# A balance with three pans, detecting the lightest pan (find the one lighter ball)

The following puzzle was told to me by a friend, Markus Götz, who put it online here: Deviating Ball Puzzles (pdf). After some searching, I did not find this puzzle anywhere else online. The only similar (but ultimately different) puzzle I found is Three Pan Balance, 15 Coins.

Are there any other of three-pan balance puzzles online?

The three-pan balance

Imagine a balance with not two, but three pans. Weightings using the balance follow these rules:

• If there exists a pan that is lighter than each of the other two pans, then this pan goes up and the other two pans go down to a stop. (Note that one cannot see which of the two heavier pans, if any, is the heaviest.)
• If there is no single lightest pan, then nothing happens. (This includes the case of two equally light pans and one heavier pan.)

Let's call this the "lightest-pan-detection-rule" (LPDR).

The problems

The puzzle consists of several related problems. (The original contains 12 problems, with different kinds of ball deviation. I will split them across several questions.)

In each problem you are given n balls. They are all of the same weight, exept as stated in each problem. You are to identify the deviating ball(s) by using the new balance a maximum number of weighings stated in the puzzle. You are also to present a method to identify the deviating ball(s).

1) You are given n balls, one of which is lighter. What is the largest n, so that the lighter ball can be identified with 1 weighting?

2) You are given n balls, one of which is lighter. What is the largest n, so that the lighter ball can be identified with 2 weightings?

3) You are given n balls, one of which is lighter. What is the largest n, so that the lighter ball can be identified with k weightings?

• If you are planning on asking more questions to obtain answers for all 12 of your questions, I would recommend (possible after this question) asking a more abstract version such that the answer can be used for the remaining questions (and any other similar variants) [Assuming one is not provided in this question]. This has been done for many problems such as: 'coin weighing', 'guessing hat colour' etc.. Jun 4 '15 at 20:16
• @MarkN The answers for one lighter ball and the answers for one heavier ball can be generalized to k weightings. For the other problems, with two lighter or two heavier balls, I know of no general answer, but had to solve each number of weightings separately. Jun 4 '15 at 20:21
• You can try asking for a generalization of those in a separate question if you like and see what people can think of, some people here might surprise you. Jun 4 '15 at 20:23
• I will ask several questions. How should I interlink them? In the question? In a comment? All from the first question? Linearly (each one to the next)? Backwards? Not at all (relying on SO's "related" feature)? Jun 4 '15 at 20:36
• (My opinion - so not the rules) I would recommend that you wait until some are solved before posting the next, so that you could easily link linearly. But otherwise I would recommend linking them based on their category (i.e group finding n heavy balls, group finding n light balls etc.). Just make sure that each question is unique enough such that it wont be too close to a duplicate or such that each general answer is contained within its own question (and doesn't answer your possible other already posted question). Jun 4 '15 at 20:41

1) Four. This is clearly the maximum, as there are four possible outcomes from a single weighing.

Put one ball on each pan, and the fourth ball aside. If one pan rises, that is the light one. If none do, the fourth one is lightest.

2) Sixteen.

Put four balls on each pan and put four aside. This allows you to identify a group of four that contains the light ball. Put the others aside, and weigh those four as in case 1.

3) $4^k$

Put $1/4$ of the balls on each pan, and the other $1/4$ aside. This identifies one group with the lightest ball. Iterate through all $k$ weighings until you are left with 1 ball.

• Perfect. Soon you'll get to see the tough variants. :-) Jun 5 '15 at 6:32