[Guest]

You have 240 barrels of wine, one of which has been poisoned. After drinking the poisoned wine, one dies within 24 hours. You have 5 slaves whom you are willing to sacrifice in order to determine which barrel contains the poisoned wine. How do you achieve this in 48 hours?

[Guest]

Does the person drinking the wine die within 24 hours, or do they die exactly 24 hours after drinking the wine? Just wanted to clarify...

I'm thinking...

3 of the slaves a, b and c drink out of 120 barrels, of which a and b shares 30, b and c shares 30, a and c shares 30, and a, b and c shares 30.

This will also leave 30 not drunk.

So depending on which slaves are dead you can whittle it down to 30 barrels, but I am a bit stumped after that.

EDIT - attached image of split

Edited July 21, 2009 by d31b0y

[Guest]

You have 240 barrels of wine, one of which has been poisoned. After drinking the poisoned wine, one dies within 24 hours. You have 5 slaves whom you are willing to sacrifice in order to determine which barrel contains the poisoned wine. How do you achieve this in 48 hours?

all 5 slaves drink from 15 barrels and then each pair share drinks from 15 barrels, and 15 barrels remain undrunk. this should narrow it 15 possible poisoned barrows and worse case is you lose 2 slaves. if you only lose one (or zero) slave you can then repeat this with each slave drinking from one barrel, each pair drinking another, each threesome drinking from the rest. that will find your poisoned barrel. Not sure what you do if more slaves die. i will have to think a little longer.

[Guest]

If you die in exactly 24 hours...

You only need 1 slave. He drinks 1 glass (or actually a tiny amount so that he doesnt pass out) every minute. This will take 4 hours to test them all. Then you time how long it takes for him to die. From this you find which minute he drank the poison and hence which barrel was poisoned.

[Guest]

Yeah, that's why I asked about it but I think it is meant to be taken as "within" 24 hours since that is what the OP says...

[Guest]

If you die in exactly 24 hours...

You only need 1 slave. He drinks 1 glass (or actually a tiny amount so that he doesnt pass out) every minute. This will take 4 hours to test them all. Then you time how long it takes for him to die. From this you find which minute he drank the poison and hence which barrel was poisoned.

it does say "within 24 hours"...

[Guest]

OK, I think I found a way to do it with a 14/15 success rate. At worst it will narrow it down to 2 but will normally get it down to 1 depending on how lucky you are. I'll post it up later.

[soop]

divide them into 48. Then the lettal batch gets divided into 12, 4, 2, and then 1 each.

[Guest]

all 5 slaves drink from 15 barrels and then each pair share drinks from 15 barrels, and 15 barrels remain undrunk. this should narrow it 15 possible poisoned barrows and worse case is you lose 2 slaves. if you only lose one (or zero) slave you can then repeat this with each slave drinking from one barrel, each pair drinking another, each threesome drinking from the rest. that will find your poisoned barrel. Not sure what you do if more slaves die. i will have to think a little longer.

I'll help you along cause I don't have time to solve (at work).

If you don't lose any slaves, you can still have up to 32 barrels after the first 24 hours.

[Guest]

I'll help you along cause I don't have time to solve (at work).

If you don't lose any slaves, you can still have up to 32 barrels after the first 24 hours.

not sure that is helpful. I have already solved for losing no slaves and losing one slave. it is the losing 2 slaves that is the problem, this narrows it to 2 barrels, not 1.

[Guest]

Great puzzle!

Call the slaves a,b,c,d,& e

Each slave (5 - a,b,c,d,e) drinks from 16 barrels - 80

Each unique pair (10 - ab,ac,ad,ae,bc,bd,be,cd,ce,de) drinks from 8 barrels - 80

Each unique trio (10 - abc,abd,abe,acd,ace,ade,bcd,bce,bdc,cde) drinks from 4 barrels - 40

Each unique quartet (5 - abcd,abce,abde,acde,bcde) drinks from 2 barrels - 10

Everyone (5 - abcd) drinks from 1 barrel - 1

Nobody drinks from remaining 9 barrels - 9

If 1 slave dies we have 16 suspect barrels

Each remaining slave (4), pair (6), trio (4) and quartet (1) drink from 1 barrel - 15

If 2 slaves die we have 8 suspect barrels

Each remaining slave (3), pair (3), and trio (1) drink from 1 barrel - 7

If 3 slaves die we have 4 suspect barrels

Each remaining slave (2), pair (1) drink from 1 barrel - 3

If 4 slaves die we have 2 suspect barrels

The remaining slave drinks from 1 barrel

If 5 slaves die we have a positive ID on the barrel

If 0 slaves die we have 9 suspect barrels

Each remaining slave (5) and 3 pairs drink from 1 barrel - 8

Edited July 21, 2009 by SteveK

[Guest]

Great puzzle!

Call the slaves a,b,c,d,& e

Each slave (5 - a,b,c,d,e) drinks from 16 barrels - 80

Each unique pair (10 - ab,ac,ad,ae,bc,bd,be,cd,ce,de) drinks from 8 barrels - 80

Each unique trio (10 - abc,abd,abe,acd,ace,ade,bcd,bce,bdc,cde) drinks from 4 barrels - 40

Each unique quartet (5 - abcd,abce,abde,acde,bcde) drinks from 2 barrels - 10

Everyone (5 - abcd) drinks from 1 barrel - 1

Nobody drinks from remaining 9 barrels - 9

If 1 slave dies we have 16 suspect barrels

Each remaining slave (4), pair (6), trio (4) and quartet (1) drink from 1 barrel - 15

If 2 slaves die we have 8 suspect barrels

Each remaining slave (3), pair (3), and trio (1) drink from 1 barrel - 7

If 3 slaves die we have 4 suspect barrels

Each remaining slave (2), pair (1) drink from 1 barrel - 3

If 4 slaves die we have 2 suspect barrels

The remaining slave drinks from 1 barrel

If 5 slaves die we have a positive ID on the barrel

If 0 slaves die we have 9 suspect barrels

Each remaining slave (5) and 3 pairs drink from 1 barrel - 8

what about the other 20 barrels?

[Guest]

Great puzzle!

Call the slaves a,b,c,d,& e

Each slave (5 - a,b,c,d,e) drinks from 16 barrels - 80

Each unique pair (10 - ab,ac,ad,ae,bc,bd,be,cd,ce,de) drinks from 8 barrels - 80

Each unique trio (10 - abc,abd,abe,acd,ace,ade,bcd,bce,bdc,cde) drinks from 4 barrels - 40

Each unique quartet (5 - abcd,abce,abde,acde,bcde) drinks from 2 barrels - 10

Everyone (5 - abcd) drinks from 1 barrel - 1

Nobody drinks from remaining 9 barrels - 9

If 1 slave dies we have 16 suspect barrels

Each remaining slave (4), pair (6), trio (4) and quartet (1) drink from 1 barrel - 15

If 2 slaves die we have 8 suspect barrels

Each remaining slave (3), pair (3), and trio (1) drink from 1 barrel - 7

If 3 slaves die we have 4 suspect barrels

Each remaining slave (2), pair (1) drink from 1 barrel - 3

If 4 slaves die we have 2 suspect barrels

The remaining slave drinks from 1 barrel

If 5 slaves die we have a positive ID on the barrel

If 0 slaves die we have 9 suspect barrels

Each remaining slave (5) and 3 pairs drink from 1 barrel - 8

Actually this will work if you have all 29 barrels in the first day that no one drinks from as all 5 slaves can test them in the various combinations.

Well done.

[bushindo]

I think this can save up to 243 prisoners.

Let's work backwards, and assume that we are on the second day. If we have n prisoners remaining, then we can always uniquely determine the poisoned wine in 2ⁿ barrels.

The trick is to divide the wine in the first day in such a way that the number of suspected barrels match the uniquely determinable number of barrels given the remaining prisoners. So, on the first day, divide the barrels in such a way that

There are 2⁵= 32 barrels not drunk by any of the prisoners. (There is C(5,0) = 1 way of this happening)

There are 2⁴ = 16 barrels drunk ONLY by 1 prisoner (There is C(5,1) = 5 way of this happening)

A subset of any two prisoners will uniquely share 2³ barrels between them. (There is C(5,2) = 10 way of this happening)

A subset of any three prisoners will uniquely share 2² barrels between them. (There is C(5,3) = 10 way of this happening)

A subset of any four prisoners will uniquely share 2¹ barrels between them. (There is C(5,4) = 5 way of this happening)

A subset of any five prisoners will uniquely share 2⁰ = 1 barrels between them. (There is C(5,5) = 1 way of this happening)

Multiply the number of uniquely determinable barrels by the ways of it happening, we get

32*1 + 16*5 + 8*10 + 4*10 + 2*5 + 1*1 = 243

I don't know what's wrong with the spoiler function. This is weird. I'm sure the tags are right.

Edited July 21, 2009 by bushindo

[Guest]

Label the barrels 1-240 and the slaves A-E

Split the barrels into 8 sets that contains 27 barrels and the ninth set which is the remainder. 1-27, 28-54, 55-81... etc.

The slaves should drink the following from the first set and similar for the other 7 sets (not the remainder set).

A: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 23

B: 1, 11, 12, 13, 5, 6, 7, 14, 15, 16, 24

C: 17, 2, 21, 13, 5, 19, 20, 8, 9, 16, 25

D: 17, 11, 3, 13, 18, 6, 20, 8, 15, 10, 26

E: 17, 11, 21, 4, 18, 19, 7, 14, 22, 10, 27

At most 3 will die and you can identify which ONE would be the poison in each set, which will leave you with 8 after the first 24 hours. As long as you have 3 left you can then find the poison out of the last 8. However, if you only have 2 slaves left then you can only narrow it down to 2.

If none of them die you will have 24 left with all 5 slaves which can easily be found.

[Guest]

*********WARNING SPOILER NOT WORKING SOLUTION POSTED BELOW*********

Start by having each slave drink from every barrel from a different set of 16 barrels. This means 80 barrels have been drunk from (5x16), so 160 remain.

In the same day, have every possible distinct pair from the five slaves (10 total pairs) drink from every barrel from a different set of 8 barrels. The pairs will have collectively drank from 80 barrels (10x8), so 80 remain.

In the same day (still in the first 24 hours) have every distinct possible set of 3 from the 5 slaves (10 groups of 3 slaves) drink from every barrel from a different set of 4 barrels. The groups of three will have collectively drank from 40 barrels, so 40 remain undrunk (undrank?).

In the same day, have every distinct possible group of 4 slaves (5 groups of 4 slaves) drink from every barrel from a different set of 2 barrels. The 5 groups will have collectively drank from 10 barrels, so 30 remain undrunk from.

Now we wait . . .

If no slaves die, then we know the poisoned barrel is in the set of 30 that no slave drank from. Then we follow a similar procedure as above only we have every "group" drink from 1 barrel of wine. So each slave drinks from a different barrel (5 barrels drank from), the 10 groups of 2 drink from different barrels (15 barrels drank from), the 10 groups of 3 drink from different barrels (25 barrels drank from), and the 5 groups of 4 drink from different barrels (30 barrels drank from). Wait again for 24 hours and depending on the number of slaves that die, we know which barrel was poisoned.

The procedure is the same for the different scenarios, sorry I don't have time to go through them all.

*******WARNING SPOILER NOT WORKING SOLUTION POSTED ABOVE******

[Guest]

Oops! My original solution was based on 220 (don't know how I ended up doing that

Bushindo is correct that you can have up to 243 barrels.

Call the slaves a,b,c,d,& e

Each slave (5 - a,b,c,d,e) drinks from 16 barrels - 80

Each unique pair (10 - ab,ac,ad,ae,bc,bd,be,cd,ce,de) drinks from 8 barrels - 80

Each unique trio (10 - abc,abd,abe,acd,ace,ade,bcd,bce,bdc,cde) drinks from 4 barrels - 40

Each unique quartet (5 - abcd,abce,abde,acde,bcde) drinks from 2 barrels - 10

Everyone (5 - abcd) drinks from 1 barrel - 1

Nobody drinks from remaining 9 barrels - 32

If 1 slave dies we have 16 suspect barrels

Each remaining slave (4), pair (6), trio (4) and quartet (1) drink from 1 barrel - 15

If 2 slaves die we have 8 suspect barrels

Each remaining slave (3), pair (3), and trio (1) drink from 1 barrel - 7

If 3 slaves die we have 4 suspect barrels

Each remaining slave (2), pair (1) drink from 1 barrel - 3

If 4 slaves die we have 2 suspect barrels

The remaining slave drinks from 1 barrel

If 5 slaves die we have a positive ID on the barrel

If 0 slaves die we have 32 suspect barrels

Each remaining slave (5), pair (10), trio (10), quartet (5), quintet (1) drink from 1 barrel - 31

[Guest]

Each slave takes a drink from a new barrel every 30 minutes for the first 24 hours, marking the time on each barrel as it is used (48 barrels x 5 slaves = 240 barrels)....the next day when the poisoned slave dies go back exactly 24 hours and the barrel he/she drank from is poisoned.

[Guest]

JUST REALIZED; LOL!! I'm sitting here thinking that you'd need to find the poisoned barrel specifically, when of course you are only concerned that it isn't one of the barrels you are left with.....BEEN SCRATCHIN HEAD 4 AGES 4 NO REASON!!!

I NEVER read the prior posts and VERY seldom the spoiler - those of you who have read my answers before could testify to that fact!! and admit it when I have, so I'LL BE BACK as old arnie would say (HOPEFULLY WITH A [sEMI] INTELLIGENT guess!xx d^ o ^b = RQ on the ipod just now xxx

[Guest]

The first 24 hours:

a)Each slave drinks 16 barrels himself = 80 barrels down

b)Each pair drinks 8 barrels together = 80 barrels down (10 possible pairs)

c)Each set of 3 slaves drink 4 barrels together = 40 barrels down (10 possible sets)

d)Each set of 4 slaves drink 2 barrels together = 10 barrels down (5 possible sets)

e)All 5 slaves drink 1 barrel together = 1 barrel down

f)29 barrels go untouched

The second 24 hours:

After finding out who dies, if the poisoned barrel is in group....

a) 4 slaves are left with 15 barrels, 4 slaves = 16 combinations (with the 16th being a barrel that nobody drinks)

b) 3 slaves are left with 8 barrels, 3 slaves = 8 combinations

c) 2 slaves are left with 4 barrels, 2 slaves = 4 combinations

d) 1 slave is left with 2 barrels (50/50 chance of dying!)

e) If this happens you have your barrel after 24 hours

f) 5 slaves are left with 29 barrels, 5 slaves = 32 combinations

Have each combination in each group drink one barrel and after 48 hours you should be able to tell which barrel is poisoned.

[Guest]

So I got the answer with 243 people or whatever but at first i was trying a staggering method such as if a prisoner died in this 24 hour period it limited it to x barrels he drank in the past 24 hours, so then if another prisoner died then you look in his past 24 also. This way we may be able to squeak out a few more barrels. But then of course no barrels could be drank in the last 24 because if someone didnt die you wouldnt know if it was impending doom or no one had drank it. So you gain no time. But then any overlap couldn't be greater then the possible summation(x choose y) or 2^x able because this would make it impossible if someone died in that time. So it would have to be less or equal. Equal would mean the greatest you can hope for is using all the time so it must be equal and maximized at the beginning. This gives you the original answer. Obviously if it was ever less then this it would be less efficient. Anyway this started as a anyone have any thoughts about this, but became me explaining to myself why it is the only answer. but im posting it anyway.

if anyone understands this round about logic you may want to get yourself checked for the crazies

[Guest]

If its exactly 24 hours til they die drinking 10 barrels an hour would get you to 240 in your 24 hours leaving the last 24 hours for if the last barrel was the poisoned one. 5 slaves (a,b,c,d,e)

a,b - 1

a,c - 2

a,d - 3

a,e - 4

b,c - 5

b,d - 6

b,e - 7

c,d - 8

c,e - 9

d,e - 10

what ever 2 slaves die you just look at what barrel they shared on that hour the day before.

[Guest]

First give each slave a sip of wine from 16 barrels. Then give every combination of two slaves a sip from 8 barrels; every combination of three a sip from 4 barrels, every combination of 4 a sip from 2 barrels and give all five slaves a sip from one barrel. the remaining barrels (there should be 29 if my math is correct) give to no one. This way, you can narrow it down to whichever group has whichever combination of slaves die, and that group should have just enough barrels for every combination of the remaining slaves to try one barrel (plus one for none to try). The exception is if all die (in which case it was the one they all tried) or if none of them die (in which case there are more combinations of slaves than barrels). So the poisoned barrel can be found out within the second 24 hours.

[Guest]

We test it for 24hours, every hour we test wine from 10 different barrels. Each slave taste wine from 4 different barrels, wine from each barrel should be tasted by two different slaves. So by the end of 24 hours all 240 barrls of wine is tasted. If two slaves die after completion of 24th hour, then first sample has poison. We should supply the wine from 10 barrels say A to J, in the following way.

Slave1 Slave2 Slave3 Slave4 Slave5

A B C D E

F G H I J

B C D E A

I J F G H

If barrel A has poison slave 1&5 will die, B - 1&2, C - 2&3, D - 3&4, E - 4&5, F - 1&3, G - 2&4, H - 3&5, I - 1&4, J - 2&5.

[Guest]

Each would drink from 72 barrels

firstly, since we have 5 slaves, we divide this in 6 groups 240/6 = 40

now, each slave would own 1 gp,i.e.40 and 8 from 6th gp and additionally would drink from 8 of other 4 gps;the 8 other would be mutually exclusive for each slave.

it would look like

40 40 40 40 40 40

a 40+8 8 8 8 8 -

b 8 40+8 8 8 8 -

c 8 8 40+8 8 8 -

d 8 8 8 40+8 8 -

e 8 8 8 8 40+8 -

eg; if 'a' alone dies; its 8 barrels from the a's 40+8 untouched by b,c,d,e; else a dies with be its gp of 8 both a n b drank...likewise.

if in case god forbid , no one dies. then v have 32 left.

here, important point is

if none dies v have 32 to check by 5; it could be checked since 2^5=32

if 1 dies v have 16 to check by 4;it could be checked since 2^4=16

if 2 die v have 8 to check by 3; it could be checked since 2^3=8

---I tried but can't help spaces, so appologies for formatting

Edited July 22, 2009 by Vikram@den

[Guest]

We test it for 24hours, every hour we test wine from 10 different barrels. Each slave taste wine from 4 different barrels, wine from each barrel should be tasted by two different slaves. So by the end of 24 hours all 240 barrls of wine is tasted. If two slaves die after completion of 24th hour, then first sample has poison. We should supply the wine from 10 barrels say A to J, in the following way.

Slave1 Slave2 Slave3 Slave4 Slave5

A B C D E

F G H I J

B C D E A

I J F G H

If barrel A has poison slave 1&5 will die, B - 1&2, C - 2&3, D - 3&4, E - 4&5, F - 1&3, G - 2&4, H - 3&5, I - 1&4, J - 2&5.

Anantham, ur approach is splendid, and this will help me in approaching other probs in future.

For this particular problem, I think we should not use this since, it does not say 24hr exact, but within 24 hrs, and even num of hrs are a constant figure below 24th, it remains inconclusive as to which hr barrel caused the death.

But surely your approach makes this otherwise difficult looking problem so easy. thanks man.