Hey me and my buddy have been arguing about this logic's question
''you have 12 Golf Balls within this set of golf balls one is defective it can be heavier or lighter than the non-defective ball the only tools you have to use is a balancing scale, this will evaluate equal weight. Also for your convenience the golf balls are numbered. Each time you use the scale there is a cost associated so you want to minimize the number of times you use the scale.''
Can some one help me ?
Trivial upper bound is 11 (weight ball 1 against all balls)
Good upper bound is 4:
Weight 3v3. Then you know which half has a defective one (the one being weighted, the one not being weighted). 6 balls left. Repeat, 2v2. If they're equal, you got 2 where the problem can be. Weight one good ball against one of them and you got it. Otherwise, 4 balls left, you know AB weights more than CD. Take A and C out. Weight BvD. Then weight AvB and you know.
Can we do better? I can't think of how to get it on 3, but we do get more information when weighting 4v4 at the start. If they're equal, we have 4 balls left and can find the odd one in 2 moves. Otherwise, we have 8 balls + weight info. Can this be done in two?
R = right side of scale
L = left side of scale
O = not weighed
where you put the coin on each weighing also determines which coin is the defective one. For example, if the scale tips to the right on the first weighing, is even on the second weighing, and tips to the left on the third, this corresponds to the 4th coin being heavier.
>Step 1: 4v4
Case 1 : 1/3 chance of equal weigh -> defective is one of the 4 other balls
>Step 2 : 1v1 with two of the balls, if there is a weight difference it's one of those 2, if not it's one of the other two
>Step 3 : weight one of the two possible defective balls with a normal ball. If there is a weight difference it's this ball, if not it's the other
Case 2 : 2/3 chance of different weight on step 1 -> defective is one of the 8 balls weighted
>Step 2 : 2v2 with four of the balls, if there is a weight difference it's one of those 4, if not it's one of the other 4
>Step 3 and 4 : See Step 2 and 3 of Case 1
The average number of steps is 3.66 .
>Can this be done in two?
Only if you are really really lucky. If you weight 1v1 and have a difference on the first try, you can weight one of them with another ball. If the weight is different on the second step, the odd one is the one you kept for the second try. If the weight is the same, it's the one you put away.