doesn't work... You only have 2 tests. Anza's method requires up to 6 tests. Or is it 2 tests per mouse, taken whenever you want?

[Guest]

When I first saw this I immediately thought of the weighing puzzles. It's much like a weighing puzzle, except the scale has a chance to be broken after weighing.

In any case, I like Tuckleton's logic

Given two total tests which may each include up to five living mice, they could run something like this:

On the first test, mice sample from a mixture of the following bottles

Mouse 1 - 1,2,3,4,5,6,7,8,9,      12,13,14,15,16,17,18,19,20,21,22,23,            28,29,30,31,32,33,34,35,            40,41,42,43,                        52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,                                                                                                                                                                                132,133,134,135,136,137,138,139,140,141,142,143,144,145,146,147

Mouse 2 - 1,2,3,4,5,6,7,    10,11,12,13,14,15,16,17,18,19,            24,25,26,27,28,29,30,31,            36,37,38,39,            44,45,46,47,            52,53,54,55,56,57,58,59,                                                                        84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,                                                                                                                                                                148,149,150,151,152,153,154,155,156,157,158,159,160,161,162,163

Mouse 3 - 1,2,3,4,5,    8,9,10,11,12,13,14,15,            20,21,22,23,24,25,26,27,            32,33,34,35,36,37,38,39,                        48,49,50,51,                        60,61,62,63,64,65,66,67,                                                84,85,86,87,88,89,90,91,                                                        108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,                                                                                                                                                                164,165,166,167,168,169,170,171,172,173,174,175,176,177,178,179

Mouse 4 - 1,2,3,    6,7,8,9,10,11,            16,17,18,19,20,21,22,23,24,25,26,27,                                    40,41,42,43,44,45,46,47,48,49,50,51,                                                68,69,70,71,72,73,74,75,                                                92,93,94,95,96,97,98,99,                                108,109,110,111,112,113,114,115,                                124,125,126,127,128,129,130,131,                                                                                                                                                                                                180,181,182,183,184,185,186,187,188,189,190,191,192,193,194,195

Mouse 5 - 1,    4,5,6,7,8,9,10,11,                                                28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,                                                                        76,77,78,79,80,81,82,83,                                                100,101,102,103,104,105,106,107,                                116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,                                                                                                                                                                                                                                                                196,197,198,199,200,201,202,203,204,205,206,207,208,209,210,211

You'll see this follows the logic that for each possible number of remaining living mice, there is a subset of possibilities that can be tested with only those mice. For example, if only one mouse lives (say number 5) then only bottles tested by mice 1-4 need be checked (bottles 2 & 3) and so test mouse 5 on either. If 2 mice live (say 4 and 5) that leaves 4 bottles (12-15) to be checked, giving

Mouse 4 - 12,13

Mouse 5 - 12,   14

If 3 mice live (say 3, 4, and 5) that leaves 8 bottles (52-59) to be checked

Mouse 3 - 52,53,54,   56

Mouse 4 - 52,53,   55,   57

Mouse 5 - 52,   54,55,      58

If 4 mice live (say 2, 3, 4, and 5) that leaves 16 bottles (132-147) to be checked

Mouse 2 - 132,133,134,135,    137,138,    140,        143

Mouse 3 - 132,133,134,    136,137,    139,    141,        144

Mouse 4 - 132,133,    135,136,    138,139,        142,        145

Mouse 5 - 132,    134,135,136,            140,141,142,            146

And of course, if all 5 mice live, that leaves the untested bottles (212-240) to be checked

Mouse 1 - 212,213,214,215,216,    218,219,220,    222,223,    225,        228,229,    231,        234,            238

Mouse 2 - 212,213,214,215,    217,218,219,    221,222,    224,    226,    228,    230,    232,        235,            239

Mouse 3 - 212,213,214,    216,217,218,    220,221,    223,224,        227,    229,230,        233,        236,            240

Mouse 4 - 212,213,    215,216,217,    219,220,221,            225,226,227,            231,232,233,            237

Mouse 5 - 212,    214,215,216,217,                222,223,224,225,226,227,                        234,235,236,237

This can all be done in two well defined (if complicated) tests.

[Guest]

If the tests can be done at seperate times.

First use mouse A to taste bottles 123. If mouse A dies then use mouse B to test bottle 1. If he survives test bottle 2. If he survives bottle 3 is poision.

If mouse A does not die after tasting bottles 123 then use mouse A to test bottles 456. If mouse A dies then Mouse B tests bottle 4. If he survives test bottle 5. If he survives bottle 6 is poision.

If mouse A does not die after tasting bottles 123 or 456 then use mouse B to test bottle 7. If he survives test bottle 8. If he survives the poision is in bottle 9.

This method uses 4 tastes maximum with 2 tastes for each mouse but must be done in sequence ie you cannot do two tastes at once

[Guest]

Given 5 vessels, one for each mouse, one can pour in a mix that can test for the poison.

For the following, where I present a binary representation, the binary digit 1 represents a vessel in which wine from n distinct bottles may be poured. i.e., 01001 would represent a sample poured from a wine bottle into the second and fifth vessel. The number n of such bottles is also provided by the quantity that may be tested in a second test.

If any mice die in the first test the poison is in one of the bottles sampled from in the first test.

If 5 mice die after the first test, the largest quantity that can be tested in a second test is 0. But, then the poison would have been found if all 5 mice died. This can only occur if all mice sampled from the same one bottle, thus a sample for each vessel from 1 and only 1 bottle can be added.

1

If 4 mice die after the first test, the largest quantity that can be tested in a second test is 2. There are 5 combinations where 4 mice may die: {00001, 00010, 00100, 01000, 10000}. Therefore, we can add samples from 2 bottles for each of the 5 mice.

10 = 5x2

If 3 mice die after the first test, the largest quantity that can be tested in a second test if 4. There are 10 combinations where 3 mice may die: {00011, 00101, 00110, 01001, 01010, 01100, 10001, 10010, 10100, 11000}. Therefore, we can add samples from 4 bottles for each pair of 5 mice.

40 = 10x4

If 2 mice die after the first test, the largest quantity that can be tested in a second test is 8. There are 10 combinations where 2 mice may die: {11100, 11010, 11001, 10110, 10101, 10011, 01110, 01101, 01011, 00111}. Therefore, we can add samples from 8 bottles for each triad of 5 mice.

80 = 10x8

If 1 mouse dies after the first test, the largest quantity that can be tested in a second test is 16. There are 5 combinations where 1 mouse may die: {11110, 11101, 11011, 10111, 01111}. Therefore, we can add samples from 16 bottles for each quad of 5 mice.

80 = 5x16

Total bottles in first and second test with poison in first test batch: 1 + 10 + 40 + 80 + 80 = 211.

If 0 mice die after the first test, the poison must be in the remaining unsampled bottles. The maximum number of bottles that can be tested with 5 mice is 32. {00000, 00001, 00010, 00011, 00100, 00101, 00110, 00111, 01000, 01001, 01010, 01011, 01100, 01101, 01110, 01111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111}.

Total bottles that can be tested in a second test with the poison in the second test batch: 32.

Total bottles that can be tested, therefore, is 243. The quantity of 240 is within our means of testing.

[Guest]

I will be naming the mice A,B,C,D,E.

Mouse E, the last mouse, cannot find the poision if there are more than three bottles left (except by chance). He will test 2 of the bottles, 1 at a time, if he dies on either 1 you know the poision and if he lives the last bottle is the poision.

E - 1 1 1

Since mouse E can only have 3 bottles left to find the poision we must use mouse D to test 3 bottles at a time. He will test 6 of the bottles, 3 at a time, if he dies on either test you know the poision is in that group of 3 and if he lives the poision is in the last group of 3 bottles.

D - 3 3 3

Since mouse D can test no more than 9 bottles then mouse C must test nine bottles at a time. He will test 18 of the bottles, 9 at a time, if he dies on either test you know the poision is in that group of 9 and if he lives the poision is in the last group of 9 bottles.

C - 9 9 9

Since mouse C can test no more than 27 bottles mouse B must test 27 bottles at a time. He will test 54 of the bottles, 27 at a time, if he dies on either test you know the poision is in that group of 27 and if he lives the poision is in the last group of 27 bottles.

B - 27 27 27

Since mouse B can test no more than 81 bottles mouse A must test 81 bottles at a time or 80 in this case. He will test 162 of the bottles, 81 at a time, if he dies on either test you know the poision is in that group of 81 and if he lives the poision is in the last group of 81 bottles.

A - 81 81 81

Using this method the max number of bottles that 5 mice can search for one bottle of wine is 81 x 3 = 243 bottles of wine.