[Guest]

You are provided 13 identical looking balls. However one of the ball's weight is either less or more than all other balls. Rest balls weigh the same. You are provided a Beam Balance. By weighing just 3 times, find out which is the odd ball and also find out whether it is heavier or lighter than rest of the balls.

[Gyvven]

first weigh you put 6 balls on each side, if they stay even the ball left over is the odd one out.

second weigh you divide the 6 balls that are on the heavy side of the scale.

third weigh you divide the 3 balls that are one the heavy side , 1 on each side, if it is even, the third one is it.

otherwise the heavy side is the odd one out

[Guest]

first weigh 6 and 6. if they are the same, the other one is the odd ball. If not, take the lighter set and measure 3 and 3 (it doesnt matter which ones). take the lighter set of 3 and weigh any 2 balls. if they are the same, the third ball is the odd one. If not, the lighter one is odd.

[Guest]

first weigh 6 and 6. if they are the same, the other one is the odd ball. If not, take the lighter set and measure 3 and 3 (it doesnt matter which ones). take the lighter set of 3 and weigh any 2 balls. if they are the same, the third ball is the odd one. If not, the lighter one is odd.

If you knew whether the odd ball is heavier or lighter. In both the last 2 posts it is assumed that there is a difference between the 3 and 3 of the 6 selected for re-weighing.

It seems you might need a 4th measurement.

[Guest]

ok... the actual problem is impossible, but this is as close as you can get.

1. Weight 4 vs 4. If they are equal go to step 3, if not then go to step 2.

2. One of these 8 balls is the odd one. Name the balls on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the balls on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one ball from the remaining 5 balls in intial weighing.

If they are equal then weight one of L2, L3 and L4 is lighter. Weigh L2 against L3. Which ever is lighter is the odd one, if they are equal then L4 is the lighter one.

If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2 and the heavier ball is the heaviest.

If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4;

If both are equal, L1 is the odd ball and is lighter.

If H3 is heavy, H3 is the odd ball and is heavier.

If H4 is heavy, H4 is the odd ball and is heavier.

3.One of the remaining 5 balls is the odd one. Name the balls as C1, C2, C3, C4, C5. Weight (C1, C2, C3) against (X1, X2, X3) where X1, X2, X3 are any three balls from the first weighing of 8 balls.

If both are equal, C5 is the odd ball. But you can not tell whether it is heavier or lighter.

If C4 is heavy, C4 is the odd ball and is heavier. If C4 is light, C4 is the odd ball and is lighter.

If (C1, C2, C3) is heavier side, one of C1, C2, C3 is the odd ball and is heavier. Weigh C1 and C2.

If both are equal, C3 is the odd ball and is heavier.

If C1 is heavy, C1 is the odd ball and is heavier.

If C2 is heavy, C2 is the odd ball and is heavier.

If (C1, C2, C3) is lighter side, one of C1, C2, C3 is the odd ball and is lighter. Weigh C1 and C2.

If both are equal, C3 is the odd ball and is heavier.

If C1 is light, C1 is the odd ball and is lighter.

If C2 is light, C2 is the odd ball and is lighter.

[Prime]

You are provided 13 identical looking balls. However one of the ball's weight is either less or more than all other balls. Rest balls weigh the same. You are provided a Beam Balance. By weighing just 3 times, find out which is the odd ball and also find out whether it is heavier or lighter than rest of the balls.

This riddle has already been given a full treatment here.

You can find one different ball out of 13 in 3 weighings. But I don't see how you can always determine whether it is heavier or lighter.

Here is the variation, which does not allow that:

1. Weigh 4 against 4 -- they weigh the same.

2. Out of the 5 on the side, weigh 3 against 3 good balls from the previous step. They weigh the same.

3. Weigh 1 of the 2 remaining possibly different balls against one of the good balls. It weighs the same.

Then you know the different ball is the last one on the side, but you have used up all 3 allowed weighings and cannot determine whether the faulty ball is heavier, or lighter.

If you used more than 4 vs. 4 in the 1st weighing, and they weighed different, then the remaining 2 weighings wouldn't be enough to find different ball out of more than 9.

[Guest]

I also thought that with 13 balls its not possible to tell whether the odd ball is heavy or light, but couldn't give up too. So thanks prime for the proof. But we can with 12 balls. So,

Where is the solution for 12 balls Prime ?

[bonanova]

Thirteen balls, one is heavier or lighter = 26 possibilities.

Each weighing has three possible outcomes, and 3³ = 27 distinguishable cases.

Are we sure this is impossible?

[Guest]

Oh! I din't think like that, bonanova. Oh yes, there must a solution as bonanova has shown that there 27 possible outcomes of 3 weighings,(Equal Equal Equal, Equal Equal A side Down, Equal Equal B side down, Equal A-side down Equal .... etc) And we have 26 possible combination with our balls( No1, heavy, No 2. Heavy, .... No. 13 Heavy, No 1. Light, No.2 Light, ...... No. 13 light) So, there must be a method of linking each outcome to say which ball is the odd and whether its heavy or light.

[Guest]

Error: View Next post.

Edited March 9, 2009 by the-genius

[Guest]

But you can't weigh 4---4 at the first weighing, because, if they weigh the same then, you are left with 5 unknown balls with 10 possible combination. You hav only 3*3 = 9 possible outcome of weighing, so it isn't solvable.

But even if you weigh 5 (1,2,3,4,5)----5(6,7,8,9,10) at the first wighing, of-couse you would be very happy if they weigh the same, but if they not, then You can now have 10 possible combination of unknown balls (supposing (1,2,3,4,5) side went down: 1Heavy or 2 Heavy ... 5Heavy or 6 Light or 7light ...10 light) so you can't find the solution in just 2 weighing that gives only 9 outcomes.

So either way, you fall short of one outcome from the weighing. However at first glance, total 13 balls have only 26 combination but 3^3 = 27 is the total wighing results. This aparent paradox can be understood by that you have to divide Ball combination and weighing combination into at least 2 parts so that in each part, the weighing combination exceed ball combination by 1. And we know its not possible.

In general terms, the problem is that we cannot, bisect the ball.

So I declare this puzzle as UNSOLVABlE. Credits to Prime and Bonovoa.

[bonanova]

But you can't weigh 4---4 at the first weighing, because, if they weigh the same then, you are left with 5 unknown balls with 10 possible combination. You hav only 3*3 = 9 possible outcome of weighing, so it isn't solvable.

But even if you weigh 5 (1,2,3,4,5)----5(6,7,8,9,10) at the first wighing, of-couse you would be very happy if they weigh the same, but if they not, then You can now have 10 possible combination of unknown balls (supposing (1,2,3,4,5) side went down: 1Heavy or 2 Heavy ... 5Heavy or 6 Light or 7light ...10 light) so you can't find the solution in just 2 weighing that gives only 9 outcomes.

So either way, you fall short of one outcome from the weighing. However at first glance, total 13 balls have only 26 combination but 3^3 = 27 is the total wighing results. This aparent paradox can be understood by that you have to divide Ball combination and weighing combination into at least 2 parts so that in each part, the weighing combination exceed ball combination by 1. And we know its not possible.

In general terms, the problem is that we cannot, bisect the ball.

So I declare this puzzle as UNSOLVABlE. Credits to Prime and Bonovoa.

I agree.

Until we've determined whether the odd ball is heavy or light, the number of possible outcomes must be even: that is, 2 times the number of suspect balls.

To render 26 outcomes to a manageable level with two weighings to go, the first weighing would have to present three cases of 9, 9, and 8 possible outcomes.

One weighing cannot determine the relative weight of the odd ball. So it can't yield a division of cases containing an odd number.

[Prime]

The 12 balls puzzle has been there all along. And we gave this puzzle quite a work over in my Ultimate weighing problem.

The power of 3 rule does not work on the first weighing when one of the balls has unknown fault. (Try 4 balls, one with unknown fault in 2 weighings, for a simple case.)

Balance beam can potentially cut the number of variations to 1/3, provided we can arrange the balls so. With known fault we just toss 1/3 of all balls into one pan, 1/3 -- into another, and leave 1/3 on the side. Where we have unknown fault, for x balls we have 2x variations. On the first weighing we must use equal number of unknown balls in each pan of the balance.

So if we use w balls in each pan and leave s on the side (2w+s=total balls; 4w+2s=total variations), the two distinct outcomes are:

1) The 2w balls on the balance weigh different. Then we have excluded 2w + 2s variations, leaving 4w+2s-2w-2s=2w variations.

2) The 2w balls balance. Then we have excluded 4w variations, leaving 4w+2s-4w=2s variations.

In the case of 13 balls, the largest number of balls we could put on the balance is 8. (10 could leave 10 variations for the remaining 2 weighings.) That would leave 5 balls of unknown fault on the side with possibility of leaving 10 variations for the remaining 2 weighings.

On consecutive weighings, the constraint of having even number of unknown balls on the balance is removed. That is because at that point we have some "good" reference balls, which we could throw on the balance to even number of balls in two pans.

Edited March 9, 2009 by Prime

[bonanova]

Of course, it's a different story if we only have to identify the odd ball.