Awesome Thinker Problem!!

[Backglass]

This reminds me of an old nursery rhyme from my childhood:

As I was going to St. Ives, I met a man with Seven Wives.
Each Wive had Seven Sacks,
Each Sack held Seven Cats,
Each Cat had Seven "Kits".

Kits, Cats, Sacks, Wives...how many went to St. Ives?

[e_i_pi]

This reminds me of an old nursery rhyme from my childhood:

As I was going to St. Ives, I met a man with Seven Wives.
Each Wive had Seven Sacks,
Each Sack held Seven Cats,
Each Cat had Seven "Kits".

Kits, Cats, Sacks, Wives...how many went to St. Ives?

Depends how many got ran over by the bus on the way

[guylian]

This reminds me of an old nursery rhyme from my childhood:

As I was going to St. Ives, I met a man with Seven Wives.
Each Wive had Seven Sacks,
Each Sack held Seven Cats,
Each Cat had Seven "Kits".

Kits, Cats, Sacks, Wives...how many went to St. Ives?

1

[Iliad]

I like this one
You are the ruler of a medieval empire and you are about to have a celebration tomorrow. The celebration is the most important party you have ever hosted. You've got 1000 bottles of wine you were planning to open for the celebration, but you find out that one of them is poisoned.

The poison exhibits no symptoms until death. Death occurs within ten to twenty hours after consuming even the minutest amount of poison.

You have over a thousand slaves at your disposal and just under 24 hours to determine which single bottle is poisoned.

You have a handful of prisoners about to be executed, and it would mar your celebration to have anyone else killed.

What is the smallest number of prisoners you must have to drink from the bottles to be absolutely sure to find the poisoned bottle within 24 hours?

[MeDeFe]

With as little as 32 prisoners you can find the poisoned bottle in 2 "test runs", though it's not guaranteed that you will have indentified the bottle within 24 hours. Assign 31 or 32 bottles to each of the 32 prisoners, give each prisoner a drop from each of the bottles assigned to him, see which one dies. Reassign the bottles he drank from, one to each prisoner and give them a drop and you'll have the bottle. If the prisoner who died first had 32 bottles and you have one bottle more than you have prisoners, keep it and if noone dies that's the poisoned bottle. If the poison works within 10 hours you'll have it, if it takes 20 you'll at least have 968 bottles for your celebration.
The variable time it takes for the poison to work messes things up a lot. If it were always the same length of time before a person dies from the poison this would be fairly easy because you could just let them drink every 30 minutes or so in a system and find out which bottle is poisoned that way.

[TheProwler]

With as little as 32 prisoners you can find the poisoned bottle in 2 "test runs", though it's not guaranteed that you will have indentified the bottle within 24 hours. Assign 31 or 32 bottles to each of the 32 prisoners, give each prisoner a drop from each of the bottles assigned to him, see which one dies. Reassign the bottles he drank from, one to each prisoner and give them a drop and you'll have the bottle. If the prisoner who died first had 32 bottles and you have one bottle more than you have prisoners, keep it and if noone dies that's the poisoned bottle. If the poison works within 10 hours you'll have it, if it takes 20 you'll at least have 968 bottles for your celebration.
The variable time it takes for the poison to work messes things up a lot. If it were always the same length of time before a person dies from the poison this would be fairly easy because you could just let them drink every 30 minutes or so in a system and find out which bottle is poisoned that way.

The problem also is that he didn't say how many bottle were needed for the celebration. One would assume he meant he wanted to identify the specific bottle that was poisoned. Which, given the possibility that it may take 20 hours to complete the first round of tests, and that there is no time to complete a second round of tests before 24 hours, would mean you would have to test 999 bottles in the first round.

[PLAYER57832]

HINT: 7 girls..........they each hold 7 bags apiece, and within every single damned bag there are 7 cats and 7 kittens. They all have legs. This should be your thought process.

NOT

7 girls * 7 cats * 7 kittens * 2 legs for the girls * 4 legs for the cats and kittens.

You said they each had A bad ... singular, not 7 bags. Also, the kitten's location was not specified.

If each girl has 1 bag, then the formula is
# girls =7
# girls legs= #girls * 2= 7 *2 = 14
# bags= #girls *1 = 7 * 1= 7
#cats # bags * 7 = 7 * 7 = 49
# cats legs # cats * 4 = 49 * 4 = 196
# kittens -- not specified, but if as later stated they were all on the bus, then
# kittens = # cats * 7= 49 * 7 = 343
# kittens legs = # kittens * 4 = 343 * 4
# bus driver legs = 2

total = 2 + 14 + 196 + 2401= 2613


If each girl has 7 bags, then change # bags above to # girls * 7 = 7 * 7 =49
= 2 + 14 + 196*7 + 2401*7 =


The BASE forumula for # legs is

(# Girls * 2) + (* girls *#bags* # cats * 4) + (#girls*#bags*#cats*#kittens*4) + bus driver * 2 = number of legs on the bus.


My addition might be off ... I did it in my head, but the forumula IS correct.

[john9blue]

# kittens legs = # kittens * 4 = 343 * 7

Heh, you multiplied the number of kittens by 7 instead of 4.

But you were on the right track.

7 girls = 14 legs

7 girls * 7 cats = 49 cats = 196 legs

49 cats * 7 kittens = 343 kittens = 1372 legs

Total = 1582 legs.

About a minute.

Here's a classic.

You are testing four switches, one of which is connected to an incandescent light bulb. The bulb is on a table, and is covered by a box, so you cannot tell whether it is on or off. How can you flip the switches so that, when you remove the box, you can instantly tell which switch is connected to the lightbulb?

SMARTASS NOTE: You cannot see the wires. You cannot shoot the box or remove it from a distance. The switches can only turn on and off. This is not Schrodinger's lightbulb.

[TheProwler]

You are testing four switches, one of which is connected to an incandescent light bulb. The bulb is on a table, and is covered by a box, so you cannot tell whether it is on or off. How can you flip the switches so that, when you remove the box, you can instantly tell which switch is connected to the lightbulb?

On the first try? It can't be done with 100% accuracy.

[john9blue]

On the first try? It can't be done with 100% accuracy.

That is baloney, sir. It can be done. Think outside the box. Pun intended.

[TheProwler]

On the first try? It can't be done with 100% accuracy.

That is baloney, sir. It can be done. Think outside the box. Pun intended.

I could set up a test that fully adhered to the conditions you stipulated in the description of the problem, but would make it impossible to find a solution with 100% accuracy on the first try. So there is a catch or a trick that was not fully explained in the question.

Quick edit: The word "instantly" was used in the question, which would imply that you are using light as the indication of whether the bulb is on or off. Use of temperature. or heat of the bulb, would not be instant.

Also, it is implied that you would instantly know at the precise time that the box is removed. Not before or after.

[john9blue]

I could set up a test that fully adhered to the conditions you stipulated in the description of the problem, but would make it impossible to find a solution with 100% accuracy on the first try. So there is a catch or a trick that was not fully explained in the question..

Well, I'm not going to tell you the trick during the question, am I?

The information is right there. You remove the box, and you can tell which switch is connected to the light bulb using the bulb alone, without taking it out of its socket.



EDIT: I said "instantly" so that it was clear you could not go back to the switches. Let's say you have one second.

[Iliad]

With as little as 32 prisoners you can find the poisoned bottle in 2 "test runs", though it's not guaranteed that you will have indentified the bottle within 24 hours. Assign 31 or 32 bottles to each of the 32 prisoners, give each prisoner a drop from each of the bottles assigned to him, see which one dies. Reassign the bottles he drank from, one to each prisoner and give them a drop and you'll have the bottle. If the prisoner who died first had 32 bottles and you have one bottle more than you have prisoners, keep it and if noone dies that's the poisoned bottle. If the poison works within 10 hours you'll have it, if it takes 20 you'll at least have 968 bottles for your celebration.
The variable time it takes for the poison to work messes things up a lot. If it were always the same length of time before a person dies from the poison this would be fairly easy because you could just let them drink every 30 minutes or so in a system and find out which bottle is poisoned that way.

You can do it with less prisoners and in only one "test run"

[john9blue]

You can do it with less prisoners and in only one "test run"

You can do it with no prisoners if you use slaves.

Amirite?

[MeDeFe]

You can do it with less prisoners and in only one "test run"

You can do it with no prisoners if you use slaves.

Amirite?

That would impair the quality of the celebrations.

[john9blue]

That would impair the quality of the celebrations.

Lol, I was being semi-sarcastic.

I think I know what it is, though. If nobody else has any idea, then I'll post the answer. It's a tricky riddle.



EDIT: I'll just post the answer now. Don't look if you want to figure it out yourself (which I suggest you do).

So, suppose you only had 1 prisoner. The max number of bottles to test in 1 trial is 2 (have him drink one).

Now, suppose you have 2 prisoners. You have them both drink one, and leave a third aside. But, if they BOTH drank from a bottle, and it was poisoned, they would both die. So, you can test 4 bottles with 2 prisoners.

Same with three prisoners. Have P1 drink B1, P2 drink B2, and P3 drink B3. Then, have P1 and P2 drink B4, P1 and P3 drink B5, and P2 and P3 drink B6. Then, have all of them drink B7, and leave B8 aside. You can test 8 bottles with 3 prisoners.

Obviously, these are powers of two. So, to test 1000 bottles, you would need ceil(log2(1000)) = 10 prisoners.

Thinking further, the alive/dead status of a prisoner after one trial can be represented with a bit (1 or 0). 10 bits (or a 10 digit binary number) can go up to 1023, which is more than enough to test every bottle.

That was a good one, Iliad.

[Iliad]

I know the answer to this one.

[PLAYER57832]

.

Here's a classic.

You are testing four switches, one of which is connected to an incandescent light bulb. The bulb is on a table, and is covered by a box, so you cannot tell whether it is on or off. How can you flip the switches so that, when you remove the box, you can instantly tell which switch is connected to the lightbulb?

SMARTASS NOTE: You cannot see the wires. You cannot shoot the box or remove it from a distance. The switches can only turn on and off. This is not Schrodinger's lightbulb.

first you turn 2. If the light goes on, you turn one of those switches off (does not matter which one). If not, you turn on one added switch. Either way, you know which is which.


There are quite a few variations on this.

[Iliad]

That would impair the quality of the celebrations.

Lol, I was being semi-sarcastic.

I think I know what it is, though. If nobody else has any idea, then I'll post the answer. It's a tricky riddle.



EDIT: I'll just post the answer now. Don't look if you want to figure it out yourself (which I suggest you do).

So, suppose you only had 1 prisoner. The max number of bottles to test in 1 trial is 2 (have him drink one).

Now, suppose you have 2 prisoners. You have them both drink one, and leave a third aside. But, if they BOTH drank from a bottle, and it was poisoned, they would both die. So, you can test 4 bottles with 2 prisoners.

Same with three prisoners. Have P1 drink B1, P2 drink B2, and P3 drink B3. Then, have P1 and P2 drink B4, P1 and P3 drink B5, and P2 and P3 drink B6. Then, have all of them drink B7, and leave B8 aside. You can test 8 bottles with 3 prisoners.

Obviously, these are powers of two. So, to test 1000 bottles, you would need ceil(log2(1000)) = 10 prisoners.

Thinking further, the alive/dead status of a prisoner after one trial can be represented with a bit (1 or 0). 10 bits (or a 10 digit binary number) can go up to 1023, which is more than enough to test every bottle.

That was a good one, Iliad.

Yep, you only needed 10 prisoners.

[TheProwler]

I could set up a test that fully adhered to the conditions you stipulated in the description of the problem, but would make it impossible to find a solution with 100% accuracy on the first try. So there is a catch or a trick that was not fully explained in the question..

Well, I'm not going to tell you the trick during the question, am I?

The information is right there. You remove the box, and you can tell which switch is connected to the light bulb using the bulb alone, without taking it out of its socket.



EDIT: I said "instantly" so that it was clear you could not go back to the switches. Let's say you have one second.

Well, if it is not instantly, then you can do it with heat...

Well, with heat you could know before you even lift the box....turn one switch on...feel for heat...if no heat, turn it off and try the next one...until you find the one giving heat through the box...if you could only feel the box for heat once, try 2 switches...if there is heat, it is one of those, if not, it is one of the other 2. Once you have it narrowed down to 2, turn one on and lift the box. Light, it is that one...no light, the other.

But lets assume the box is insulated so no heat is detectable through the box.

Call the switches A, B, C, D. So, turn 2 switches on - A and B. Wait one minute or so (enough time for the bulb to get very hot if it is on). Now, quickly turn off B and turn on C and lift the lid. Touch the bulb. Is it very hot?

A - bulb is lit and very hot
B - bulb is out and very hot
C - bulb is lit and cool
D - bulb is out and cool

But I have to argue that this is not instantly. Sorry for being so picky.

[TheProwler]

Thinking further, the alive/dead status of a prisoner after one trial can be represented with a bit (1 or 0). 10 bits (or a 10 digit binary number) can go up to 1023, which is more than enough to test every bottle.

I am humbled.

The binary idea helps to visualize...essentially assign a unique number to each of the bottles and use the binary form of that number to determine what prisoners need to drink it.

[pmchugh]

nice one illy I got the 10 but never could have explained as well as john.

However if the heat thing is correct i don't like the question, its cheating.

[TheProwler]

nice one illy I got the 10 but never could have explained as well as john.

However if the heat thing is correct i don't like the question, its cheating.

Would it be cheating to use a glass table and look underneath as your flipped the switches?

[john9blue]

I tried to reinforce the phrasing against the onslaught of loopholes, but apparently failed...

[Snorri1234]

What if I "accidentally" fell against the table after turning each switch?