logic?

[phanbachloc]

there are 13 stones ,12 stones have the same weight,and the rest one has different weight.
How can we find out the stone that has different weight by using 3 times of balancing?

[Acidic]

you need to know if the "odd" stone is lighter or heavier. let's assume it is lighter.

1.weight 4 stones and 4 stones. if one group is lighter, chuck the rest of the stones away. if not, goto 5

3. weigh two and two of those four stones. throw away the two heavier ones.

4. weigh the two remaining stones, the lighter one is it.

5. take the 5 which haven't been weighed, add one (any one) from those that have. wiegh 3 and 3 of them. throw away the heavier group

6. from the 3 left, weigh 1 against another. if one weighs less, thats the one else the one which you didn't weigh is it.

[twanvl]

I'll take a bottom up approach.

a. Given enough normal stones, you can find the odd one in 2 stones by weighing one of them against a normal one.

b. If you know the 'direction' (wether the odd stone is lighter or heavier) you can find it in 3 stones by weighing two of them against each other

c. Given 5 stones and unknown direction you can end up with one of {a,b} by weighing 3 of them against 3 normal stones.

d. Given two groups (gl,gh) of 1 and 2 stones respectivly, and the information that if the odd stone is in gl it is lighter then a normal one, and if it is in gh is is heavier then a normal one:
weigh the two gl stones against each other, the odd stones is the lighter one, if they weigh the same the odd stone is the one from gh.

e. Given two groups (gl,gh) of 4 stones:
weigh 3 stones from gl and 2 from gh against 5 normal ones (6 would be easier).
- if the unknown stones are lighter we are left with the 3 from gl (case b)
- if they are heavier the 2 from gh (case b or a),
- otherwise the 2 remaining stones (case d)

f. Given 13 stones weigh 4 of them against 4 other. You either end up in case e or in case c with the 5 remaining stones.

[Acidic]

we basically have the same answer.

[alkatran]

Here's a picture.

[DiGiTaIErRoR]

You can find this solution on-line. It's typically used with coins though. "There are 12 coins, all identical, except weight. Find the odd coin and determine whether it weighs more or less in 3 balances."

It's very possibly.

Basically:

It's breaking up the 'stones' into small groups, identifying each, then using 'stones' you know are good mixed with ones you don't to find the 'odd' one.

[edit]13?! WHAT WAS I THINKING?! [/edit]

[bugzpodder]

actually either i am completely out of it, or both of you are wrong. you do not know if the fake stone is heavier or lighter

[alkatran]

If there are 12 objects, you can determine heavier/lighter. When there are 13 there's a 1/13 chance you won't get the weight.

[DiGiTaIErRoR]

Originally posted by alkatran
If there are 12 objects, you can determine heavier/lighter. When there are 13 there's a 1/13 chance you won't get the weight.

And by the time you figure out which one it is with your balance, I'll use a scale and be done quick!

That's logic and reasoning for you!

[phanbachloc]

but the problem is that we dont know if the rest stone is lighter or heavier

[Acidic]

do you know the answer?

[kedaman]

here's my solution
L = light/normal
H = heavy/normal
U = unknown
N = normal

1 UUUU UUUU UUUUU
1A LLLL HHHH NNNNN
1A2 LLHHH NNNNN LLH
1A2A NNHHH NNNNN NNN (Heavy)
1A2A3 HH HNNNNNNNNNN
1A2A3A HN NNNNNNNNNNN
1A2A3B NN HNNNNNNNNNN
1A2B LLNNN NNNNN NNN (Light)
1A2B3 L N LNNNNNNNNNN
1A2B3A L N NNNNNNNNNNN
1A2B3B N N LNNNNNNNNNN
1A2C NNNNN NNNNN LLH
1A2C3 L L HNNNNNNNNNN
1A2C3A L N NNNNNNNNNNN
1A2C3B N N HNNNNNNNNNN
1B NNNN NNNN UUUUU
1B2 UUU NNN NNNNNUU
1B2A HHH NNN NNNNNNN
1B2A3 H H HNNNNNNNNNN
1B2A3A H N NNNNNNNNNNN
1B2A3B N N HNNNNNNNNNN
1B2B LLL NNN NNNNNNN
1B2B3 L L LNNNNNNNNNN
1B2B3A L N NNNNNNNNNNN
1B2B3B N N LNNNNNNNNNN
1B2C NNN NNN NNNNNUU
1B2C3 U N NNNNNNNNNNU
1B2C3A H N NNNNNNNNNNN
1B2C3B L N NNNNNNNNNNN
1B2C3C N N NNNNNNNNNNU

[phanbachloc]

i do not know the answer,is it possible to solve ?

[fundu]

phanbachloc, IT CANNOT BE DONE.

If we don't have the extra information that the odd ball is heavier or lighter, in three weighings, we can only get the odd ball out, if the number of balls is less than or equal to 8.

Doing so, means not only we will find the odd ball, but will also find what is the defect.


In case, if we assume that we have this information, in three weighings we can find the odd ball if the number of balls is less than or equal to 24.


If anyone, wants I can explain.


phanbachloc, I hope this helps you.

[alkatran]

Excuse me, but you can determine wether or not it is heavier/lighter with 12 stones assuming you have 3 weighings (see my picture, do the last logical step for yourself, the 13th stone is only a problem when the first weighing is equal).

and if you know the stone is heavier, you can:

Last weighing, 3 unknown stones, compare 2. If different take heavier, else is remaining. before last gives 3, so 9.

3^x stones can be sorted where x is the number of weighings. So with 3 weighs we can actually find the odd stone out of 3^3 = 3*3*3 = 3*9 = 27 stones.

[kedaman]

I noticed something.. nobody reads anyone's posts here I just read trough everyones solutions and twanvl seems to have got it long ago already

[alkatran]

Sometimes finding the solution for yourself is fun, too.

[wossname]

Originally posted by twanvl
I'll take a bottom up approach.

I did hear rumours. j/k

[Ecniv]

Originally posted by phanbachloc
there are 13 stones ,12 stones have the same weight,and the rest one has different weight.
How can we find out the stone that has different weight by using 3 times of balancing?

I take this to mean:
1 stone of a different weight
12 stones of same weight
You have 3 balances MAX to get the stone.

If that is correct, then the lovely graphc is nice, but uses too many balances.

If you know if its heavier or lighter then you can possibly get it in the 3 balances :

1) 1st 6st-6st
If balanced the odd stone is the difference
Otherwise take the 6st from the unbalanced side

2) 3st-3st
Take the 3st from the unbalanced side

3) 1 1st-1st
If balanced its the removed stone
If unbalanced its that stone

However, if you don't know if the stone thats different, its going to be more tricky. And if the original poster now has the answer from a school book - please post


Vince

[alkatran]

You don't know if the stone is heavier or lighter, so there is no way to eliminate a group of stones in one weighing unless it returns equal. (You don't know if it's on the heavy or light side)

If you knew the stone was heavier, it would make more sense to take 2/3 of your stones, and compare them (1/3 on each side) instead of putting them all on.