Can You Solve the Riddle of the 9 Balls with Only 3 Weighings?

[STAii]

Ok, here is a good riddle (waiting for a Riddles thread to be made).
You have 9 balls that look totally identical.
They all weight the same, except for one of them (that weight either less or more, we don't know).
You are supposed to get the odd one out by using a two-sided weight scale only for 3 times .
Think about it, it is a bit tricky

 

[Zero]

I know the first step is three on each side, logically...

 

[drag]

The next step is either 4 - 4 or 1 - 1,
not too hard to guess the last...

Live long and prosper.

 

[ermines]

how about if there were twelve billiard balls? that was the question asked to me when i was in 3rd year high school? still can't think of the answer.

 

[Integral]

Here is my solution to the 12 ball problem

separate the 12 balls into 3 sets, A, B, & C, of 4 balls each
1st measurement: compare sets A and B.

IF Sets A and B balance
the oddball is in set C, we also know that all of the balls from sets A and B are the same and may use them for comparison, any of these balls may be considered standard.

2nd measurement
3 balls from C against 3 balls (standards) from A or B.
Balance
The odd ball is the remaining 1 of C

3rd measure
compare 1 of the remaining balls from C to 1 of the balls from A or B to
determine relative weight
Not Balance
The oddball is one of the 3 from set C on the scale. We know wethere it is light
or heavy

3rd measure
compare 2 of the 3 balls in question
Balance
The remaining ball is the oddball and we know its relative wight.

Else IF Sets A and B do not Balance
The oddball is on the scale so the balls of set C are standard
2nd Measure
1. Remove 3 balls from the low side (leave one behind).
2. Move 3 balls from the high side to low side(leave one behind).
3. Place 3 balls from set C (standard) on the high side.
If Balance then
Oddball is heavy and is one of the 3 removed from low side.
3rd measure
Compare 2 of the balls removed from the heavy side.
Balance => the other ball is the oddball
no Balance => the oddball is the one on the low side of the scale
Else IF The Balance Shifts (previous high side is now low)
The oddball was moved and is light.
3rd measure
Compare 2 of the moved balls.
Balance => other is oddball
No Balance => oddball is on the high side of scale
Else if The Balance does not Shift (previous high side remains high)
The oddball is one of the 2 balls which were not moved
3rd measure
compare one of these balls with a standard.
Balance => other ball is oddball
No Balance => oddball is on the scale.

Done

 

[bogdan]

no offence...but we were solving problems like this in primary school...
Integral -> wasn't it a real challenge for your intelligence ?
Try this way :
1st : 5-5;2 remain...
2nd:2-2(from the odd 5-group)
3rd:1-1(from the odd 2-group)
[zz)]

 

[STAii]

Originally posted by Zero
I know the first step is three on each side, logically...

Good logic, but try to continue .

Originally posted by Drag
The next step is either 4 - 4 or 1 - 1,
not too hard to guess the last...

Well if not, try to do it , i don't think you will be able to continue from here (i was unable to continue when taking a group of 4).

Originally posted by Integral
Here is my solution to the 12 ball problem

Can you addapt it for 9 balls ?

Originally posted by bogdan
no offence...but we were solving problems like this in primary school...
Integral -> wasn't it a real challenge for your intelligence ?
Try this way :
1st : 5-5;2 remain...
2nd:2-2(from the odd 5-group)
3rd:1-1(from the odd 2-group)

Sorry, but i don't think this is quite right.
It seems that you forgot that we do not know wether the odd ball is lighter or heavier than the other balls.

 

[bogdan]

I'm so sorry...I was wrong...can you forgive me ?

As for the 9 balls...divide them into 3 groups of 3 and the rest is quite the same...
sorry again...

 

[drag]

Originally posted by bogdan
[no offence...but we were solving problems like this in primary school...

Must've been awfully hard... (Sorry, just couldn't
resist. )

BTW, the problem can also be solved for 13
balls. (But not 14+... )

Peace and long life.

 

[kyleb]

that is a cool one STAii, i have never thought about it before. the best i have been able to manage is three steps but could take between two and four uses of the scale.


9 balls, 3 groups of 3; a, b and c

1)
a~b
if a=b; c=x
2)
if a!=b; a~c
if a=c; b=x
else a=x
3)
repeate process for x, 3 balls; a, b and c
x = the odd ball

so how do you improve upon that[?]

 

[drag]

Greetings !

If you'd like, here's the solution for 13 balls :

1. measure 4 and 4.

2.1 If equal - measure 3 of the remaining with
3 measured (normal).

2.2 If not equal - measure 3 of one side + 1
from the other side with the 3 normal balls +
1 from the other side.

3.1 Following from 2.1 if the balls are equal
then measure 1 of two yet unknown balls with
a normal ball.

3.2 Following from 2.1 if the balls are not equal
measure 1 - 1 (2 balls) from the 3 "unnormal".
Since from 2.1 we know that the ball is either
heavier or lighter this 1-1 can allow us to find
it, or if they're equal it's the remaining 1.

3.3 If 2.2 turns out eual then we have 3 balls
left and we know we're looking for a certain -
heavier or lighter ball. We now do it like 3.2.

3.4 If 2.2 turns out to have changed previous
heavier/lighter sides then it's either the one we
transfered to the normal 3 or the one transfered
sides to the other unknown 3. So, we measure
one of them with a normal ball and find out.

3.5 If the heavier/lighter after 2.2 stays the same
then we know it's one of the 3 unknown on the
left and we know it's either heavier or lighter.
So, we do it like in 3.2.

Live long and prosper.

 

[kyleb]

bump for the answer.


and i mean the one for 9 balls in 3 trys, none of this off topic junk!

 

[STAii]

Well kyleb, good answer, but not good enough.
You see if when you repeat the process, you go to the point (2) again, then you will use the scale 4 times, and you are supposed to use it 3 times in the maximum.
Here if the solution.
Make the 9 balls into 3 sets, a, b and c.
The balls are also numbered from 1 to 9, (iow set a containst balls 1,2 and 3 ... etc)
->a&b on scale
-->if a=b, then c is odd group, put ball7 and ball8 on the scale.
--->if ball7=ball8 then ball9 is the odd one (used the scale only twice)
--->if ball7[x=]ball8, then leave the heavier on the scale, and put instead of the other one the last of the balls in set c, if the scale is still not in equilibrum, then the heavier ball is the odd one, otherwise the one you took out (between ball7 and ball8) is the heavier. (founded using the scale 3 times)
-->if a[x=]b, either a or b is heavier, suppose a is heavier. Put a and c on the scale.
--->if a=c then b is odd set, and the odd ball is lighter than all other balls (since a is heavier than b), if a[x=]c then a is the odd set, and the odd ball is heavier than all other balls.
---->Put any two balls of the odd set onto the scale, if the scale stays in equilibrum then the third ball is the odd one, if not then you will know which one is the odd one (from the last step you know if the odd ball is heavier or lighter, so if at this step the scale is not in equilibrum, then the heavier ball is the odd one if the odd ball is heavier than the other balls (From the last step), or the lighter ball is the odd one if the odd ball is lighter than the other balls (From the last step too)).

Maybe it is not clear enough, so feel free to ask any questions on it.

 

[kyleb]

than you STAii, it makes sense now. i just failed to realize that pervious measurements could be used to determine if the odd ball was heavier or lighter.

 

[STAii]

Well, this is the tricky part in it