How pizza can solve this unusual math puzzle

This article is from Proof Positive, our friendly newsletter that explores the joys and peculiarities of math. Sign up today for a weekly math essay and puzzle in your email inbox.

How do you think the number sequence 1, 2, 4, 8, 16... continues? Most people would guess 32 because all the listed values so far have been the previous number doubled. But this sequence, which was devised by number theorist Leo Moser in 1949, takes a surprising turn: 16 is followed by 31.

On supporting science journalism

If you're enjoying this article, consider supporting our award-winning journalism by subscribing. By purchasing a subscription you are helping to ensure the future of impactful stories about the discoveries and ideas shaping our world today.

In fact, this sequence, designated as “A000127” in The On-line Encyclopedia of Integer Sequences (OEIS), continues with the numbers 31, 57 and 99. So even though it begins innocently enough with doubling, it suddenly deviates from that obvious pattern at the sixth term. Moser used this example to warn against drawing hasty conclusions from supposed patterns.

But where did this sequence, now known as Moser’s circle area problem, come from? It’s not random; the underlying logic can be derived using the example of cutting a pizza.

Imagine two people want to cut a circular pizza into many small slices. One person marks points on the edge of the dough, while the other makes a straight cut between all the marked points. They use a pizza cutter to cut from point to point along the edge. The question Moser posed was, essentially: What is the maximum number of pizza slices that can be created in this way?

Strange Pizza Patterns

To think of this situation as a sequence, let’s assume the first person adds one point at a time to the outer edge. With only one point, the pizza cannot be divided according to the rules mentioned, so there is only one slice: the whole pie.

With two points, the surface can be cut in two by cutting through both of them. With three points, three cuts can be made, thus dividing the surface into four. With four points, there are six possible cuts, resulting in eight pizza slices. The pattern continues as expected until there are six points, whereupon the expected pattern breaks down, and the result is 31 slices. But why?

There’s a formula that will tell you how many slices, or areas, are created for n points on a boundary, such as a pizza’s edge. We’ll start by thinking about a simpler relationship: the intersecting lines that result from connecting any two of n points with cuts from a pizza cutter. If you ever studied combinatorics in school, you may know there’s a handy quantity for this: the binomial coefficient, written as B(n, 2) = n!/(2!(n – 2)!). This can be quickly verified using the example with n = 5 boundary points; in this case, there are B(5, 2) = 10 intersections.

A similar approach can be used to determine how often the cut lines intersect. If you consider the pizza with one, two or three points on the edge, the cuts do not intersect as they crisscross the pizza. When there are four or more cuts, however, there is at least one intersection point. Therefore, for every four points, there is one intersection point. The number of these intersections can be expressed using a binomial coefficient as well: B(n, 4), where n points are marked at the edge. For the example with n = 5 edge points, this results in B(5, 4) = 5 intersection points on the pizza.

To keep things simple, we can disregard certain slicing scenarios. Moser’s initial question was: What is the maximum number of separate pizza slices that can be created? To maximize the number of slices, we can disregard cases that involve the intersection of more than two cuts through the pizza.

If you look closely at one of the sliced pizzas, you might notice that it resembles a graph: it consists of a jumble of points connected by lines. It turns out we can use Euler’s polyhedron formula (which I wrote about in last week's newsletter about perfect shapes) to think about this problem.

According to this formula, for any graph where all points are connected, there is a relationship between the number of points, V, edges, E, and faces, F. Specifically: V – E + F = 1. Applying the formula to our pizza, F will correspond to the total number of pizza slices, so if we know how many points along the crust and edges of individual slices there are in the graph, we can calculate our total slices of pizza pie.

The only problem is that, in our example, the number of cuts doesn’t correspond to the number of edges in the formula because each segment of a cut made by intersecting cuts counts as an independent edge of the graph.

To determine the total number of edges, we should recognize that there are three basic types of edges. The first type connects two points on the boundary or crust without being interrupted by other cuts. For n boundary points, there are also n of this type of edge.

Then there is the second type of edge, which connects interior intersection points with boundary points. Each of the B(n, 2) intersections is bounded by two boundary points. This means that a total of 2B(n, 2) edges emanate from all boundary points. This also includes edges of the first type, however, meaning any edges that directly connect two boundary points to each other must, of course, be subtracted to avoid counting them twice. Thus, there are 2(B(n, 2) – n) edges of the second type, which connect outer boundary points with interior intersection points. For n = 5 boundary points, for example, this results in 2(10 – 5) = 10.

Finally, there is the third type of edge, which connects internal intersection points. Four edges terminate at each internal intersection point, of which there are B(n, 4). Therefore, each internal intersection point is connected by four edges, resulting in a total of 4 × B(n, 4) edges. To avoid including all edges connecting internal points and boundary points, we need to do some subtraction and division: 1/2 × 4 × B(n, 4) – 2 × (B(n, 2) – n) = 2 × B(n, 4) – (B(n, 2) – n). Again, the result can be tested for n = 5 points: 2 × 5 – (10 – 5) = 5 edges connecting interior points.

Now we only need to add up all three edge types to get the total number: n + 2 × (B(n, 2) – n) + 2 × B(n, 4) – (B(n, 2) – n) = 2 × B(n, 4) + B(n, 2). With that, we’re almost ready to calculate the number of pizza slices.

Only one small detail is missing: because we abstracted the pizza into a graph to use Euler’s formula, the pizza’s crust also represents edges. Therefore, the term n is added to the previously calculated total. This gives us the E from Euler’s formula: E = 2 × B(n, 4) + B(n, 2) + n.

We already know V, the number of intersection points, and the number of boundary points, n, so: V = B(n, 4) + n.

From Euler’s formula, we obtain the following formula for the number of pizza slices: F = 1 – V + E = 1 – B(n, 4) – n + 2 × B(n, 4) + B(n, 2) + n = 1 + B(n, 4) + B(n, 2). So if you have a pizza in front of you, and you mark n points and divide it by B(n, 2) slices, you get 1 + B(n, 4) + B(n, 2) pizza slices. And as it turns out, for n = 1, 2,..., 5, this always results in the power of 2n – 1, and deviates from this starting at n = 6.

So now you know the logic of Moser’s circle problem—and have learned how to divide a pizza into 31 slices, albeit not all the same size. Fortunately, mathematics can also help us find the fairest way of divvying up pizza pie—but that is a different story.

This article originally appeared in Spektrum der Wissenschaft and was reproduced with permission. It was translated from the original German version with the assistance of artificial intelligence and reviewed by our editors.