Asymptotics of Largest Components in Combinatorial Structures

Abstract

Given integers m and n, we study the probability that structures of size n have all components of size at most m. The results are given in term of a generalized Dickman function of n/m. 1. Introduction. We consider a class of decomposable combinatorial objects A such that each object has a unique decomposition over a sub-class of the original class C, called irreducible or connected components. We examine structures such as general graphs or graphs with certain properties and their components, monic polynomials over finite fields viewed as the product of irreducible polynomials, permutations viewed as set of cycles, and so on. There is a notion of size related to these combinatorial objects and their components. We let Ai denote the number of structures of size i and let Ci be the number of those that are connected. We let A(x) denote the generating function for the objects and C(x) denote the generating function for the connected objects or components. It is well known that if the objects are labelled one uses the exponential generating function and A(x) = exp(C(x)), and furthermore if the objects are unlabelled the ordinary generating function is appro- priate and

Full text

DOI: 10.1007/s00453-006-0103-y
Algorithmica (2006) 46: 493–503 Algorithmica
©2006 Springer Science+Business Media, Inc.
Asymptotics of Largest Components
in Combinatorial Structures
Mohamed Omar,1Daniel Panario,2Bruce Richmond,1and Jacki Whitely3
Abstract. Given integers mand n, we study the probability that structures of size nhave all components of
size at most m. The results are given in term of a generalized Dickman function of n/m.
Key Words. Decomposable combinatorial structures, Largest components, Permutations, Polynomials over
finite fields, Dickman function.
1. Introduction. We consider a class of decomposable combinatorial objects Asuch
that each object has a unique decomposition over a sub-class of the original class C, called
irreducible or connected components. We examine structures such as general graphs or
graphs with certain properties and their components, monic polynomials over finite fields
viewed as the product of irreducible polynomials, permutations viewed as set of cycles,
and so on.
There is a notion of size related to these combinatorial objects and their components.
We let Aidenote the number of structures of size iand let Cibe the number of those
that are connected. We let A(x)denote the generating function for the objects and C(x)
denote the generating function for the connected objects or components. It is well known
that if the objects are labelled one uses the exponential generating function and
A(x)=exp(C(x)),
and furthermore if the objects are unlabelled the ordinary generating function is appro-
priate and
A(x)=exp 
k≥1
C(xk)
k.
We let An,mbe the number of structures of size nwhose largest component has size
at most m. For a general discussion over both labelled and unlabelled structures let
an=


An
n!,if the structures are labelled,
An,if the structures are unlabelled,
1Department of Combinatorics and Optimization, University of Waterloo, Waterloo, Ontario, Canada N2L
3G1. momar@student.math.uwaterloo.ca; lbrichmond@math.uwaterloo.ca.
2School of Mathematics and Statistics, Carleton University, Ottawa, Ontario, Canada K1S 5B6.
daniel@math.carleton.ca.
31944 Saint Margaret’s Bay Road, Timberlea, Nova Scotia, Canada B3T 1C3. jackiw@gmail.com.
Received November 30, 2004; revised January 19, 2006. Communicated by P. Jacquet, D. Panario, and
W. Szpankowski. Online publication October 28, 2006.

494 M. Omar, D. Panario, B. Richmond, and J. Whitely
cn=


Cn
n!in the labelled case,
Cn,in the unlabelled case,
an,m=


An,m
n!,in the labelled case,
An,m,in the unlabelled case.
We suppose throughout this paper that the objects satisfy the following asymptotic
conditions:
ci∼a
iRi,ai∼ia−1KRi,
where a,K, and Rare positive constants. These conditions are satisfied by many struc-
tures, the best known labelled example is permutations (the components are cycles), and
the best known unlabelled example is polynomials over a finite field (the components
are the irreducible polynomials). See Section 3 for a discussion of these examples and
others. We also suppose that m≥εnfor εa positive constant in our main theorem
(Theorem 1 below).
In Section 2 we first show that under certain restrictions almost all objects of size n
have only one component of size iwhen nand itend to infinity (Lemma 1). We then
study the proportion of objects of size nwith the largest component equal to mwhen
m≥εn, for εa positive constant (Theorem 1). Our results are expressed in terms of
a generalized Dickman function ρa(u)(the Dickman function corresponds to ρ1(u)).
This problem has been considered by Gourdon [5]. In the range of applicability of our
theorem, our results simplify and extend Gourdon’s results. Our proof is elementary and
in line with the Bender et al. [1] study of smallest components. We also give bounds
for the generalized Dickman function (Theorem 2). Finally, in Section 3 we give several
examples, including one where Gourdon’s results do not apply.
2. Results and Proofs. We start by giving the following auxiliary result, of indepen-
dent interest, which covers many important combinatorial problems like permutations,
polynomials over finite fields, and square-free polynomials over finite fields.
LEMMA 1. Suppose a =1. Almost all objects of size n have at most one component
of size i as n,i→∞.Moreover,the fraction with more than one component of size i
is O(i−2).
PROOF. In the labelled case the number of objects, An,i, with no components of size i
has the exponential generating function
exp ∞

=1
cx−cixi=e−cixiA(x).
So
An,i
n!=an−an−ici+an−2ic2
i
1
2! −an−3ic3
i
1
3! +···.

Asymptotics of Largest Components in Combinatorial Structures 495
As n−i→∞,wehavefor≥3,
an−ic
i
!∼KRn−i1
iRi
!∼an(1/i)
!
≤an(1/i)3(1/i)−3
( −3)!.
So as i→∞,
¯
An,i
n!=KRn−KRn1
i+1
2! 1
i2
KRn−···
=an1−1
i+1
2! 1
i2
+Oe1/i
i3
=an1−1
i+O1
i2+Oe1/i
i3.
The number of objects of size nwith two or more components of size iis smaller than
or equal to
n
iCin−i
iCiAn−2i/2∼n!c2
ian−2i
2
∼n!K1
i2
R2iRn−2i/2∼n!KRn
2i2=n!anO(i−2).
If n−iis bounded, we must replace the asymptotic equalities in the last two equations
by O-estimates; the arguments then hold. It follows that the number of objects of size n
with one component of size iis
an1
i+O(i−2)=an
1
i1+O1
i=cn1+O1
i.
The unlabelled case is similar; the radius of convergence of ∞
k=2C(xk)/kis easily
seen to be √R, so from Schur’s lemma (see page 62 of [2]), we have
[xn]A(x)=[xn]exp(C(x)) ·exp 
k≥2
C(xk)
k∼α·[xn]exp(C(x)),
where αis a constant. We may now follow the proof for the labelled case, the difference
is only the factor α.
This lemma is one way to see that frequently the combinatorial objects we are con-
sidering behave like the primes even though ciis large while there is at most one prime
of size i.

496 M. Omar, D. Panario, B. Richmond, and J. Whitely
We now suppose m≥εn, for a positive number ε. The proof of our main result uses
a modification of the Buchstab Identity (see [6]). Let
ψ(x,y)=|{n≤x,where the largest prime factor of xis ≤y}|.
Buchstab’s identity is defined as
ψ(x,y)=ψ(x,z)−
y<p≤z
ψx
p,p,
where the summation is over primes. We use the following modifications in the labelled
case since n
iAn−i,iCiovercounts structures with largest components:
An,m≥An,M−
m<i≤Mn
iAn−i,iCi=An,M−
m<i≤M
n!an−i,ici
or
an,m≥an,M−
m<i≤M
an−i,ici.
Furthermore,
An,m≤An,M−
m<i≤Mn
iCi¯
An−i,i+1
since this sum only counts structures with exactly one component of size i. Thus
an,m≤an,M−
m<i≤M
cian−i,i+11+O1
i.
(These inequalities for an,mhold in both the labelled and unlabelled cases.)
We wish, by analogy with the integers ≤nwhich have largest prime factor ≤m,to
define a function ρa(n/m)so that
an,m∼anρan
m+O1
m.(1)
Thus ρa(n/m)is the probability that the largest component of the structures of size n
is ≤m. It is well known in number theory that the probability that an integer ≤nhas
largest prime factor ≤mis ∼ρ1(log n/log m)where ρa(u)is defined next.
DEFINITION 1. Let ρa(u)=1 for 0 <u≤1. We recursively define ρa(u)for k<u≤
k+1, k≥1by
ρa(u)=ρa(k)−au
k
ρa(v −1)(v −1)a−1
vadv.

Asymptotics of Largest Components in Combinatorial Structures 497
We now derive a few properties of ρa(u)which satisfies the differential-difference
equation
uaρ
a(u)+aρa(u−1)(u−1)a−1=0,
so
uaρa(u)=aua−1ρa(u)−a(u−1)a−1ρa(u−1),
and we have
uaρa(u)=au
u−1
va−1ρa(v) dv.(2)
It now follows that ρa(u)>0, and hence ρ
a(u)<0.Furthermore, for u>1,
ρa(u+δ) −ρa(u)=−au+δ
u
ρa(v −1)(v −1)a−1
vadv,
and since ρa(u)is decreasing and <1wehaveforδ≥0,
0≥ρa(u+δ) −ρa(u)≥−au+δ
u
(1−v−1)a−1
vdv.
It follows that for u≥2 there is an absolute constant Cso that
|ρa(u+δ) −ρa(u)|≤C|δ|.(3)
Note that ρais infinitely differentiable except at u=1 where its right and left derivatives
of all orders exist.
Furthermore, if a=1 then for 1 ≤u<2, ρ1(u)=1−ln u, for 2 ≤u<3,
ρ1(u)=1−ln 2 −u
2
1−ln v
vdv
and so on. The case a=1 gives the classical Dickman’s function (for example, see [6]).
THEOREM 1. Let m ≥εn,where εis a positive constant.Let ρa(u)be defined as in
Definition 1. Then
an,m
an=ρan
m+O1
m.
PROOF. The proof that (1) holds is an induction on kwhere
n
k+1<m≤n
k.
If m>nthen an,m=anso
an,m
an=1=ρn
m,(4)

498 M. Omar, D. Panario, B. Richmond, and J. Whitely
for m>n, that is, n/m<1. If n/2<m≤n, that is, 1 ≤n/m<2, then
an,m≥an,n−
m<i≤n
an−i,ici
and
an,m≤an,n−
m<i≤n
an−i,i+1ci.(5)
Now n−i<n/2so(n−i)/i=−1+n/i≤1. Also (n−i)/(i+1)<(n−i)/i≤1,
so an,m
an≥1−
m<i≤n
an−i,i
an
ci
and by (1) this is equal to
1−
m<i≤n
ρan
i−1cian−i
an1+O1
i.
Our assumptions on anand ciimply that this is equal to
1−a
m<i≤n
ρn
i−1(1−i/n)a−1
i1+O1
i.
We estimate this sum by an integral. We may use the estimate derived in [1] to do this or
we may routinely apply the Euler–Maclaurin summation estimates. Either way the error
in estimating the sum by an integral is O(1/m).Wefind

m<i≤n
ρan
i−1(1−i/n)a−1
i=n
m
ρan
x−1(1−x/n)a−1
xdx +O1
m.(6)
If we let v=n/xor x=n/v,wefinddv=−n/x2dx,dx =−x2/ndv=−ndv/v2
so this is
1
n/m
ρa(v −1)1−1
va−1v
n−n2
v21
ndv+O1
m
=−
1
n/m
ρa(v −1)(v −1)a−1
vadv+O1
m
=n/m
1
ρa(v −1)(v −1)a−1
vadv+O1
m.
Now since n/m<2, 0 <v−1<1, ρa(v −1)≤1, and v−a≤1 so this expression is
smaller than 1 +O(1/m).
Thus using (6) in our first inequality for an,m/angives
an,m
an≥1−an/m
1
ρa(v −1)(v −1)a−1
vadv+O1
m.

Asymptotics of Largest Components in Combinatorial Structures 499
Furthermore, from the second inequality, (5), we get
an,m
an≤1−
m<i≤n
ρan
i+1−1+1
i+1(n−i)a−1
na−1
a
i.
Now n/i−n/(i+1)=n/i(i+1)≤n/(εn)(εn+1)≤(1/ε)/m. Thus n/(i+1)−1+
1/(i+1)=n/i−1+O(1/m)and from (3) it follows that ρa(n/(i+1)−1+1/(i+1)) =
ρa(−1+n/i)+O(1/m)and we can apply the same reasoning with an extra term O(1/m)
in (6). Thus
an,m
an=1−an/m
1
ρa(v −1)(v −1)a−1
vadv+O1
m.
We have our asymptotic estimate for n/2<m≤n. If we suppose (1) holds for k−1<
n/m≤kthen since (n−i)/i<k+1−1=kand since (n−i)/(i+1)<kthe
argument above for the case k=2 is easily modified to apply and one obtains that for
n/(k+1)<m<n/k,
an,m
an=ρa(k)−an/m
k
ρa(v −1)(v −1)a−1
vadv+O1
m.
Thus by induction (1) holds for m≥εn.
REMARK 1. When a=1, ρ1(u)is the original Dickman function.
REMARK 2. Theorem 1 may be compared with one of Gourdon’s results in his Ph.D.
thesis on the largest component of a random structure [5]. Our assumptions here are
slightly weaker as we show by example later, and so we extend his results. However,
when the Flajolet–Odlyzko singularity analysis applies, we have the same results as
Gourdon. On the other hand, our elementary results only hold for m≥εnwhereas
Gourdon proves that
an,m
an=ρan
m+O1
mfor 2 ≤m≤n.
Gourdon does not determine the asymptotic behaviour of ρa(n/m). However, ρ1(u)
is the classical Dickman function and it is well known that ρ1(u)=O(1/(u)).It
follows that when a=1, Gourdon’s leading term is large compared with O(1/m)when
m>n/(log n). Our result only holds for m>εn, of course, but perhaps it is surprising
that our elementary result is so close to the best known result as far as the range of mis
concerned. An analysis similar to that of Tenenbaum [6] gives the asymptotic behaviour
of ρa(u)for all a. As far as we know, such an analysis has not been published and it
would be much more intricate than the one we now give.
We now bound ρa(u).
THEOREM 2. There is a constant C1such that for u >1and a <1,
ρa(u)≤C1au,

500 M. Omar, D. Panario, B. Richmond, and J. Whitely
while for a ≥1there is a constant C2so that
ρa(u)≤Cu
2
(u+1).
PROOF. Let u>1. If 0 <a<1, (u−1)a−1≥ua−1so (u−1)a−1ρa(u−1)≥
ua−1ρa(u−1)≥ua−1ρa(u). We have already seen that
uaρa(u)=aua−1ρa(u)−a(u−1)a−1ρa(u−1)
so uaρa(u)is a decreasing function. Furthermore, from (2)
uaρa(u)=au
u−1
va−1ρa(v) dv=au
u−1
1
vvaρa(v) dv
<au
u−1
vaρa(v) dv<a(u−1)aρa(u−1),
so
ρa(u)< a
ua(u−1)aρa(u−1).
Therefore, it follows by induction that
ρa(u)≤C1au,
for some constant C1and the first part of the theorem is proved.
Suppose now a≥1. We have
uaρa(u)≤ρa(u−1)u
u−1
ava−1dv
=ρa(u−1)(ua−(u−1)a)≤ρa(u−1)C2ua−1,
where C2is a constant, so
ρa(u)≤C2
uρa(u−1).
By induction therefore
ρa(u)≤Cu
2
(u+1)
and the second part of the theorem is proved.
3. Examples. The following examples are from [4] and appear also in [5].
•Permutations. The exponential generating function for cycles in permutations is
C(x)=log(1/(1−x)). Therefore Cn=(n−1)! and hence cn=(n−1)!/n!=1/n.
On the other hand, the radius of convergence of the function log(1/(1−x)) is 1.
Consequently, the theorem applies with a=K=1. This gives rise to ρ1.

Asymptotics of Largest Components in Combinatorial Structures 501
•Polynomials. The ordinary generating function for monic polynomials over a finite
field Fqis 1/(1−qx)and we have the well-known approximation for Cn, the number
of irreducible polynomials of degree n,
Cn=qn
n+O(qn/2).
From this approximation we can find R, the radius of convergence of C(x), as follows:
R=lim sup
n→∞ |Cn|−1/n=lim sup
n→∞ 
qn
n+O(qn/2)
−1/n
=q−1.
From our definition cn=Cn=qn/n+O(qn/2)in the case of unlabelled objects. We
now compute the term 1/(nRn):
1
nRn=R−1n
n=qn
n.
Hence a=K=1 and the theorem applies. The case of squarefree polynomials has
the same asymptotics for an,m/an.
•2-regular graphs. The exponential generating function for labelled 2-regular graphs
is
e−x/2−x2/4
(1−x)1/2=exp 1
2log 1
1−x−x
2−x2
4.
Hence the exponential generating function for the components is
1
2log 1
1−x−x
2−x2
4.
Extracting coefficients yield in C1=C2=0 and
Cn=(n−1)!
2.
Hence cn=1/2nand this gives rise to ρ1/2.
The unlabelled case has the same asymptotics for an,m/an.
•Labelled functional digraphs.IfT(x)denotes the exponential generating function
for labelled rooted trees (defined by T(x)=xexp(T(x))), then we have for cycles
of labelled trees the generating function
log 1
1−T(x)=1
2log 1
1−ex +O(√1−ex).
This is the generating function for labelled funtional digraphs. This gives an example
with a=1
2,K=1.
•Unlabelled functional digraphs. This is also an example with a=1
2,K=1. If
one lets A(x)denote the ordinary generating function for unlabelled rooted trees then
A(x)=xexp(A(xk/k)) and the generating function for cycles of these trees is
∞

k=1
ϕ(k)
klog 1
1−A(xk),

502 M. Omar, D. Panario, B. Richmond, and J. Whitely
where ϕ(k)is Euler’s function. This is the generating function for unlabelled functional
digraphs.
•Coloured permutations. Now we give an example where Gourdon’s results do not
apply but ours do. Colour the integers {1,2,...,n}red and green and consider per-
mutations of them where the colours do not change under the mapping. We can think
of these permutations as being composed of red and green cycles. Let us add the
following refinement; consider a cycle
12··· n
i1i2··· 1,
where as usual is mapped to iand add the restriction that i1=2. There are
clearly (n−2)! such cycles. Let us now consider permutations such that when n=
4,8,...,2i,...,i≥2, no cycles
12··· n
2i2··· 1,
are allowed. Let cndenote the number of such cycles and let anbe the number of
permutations constructed from these cycles. Then
∞

n
anxn=exp 
n≥0
cnxn
n!=exp 2
i≥1
xi
i−2
i≥2
x2i
2i(2i−1).
Clearly ci∼2(i−1)!. We have that
ai∼iexp −2
i≥2
1
2i(2i−1)=Ci,
where Cis a constant. We have
∞

n=0
anxn=1
(1−x)2exp −2
i≥2
x2i
2i(2i−1).
An elementary Abelian argument shows that
∞

n=0
anxn∼C
(1−x)2as x→1.
Now
exp(−2i≥1(x2i/2i(2i−1)))
1−x=exp 
i≥1
xi
i−2
i≥2
x2i
2i(2i−1)=
i≥2
bixi,
where bi≥0. Thus the anabove are monotonic increasing. From the Hardy–
Karamata–Littlewood Tauberian theorem [3]
an∼Cn.

Asymptotics of Largest Components in Combinatorial Structures 503
We observe that singularity analysis does not apply since

i≥2
x2i
2i(2i−1)
is a lacunary series, so |x|=1 is a natural boundary for it and therefore for ∞
n=0anxn.
We have that k-coloured permutations give examples where a=k,K=1/(k−1)!.
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