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Thierry Monteil and Khaydar Nurligareev 11institutetext: LIPN, CNRS (UMR 7030), Université Paris 13, F-93430 Villetaneuse, France,
11email: [email protected],
https://monteil.perso.math.cnrs.fr/
22institutetext: LIPN, CNRS (UMR 7030), Université Paris 13, F-93430 Villetaneuse, France,
22email: [email protected],
https://lipn.univ-paris13.fr/~nurligareev/

Asymptotics for connected graphs and irreducible tournaments

Thierry Monteil 11    Khaydar Nurligareev 22
Abstract

We compute the whole asymptotic expansion of the probability that a large uniform labeled graph is connected, and of the probability that a large uniform labeled tournament is irreducible. In both cases, we provide a combinatorial interpretation of the involved coefficients.

1 Introduction

Let us consider the Erdös-Rényi model of random graphs G(n,1/2)G(n,1/2), where for each integer n0n\geqslant 0, we endow the set of undirected simple graphs on the set {1,,n}\{1,\ldots,n\} with the uniform probability: each graph appears with probability 1/2(n2)1/2^{n\choose 2}. The probability pnp_{n} that such a random graph of size nn is connected goes to 11 as nn goes to \infty. In 1959, Gilbert [2] provided a more accurate estimation and proved that

pn=12n2n+O(n223n/2).p_{n}=1-\dfrac{2n}{2^{n}}+O\left(\dfrac{n^{2}}{2^{3n/2}}\right).

In 1970, Wright [8] computed the first four terms of the asymptotic expansion of this probability:

pn=1(n1)12n12(n3)123n624(n4)124n10+O(n525n).p_{n}=1-{n\choose 1}\dfrac{1}{2^{n-1}}-2{n\choose 3}\dfrac{1}{2^{3n-6}}-24{n\choose 4}\dfrac{1}{2^{4n-10}}+O\left(\dfrac{n^{5}}{2^{5n}}\right).

The method can be used to compute more terms, one after another. However, it does not allow to provide the structure of the whole asymptotic expansion, since no interpretation is given to the coefficients 1,2,24,1,2,24,\dots.

The first goal of this paper is to provide such a structure: the kkth term of the asymptotic expansion of pnp_{n} is of the form

ik 2k(k+1)/2(nk)12kn,i_{k}\ 2^{k(k+1)/2}\ {n\choose k}\ \dfrac{1}{2^{kn}},

where iki_{k} counts the number of irreducible labeled tournaments of size kk. A tournament is said irreducible if for every partition ABA\sqcup{B} of the set of vertices there exist an edge from AA to BB and an edge from BB to AA. Equivalently, a tournament is irreducible if, and only if, it is strongly connected [6][7].

Theorem 1.1 (Connected graphs).

For any positive integer rr, the probability pnp_{n} that a random graph of size nn is connected satisfies

pn=1k=1r1ik(nk)2k(k+1)/22nk+O(nr2nr),p_{n}=1-\sum\limits_{k=1}^{r-1}i_{k}{n\choose k}\dfrac{2^{k(k+1)/2}}{2^{nk}}+O\left(\dfrac{n^{r}}{2^{nr}}\right),

where iki_{k} is the number of irreducible labeled tournaments of size kk.

In particular, as there are no irreducible tournament of size 22, this explains why there is no term in (n2)122n{n\choose 2}\dfrac{1}{2^{2n}} in Wright’s formula. This result might look surprising as it relates asymptotics of undirected objects with directed ones.

A similar development happened for irreducible tournaments. For n0n\geqslant 0, we endow the set of tournaments on the set {1,,n}\{1,\ldots,n\} with the uniform probability: each tournament appears with probability 1/2(n2)1/2^{n\choose 2}. In 1962, Moon and Moser [5] gave a first estimation of the probability qnq_{n} that a labeled tournament of size nn is irreducible, which was improved in [4] into

qn=1n2n2+O(n222n).q_{n}=1-\dfrac{n}{2^{n-2}}+O\left(\dfrac{n^{2}}{2^{2n}}\right).

In 1970, Wright [9] computed the first four terms of the asymptotic expansion of the probability that a labeled tournament is irreducible:

qn=1(n1)22n+(n2)242n(n3)283n(n4)2154n+O(n525n).q_{n}=1-{n\choose 1}2^{2-n}+{n\choose 2}2^{4-2n}-{n\choose 3}2^{8-3n}-{n\choose 4}2^{15-4n}+O\left(n^{5}2^{-5n}\right).

Here again, we provide the whole structure of the asymptotic expansion, together with a combinatorial interpretation of the coefficients (they are not all powers of two):

Theorem 1.2 (Irreducible tournaments).

For any positive integer rr, the probability qnq_{n} that a random labeled tournament of size nn is irreducible satisfies

qn=1k=1r1(2ikik(2))(nk)2k(k+1)/22nk+O(nr2nr),q_{n}=1-\sum\limits_{k=1}^{r-1}\big{(}2i_{k}-i_{k}^{(2)}\big{)}{n\choose k}\dfrac{2^{k(k+1)/2}}{2^{nk}}+O\left(\dfrac{n^{r}}{2^{nr}}\right),

where ik(2)i_{k}^{(2)} is the number of labeled tournaments of size kk with two irreducible components.

We can notice that the coefficients cannot be interpreted as counting a single class of combinatorial objects, since the coefficient 2i2i2(2)=022i_{2}-i_{2}^{(2)}=0-2 is negative.

2 Notations, strategy and tools

Let us denote, for every integer nn, gng_{n} the number of labeled graphs of size nn, cnc_{n} the number of connected labeled graphs of size nn, tnt_{n} the number of labeled tournaments of size nn, and ini_{n} the number of irreducible labeled tournaments of size nn. We have pn=cn/gnp_{n}=c_{n}/g_{n} and qn=in/tnq_{n}=i_{n}/t_{n}.

Looking for a proof of Theorem 1.1, we see two issues: finding a formal relation between connected graphs and irreducible tournaments, and proving the convergence. A tool to settle the first issue is the symbolic method: we associate to each integer sequence its exponential generating function:

G(z)=n=0gnznn!,C(z)=n=0cnznn!,T(z)=n=0tnznn!,I(z)=n=0inznn!.G(z)=\sum\limits_{n=0}^{\infty}g_{n}\dfrac{z^{n}}{n!},\ \ C(z)=\sum\limits_{n=0}^{\infty}c_{n}\dfrac{z^{n}}{n!},\ \ T(z)=\sum\limits_{n=0}^{\infty}t_{n}\dfrac{z^{n}}{n!},\ \ I(z)=\sum\limits_{n=0}^{\infty}i_{n}\dfrac{z^{n}}{n!}.

Since gn=tn=2(n2)g_{n}=t_{n}=2^{n\choose 2}, we have

G(z)=T(z)=n=02(n2)znn!.G(z)=T(z)=\sum\limits_{n=0}^{\infty}2^{n\choose 2}\dfrac{z^{n}}{n!}. (1)

Note that, while the number of labeled tournaments of size nn is equal to the number of labeled graphs of size nn, their associated species are not isomorphic: for n=2n=2, the two labeled tournaments are isomorphic (by swapping the vertices), while the two labeled graphs are not, so this equality is somewhat artificial.

Since every labeled graph can be uniquely decomposed as a disjoint union of connected labeled graphs, we have

G(z)=exp(C(z)).G(z)=\exp(C(z)). (2)

It remains to find a relation between tournaments and irreducible tournaments.

Lemma 2.1.

Any tournament can be uniquely decomposed into a sequence of irreducible tournaments.

In terms of generating functions, Lemma 2.1 translates to

T(z)=11I(z).T(z)=\dfrac{1}{1-I(z)}. (3)

Hence, part of the work will be to let those expressions interplay.

Regarding asymptotics, we will rely on Bender’s Theorem [1]:

Theorem 2.2 (Bender).

Consider a formal power series

A(z)=n=1anznA(z)=\sum\limits_{n=1}^{\infty}a_{n}z^{n}

and a function F(x,y)F(x,y) which is analytic in some neighborhood of (0,0)(0,0). Define

B(z)=n=1bnzn=F(z,A(z))andD(z)=n=1dnzn=Fy(z,A(z)),B(z)=\sum\limits_{n=1}^{\infty}b_{n}z^{n}=F(z,A(z))\qquad\mbox{and}\qquad D(z)=\sum\limits_{n=1}^{\infty}d_{n}z^{n}=\dfrac{\partial F}{\partial y}(z,A(z)),

Assume that an0a_{n}\neq 0 for all nn\in\mathbb{N}, and that for some integer r1r\geqslant 1 we have

 (i) an1an0 as n; (ii) k=rnr|akank|=O(anr) as n.\mbox{ (i) }\quad\dfrac{a_{n-1}}{a_{n}}\to 0\,\mbox{ as }\,n\to\infty;\qquad\qquad\mbox{ (ii) }\quad\sum\limits_{k=r}^{n-r}|a_{k}a_{n-k}|=O(a_{n-r})\,\mbox{ as }\,n\to\infty.

Then

bn=k=0r1dkank+O(anr).b_{n}=\sum\limits_{k=0}^{r-1}d_{k}a_{n-k}+O(a_{n-r}).

3 Proofs

Proof 3.1 (Proof of Lemma 2.1).

Let TT be a tournament. It is either irreducible, and all is done, or it consists of two nonempty parts AA and BB such that all edges between AA and BB are directed from AA to BB. Applying the same argumentation recursively to AA and BB, we obtain a decomposition of TT into a sequence of subtournaments T1,,TkT_{1},\ldots,T_{k}, such that each TiT_{i} is irreducible and for every pair i<ji<j, all edges go from TiT_{i} to TjT_{j} (see Figure 1). Since TiT_{i} are also the strongly connected components of TT, the decomposition is unique.

112233445566\leadsto445533112266
Figure 1: Decomposition of a tournament as a sequence of irreducible components.
Proof 3.2 (Proof of Theorem 1.1).

Let us apply Bender’s theorem (Theorem 2.2) the following way: take

A(z)=G(z)1andF(z,w)=ln(1+w).A(z)=G(z)-1\qquad\qquad\mbox{and}\qquad\qquad F(z,w)=\ln(1+w).

Then, in accordance with formulas (1), (2) and (3),

B(z)=ln(G(z))=C(z)andD(z)=1G(z)=1T(z)=1I(z).B(z)=\ln(G(z))=C(z)\qquad\mbox{and}\qquad D(z)=\dfrac{1}{G(z)}=\dfrac{1}{T(z)}=1-I(z).

Check the conditions of Theorem 2.2. In the case at hand, condition (i) has the form:

an1an=2(n12)(n1)!n!2(n2)=n2n10,asn.\dfrac{a_{n-1}}{a_{n}}=\dfrac{2^{n-1\choose 2}}{(n-1)!}\dfrac{n!}{2^{n\choose 2}}=\dfrac{n}{2^{n-1}}\to 0,\qquad\mbox{as}\quad n\to\infty.

To establish condition (ii), consider xk=n!akank=(nk)2(k2)+(nk2)x_{k}=n!a_{k}a_{n-k}={n\choose k}2^{{k\choose 2}+{n-k\choose 2}}, where rknrr\leqslant k\leqslant n-r. Then (xk)(x_{k}) decreases for rkn/2r\leqslant k\leqslant n/2 and increases symmetrically for n/2knrn/2\leqslant k\leqslant n-r. Bounding each summand by the first term is not enough, but bounding each summand (except for the first and last) by the second term gives the following:

k=rnrakank1n!(nr)2(r2)+(nr2)+1+n2r1n!(nr+1)2(r+12)+(nr12)=O(2(n2(2r+1)n)/2(nr)!)+O(2(n2(2r+3)n)/2(nr2)!)=O(anr).\begin{split}\sum\limits_{k=r}^{n-r}a_{k}a_{n-k}&\leqslant\dfrac{1}{n!}{n\choose r}2^{{r\choose 2}+{n-r\choose 2}+1}+\dfrac{n-2r-1}{n!}{n\choose r+1}2^{{r+1\choose 2}+{n-r-1\choose 2}}\\ &=O\left(\dfrac{2^{\big{(}n^{2}-(2r+1)n\big{)}/2}}{(n-r)!}\right)+O\left(\dfrac{2^{\big{(}n^{2}-(2r+3)n\big{)}/2}}{(n-r-2)!}\right)=O(a_{n-r}).\end{split}

Hence, Bender’s theorem implies

bn=cnn!=2(n2)n!k=1r1ikk!2(nk2)(nk)!+O(2(nr2)(nr)!).b_{n}=\dfrac{c_{n}}{n!}=\dfrac{2^{n\choose 2}}{n!}-\sum\limits_{k=1}^{r-1}\dfrac{i_{k}}{k!}\dfrac{2^{n-k\choose 2}}{(n-k)!}+O\left(\dfrac{2^{n-r\choose 2}}{(n-r)!}\right).

Dividing by gn/n!=2(n2)/n!g_{n}/n!=2^{n\choose 2}/n!, we get

pn=cngn=1k=1r1ik(nk)2(nk2)2(n2)+O(nr2nr).p_{n}=\dfrac{c_{n}}{g_{n}}=1-\sum\limits_{k=1}^{r-1}i_{k}{n\choose k}\dfrac{2^{n-k\choose 2}}{2^{n\choose 2}}+O\left(\dfrac{n^{r}}{2^{nr}}\right).
Proof 3.3 (Proof of Theorem 1.2).

Let us apply Bender’s theorem (Theorem 2.2) for

A(z)=T(z)1andF(z,w)=11+w.A(z)=T(z)-1\qquad\qquad\mbox{and}\qquad\qquad F(z,w)=-\dfrac{1}{1+w}.

Then, in accordance with formula (3),

B(z)=1T(z)=1+I(z)andD(z)=1(T(z))2=(1I(z))2.B(z)=-\dfrac{1}{T(z)}=-1+I(z)\qquad\qquad\mbox{and}\qquad\qquad D(z)=\dfrac{1}{\big{(}T(z)\big{)}^{2}}=\big{(}1-I(z)\big{)}^{2}.

Since (I(z))2\big{(}I(z)\big{)}^{2} is the generating function for the class of labeled tournaments which can be decomposed into a sequence of two irreducible tournaments, we can rewrite the latter identity in the form

D(z)=1n=1(2ikik(2))znn!.D(z)=1-\sum\limits_{n=1}^{\infty}\big{(}2i_{k}-i_{k}^{(2)}\big{)}\dfrac{z^{n}}{n!}.

In the case at hand, the conditions that are needed to apply Theorem 2.2 are the same as in the proof of Theorem 1.1, since the sequence (an)(a_{n}) is the same. Hence,

bn=inn!=2(n2)n!k=1r12ikik(2)k!2(nk2)(nk)!+O(2(nr2)(nr)!).b_{n}=\dfrac{i_{n}}{n!}=\dfrac{2^{n\choose 2}}{n!}-\sum\limits_{k=1}^{r-1}\dfrac{2i_{k}-i_{k}^{(2)}}{k!}\dfrac{2^{n-k\choose 2}}{(n-k)!}+O\left(\dfrac{2^{n-r\choose 2}}{(n-r)!}\right).

Dividing by tn/n!=2(n2)/n!t_{n}/n!=2^{n\choose 2}/n!, we get

qn=intn=1k=1r1(2ikik(2))(nk)2(nk2)2(n2)+O(nr2nr).q_{n}=\dfrac{i_{n}}{t_{n}}=1-\sum\limits_{k=1}^{r-1}\big{(}2i_{k}-i_{k}^{(2)}\big{)}{n\choose k}\dfrac{2^{n-k\choose 2}}{2^{n\choose 2}}+O\left(\dfrac{n^{r}}{2^{nr}}\right).

4 Further results

With a bit more work, we can compute the probability that a random graph of size nn has exactly mm connected components, and the probability that a random tournament of size nn has exactly mm irreducible components as nn goes to \infty.

In another direction, we can also generalize Theorem 1.1 to the Erdös-Rényi model G(n,p)G(n,p), where the constant 22 in the formulas is replaced by ρ=1/(1p)\rho=1/(1-p) and the sequence (ik)(i_{k}) is replaced by a sequence of polynomials (Pk(ρ))=1,ρ2,ρ36ρ+6,ρ68ρ36ρ2+36ρ24,(P_{k}(\rho))=1,\rho-2,\rho^{3}-6\rho+6,\rho^{6}-8\rho^{3}-6\rho^{2}+36\rho-24,\dots with an explicit combinatorial interpretation.

The methods presented here can also be extended to some geometrical context where connectedness questions appear. In particular, we will provide asymptotics for combinatorial maps, square tiled surfaces, constellations, random tensor model [3]. In some of the models, the coefficients in the asymptotic expansions show connections with indecomposable tuples of permutations and perfect matchings.

References

  • [1] E. A. Bender. An asymptotic expansion for the coefficients of some formal power series. J. Lond. Math. Soc., II. Ser., 9:451–458, 1975.
  • [2] E. N. Gilbert. Random graphs. Ann. Math. Stat., 30:1141–1144, 1959.
  • [3] T. Monteil and Kh. Nurligareev. In preparation.
  • [4] J. W. Moon. Topics on tournaments. New York etc: Holt, Rinehart and Winston VIII,104 p 56 s, (1968)., 1968.
  • [5] J. W. Moon and L. Moser. Almost all tournaments are irreducible. Can. Math. Bull., 5:61–65, 1962.
  • [6] R. Rado. Theorems on linear combinatorial topology and general measure. Ann. Math. (2), 44:228–270, 1943.
  • [7] B. Roy. Sur quelques propriétés des graphes fortement connexes. C. R. Acad. Sci., Paris, 247:399–401, 1958.
  • [8] E. M. Wright. Asymptotic relations between enumerative functions in graph theory. Proc. Lond. Math. Soc. (3), 20:558–572, 1970.
  • [9] E. M. Wright. The number of irreducible tournaments. Glasg. Math. J., 11:97–101, 1970.