In my spam folder, I just found an e-mail from Sam Pottle, who sent his solution on Feb. 27 (before I posted mine).
Sam's solution was not only correct, but it was mathematically thorough. I am confident that he could get quite wealthy off of King Sharuyar's puzzles at higher weigh levels, given the opportunity.
Sam closed with
I want to offer a word in praise of this fair-minded monarch. We have all of us known despots who would pose a problem with no solution, thereby giving himself a freeroll. But King Shahryar has twice offered his valued subject the opportunity to earn a fortune, if he has both wit and courage enough. We should all have such playing companions.
As for Mrs. A, perhaps she will be less nervous about the outcome this time around!
When you know whether the counterfeit is light or heavy, the formula for the maximum number of coins, where "n" is the number of weighs, is 3^n. For example, for three weighs, you can solve for 27 coins. You simply divide them into three subsets (left side, right side, off scale) and repeat.
When you don't know if the counterfeit is light or heavy, the formula is (3^n - 1) / 2. I worked this out by trial and error, but learned later it was already known (albeit not widely). [Sam figured it out faster than I did, clearly having a superior mathematical background, and perhaps a superior mind.]
This formula generates 4 coins for two weighs, 13 for three, 40 for four, 121 for five, etc. Perhaps one can look it up somewhere, but the fun (for me, anyway) was working out the method and figuring out the maximum number of coins (as I was not provided with that information). At the three-weigh level I conjectured the formula, at the four-weigh level I confirmed the formula and could see why it worked, and by the five-weigh level I understood the puzzle thoroughly.
I discovered that by introducing a known neutral coin at the start, it is possible to add a coin to the (3^n - 1) / 2 pool. With it, one can solve for 5, 14, 41, 122, etc., which becomes 6, 15, 42, 123, etc., when counting the neutral coin itself (the one with the X on it), thereby tweaking the formula to (3^n + 3) / 2). That's my contribution to coin puzzle theory; while I suppose non-trivial, it is not exactly path-breaking.
When I conceived this expanded version of the puzzle, I thought of relating it to backgammon because a player has 6 points in a quadrant and 15 checkers total. While obviously of the coin genre, maybe it could be presented as a checker problem. Checker weigh rhymes with checker play, and that might subliminally catch on.
I'm glad I wasn't tasked with writing Part 3. I'm not sure I have the imagination to make 42 a meaningful number of checkers. I suppose there could be a variant with 21 checkers on each side... hmm but silver is lighter than gold. Maybe the silver checkers could be changed to platinum... but can anyone tell me: which weighs more, a pound of gold or a pound of platinum?
Nack
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