[bonanova]

One of the site's classic weighing problems has three pairs of balls colored red white and blue.

In each pair, one ball is slightly heavier.

All heavy balls weigh the same, and all light balls weigh the same.

In just two weighings on a balance scale, one may identify the three heavy balls!

How many weighings are required if the balls are all white?

[Guest]

One of the site's classic weighing problems has three pairs of balls colored red white and blue.

In each pair, one ball is slightly heavier.

All heavy balls weigh the same, and all light balls weigh the same.

In just two weighings on a balance scale, one may identify the three heavy balls!

How many weighings are required if the balls are all white?

Three

Lay one ball on each pan:

- If there is equilibrium then the two balls are either heavy or light. Take then one of the already weighed balls and compare it to a third one.

* If there is equilibrium then the three weighed balls are either heavy or light and one more comparision solves the problem.

- If at first weighing, one side is heavier. Than the heavier side is a heavy ball, the other is a light one. Take then two different balls:

* If there is equilibrium then the two balls are either heavy or light. Take then one of the already weighed balls and compare it to a third one, the problem is solved.

* If at second weighing, one side is heavier. Than the heavier side is a heavy ball, the other is a light one. Do a third comparision and the problem is solved.

Edit: explenation

Edited June 26, 2009 by mojail

[Guest]

One of the site's classic weighing problems has three pairs of balls colored red white and blue.

In each pair, one ball is slightly heavier.

All heavy balls weigh the same, and all light balls weigh the same.

In just two weighings on a balance scale, one may identify the three heavy balls!

How many weighings are required if the balls are all white?

you have 6 balls. You way the first ball and it is heavy. You weigh the second, it is light. The third is light. Now we know the first three is heavy, light, light. For this we still cant conclude the weight of the last three. So we way the fourth. It is light. We now know the first four is heavy, light, light, light. from this we can assume the last are heavy. but if the fourth ball was heavy then the four would be heavy, ligh, light, heavy. two heavy and two light. from this we can not conclude the weight of the last two. so we have to wiegh a fith time. Answer is you will have to do between 3-5 weighings. depending on the outcome of each weighing.

[Guest]

Using a two pan balance or feel if the weight difference is significant, put three in each pan or hand. The heavy side eliminates three identical balls. Discard them. Place one ball in each pan or hand. If the two balance then the third is heavier. If the ball is in a pan or hand then it is identified. Total of three weighings.

Edited June 26, 2009 by decal-last

[Guest]

Edit: explenation

Three

Lay one ball on each pan:

- If there is equilibrium then the two balls are either heavy or light. Take then one of the already weighed balls and compare it to a third one.

* If there is equilibrium then the three weighed balls are either heavy or light and one more comparision solves the problem.

- If at first weighing, one side is heavier. Than the heavier side is a heavy ball, the other is a light one. Take then two different balls:

* If there is equilibrium then the two balls are either heavy or light. Take then one of the already weighed balls and compare it to a third one, the problem is solved.

* If at second weighing, one side is heavier. Than the heavier side is a heavy ball, the other is a light one. Do a third comparision and the problem is solved.

Three is correct, and I agree with your method, except you don't clarify what to do if, in the initial equilibrium scenario, you do NOT have equilibrium in the second measure. If Measure #1 gives you equilibrium and Measure #2 does not, you then know that those three balls are either HHL (heavy, heavy, light) or LLH. With Measure #3, measure two of the remaining balls against each other. If they weigh the same, they must match the odd ball out; if they are different, you will know which group to add each ball to.

[Guest]

3 vs 3. take the lighter side and weigh 1 vs 1. if they are equal the reminder is one of the heavy balls. then Take the heavier side from the first weighing and do the same thing. So four times..??

[Guest]

let's say the 3 colors are: red ( r ) green( g ) and blue( b )

Step 1:

Compare r1 and g1 (left side) to g2 and b1 on the right side

We have 3 situations

1) the balance is equal: Then r1 diff from g1 and g2 diff from b1 => step two measure g1 left g2 right. we find out the heavy g => we know then the heavy read and blue

2) r1+g1>g2+b1 => both r1 and g1 are heavy. Step 2 compare b1 to b2 and we know all 3

3) r1+g1<g2+b1 => both r1 and g1 are ligth. Step 2 compare b1 to b2 and we know the heavy

let call the balls w1,w2,w3,w4,w5,w6

compare

w1-w2

w3-w4

w5-w6

al least one measurement is not equal so we know a heavy ball. If there are equal comparasions compare one ball from that to the heavy ball.

worst case 5 measurements

Edited June 26, 2009 by Gaspar Zoltan

[dyalDragon]

notation(H heavy ball, L light ball)

first, label the balls 1,2,3,4,5, and 6

compare balls 1 and 4, there are two possible results: either they are different or the same...

if different, then move on to 2 and 5, same results: either same or different...

if different then weigh 3 and 6 to find their weights and you are done.

if 2 and 5 are the same, then 3 and 6 are also the same and you just need to compare 2 and 3 to find which pair is HH and which is LL...done

now back to 1 and 4 being the same...

compare 1+4 against 2+5... you know 1 and 4 are the same, and this step will tell you whether they are HH or LL, you now know the weights of 1 and 4

next compare 3 and 6, if they are different then you are done, because 2 and 5 will then be the same and the opposite of 1 and 4, but if 3 and 6 are the same, then they are the opposite of 1 and 4 and you must weigh 2 and 5 separately to find out their weights

sorry if this is a little hard to follow, but basically you have 3 steps best case and 4 steps worst case

put one red with one of the whites and the other white with one of the blues

if both sides are the same then you know the blue red are opposites because we already know the whites are opposites, now you just have to weigh the whites to tell which is heavy/light and that will tell you which of the red/blue is heavy/light, and from there it is simple

Edited June 26, 2009 by dyalDragon

[Guest]

One of the site's classic weighing problems has three pairs of balls colored red white and blue.

In each pair, one ball is slightly heavier.

All heavy balls weigh the same, and all light balls weigh the same.

In just two weighings on a balance scale, one may identify the three heavy balls!

How many weighings are required if the balls are all white?

Assuming the 3 heavier balls are all of equal weight and the 3 lighter balls are all of equal weight. Both best and worst case is 3 weighings:

Weigh balls 1 and 2. You have determined that (one is heavy and one is light) or you have determined that they are both of the same class (but not which one).

Case 1: balls 1 and 2 are different - this also tells you which is heavier.

you know that the remaining four balls contain 2 each of heavy and light. You also know the class of the two balls (since you know which is heavier) Separate balls 1 and 2 and weigh balls 3 and 4.

Case 1a: balls 3 and 4 are different. You know the classes of 3 and 4, so separate into the matching class of 1 and 2. You know 5 and 6 are different, so Weigh them and separate into appropriate class Total three weighings

Case 1b: balls 3 and 4 are the same. This means that both 5 and 6 are also of the same class. Weigh balls 3 and 4 against 5 and 6, and you can tell which of (5,6) and (3,4) are heavier. Place the pairs into the appropriate class, since you know which of 1 and 2 are heavier. Total three weighings

Case 2: balls 1 and 2 are the same.

You know balls 1 and 2 are of the same class, and that 3-6 are 3 of the other class and 1 of the same class. Weigh ball 1 against ball 3

Case 2a: balls 1 and 3 are the same. You can now separate the different class balls (ball 3 belongs with 1 and 2). Weigh 1, 2, 3 against the remaining 3. You now know which are heavier. Three weighings. Done.

Case 2b: balls 1 and 3 are different. You now know that balls 1 and 3 are of different classes, AND you know to which class each belongs. You also know 2 out of 4,5,6 belongs with 3 and the other belongs with 1 and 2. Weigh balls 4,5. If they are the same put them with ball 3. If they are different, put them with their matching class (since you know whether the balls in 1,2 are heavier than the ones in 3). Put ball 6 into the class that currently has 5. Three weighings. Done.

There's probably some generic method for 2n number of balls (for some n) ...but i'm too lazy to do it

[bonanova]

mojail [with glib] and tpaxatb have it. Nice going.

sands: we use a balance scale, comparing two balls or groups of balls.

decal-last: How do you distinguish HHH from HHL in the heavy pan?

overide and dvalDragon: close. [dD very close.]

Gaspar Zoltan [gotta love that handle] rethink it.