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Weighing Pool Balls - Solution
Puzzles -> Measuring Puzzles
 The Puzzle: 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? |
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(Scroll down if you really want
to see the Solution ...)
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The Solution . . .
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.
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! |
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