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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
1235 4678
1236 4578
1237 4568
1238 4567
1245 3678
1246 3578
1247 3568
1248 3567
1256 4378
1257 4368
1258 4367
1267 4358
1268 4357
1278 4356
1345 2678
1346 2578
1347 2568
1348 2567
1356 2478
1357 2468
1358 2467
1367 2458
1368 2457
1378 2456
1456 2378
1457 2368
1458 2367
1467 2358
1468 2357
1478 2356
1567 2348
1568 2347
1578 2346
1678 2345


Table 2. RANDOMLY CHOSEN PAIRINGS

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.

1

1378

2456

2

1258

4367

3

1347

2568

4

1358

2467

5

1237

4568

6

1468

2357

7

1368

2457

8

1256

4378

9

1267

4358

10

1458

2367

11

1346

2578

12

1457

2368

13

1578

2346

14

1238

4567

15

1268

4357

16

1248

3567

17

1235

4678

18

1456

2378

19

1478

2356

20

1247

3568

21

1234

5678

22

1348

2567

23

1567

2348

24

1345

2678

25

1246

3578

26

1236

4578

27

1367

2458

28

1357

2468

29

1568

2347

30

1356

2478

31

1678

2345

32

1245

3678

33

1467

2358

34

1278

4356

35

1257

4368

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.

1 1   2 11 1   2
3   4   3   5
7   5   4   7
8   6   6   8
             
2 1   4 12 1   2
2   3   4   3
5   6   5   6
8   7   7   8
             
3 1   2 13 1   2
3   5   5   3
4   6   7   4
7   8   8   6
             
4 1   2 14 1   4
3   4   2   5
5   6   3   6
8   7   8   7
             
5 1   4 15 1   4
2   5   2   3
3   6   6   5
7   8   8   7
             
6 1   2 16 1   3
4   3   2   5
6   5   4   6
8   7   8   7
             
7 1   2 17 1   4
3   4   2   6
6   5   3   7
8   7   5   8
             
8 1   4 18 1   2
2   3   4   3
5   7   5   7
6   8   6   8
             
9 1   4 19 1   2
2   3   4   3
6   5   7   5
7   8   8   6
             
10 1   2 20 1   3
4   3   2   5
5   6   4   6
8   7   7   8
             
             
21 1   5 31 1   2
2   6   6   3
3   7   7   4
4   8   8   5
             
22 1   2 32 1   3
3   5   2   6
4   6   4   7
8   7   5   8
             
23 1   2 33 1   2
5   3   4   3
6   4   6   5
7   8   7   8
             
24 1   2 34 1   4
3   6   2   3
4   7   7   5
5   8   8   6
             
25 1   3 35 1   4
2   5   2   3
4   7   5   6
6   8   7   8
             
26 1   4     1 Carlson
2   5     2 Huisinga
3   7     3 Linde
6   8     4 Krock
          5 MacDonald
27 1   2     6 Moore
3   4     7 McAlister
6   5     8 Christy
7   8        
             
28 1   2        
3   4        
5   6        
7   8        
             
29 1   2        
5   3        
6   4        
8   7        
             
30 1   2        
3   4        
5   7        
6   8        


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