Congressional seating for SOTU and Discrete Math

Remember when in elementary school when teachers would instruct you to sit “boy-girl-boy-girl?” Well, it appears as if the United States Congress needs to be reprimanded for their childish behavior and must sit “Democrat-Republican-Democrat-Republican.” You see, Colorado Senator Mark Udall made a proposal that during the 2011 State of the Union (SOTU) speech members of congress would sit amongst members of the opposite party. This is in contrast to previous SOTU addresses where the house of congress was firmly divided and after each line half of the house would cheer like their favorite team is trying to make a crucial 3rd down stop and the other half would sit on their hands and scowl. So members of congress accepted Udall’s proposal and immediately members of congress began trying to decide whom to sit next to so they could appear bi-partisan but so as not to sit next to anyone that’s an anathema to that politician’s base supporters (Seriously, they are children).

Anyway, as you may also know, Democrats lost a lot of seats in the 2010 mid-term elections after four years of steady gains. For the 2011 SOTU there were 242 Republicans (Rs) and 193 Democrats (Ds) in the House of Representatives; 53 Ds (and Independents who caucus with the Democrats) and 47 Rs in the Senate. As far as I know, the Senate and Congress sit together during the SOTU, so that’s a total of 289 Rs and 246 Ds.

So upon hearing that Ds and Rs were to sit together at the 2011 SOTU, I wondered a couple things:

  • Is it possible for every member of congress to sit next to a member of the opposite party?
  • If it is possible, how lopsided would the party split have to be before it becomes impossible?

I was hoping this wouldn’t be a simple discrete math problem. And as luck would have it, it isn’t (at least, I don’t think it is). Behold the floor-plan of the House of Representatives.

(note: not actually the floor plan of House of Representatives. At least I don’t think it is. And it certainly isn’t the seats used for the SOTU seeing as there are more members of congress than seats in this diagram. If anyone out there can find a better, more-correct diagram of the seats of Congress during a SOTU please email me. Below, I’ll start out using the entire congress, but eventually I’ll switch to just the House, and then I’ll declare that it doesn’t really matter.)

What a wonderful disparity of seats! There are rows of 15 and rows of two! There are things called the “Republican and Democratic Committee Tables”. And I’ve always wanted to know where the Tally Clerk sits.

So here’s how I would produce this to students.

Artifact

Provide students the floor-plan of the House of Representatives. Printouts or splash it on the screen.

Say,

“There are currently 289 Republicans and 246 Democrats in Congress. And they’re supposed to sit together and play nice. Any questions?”

Guiding Questions (GQs)

Hopefully the students will come up with similar questions I posted above. Namely,

  • Is it possible for every member of congress to sit next to a member of the opposite party?
  • If it is possible, how lopsided would the party split have to be before it becomes impossible?

And I guess I’ll include a couple more questions that I’m not totally certain I know the answer to right now.

  • Is this a math problem?
  • Tangentially, is there some mathematical formula we can apply here?

Not having worked the problem, my gut is telling me the answer to these two questions is “sort of.”

Suggested Activities

  • Once students have put forth the pertinent questions (as determined by the teacher), hand student groups a printout of the floor plan, perhaps some red and blue colored writing utensils and have them go to town.

Solutions

Let’s start with the current breakdown of congress at 298-246, Rs-Ds. Now, not all members of congress attend. So there’s another level of complexity we could deal with. As noted above, this diagram, which is crucial to our problem, probably isn’t accurate. There are only 417 seats shown for 435 members of the House (note: the Speaker of the House sits behind the President.)

In this task of assigning seats, Ds are “more valuable”. That is, given a row with only three seats, we’d want 2 Rs and 1 D so we can “save” the D for later. With that in hand, we can immediately start with the rows with two or three seats, assigning 1 D and 1 or 2 Rs:

We can also go ahead and mark up the rows of four as 2 Rs and 2 Ds.

At this point, we have used 14 Rs and 12 Ds, leaving us 284 and 234, respectively. Now for the rows of 5 seats. Those will either have to go R-D-R-R-D, or R-D-R-D-R, etc. The point is clear though, we’ll need 3 Rs and 2 Ds for those rows, no matter what. So let’s just do those.

That used up another 18 Rs and 12 Ds, down to 266 Rs and 222 Ds. So for every row of 5, you’re able to “use up” one more R versus D. (R=D+1. See where we’re headed? Possibly??)

Now it gets a tad bit trickier. For a row of 6 we can either alternate R-D-R-D-R-D, using 3-and-3 or we can go R-D-R-R-D-R, using 4 Rs and 2 Ds. But now that we’ve come this far, consider this. If we continue playing this game and we ever get to an equal number of Ds and Rs, we win: we know it’s possible to accommodate Mark Udall’s sense of decency. So let’s keep going with rows of six, knocking out 4 Rs and 2 Ds each step of the way.

That took care of another 12 Ds and 24 Rs. So we’re at 242 Rs and 210 Ds.

But you know what? This is getting exhausting! Mr. Krall, there must be an easier way of doing this!

Now do you see where I’m headed with this, dear student??

We could continue filling it out with rows of 7, 8, 9, etc. But it’s pretty clear from this point, that it would be possibly to assign seats for every D and R so they’re sitting next to an R or D, respectively.

But that brings us to our second question: how lopsided would congress have to be before it becomes impossible to seat everyone next to a member of the opposite party?

Consider the varying rows and tallies:

For every row of 2, we end up even with R=D.

For every row of 3, we could swing the balance by 1 (R=D+1).

For every row of 4, it remained even again (D=R).

For every row of 5, we have R=D+1 again.

For every row of 6, we have the potential for R=D+2, a swing of two. Let’s keep going, trying to maximize the number of Rs and minimize the number of Ds for each row.

Rows of 7: R-D-R-R-D-R-D. R=D+1, another swing of 1.

Rows of 8: R-D-R-R-D-R-R-D. R=D+2, another swing of 2.

Rows of 9: R-D-R-R-D-R-R-D-R. That’s 3 Ds and 6 Rs. A swing of three! (R=D+3)

Rows of 10: R-D-R-R-D-R-R-D-R-D. Swing of 2 (R=D+2).

Rows of 11: R-D-R-R-D-R-R-D-R-R-D. Swing of 3 (R=D+3)

Rows of 12: R-D-R-R-D-R-R-D-R-R-D-R. Swing of 4 (R=D+4). It’s getting larger!

Rows of 13: R-D-R-R-D-R-R-D-R-R-D-R-D. Swing of 3 (R=D+3). Hmm. Back down to 3.

Rows of 14: R-D-R-R-D-R-R-D-R-R-D-R-R-D. Swing of 4 (R=D+4).

Rows of 15: R-D-R-R-D-R-R-D-R-R-D-R-R-D-R. Swing of 5 (R=D+5).

Whew! Counting the seats in each row:

Now, excuse me while I get my trusty Excel spreadsheet out.

(A)Number of seats in row (B)Potential “swing” for the minority party, each row | and corresponding equation (C)Number of rows (D)Potential “swing” for the minority party, all rows of size (B) x (C)
2 0 | or, R=D 4 0
3 1 | R=D+1 3 3
4 0 | R=D 2 0
5 1 | R=D+1 6 6
6 2 | R=D+2 4 8
7 1 | R=D+1 6 6
8 2 | R=D+2 6 12
9 3 | R=D+3 4 12
10 2 | R=D+2 4 8
11 3 | R=D+3 6 18
12 4 | R=D+4 4 16
13 3 | R=D+3 0 0
14 4 | R=D+4 2 8
15 5 | R=D+5 2 10
Total = 107

So we could play this game for a disparity of house seats up to 107 members of the seated house. There are 417 members of the House as shown in the diagram (again, note: there are actually 434 members of the House and 100 members of the Senate). So for 434 seated members, there could be a lopsidedness of 270-164. But a split of 271-163 and sadly, there would not be enough house members in the minority party to satisfy Mark Udall’s utopian dream.

A couple final thoughts to take away from this exercise, some obvious, some not so obvious:

  • In general, the more seats in a row, the more wiggle room we had to adjust our seating numbers.
  • We could easily extend this same thought process to, say, school assemblies and boy-girl seating. Or shoot, your own classroom.
  • We could also use these formulas to stricter a “more bi-partisan seating friendly” congress. There would be more 15 seat rows and fewer three seaters.
  • After the 2010 Census there will be a few additional seats allotted (mostly to Republicans), where should we put these seats to better allow for Mark Udall’s rule?
  • Maybe the 435-417 disparity between reality and the picture we have would provide an interesting extension along those same lines: We have to bring in a bunch of folding chairs for the SOTU for the rest of Congress: where should we put these seats to better allow for Mark Udall’s rule?
  • The pattern of swings for the minority party was interesting (0-1-0, 2-1-2, 3-2-3, etc.). Not sure if it means anything, but it was interesting.
  • Probably the most damning puncture in this exercise is the simple fact that I can’t find a seating arrangement that seats 435 members of the House. But frankly, that’s not terribly important. It’s the disparity that matters. “Sort of”.
This entry was posted in combinations, discrete math, tasks. Bookmark the permalink.

One Response to Congressional seating for SOTU and Discrete Math

  1. Pingback: “Isn’t Problem Based Learning easier than Project Based Learning?” and 10 other myths about PrBL, (“Real or not real”) | emergent math

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