Exact Largest and Smallest Size of Components
ABSTRACT Golomb and Gaal [15] study the number of permutations on n objects with largest cycle length equal to k . They give explicit expressions on ranges n/(i+1) < k ≤ n/i for i=1,2, \ldots, derive a general recurrence for the number of permutations of size n with largest cycle length equal to k , and provide the contribution of the ranges (n/(i+1),n/i] for i=1,2,\ldots, to the expected length of the largest cycle.
We view a cycle of a permutation as a component. We provide exact counts for the number of decomposable combinatorial structures
with largest and smallest components of a given size. These structures include permutations, polynomials over finite fields,
and graphs among many others (in both the labelled and unlabelled cases). The contribution of the ranges (n/(i+1),n/i] for i=1,2,\ldots, to the expected length of the smallest and largest component is also studied.
Key words. Largest and smallest components, Random decomposable combinatorial structures, Exponential class.
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Conference Paper: Probabilistic algorithms in finite fields
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ABSTRACT: Not AvailableFoundations of Computer Science, 1981. SFCS '81. 22nd Annual Symposium on; 11/1981  Discrete Mathematics  DM. 01/1975;
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ABSTRACT: This paper begins with the observation that half of all graphs containing no induced path of length 3 are disconnected. We generalize this in several directions. First, we give necessary and sufficient conditions (in terms of generating functions) for the probability of connectedness in a suitable class of graphs to tend to a limit strictly between zero and one. Next we give a general framework in which this and related questions can be posed, involving operations on classes of finite structures. Finally, we discuss briefly an algebra associated with such a class of structures, and give a conjecture about its structure. 1 1 Introduction The class of graphs containing no induced path of length 3 has many remarkable properties, stemming from the following wellknown observation. Recall that an induced subgraph of a graph consists of a subset S of the vertex set together with all edges contained in S. Proposition 1.1 Let G be a finite graph with more than one vertex, containin...Combinatorics Probability and Computing 06/1998; · 0.61 Impact Factor
Page 1
DOI: 10.1007/s0045300100471
Algorithmica (2001) 31: 413–432
Algorithmica
© 2001 SpringerVerlag New York Inc.
Exact Largest and Smallest Size of Components1
D. Panario2and B. Richmond3
Abstract.
equal to k. They give explicit expressions on ranges n/(i + 1) < k ≤ n/i for i = 1,2,..., derive a general
recurrence for the number of permutations of size n with largest cycle length equal to k, and provide the
contribution of the ranges (n/(i + 1),n/i] for i = 1,2,..., to the expected length of the largest cycle.
Weviewacycleofapermutationasacomponent.Weprovideexactcountsforthenumberofdecomposable
combinatorial structures with largest and smallest components of a given size. These structures include per
mutations, polynomials over finite fields, and graphs among many others (in both the labelled and unlabelled
cases). The contribution of the ranges (n/(i +1),n/i] for i = 1,2,..., to the expected length of the smallest
and largest component is also studied.
Golomb and Gaal [15] study the number of permutations on n objects with largest cycle length
KeyWords.
class.
Largestandsmallestcomponents,Randomdecomposablecombinatorialstructures,Exponential
1. Introduction.
with finding the largest or smallest “prime” component. The algorithms we have in mind
here are structurally similar to the school method of decomposing a number into primes.
For instance, the factorization of polynomials over finite fields can be performed by
finding irreducible factors of degree 1,2,..., the socalled distinctdegree factoriza
tion. This process halts when the irreducible factor of largest degree is found (see [12,
Chapter 14] and [13] for uptodate accounts on factorization algorithms). The average
case analysis of this algorithm involves the study of the expected largest degree among
the irreducible factors of a random polynomial (see [6]).
The same algebraic property used in the distinctdegree factorization stage of the
factoring process can be used to derive a fast algorithm for finding irreducible poly
nomials over finite fields (see [1] and [12, Chapter 14]). This is an important problem
when implementing extension fields (a crucial step for cryptographical applications).
For the analysis of this algorithm one should find the expected smallest degree among
the irreducible factors of a random polynomial (see [20]).
We would like to have a better understanding of how often these large and small
degreeirreduciblefactorsappear.Inaddition,wehavesituationswherenoconcentration
results occur, and thus, averagecase analysis may not provide an accurate picture of
Some algebraic algorithms have stopping conditions that are related
1The work of the first author was done while he was with the Department of Computer Science, University
of Toronto.
2School of Mathematics and Statistics, Carleton University, 1125 Colonel By Drive, Ottawa, Canada K1S
5B6. daniel@math.carleton.ca.
3Department of Combinatorics and Optimization, University of Waterloo, Waterloo, Canada N2L 3G1.
lbrichmond@watdragon.uwaterloo.ca.
Received June 27, 2000; revised October 8, 2000. Communicated by R. Kemp and H. Prodinger.
Online publication August 9, 2001.
Page 2
414D. Panario and B. Richmond
the behaviour of the algorithm. For instance, the expected smallest degree in a random
polynomial is c1logn but its variance is c2n, for constants c1,c2. Therefore, the exact
number of polynomials over finite fields of degree n with largest or smallest irreducible
factor of degree k provides additional interesting information.
The similar problem of counting permutations on n objects with largest cycle length
equal to k was analysed by Golomb and Gaal [15]. They provide:
• explicit expressions for the number of permutations on n objects with largest cycle
length equal to k on the ranges n/(i + 1) < k ≤ n/i for i = 1,2,...;
• ageneralrecurrenceforthenumberofpermutationsofsizen withlargestcyclelength
equal to k; and
• the contribution of the ranges (n/(i +1),n/i] for i = 1,2,..., to the expected length
of the largest cycle.
In this paper we generalize Golomb and Gaal’s work to provide exact extremal prop
erties in random decomposable combinatorial structures. We have in mind objects that
decompose into “primes”. Typical examples are the decomposition of permutations into
a set of cycles, of polynomials over finite fields into a multiset of irreducible factors, of
graphsintoasetofconnectedcomponents,etc.Weareinterestedinthesamekindofprop
ertiesasGolombandGaalbutforthenumberofobjectsofsizen withlargestsizeofcom
ponents exactly equal to k. We also derive equivalent results for the smallest component.
Unfortunately, our motivation in the case of polynomials over finite fields does not
seem to apply to other decomposable structures like permutations and graphs. There are
efficient methods for decomposing a permutation into cycles or a graph into connected
components that do not involve a similar process to the one in factoring a number into
its primes. In any case, the problem of exact counting extremal properties for these
decomposable structures deserves to be studied for its intrinsic interest.
We now comment on the contents of the paper. Section 2 briefly introduces the
exponential class that covers a large number of combinatorial structures of interest. Our
studyoflabelledstructuresisinSection3.Westateexplicitexpressionsforthenumberof
labelledobjectsofsizen withlargest(andsmallest)sizeofcomponentsexactlyequaltok
fortherangesn/(i+1) < k ≤ n/i,i = 1,2,....Thenwederiveageneralrecurrencefor
the number of labelled objects of size n with largest (and smallest) size of components
exactly equal to k in terms of smaller (respectively larger) sizes. Similar results for
unlabelled objects are obtained in Section 4. The application of our theorems to several
combinatorial structures included in the exponential class is presented in Section 5. We
also provide the contribution of the ranges (n/(i + 1),n/i] for i = 1,2,..., to the
expected size of the largest and smallest component for both labelled and unlabelled
structures. These contributions give useful information related to how often the largest
and smallest components occur.
We finish this section by fixing the notation that we use throughout the paper. The
generating functions of objects with largest irreducible component of size k are denoted
by L?
[zn]L?
weuse Ls
Moreover, for labelled structures we use exponential generating functions while for
unlabelled structures we use ordinary generating functions.
k(z) and U?
k(z) = L?
k(z),Us
k(z), for the labelled and unlabelled case, respectively, with coefficients
k,nand [zn]U?
k(z), Ls
k(z) = U?
k,nandUs
k,n. For smallest irreducible components of size k
k,n,againforlabelledandunlabelledcases,respectively.
Page 3
Exact Largest and Smallest Size of Components 415
2. Exponential Class.
intoirreduciblecomponents.Forinstance,polynomialsoverfinitefieldsdecomposeinto
irreduciblefactors,permutationsintocycles,graphsintoconnectedcomponents,etc.We
consider these problems for labelled and unlabelled objects.
We consider exponential generating functions when the structures are labelled, and
thus we have for the components C(z) =?
?
Then the exponential generating function of objects formed with components of size at
most k is
?
i=1
We immediately obtain the exponential generating function for the objects formed with
components of largest size exactly equal to k:
?
i=1
?k−1
i=1
In the same way we can obtain the exponential generating function of objects formed
with components of size at least k:
??
The exponential generating function for the objects formed with components of smallest
size exactly equal to k is then
??
??
When the structures are unlabelled, using ordinary generating functions, we have for
the components C(z) =?
?
In many situations combinatorial objects decompose uniquely
kckzk/k!. It is well known [10] that the
exponential generating function L(z) for the labelled objects satisfies
L(z) =
n
?n
zn
n!= exp(C(z)).
(1)
exp
k ?
cizi/i!
?
.
L?
k(z) = exp
k ?
?
cizi/i!
?
??
− exp
?k−1
i=1
?
cizi/i!
?
(2)
= exp
cizi/i!
eckzk/k!− 1
?
.
exp
i≥k
cizi/i!
?
.
Ls
k(z) = exp
i≥k
cizi/i!
?
??
− exp
??
i≥k+1
cizi/i!
?
(3)
= exp
i≥k
cizi/i!
1 − e−ckzk/k!?
.
kckzk. In this case, it is also well known (see for instance
[21]) that the ordinary generating function U(z) for the unlabelled objects satisfies
?
U(z) =
n
unzn= exp
C(z) +C(z2)
2
+C(z3)
3
+ ···
?
.
(4)
Page 4
416D. Panario and B. Richmond
Another expression for the ordinary generating function of unlabelled objects is well
known to be
?
In the same vein, the ordinary generating function of objects formed with components
of size at most k is
?
The generating function for the objects formed with components of largest size exactly
equal to k is
U(z) =
∞
i=1
?
1
1 − zi
?ci
.
k ?
i=1
1
1 − zi
?ci
.
U?
k(z) =
k ?
∞
?
i=1
?
?
1
1 − zi
1
1 − zi
?ci
?ci ?
−
k−1
?
i=1
?
?
1
1 − zi
1
1 − zi
?ci
?−ci
=
k ?
(1 − (1 − zk)ck).
i=1
?
1
1 − zi
?ci
(1 − (1 − zk)ck)
(5)
=
i=1
i≥k+1
The ordinary generating function of objects formed with components of smallest size
exactly equal to k is
?
?
When no component is allowed to repeat and Q(z) denotes the ordinary generating
function, we obtain
?
We note that the objects we are considering in both the labelled and unlabelled case
have generating functions that can be expressed in terms of the exponential function.
Thus, we now define the exponential class for generating functions.
Us
k(z) =
?
i≥k
1
1 − zi
1
1 − zi
?ci
?ci
−
?
i≥k+1
?
1
1 − zi
?ci
(6)
=
i≥k
?
(1 − (1 − zk)ck).
Q(z) =
?
n
unzn= exp
C(z) −C(z2)
2
+C(z3)
3
− ···
?
.
(7)
DEFINITION 1.
irreducible components. If one of (1), (4) or (7) holds, then the objects are said to be in
the exponential class.
Let C(z) be the (ordinary or exponential) generating function for the
Many authors give definitions for classes of combinatorial objects. Flajolet and Soria
[10] define a similar combinatorial class, the explog class. The main difference with
our definition is that they are interested in making explicit the type of singularity of
the generating function of the components in order to use asymptotic techniques like
Page 5
Exact Largest and Smallest Size of Components 417
singularity analysis (see [8]). On the other hand, we are not concerned with singularities
in this paper since our focus is on exact combinatorial counting rather than asymptotic
analysis; thus, we drop the logarithmic type of the components. However, since our
results also apply to the explog class, several examples of configurations in that class
are studied in Section 5.
The same authors also define the alglog class (see [11]). Other authors presenting
settingsinwhichgeneratingfunctionsoftheform A(C(z))arise,withC(z)thegenerating
function for the irreducible components, are Cameron [4] (see also [3]) and Labelle and
Leroux [19]. In this paper we only consider the exponential class as defined above.
3. Labelled Objects.
fore all generating functions are of exponential type. The connection between objects
and components given in (1) is used extensively.
In this section we concentrate on labelled structures and there
3.1. Largest Labelled Case
THEOREM 2.
size n having largest size of irreducible components equal to k. Then if n/2 < k ≤ n we
have
Let L?
k(z) be the exponential generating function for labelled objects of
n![zn]L?
k(z) =
?n
k
?
ck?n−k;
(8)
in the range n/3 < k ≤ n/2 we have
n![zn]L?
k(z) = n!ck
k!
?
?n−k
(n − k)!−
ck
2k!
?n−2k
(n − 2k)!−
n−k
?
i=k+1
ci
i!
?n−k−i
(n − k − i)!
?
;
(9)
in the range n/4 < k ≤ n/3 we have
n![zn]L?
k(z) = n!ck
k!
?
?n−k
(n − k)!−
n−k
?
+1
2
i=2k
ck
2k!
?n−2k
(n − 2k)!−
c2
k
3(k!)2
?n−3k
(n − 3k)!
(10)
−
i=k+1
n−k
?
ci
i!
?i−k
j=k
?n−k−i
(n − k − i)!−
?
n−2k
?
?
(n − k − i)!
i=k+1
ci
i!
ck
2k!
?n−2k−i
(n − 2k − i)!
?
cj
j!
ci−j
(i − j)!
?n−k−i
;
and so on, where cnand ?nare the coefficients of znin n!C(z) and n! L(z), respectively.
For the coefficients of L?
k(z), we have the recurrence
L?
k,n=
?n/k?
?
i=1
ci
i!
k
n!
(k!)i(n − ki)!
min{k−1,n−ki}
?
j=1
L?
j,n−ki,
(11)
for k > 1, with L?
1,n= cn
1.
Page 6
418 D. Panario and B. Richmond
PROOF.Recall (2). Using (1) we obtain
L?
k(z) = exp
?
∞
?
i=1
cizi/i! −
?
i=k
∞
?
∞
?
i=k
cizi/i!
??
eckzk/k!− 1
?
= L(z)exp
−
∞
?
cizi/i!
??
eckzk/k!− 1
?
= exp
?
−
i=k
cizi/i!
??
eckzk/k!− 1
???
j≥0
?j
zj
j!
?
.
Consider first the range n/2 < k ≤ n. Extracting coefficients we obtain
k(z) = [zn]ckzk
[zn]L?
k!
?
j≥0
?j
zj
j!=ck
k!
?n−k
(n − k)!.
Since we have an exponential generating function, we finally have
n![zn]L?
k(z) =
?n
k
?
ck?n−k,
with the obvious interpretation as an irreducible component of size k and an object of
size n − k. This proves (8).
We now consider the range n/3 < k ≤ n/2. Extraction of coefficients provides
??
i≥k
ck
k!2(k!)2
[zn]L?
k(z) = [zn]1 −
?
cizi/i!
??ckzk
k!
+c2
kz2k
2(k!)2
??
ck
k!
j≥0
?j
zj
j!
?
=
?n−k
(n − k)!+
c2
k
?n−2k
(n − 2k)!−
n−k
?
i=k
ci
i!
?n−k−i
(n − k − i)!.
We conclude that in this range
n![zn]L?
k(z) = n!ck
k!
?
?n−k
(n − k)!−
ck
2k!
?n−2k
(n − 2k)!−
n−k
?
i=k+1
ci
i!
?n−k−i
(n − k − i)!
?
.
This expression can be interpreted as an inclusion–exclusion formula when n/3 < k <
n/2. Indeed, the number of objects of size n with largest irreducible component of size
k with n/3 < k < n/2 can be obtained by considering all objects with an irreducible
component of size k (first term) minus all objects with another irreducible component
with size from k up to n − k (the other two terms). In the case k = n/2, this formula
is not an inclusion–exclusion formula due to the possibility of two components of size
n/2. We have proven (9).
We consider one more range, that is a k such that n/4 < k ≤ n/3. Extracting
coefficients we obtain
i≥k
[zn]L?
k(z) = [zn]
1 −
?
cizi/i! +1
2
??
i≥k
cizi/i!
?2
Page 7
Exact Largest and Smallest Size of Components419
×
?ckzk
k!
+c2
kz2k
2(k!)2+c3
kz3k
6(k!)3
??
c2
k
3(k!)2
j≥0
?j
zj
j!
?
=
ck
k!
?
?n−k
(n − k)!−
n−k
?
+1
2
i=2k
ck
2k!
?n−2k
(n − 2k)!−
?n−3k
(n − 3k)!
−
i=k+1
n−k
?
ci
i!
?i−k
j=k
?n−k−i
(n − k − i)!−
?
ck
2k!
?
n−2k
?
?n−k−i
(n − k − i)!
i=k+1
ci
i!
?n−2k−i
(n − 2k − i)!
?
cj
j!
ci−j
(i − j)!
.
Equation (10) follows after multiplying the last expression by n!.
Asweadvanceoni andconsiderrangesn/(i+1) < k ≤ n/i theinclusion–exclusion
formulas get more and more involved. Therefore, we derive a recurrence relation for
L?
components of largest size equal to k, and consider first the case k = 1. Since the largest
componenthassize1,theobjectisformedbyn componentseachofsize1.Inourgeneral
context, the number of such objects is given by
?n
since we have to choose the n fixed points, for each one we have c1possibilities, and we
have to divide by n! due to the fact of being labelled objects. Therefore, L?
usefultobearinmindthepermutations.Inthatcase,wewantthenumberofpermutations
with all points fixed; clearly, c1= 1 and that quantity is 1.
For the case k > 1, we can have i components of size k, 1 ≤ i ≤ ?n/k?. The number
of objects is
k,n, 1 ≤ k ≤ n. Recall that L?
k,ndenote the number of labelled objects of size n with
cn
n!
1
1
??n − 1
1
?
···
?1
1
?
,
1,n= cn
1. It is
ci
i!
k
?n
k
??n − k
k
?
···
?n − ki + k
k
?
=ci
k
i!
n!
(k!)i(n − ki)!.
We combine this last expression with the number of objects with the remaining n − ki
elements not considered in the i components of size k. For that we use largest size j
with j = min{k − 1,n − ki}. This proves (11).
3.2. Smallest Labelled Case.
structures.
We focus now on the exact smallest size of labelled
THEOREM 3.
size n having smallest size of irreducible components equal to k. Then if n/2 < k ≤ n
we have
?
0
Let Ls
k(z) be the exponential generating function for labelled objects of
n![zn]Ls
k(z) =
ck
if
if
k = n;
n/2 < k < n;
(12)
Page 8
420 D. Panario and B. Richmond
in the range n/3 < k ≤ n/2 we have
n![zn]Ls
k(z) =
?1
2
?n
?n
k
k
?c2
k
if
if
n = 2k;
n/3 < k < n/2;
?cn−kck
(13)
in the range n/4 < k ≤ n/3 we have
n![zn]Ls
(14)
and so on, where cnand ?nare the coefficients of znin n!C(z) and n! L(z), respectively.
For the coefficients of Ls
?
i!
(k!)i(n − ki)!
where the notation [S] means 1 if S is integral and 0 otherwise.
k(z)
?
?
=
n!
c3
n/3
6((n/3)!)3+
cn/3
(n/3)!
c2n/3
(2n/3)!
?
ifn = 3k;
n!
cn−k
(n − k)!
ck
k!+1
2
ck
k!
n−2k
?
j=k+1
cj
j!
cn−k−j
(n − k − j)!
?
ifn/4 < k < n/3;
k(z), we have the recurrence
Ls
k,n=
?n/k?
?
i=1
ci
k
n!
n−ki
?
j=k+1
Ls
j,n−ki
?
+ [n/k]
cn/k
k
(n/k)!
n!
(k!)n/k,
(15)
PROOF.Consider (3). In the range n/2 < k ≤ n we have
??
i≥k
Since we have an exponential generating function, we finally have, for k = n,
n![zn]Ls
[zn]Ls
k(z) = [zn]1 +
?
ci
zi
i!
??ckzk
k!
??
=
?
ck/k!
0
if
if
k = n;
n/2 < k < n.
k(z) = n!cn
n!= cn,
that is, the number of objects of size n with smallest size of component equal to n is cn.
Note that we cannot have objects of size n with smallest size of irreducible component
equal to k in the range n/2 < k < n, and thus the number of these objects should be
zero. This proves (12).
We now consider the range n/3 < k ≤ n/2. Extracting coefficients we obtain
??
i≥k
Aftermultiplicationbyn!weobtain(13).Theseexpressionscanbeinterpretedasfollows.
For an object of size n with smallest irreducible component of size k, n/3 < k < n/2,
[zn]Ls
k(z) = [zn]1 +
?
cizi/i!
??ckzk
k!
−c2
kz2k
2(k!)2
??
=
c2
k
2(k!)2
cn−k
(n − k)!
if
n = 2k;
n/3 < k < n/2.
ck
k!
if
Page 9
Exact Largest and Smallest Size of Components421
the only possibility is to be formed by two irreducible components, one of size k and
the other of size n − k. On the other hand, if the smallest size is k = n/2, it can only
be formed by two irreducible components of size n/2, and since we are double counting
we have to divide by two.
We consider one more range, that is a k such that n/4 < k ≤ n/3. Extracting
coefficients we obtain
i≥k
?ckzk
where we have collected appropriate terms. The interpretation of this last formula is
clear: an object of size n with smallest component of size k, n/4 < k ≤ n/3, can be
formed by one component of size k and another of size n −k, or by three components at
least one of them having size k and the other two with a compound size of n −k and no
one with size smaller than k. Equation (14) follows after multiplying the last expression
by n!.
We now derive a recurrence relation for Ls
number of labelled objects of size n with components of smallest size equal to k. Then,
in a similar way to (11), we have
?
i!
(k!)i(n − ki)!
where the notation [n/k] means 1 if n is a multiple of k and 0 otherwise. The basis in
this case is when k = n, where we obtain Ls
[zn]Ls
k(z) = [zn]
1 +
?
cizi/i! +1
2
??
i≥k
cizi/i!
?2
×
k!
−c2
kz2k
2(k!)2+c3
kz3k
6(k!)3
?
=
c3
n/3
6((n/3)!)3+
cn−k
(n − k)!
cn/3
(n/3)!
ck
k!
c2n/3
(2n/3)!
n−2k
?
if
n = 3k;
ck
k!+1
2
j=k+1
cj
j!
cn−k−j
(n − k − j)!
if
n/4 < k < n/3,
k,n, 1 ≤ k ≤ n. Recall that Ls
k,ndenote the
Ls
k,n=
?n/k?
?
i=1
ci
k
n!
n−ki
?
j=k+1
Ls
j,n−ki
?
+ [n/k]
cn/k
k
(n/k)!
n!
(k!)n/k,
n,n= cn, as expected.
4. Unlabelled Objects.
dinary generating functions. We recall that U?
atingfunctionsforunlabelledobjectswithlargestandsmallestirreduciblecomponentof
size k, respectively. Consider (5) and (6) for the generating functions U?
respectively.
We concentrate now on unlabelled structures so we have or
k(z) and Us
k(z) denote the ordinary gener
k(z) and Us
k(z),
4.1. Largest Unlabelled Case
THEOREM 4.
size n having largest size of irreducible components equal to k. Then if n/2 < k ≤ n
Let U?
k(z) be the ordinary generating function for unlabelled objects of
Page 10
422 D. Panario and B. Richmond
we have
[zn]U?
k(z) = ckun−k;
(16)
in the range n/3 < k ≤ n/2 we have
[zn]U?
k(z) = ckun−k−
?ck
2
?
un−2k− ck
n−k
?
i=k+1
ciun−k−i;
(17)
in the range n/4 < k ≤ n/3 we have
[zn]U?
k(z) = ckun−k−
?ck
2
?
un−2k− ck
n−k
?
i=k+1
ciun−k−i+
?ck
3
?
un−3k
(18)
+
?ck
2
?n−2k
?
?
i=k+1
n−k
ciun−2k−i−ck
?i−(k+1)
j=k+1
2
?
?(n−k)/2?
?
i=k+1
ciun−k−2i
+ck
2
i=2k+2
?
cjci−j
un−k−i;
and so on, where cnand unare the coefficients of znin C(z) and U(z), respectively.
PROOF.
of (5), we obtain
Recall (5). Using (4) for the restricted range i ≥ k + 1 in the second product
U?
k(z) = U(z)exp
?
−
?
i≥k+1
cizi−1
2
?
i≥k+1
ciz2i− ···
?
(1 − (1 − zk)ck).
Consider first the range n/2 < k ≤ n. Extracting coefficients we obtain
[zn]U?
k(z) = ckun−k.
The interpretation here is: an object of size n with an irreducible component of size k,
n/2 < k ≤ n, is formed by an irreducible component of size k and an object of size
n − k. This proves (16).
We now consider the range n/3 < k ≤ n/2. Extraction of coefficients provides
??
i≥k+1
?ck
[zn]U?
k(z) = [zn]1 −
?
?
cizi
??
ckzk−
?ck
2
?
z2k
??
j≥0
ujzj
?
= ckun−k−
2
un−2k− ck
n−k
?
i=k+1
ciun−k−i.
We bring the attention of the reader to (1) and (4). The difference between labelled
and unlabelled structures given by the extra exponential terms results in expressions
somewhat simpler in the labelled cases. Indeed, we now show this effect in the first
Page 11
Exact Largest and Smallest Size of Components423
range where it appears. Extracting coefficients for the range n/4 < k ≤ n/3, we
obtain
[zn]U?
k(z) = [zn]
1 −
?
?
i≥k+1
cizi+1
2
??
?ck
i≥k+1
cizi
?2
−1
2
?
i≥k+1
ciz2i
×
ckzk−
?ck
2
?
z2k+
3
?
z3k
??
j≥0
ujzj
?
= ckun−k−
?ck
2
?
un−2k− ck
n−k
?
i=k+1
ciun−k−i+
?ck
3
un−3k
+
?ck
2
?n−2k
?
?
i=k+1
n−k
ciun−2k−i−ck
?i−(k+1)
j=k+1
2
?
?(n−k)/2?
?
i=k+1
ciun−k−2i
+ck
2
i=2k+2
?
cjci−j
un−k−i.
As in the largest labelled case, an inclusion–exclusion formula emerges. Indeed, the
numberofobjectsofsizen withlargestirreduciblecomponentofsizek,n/4 < k ≤ n/3,
can be computed by considering all objects with an irreducible component of size k, first
term in (18), minus the objects with two irreducible components, at least one of them
of size k, second and third terms, plus the objects with three irreducible components (at
least one of them of size k), the remaining terms in (18).
Note that in the term involving ciun−k−2i we have ci components of size i occur
ring exactly twice. This is the essential difference between labelled and unlabelled
counting.
4.2. Smallest Unlabelled Case
THEOREM 5.
size n having smallest size of irreducible components equal to k. Then if n/2 < k ≤ n
we have
?
0
Let Us
k(z) be the ordinary generating function for unlabelled objects of
[zn]Us
k(z) =
ck
if
if
k = n;
n/2 < k < n;
(19)
in the range n/3 < k ≤ n/2 we have
[zn]Us
k(z) =
?ck+ 1
cn−kck
2
?
ifn = 2k;
n/3 < k < n/2;
if
(20)
Page 12
424D. Panario and B. Richmond
in the range n/4 < k ≤ n/3 we have
[zn]Us
k(z) =
cn/3c2n/3+
cn−kck+ck
?ck+ 2
n−2k
?
?ck
3
?
ifn = 3k;
2
j=k
cjcn−k−j
?
−cn−2k
2
+ck
2c(n−k)/2
ifn/4 < k < n/3;
(21)
and so on, where cnand unare the coefficients of znin C(z) and U(z), respectively, and
c(n−k)/2is zero if n − k is an odd integer.
PROOF. Again we start from (6). Using (4) for the restricted range i ≥ k, we obtain
??
Consider first the range n/2 < k ≤ n. Extracting coefficients we obtain
[zn]Us
0
Us
k(z) = exp
i≥k
cizi+1
2
?
i≥k
ciz2i+ ···
?
(1 − (1 − zk)ck).
k(z) =
?ck
if
if
k = n;
n/2 < k < n.
As we commented in the labelled case, we cannot have objects of size n with smallest
size of irreducible component equal to k in the range n/2 < k < n, and thus the number
of these objects should be zero. This proves (19).
We now consider the range n/3 < k ≤ n/2. Extraction of coefficients provides
??
i≥k
i≥k
?
cn−kck+ck
2
j=k
−cn−2k
2
[zn]Us
k(z) = [zn]1 +
?
cizi
??
ckzk−
?ck
2
?
z2k
??
=
?ck+ 1
cn−kck
2
?
if
n = 2k;
n/3 < k < n/2.
if
We consider the range n/4 < k ≤ n/3. Extracting coefficients we obtain
[zn]Us
k(z) = [zn]
1 +
?
cizi+1
2
??
i≥k
cizi
?2
+1
?
2
?
i≥k
ciz2i
×
ckzk−
?ck+ 2
n−2k
?
?ck
?ck
2
?
?
z2k+
?ck
3
?
z3k
=
cn/3c2n/3+
3
if
n = 3k;
cjcn−k−j
?
+ck
2c(n−k)/2
if
n/4 < k < n/3.
Page 13
Exact Largest and Smallest Size of Components425
Here, c(n−k)/2is zero if n − k is an odd integer. Again, as in the largest unlabelled case,
the extra exponential terms in unlabelled situations start appearing in this range.
4.3. Recurrences for Unlabelled Objects.
currence relations for unlabelled objects using the labelled recurrences obtained in the
previous sections. We only do it for smallest components of size k.
In this section we show how to obtain re
THEOREM 6.
belled objects of size n having smallest size of irreducible components equal to k, we
have the recurrence
?
i!
(k!)i(n − ki)!
For the coefficients of Us
k(z), the ordinary generating function for unla
Us
k,n=
?n/k?
?
i=1
¯ ci
k
n!
n−ki
?
j=k+1
Us
j,n−ki
?
+ [n/k]
¯ cn/k
(n/k)!
k
n!
(k!)n/k,
(22)
where the notation [S] means 1 if S is integral and 0 otherwise, and
¯ ci= i!
?
j≥1
ci/j
j
,
with ci/j= 0 if j ? i.
PROOF.
irreducible components at least k is
The generating function of unlabelled objects of size n having smallest size of
exp
??
i≥k
cizi+1
2
?
i≥k
ciz2i+1
3
?
i≥k
ciz3i+ ···
?
.
This expression can be rewritten as
exp
??
i≥k
¯ cizi
i!
?
,
where
¯ ci= i!
?
k(z), that is,
?
??
j≥1
ci/j
j
,
with ci/j= 0 if j ? i. We are interested in Us
Us
k(z) = exp
??
??
i≥k
¯ cizi
i!
− exp
??
i≥k+1
¯ cizi
i!
?
= exp
i≥k
¯ cizi
i!
1 − e−¯ ck(zk/k!)?
.
Our recurrence for Ls
since their generating function then satisfies the same equation as Ls
k,ngiven in Theorem 3 holds for the Us
k,nwith cireplaced by ¯ ci
k(z).
Page 14
426D. Panario and B. Richmond
We note that a recurrence for the ¯ ci gives a recurrence for the ci. Indeed, if ¯ ci =
i!?
cj =
dj
j≥1(ci/j/j), then
?
µ(d)
d
¯ cj/d
(j/d)!,
where µ is the wellknown Moebius function.
5. Examples.
structures. We include a section about polynomials over finite fields. These data may be
of interest for practitioners in cryptography (see some possible applications for instance
in [6] and [20]).
This section exemplifies our results for several decomposable random
5.1. Labelled Structures in the explog Class.
explog class when studying the limit distributions of the number of components in
combinatorialstructures.Thisclassissimilartoourexponentialclasswiththeadditional
requirement that the generating function for the basic components have a singularity
of the logarithmic type. More precisely, we say that C(z) is of logarithmic type with
multiplicity constant a > 0 if
Flajolet and Soria [10] introduced the
C(z) = a log
?
1
1 − z/ρ
?
+ R(z),
where R(z) is analytic (see [21] for details).
The simplest possible example in this class is permutations viewed as a product of
cycles. The generating functions of permutations is well known to be
L(z) =
1
1 − z= exp
?
log
1
1 − z
?
.
Thus, L(z)isintheexponentialclass.Inthiscase,?k= k!andck= (k−1)!.Substituting
these quantities in our formulas of Theorem 2, we obtain the same results as in [15] for
the largest permutation.
In the following, we derive new results for permutations. If n/2 < k ≤ n we have
?
0
n![zn]Ls
k(z) =
(k − 1)!if
if
k = n;
n/2 < k < n;
in the range n/3 < k ≤ n/2 we have
n![zn]Ls
k(z) =
n!
2
1
(n/2)2
1
k(n − k)
if
n = 2k;
n!
if
n/3 < k < n/2;
Page 15
Exact Largest and Smallest Size of Components 427
in the range n/4 < k ≤ n/3 we have
The recurrence for the coefficients of Ls
n![zn]Ls
k(z) =
n!
?
?
1
6(n/3)3+
1
n/3
1
2n/3
?
k(z) is in this case
?
1
j
if
n = 3k;
n!
1
n − k
1
k+
1
2k
n−2k
j=k+1
1
n − k − j
?
if
n/4 < k < n/3.
Ls
k,n=
?n/k?
?
i=1
?
1
i!(k)i
n!
(n − ki)!
n−ki
?
j=k+1
Ls
j,n−ki
?
+ [n/k]
1
kn/k
n!
(n/k)!.
The expected value of the largest component is asymptotically λn, where λ =
0.62432965... is Golomb’s constant. This was proven for permutations by Golomb
[14]. Permutations are in the explog class with multiplicative constant a = 1. In his
thesis, Gourdon [16] generalized Golomb’s result. More precisely, he proved that the
expected largest component in the explog class with multiplicative constant a = 1 is
asymptotic to λn.
From Theorem 2, the contribution to λ from those components of size k in the range
n/2 < k ≤ n is
1
n?n
k=n/2+1
If the objects are in the explog class with multiplicative constant a = 1 and radius of
convergence ρ (see [21] for details), an application of singularity analysis [8] provides
n ?
k
?n
k
?
ck?n−k=
n!
n?n
n ?
k=n/2+1
kck
k!
?n−k
(n − k)!.
ck
k!∼
1
kρk,
?k
k!∼
1
ρk.
It implies that this contribution is asymptotic to
1
n?n
n ?
k=n/2+1
k
?n
k
?
ck?n−k∼1
n
n ?
k=n/2+1
k1
k=1
2.
This is exactly the asymptotic contribution Golomb and Gaal [15] got for the largest
cycle of a permutation.
Furthermore, if we consider components on the range n/3 < k ≤ n/2, our theorem
gives an asymptotic contribution to λ of
?
2k!
?
i=k+1
n!
1
n?n
n/2
?
k=n/3+1
∼1
n
kck
k!
?n−k
(n − k)!−
ck
?n−2k
(n − 2k)!−
?
n−k
?
i=k+1
ci
i!
?n−k−i
(n − k − i)!
?
n/2
?
k=n/3+1
1 −
1
2k−
n−k
?
1
i
.