Solve the Mystery: Identify the Odd Weight Snooker Ball

[Ian Rumsey]

We have 12 red snooker balls and a scale balance.
One of the snooker balls has a different weight but we do not know whether it is heavier or lighter.
The scale balance may be only used three times by putting an equal number of balls on each side and noting the balancing result.
What 3 balancing arrangements would be required to ensure the identity of the odd weight ball and whether it was heavy or light.

 

[DarkForXe]

hmm. I am not trying to be rude here.
But this q is asked before.

 

[TenaliRaman]

i don't know if this is asked before but nice *twist* on the old *ionc* problem

 

[K.J.Healey]

I don't think this is solveable with the alotted measurements and information.

 

[Rogerio]

I don't think this is solveable with the alotted measurements and information.

Surely it is!
Consider 1234 x 5678 at first...

 

[K.J.Healey]

youre sure you read right? that we don't know if the unique ball is heaver OR lighter? I think that creates a problem.

 

[TenaliRaman]

Healey,
It can be solved ...
infact it can be shown that
In n weighings, [(3^n-3)/2 + 1] snooker balls can be weighed and the odd ball can be found.

Placing n = 3, we get 13 .. so ofcourse it is possible even for 12.

I won't spoil this one for anyone ... i already received too many insinuating glares from Gokul already... :D

-- AI

P.S -> So what does this post mean? It means that once the original problem is solved some followers will be,
Find a solution when we have 13 snooker balls?
Show that the general statement of n weighings given by me is true?

 

[NateTG]

Show that the general statement of n weighings given by me is true?

Odd -- according to your formula, I can't find it in 1 ball with 0 weighings, when the number for n=0 should be 1.

 

[Rogerio]

Placing n = 3, we get 13 .. so ofcourse it is possible even for 12.

Unfortunately, in just 3 weighings it's not possible for 13 balls...
(12 is the max)

 

[NateTG]

Unfortunately, in just 3 weighings it's not possible for 13 balls...
(12 is the max)

Really, I'm pretty sure it's possible to find the odd ball out with 13 balls - you may not be able to tell whether it's light or heavy though.

 

[Hurkyl]

Odd -- according to your formula, I can't find it in 1 ball with 0 weighings, when the number for n=0 should be 1.

Then you're not looking hard enough.

If you have one ball, and you know one ball has the incorrect weight, then you don't need to do any weighings to find it!

The impossible case is where you have two balls.

(Oh, and incidentally, the formula says you can find 0 balls in 1 weighing)

 

[K.J.Healey]

Someone post the answer in white. I've tried it numerous ways for a few minutes, and i can get pretty close, but for every possibility there's one where I can't determine if the odd ball was heavier or lighter than the others, but just different.


Hmm by that equation you should be able to find the odd ball out of 4 in 2 tries, and tell me if its heavier or lighter?? Please show me (its a lot simpler than the stated riddle).

EDIT : nevermind, i got it. But my simple question still stands.

 

[NateTG]

Then you're not looking hard enough.

If you have one ball, and you know one ball has the incorrect weight, then you don't need to do any weighings to find it!

The impossible case is where you have two balls.

(Oh, and incidentally, the formula says you can find 0 balls in 1 weighing)

You're usually so on the ball (no pun intended) that I'm guessing you misread, or I wasn't clear enough, since we appear to (in principle) agree that the formula is incomplete.

However, the formula indicates that one ball can be identified in one weighing:

In the formula n is the number of weighings, and it gives the number of balls.
My post indicated that the formula was too low in the case of 0 weigings.

\frac{3^n-3}{2}+1 = \frac{3^1-3}{2}+1 = \frac{0}{2}+1 = 1

But 0 balls in 0 weighings.

 

[Hurkyl]

Yes, I meant 0 in 0.

 

[NateTG]

Someone post the answer in white. I've tried it numerous ways for a few minutes, and i can get pretty close, but for every possibility there's one where I can't determine if the odd ball was heavier or lighter than the others, but just different.

My post in the thread
https://www.physicsforums.com/showthread.php?t=22391&highlight=Brain+Teaser+#90

Should give you a good idea on how to find a solution to this problem. Notably, using that method, the 10 coin case might be more difficult that the 12 coin one.

 

[Gokul43201]

How about , <select to see>



first : L = 5, 6, 8, 10 R = 7, 9, 11, 12

second : L = 2, 3, 4, 7 R = 5, 6, 11, 12

third : L = 1, 4, 10, 11 R = 2, 5, 7, 8 ? [/color]

I think this gives unique outcomes for each number.

Defining outcomes as <, > or = on the basis of the L pan being heavier, R pan being heavier, or both equal.


1 : ==< or ==>
2 : =<> or =><
3 : =<= or =>=
4 : =<< or =>>
5 : <>> or ><<
6 : <>= or ><=
7 : ><> or <><
8 : <=> or >=<
9 : >== or <==
10: <=< or >=>
11: >>< or <<>
12: >>= or <<=

All look different to me...

 

[TenaliRaman]

Then you're not looking hard enough.

If you have one ball, and you know one ball has the incorrect weight, then you don't need to do any weighings to find it!

The impossible case is where you have two balls.

(Oh, and incidentally, the formula says you can find 0 balls in 1 weighing)

Really, I'm pretty sure it's possible to find the odd ball out with 13 balls - you may not be able to tell whether it's light or heavy though.

Both are related observations and well observed indeed ...

Ofcourse it was my mistake that i forgot to mention one tiny 'precursor' reqd to relax the problem...

Hurkyl,
Now with 1 ball u know which is the odd one but u don't know whether it is heavier or lighter so u need one weighing (but weighed against what ? that will be ur question ... wait i will address it shortly)

NateTG,
You are pretty correct indeed and we will be able to find the lighter or heavier condition too ...

The relaxation given to the problem,
"You can add one true weight ball to the existing set of balls"
(Hurkyl and NateTG would have got the entire general solution by now )

-- AI

 

[Rogerio]

The relaxation given to the problem,
"You can add one true weight ball to the existing set of balls"
-- AI

Now it's easy (and possible): compare 12345 x 6789T at first...:-)

 

[BobG]

We have 12 red snooker balls and a scale balance.
One of the snooker balls has a different weight but we do not know whether it is heavier or lighter.
The scale balance may be only used three times by putting an equal number of balls on each side and noting the balancing result.
What 3 balancing arrangements would be required to ensure the identity of the odd weight ball and whether it was heavy or light.

1. Left scale= 1, 2, 3, 4 [/color]
Right scale = 5, 6, 7, 8 [/color]

2. Left scale = 1, 4, 7, 10 [/color]
Right scale = 2, 5, 8, 11 [/color]

3. Left scale = 3, 6, 12 [/color]
Right scale = 1, 8, 9 [/color]

 

[BobG]

Revised solution (forgot to check opposites)

1. Left scale= 1, 2, 3, 4[/color]
Right scale = 9, 10, 11, 12 [/color]

2. Left scale = 1, 4, 7, 10 [/color]
Right scale = 2, 5, 8, 11 [/color]

3. Left scale = 1, 3, 9, 11, 12[/color]
Right scale = 2, 5, 6, 7, 10[/color]

 

[Rogerio]

Revised solution (forgot to check opposites)

1. Left scale= 1, 2, 3, 4[/color]
Right scale = 9, 10, 11, 12 [/color]

2. Left scale = 1, 4, 7, 10 [/color]
Right scale = 2, 5, 8, 11 [/color]

3. Left scale = 1, 3, 9, 11, 12[/color]
Right scale = 2, 5, 6, 7, 10[/color]

Bob, this is not ok, yet .
Depending on each result, you need to choose different balls for the next step.

 

[NateTG]

Depending on each result, you need to choose different balls for the next step.

Not at all. It's simpler to create solutions where that doesn't happen.

 

[K.J.Healey]

Yea, that's a really neat solution. Is it something you have learned in the past to do for these problems?

 

[Rogerio]

Bob, this is not ok, yet .
Depending on each result, you need to choose different balls for the next step.

My mistake, Bob's solution seems ok!

Another way (select to see):

1,2,3,4 X 5,6,7,8

L>R -> 1,2,5,6 X 3,9,10,11
L>R -> 1 X 2
L=R -> 7 X 8
L<R -> 5 X 6

L=R -> 1,2,3 X 9,10,11
L=R -> 1 X 12
L>R -> 9 X 10
L<R -> 9 X 10

L<R -> 1,2,5,6 X 3,9,10,11
L>R -> 5 X 6
L=R -> 7 X 8
L<R -> 1 X 2

 

[TenaliRaman]

Now it's easy (and possible): compare 12345 x 6789T at first...:-)

Nice work!

 

[NateTG]

Yea, that's a really neat solution. Is it something you have learned in the past to do for these problems?

I'm not sure whether you're talking to me, or someone else. I had this problem as homework, and realized that it would be a PITA to write out a strategy that uses different weigings as it goes through, and to show that it worked. That led me to take a more systematic approach - which turned out to be much easier than the ad-hoc one that I used to get the solution the first time I did it.

Someone posted a more involved example with 81 bags of 5 coins here which involved some more sophistication and which I had some trouble with.