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Crossings in Randomly Embedded Graphs

Santiago Arenas-Velilla
Centro de Investigación en Matemáticas,
Guanajuato, Gto. 36000, Mexico
[email protected]
   Octavio Arizmendi
Centro de Investigación en Matemáticas,
Guanajuato, Gto. 36000, Mexico
[email protected]
Abstract

We consider the number of crossings in a graph which is embedded randomly on a convex set of points. We give an estimate to the normal distribution in Kolmogorov distance which implies a convergence rate of order n1/2n^{-1/2} for various families of graphs, including random chord diagrams or full cycles.

keywords: crossings, random graphs, normal approximation.

1 Introduction

Let G=(V,E)G=(V,E) be a graph with vertex set VV and edge set EV×VE\subset V\times V, which is embedded randomly in a convex set of points. We are interested in the random variable counting the number of crossings under this embedding.

Formally, for a graph G=(V,E)G=(V,E) with vertex set [n]={1,,n}[n]=\{1,\ldots,n\}, an embedding given by the permutation π:[n][n]\pi:[n]\to[n], is the graph isomorphism induced by the permutation π\pi. The crossings of such embedding are given by the set {(a,b,c,d)|{a,b},{c,d}E,π(a)>π(c)>π(b)>π(d)}\{(a,b,c,d)|\{a,b\},\{c,d\}\in E,\pi(a)>\pi(c)>\pi(b)>\pi(d)\}. Figure 1 shows graphical representation of a couple embeddings of a path graph P20P_{20}. The first one having 4040 crossing and the second one having 6060 crossings.

Refer to caption
(a) P20P_{20} with 40 crossings
Refer to caption
(b) P20P_{20} with 60 crossings
Figure 1: Examples of an embedding of a path with 20 vertices

To our best of our knowledge there is not much work about general graphs. The paper by Flajolet and Noy [4] considers the case where GG is a union of disjoint edges (is called a matching, a pairing or a chord diagram) and proves a central limit theorem. This result is also proved with the use of weighted dependency graphs, in [3]. More important to us the recent paper by Paguyo [5] gives a rate of convergence in that case. Another related paper is [1], where the authors consider a uniform random tree.

In this paper, we will show that under some asymptotic behaviour of very precise combinatorial quantities of the graph, the random variable counting the number of crossings in a random embedding approximates a normal distribution with mean μ\mu and variance σ2\sigma^{2} which can be calculated precisely (see Lemmas 1 and 2). Moreover, we give a convergence rate in this limit theorem.

Theorem 1.

Let GG be a graph with maximum degree Δ\Delta, and let XX be the number of crossings of a random uniform embedding of GG. Let μ\mu and σ2\sigma^{2} the mean and the variance of XX. Then, with W=(Xμ)/σW=(X-\mu)/\sigma,

dKol(W,Z)4m2(G)Δm3σ2(6Δmσ+16m4(G)m2(G)2+m(Δ1)2m2(G)),d_{Kol}(W,Z)\leq\frac{4m_{2}(G)\Delta m}{3\sigma^{2}}\left(\frac{6\Delta m}{\sigma}+\sqrt{1-\frac{6m_{4}(G)}{m_{2}(G)^{2}}+\frac{m(\Delta-1)^{2}}{m_{2}(G)}}\right), (1.1)

where mm is the number of edges of GG, mr(G)m_{r}(G) is the number of rr-matching of GG, and ZZ is a standard Gaussian random variable.

Examples of families of graphs that satisfy such a normal approximation, with a rate proportional to 1/n1/\sqrt{n} are pairings, cycle graphs, path graphs, union of triangles, among others. We explain in detail these examples in Section 4.

We should mention that our method of proof resembles the one used by Paguyo in [5], for the case of pairings. The main idea is to write the number of crossing as a sum of indicators variables and then consider the size biased transform in the case of sums of Bernoulli variables. However, there is a crucial difference between Paguyo’s way to write such variables and how we do it, which in our opinion is more flexible. To be precise, Paguyo considers for each four points a<b<c<da<b<c<d in the circle the indicator that there is a crossing formed by the edges {a,c}\{a,c\} and {b,d}\{b,d\}. This random variable is easy to handle for the case of a pairing but for a general given graph, even calculating the probability of such indicator to be 11 can be very complicated. Our approach instead looks at a given 22-matching in the graph GG and consider the indicator random variable of this 22-matching, when embedded randomly, to form a crossing.

2 Preliminaries

In this section we establish some notations for graphs and remind the reader about the main tool that we will use to quantify convergence to a normal distribution: the size bias transform.

2.1 Notation and Definitions on Graphs

A graph is pair G=(V,E)G=(V,E), where E{{v,w}|v,wV}E\subset\{\{v,w\}|v,w\in V\}. Elements in VV are called vertices and elements in EE are called edges. An edge {v,w}\{v,w\} is sometimes written as vGwv\sim_{G}w or vwv\sim w if the underlying graph GG is clear. The number or vertices |V||V|, will be denoted by nn, while the number of edges, |E||E|, will be denoted be mm.

For a vertex vv, we say that ww is a neighbour of vv, if {v,w}E\{v,w\}\in E. The number of neighbours of vv is called the degree of vv, denoted by deg(v)deg(v). The largest degree among all vertices in a graph will be denoted by Δ\Delta.

A subgraph of GG, is a graph, H=(W,F)H=(W,F), such that WVW\subset V and FGF\subset G. We say that G=(V,E)G=(V,E) and G=(V,E)G^{\prime}=(V^{\prime},E^{\prime}) are isomorphic if there exist a bijection φ:VV\varphi:V\to V^{\prime} such that {u,v}E\{u,v\}\in E if and only if {φ(u),φ(v)}E\{\varphi(u),\varphi(v)\}\in E^{\prime}, for all u,vVu,v\in V.

An rr-matching in a graph GG is a set of rr edges in GG, no two of which have a vertex in common. We denote by Mr(G)M_{r}(G) the set rr-matchings of GG and by mr(G)m_{r}(G) their cardinality. Note the m1=mm_{1}=m corresponds to the number of edges of the graph GG.

2.2 Size Bias Transform

Let XX be a positive random variable with mean μ\mu finite. We say that the random variable XsX^{s} has the size bias distribution with respect to XX if for all ff such that 𝔼[Xf(X)]<\mathbb{E}[Xf(X)]<\infty, we have

𝔼[Xf(X)]=μ𝔼[f(Xs)].\mathbb{E}[Xf(X)]=\mu\mathbb{E}[f(X^{s})].

In the case of X=i=1nXiX=\sum_{i=1}^{n}X_{i}, with XiX_{i}’s positive random variables with finite mean μi\mu_{i}, there is a recipe to construct XsX^{s} (Proposition 3.21 from [6]) from the individual size bias distributions of the summands XiX_{i}:

  1. 1.

    For each i=1,,ni=1,\ldots,n, let XisX_{i}^{s} having the size bias distribution with respect to XiX_{i}, independent of the vector (Xj)ji(X_{j})_{j\neq i} and (Xjs)ji(X_{j}^{s})_{j\neq i}. Given Xis=xX_{i}^{s}=x, define the vector (Xj(i))ji(X_{j}^{(i)})_{j\neq i} to have the distribution of (Xj)ji(X_{j})_{j\leq i} conditional to Xi=xX_{i}=x.

  2. 2.

    Choose a random index II with (I=i)=μi/μ\mathbb{P}(I=i)=\mu_{i}/\mu, where μ=μi\mu=\sum\mu_{i}, independent of all else.

  3. 3.

    Define Xs=jIXj(I)+XIsX^{s}=\sum_{j\neq I}X_{j}^{(I)}+X_{I}^{s}.

It is important to notice that the random variables are not necessarily independent or have the same distribution. Also, XX can be an infinite sum (See Proposition 2.2 from [2]).

If XX is a Bernoulli random variable, we have that Xs=1X^{s}=1. Indeed, if (X=1)=p\mathbb{P}(X=1)=p, 𝔼(X)=p=μ\mathbb{E}(X)=p=\mu and then

𝔼[Xf(X)]=(1p)(0f(0))+p(1f(1))=pf(1)=μf(1)=μ𝔼[f(1)].\mathbb{E}[Xf(X)]=(1-p)(0f(0))+p(1f(1))=pf(1)=\mu f(1)=\mu\mathbb{E}[f(1)].

Therefore, we have the following corollary (Corollary 3.24 from [6]) by specializing the above recipe.

Corollary 1.

Let X1,X2,,XnX_{1},X_{2},\ldots,X_{n} be Bernoulli random variables with parameter pip_{i}. For each i=1,,ni=1,\ldots,n let (Xj(i))ji(X_{j}^{(i)})_{j\neq i} having the distribution of (Xj)ji(X_{j})_{j\neq i} conditional on Xi=1X_{i}=1. If X=i=1nXiX=\sum_{i=1}^{n}X_{i}, μ=𝔼[X]\mu=\mathbb{E}[X], and II is chosen independent of all else with (I=i)=pi/μ\mathbb{P}(I=i)=p_{i}/\mu, then Xs=1+jIXj(I)X^{s}=1+\sum_{j\neq I}X_{j}^{(I)} has the size bias distribution of XX.

The following result (Theorem 5.3 from [2]) gives us bounds for the Kolmogorov distance, in the case that a bounded size bias coupling exists. This distance is given by

dKol(X,Y):=supz|FX(z)FY(z)|,d_{Kol}(X,Y):=\sup_{z\in\mathbb{R}}|F_{X}(z)-F_{Y}(z)|,

where FXF_{X} and FYF_{Y} are the distribution functions of the random variables XX and YY.

Theorem 2.

Let XX be a non negative random variable with finite mean μ\mu and finite, positive variance σ2\sigma^{2}, and suppose XsX^{s}, have the size bias distribution of XX, may be coupled to XX so that |XsX|A|X^{s}-X|\leq A, for some AA. Then with W=(Xμ)/σW=(X-\mu)/\sigma,

dKol(W,Z)6μA2σ3+2μΨσ2,d_{Kol}(W,Z)\leq\frac{6\mu A^{2}}{\sigma^{3}}+\frac{2\mu\Psi}{\sigma^{2}}, (2.1)

where ZZ is a standard Gaussian random variable, and Ψ\Psi is given by

Ψ=Var(𝔼[XsX|X])\Psi=\sqrt{\mathrm{Var}(\mathbb{E}[X^{s}-X|X])} (2.2)

3 Mean and Variance

Let SnS_{n} be the set of permutation of nn elements. For a permutation πSn\pi\in S_{n}, and a graph GG, let GπG_{\pi} be the graph whose edges are given by

vGwπ(v)Gππ(w),v,wV.v\sim_{G}w\Longleftrightarrow\pi(v)\sim_{G_{\pi}}\pi(w),\qquad\forall v,w\in V.

For a random uniform permutation π\pi, let X:=X(Gπ)X:=X(G_{\pi}) be the random variable that counts the number of crossings of GπG_{\pi}, that is

X=iM2(G)𝕀{i is a crossing in Gπ}=iM2(G)YiX=\sum_{i\in M_{2}(G)}\mathbb{I}_{\{i\text{ is a crossing in }G_{\pi}\}}=\sum_{i\in M_{2}(G)}Y_{i} (3.1)

where M2(G)M_{2}(G) is the set of 2-matching of GG.

In this section we give a formula for the mean and variance of the random variable XX in terms of the number of subgraphs of certain type.

Lemma 1.

For a graph GG, if XX denote the number of crossings in a random embedding on a set of nn points in convex position, then its expectation is given by

μ:=μ(G)=𝔼(X)=13m2(G),\mu:=\mu(G)=\mathbb{E}(X)=\frac{1}{3}m_{2}(G),

where m2(G)m_{2}(G) denotes the number of 2-matching of GG.

Proof.

For each iM2(G)i\in M_{2}(G), we notice that YiBernoulli(1/3)Y_{i}\sim\mathrm{Bernoulli}(1/3). Indeed, if ii consist of two edges {v1,v2}\{v_{1},v_{2}\} and {v3,v4}\{v_{3},v_{4}\}, the probability to obtain a crossing only depends on the cyclic orders in which v1,v2,v3v_{1},v_{2},v_{3} and v4v_{4} are embedded in {1,,n}\{1,\dots,n\}, not in the precise position of them. From the 66 possible orders, only 1/31/3 of them yield a crossing. See Figure 2, for the 66 possible cyclic orders of v1,v2,v3v_{1},v_{2},v_{3} and v4v_{4}.

v1v_{1}v2v_{2}v4v_{4}v3v_{3}v1v_{1}v2v_{2}v3v_{3}v4v_{4}v1v_{1}v2v_{2}v4v_{4}v3v_{3}v1v_{1}v2v_{2}v3v_{3}v4v_{4}v1v_{1}v2v_{2}v3v_{3}v4v_{4}v1v_{1}v2v_{2}v4v_{4}v3v_{3}
Figure 2: Possibles cyclic orders for a 22-matching.

Summing over all jj, the expected value of XX is

𝔼X=iM2(G)(i is a crossing)=13m2(G),\mathbb{E}X=\sum_{i\in M_{2}(G)}\mathbb{P}(i\text{ is a crossing})=\frac{1}{3}m_{2}(G),

as desired. ∎

For the the second moment it is necessary to expand X2X^{2} to get

𝔼X2=𝔼i,jM2(G)𝕀{i,j are crossings}=i,jM2(G)(i,j are crossings).\mathbb{E}X^{2}=\mathbb{E}\sum_{i,j\in M_{2}(G)}\mathbb{I}_{\{i,j\text{ are crossings}\}}=\sum_{i,j\in M_{2}(G)}\mathbb{P}(i,j\text{ are crossings}).

The analysis for 𝔼X2\mathbb{E}X^{2} is significantly more complicated, since (i,j are crossings)\mathbb{P}(i,j\text{ are crossings}) depends of how many edges and vertices the two 22-matchings, ii and jj, share. Thus the previous sum can be divided in 88 types, depending of how the 22-matchings, ii and jj, share edges and vertices as is shown in Figure 3. We call such different configuration the “kind of pair of 22-matching” .

C1C_{1}C2C_{2}C3C_{3}C4C_{4}C5C_{5}C6C_{6}C7C_{7}C8C_{8}
Figure 3: Kinds of pair of 22-matchings in the sum of the second moment of XX.

The probabilities of that both ii and jj are crossing for each type of double 22-matching are the following (with the obvious abuse of notation):

(C1)=19,(C2)=19,(C3)=215,(C4)=760,\displaystyle\mathbb{P}(C_{1})=\frac{1}{9},\quad\mathbb{P}(C_{2})=\frac{1}{9},\quad\mathbb{P}(C_{3})=\frac{2}{15},\quad\mathbb{P}(C_{4})=\frac{7}{60},
(C5)=110,(C6)=112,(C7)=16,(C8)=13.\displaystyle\mathbb{P}(C_{5})=\frac{1}{10},\quad\mathbb{P}(C_{6})=\frac{1}{12},\quad\mathbb{P}(C_{7})=\frac{1}{6},\quad\mathbb{P}(C_{8})=\frac{1}{3}.

Thus, we arrive to the following.

Lemma 2.

The second moment of XX is given by the formula,

𝔼X2=69m4(G)+45m3(G)+13m2(G)+49S2+715S4+15S5+16S6+13S7\mathbb{E}X^{2}=\frac{6}{9}m_{4}(G)+\frac{4}{5}m_{3}(G)+\frac{1}{3}m_{2}(G)+\frac{4}{9}S_{2}+\frac{7}{15}S_{4}+\frac{1}{5}S_{5}+\frac{1}{6}S_{6}+\frac{1}{3}S_{7} (3.2)

where SiS_{i} is the number of subgraphs of GG of type CiC_{i}.

Before proving our main result we will apply the above lemmas for a few examples.

Example 1 (Pairing).

Consider GG to be a disjoint union of nn copies of K2K_{2} graphs. The expectation is given by

𝔼X=13m2(G)=n(n1)6.\mathbb{E}X=\frac{1}{3}m_{2}(G)=\frac{n(n-1)}{6}.

For the variance, we only need to consider, m2(G)m_{2}(G), m4(G)m_{4}(G) and m3(G)m_{3}(G), since the other types of subgraphs are not present in GG. The number of rr-matchings is given by mr(G)=(nr)m_{r}(G)=\binom{n}{r}, since any choice of rr different edges is an rr-matching, then we obtain

𝔼X2=69(n4)+1215(n3)+13(n2)=n436n330+13n2180n15,\mathbb{E}X^{2}=\frac{6}{9}\binom{n}{4}+\frac{12}{15}\binom{n}{3}+\frac{1}{3}\binom{n}{2}=\frac{n^{4}}{36}-\frac{n^{3}}{30}+\frac{13n^{2}}{180}-\frac{n}{15},

and thus the variance is

Var(X)=𝔼X2(𝔼X)2=n(n1)(n+3)45.\mathrm{Var}(X)=\mathbb{E}X^{2}-(\mathbb{E}X)^{2}=\frac{n(n-1)(n+3)}{45}.
Example 2 (Path).

Let PnP_{n} be the path graph on nn vertices. In this case, the number of rr-matchings is mr(Pn)=(nrr)m_{r}(P_{n})=\binom{n-r}{r}, so we obtain that

m4=(n44),m3=(n33),m2=(n22),S2=3(n43),\displaystyle m_{4}=\binom{n-4}{4},\quad m_{3}=\binom{n-3}{3},\quad m_{2}=\binom{n-2}{2},\quad S_{2}=3\binom{n-4}{3},
S4=(n42),S5=2(n42),S6=n4,S7=2(n32).\displaystyle S_{4}=\binom{n-4}{2},\quad S_{5}=2\binom{n-4}{2},\quad S_{6}=n-4,\quad S_{7}=2\binom{n-3}{2}.

Then,

𝔼X2=n43623n390+35n23686n4553,\mathbb{E}X^{2}=\frac{n^{4}}{36}-\frac{23n^{3}}{90}+\frac{35n^{2}}{36}-\frac{86n}{45}-\frac{5}{3},

and thus the variance is given by

Var(X)=𝔼X2(𝔼X)2=n345n21811n45+23.\mathrm{Var}(X)=\mathbb{E}X^{2}-(\mathbb{E}X)^{2}=\frac{n^{3}}{45}-\frac{n^{2}}{18}-\frac{11n}{45}+\frac{2}{3}.
Example 3 (Cycle).

Let CnC_{n} be the cycle graph on nn vertices. The number of rr-matchings is given by mr(Cn)=nr(nr1r1)m_{r}(C_{n})=\frac{n}{r}\binom{n-r-1}{r-1}, then

𝔼X=m23=n(n3)6.\mathbb{E}X=\frac{m_{2}}{3}=\frac{n(n-3)}{6}.

On the other hand, the number of subgraphs is given by

m4=n4(n53),m3=n3(n42),m2=n(n3)2,S2=n(n52),\displaystyle m_{4}=\frac{n}{4}\binom{n-5}{3},\quad m_{3}=\frac{n}{3}\binom{n-4}{2},\quad m_{2}=\frac{n(n-3)}{2},\quad S_{2}=n\binom{n-5}{2},
S4=n(n5)2,S5=n(n5),S6=n,S7=n(n4),\displaystyle S_{4}=\frac{n(n-5)}{2},\quad S_{5}=n(n-5),\quad S_{6}=n,\quad S_{7}=n(n-4),

from where the second moment is

𝔼X2=n43613n390+47n2180n3,\mathbb{E}X^{2}=\frac{n^{4}}{36}-\frac{13n^{3}}{90}+\frac{47n^{2}}{180}-\frac{n}{3},

and thus the variance is given by

Var(X)=𝔼X2(𝔼X)2=n345n290n3.\mathrm{Var}(X)=\mathbb{E}X^{2}-(\mathbb{E}X)^{2}=\frac{n^{3}}{45}-\frac{n^{2}}{90}-\frac{n}{3}.
Example 4 (Triangles).

Let GG be the disjoint union of nn copies of K3K_{3}. In this case, the subgraphs of type S5S_{5} and S6S_{6} are not present in GG. In order to obtain an rr-matching for r2r\geq 2, we need to choose rr triangles and for each one we have 3 options to form the matching, so the number of rr-matching is mr(G)=3r(nr)m_{r}(G)=3^{r}\binom{n}{r}. Similarly, the number of other types of subgraphs is given by

m4=34(n4),m3=33(n3),m2=32(n2),\displaystyle m_{4}=3^{4}\binom{n}{4},\quad m_{3}=3^{3}\binom{n}{3},\quad m_{2}=3^{2}\binom{n}{2},
S2=34(n3),S4=32(n2),S7=232(n2).\displaystyle S_{2}=3^{4}\binom{n}{3},\quad S_{4}=3^{2}\binom{n}{2},\quad S_{7}=2\cdot 3^{2}\binom{n}{2}.

Then, the expectation and the second moment are

𝔼X=13m2(G)=3n(n1)2,𝔼X2=9n4439n310+51n2209n10,\mathbb{E}X=\frac{1}{3}m_{2}(G)=\frac{3n(n-1)}{2},\qquad\mathbb{E}X^{2}=\frac{9n^{4}}{4}-\frac{39n^{3}}{10}+\frac{51n^{2}}{20}-\frac{9n}{10},

and thus the variance is

Var(X)=𝔼X2(𝔼X)2=3n35+3n2109n10.\mathrm{Var}(X)=\mathbb{E}X^{2}-(\mathbb{E}X)^{2}=\frac{3n^{3}}{5}+\frac{3n^{2}}{10}-\frac{9n}{10}.

4 Proof of the Main Theorem

4.1 Construction of Size Bias Transform

Let π\pi be a fixed permutation and let I=(e,f)I=(e,f) be a random index chosen with probability (I=i)=1/m2(G)\mathbb{P}(I=i)=1/m_{2}(G) , which corresponds to a 2-matching (in this way e,fe,f are edges), and which is independent of π\pi. For said fixed π\pi, we have a fit GπG_{\pi}. We construct πs\pi^{s} as follows

  • If π\pi is such that GπG_{\pi} has a crossing at II, we define πs=π\pi^{s}=\pi.

  • Otherwise, we perform the following process to obtain a permutation with a cross on II. Suppose e={u1,u2}e=\{u_{1},u_{2}\} and f={v1,v2}f=\{v_{1},v_{2}\}, then under π\pi these edges are π(e)={π(u1),π(u2)}\pi(e)=\{\pi(u_{1}),\pi(u_{2})\} and π(f)={π(v1),π(v2)}\pi(f)=\{\pi(v_{1}),\pi(v_{2})\}. Since they do not cross, without loss of generality we can assume that π(u1)<π(v1)<π(v2)<π(u2)\pi(u_{1})<\pi(v_{1})<\pi(v_{2})<\pi(u_{2}) satisfies. Now, we choose a random vertex uniformly among the vertices of the edges of II. In case the vertex is u1u_{1} or v1v_{1}, we leave these fixed and swap the positions between π(v2)\pi(v_{2}) and π(u2)\pi(u_{2}) and define πs\pi^{s} as the resulting permutation. In case of choosing u2u_{2} or v2v_{2}, we do the same process leaving these vertices fixed and exchanging π(v1)\pi(v_{1}) and π(u1)\pi(u_{1}). In this way we obtain a permutation πs\pi^{s} such that it has a crossing at II

Note that πs\pi^{s} is a uniform random permutation conditional on the event that π(ui)\pi(u_{i})’s and π(vi)\pi(v_{i})’s are in alternating in the cyclic order. This in turn means that GπsG_{\pi^{s}}, is a uniform random embedding conditioned on the event that the event II has a crossing.

In summary, we obtain that

X(Gπs)=iM2(G)𝕀{πs(i) is a crossing in Gπs}=iM2(G)Yπs(i),X(G_{\pi^{s}})=\sum_{i\in M_{2}(G)}\mathbb{I}_{\{\pi^{s}(i)\text{ is a crossing in }G_{\pi^{s}}\}}=\sum_{i\in M_{2}(G)}Y_{\pi^{s}(i)},

satisfies {Yπs(j)}jI\{Y_{\pi^{s}(j)}\}_{j\neq I} has the distribution of {Yπ(j)}jI\{Y_{\pi(j)}\}_{j\neq I} conditional on Yπ(I)=1Y_{\pi(I)}=1. Then, by Corollary 1, we get that Xs=X(Gπs)X^{s}=X(G_{\pi^{s}}), has the size bias distribution of XX.

Define Xs=X(Gπs)X^{s}=X(G_{\pi}^{s}) as the size bias transform of X(Gπ)X(G_{\pi}). By construction, |XsX|2Δ(m1)|X^{s}-X|\leq 2\Delta(m-1). Indeed, because in the worst case each of the edges incident to each vertex creates a new crossing.

4.2 Bounding the Variance of 𝔼[XsX|X]\mathbb{E}[X^{s}-X|X]

In order to use Theorem 2 we need a bound for the variance of a conditional expectation, which depends of XX and its size bias transform XsX^{s}. A bound for that variance is given in the next lemma, which is one of the main results in this paper.

Lemma 3.

Let G=(V,E)G=(V,E) a graph with maximum degree Δ\Delta, and let XX be the number of crossings of a random uniform embedding of GG. Then

Var(𝔼[XsX|X])4Δ2(m1)2(16m4(G)m2(G)2+(Δ1)2(m4)2m2(G)),\mathrm{Var}(\mathbb{E}[X^{s}-X|X])\leq 4\Delta^{2}(m-1)^{2}\left(1-\frac{6m_{4}(G)}{m_{2}(G)^{2}}+\frac{(\Delta-1)^{2}(m-4)}{2m_{2}(G)}\right), (4.1)

where XsX^{s} is the size bias transform of XX, mm is the number of edges of the graph GG and mr(G)m_{r}(G) is the number of rr-matchings of the graph GG.

Proof.

Note that

𝔼[XsX|X]\displaystyle\mathbb{E}[X^{s}-X|X] =iM2(G)𝔼[XsX|X,I=i](I=i)\displaystyle=\sum_{i\in M_{2}(G)}\mathbb{E}[X^{s}-X|X,I=i]\mathbb{P}(I=i)
=1m2(G)iM2(G)(X(i)X),\displaystyle=\frac{1}{m_{2}(G)}\sum_{i\in M_{2}(G)}(X^{(i)}-X),

where X(i)X^{(i)} denote XsX^{s} conditioned to have a crossing in the index ii. This gives that

Var(𝔼[XsX|X])=1m2(G)2i,jM2(G)Cov(X(i)X,X(j)X).\mathrm{Var}(\mathbb{E}[X^{s}-X|X])=\frac{1}{m_{2}(G)^{2}}\sum_{i,j\in M_{2}(G)}\mathrm{Cov}(X^{(i)}-X,X^{(j)}-X). (4.2)

We identify two kinds of terms in the summation of the covariance, when the indices satisfy V(i)V(j)V(i)\cap V(j)\neq\emptyset or when they satisfy V(i)V(j)=V(i)\cap V(j)=\emptyset, where V(i)V(i) denote the set of vertices of the 2-matching ii.

Case V(i)V(j)V(i)\cap V(j)\neq\emptyset: In this case, we have that

|{i,jM2(G):V(i)V(j)}|=m2(G)2(42)m4(G)=m2(G)26m4(G).|\{i,j\in M_{2}(G):V(i)\cap V(j)\neq\emptyset\}|=m_{2}(G)^{2}-\binom{4}{2}m_{4}(G)=m_{2}(G)^{2}-6m_{4}(G).

From the construction of the size bias transform, we have that |X(i)X|2Δ(m1)|X^{(i)}-X|\leq 2\Delta(m-1). Indeed, if XsX^{s} have a crossing in the matching ii, there are two options for the matching ii in XX, that is, it is a crossing or not. If ii is a crossing, then X(i)=XX^{(i)}=X, because we don’t need to crossing any edges. On the other hand, if ii is not a crossing of XX, then to obtain X(i)X^{(i)} it is necessary crossing the edges of the matching ii, which can be generate at least a new crossing for each of the edges incidents to the four vertices of ii. Then, an upper bound for the variance given by

Var(X(i)X)𝔼[(X(i)X)2]4Δ2(m1)2.\mathrm{Var}(X^{(i)}-X)\leq\mathbb{E}[(X^{(i)}-X)^{2}]\leq 4\Delta^{2}(m-1)^{2}. (4.3)

Then, the contribution in the sum (4.2) of the 2-matchings such that V(i)V(j)V(i)\cap V(j)\neq\emptyset is bounded by

m2(G)26m4(G)m2(G)24Δ2(m1)2=(16m4(G)m2(G)2)4Δ2(m1)2.\frac{m_{2}(G)^{2}-6m_{4}(G)}{m_{2}(G)^{2}}4\Delta^{2}(m-1)^{2}=\left(1-\frac{6m_{4}(G)}{m_{2}(G)^{2}}\right)4\Delta^{2}(m-1)^{2}.

Case V(i)V(j)=V(i)\cap V(j)=\emptyset: In this case, we have that

|{i,jM2(G):V(i)V(j)=}|=(42)m4(G)=6m4(G).|\{i,j\in M_{2}(G):V(i)\cap V(j)=\emptyset\}|=\binom{4}{2}m_{4}(G)=6m_{4}(G).

Let N(i)N(i) be the set of edges incidents to the vertices of the 2-matching ii. We can divide the sum over the 2-matching with V(i)V(j)=V(i)\cap V(j)=\emptyset. In the case of N(i)N(j)=N(i)\cap N(j)=\emptyset, we obtain that the random variables X(i)XX^{(i)}-X and X(i)XX^{(i)}-X are independent. Indeed, from the construction X(i)XX^{(i)}-X only depends on the edges incident to the 22-matching ii, and similarly X(j)XX^{(j)}-X only depends on the edges incident to the 22-matching jj. Hence, in this case we obtain that Cov(X(i)X,X(j)X)=0\mathrm{Cov}(X^{(i)}-X,X^{(j)}-X)=0.

So we are interested in the pairs of 22- matchings such that V(i)V(j)=V(i)\cap V(j)=\emptyset, but N(i)N(j)N(i)\cap N(j)\neq\emptyset. An upper bound for the number of such pairs of 22-matchings is given by m2(G)(Δ1)2(m4)/2m_{2}(G)(\Delta-1)^{2}(m-4)/2.

Indeed, in this case, there exists at least one edge between V(i)V(i) and V(j)V(j). So, to obtain such configuration one may proceed as follows. First, one chooses a 22-matching, ii, and one considers one of the 44 vertices in ii, say vv, and looks for a neighbour of vv, say ww, which should be in the 22-matching jj. There are at most Δ1\Delta-1 choices for ww. Now, to construct jj, we need to find a neighbour of ww which is not vv for one of the edges forming jj, giving at most Δ1\Delta-1 possibilities and another edge which cannot be in ii, or contain ww, giving at most m4m-4 possibilities. Putting this together and considering double counting, we obtain the desired m2(G)(Δ1)2(m4)/2m_{2}(G)(\Delta-1)^{2}(m-4)/2. See Figure 4 for a diagram explaining this counting.

wwiijj
Figure 4: A generic pair of 22-matchings sharing a vertex.

Finally, using the upper bound for the variance given in (4.3), we obtain

1m2(G)2i,jM2(G)V(i)V(j)=Cov(X(i)X,X(j)X)\displaystyle\frac{1}{m_{2}(G)^{2}}\sum_{\begin{subarray}{c}i,j\in M_{2}(G)\\ V(i)\cap V(j)=\emptyset\end{subarray}}\mathrm{Cov}(X^{(i)}-X,X^{(j)}-X) =1m2(G)2i,jM2(G)N(i)N(j)Cov(X(i)X,X(j)X)\displaystyle=\frac{1}{m_{2}(G)^{2}}\sum_{\begin{subarray}{c}i,j\in M_{2}(G)\\ N(i)\cap N(j)\neq\emptyset\end{subarray}}\mathrm{Cov}(X^{(i)}-X,X^{(j)}-X)
1m2(G)2i,jM2(G)N(i)N(j)4Δ2(m1)2\displaystyle\leq\frac{1}{m_{2}(G)^{2}}\sum_{\begin{subarray}{c}i,j\in M_{2}(G)\\ N(i)\cap N(j)\neq\emptyset\end{subarray}}4\Delta^{2}(m-1)^{2}
1m2(G)2Δ2(Δ1)2(m1)2(m4).\displaystyle\leq\frac{1}{m_{2}(G)}2\Delta^{2}(\Delta-1)^{2}(m-1)^{2}(m-4).

Thus, the contribution of the pairs of 22-matchings such that V(i)V(j)=V(i)\cap V(j)=\emptyset in the covariance sum (4.2) is bounded by

2Δ2(Δ1)2(m1)2(m4)m2(G).\frac{2\Delta^{2}(\Delta-1)^{2}(m-1)^{2}(m-4)}{m_{2}(G)}.

Therefore,

Var(𝔼[XsX|X])4Δ2(m1)2(16m4(G)m2(G)2+(Δ1)2(m4)2m2(G)).\mathrm{Var}(\mathbb{E}[X^{s}-X|X])\leq 4\Delta^{2}(m-1)^{2}\left(1-\frac{6m_{4}(G)}{m_{2}(G)^{2}}+\frac{(\Delta-1)^{2}(m-4)}{2m_{2}(G)}\right).

4.3 Kolmogorov Distance

Using the previous results, we are in position to apply Theorem 2. Therefore, we can obtain a bound for the Kolmogorov distance of the (normalized) number of crossings and a standard Gaussian random variable.

Theorem 3.

Let GG be a graph with maximum degree Δ\Delta, and let XX be the number of crossings of a random uniform embedding of GG. Let μ\mu and σ2\sigma^{2} the mean and the variance of XX. Then, with W=(Xμ)/σW=(X-\mu)/\sigma,

dKol(W,Z)4m2(G)Δm3σ2[6Δmσ+16m4(G)m2(G)2+(Δ1)2m2m2(G)],d_{Kol}(W,Z)\leq\frac{4m_{2}(G)\Delta m}{3\sigma^{2}}\left[\frac{6\Delta m}{\sigma}+\sqrt{1-\frac{6m_{4}(G)}{m_{2}(G)^{2}}+\frac{(\Delta-1)^{2}m}{2m_{2}(G)}}\right],

where mm is the number of edges of GG, mr(G)m_{r}(G) is the number of rr-matchings of GG, and ZZ is a standard Gaussian random variable.

Proof.

By Lemma 1 we have that μ=m2(G)/3\mu=m_{2}(G)/3, also by Lemma 3, Ψ\Psi defined in (2.2) is bounded as follows,

Ψ2Δm16m4(G)m2(G)2+(Δ1)2m2m2(G).\Psi\leq 2\Delta m\sqrt{1-\frac{6m_{4}(G)}{m_{2}(G)^{2}}+\frac{(\Delta-1)^{2}m}{2m_{2}(G)}}.

Then, using Theorem 2, and the fact that |XsX|A=2Δm|X^{s}-X|\leq A=2\Delta m, we obtain

dKol(W,Z)\displaystyle d_{Kol}(W,Z) 6μA2σ3+2μΨσ2\displaystyle\leq\frac{6\mu A^{2}}{\sigma^{3}}+\frac{2\mu\Psi}{\sigma^{2}}
8Δ2m2(G)m2σ3+4m2(G)Δm3σ216m4(G)m2(G)2+(Δ1)2m2m2(G)\displaystyle\leq\frac{8\Delta^{2}m_{2}(G)m^{2}}{\sigma^{3}}+\frac{4m_{2}(G)\Delta m}{3\sigma^{2}}\sqrt{1-\frac{6m_{4}(G)}{m_{2}(G)^{2}}+\frac{(\Delta-1)^{2}m}{2m_{2}(G)}}
=4m2(G)Δm3σ2(6Δmσ+16m4(G)m2(G)2+(Δ1)2m2m2(G))\displaystyle=\frac{4m_{2}(G)\Delta m}{3\sigma^{2}}\left(\frac{6\Delta m}{\sigma}+\sqrt{1-\frac{6m_{4}(G)}{m_{2}(G)^{2}}+\frac{(\Delta-1)^{2}m}{2m_{2}(G)}}\right)

5 Some Examples

In this section we provide various examples for which Theorem 3 can be applied directly. To show its easy applicability, we give explicit bounds on the quantities appearing in (1.1).

5.1 Pairing

Let GG be a pairing or matching graph on 2n2n vertices, that is, a disjoint union of nn copies of K2K_{2}, as in Example 1. In particular, m=nm=n, m2(G)=(n2)m_{2}(G)=\binom{n}{2} and m4(G)=(n4)m_{4}(G)=\binom{n}{4} and the variance is given by

σ2=n(n1)(n+3)45\sigma^{2}=\frac{n(n-1)(n+3)}{45}

which is bigger that n3/45n^{3}/45 for n>3n>3. On the other hand, since Δ=1\Delta=1, we see that, for n>3n>3,

16m4(G)m2(G)2+m(Δ1)22m2(G)=16m4(G)m2(G)2=16(n4)(n2)2=4n6n2n=4n2n2n<4n.1-\frac{6m_{4}(G)}{m_{2}(G)^{2}}+\frac{m(\Delta-1)^{2}}{2m_{2}(G)}=1-\frac{6m_{4}(G)}{m_{2}(G)^{2}}=1-\frac{6\binom{n}{4}}{\binom{n}{2}^{2}}=\frac{4n-6}{n^{2}-n}=\frac{4}{n}-\frac{2}{n^{2}-n}<\frac{4}{n}.

Thus,

dKol(W,Z)445n332n3(645nn3/2+2n)1268n.d_{Kol}(W,Z)\leq\frac{4\cdot 45n^{3}}{3\cdot 2n^{3}}\left(\frac{6\cdot\sqrt{45}n}{n^{3/2}}+\frac{2}{\sqrt{n}}\right)\leq\frac{1268}{\sqrt{n}}.

5.2 Path Graph

Let PnP_{n} be the path graph on nn vertices. From Example 2, m2(Pn)=(n22)m_{2}(P_{n})=\binom{n-2}{2} and m4(Pn)=(n44)m_{4}(P_{n})=\binom{n-4}{4}, and we obtain that

16m4(Pn)m2(Pn)2=16(n44)(n22)2=2(6n371n2+289n402)n410n3+37n260n+36=12n+o(n1)1-\frac{6m_{4}(P_{n})}{m_{2}(P_{n})^{2}}=1-\frac{6\binom{n-4}{4}}{\binom{n-2}{2}^{2}}=\frac{2(6n^{3}-71n^{2}+289n-402)}{n^{4}-10n^{3}+37n^{2}-60n+36}=\frac{12}{n}+o(n^{-1})

On the other hand, Δ=2\Delta=2, and then, one easily sees that,

4m2(Pn)Δm4n3,6Δm12n, and m(Δ1)22m2(Pn)=1(n2).4m_{2}(P_{n})\Delta m\leq 4n^{3},\qquad 6\Delta m\leq 12n,\quad\text{ and }\quad\frac{m(\Delta-1)^{2}}{2m_{2}(P_{n})}=\frac{1}{(n-2)}.

Finally, since the variance is given by σ2=n3/45n2/1811n/452/3>n3/60\sigma^{2}=n^{3}/45-n^{2}/18-11n/45-2/3>n^{3}/60 , for n14n\geq 14, we get

dKol(W,Z)460n33n3(1260nn3/2+4n)7757n.d_{Kol}(W,Z)\leq\frac{4\cdot 60n^{3}}{3n^{3}}\left(\frac{12\cdot\sqrt{60}n}{n^{3/2}}+\frac{4}{\sqrt{n}}\right)\leq\frac{7757}{\sqrt{n}}.

5.3 Cycle Graph

Let CnC_{n} be the cycle graph on nn vertices. In this case m=nm=n and Δ=2\Delta=2, and from Example 3, m2(Cn)=n(n3)2m_{2}(C_{n})=\frac{n(n-3)}{2} and m4(Cn)=n4(n53)m_{4}(C_{n})=\frac{n}{4}\binom{n-5}{3}, then

16m4(Cn)m2(Cn)2+(Δ1)2m2m2(Cn)=16n4(n53)(n(n3)2)2+nn(n3)=13n2101n+210n(n3)213n, for n5.1-\frac{6m_{4}(C_{n})}{m_{2}(C_{n})^{2}}+\frac{(\Delta-1)^{2}m}{2m_{2}(C_{n})}=1-\frac{6\frac{n}{4}\binom{n-5}{3}}{\left(\frac{n(n-3)}{2}\right)^{2}}+\frac{n}{n(n-3)}=\frac{13n^{2}-101n+210}{n(n-3)^{2}}\leq\frac{13}{n},\text{ for }n\geq 5.

Also, 4m2(Cn)Δm4n34m_{2}(C_{n})\Delta m\leq 4n^{3}, and 6Δm=12n6\Delta m=12n. Since the variance is σ2=n3/45n2/90n/3>n3/50,\sigma^{2}=n^{3}/45-n^{2}/90-n/3>n^{3}/50, for n15n\geq 15, we obtain

dKol(W,Z)450n33n3(1250nn3/2+13n)5898n.d_{Kol}(W,Z)\leq\frac{4\cdot 50n^{3}}{3n^{3}}\left(\frac{12\sqrt{50}n}{n^{3/2}}+\frac{\sqrt{13}}{\sqrt{n}}\right)\leq\frac{{5898}}{\sqrt{n}}.

5.4 Disjoint Union of Triangles

Consider nn copies of K3K_{3} and let GG be the disjoint union of them. From Example 4 we have that GG is a graph with 3n3n vertices, m=3nm=3n edges, maximum degree Δ=2\Delta=2, m2(G)=32(n22)m_{2}(G)=3^{2}\binom{n-2}{2} and m4(G)=34(n44)m_{4}(G)=3^{4}\binom{n-4}{4}, then

16m4(G)m2(G)2+(Δ1)2m2m2(G)=16(34(n4))(32(n2))2+3n232(n2)=13n183n(n1)133n.1-\frac{6m_{4}(G)}{m_{2}(G)^{2}}+\frac{(\Delta-1)^{2}m}{2m_{2}(G)}=1-\frac{6\left(3^{4}\binom{n}{4}\right)}{\left(3^{2}\binom{n}{2}\right)^{2}}+\frac{3n}{2\cdot 3^{2}\binom{n}{2}}=\frac{13n-18}{3n(n-1)}\leq\frac{13}{3n}.

On the other hand, we can obtain that, 4m2(G)Δm108n34m_{2}(G)\Delta m\leq 108n^{3} and 6Δm=36n6\Delta m=36n. Finally, since the variance is σ2=3n3/5+3n2/109n/10>3n3/5,\sigma^{2}=3n^{3}/5+3n^{2}/10-9n/10>3n^{3}/5, for n3n\geq 3, then we get

dKol(W,Z)1085n39n3(365n3n3/2+132n)2942n.d_{Kol}(W,Z)\leq\frac{108\cdot 5n^{3}}{9n^{3}}\left(\frac{36\sqrt{5}n}{\sqrt{3}n^{3/2}}+\frac{\sqrt{13}}{\sqrt{2n}}\right)\leq\frac{2942}{\sqrt{n}}.

6 Another Possible Limit

The following shows that not every sequence of graphs satisfies a central limit theorem for the number of crossings, even if the variance is not always 0 and that having m2m_{2} going to infinity is not enough. Moreover, it shows that we can have another type of limit for the number of crossings.

Consider the graph GnG_{n} which consists of a star graph with n1n-1 vertices for which an edge is attached at one of the leaves, as in Figure 5a.

Note that in this case m2=n3m_{2}=n-3 and m4=0m_{4}=0 and the only other term appearing in (3.2) is S7S_{7}, which for this graph equals (n32)\binom{n-3}{2} . This implies that

𝔼(X)=n33,𝔼(X2)=(n2)(n3)6,\mathbb{E}(X)=\frac{n-3}{3},\quad\mathbb{E}(X^{2})=\frac{(n-2)(n-3)}{6},

from where σ2=n(n3)/18\sigma^{2}=n(n-3)/18, 16m4/m22=11-6m_{4}/m_{2}^{2}=1 and

(Δ1)2m2m2=(n2)2(n1)2(n3)n22.\frac{(\Delta-1)^{2}m}{2m_{2}}=\frac{(n-2)^{2}(n-1)}{2(n-3)}\approx\frac{n^{2}}{2}.

Thus the right hand of (1.1) does not approximate 0 as nn\to\infty.

One can calculate explicitly the probability of having kk crossings. Indeed, lets us denote by v0v_{0} is the center and by vnv_{n}, the tail (the only vertex at distance 2 from v0v_{0}) and by vn1v_{n-1} the vertex which has v0v_{0} and vnv_{n}. The number of crossings in an embedding of GnG_{n} depends only on the position of this three vertices. More precisely, there will be exactly kk crossing if the following two conditions are satisfied (see Figure 5b for an example):

  1. 1.

    There are exactly kk and n2kn-2-k vertices in the two arcs that remain when removing vnv_{n} and vn1v_{n-1}.

  2. 2.

    v0v_{0} is in the arc with n2kn-2-k vertices.

a) G12G_{12}
v0v_{0}vn1v_{n-1}vnv_{n}b) G10G_{10}
Figure 5: a) G12G_{12} drawn without crossing (left). b) G10G_{10} drawn in convex position with 33 crossings formed by the edge (vn1,vn)(v_{n-1},v_{n}) and the edges (v0,vi)(v_{0},v_{i}) where viv_{i} are between vn1v_{n-1} and vnv_{n} in the cyclic order.

By a simple counting argument, since all permutations have the same probability of occurrence, one sees that such conditions are be satisfied with probability

(Xn=k)=2(n2k)(n1)(n2),k=0,,n2.\mathbb{P}(X_{n}=k)=\frac{2(n-2-k)}{(n-1)(n-2)},\qquad k=0,\ldots,n-2.

Finally, dividing by nn, the random variable Yn=Xn/nY_{n}=X_{n}/n satisfies that

(Yn=kn)=2(n2k)(n1)(n2)2n(1kn),k=0,,n2,\mathbb{P}\left(Y_{n}=\frac{k}{n}\right)=\frac{2(n-2-k)}{(n-1)(n-2)}\approx\frac{2}{n}\left(1-\frac{k}{n}\right),\qquad k=0,\ldots,n-2,

which implies that YnYY_{n}\to Y, weakly, where YY is a random variable supported on (0,1)(0,1) with density fY(x)=2(1x).f_{Y}(x)=2(1-x).

Acknowledgement

We would like to thank Professor Goldstein for pointing out the paper [5] and for various communications during the preparation of this paper. OA would like to thank Clemens Huemer for initial discussions on the topic that led to work in this problem.

Santiago Arenas-Velilla was supported by a scholarship from CONACYT.

Octavio Arizmendi was supported by CONACYT Grant CB-2017-2018-A1-S-9764.
[Uncaptioned image] This project has received funding from the European Union’s Horizon 2020 research and innovation programme under the Marie Skłodowska-Curie grant agreement No 734922.

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