I like to assign peer reviews for my 1L Legal Writing class. Generally, the course’s collaboration rules don’t let students help each other with their final graded assignments. But I make an exception for in-class peer review.

Anyway, this quarter I wanted to try something different: I wanted each student to have ** two **peer review partners, instead of just one. So at the end of the class before the peer review I randomly assigned each student two different partners. Everyone exchanged papers with their two partners and read their two partners’ papers before the next class. My plan was to then break the peer-review class into two separate periods. During the first half of class each student would meet with one partner and discuss each other’s papers. And then during the second half of class each student would meet with the other partner and discuss each other’s papers.

The idea was that everyone would be discussing papers during the entire class, and no student would need to meet with their partner after class. Nice and easy! But perhaps you can see the problem. I’d like to think I would have seen it if I’d thought about it for a second. Oh well.

Anyway, on the peer-review day the problem became apparent when I said “Okay, meet with your first partner for 25 minutes.” But who was the “first” partner? I’d just randomly assigned everyone two partners. Hmm. Good question. So I just started naming off pairs until I ran into a problem. I got to a person for whom ** both** partners were already paired up with other people. Disaster! My random assignment of partners did not allow for everyone to meet with their partners at the same time during class. Some students had no partner for half the class, and then had to meet with a partner after class.

One student pointed out an easy way to ensure this didn’t happen: I could have just numbered all the students 1 to 24 (I have 24 students in this class) and then partnered student 1 with student 2 and student 24, partnered 2 with 1 and 3, partnered 3 with 2 and 4, etc. With that division, everyone could meet with their partners at the same time. You’d have (S1,S2), (S3,S4), (S5,S6) . . . (S23,S24) for the first session. And then (S2,S3), (S4,S5) . . . (S24,S1) for the second session.

And that certainly works. I think of that as the “everyone hold hands in a circle” solution. And that will always work for any even-numbered class. (There is never going to be a solution for a class with an odd number of students, since you can’t pair them all off at the same time using any pairing method.)

But is that the only way to make it work? I didn’t think so. And despite finding one solution, I wanted to figure out ** every** solution to the problem of how to pair off a group of students in a way that would allow for (1) each student to have two different partners and (2) all students to meet with both of their partners during the two in-class sessions. To figure this out I crowd sourced some help from Facebook. Lots of folks explained the math to me in a number of different ways, but the way that made the most sense to me was a visual/graphic depiction of the problem. Here it is:

Picture each student as a point on a page. Two points connected by a line means that those two students are partners for one of the periods. And since each student needs two different partners, each point must have two lines coming out of it—not one, not three, just two.

So, to make it simple, if we had a class of just four students, this arrangement would ** not** work:

Student 4 has two partners: Student 2 and Student 3. And Student 2 has two partners: Student 4 and Student 3. But S1 only has one partner: S3. And S3 has three partners: S1, S2, and S4. So that’s no good.

But there are a variety of ways that would work for a four-person class:

Though these look different, you can see they are all basically versions of the same “hold hands” solution. Each solution is the same because even though they look different when depicted as lines and shapes, I’ve randomly assigned the numbers S1, S2, S3, and S4. If you just change the labels, it doesn’t matter whether the “hold hands” circle goes S1, S2, S3, S4, or S1, S2, S4, S3, or S1, S3, S2, S4. Any order works as long as everyone is holding hands. For each of the above solutions, you can follow the line continuously through all the points in the same manner. That works and there’s no other way for it to work.

But that’s just for a four-person class. What if we go to eight students? The “hold hands” solution will still work:

And just like the four-person solutions, the eight-person “hold hands” solution works no matter the order in which the students hold hands. So these would also work:

That’s a hand-holding order of S1, S4, S7, S2, S5, S8, S3, S6 on the right. And then on the left you’ve got S1, S3, S4, S6, S2, S7, S5, S8. They’re all essentially the same because the numbering is random.

But for an eight-person class, the “hold hands” solution is not the only solution! We can also do something like this:

Instead of doing a single eight-person “hold hands” solution, we can do two separate four-person “hold hands” solutions. Again, since all the different four-person solutions are essentially the same, these different versions of two four-person solutions are essentially the same as well. But the separate four-person “hold hands” solutions are different from the single eight-person “hold hands” solution because you can’t just trace around person to person and complete the entire class. To follow through with the metaphor, you’d need to pick up your pencil and make a separate loop.

And now we’ve got a way to think about all possible solutions. Because for the eight-person class, we can do a complete eight-person “hold hands” solution, or two smaller four-person “hold hands” solutions. That’s it. We can’t do a three-, five-, or seven-person small group because those are odd, which makes the pairing impossible. And we can’t do a six-person “hold hands” solution because, though that would work for the six-person group, it would leave two students left over which means those two students would only have one partner. That doesn’t work either.

So for our twenty-four person class, every solution must either be (1) the entire group in a “hold hands” solution, or (2) completely separate subgroups of the class in “hold hands” solutions, such that each subgroup contains an even number of students greater than two. We could do a single 24-person “hold hands” circle. Or four groups of six students each. Or three groups of eight. Or six groups of four. Or two groups of six, a group of eight, and a group of four. Whatever. As long as all the subgroups can be even and greater than two, it’ll work!

And that’s it for all the different ways I could have broken up my class into partners for peer review. Also, I learned about a new area of math that I never learned in college: graph theory.

And now, back to grading papers.