Broken Billiard Balls (August 2021 Challenge Question)

You put four balls on each side and weigh them, one side will be heavier, you remove one ball from each side, and continue until both sides weigh the same, now you know that the last ball removed from the heaviest side is the defective ball...worst case scenario is 4 weighs...best case is 2...

A little more explanation. For example, suppose the 6th ball (101) is heavier.

1st weighing: 000, 001, 010, 011 would be lighter than 100, 101, 110, 111; "1" side is heavier.

2nd weighing: 000, 001, 100, 101 would be heavier than 010, 011, 110, 111; "0" side is heavier.

3rd weighing: 000, 010, 100, 110 would be lighter than 001, 011, 101, 111; "1" side is heavier.

The heavier ball is "101".

That's even better than the three weighs - which I came to the as the first solution as you did...I didn't think of the two weigh trick - nice logic!

Winner winner chicken dinner.

Three .

Four balls to each side. If balanced then no ball is less than others.

If not balanced take lighter pan and split two balls to each pan.. lighter pan again.. split one ball to each pan.. lighter pan is ball that weighed light.

Yes, it does work with both

(two-groups-of-3-balls-coupled-with-one-group-of-2-balls)

as well as

(two-groups-of-4-balls-coupled-with-one-group-of-1-ball).

Huh?

Just weigh them individually on a good scale!

With eight balls, the answer is three as most have already replied:

1st: weight four and four, opposite each other. The heavier side will contain the defective ball. Thus the opposite four can be temporarily discarded.

2nd: Splitting the four heavier balls to two and two, repeat the above step. Again, discard the lighter balls and split the two remaining balls.

3rd: Weigh the two remaining balls, opposite. The heavier ball is your defect!

Why on earth would anyone have 8 billiard balls and you represent them with a black 8 ball? There is no black ball in billiards and certainly not 8 balls. Three is all that is needed for billiards. Snooker needs 22 balls.

1 Weigh 6 balls split between two pans. If they balance the heavy ball is one of the 2 on the desk, not yet weighed.

2. If they don't balance: Remove 3 balls from the lighter pan. The remaining 3 balls on the pan, remove one to the empty pan, add one ball from the desk giving 2 balls in each pan.

3. If it balances the heavy ball is on the desk, the last un-weighed ball. If the balance is out of balance then the heavy ball is in the drooped pan.

4 Remove the balls from the raised pan, split the remaining two balls from the heavier pan and determine which one is heavier on the balance.

Billiards is a "catch all" term for all "cue, ball and table" games. The original "billiards" is now commonly known as "English billiards".

Some games like croquet have even been called ground billiards.

One!

Just weigh one ball against another and you stand a 1 in four chance of finding the duff one.

However to be sure of finding the right one you need two.

Weigh 1, 2 and 3 against 4, 5 and 6

If they balance weigh 7 against 8 to find the duff one.

If they don't balance: suppose without loss of generality that 1, 2, and 3 are heavier. Weigh 1 against 2: if they balance the duff one is 3; if they don't its obvious which is heavier.

But Rixter has already given the correct answer, so, we should try a more difficult problem. You've got twelve balls: one of them is duff but you don't know if it's heavier or lighter. Can you find out which one is duff, and, if it's heavier or lighter in three weighings?

The title of this months challenge reminds me of a trick I played on a friend at work. We had a couple of pool tables, when one of the guys who used to manage the tables left his boss, who knew I played at lunch times, gave me a set of brand new balls he found in his desk. Although the balls were new the black ball was broken in half: I can't imagine how that happened, the break was rough and the two halves were roughly equal, so you could push them together and it would just hold. I challenged my friend to a game and set them up with the broken ball secretly substituted for the old one in the middle of the pack. My friend was a very good player: on his second visit to the table he had cleared all his balls, and, remarkably the black was still in its original position. So he was actually playing to win the game when the cue ball hit the black it simply fell apart on the spot.

I can't imagine how many times you would need to repeat that experiment to get to the point where a player was actually playing the black ball before the fault was discovered.

You've got twelve balls: one of them is duff but you don't know if it's heavier or lighter. Can you find out which one is duff, and, if it's heavier or lighter in three weighings?

The answer is probably yes, the question is how.

In the original problem, the scale gave you one of 3 indications - bad ball on left, right, or neither. Not knowing if the bad ball is heavier or lighter means that the scale can only tell whether the bad ball is on the scale, not on which side - 2 indications.

1. Take half the balls, three on each side. If it doesn't balance, the bad ball is among these six (bad-six). If it does balance, these six are good (good-six) the bad ball is among the other six (bad-six).

2. Balance 3 balls from bad-six with 3 balls from "good-six". If it doesn't balance, they are "bad-three", else the other three are "bad-three". You now know whether the bad ball is lighter or heavier!

3. You now have 3 balls from "bad-three". Put one on each side. If they balance, the third one is bad. If they don't, you know which one is bad because you know from (2) whether it is lighter or heavier.

Nice puzzle!

You seem to be interested in variations. If you like, next level of challenge is: How many trials are required to find odd ball (don't know lighter/heavier) for 'n' balls ( n>2 & natural number)

a) with single pan balance (like we see in stores; which displays weight)

b) with two pan balance

?

"Balance 3 balls from bad-six with 3 balls from "good-six". If it doesn't balance, they are "bad-three", else the other three are "bad-three". You now know whether the bad ball is lighter or heavier!"

Unfortunately: if the "bad six" were the six not in the first weighing, and, if it's the other three, you don't know if the ball is lighter or heavier.

You are absolutely correct. Back to the drawing board...

Let's try this again.

You've got twelve balls: one of them is duff but you don't know if it's heavier or lighter. Can you find out which one is duff, and, if it's heavier or lighter in three weighings?

1. Take half the balls, three on each side. If it doesn't balance, the bad ball is among these six (bad-six). If it does balance, these six are good (good-six) the bad ball is among the other six (bad-six).

2. Balance 3 balls from "bad-six" with 3 balls from "good-six". If it doesn't balance, they are "bad-three" and you know whether the bad ball is heavier or lighter.

If it does balance, the other 3 balls are "bad-three". Weigh them against 3 balls from "good-six" to see whether the bad ball is heavier or lighter.

3. You now have 3 balls from "bad-three". Put one on each side. If they balance, the third one is bad. If they don't, you know which one is bad because you know from (2) whether it is lighter or heavier.

So, the best I can do is 3.5 weighings. Half the time requires a 4th weighing to determine heavier/lighter.

I'm sure you can do better: I assure you that it's possible.

Divide the 12 balls into 3 groups of 4, call them A(1-4), B(1-4), and C(1-4). Weigh A(1-4) against B(1-4).

• If A < B, A contains a lighter ball or B contains a heavier ball.
• If A > B, A contains a heavier ball or B contains a lighter ball.
• If A=B, C contains the bad ball, heavier or lighter, we don't know.

In the first 2 cases, C can be used to supply balls known to be good. In the last case, A or B can be used to supply good reference balls.

If A < B (A contains lighter ball or B contains heavier ball)

• Weigh B1, A1, A2 against A3, A4, C1.
• If (B1,A1,A2 = A3,A4,C1), all A is good and B1 is good. B2, B3, or B4 is heavy. Weigh B2 vs B3. If they balance, B4 is heavy, else the heavier of B2 and B3 is the faulty ball
• If (B1,A1,A2 > A3,A4,C1) B1 is heavy or A3 or A4 is light. Weigh A3 and A4. If equal, B1 is heavy, else the lighter of A3, A4 is faulty.
• If (B1,A1,A2 < A3,A4,C1) A1 or A2 must be light. Weigh A1 vs A2 and the lighter one is faulty.

If A > B (A contains heavier ball or B contains lighter ball)

• Weigh A1, B1, B2 against B3, B4, C1.
• If A1,B1,B2 = B3,B4,C1, all B is good and A1 is good. A2, A3, or A4 is heavy. Weigh A2 vs A3. If they balance, A4 is heavy, else the heavier of A2 and A3 is the faulty ball
• If A1,B1,B2 > B3,B4,C1, A1 is heavy or B3 or B4 are light. Weigh B3 and B4. If equal, A1 is heavy, else the lighter of B3, B4 is faulty.
• If A1,B1,B2 < B3,B4,C1, B1 or B2 must be light. Weigh B1 vs B2 and the lighter one is faulty.

If A and B balance, C contains the faulty ball, balls in A and B are good

• Weigh A1,A2,A3 vs C1,C2,C3.
• If (A1,A2,A3 = C1,C2,C3) C4 is faulty, weigh A1 vs C4 to determine if C4 is heavy or light
• If (A1,A2,A3 > C1,C2,C3) weigh C1 vs C2. If they are equal, C3 is light. Otherwise, the lighter of C1, C2 is the faulty ball.
• If (A1,A2,A3 < C1,C2,C3) weigh C1 vs C2. If they are equal, C3 is heavy. Else the faulty ball is the heavier of C1 and C2.

Good Answer: I can't see any problems with that.

The way I know is to weigh four against four. If they balance use the same method as you.

If they don't: move three from the right scale to a holding area; three from the left scale to the right scale and three good to the left scale. If they balance the dud is in the three in the holding area, and you already know if it's heavy or light: weigh one against one etc.

If they stay the same you know that dud is one of the two balls which didn't move, and you know which might be light and which might be heavy. Weigh either against a good ball to resolve.

If the scale tips the other way then you know that the dud is in the three which moved from one side to the other, and, if its heavy or light. Weigh one against one etc.

Three groups, of 4 balls in each, would solve that puzzle in three ''weighs''. if the first ''weigh'' was any four against any other four.

Can you expand on that a bit?

OK, adding 4 more balls to make a total of 12 billiards enables that total to be evenly divided into three groups of 4 per each.

Weighing any 1 group of 4 against either of the other 2 groups will reveal which is the heaviest group of in only one weighing.

Breaking down that one group of four into two groups of two per each, will reveal which pair is the heaviest pair, after the second weighing.

Breaking down that last pair for a third weighing will reveal which single billiard is the heaviest one.

QED.

"You've got twelve balls: one of them is duff but you don't know if it's heavier or lighter."

3 weighings only

4:4 weigh all 8

2:2 Re weigh lighter 4

1:1 Re weigh lighter 2

Bob's your ankle!

GA for creativity - but with three provisos: You are assuming the pan is large enough to accommodate 7 balls; it might require a small hammer and punch to make the required dent in the centre of the bottom of the pan; and the use of double-sided tape introduces another material aid that does not violate the letter of the puzzle, but probably the spirit.

Thanks!

You are correct: in such a puzzle some assumptions must be made - but that holds for any of the solutions above. I must also assume that the pan is round :-)

For example, for all solutions above it must be assumed that the balance is "good enough", which basically means that the weight difference of the deviating ball is sufficient to overcome the static friction of the balance.

When solving such a puzzle, I think it is good to think about the 'silent assumptions' one makes when thinking about the solutions. Do not let yourself be limited by restrictions in your mind, but do keep the spirit of the question in mind :-)

The hammer/punch and the double sided tape are not necessary for the theoretical solution but could make it more practical.

Put the balls in a centrifuge lined with clay, the ball that makes the deepest impression is the heaviest ball....1 weigh