Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: notesfiles Path: utzoo!linus!decvax!decwrl!amd!dual!zehntel!hplabs!hp-pcd!hpfcrs!lief From: lief@hpfcrs.UUCP (lief) Newsgroups: net.puzzle Subject: Re: Orphaned Response Message-ID: <15400002@hpfcrs.UUCP> Date: Sun, 14-Oct-84 18:57:00 EDT Article-I.D.: hpfcrs.15400002 Posted: Sun Oct 14 18:57:00 1984 Date-Received: Thu, 18-Oct-84 19:23:04 EDT References: <-26200@iwu1c.UUCP> Lines: 73 Nf-ID: #R:iwu1c:-26200:hpfcrs:15400002:37777777600:3834 Nf-From: hpfcrs!lief Oct 9 14:57:00 1984 PROBLEM: Given 12 golf balls, one being odd (either heavier or lighter). Determine which one is heavier or lighter with only 3 weighings on a balance scale. I don't know what the solution is supposed to be, but here is how I would do it (same technique as Hamming codes are done): (1) The balance scale can be thought of as a 3-state digit. In other words: a. If the left is heavier than the right, the value is 0 (arbitrary). b. If the left is ligther than the right, the value is 1 c. If the left is the same as the right, the value is 2 (2) Because of the 3-state nature of the balance scale, if we perform 3 seperate weighings, then that gives us 3^3 or 27 possibilities, which should be sufficient. For our situation there exists 24 possibilities, either 1 of the 12 balls will be heavier or 1 of the 12 balls will be lighter. (3) I will assign a unique 3 digit value to each golf ball (using 3-state digits). This to is somewhat arbitrary, except that we would like the heavy ball to be the complement of the light ball. Thus we have: BALL ID HEAVY VALUE LIGHT VALUE ======= =========== =========== a 000 111 b 001 110 c 002 112 d 010 101 e 102 012 f 120 021 g 121 020 h 122 022 i 210 201 j 211 200 k 212 202 l 221 220 Note that I did not use 011, 100, or 222. Obviously, there is no complement for 2 since that is when both sides of the scale are equal. Thus, one would not use 222 as it could not be unique. (4) Divide the 12 golf balls into 3 groups of 4 and perform the 3 weighings in the following order (determined by the values assigned above in (3)): WEIGHING NUM LEFT SIDE RIGHT SIDE OFF SCALE ============ ========== ========== ========== 0 a b c d e f g h i j k l 1 a b c e d i j k f g h l 2 a d f i b g j l c e h k Note that the left side corresponds to digit 0, the right side corresponds to digit 1 and off scale corresponds to digit 2. Weighing number 0 corres- ponds to the 1st digit of the ball value, weighing number 1 corresponds to the 2nd digit of the ball value, and weighing number 3 corresponds to the 3rd digit of the ball value (1st digit is the left-most digit). Just in case you don't understand this table, what I am saying is that in the 2nd weighing, place balls a, b, c, and e on the left side, balls d, i, j, and k on the right side, and balls f, g, h, and l off the scale. (5) Example: Suppose on the first weighing, the sacles are equal. On the second weighing the left side is heavier. On the third weighing, the scales are equal again. This gives me a result of 202 (where 2 is assigned for equality and 0 is assigned when the left side is heavier, as defined above). Looking at the table in (3) shows that 202 corresponds to ball k being lighter than all the other balls. Example: Suppose on the first weighing, the left side is heavier. On the second weighing the left side is heavier. On the third weighing, the right side is heavier. This gives a result of 001. Looking at the table in (3) shows that this corresponds to ball being heavier than all the other balls. Lief Sorensen