Assume an old-fashioned scale balance in which weights can be placed on either side. The associated set of weights (each of which is greater than zero) is 'complete' for some W if it is capable of measuring all integer weights from 1 to W.

Clearly it is possible for such sets to exist even if no combination of the weights themselves can be balanced against any combination of the remaining weights - the set {1,2,4,8..} is just one such example.

On the other hand, the set {1,1,2,4} is also 'self-measuring', because, assuming that one of the weights were unmarked, a stranger could neverthless establish its value by weighing it in the scales with the others.

Question: You are allowed 5 weights. They must form a set which is complete, and also self-measuring.

What is the largest possible value of W, given these constraints?

Bonus: What is the largest possible value of W, if 8 weights are allowed?

I assume the solution rests with manipulation of consecutive Fibonacci numbers, but I'll leave the detailed exploration to others.

The value for W = 29 = (2,3,5,8,11) (the first sequence of five after value 1) allows representing all weighings from 1 thru 29, by appropriate combinations of known weights, placed on the same pan and/or the opposite pan from an unknown of integer weight in the range, hence is both "complete" and "self-measuring" as required.  (The value 28 is discerned by two weighings, i.e. heavier than (11+8+5+3)=27 and lighter than (11+8+5+3+2)=29, since all weights are in integer units.)

The next set of five would be 3,5,8,11,19 or 46.  I haven't worked on that yet.  If the principle is sound, then presumably some exploration of eight consecutive Fibs might answer the Bonus question.  The exploration of the set of powers of 2, including 2**0=1, is not as extensive. 

 

Posted by ed bottemiller on 2010-11-23 22:34:49