 # Bounding the number of edges of matchstick graphs

We show that a matchstick graph with n vertices has no more than 3n-c√(n-1/4) edges, where c=1/2(√(12) + √(2π√(3))). The main tools in the proof are the Euler formula, the isoperimetric inequality, and an upper bound for the number of edges in terms of n and the number of non-triangular faces. We also find a sharp upper bound for the number of triangular faces in a matchstick graph.

## Authors

##### This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

## 1. Introduction

Matchstick graphs, first introduced by Harborth in 1981 [6, 7], are graphs drawn in the plane with each edge a straight-line segment of unit length, such that no two edges have a point in common, unless the common point is an endpoint of both edges (Figure 1).

Harborth posed various problems about matchstick graphs. The one that drew most attention in the literature is that of finding -regular matchstick graphs with the smallest number of vertices. For example, it is known that there are no -regular matchstick graphs [2, 8], and the currently smallest known -regular matchstick graph, described in  and known as the Harborth graph, has vertices and edges. Another of Harborth’s problems in  is to find the maximum number of edges in a matchstick graph on vertices, for which he conjectured the following.

###### Conjecture 1 (Harborth ).

For each , the maximum number of edges in a matchstick graph on vertices is .

In  Harborth proved that this is indeed the maximum number of edges if we furthermore assume that any two vertices are separated by a distance of at least (the so-called penny graphs). This maximum is attained by an appropriately chosen set of points on the triangular lattice (Figure 2).

In [3, p. 225] this problem is again mentioned, where it is stated that it “seems very likely that the maximum number of edges in a crossing-free unit-distance graph is again …” It is not hard to show from the Euler formula and the isoperimetric inequality that the number of edges is at most . Our main result is the following improvement.

###### Theorem 1.

In a matchstick graph with vertices, the number of edges satisfies

 e⩽3n−c√n−1/4,

where .

As a calculation shows, this bound turns out to be strong enough that it settles Conjecture 1 for all , as well as for

 n =16,18,19,20,21,23,24,26,27,29,30,32,33,36,37,39,40, 43,44,47,48,51,52,55,56,60,61,65,69,70,74,75,79,80,85, 90,91,96,102,108,114,120,127.

The proof of Theorem 1 uses the isoperimetric inequality (Lemma 5 below), as well as the following result which bounds the number of edges in terms of the number of bounded non-triangular faces of the matchstick graph. Its proof is based on Harborth’s induction proof  of Conjecture 1 for penny graphs.

###### Theorem 2.

In a matchstick graph with vertices, edges, and bounded non-triangular faces, we have

 e⩽3n−√12n−3+g.

We do not assume that the graph is -connected or even connected in Theorem 2. This is not just for the sake of generality, as this general statement is needed when the induction hypothesis is applied in the proof.

Even though Theorem 2 does not seem to be sharp if there are bounded non-triangular faces (), when combined with the Euler formula, it gives the following sharp upper bound for the number of triangular faces of a matchstick graph.

###### Corollary 3.

In a matchstick graph with vertices, the number of bounded triangular faces is at most .

Note that for each there is a matchstick graph on the triangular lattice with vertices and triangular faces (Figure 2).

The above results are proved in Section 3. In the next section, we establish our terminology and introduce the fundamental tools we’ll need: the Euler formula, a double-counting identity, and the isoperimetric inequality.

## 2. Plane graphs and matchstick graphs

A plane graph is defined to be a drawing of a graph in the plane such that each vertex is a different point in the plane, and each edge is represented by a simple arc joining and , in such a way that two arcs only intersect in a common endpoint. The faces of a plane graph are the connected components of the complement of the plane graph in the plane. One of the faces is unbounded. Throughout this paper, we will denote the number of vertices by , the number of edges by , and the number of bounded faces by .

By the Euler formula, whenever is connected, we have . If is furthermore -connected, then each face is bounded by a cycle. Denote the number of vertices of the unbounded face by , and the number of bounded faces with exactly boundary vertices by . We have the following well-known relation.

###### Lemma 4.

For any -connected plane graph with vertices, edges, boundary vertices, and bounded faces with vertices, , we have .

###### Proof.

If we add up the number of vertices of each face, including the unbounded face, we count each edge twice, thus obtaining . By Euler’s formula, . Subtracting these two identities, we obtain the result. ∎

A matchstick graph is a plane graph in which each edge is represented by a straight-line segment of unit length. A matchstick graph is called a penny graph if the distance between any two vertices is at least , and there is an edge between all pairs of vertices at distance .

In the proof of Theorem 1 we will need the following consequence of the isoperimetric inequality that asserts that among all simple closed curves in the plane of a fixed length, the circle is the unique curve that encloses the largest area .

###### Lemma 5.

Let be a -connected matchstick graph with vertices on the outer boundary and bounded triangular faces. Then .

###### Proof.

The polygon bounding the unbounded face has edges and encloses equilateral triangles of unit side length. Each of these triangles has area . Thus the polygon has area and perimeter . By the isoperimetric inequality, any region of area bounded by a simple closed curve of length satisfies , with equality only if the curve is a circle. It follows that . ∎

## 3. Proofs

We will repeatedly use the following inequality involving sums of square roots.

###### Lemma 6.

Let be non-negative real numbers with and . Then

 √β+√γ⩽√α+√δ,

with equality iff .

###### Proof.

From

 0⩽(α−β)(γ−α)=αδ−βγ

we obtain

 (√β+√γ)2=β+γ+2√βγ⩽α+δ+2√αδ=(√α+√δ)2.\qed
###### Lemma 7.

If are integers such that , , then

 ⌊3n−√12n−3⌋+1⩾⌊3n1−√12n1−3⌋+⌊3n2−√12n2−3⌋.
###### Proof.

The required inequality is equivalent to

 √12n−3+5⩽⌈√12n1−3⌉+⌈√12n2−3⌉. (1)

Without loss of generality, . If , then we set , , , in Lemma 6 and obtain

 √69+√12n−51⩽√12n1−3+√12n2−3. (2)

Apply Lemma 6 again with , , , (since ), we get

 √12n−3+√33⩽√12n−51+9. (3)

It follows from (2) and (3) that

 √12n−3+√69+√33−9⩽√12n1−3+√12n2−3.

Finally, a calculation shows that , and (1) follows.

For the remaining cases , we need to round up to obtain (1). For instance, if , then we need to show , which follows from Lemma 6 by setting , , , (since ), and noting that . Similarly, when then , and we obtain by setting , , , , and when then , and follows by setting , , , . ∎

###### Proof of Theorem 2.

We use induction on the number of vertices . The theorem clearly holds when or . We now assume that and that the theorem holds for all smaller values of as induction hypothesis.

If the matchstick graph is not connected, let be a connected component of . If is in a bounded face of , we can move to the unbounded face of . Note that this does not change the number of non-triangular faces, unless was originally inside a triangular face of . However, then cannot have any edges, and we are done by applying induction to . Thus we may assume that neither nor lies in a bounded face of the other graph. Then it is easy to move so that one of its vertices is at distance from a vertex of , while keeping and disjoint. This creates a new edge that joins two connected components of . This process can be repeated until is connected, without decreasing the number of edges.

We now assume without loss of generality that is connected. If is not -connected, then there is a vertex such that is disconnected (Figure 3(a)). We can then decompose into two induced subgraphs and having only in common. If has vertices, edges, and non-triangular faces (), then , , and . It is clear that if lies in the unbounded face of and in the unbounded face of . Suppose that (say) lies in a bounded face of . Then this face cannot be a triangle, as then would not have any edges, contradicting the connectedness of . It follows that in this case too. By induction,

 e=e1+e2 ⩽3n1−√12n1−3+g1+3n2−√12n2−3+g2 =3n+3−√12n1−3−√12n2−3+g ⩽3n−√12n−3+g,

by Lemma 6 with , , , .

For the remainder of the proof we assume without loss of generality that is -connected. In particular, the boundary of the unbounded face is a cycle.

Suppose that the boundary cycle has a chord (Figure 3(b)). Let and be two induced subgraphs covering such that the chord is their only common edge and the endpoints of the chord their only two common vertices. Using the same notation as before, with having vertices, edges, and non-triangular faces (), we now have , , , and . Again we use induction on these two subgraphs to obtain

 e=e1+e2−1 ⩽⌊3n1−√12n1−3⌋+g1+⌊3n2−√12n2−3⌋+g2−1 ⩽3n−√12n−3+gby Lemma~{}???.

From now on we assume without loss of generality that the boundary cycle does not have a chord.

We next show that if a bounded non-triangular face shares more than one vertex with the unbounded face, then we can assume without loss of generality that only two of its vertices are on the boundary cycle, they are adjacent, and at most one of the two interior angles of the face at these two vertices is smaller than .

First suppose that a non-triangular bounded face has two non-adjacent vertices on the boundary (Figure 3(c)). We decompose into two induced subgraphs and such they only have these two vertices in common, with no common edge. Again using the notation with having vertices, edges, and non-triangular faces (), we have , , , and . By induction,

 e=e1+e2 ⩽⌊3n1−√12n1−3⌋+g1+⌊3n2−√12n2−3⌋+g2 ⩽3n−√12n−3+gby Lemma~{}???.

From now on we assume without loss of generality that whenever a non-triangular bounded face has more than one vertex on the boundary, it has only two vertices on the boundary and they are adjacent. If both angles at these boundary vertices are , then two of the edges of the face will be forced to intersect. It follows that at most one angle of a non-triangular face is less than .

Let denote the number of non-triangular bounded faces that share a vertex with the unbounded face. At each vertex of of the boundary cycle of degree there are angles interior to bounded faces. If we denote the number of boundary vertices of degree by , then the total number of boundary vertices is , the number of angles is , and the sum of these angles equals . By the previous paragraph we have that at most of these angles are smaller than . Therefore, , hence

 ∑i⩾2(i−1)bi−gb⩽3b−6. (4)

Since the boundary cycle has no chord, when we remove the vertices on the boundary together with their incident edges, we remove exactly edges. We also remove non-triangular faces and vertices. Without loss of generality, , otherwise is just the boundary cycle, hence since . We use induction on the remaining graph of vertices to obtain

 e−∑i⩾2(i−1)bi⩽3(n−b)−√12(n−b)−3+g−gb,

hence

 e ⩽3(n−b)−√12(n−b)−3+g−gb+∑i⩾2(i−1)bi ⩽3(n−b)−√12(n−b)−3+g+3b−6by (???) =3n−√12(n−b)−3+g−6,

and in order to conclude that , we need to show that

 √12n−3⩽√12(n−b)−3+6. (5)

If , then

 √12(n−b)−3⩾√12(n−√12n−3+3)−3=√(√12n−3−6)2,

which gives (5), since , hence and . Otherwise, , and by Lemma 4, . ∎

###### Remark 8.

In the last step of the above proof, there is some slack, as we actually have from Lemma 4. By taking this into account, it is possible to prove the slightly stronger inequality . However, there are then more boundary cases to deal with, and as this is not much of an improvement, we settled for the weaker inequality in Theorem 2.

###### Remark 9.

When we removed the outer boundary cycle in the above proof, we needed this cycle not to have a chord in order to count the number of edges that are removed. This point is overlooked in Harborth’s original proof  on which this proof is based.

###### Proof of Corollary 3.

By the Euler formula, , and by Theorem 2, . It follows that . ∎

###### Proof of Theorem 1.

As in the proof of Theorem 2 we use induction on . The theorem is easy to verify for , and as in the proof of Theorem 2, we can assume that is connected.

To show that we can furthermore assume that is -connected, we also proceed as in the proof of Theorem 2. If is not -connected, we can decompose into two induced subgraphs and having only a single vertex in common. Let have vertices and edges . Then , and . By induction and using the shorthand we have

 e=e1+e2 ⩽3n1−c√n1−1/4+3n2−c√n2−1/4 =3n+3−c√n1−1/4−c√n2−1/4.

To conclude that we need to show that

 3c+√n−1/4⩽√n1−1/4+√n2−1/4.

Since , we can apply Lemma 6 with , , , to get

 √3/2+√n−1⩽√n1−1/4+√n2−1/4.

Since , we can again apply Lemma 6 with , , , , to obtain

 √5/2+√n−1/4⩽√n−1+√2.

Combining these two equations together gives

 √n−1/4+√5/2−√2+√3/2⩽√n1−1/4+√n2−1/4

which shows the required inequality since .

We now assume that is 2-connected. Thus the unbounded face is bounded by a cycle with edges. As before, denote the number of bounded faces with vertices by (), and the number of non-triangular faces by . By Lemma 4, noting that , we have

 e⩽3n−3−b−g. (6)

Suppose for the sake of contradiction that

 e>3n−c√n−1/4. (7)

Then (6) and (7) give the following upper bound for :

 g

We obtain the following lower bound for from (7), (8), and the Euler formula:

 f3 =e−n−g+1 >3n−c√n−1/4−n−c√n−1/4+3+b+1 =2n−2c√n−1/4+4+b.

Substitute this into the inequality from Lemma 5 to obtain

 b2−π√3b >2π√3(n−c√n−1/4+2).

By completing the square,

 (b−π√32)2 >3π24+2π√3(n−c√n−1/4+2),

we get the following lower bound for :

 b

We would like to deduce from this that . It is sufficient to show the following:

 √3π24+2π√3(n−c√n−1/4+2)⩾√2π√3(n−1/4)−π√32−3. (9)

Since the left-hand side is non-negative, we can assume without loss of generality that

 √2π√3(n−1/4)⩾π√32+3. (10)

Then we can square both sides of (9) and rearrange to obtain the equivalent

 (π√3+6−c√{}2π√3)√2π√3(n−1/4)⩾9−3π√32.

This follows from (10), upon checking that and

 (π√3+6−c√{}2π√3)(π√32+3)⩾9−3π√32.

So we have shown that , which, together with (6) gives

 e ⩽3n−√{}2π√3(n−1/4)−g.

By Theorem 2 we also have . Adding these two bounds, we obtain , which contradicts the assumption (7). Thus the assumption (7) is false and the theorem follows. ∎

###### Remark 10.

Eppstein  uses the isoperimetric inequality to find an upper bound of the form for the number of edges in a triangle-free penny graph on vertices. To show an upper bound of this form for triangle-free matchstick graphs will need a new idea, as there is no obvious way to bound the area of the bounded faces from below.

## References

•  Blåsjö, V. (2005). The Isoperimetric Problem. Amer. Math. Monthly 112, 526–566.
•  Blokhuis, A. (1982). Regular finite planar maps with equal edges. arXiv preprint arXiv:1401.1799
•  Brass, P., W. O. J. Moser, and J. Pach (2005). Research Problems in Discrete Geometry. Springer-Verlag, New York.
•  Eppstein, D. (2018). Edge bounds and degeneracy of triangle-free penny graphs and squaregraphs. Journal of Graph Algorithms and Applications 22, 483–499.
•  Harborth, H. (1974). Problem 664A. Elemente der Mathematik 29, 14–15.
•  Harborth, H. (1981). Point sets with equal numbers of unit-distant neighbors (Abstract), Discrete Geometry, 12–18 July 1981, Oberwolfach, Tagungsbericht 31/1981, Mathematisches Forschungsinstitut Oberwolfach. pp. 11–12.
•  Harborth, H. (1986). Match sticks in the plane. In: The Lighter Side of Mathematics, edited by R. K. Guy and R. E. Woodrow, 281–288. Mathematical Association of America, Washington, D.C.
•  Kurz, S. and R. Pinchasi (2011). Regular matchstick graphs. Amer. Math. Monthly 118, 264–267.