How can you solve the 12 ball puzzle with a twisted balance scale?

[lugita15]

There's an well-known puzzle that goes as follows: You have 12 balls. They look identical, and 11 of them have the same weight but one of them, you don't know which, has a different weight. You don't even know whether the odd ball is heavier or lighter than the others. You have an old-fashioned balance scale (they look like this if you're not familiar with them), and you can put balls on the two sides of the scale and weigh them against each other. The scale is really rusty, so it will break after using it three times. So how can you use the three weighings to determine which of the 12 balls has a different weight than the others?

You might want to try your hand at it, but I have just thought of the following twist on the traditional puzzle: Suppose that you don't have a regular balance scale that tells you which side is heavier and which side is lighter, but instead you had some kind of crippled balance scale that would only tell you whether the two sides are equal or unequal, not which side has a greater weight. Given this modification, how many weighings would suffice to solve the problem, and how would you solve it? I'm guessing that 4 or 5 weighings would suffice, but I'm not sure.

I look forward to hearing people's thoughts on this.

 

[mfb]

As you have to separate 12 cases and each test gives you a single bit of information, you need at least 4 weightings (in the worst case). To fully exploit the power of the broken scale, we can use 16 balls.

Solution:

Spoiler

1) ABCD vs EFGH
2) ABCD vs IJKL
both unequal -> wrong one is in ABCD
1 equal, 2 unequal -> wrong one is in IJKL
1 unequal, 2 equal -> wrong one is in EFGH
both equal -> wrong one is in MNOP
In all cases, you have 4 balls to check and enough correct balls. I'll restrict the analysis to a wrong ball in ABCD:
3) AB vs EF
4) AC vs EF

This is easy to expand to arbitrary numbers of balls.

There's an well-known puzzle that goes as follows: You have 12 balls.

Try the same puzzle with 13 balls. You just have to find the ball with a different weight, you do not need to determine whether it is heavier or lighter.

 

[harishm18]

1st mesurement -12 divide into 6 and 6 - you can eliminate 6 from it.
2nd - from 6 mesure 3 and 3
3rd mesure - pick any two if it is same then the left one is different . otherwise we can get the answer from mesurement

 

[mfb]

1st mesurement -12 divide into 6 and 6 - you can eliminate 6 from it.

This does not work for the new problem, and it does not work for the well-known puzzle either. How can you eliminate 6 balls, if you don't know whether the wrong one is lighter or heavier?

 

[harishm18]

Think carefuly man

pick 6 randomly from 12 and measure . out of 6 and 6 one bunch must be different weigh high . so remove the other 6.

 

[mfb]

Measure what?
You have to fill both sides of the balance scale, and you have no absolute weight (or reference balls) to compare the balls with.

 

[Maui]

1st mesurement -12 divide into 6 and 6 - you can eliminate 6 from it.
2nd - from 6 mesure 3 and 3
3rd mesure - pick any two if it is same then the left one is different . otherwise we can get the answer from mesurement

Good.

 

[D H]

Good.

Good? No. It's completely wrong. Splitting the twelve balls into two groups of six and weighing one group against the other yields zero information, even in the original formulation of the problem. In both formulations, eleven of the balls have identical weights and the remaining ball is different. We don't know if it's heavier or lighter than the others. Weighing six versus six will tell us (original formulation) that one group is heavier than the other, or (new formulation) that the two groups differ in mass. Then what? There's no information here.

Compare that with what happens if we split the twelve balls into three groups of four and weigh one of those groups against another. Now the result is not preordained. If the scale balances it means the odd ball is in the group that wasn't involved in the weighing; we've narrowed the problem down to four balls. If the scale doesn't balance we can rule out that unweighed group of four, narrowing the problem down to eight balls.

 

[micromass]

Good? No. It's completely wrong. Splitting the twelve balls into two groups of six and weighing one group against the other yields zero information, even in the original formulation of the problem. In both formulations, eleven of the balls have identical weights and the remaining ball is different. We don't know if it's heavier or lighter than the others. Weighing six versus six will tell us (original formulation) that one group is heavier than the other, or (new formulation) that the two groups differ in mass. Then what? There's no information here.

Compare that with what happens if we split the twelve balls into three groups of four and weigh one of those groups against another. Now the result is not preordained. If the scale balances it means the odd ball is in the group that wasn't involved in the weighing; we've narrowed the problem down to four balls. If the scale doesn't balance we can rule out that unweighed group of four, narrowing the problem down to eight balls.

His point was not to weigh 6 balls versus 6 balls. His point was to take 6 balls from the 12 and weigh them 3 versus 3. If there is a weight difference, then the different ball must be among the 6. Otherwise, the different ball must be among the other 6.

 

[thelevster]

solution

so I'm not sure if anybody else posted this but...

I did this in 3-4 steps

1)I split it into 4 (3 balls each) groups (A,B,C,D)
2)then I weighed A vs. B
-if they're equal then both have equal weight balls, if they're unequal then one has the unequal ball
3)either way, the second time you weigh it is A vs. C
if A = B, and A = C, then D has the unequal ball
if A = B, but A =/= C then C has the unequal ball
if A = C, but A =/= B then B has the unequal ball
if A =/= B, and A =/= C then A has the unequal ball
4) finally the third time weighing you try 2 balls from the unequal group
if they equal each other then the last ball is the odd one out and your done in 3 steps
5) if they're unequal then your last weigh is against the ball you didn't weigh in the last step
if they're equal then the ball which is left is the odd one
if they're unequal then the ball weighed in both the 3rd an 4th time is the odd one.

correct me if someone else already posted it or if I am just wrong.