Suppose you have 8 balls. One of them is slightly heavier, but the only way to tell is by putting it on a scale against others. Whats the fewest number of times you'd have to use the scale to find the ball?


Almost everyone i know (including me) gave the answer 3.

But the correct answer is supposed to be 2 . But how???? :?

This is where i got the problem from. Click your way to the 4th page in the excerpt' and you will find the problem there?

So can anyone find the solution?
Posted on 2004-11-14 06:03:55 by clippy
Just a hint: The key to the puzzle is not only getting information by what you measure but also gaining information about what you do not measure... (according to google)
Posted on 2004-11-14 06:33:19 by lifewire
oh, and just a little anecdote about my unmeasurable stupidity: in my native language, billiard does also mean 10^5 (one billion). (well, you write it slightly different). So I imagined that the question was: you have 8.10^5 balls to measure... that are many balls.
Posted on 2004-11-14 06:37:44 by lifewire
I thought so too :oops:

wouldn't the fewest be 1 (luck and all that)
Posted on 2004-11-14 07:41:07 by Hiroshimator
Well i found the answer and yes it is 2 only.
Btw, Ms gives you only 3 mins to solve this in an interview.
I guess i will never be able to get into MS ;)

I have edited the post so now it says '8 balls' instead of '8 billiard balls'.
Posted on 2004-11-14 07:50:17 by clippy
yes but if you pick 2 balls and put them on and 1 is heavier then that's only 1 try, no?

edit: I guess it'll depend on the actual type of scale used....
Posted on 2004-11-14 07:56:16 by Hiroshimator
I love these kind of problems :) Thanks for the link!

Btw, I got the answer in just under two minutes :P (Okay, I did read lifewire's hint)
Posted on 2004-11-14 08:10:32 by Qweerdy
I don't know the answer to the question but you would sure have a very deep voice with that many balls. :P
Posted on 2004-11-14 08:12:08 by hutch--
qweerdy: explain please :s
Posted on 2004-11-14 08:12:27 by Hiroshimator
the only way I can think off is a cheat :grin:


you need a 2 hold scale as well

put on 4 each, determine largest group (step 1)
seperate heaviest group over scales (step 2)

while removing 1 ball from each scale (your mother told you to clean up after playing didn't she? ;) ) you'll notice wether they stay the same or not (if they do you just picked the largest ball else it's still on there)

but technically I think it's a cheat
Posted on 2004-11-14 08:19:14 by Hiroshimator
I think you mean 8*10^15 (English quadrillion).

This is very easy. Using the scale twice you get 9 possible outcomes, which should be more than enough for most of you.

First, put 3 balls on each side of the scale. If they are equal, weigh the two remaining balls and you're done. Otherwise, take two of the balls from the heavier side and weigh them. If they're equal too, the last ball is heavier. If they aren't, the heaviest ball is the heavier, obviously. I was expecting more from a board of computer programmers. No wonder all programs made these days are more than 30MB in size.
Posted on 2004-11-14 08:44:21 by Sephiroth3
Ok i got another one for you guys.



Put a coin in a bottle, and a cork in the neck. Take the coin out without breaking the bottle or pulling out the cork


Before some of you say impossible think about it for a sec. If i tell you the answer you would laugh at its simplicity.
Posted on 2004-11-14 09:13:36 by clippy
Here two more along the same lines, first is very similar;

9 coins, all look identical but one is heavier you only have 2 weighings to figure it out.


Second one is alot harder :twisted:

12 coins, one is different weight but you don't know if its heavier or lighter. 3 weighings to figure out which coin is the different one and also find out if its heavier or lighter than the rest.


The fact that you don't know if its heavier or lighter is what makes it so difficult. A clue for it is: Posted on 2004-11-14 10:00:39 by Eóin
Ok i got another one for you guys.



Put a coin in a bottle, and a cork in the neck. Take the coin out without breaking the bottle or pulling out the cork


Before some of you say impossible think about it for a sec. If i tell you the answer you would laugh at its simplicity.


my idea:
if you put it like that i would say push the cork in the bottle (didnt say anything about touching the cork) hold it upside down, shake a bit and make the coin fall out

EDIT: YES, found the solution to the 8 balls in like 1 minute (using the hint and 8 marbels for the visual effect).
btw, your way is cheating because you look at the scale 3 times ;)

OH i just see that this idea also works for the 9 coins :P (muahahaha)

!!!SPOILER!!!
Put two groups of balls on the scale, that leaves 2 balls on the table.
IF the scale is balaced, empty the scale and put the 2 from the table on the scale, which side goes down is the heavy ball.

IF 1 side of the scale goes down it means the heavy ball is on of that 3 balls.
take the 3 balls, put 1 on the table and the other 2 on the scale
if 1 side goes down that side has the heavy ball, if the scale is balaced the ball on the table is heavyest.

/SPOILER

now i will try the hardest problem (solving this probs is a HUGE ego-boost btw :D)
Posted on 2004-11-14 10:12:26 by Scorpie
Just shake the bottle around so that the coin rotates at the bottom of the bottle.
The coin will cut the glass and then the bottom will be separated from the bottle.
Now you can get the coin. :-D
Posted on 2004-11-14 11:49:45 by Siekmanski
Posted on 2004-11-14 11:57:56 by Hiroshimator
Scorpie,
Yeah you are correct.
Posted on 2004-11-14 12:25:24 by clippy
Yeah, but I don't see how that is any different from what I said above.

The 12 coin problem can be solved by the following method:
We'll call the balls A-L and use this diagram. We will look up which balls to weigh in the first column and find the result in the next one. Then we advance to the cell to the right and repeat until we get at the result column.


Step 1 | Step 2 | Step 3|
-----+---+-----+---+---+---|
ABCD | = | ABC | = | A | > | L is smaller
- | | - | | - |---|
EFGH | | IJK | | L | < | L is larger
| | |---|---|---|
| | | > | I | = | K is larger
| | | | - |---|
| | | | J | > | I is larger
| | | | |---|
| | | | | < | J is larger
| | |---|---|---|
| | | < | I | = | K is smaller
| | | | - |---|
| | | | J | > | J is smaller
| | | | |---|
| | | | | < | I is smaller
|---|-----|---|---|---|
| < | ABE | = | F | = | H is larger
| | - | | - |---|
| | CDL | | G | > | F is larger
| | | | |---|
| | | | | < | G is larger
| | |---|---|---|
| | | > | C | = | E is larger
| | | | - |---|
| | | | D | > | D is smaller
| | | | |---|
| | | | | < | C is smaller
| | |---|---|---|
| | | < | A | > | B is smaller
| | | | |---|
| | | | | < | A is smaller
|---|-----|---|---|---|
| > | ABE | = | F | = | H is smaller
| | - | | - |---|
| | CDL | | G | > | F is smaller
| | | | |---|
| | | | | < | G is smaller
| | |---|---|---|
| | | > | C | = | E is smaller
| | | | - |---|
| | | | D | > | D is larger
| | | | |---|
| | | | | < | C is larger
| | |---|---|---|
| | | < | A | > | B is larger
| | | | |---|
| | | | | < | A is larger
Posted on 2004-11-14 13:02:31 by Sephiroth3
Nice approach, my way of solving it is here if you're interested :) .
Posted on 2004-11-14 14:01:45 by Eóin
I was kinda hoping for a "clean" solution to the 12 coins problem instead of brute-forcing it, but after bothering to write down my solution I saw it wasn't going to happen and gave up....

As for the coin in the bottle, that's a neat problem but as soon as I imagine the cork being pushed into the bottle, I imagine it going too far in and falling in the bottle :)
Posted on 2004-11-16 16:13:54 by Qweerdy