I’m back after a little break to spend time with family who, coincidentally, showed up a few weeks after the birth of baby #3. In a couple of weeks, I start teaching precalculus again and the pseudo-finite-mathematics class. Last night, while reading a sample problem prepared for the class by the course coordinator, I decided that I don’t draw circles well—especially several that are supposed to intersect—and that, while I understand what Venn diagrams represent and how they are put together, Venn diagrams do not illuminate raw data for me. Inspired by this TED talk of David McCandless, I tried to redraw the problem. Ah, perhaps you would like to see the problem.
A survey of 100 students shows that: 48 take English (E), 49 take History (H), 38 take Spanish (S), 17 take E and H, 15 take E and S, 18 take H and S, and 7 take all three. How many students:
- take only S
- take S and H but not E
- take only one of those courses
- take none of those courses
The solution is not difficult to find without the aid of a diagram, but for the type of students in this class, every little picture helps. The book suggests the following Venn diagram will be helpful:
Once the students draws this (by hand, I always seem to be to impatient and screw it up, but maybe the students are better at this) he or she might then argue that the middle region is 7 and that the three regions surrounding the middle are (in clockwise order from the top) 10, 11, and 8. Starting from the left and working clockwise, the three missing regions are then 23, 21 and 12. At this point, they would be able to answer the four questions.
This is a best case scenario, I think. Let’s assume that they are taught to always throw the 7 in the middle. My guess is that many of the students will then toss the numbers 17, 18 and 15 into the three cells around the middle. Why? Well, duh! Those three cells represent E and H, H and S and E and S. Obviously. At this point, I don’t think it matters what they do, they will get the wrong answer. If they get to the next step, however, I’m still not sure they will see how to compute the outer ring. If they get that far, will they really get how to read the picture to answer the questions? I don’t know; I’ve never had to teach Venn diagrams. Perhaps they will find this very easy. On the other hand, maybe they won’t.
An Alternative to the Venn Diagram
Here is an alternative way to solve this problem, one that mathematicians should appreciate for its passing reference to Euler’s formula (Not that one; the other one 😉 ).
First, we draw a triangle. The type is not particularly relevant, only draw it sufficiently regular in order to fit numbers. Label the edges with E, H and S.
We now the label the vertices, edges and faces as naturally as possible. By naturally, I mean as a typical student would do it. The vertices are the numbers given in the problem for the individual subjects, the edges are the numbers given for combinations of two subjects and the middle (face) is for all three subjects.
This is a graphical representation of the given data—and I understand it. This is what the students want to write, so let them write it down. Now, we draw a new triangle. I don’t want to give it a name. If I must, I might call it the companion triangle, but there is probably a better name for it.
How will we label this new companion triangle? Euler’s formula, of course. Half of it, anyway.
Euler proved for any convex polyhedron that the number of vertices plus the number of faces is two more than the number of edges, i.e., V + F = 2 + E.* Traditionally, this is written V – E + F = 2. An interesting thing is that this formula fails in higher dimensions, but the idea of alternating between plus and minus as you increase the dimension of the object is still useful.
* Note that the E in the formula V – E + F = 2 stands for edges, not English. This ambiguity will continue below.
Back to our triangle
With a bit of thought, one sees that to put the “right” data into our new triangle, we simply need to compute V – E + F for the vertices, E – F for the edges and F is still the face. Let me explain what I mean. In our new triangle, we want the E vertex number to represent the number of people who took only English. The original E double counts the numbers coming from the EH and ES edges, so we should subtract those. But this double subtracts things that were on both edges, i.e., are in the middle, so we need to add those back in. So, V – E + F. The EH edge should represent the people who took both English and History and not Spanish. As the original edge number counts both those who took just English and History as well as those who took all three, we need to subtract those in the middle. So E – F. The middle represents those who took all three and so does not change from the original. Let’s compute the new E vertex.
V – E + F
The original E vertex says 48
The edges going out of E (the EH and ES edges) add up to 32, so we subtract them from 48 to get 16.
The face is 7, so we add that to get 23—our new E vertex is 23.
The original S vertex says 38
The edges going out of S (the ES and HS edges) add up to 33, so we subtract them from 38 to get 5.
The face is 7, so we add that to get 12—our new E vertex is 12.