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Las Posas Country Club Flubbers rotation Problem: Eight golfers, divided into two foursomes, are to play golf with each other an equal number of times over an unspecified period of time. How many permutations are necessary? Answer: Each golfer plays with each other golfer fifteen times (3/7 of the time) over thirty-five matches. Each golfer plays against each other golfer twenty times (4/7 of the time). Solution: Assign the digits 1 through 8 to the players, i.e., 12345678. The number 1234 represents foursome number one, players 1, 2, 3 and 4. Now count from 1234 to 1678 using base system 8 since there are 8 golfers in the rotation. The numbers for foursome two are axiomatic. The problem reduces to counting in octal from 0123 to 0567, the four columns, for the purposes of the discussion, being represented by the letters a, b, c, d, respectively. Instead of the numeral 0, we start with the numeral 1 for the sake of simplicity. Start counting by increasing the digits in column d (e.g., 1234, 1235, 1236, 1237, 1238, 1245...). When column c increases, the new number in column d must be one greater than the new number in column c. Likewise, when column b increases, the new number in column c must be one greater than the new number in column b, and the new number in column d must be one greater than the new number in column c. This ensures that none of any four digits repeats, regardless of a single digit's order in a column. In the last case, the sequence 1278 must be followed by 1345, to ensure that the rules for the columns are obeyed. Let's count now starting with 1234 instead of 0123, e.g., 1234, 1235, 1236, 1238, 1245, 1246, 1247, 1248, 1256...etc. Between the numbers 1238 and 1245, the numbers 1241, 1242, 1243, 1244 wouldn't make sense; hence, the rules for columns b, c, and d. Problem: Take four golfers, pair them into twosomes and produce player permutations. Hint: The problem reduces to counting in a numbering system based on 4 digits without repeating any two digits regardless of order, i. e, counting from 11 to 44 in base 4. Along a true counting, without regards to ascending order, the numbers 11, 21, 22, 31, 32, 33, 41, 42, 43 and 44 would be illegal as some of them would be repeating other numbers counted, while others would have players partnering with themselves. For our purposes, the numbers 12, 13, 14 solve the twosome problem, the other numbers being axiomatic. Answer: 12 34 13 24 14 23 In the first row above, players 1 and 2 are matched against players 3 and 4. Proof of Algorithm: Find a combination of player permutations that does not hold for the solution. Formula for permutations P is: P(x) = 5 * y, where x is equal to the number of golfers and y = n(x-2) * 3 + 1, starting with n(6) = 2. Hence, n(8) = 2 * 3 + 1; n(10) = 7 * 3 + 1. Example: To find P(12) for 12 golfers: take n(10) which is 22 and multiply it by 3 and add 1, multiplying that quantity by 5, i.e., P (12) = (n(10)*3+1)*5 = (22*3+1)*5 = 335. Pairings: Using a random number generator, assign the two foursomes to a time slot. Table 1. Algorithmic parings (8 golfers) obtained by counting from 1234 to 1678 in ascending order, foursome one only. 1 2 1234
5678
Using the random numbers below (Table 3), assign the foursomes to a time slot. For example, pairing number one (1234 5678) is assigned to slot 21.
Table 3. Random numbers for assigning slots in Table 2. Random numbers: 21, 17, 26, 5, 14, 32, 25, 20, 16, 8, 6, 35, 2, 9, 15, 34, 24, 11, 21, 22, 30, 28, 21, 15, 1, 27, 12, 27, 10, 26, 6, 14, 12, 35, 13, 10, 27, 33, 23, 12, 7, 20, 3, 13, 9, 10, 29, 11, 19 Table 4. Creating a spread sheet showing the pairings Create a spread sheet using the numbers shown in Table 2. Then for each number, use find-and-replace-all to replace it with a golfer's name. For example, golfer Linde is number 3. Find and replace all 3's with the name Linde, using the edit function--and so on for each golfer's name and number. Enter the time slots 1 through 35 on the spread sheet.
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