28 Aug 2022 2:06 am
It's 4x4 on the first and 4x4 on the second. You do have to slide 3 marbles on the 2nd turn and use your control marbles. you're so close!Cannonpointer » 27 Aug 2022, 12:18 pm » wrote: ↑ @Jantje_Smit
I think that did the trick. Gonna use hypothetical colors for clarity, so the mind's eye can track the moves.
We have 4 side a light-side suspects (reds), 4 side b heavy-side suspects (blues) and 4 controls (yellow). In two weighings, we have to find the culprit.
You pull 3 reds from light side a and replace with 2 blues and 1 yellow control, replacing the 2 blues on heavy side b with 2 yellow controls. If even, you know the odd marble is lighter and is one of the 3 sequestered reds. You weigh two of them against each other. The lighter one is the culprit - if same, the remaining sequestered red is the odd light marble.
If odd, and side a is still lighter, you know the culprit is EITHER the 1 red OR one of the the two blues from side b. Weigh the two blues against each other. The heavier blue is the culprit. If even, the lighter red is the culprit.
If odd, and side a is now heavier, it's the heavier of the 2 blues on side a.
Solved. That was a bear.
Thank you for standing up for me and giving me the opportunity to be a smart-***!
PS - I have a hunch that it did not occur to you to use colors in the explanation you gave to Jantje Smit, since you appear to have been somewhat frustrated by the arduousness of getting the solution across.
I believe that the idea to assign the 3 groups of marbles an arbitrary color actually helped me solve the puzzle, as it gave me a short-hand way of tracking and talking about the marbles to myself (yes, this puzzle had me talking to myself).
The rest of your control marbles will solve the puzzle.