Here is a truth table for my proposed solution of this week’s Puzzler. An “S” means that rat samples a small amount of wine from the bottle. It’s a simple solution with only four bottles of wine and four rats.

Decimal 8 4 2 1

Rat 1 Rat 2 Rat 3 Rat 4

Bottle #1 S

Bottle #2 S

Bottle #4 S

Bottle #8 S

But there are 13 bottles of wine. So the binary numbering sequence continues. Make sure Rat 1 only samples bottles with an S in the far left column, Rat 2 only samples bottles with an S in the next column to the right, and so on.

Bottle #3 S S

Bottle #5 S S

Bottle #6 S S

Bottle #7 S S S

Bottle #9 S S

Bottle #10 S S

Bottle #11 S S S

Bottle #12 S S

Bottle #13 S S S

Rat 1 samples wine from bottles 8 through 13.

Rat 2 samples wine from bottles 4 -7 and 12-13.

Rat 3 samples wine from bottles 2, 3, 6, 7, 10 and 11.

Rat 4 samples wine from all odd numbered bottles.

# Puzzler Solution (2-13-10)

Actually, you can set bottle 13 aside. I would renumber the rats but essentially do the same thing.

jugs tested

Rat 1: 1,3,5, 7,9, 11

Rat 2: 2,3,6,7,10,11

Rat 3: 4,5,6,7 12

Rat 4: 8,9,10,11,12

Looking at the chart above, if rats 1 and 3 die, the bad wine is in jug 5 because both 1 and 3 sampled from this jug. If only rat 2 dies, the bad wine is in jug 2, because only rat 2 sampled from this jug, etc. If no rat dies, the poison is in jug 13.

Writing the values in binary from 1 to 12 explains the whole thing: 1,10,11,100,101, 110,111, 1000,1001 1010 1011, and 1100. We do the same thing in computer science to find a one bit error in a byte of 8 bits. We add 4 parity bits and make the parity bits powers of 2: 1,2,4,8. Then each parity bits check the bits that have a 1 bit in the power of 2 represented by that bit.

You guys are making this way too complicated…

We can make the following unique groups from the set of four test subjects {1,2,3,4}:

#1: {1}

#2: {2}

#3: {3}

#4: {4}

#5: {1,2}

#6: {1,3}

#7: {1,4}

#8: {2,3}

#9: {2,4}

#10: {3,4}

#11: (1,2,3}

#12: {1,2,4}

#13: {1,3,4}

#14: {2,3,4}

#15: {1,2,3,4}

Leave one bottle sealed–if all the subjects live, it’s poisoned. If subject 1 dies, we know the poison is in bottle 1. If 1 and 2 die, it’s in bottle 5. If 1, 2, and 3 die, it’s in bottle 11. And so on. The unique bottle assigned to each group is arbitrary, of course.

You can see that with four test subjects we could have tested 16 bottles of wine, provided we have no qualms about possibly killing all four of them. The choice of 13 bottles, it seems, was a “red herring”.

This is the best answer. I like the fact you took the humane approach as well. Taking away the most deadly scenerios is very kind of you.

I also like that the first and second rat have a 50% survival rate while rats 3 and 4 have slightly better odds (58%).

Actually all these answers are all essentially the same.

Leaving on bottle unopened equates to the binary 0000 or all rats live (assuming 0 for live, 1 for death)

As you correctly point out we could create 16 unique combinations. Expressed as the previous answers point out 4 rats, 2 possible states. 2^4 = 16

@meroche – the odds are the same in every solution!

I solved the thing by using logic. Explain how writing the binary values explains the whole thing. I understand base two, and I wrote out the numbers. I couldn’t make a connection between the binary system and my solution.

Look at the unit’s column. The values that have a 1 in the unit’s column are 1, 3, 5, 7, 9, 11 (binary (1, 11, 101, 111, 1001, 1011). These are the samples of wine tested by rat number 1. The second column from the right is the two’s column. The values 2,3,6 7, 10, 11 (binary 10, 11, 110, 111, 1010, 1011) have a 1 in the two’s column. These are the samples of wine tested by rat 2. Look ath the four’s column. These values are 4,5,6,7,12 (binary 100, 101, 110, 111, 1100). These samples are tested by rat 3. Finally, look at the values that have a 1 in the eight’s column which are 8, 9, 10, 11, 12 (binary 1000, 1001, 1010, 1011 and 1100). These values are tested by rat 4. If I list the binary values in a column, then you have

0001

0010

0011

0100

0101

0110

0111

1000

1001

1010

1011

1100

One’s in the first column represent wines tested by rat 1, One’s in the two’s column are samples tested by rat 2, and so on.