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cushioncrawler
09-11-2012, 05:18 AM
Weighing Pool Balls
Puzzles -> Measuring Puzzles

One of twelve pool balls is a bit lighter or heavier (you do not know) than the others.

At least how many times do you have to use an old balance-type pair of scales to identify this ball?

Rich R.
09-11-2012, 07:00 AM
3 /forums/images/%%GRAEMLIN_URL%%/confused.gif

Rich R.
09-11-2012, 07:02 AM
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SpiderMan
09-11-2012, 09:01 AM
I'd only have to do it once. After that, I'd remember which was the oddball.

BTW, congratulations on your improved writing skills!

SpiderMan

Fran Crimi
09-11-2012, 03:58 PM
Yea! Mac's writing the Queen's English!! /forums/images/%%GRAEMLIN_URL%%/smile.gif

Twice if you only have 12 pool balls.

I think 4 times if you have 15 or 16 pool balls and one out of every 12 is off.

cushioncrawler
09-11-2012, 04:40 PM
DONT READ THIS UNLESS U WANT SOLUTION NO1.
It is enough to use the pair of scales just 3 times. We know of two possible solutions: Solution 1

Let's mark the balls using numbers from 1 to 12 and these special symbols:
x? means I know nothing about ball number x;
xL means that this ball is maybe lighter then the others;
xH means that this ball is maybe heavier then the others;
x. means this ball is "normal".

At first, I lay on the left pan balls 1? 2? 3? 4? and on the right pan balls 5? 6? 7? 8?.

If there is equilibrium, then the wrong ball is among balls 9-12. I put 1. 2. 3. on the left and 9? 10? 11? on the right pan.

If there is equilibrium, then the wrong ball is number 12 and comparing it with another ball I find out if it is heavier or lighter.

If the left pan is heavier, I know that 12 is normal and 9L 10L 11L. I weigh 9L and 10L.

If they are the same weight, then ball 11 is lighter then all other balls.

If they are not the same weight, then the lighter ball is the one up.

If the right pan is heavier, then 9H 10H and 11H and the procedure is similar to the former text.

If the left pan is heavier, then 1H 2H 3H 4H, 5L 6L 7L 8L and 9. 10. 11. 12. Now I lay on the left pan 1H 2H 3H 5L and on the right pan 4H 9. 10. 11.

If there is equilibrium, then the suspicious balls are 6L 7L and 8L. Identifying the wrong one is similar to the former case of 9L 10L 11L

If the left pan is lighter, then the wrong ball can be 5L or 4H. I compare for instance 1. and 4H. If they weigh the same, then ball 5 is lighter the all the others. Otherwise ball 4 is heavier (is down).

If the left pan is heavier, then all balls are normal except for 1H 2H and 3H. Identifying the wrong ball among 3 balls was described earlier.

cushioncrawler
09-11-2012, 04:41 PM
DONT READ THIS UNLESS U WANT SOLUTION NO2.
Solution 2 This solution was provided by Charles Naumann. His method also solves it with just three weighings:
Label the balls 1-12

First Weighing:
Left: 1 2 3 4
Right: 5 6 7 8
Off: 9 10 11 12
Record the heavier side (L, R, or B)

Second Weighing:
Left: 1 2 5 9
Right: 3 4 10 11
Off: 6 7 8 12
Record the heavier side (L, R or B)

Third Weighing:
Left: 3 7 9 10
Right: 1 4 6 12
Off: 2 5 8 11
Record the heavier side (L, R, B)

There are 27 (3^3) possible combination of scale readings. A complete sorted list of the scale reading appears below. Note that only 24 of the 27 readings should be possible given the original problem statement. The algorithm was designed so that if all three scale readings are the same, an error is flagged indicating that the scale is stuck.

BBB Error! There is not a single light or heavy ball (or scale is stuck).
BBL Ball #12 is light
BBR Ball #12 is heavy
BLB Ball #11 is light
BLL Ball #9 is heavy
BLR Ball #10 is light
BRB Ball #11 is heavy
BRL Ball #10 is heavy
BRR Ball #9 is light
LBB Ball #8 is light
LBL Ball #6 is light
LBR Ball #7 is light
LLL Error! Scale is stuck!
LLB Ball #2 is heavy
LLR Ball #1 is heavy
LRB Ball #5 is light
LRL Ball #3 is heavy
LRR Ball #4 is heavy
RBB Ball #8 is heavy
RBL Ball #7 is heavy
RBR Ball #6 is heavy
RLB Ball #5 is heavy
RLL Ball #4 is light
RLR Ball #3 is light
RRB Ball #2 is light
RRL Ball #1 is light
RRR Error! Scale is stuck!

cushioncrawler
09-11-2012, 04:44 PM
I havnt wrapped my brain around solution 2 yet, i will later.
I remember doing solution 1 about 40yrs ago, and i think the original question did inklood having to identyfy whether light or heavy, which solution 1 duz anyhow.
mac.

cushioncrawler
09-11-2012, 04:46 PM
deleted

Fran Crimi
09-13-2012, 06:29 AM
Why do I think it's much simpler than your solution? What am I missing?

You asked for the least number of times. That would translate into one of the first two balls being the odd ball. The only thing you didn't clarify was how many balls in total -- This assumes there were 12 balls:

1. Weigh one ball against another. One ball weighs differently. You know that one of the two is the odd one.

2. Remove one and keep the other on the scale. Add another to the other side. If they weigh the same, then you know that the one you removed from the scale is the odd ball. If they are different, then you know that the odd ball is the one you left on the scale.

cushioncrawler
09-13-2012, 07:21 AM
Yea! Mac's writing the Queen's English!!
Twice if you only have 12 pool balls.
I think 4 times if you have 15 or 16 pool balls and one out of every 12 is off.

YES I SEE WHAT U MEAN FRAN, IT WOULD BE POSSIBLE TO IDENTIFY A FAULTY BALL WITH JUST 2 WEIGHINGS, IF U WEIGHED JUST 2 BALLS, AND IF IT WOZ ONE OF THEZE FIRST TWO.
BUT THAT WOULD BE A SORT OF TRICK QUESTION AND TRICK ANSWER.
ANYHOW 3 DUZZ THE TRICK.
MAC.

Fran Crimi
09-14-2012, 08:09 AM
Rats, you're back to your old spelling. Well, it was nice while it lasted.

As far as tricks go, I think you're question probably left the door open for the assumption I made that the least number of times would imply that the first ball was the odd one.

enjoydgame
09-14-2012, 08:24 AM
You only need to use the pair of scales three times.
The trick is to divide the balls into three groups: [1, 2, 3 and 4], [5, 6, 7 and 8] and [9,10,11 and 12].
You then need to weigh each group and find out the odd group. You can find out the odd ball depending on the odd group.

cushioncrawler
09-14-2012, 04:58 PM
The camel puzzle iz gooder.
The dropping balls puzzle iz even gooderer.
mac.

Fran Crimi
09-14-2012, 06:41 PM
Sorry, Mac. You're back to your old spelling. Count me out again.

Sid_Vicious
09-14-2012, 06:59 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Fran Crimi</div><div class="ubbcode-body">Sorry, Mac. You're back to your old spelling. Count me out again. </div></div>

/forums/images/%%GRAEMLIN_URL%%/confused.gif

sid

Rich R.
09-14-2012, 07:04 PM
<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: enjoydgame</div><div class="ubbcode-body">
You only need to use the pair of scales three times.
The trick is to divide the balls into three groups: [1, 2, 3 and 4], [5, 6, 7 and 8] and [9,10,11 and 12].
You then need to weigh each group and find out the odd group. You can find out the odd ball depending on the odd group.
</div></div>
But you have to continue weighing the balls within the odd group to determine the odd ball.

enjoydgame
09-20-2012, 06:33 AM
But you have to continue weighing the balls within the odd group to determine the odd ball. [/quote]

yes Rich, i agree.. i think i had missed that out.. thanks!