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sleepsleep
6 vs 6
Code: 1M measure (A)[1][2][3][4][5][6] vs (B)[7][8][9][10][11][12] 2M measure (A)[1][2][3][10][11][12] vs (B)[7][8][9][4][5][6] 3M measure (A)[1][2][3][7][8][9] vs (B)[4][5][6][10][11][12] assume oddball is (A.H) 1M = (A.H) (B.L) 2M = (A.H) means ball in [1][2][3] (A.L) means ball in [4][5][6] assume oddball is (A.L) 1M = (A.L) (B.H) 2M = (A.L) means ball in [1][2][3] (A.H) means ball in [4][5][6] assume oddball is (B.H) 1M = (A.L) (B.H) 2M = (A.L) means ball in [7][8][9] (A.H) means ball in [10][11][12] assume oddball is (B.L) 1M = (A.H) (B.L) 2M = (A.H) means ball in [7][8][9] (A.L) means ball in [10][11][12] i guess i am near, but without knowing it is heavier or lighter seems like a problem to solve it in 3 measures. 

08 Feb 2014, 16:41 

sleepsleep
tthsqe wrote: if you want to make a decision tree, it could start like this Code: (A)[1][2][3][4] vs (B)[5][6][7][8] same equal odd ball in [9][10][11][12] different = which set? A or B, oddball is heavy or light? 

08 Feb 2014, 16:47 

TmX
sleepsleep wrote:
That is my point, exactly. Actually this has been the problem since the beginning. Let's say first you want to weigh 6 vs 6 balls. Sure it will not be balanced. Unfortunately we don't know which ball is the culprit, which is either lighter or heavier. 

08 Feb 2014, 16:49 

tthsqe
guys, the problem is solvable in 3 weighings even if you don't know if the oddball is overweight or underweight, or even there is no oddball. I don't want to give it away.
(A)[1][2][3][4] vs (B)[5][6][7][8] if they are different then you know that every ball in [9][10][11][12] is a good ball. I would then try the comparisons [1][5][9] vs [2][3][7] [2][5] vs [1][6] this is enough info to find the possible odd ball. 

08 Feb 2014, 17:01 

sleepsleep
ok, i found some insight using this method
Code: 1M (A)[1][2][3] vs (B)[4][5][6] 2M (A)[7][8] vs (B)[9][10] 3M (A)[11] vs (B)[1] 1M if same means oddball in [7][8][9][10][11][12] 2M if same means oddball in [11][12] 3M if same means oddball in [12] if different means oddball is [11] i guess this pretty much could solve =) could revolution gives me a star? 

08 Feb 2014, 17:02 

MHajduk
I claim that you can point out which of tennis balls is lighter or heavier than others only in one weighing, doesn't matter how many balls you have (a kind of pun, lol).
You can drop all balls into a big aquarium filled with water and after some time you will see which of the balls is submerged differently than others (lighter ball will be less submerged, heavier will be more submerged). 

08 Feb 2014, 17:18 

revolution
sleepsleep wrote: ok, i found some insight using this method 

08 Feb 2014, 19:20 

revolution
MHajduk wrote: I claim that you can point out which of tennis balls is lighter or heavier than others only in one weighing, doesn't matter how many balls you have (a kind of pun, lol). 

08 Feb 2014, 19:22 

revolution
tthsqe wrote: Maybe our methods are different? 

08 Feb 2014, 19:31 

MHajduk
revolution wrote: Also, what if a heavier ball is also slightly larger and thus the amount the ball submerses is the same as the others? Wikipedia about tennis ball wrote: A tennis ball is a ball designed for the sport of tennis, approximately 6.7 cm (2.63 in.) in diameter. Tennis balls are yellow at major sporting events, but in recreational play can be virtually any color. 

08 Feb 2014, 20:08 

sleepsleep
Code: 1M (A)[1][2][3] vs (B)[4][5][6] 2M (A)[7][8] vs (B)[9][10] 3M (A)[11] vs (B)[1] 1M if same means oddball in [7][8][9][10][11][12] 2M if same means oddball in [11][12] 3M if same means oddball in [12] if different means oddball is [11] Quote:
if [11] is balance with [1], then no doubt, [12] is oddball. the above solution is with the assumption, they are balance in 1M, 2M and 3M, for unbalance approach, Code: 1M (A)[1][2][3] vs (B)[4][5][6] balance = oddball inside [7] to [12] unbalance = oddball inside [1] to [6] but we obtain information (A) is heavier or (B) is heavier 2M (A)[1][4] vs (B)[2][5] balance = oddball inside [3] or [6] unbalance = oddball inside [1][4] or [2][5] assume (A) < (B) in 1M assume (A) < (B) in 2M [1][4] < [2][5] means [4] and [2] are normal ball. odd ball [1] or [5] 3M (A)[1] vs (B)[2] assume (A) < (B) in 3M then [1] is oddball assume (A) == (B) in 3M then [5] is oddball 

08 Feb 2014, 21:30 

tthsqe
revolution, if you go with the conditional (decision tree) approach, you don't nec. have to put each ball on the scale in each path through the decision tree. If you use an unconditional approach, I would assert that you do need to put each ball on the balance at least once to determine if there is an oddball and if the oddball is underweight/overweight.
Here is another fundamental question: It should be plain that the conditional approach can correctly decide upon just as many balls as an unconditional approach; can an conditional approach decide upon more balls than an unconditional approach (assuming the total number of weighings is the same)? 

08 Feb 2014, 22:02 

revolution
Okay the puzzle here is slightly easier than the one I did with coins.
For the coin puzzle it states: That all coins might be the same weight. And also that you must determine if a coin is heavier or lighter (or same) so you can't assume anything about a coin that has not been on the scale. So in this case okay, not all balls must go on the scale. But I would encourage you to think of a general solution that produces a stronger result than just assuming the nature of the last ball. tthsqe wrote: Here is another fundamental question: It should be plain that the conditional approach can correctly decide upon just as many balls as an unconditional approach; can an conditional approach decide upon more balls than an unconditional approach (assuming the total number of weighings is the same)? 

09 Feb 2014, 04:41 

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