Find the off weight golf ball

[TakenItSeriously]

There are 12 golf balls in a box. You know that one flawed ball accidentally got mixed into the batch.

 

The flaw is that the weight is off but you don't know if it is too heavy or too light. You can only detect the flaw through the use of a balance (not a scale) that can only detect which of two trays contents is heavier.

 

Describe the most efficient method (fewest number of balances) to identify the flawed ball.

 

To help you describe the method use the following codes:

U = Untested

L = golf ball from a Lighter group

H = golf ball from a Heavier group

E = golf ball from an equal or balanced result (good golf ball).

Example:

After balancing 6U vs 6U you get 6L & 6H

 

Please put solutions within spoiler tags

1) find the most efficient method

 

P.S. I believe It should be possible to prove if a method is optimal but I haven't validated my method for testing optimal conditions yet so 2) comes with no guarantees for an answer.

2) prove it's the optimal method

Edited July 18, 2016 by TakenItSeriously

[Daecon]

Split the 12 into two groups of 6 and put them on the balance.

 

Now take 1 ball from each group (you now have B1 & B2 in your hands) and see if the weights balance.

 

If the scales are still unbalanced, discard B1 & B2 (because you know they can't be the faulty ones) and repeat the previous step.

 

When the scales eventually balance, (they'll either both have x number of balls on them or they'll both be empty) you can discard all other balls because you know that either B1 or B2 is the faulty ball. Balance B1 and B2 against one of the previously discarded balls and see which of the two don't balance and you've found the faulty ball.

 

Whether or not it's the fastest method is all down to how lucky you are with your choice of balls to remove from each side of the balance.

[Raider5678]

I believe the fewest possible weigh ins would be 3, but the method can only guarantee 4.

S = weigh ins

Firs one is as follows

6U vs 6U

Result

6L vs 6H

 

You would then weigh 3 vs 3 on either side. If they are the same, you would know its the opposite weight,lighter rather then heavier for example.

 

3H vs 3H

Result

3E vs 3E

E = equal.

You now know that the golf ball is too light, not too heavy, because all 6 heavier golf balls weigh the same, and there's only 1 flawed golf ball. Should there be 2, this problem would be a lot harder.

 

3L vs 3L

Result

3L vs 3H

 

You know the golf ball is too light, so you weigh the lighter batch.

Note you now only have 3 balls. All the others have been ruled out.

1L vs 1L

Result 1.

1E vs 1E

 

You now know it was the other ball, that was not weighed.

 

Result 2

1L vs 1H

 

You now know its the lighter golf ball.

Now going back to step 2, should the golf bald have been different weights, 2 things would be different. You know the golf ball is too heavy, rather then light, and you get to skipma step, bringing the total number of steps to 3.

 

I believe this is the most efficient method. Anybody got this beat?

[TakenItSeriously]

 

I believe the fewest possible weigh ins would be 3, but the method can only guarantee 4.

S = weigh ins

Firs one is as follows

6U vs 6U

Result

6L vs 6H

You would then weigh 3 vs 3 on either side. If they are the same, you would know its the opposite weight,lighter rather then heavier for example.

3H vs 3H

Result

3E vs 3E

E = equal.

You now know that the golf ball is too light, not too heavy, because all 6 heavier golf balls weigh the same, and there's only 1 flawed golf ball. Should there be 2, this problem would be a lot harder.

3L vs 3L

Result

3L vs 3H

You know the golf ball is too light, so you weigh the lighter batch.

Note you now only have 3 balls. All the others have been ruled out.

1L vs 1L

Result 1.

1E vs 1E

You now know it was the other ball, that was not weighed.

Result 2

1L vs 1H

You now know its the lighter golf ball.

Now going back to step 2, should the golf bald have been different weights, 2 things would be different. You know the golf ball is too heavy, rather then light, and you get to skipma step, bringing the total number of steps to 3.

I believe this is the most efficient method. Anybody got this beat?

 

Not bad, 50% 4bal, 50% 3bal depending on 2nd result.

 

Not the best though.

 

All future solutions should be placed within spoiler tags, thanks.

Edited July 18, 2016 by TakenItSeriously

[imatfaal]

three weighings is enough - too long to post so here is a pdf.

 

FYG I have not double checked the details and I am likely to have screwed up somewhere; the method works it is the dood at the keyboard that needs fixing

 

to read chart note that after weighing balls are considered as

G - Known as Good,

U - Still no knowledge

H - in a group one of which is heavy

L - in a group one of which is light

 

3 weighings.pdf

 

Didn't know if I could put a screen grab in spoiler - turns out you can

 

 

 

 

 

 

[TakenItSeriously]

three weighings is enough - too long to post so here is a pdf.

 

FYG I have not double checked the details and I am likely to have screwed up somewhere; the method works it is the dood at the keyboard that needs fixing

 

to read chart note that after weighing balls are considered as

G - Known as Good,

U - Still no knowledge

H - in a group one of which is heavy

L - in a group one of which is light

 

3 weighings.pdf

 

Didn't know if I could put a screen grab in spoiler - turns out you can

 

 

 

Capture.JPG

 

 

Very close to the solution I had chosen to show. Their are a few minor variations possible. But they will always be solved in 3 tests.

 

Tips:

• always try to work with 3 groups, two to test and one set aside in case they balance. This will max the utility of each test leveraging all three results. H vs L, L vs H, or Balanced
• Make sure you leave only three balls to for the last test. Leaving less than three to test is less efficient.
• 1U is roughly the same as 2 partially known HH, HL, or LL

  So 4U is the same amt. of work to solve as 4H + 4L.

• If all tips are optimized, it should result in an optimal solution.

Start with 12U

Test 1) 4U vs 4U || 4U

Split into 3 groups of 4 and test 4U vs 4U giving you two possible results

L vs H: a) 4L,4H,4G

H vs L: symmetrical

balance: b) 4U,8G

 

Start with 4U,8G

Test 2a) 2U vs 1U1G || 1U,7G

Objective is to make all results have 3 or fewer balls that havent tested as good (G)

L vs H: 2L,1H,9G

H vs L: 2H,1L,9G

Balance: 1U,11G

 

Start with 4H,4L,4G

Test 2b) 2L, 1H vs 1H, 2L || 2H, 4G

Objective is to make all results have 3 or fewer balls that havent tested as good (G)

L vs H: 2L, 1H, 9G

H vs L: 1H, 2L, 9G

Balance: 2H, 10G

 

Start with 2L,1H,9G

Test 3) 1L,1H vs 2G || 1L,7G

L vs H: 1L

H vs L: 1H

Balance; 1L

Done: any test with three balls or less can be resolved in one test, with the exception of 3U

 

 

Edited July 18, 2016 by TakenItSeriously