Okay - so I posted this puzzle to reddit a few weeks ago and one guy had the following comment:

*Same problem, but your solution must be non-adaptive. IOW, specify the weighings that you will make up front. You are not allowed to change them based on the results of previous weighings*

So the whole problem becomes: Identify the odd (heavy or light) ball among 12 identical balls **with a static plan (i.e. no conditional decisions)** that uses at most 3 weighings of a balance scale.

The original puzzle is hard enough, but this variant has been killing me. I think I could come up with a solution if the original problem specified whether we expect the odd ball to be heavy or light.

Anyone have any ideas? It feels nearly impossible.

Unsolved

## Comments

Sign uporlog in with Facebookto comment.Jared WellsOK here's a little thought, not sure either about a full solution: Might there be some tricky way to weigh the balls in a some overlapping sets in order to discern which ball it is exactly? I think that still causes problems when you don't know whether it's heavier or lighter though...

Jake GlascockLabel the balls A1-A6 and B1-B6.

LeftRightLeftRightLeftRightOffExplanation:Weigh Onegives you group A or B.Weigh Twogives you 1-3 or 4-6.Weigh Threegives you 1, 2, or 3 OR 4, 5, or 6.Larry Fennheavier or lighter means your first weighing is irrelevant

Jake GlascockYou're right. I read the problem too quickly and failed to account for heavy or light. Solution foiled!

JoshI just figured it out. It should be possible because each weighing has 3 outcomes so 3*3= 27 outcomes and there are only 24 different possibilities for the answer.

Letâ€™s say you have balls ABCDEFGHIJKL

1st weighing ABCDE v FGHIJ

2nd weighing GCJE v AIDK

3rd weighing BEI v GHL

S= same, L= left side heavier, R= right side heavier for the outcomes AL = A is lighter and AH = A is heavier (for example HHH = EH)

F, K, and L are only weighed once, so if their side is heavier or lighter and the other two weighings are the same, then you know if they are heavier or lighter.

E and I and G are weighed 3 times, LLL= EH and RRR=EL and RRL= IH and LLR = IL and LRL = GH and RLR= GL

All other balls are weighed twice. If their side is heavier or lighter both times then that ball is heaver or lighter. The key to this is that none of the balls weighed twice are weighed with any of the other twice weighed balls.

All combinations except the impossible SSS, RLL, and LRR give unique answers which is also 27 possible outcomes- 3 impossible ones = 24 total outcomes

Satya SingamsettiThought provoking..

"None of the balls weighed twice are weighed with any of the other twice weighed balls". I think you meant to type this (corrected weighing) 1st weighing ABCDE v FGHIJ 2nd weighing GCJE v AIHK 3rd weighing BEI v GDL

But neither of these produce a solution.

(your weighing) If the result is LRS => Either A or D can heavy or J can be Light. So Cant say.

(Corrected weighing) If the result is LRS => Either A can heavy or J can be Light. So Cant say.

But I still like the idea. Good luck!!

Anurag KapaleI think I have come up to a solution. Please check and try to find failing case if any (though I am pretty sure there shouldn't be any.)

^{ spoiler}don't read thisDivide the balls in 3 sets of 4 balls each: A, B, C

*First Weighing: A1, A2, A3, A4 vs B1, B2, B3, B4

*Second Weighing: A1, A2, B1, B2 vs A3, C1, C2, C3

*Third Weighing: A1, B1, B3, C1 vs A2, B2, B4, C2

Explanation:We can consider case wise: *Case 1: all weighing are equal or First two weightings are equal: C4 as it was never taken for weightings, and all other are defect-less.*Case2: First Weighting is equal; second is unequal:

Case 3: First weighting is unequal; Again without loss of generality assume that Left was heavier. One of balls As can be heavier or one of the balls in B can be lighter.Case 3.1 in second weighting left is Heavier. That can be only because of balls A1 or A2. So, one of them is odd. In the third weighting they are in opposite pans, so whichever pan is heavier that ball is odd and heavy.*Case 3.2: In second weighting right is heavier: That can either be because one of the balls B1 or B2 is ligher or ball A3 is heavier In the third weighting we dont have ball A3 and balls B1 and B2 are in opposite pans. So, whichever pan is ligher that ball is odd. If the pans are equal that means A3 is odd and heavier.

*Case 3.3 In second weighting pans are equal. That means none of the A1, A2, A3 or B1, B2 are defective. In the third weighting we have B3 and B4 in opposite pans, whichever pan is lighter that ball is defective. If the pans are equal, then A4 is odd and heavier.

This covers all the cases. Hope you enjoy the solution. :)

Kyle GilbertI think I've solved it down to a 50/50 shot.

^{ spoiler}number balls 1-12. weigh 1-4 vs 5-8. then, weigh 5-8 vs 9-12. final weighing is 3-5 vs 8-10. the first two weighings will tell you which group has the ball. if weigh 1 tilts and 2 doesn't, its 1-4. if weigh 2 tilts and 1 doesn't, its 9-12. if both tilt, its 5-8. the third weigh cuts each group in half so that if the scale tilts, you know that it is one of the two from the group already causing the scale to tilt. if it doesn't, then you know that it's one of the two that you left out from the group.IasonGalaxyThere is no solution to this problem. Let the predetermined pairs of sets of balls to be weighed be (S1,T1), (S2,T2) and (S3,T3). Now each weighing will (independently) determine a set of possible balls B1, B2, B3 which may contain the odd ball. The solution (assuming it exists) will be the single ball which lies in the intersection of B1, B2 and B3. It is key to note that Bj will either be the union of Sj and Tj or the set of of balls that are not in Sj and Tj. In other words each Bj can be one of two possible sets so there are at most 8 values for this intersection. If the odd ball isn't one of these (at most) 8 values then it obviously can't be determined by this weighing process.

Nigel ParsonsI've already given this solution in the original 12 balls problem, but here goes again: There is a solution to this which allows you to know in advance what three weighings/comparisons you will make, so that the choice of balls for the second and third weighing is not dependant on what has gone before. Label/identify each ball with a single letter: (using the letters from) "THE KP FORMULA" Weigh: TAKE against FOUR Weigh: PARK against THEM Weigh: HALF against MORE Record the results of the three weighings (Left pan heavy / left pan light / pans balance) By comparing the results with the pan contents you can identify the odd ball and whether it is heavy or light every time. eg. if the left pan always goes down then 'A' is heavy, if it always goes up 'A' is light. each ball has its own set of three unique weighings.