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Dad • 01/31/06 • 12:52 PM:

It’s easy to set it up so everyone plays with everybody else. However, I don’t think it’s possible to prevent some players from playing more than twice together. In fact, in order for everyone to play everyone at least once, 2 players may have to play each other 4 times! (Ed thinks this is fairly obvious from the problem statement and a few other logical conclusions.)

Patrick • 02/03/06 • 5:31 PM:

I love these entries where Dad poses a question and then (at least partially) answers it himself in the comments. My approach to a problem like this would be to plot out all possible scenarios. I have some math skill, I guess, since I did well in school, but mostly, I like to see examples. This is how I derived Pascal’s Triangle when Dad & I were trying to figure out the voting on Survivor statistically.

Dad • 02/03/06 • 6:36 PM:

If you love them, then try to solve them. Even if you don’t spend enough time on it to get the answer, there is usually something interesting about them, or I wouldn’t post them.

Plotting all the scenarios in this case would be pretty tedious (as was the Survivor voting analysis). Furthermore, I wasn’t able to figure out an elegant (e.g. mathematical) way of checking which guys already played together. I had to inspect all the cases and tabulate who played with whom in a separate table. I stopped well short of checking all the possibilities.

In the end, I devised a logical way of analyzing the problem to determine what the answer is, and prove that there is no other answer. This logic proof is relatively short, and something a wise logician like yourself should be able to do without much trouble.

David • 02/03/06 • 6:38 PM:

I too love it. Dad, you rule. And I concur with your answer. Speaking of Pascal - I think I want to name one of my kids (future) Blaise, after him, and after St. Blaise, whose feast day is today.

Dad • 02/06/06 • 2:32 PM:

Here I go with another partial answer. How many possible groupings? This brings back those great formulas from high school algebra for combinations. (You can also find them in Excel.) The number of ways of making a threesome from six possible players is the combinations of six things taken three at a time, which is 6!/(3!(6-3)!) = 20. In this case, however, we are picking two threesomes at a time with the second one comprised of the leftovers from the first threesome. This cuts the number of possible pairs of threesomes to 10.

Of these 10 possibilities, we are only using 1 per day for 4 days. So the number of possibilities is the number of combinations of ten things taken four at a time, 10!/(4!(10-4)!) = 210.

That’s why it would be tedious to check all of the possibilities by hand. Of course, there are lots of similar grouping, so maybe you wouldn’t have to check all of them. But there is a better way.

By analyzing the problem carefully, one can devise a 5-step (or so) logical proof that gives the answer and shows why it is the only possible answer (actually a group of equivalent answers that differ only by permutations of the players names).

Dad • 02/09/06 • 11:48 AM:

Here’s the answer. Those of you still working on the puzzle, don’t look.

On the first day, each golfer plays with 2 others. Starting with this original set of threesomes, (say, 123, 456) there is only one move for the next day—switch one player from each threesome into the other threesome. (You could switch 2 players over, but that amounts to the same thing.) Because of the symmetry of the situation, it doesn’t matter which golfers you switch. After the second round, the two switched players have now played with 4 people, while the rest have played with 3. (e.g. 124, 356)

Day three, you can either switch two who haven’t switched before, or one that has and one that hasn’t. (It’s pointless to switch back the previous switchers.) If you switch two new players, then the two previously switched players don’t see anyone new. The other four get one more pairing and now everybody is tied with four pairings each. Note that the switched golfers haven’t paired with each other, and neither have the un-switched golfers. (e.g. 154, 326)

Given this scenario, anyway you slice it, you lose. You now have 3 pairs of unmatched players (the original switchers, the second day switchers, and the un-switched) and you can at most match two of these pairs. (e.g. switch 1 and 3, and 5 and 2 are still unpaired.) So switching two new players on day three was a mistake.

Going down the other path, if you switch one of each on day three, the newly-switched golfer (say 5) pairs up with two new players, and he’s now done. The originally switched players also get one new pairing, so they are done too. (e.g. 125, 346.) At this point, the three switched players are done, while the two un-switched from one threesome (1 & 2) lack one pairing, and the single un-switched player from the other threesome (6) lacks two pairings.

Of course! The un-switched players all haven’t played with each other! Duh! So the last day, merely switch the single un-switched from the second threesome with the guy who’s already been switched from the first threesome. (e.g. 126, 345) Voila, everybody’s done, but the two un-switched guys (1 & 2) from the original threesome have been together the whole time. (Bear in mind that the numbers don’t really matter, and are given only as examples.)

So the simplest, dumbest solution anybody can come up with by inspection is the only right answer! This is may be the most inelegant problem I’ve every worked on!

Patrick • 02/09/06 • 6:20 PM:

By the way, I remember going to some St. Blaise masses, where the priest would bless your throat with two crossed candles (unlit). I thought it was kind of weird. Also, Dave, I don’t really like the name, because you’d have to pronounce it “Blaze,” and that would sound like his parents were trying to be overly “cool” with the name.

Meanwhile, one of the funniest lines I’ve recently read: “Those of you still working on the puzzle, don’t look.” I love you, Dad.

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