Talk:Puzzles/Decision puzzles/Weighings Once More/Solution

Hi, good idea to have a general proof! Here are some things I am not so happy about yet:

You say that S(n) can measure weights between 1 and |S(n)|. In standard mathematical notation |S(n)| would be just the number of elements in S(n) which is n+1 (maybe we should introduce another symbol like W(n) or so). Also, to be technically precise, we should write $$\lfloor 3^{(n+1)}/2 \rfloor$$ rather than 3^(n+1)/2 for the upper bound of weights that can be measured. Thomas

The answer (four weights: {1,3,9,27}) is correct as far as the question is posed. But, as the question does not require weights to be measured in a single weighing, the question and answer both miss an interesting aspect of measuring integral weights: that it is not necessary for the scale to be balanced to correctly identify an unknown weight. With four known weights, {2,6,18,54}, it's possible to measure up to 80, not just 40:

If x < 2, then x = 1.

If x = 2, then x = 2.

If 2 < x < 6-2, then x = 3, etc.

You could repose the question with 80 instead of 40, or turn it around and ask what range of integral weights can be measured using four known weights.

Munin