Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: $Revision: 1.6.2.14 $; site siemens.UUCP Path: utzoo!watmath!clyde!burl!ulysses!allegra!princeton!siemens!steve From: steve@siemens.UUCP Newsgroups: net.puzzle Subject: Re: Weighing Problem (solution) and new Message-ID: <23600002@siemens.UUCP> Date: Thu, 11-Apr-85 10:02:00 EST Article-I.D.: siemens.23600002 Posted: Thu Apr 11 10:02:00 1985 Date-Received: Sun, 14-Apr-85 06:52:43 EST References: <322@tekred.UUCP> Lines: 32 Nf-ID: #R:tekred:-32200:siemens:23600002:000:1739 Nf-From: siemens!steve Apr 11 10:02:00 1985 You couldn't be more wrong!! (insert the sound of a gong here) Taking 1/3 vs 1/3 in the first weighing is right, but after that you messed up! Consider that there are 24 possible cases (12 coins * heavy or light), and three weighings can discriminate 3^3 = 27 cases. That means that one might possibly be clever enough to distinguish which of 13 coins is heavy or light in only 3 weighings, and we really ought to try hard to find a way for 12. In the first weighing we must divide the 24 cases into three groups each of not more than 9 cases. 1/3 vs 1/3 divides into three sets of 8 cases each. Now cleverly find ways of weighing coins to divide each of those sets of 8 into sets of no more than 3 each -- I won't bother to say how because someone else already posted a solution (which is not unique). To prove you cannot distinguish which one is heavy or light from 13 coins, in three weighings. The first weighing must split the 26 cases into three sets of no more than 9 cases each. It must be a weighing of n coins against n coins, and will divide the 26 cases into 2n, 26-2n, and 2n cases for outcomes of <, =, and > respectively. For no integral value of n is 2n <= 9 and 26-2n <= 9. Therefore there is no way to determine which of 13 coins is heavy or light in 3 weighings. In fact, by just looking at the first weighing you can show by a generalization of the proof above that for w weighings, an upper bound on the number of coins you can distinguish is (3^w - 3)/2 (which is an integer), and the first weighing must be 1/3 vs 1/3 of those coins, which is also integral. Can you actually find the bad coin (and whether it is heavy or light) from 39 coins in 4 weighings? Steve Clark ...princeton!siemens!steve