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 Subject: Golf teams Category: Sports and Recreation > Team Sports Asked by: pippic-ga List Price: \$15.00 Posted: 04 Sep 2006 09:44 PDT Expires: 04 Oct 2006 09:44 PDT Question ID: 762106
 ```I have seen the work done by mathtalk-ga in late 2003 and have two subsequent problems: In 4 consecutive rounds of foursomes I am seeking to ensure that each 1 of 12 players only plays in the same flight as any other player once (or at least gets as close as possible to this ideal) In a round of singles (2 players) followed by a round of foursomes I am seeking to ensure that each 1 of 16 players only plays in the same flight as any other player once (or gets as close as possible to this ideal)```
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 ```Hi, pippic-ga: The first part of your Question is clear (4 consecutive rounds of foursomes implies each player has at least one of the other 11 players with whom two rounds are shared), but in the second part, are there only two rounds (eight pairs in one round and four foursomes in the other)? regards, mathtalk-ga```
 ```Hi, pippic-ga: Since I haven't heard back from you in a few days on the clarification to the second part of your Question, let me go ahead and give results about the first part. I'll also add a note on the second part at the end. Since in a group of 12 golfers, there are for each player 11 "others", any player who goes out in 4 foursomes will necessarily find at least one repetition (since there are three "others" in each foursome). While it is possible to schedule twelve foursomes so that each player experiences only one repetition, this schedule is almost certainly useless for your purpose because it does not allow three foursomes to play simultaneously on each day. In fact each pair of foursomes in a schedule of that nature will intersect in at least one player, so you couldn't even play two foursomes at a time. For the sake of completeness I will give such a schedule further down and show how it can be constructed. So I think we have to state as an important aspect of your problem that on each of four days, we schedule three mutually disjoint foursomes. By mutually disjoint I mean that no player is assigned to more than one of such three foursomes (and of course since there are exactly 12 players, this implies each player is on exactly one of the three foursomes for that day). [There is some technical terminology for this, which might be useful if you want to search the Web later for similar problems. Since two foursomes on a given day "never meet" in a common player, these various days' foursomes may be called "parallel classes". A "design" (schedule of foursomes, in this case) that can be broken ("partitioned") into parallel classes is referred to as a "resolvable design". Okay, enough technical terminology!] The first day's "flight" is arbitrary, in the sense that up to a permutation of the players, all possible schedules agree about the first day. You can just divide the 12 players into any three groups of four, and there's nothing better or worse about how those groups are chosen. Nonetheless the choices for grouping on days 2, 3, and 4 will be conditioned by the first day's choice. We can picture the first day's foursomes something like this: A1--A2--A3--A4 B1--B2--B3--B4 C1--C2--C3--C4 which shows us three rows of four corresponding to three foursomes. It is now impossible to choose _any_ foursome without getting at least two players from the same foursome of that first day. Thus after the first day, every foursome on a schedule will involve at least one "repeat". Thus we will have at least three repeats on day 2, on day 3, and on day 4, and that's only considering how many repeats may occur between the first day and each of the later days. Given that repeats will (in this sense) occur all the time, we should ask what the measure for "as close as possible to the ideal" should be. Since generally the number of distinct pairs who play together at some point is complementary to the number of repeat pairings, we could try maximize the number of distinct pairings. Unfortunately, to do so requires that a few pairings will be repeated multiple times, allowing a greater number of distinct pairings overall at the expense of some of the players. To counteract that we can impose a restriction that each player will meet the same number of "others". While this slightly reduces the maximum number of pairs who have the chance to play together, it treats all the players as equally as possible. Note that for 12 players there are C(12,2) = 66 pairs of distinct players. In the first day's schedule we will account for 6 pairings in each of three foursomes, for a first day's total of 18 pairs of distinct players. On day 2 the minimum number of repetitions is three (one for each foursome), so the maximum number of additional pairs of distinct players we can hope to account for is 15, and it is easy to attain this. On day 3 there must be three repetitions from the first day and also three repetitions from day 2. However if these repetitions were the same three pairs of distinct players, we could get as many as another 15 pairings that had not previously been acheived. If we do not allow for any "triple" repetitions at this point, we are limited to 18 - 3 - 3 = 12 new pairings. Let us assume the latter, so that after day 3 we have in total: 18 + 15 + 12 = 45 pairings of distinct players This sets the stage for day 4. We can choose the final foursomes so that in the end, each player will have played with exactly 9 of the 11 possible other players. There is a certain amount of asymmetry in how the repetitions are distributed in this case. Most players will have met three others exactly twice in their schedule (so 12 - 3 repeats = 9 distinct other players met), but two of the players do meet each other three times (and each meets another player twice). But these two players will also have met nine others (subtract 2 from 12 for the triple meeting and 1 for their "simple" repeat). With those caveats here's my schedule for the four days: First day: (A1,A1,A3,A4) (B1,B2,B3,B4) (C1,C2,C3,C4) Day 2: (A1,B1,C1,A2) (B2,C2,A3,B3) (C3,A4,B4,C4) Day 3: (A1,B2,C3,A4) (B1,C2,A3,B4) (C1,A2,B3,C4) Day 4: (A1,C2,A3,C4) (B1,A2,B2,C3) (C1,B3,B4,A4) In the end 54 of the 66 possible pairs will have played together at some point during the four days, and it is impossible for all 12 players to each meet more than nine others in such a schedule. Something of this kind is therefore my suggestion for "as close as possible to the ideal". ******************************************* Constructing Foursomes of 12 Players Where Each Distinct Pair Meets [not suitable as a solution because no two foursomes are "disjoint"] Consider the twelve hours on the dial of a clock, and starting at the top (12th hour), mark four points separated successively by intervals of 1, 2, and 4 hours: (12, 1, 3, 7) Now rotate this entire group of four points around the clock by an hour each time, until you return to the starting configuration: (1, 2, 4, 8) (2, 3, 5, 9) (3, 4, 6, 10) (4, 5, 7, 11) (5, 6, 8, 12) (6, 7, 9, 1) (7, 8, 10, 2) (8, 9, 11, 3) (9, 10, 12, 4) (10, 11, 1, 5) (11, 12, 2, 6) (12, 1, 3, 7) It will be found that every pair of distinct players (or hours) occurs at least once in the schedule. Repeats are limited to one for each player, which means there are six pairs of disinct players who meet exactly twice, and otherwise any pair meets only once. These repeats correspond to a difference of six hours (halfway around the dial) that occurs initially between 1 and 7. After shifting this "basic pattern" around by six hours, we again have a foursome that includes the same pair 1 and 7, and the same is true of every pair differing by six. ******************************************* As a final note let me comment a bit more about the second part of the Question. If all that is needed is for one complete round of pairs and one complete round of foursomes, drawn from a pool of 16 players, to avoid any repeats, this is pretty easy. Let's start with the round of 4 foursomes and label the players this way: A1--A2--A3--A4 B1--B2--B3--B4 C1--C2--C3--C4 D1--D2--D3--D4 Given that there are many ways to choose 8 nonintersecting pairs for the required round of "singles". For example we could make pairs between the top two and bottom two rows: A1--A2--A3--A4 || || || || B1--B2--B3--B4 || || || || C1--C2--C3--C4 || || || || D1--D2--D3--D4 I hope this has been helpful, though I still suspect something more was intended by the second part of the Question. regards, mathtalk-ga```
 ```Oops! I got a bit carried away in drawing the last diagram and connected the second and third rows, unnecessarily. It would be clearer to simply give this: "For example we could make pairs between the top two and bottom two rows: A1--A2--A3--A4 || || || || B1--B2--B3--B4 C1--C2--C3--C4 || || || || D1--D2--D3--D4 " Okay, now we have only eight pairings for the round of "singles". regards, mathtalk-ga```