Theorem Thursday: Kuratowski’s Reduction Theorem and Playing With Gas Pipelines


I’m doing that thing again. Sometime for my A To Z posts I’ve used one essay to explain something, but also to introduce ideas and jargon that will come up later. Here, I want to do a theorem in graph theory. But I don’t want to overload one essay. So I’m going to do a theorem that’s interesting and neat in itself. But my real interest in next week’s piece. This is just so recurring readers are ready for it.

The Kuratowski Reduction Theorem.

It starts with a children’s activity book-type puzzle. A lot of real mathematics does. In the traditional form that I have a faint memory of ever actually seeing it’s posed as a problem of hooking up utilities to houses. There are three utilities, usually gas, water, and electricity. There are three houses. Is it possible to connect pipes from each of the three utility starting points to each of the three houses?

Of course. We do it all the time. We do this with many more utilities and many more than three buildings. The underground of Manhattan island is a network of pipes and tunnels and subways and roads and basements and buildings so complicated I doubt humans can understand it all. But the problem isn’t about that. The problem is about connecting these pipes all in the same plane, by drawing lines on a sheet of paper, without ever going into the third dimension. Nor making that little semicircular hop that denotes one line going over the other.

That’s a little harder. By that I mean it’s impossible. You can try and it’s fun to try a while. Draw three dots that are the houses and three dots that are the utilities. Try drawing three lines, one from each utility to each of the houses. Or one leading into each house that comes from each of the utilities. The lines don’t have to be straight. They can have extra jogs, too. Soon you’ll hit on the possibilities of lines that go way out, away from the dots, in the quest to avoid crossing over one another. It doesn’t matter. The attempt’s doomed to failure.

You’ll be sure of this by at latest the twelfth attempt at finding an arrangement. But that leaves open the possibility you weren’t clever enough to find an arrangement. To close that possibility guess what theorem is sitting there ready to answer your question, just like I told you it would be?

This is a problem in graph theory. I’ve talked about graph theory before. It’s the field of mathematics most comfortable to people who like doodling. A graph is a bunch of points, which we call vertices, connected by arcs or lines, which we call edges. For this utilities graph problem, the houses are the vertices. The pipes are the edges. An edge has to start at one vertex and end at a vertex. These may be the same vertex. We’re not judging. A vertex can have one edge connecting it to something else, or two edges, or three edges. It can have no edges. It can have any number of edges. We’re even allowed to have two or more edges connecting a vertex to the same vertex. My experience is we think of that last, forgetting that it is a possibility, but it’s there.

This is a “nonplanar” graph. This means you can’t draw it in a plane, like a sheet of paper, without having at least two edges crossing each other. We draw this on paper by making one of the lines wiggle in a little half-circle to show it’s going over, or to fade out and back in again to show it’s going under. There are planar graphs. Try the same problem with two houses and two utilities, for example. Or three houses and two utilities. Or three houses and three utilities, but one of the houses doesn’t get one of the utilities. Your choice which. It can be a little surprise to the homeowners.

This utilities graph is an example of a “bipartite” graph. The “bi” maybe suggests where things are going. You can always divide the vertices in a graph into two groups for the same reason you can always divide a pile of change into two piles. As long as you have at least two vertices or pieces of change. But a graph is bipartite if, once you’ve divided the vertices up, each edge has one vertex in the first set and the other vertex in the second. For the utilities graph these sets are easy to find. Each edge, each pipe, connects one utility to one house. There’s our division: vertices representing houses and vertices representing utilities.

This graph turns up a lot. Graph theorists have a shorthand way of writing it. It’s written as K3,3. This means it’s a bipartite graph. It has three vertices in the first set. There’s three vertices in the second set. There’s an edge connecting everything in the first set to everything in the second. Go ahead now and guess what K2, 2 is. Or K3,5. The K — I’ve never heard what the K stands for, although while writing this essay I started to wonder if it’s for “Kuratowski”. That seems too organized, somehow.

Not every graph is bipartite. You might say “of course; why else would we have a name `bipartite’ if there weren’t such a thing as `non-bipartite’?” Well, we have the name “graph” for everything that’s a graph in graph theory. But there are non-bipartite graphs. They just don’t look like the utility-graph problem. Imagine three vertices, each of them connected to the other two. If you aren’t imagining a triangle you’re overthinking this. But this is a perfectly good non-bipartite graph. There’s no way to split the vertices into two sets with every edge connecting something in one set to something in the other. No, that isn’t inevitable once you have an odd number of vertices. Look above at the utilities problem where there’s three houses and two utilities. That’s nice and bipartite.

Non-bipartite graphs can be planar. The one with three vertices, each connected to each other, is. The one with four vertices, each vertex connected to each other, is also planar. But if you have five vertices, each connected to each other — well, that’s a lovely star-in-pentagon shape. It’s also not planar. There’s no connecting each vertex to each other one without some line crossing another or jumping out of the plane.

This shape, five vertices each connected to one another, shows up a lot too. And it has a shorthand notation. It’s K5. That is, it’s five points, all connected to each other. This makes it a “complete” graph: every set of two vertices has an edge connecting them. If you’ve leapt to the supposition that K3 is that circle and K4 is that square with diagonals drawn in you’re right. K6 is six vertices, each one connected to the five other vertices.

It may seem intolerably confusing that we have two kinds of graphs and describe them both with K and a subscript. But they’re completely different. The bipartite graphs have a subscript that’s two numbers separated by a comma: p, q. The p is the number of vertices in one of the subsets. The q is the number of vertices in the other subset. There’s an edge connecting every point in p to every point in q, and vice-versa. The points in the p subset aren’t connected to one another, though. And the points in the q subset aren’t connected to one another. That they don’t mean this isn’t a complete graph.

The others, K with a single number r in the subscript, are complete graphs, ones that aren’t bipartite. They have r vertices, and each vertex is connected to the (r – 1) other vertices. So there’s (1/2) times r times (r – 1) edges all told.

Not every graph is either Kp, q or Kr. There’s a lot of kinds of graphs out there. Some are planar, some are not. But here’s an amazing thing, and it’s Kuratowski’s Reduction Theorem. If a graph is not planar, then it has to have, somewhere inside it, K3, 3 or K5 or both. Maybe several of them.

A graph that’s hidden within another is called a “subgraph”. This follows the same etymological reasoning that gives us “subsets” and “subgroups” and many other mathematics words beginning with “sub”. And these subgraphs turn up whenever you have a nonplanar graph. A subgraph uses some set of the vertices and edges of the original graph; it doesn’t need all of them. A nonplanar graph has a subgraph that’s K3, 3 or K5 or both.

Sometimes it’s easy to find one of these. K4, 4 obviously has K3, 3 inside it. Pick three of the four vertices on one side and three of the four vertices on the other, and look at the edges connecting them up. There’s your K3, 3. Or on the other side, K6 obviously has K5 inside it. Pick any five of the vertices inside K6 and the edges connecting those vertices. There’s your K5.

Sometimes it’s hard to find one of these. We can make a graph look more complicated without changing whether it’s planar or not. Take your K3, 3 again. Go to each edge and draw another dot, another vertex, inside it. Well, now it’s a graph that’s got twelve vertices in it. It’s not obvious whether this is bipartite still. (Play with it a while.) But it hasn’t become planar, not because of this. It won’t be.

This is because we can make graphs more complicated, or more simple, without changing whether they’re planar. The process is a lot like what we did last week with the five-color map theorem, making a map simpler until it was easy enough to color. Suppose there’s a little section of the graph that’s a vertex connected by one edge to a middle vertex connected by one edge to a third vertex. Do we actually need that middle vertex for anything? Besides padding our vertex count? Nah. We can drop that whole edge-middle vertex-edge sequence and replace it all with a single edge. And there’s other rules that let us turn a vertex-edge-vertex set into a single vertex. That sort of thing. It won’t change a planar graph to a nonplanar one or vice-versa.

So it can be hard to find the K3, 3 or the K5 hiding inside a nonplanar graph. A big enough graph can have so much going on it’s hard to find the pattern. But they’ll be there, in some group of five or six vertices with the right paths between them.

It would make a good activity puzzle, if you could explain what to look for to kids.

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