Wednesday, October 5, 2016

weighing - 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?


Follow-up question: A balance with three pans, detecting the lightest pan (find the one heavier ball)



Answer



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.


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