Discrete

Two graphs that will not lie flat

Five points, every pair joined: no matter how the points are placed or how the lines are drawn, two of the lines cross. The proof is not about drawing at all — it counts edges against faces and finds one edge too many.

Worth reading first: Every corner pays for itself · Seven bridges, and the invention of throwing things away.

Four points, every pair joined by a line: the six lines can be drawn without a single crossing, and finding an arrangement that works takes about ten seconds with a pencil.

Five points, every pair joined: ten lines, and every arrangement anybody has ever tried has at least one crossing. That is a strong claim to make from failed attempts, and the reason it is true has nothing to do with attempts.

Two graphs that will not lie flat, and one that willK4, K5 and K3,3 in the best straight-line drawings a search could find. K4 has no crossings; the other two have one each, and Euler's formula shows that none can have none.K4no crossingsK51 crossing, and no drawing has noneK3,31 crossing, and no drawing has noneK4: 4 points, 6 edges, and at most 6 allowed; K5: 5 points, 10 edges, and at most 9 allowed; K3,3: 6 points, 9 edges,and at most 8 allowed6,000 layouts were tried for each; the best found is drawn, and the bound is what rules out anything better
Fig. 1 Three graphs in the best straight-line drawings a search over six thousand layouts could find. The first has no crossings. The other two have one each, and the count of edges against what a flat drawing allows is what rules out zero.

The search in that figure is honest about what it is. It tried six thousand random placements of the points for each graph, counted the crossings in every one, and kept the best. That establishes an upper bound on the crossing number and nothing else — a better arrangement might exist and the search might have missed it.

What rules out zero is a piece of arithmetic that never mentions a drawing.

Euler’s formula, turned into a ceiling

A graph drawn in the plane without crossings cuts the plane into regions. Counting corners, edges and regions — including the unbounded region outside everything — always gives the same alternating sum:

VE+F=2.V - E + F = 2.

A solid flattened into a graphThe cube pushed flat: eight vertices, twelve edges, and six faces once the region outside the drawing is counted as one. V − E + F is the same 2 it was on the solid.8 vertices, 12 edges6 faces, counting the outside — 8 − 12 + 6 = 2
Fig. 2 A cube pushed flat into the plane. The eight corners, twelve edges and six faces survive the flattening, with the outside of the drawing counted as the sixth face — which is why a formula about solids is a formula about flat drawings.

Now count edges twice. Every face of the drawing is bounded by at least three edges, since two edges cannot enclose a region in a graph without repeated edges. And every edge borders exactly two faces. So counting edge-face incidences gives at least 3F on one side and exactly 2E on the other:

3F2E.3F \le 2E.

Substituting F = 2 − V + E from Euler’s formula and rearranging gives

E3V6.E \le 3V - 6.

That is a ceiling on how many edges a graph can have and still be drawable flat, and it is arithmetic on two numbers a graph carries about with it.

The two graphs that exceed it

The complete graph on five points has 5 corners and 10 edges. The ceiling allows 3 × 5 − 6 = 9. Ten is more than nine, so no flat drawing exists — proved, from two counts, with no drawing attempted.

The ceiling on edges, and the two graphs that pass itFor each graph: how many points, how many edges, and the most edges a flat drawing could have. K5 and K3,3 exceed it; every other graph in the table does not.graphpointsedgesmost a flat drawing allowsK3333K4466K55109K661512K772115K882818K2,3566 (no triangles)K3,3698 (no triangles)K3,471210 (no triangles)a flat drawing cuts the plane into faces, every face is bounded by at least three edges,and every edge borders two faces — which turns Euler's V − E + F = 2 into a ceiling on thenumber of edgesK5 misses it by one edge and K3,3 by one; nothing about drawing appears in eithercalculation
Fig. 3 Every graph in the table with its two counts and the ceiling a flat drawing would allow. K5 exceeds it by one edge; every smaller complete graph is under it, and so are the bipartite graphs except one.

The second obstruction needs a refinement of the same argument. The graph joining three points to three other points — three houses to three utilities, in the version everybody meets as a puzzle — has 6 corners and 9 edges, and the ceiling of 3 × 6 − 6 = 12 does not rule it out.

But that graph has no triangles: every cycle in it alternates between the two sides and so has an even length of at least four. In a flat drawing of a triangle-free graph every face is bounded by at least four edges, so the count becomes 4F ≤ 2E, and the same substitution gives

E2V4.E \le 2V - 4.

For the three-and-three graph that ceiling is 2 × 6 − 4 = 8, and it has 9 edges. One too many again.

Two graphs, two counting arguments differing by one number, and in both cases the impossibility is established by an inequality that a reader can check on a napkin.

The same ceiling, counted on the solids

The inequality is worth seeing where it is tight, because that says which graphs are as dense as flatness permits.

V − E + F = 2, five timesVertices, edges and faces of the five regular solids, with the alternating sum. The edges are counted from the faces rather than listed, and the sum is 2 in every row.VEFV − E + Ftetrahedron4642cube81262octahedron61282dodecahedron2030122icosahedron1230202
Fig. 4 The five regular solids with their three counts and the alternating sum. Each is a planar graph once flattened, and the three whose faces are triangles sit exactly on the ceiling of 3V − 6 rather than under it.

Equality in 3F ≤ 2E happens exactly when every face is a triangle, so a graph achieving E = 3V − 6 is one whose flat drawing cuts the plane entirely into triangles. The tetrahedron, the octahedron and the icosahedron do it — 6 = 3 × 4 − 6, 12 = 3 × 6 − 6, 30 = 3 × 12 − 6 — and they are the reason the ceiling cannot be lowered.

That the list of regular solids stops at five is proved by a close relative of the same counting, and the two arguments sit either side of one identity. There, the counts are forced by requiring every face and every corner to look alike; here, they are bounded by requiring nothing at all. Both are Euler’s formula being asked a question about how much a flat structure can hold.

The remarkable part: those are the only two

Ruling out two graphs is a small result. What makes planarity a subject is Kuratowski’s theorem, and it is one of the most satisfying classification statements in mathematics.

A graph can be drawn in the plane without crossings if and only if it contains neither of those two graphs, in the sense of containing a subdivision of one.

A subdivision means the edges may have extra points sitting along them — a path of three edges counts as an edge that has been stretched. With that allowance, every non-planar graph in existence, however large and however complicated, has one of exactly two obstructions hiding inside it.

That is a strong statement about an infinite family, and its form is worth noticing: an infinite class characterised by a finite list of forbidden pieces. Such characterisations are rare and valuable, and the general theory of when they exist — the graph minor theorem — took two decades and hundreds of pages, with this as its first and simplest instance.

The shape of the statement deserves a moment on its own, because it is doing something an inequality cannot. The edge-count ceiling is a necessary condition: exceed it and the graph is certainly not planar. It is not sufficient, and the gap is easy to exhibit — take the two forbidden graphs and hang a long path off each, and the edge count falls well under the ceiling while the obstruction is still inside. Kuratowski’s theorem closes exactly that gap, replacing a test that can only ever say no with one that decides. Necessary conditions are cheap and common; conditions that are both necessary and sufficient are what a subject is trying to reach, and they usually arrive much later than the first bound does.

What it decides elsewhere

Planarity is not a curiosity about drawings; it is a hypothesis that other theorems need.

A 5-region wheel needs 4 coloursA hub touching 5 rim regions arranged in a ring. The rim is odd, so the whole map needs 4 colours and no fewer.the mapwho touches whom
Fig. 5 A map, which is a planar graph in disguise: one region per corner, one edge per shared border. The four-colour theorem is a statement about exactly this class of graphs, and it is false for graphs in general.

The four-colour theorem is a theorem about planar graphs and about nothing else. Five points every pair joined would need five colours, and the theorem is not contradicted because that graph is not planar. Every map on a flat sheet becomes a graph by putting a corner in each region and joining regions that share a border, and the drawing of that graph has no crossings for the geometric reason that two borders cannot cross.

The same edge-counting argument gives the four-colour theorem’s easy relative. From E ≤ 3V − 6 it follows that the average degree of a planar graph is below six, so some corner has degree at most five — and that observation, applied repeatedly, proves that six colours always suffice in a page of work. Getting from six to five takes a clever argument, and getting from five to four took a computer.

The bridges of KönigsbergA schematic of the city: a river splitting around an island, four landmasses, and the seven bridges joining them.north banksouth bankthe islandeast bank
Fig. 6 The graph that began the subject. Königsberg’s bridges are a question about traversal rather than about drawing, and the reduction is the same one: throw away the geography and keep the incidences.

Planarity also decides which circuits can be printed on one layer of a board, which is where the question is asked most often outside mathematics, and whether a graph’s dual makes sense — a construction that needs a drawing to be defined at all.

It is worth noticing which graphs on this site are planar and which are not, since the property is easy to check and rarely mentioned. The graph behind the party of six is the complete graph on six points, which contains the five-point one and is therefore about as far from planar as a small graph gets — the Ramsey argument is completely indifferent to that, because it never draws anything. The graph joining each point to its nearest neighbours in a plane divided by proximity is planar by construction, since its edges are built from a drawing rather than being fitted to one afterwards. The distinction between a graph that happens to be planar and one that arrives with a drawing attached is worth keeping, and the second kind is where the fast algorithms are used.

Give it a handle and both fit

Nothing in the argument was about the plane as such. It used Euler’s formula, and Euler’s formula depends on the surface being drawn on.

A sphere, and a sphere with handlesTwo-sided closed surfaces in order: a sphere, then one with a handle, then two, then three, each with the Euler characteristic that counts them.a sphereχ = 2a torusχ = 0a surface with 2 handlesχ = −2a sphere with g handles has characteristic 2 − 2g, so the number falls by two for everyhandleon the two-sided side of the classification, that number is the whole of the answer: twosurfaces with the same characteristic are the same surface
Fig. 7 Surfaces built by adding handles, with the alternating sum for each. Every handle costs two, so the sphere’s 2 becomes 0 on a torus and −2 on a two-holed surface — and the ceiling on edges moves with it.

On a surface where the alternating sum is χ rather than 2, the same double count gives E ≤ 3V − 3χ. For a torus, χ is 0 and the ceiling becomes 3V, which allows the complete graph on five points with room to spare — and indeed both forbidden graphs can be drawn on a torus without crossings, by routing one edge round through the hole.

The genus of a graph is the smallest number of handles a surface needs before the graph fits. It is 0 for planar graphs, 1 for the two obstructions here, and grows for larger complete graphs; every surface has its own finite list of forbidden pieces, though the lists are known explicitly only for the plane and the projective plane, where there are 35 of them.

The colour count moves too. The seven-colour bound on a torus follows from the shifted ceiling by the same averaging argument that gives six on the plane, and unlike the plane’s case it is achieved: there is a map on a torus needing all seven, and it can be drawn.

The number of crossings, when there must be some

If a graph cannot be drawn flat, the natural follow-up is how few crossings are unavoidable. That number is the crossing number, and it is a much harder quantity than planarity.

For the five-point complete graph it is 1, and both halves of that claim are visible above: the search found a drawing with one crossing, and the edge count rules out zero. For the complete graph on six points it is 3, on seven it is 9, and by ten points the value was an open problem for decades.

The general question is hard in the strongest sense — deciding whether a graph’s crossing number is below a given value is NP-complete — and even for complete graphs there is only a conjectured formula, verified up to twelve points and unproven beyond.

There is also a subtlety the figures here quietly rely on. Drawing edges as straight segments can force more crossings than drawing them as curves; the rectilinear crossing number is a different and generally larger quantity. For the two graphs above the two numbers agree at 1, so the pictures are honest, but for larger complete graphs they diverge, and a search over straight-line layouts stops being evidence about the crossing number at all.

What it costs

Deciding planarity is fast. There are algorithms that answer in time proportional to the number of edges — remarkably, no worse than reading the graph in — by attempting to build a drawing and failing in a way that exhibits an obstruction. The first such algorithm dates from 1974 and the subject has produced several since, each simpler than the last.

Finding a Kuratowski obstruction inside a non-planar graph is also linear, so the negative answer comes with a certificate: a subgraph a reader can check by hand. That is unusually good behaviour for a decision problem, and it is why planarity testing is a standard library routine rather than a research topic.

The counting bound costs nothing at all and answers the question in the direction that matters most often. If a graph has more than 3V − 6 edges it is not planar, full stop, and that test is two integers and a comparison. It gives no information in the other direction — plenty of non-planar graphs have few edges — but as a first filter it is free.

What the picture cannot show

The drawings above are three graphs in three particular layouts, and the theorem is about every layout of every graph. The search behind them can only ever report the best thing it found; the impossibility comes from the inequality, and an inequality has no picture.

Nor can a figure show a subdivision hiding inside a large graph. Kuratowski’s theorem is about containment, and finding the contained copy in a graph with a thousand edges is an algorithmic matter with no visual character.

And the pictures are all of straight-line drawings, which is a restriction the theory does not make. That every planar graph can in fact be drawn with straight edges is a separate theorem — Fáry’s — and it is what licenses these figures to be evidence about planarity at all.

The ladder from here

Rungs above: Kuratowski’s theorem proved, and the equivalent statement in terms of minors rather than subdivisions. Fáry’s theorem, that a planar graph can always be drawn with straight edges. The dual graph, and how the four-colour theorem becomes a statement about colouring faces. Graphs on surfaces other than the plane: the torus accepts both forbidden graphs, and every surface has its own finite list of obstructions. The genus of a graph, and the seven colours a map on a torus can need. Planar separator theorems, which say a planar graph can be cut into halves by removing very few corners, and are what makes planar algorithms fast. And the graph minor theorem, which says every family closed under minors has a finite list of obstructions — the most general form of what Kuratowski found first.

The shape of the idea

The argument at the centre of this essay is a double count wearing a topologist’s hat, and it is worth extracting from the graph theory.

Something is counted in two ways — edge-face incidences, once from the faces and once from the edges — and the two counts are compared. One of them is an inequality because a face needs at least three edges; the other is an equality because an edge borders exactly two faces. Feeding both into a formula that was proved elsewhere converts a topological fact into an arithmetical ceiling.

Every step is elementary and the conclusion is not: a statement about all possible drawings, which are infinite in number and impossible to enumerate, reduced to a comparison of two whole numbers. Counting the same thing twice is the oldest trick in the subject and it keeps producing results that no amount of drawing would ever settle.

Shares its objects with

Essays that name at least two of the same things, and that neither author linked.

Named objects

A dashed tag is an object no other essay names yet.

Complete graphCounting argumentCrossing numberEuler formulaGraphGraph colouringPlanar graphPlanaritySubdivisionTopological invariant