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Pastimes : Brain Teasers. -- Ignore unavailable to you. Want to Upgrade?

To: space cadet who wrote (111)	3/4/1998 2:20:00 AM
From: Mitch Blevins	Read Replies (1) \| Respond to of 136

You're right.... semi-induction. The (3^k)/2 came from the structure of the answer I posted earlier, where you have a series of if...then statements that have three branches for each weighing. The 2 divisor comes from there being 2 possible 'leafs' for each ball (heavy or light). The fact that the balls can't be divided into half-balls (or third-balls) invoked the "round_down_to_nearest_int" or "truncate". And the fact that you don't have a control ball for the first guess factored out the 3 for the final form of n = 3 * truncate((3^(k-1))/2) subbing in the numbers for k = 1,2,3... k=1 n = 3 * truncate(1/2) = 3 * 0 = 0 balls You can't tell jack with only one guess k=2 n = 3 * truncate(3/2) = 3 * 1 = 3 balls k=3 n = 3 * truncate(9/2) = 3 * 4 = 12 balls k=4 n = 3 * truncate(27/2) = 3 * 13 = 39 balls Did not verify this one by actually doing it, because I don't have a year to kill... But, you would start by putting balls 1-13 in SIDE1, and balls 14-26 in SIDE2. If it balanced, you would have 13 balls (balls 27-39) and a control ball left to get it in 3 tries (which would work). If it tipped to one side or the other, that's were it gets messy, but doable. As far a practical applications are concerned, I think that professional ball-weighers would consider this immensely practical. ;) ~Mitch