On the enumeration of rooted trees with fixed size of maximal decreasing trees

0.13 MB·English

by  Seunghyun Seo

ON THE ENUMERATION OF ROOTED TREES WITH FIXED SIZE OF
MAXIMAL DECREASING TREES
SEUNGHYUN SEO AND HEESUNG SHIN
Abstract. LetT bethesetofrootedlabeledtreeson{0,...,n}. Amaximaldecreasing
n
subtree of a rooted labeled tree is defined by the maximal subtree from the root with all
2
edges being decreasing. In this paper, we study a new refinement T of T , which is the
1 n,k n
0 set of rooted labeled trees whose maximal decreasing subtree has k+1 vertices.
2
n 1. Introduction
a
J For a nonnegative integer n, let T be the set of rooted labeled trees on [0,n] :=
n
2 {0,...,n}. For a given rooted labeled tree T, a maximal decreasing subtree of T is de-

] fined by the maximal subtree from the root with all edges being decreasing, denoted by
O
MD(T). Figure 1 illustrates the maximal decreasing subtree of a given tree T. Let T be
n,k
C
the set of rooted labeled trees in T whose maximal decreasing subtree has k +1 vertices.
n
.
h Within the scope of proven research, a maximal decreasing subtree first appeared in the
t
a paper [CDG00] of Chauve, Dulucq, and Guibert, for constructing the bijection between
m T and the set of trees in T with n being a leaf. Recently, Bergeron and Livernet [BL10]
n,0 n
[ introduced it in order to analyze the free Lie algebra based on rooted labeled trees. None

2 of them mentioned, however, the refined set T nor considered the enumeration of T .
n,k n,k
v In Section 2, we shall count the number of elements in T . We shall also introduce a set
0 n,k
of certain functions on [n], which is equinumerous to T . In Section 3, we shall decompose
9 n,k
2 a rooted labeled tree into rooted subtrees, each maximal decreasing subtree of which is a
1
single vertex. Then some formulae related to |T | are given from this decomposition. In
. n,k
6
0 Section 4, using the inverse of the matrix i+jj , Tn,k can be expressed as a linear
1 0≤i,j≤n
1 combination of {(n+1)n,(n+2)n,··· ,(2nh+(cid:0) 1)(cid:1)ni}. In the last section, we discuss bijective
: proofs of our results.
v
i
X 2. Main results
r
a
First of all, let us count the number of elements in the set T .
n,k
Theorem 1. For nonnegative integers n and k, we have
n
n+1
|T | = S(m+1,k+1)k!(n−k)n−m−1(m−k),
n,k
m+1
m=k(cid:18) (cid:19)
X
where S(n,k) is a Stirling number of the second kind.
Date: January 27, 2013.
1
2 SEUNGHYUNSEO AND HEESUNGSHIN
T 9 MD(T) 9
3 6 7 3 6 7
1 2 10 1 2
4 8 0 5
Figure 1. The maximal decreasing subtree of T from the root 9
Proof. Given a rooted labeled tree T, let V be the union of the set of vertices in MD(T)
1
and the set of children of any vertex in MD(T). Now, we will count the number of rooted
labeled trees T ∈ T with |V | = m+1.
n,k 1
First of all, the number of ways for selecting V is equal to n+1 . Make a partition
1 m+1
of V into k + 1 blocks, namely, B ,...,B . The number of such partitions is equal to
1 1 k+1
(cid:0) (cid:1)
S(m + 1,k + 1). Take the set V consisting of the minimum m of each block B . Make
0 i i
a decreasing subtree on V = {m ,...,m }. Since it is well known that the number of
0 1 k+1
(unordered) increasing trees on k + 1 nodes is k!, there are exactly k! ways of making a
decreasing subtree on V . Append vertices in V \V to this decreasing subtree such that
0 1 0
elements in B \ {m } are children of m for i = 1,...,k + 1. It is well-known that the
i i i
number of forests F on [0,n]\V such that V \V is the set of all roots of F is equal to
0 1 0
(n−k)n−m−1(m−k) (see [Sta99, Prop. 5.3.2]). Since the range of m is k ≤ m ≤ n,
n
n+1
|T | = S(m+1,k+1)k!(n−k)n−m−1(m−k).
n,k
m+1
m=k(cid:18) (cid:19)
X
(cid:3)
For a nonnegative integer n, let F be the set of functions from [n] to [n], where [n] :=
n
{1,...,n} if n is positive integer and [0] := ∅. Let F be the set of functions f ∈ F with
n,k n
[k] ⊂ f([n]), where f([n]) is the image of f.
Proposition 2. For nonnegative integers n and k, we have
n
n
|F | = S(m,k)k!(n−k)n−m (1)
n,k
m
m=k(cid:18) (cid:19)
X
k
= (−1)i (n−i)n. (2)
i
i≥0 (cid:18) (cid:19)
X
Proof. Let f−1([k]) = A and |A| = m. The number of subsets A of [n] of size m is equal
to n . The function f can be decomposed into a surjection from A to [k] with S(m,k)k!
m
ways and a function from [n]\A to [n]\[k] with (n−k)n−m ways. Since m runs through
(cid:0) (cid:1)
from k to n, the formula (1) holds.
ON THE ENUMERATION OF ROOTED TREES WITH FIXED SIZE OF MAXIMAL DECREASING TREES3
Meanwhile, defining A by the set {f ∈ F | f−1(j) = ∅}, we have
j n
F = F \(A ∪···∪A ).
n,k n 1 k
By the principle of inclusion and exclusion, we have
|F | = |F |−|A ∪···∪A | = (−1)|I|(n−|I|)n.
n,k n 1 k
I⊂[k]
X
(cid:3)
So, the formula (2) holds.
Theorem 3. For nonnegative integers n and k, we have |T | = |F |, i.e.,
n,k n,k
n k
n+1 k
S(m+1,k +1)k!(n−k)n−m−1(m−k) = (−1)i (n−i)n. (3)
m+1 i
m=k(cid:18) (cid:19) i=0 (cid:18) (cid:19)
X X
Proof. Since S(m+1,k+1)(k +1)! is the number of surjective functions from [m+1] to
[k +1], which is equal to (−1)j k+1 (k +1−j)m+1 by the principle of inclusion and
j≥0 j
exclusion, it follows that
P (cid:0) (cid:1)
n+1
|T | = [S(m+1,k +1)k!] (n−k)n−m−1(m−k)
n,k
m+1
m (cid:18) (cid:19)
X
n+1 1 k +1
= (−1)j (k +1−j)m+1 (n−k)n−m−1(m−k).
m+1 k +1 j
" #
m (cid:18) (cid:19) j≥0 (cid:18) (cid:19)
X X
Separating the term m−k to (n−k)−(n−m) and changing the order of summations,
we get
(−1)j k +1 n+1
|T | = (k +1−j)m+1(n−k)n−m
n,k
k +1 j m+1
j (cid:18) (cid:19) m (cid:18) (cid:19)
X X
(−1)j k +1 n
− (n+1) (k +1−j)m+1(n−k)n−m−1.
k +1 j m+1
j (cid:18) (cid:19) m (cid:18) (cid:19)
X X
By the binomial theorem,
(−1)j k +1
|T | = (n+1−j)n+1 −(n+1)(n+1−j)n
n,k
k +1 j
j (cid:18) (cid:19)
X (cid:2) (cid:3)
(−1)j k +1
= (−j)(n+1−j)n.
k +1 j
j (cid:18) (cid:19)
X
Substituting j = i+1 in the previous equation, the formula (3) holds.
(cid:3)
4 SEUNGHYUNSEO AND HEESUNGSHIN
3. Properties
A rooted labeled tree T is called a local minimum tree, if MD(T) consists of a single
vertex. Note that T is the set of local minimum trees on [0,n] and |T | is equal to nn
n,0 n,0
[CDG00]. Also T is the set of decreasing trees on [0,n] and |T | is equal to n!.
n,n n,n
Given T ∈ T , we can decompose T into k + 1 local minimum trees by removing k
n,k
edges in MD(T). This decomposition yields the following lemma.
Lemma 4. For nonnegative integers n and k, we have
1 n+1
|T | = n n1···n nk+1
n,k 1 k+1
k +1 n +1,...,n +1
n1+···+Xnk+1=n−k(cid:18) 1 k+1 (cid:19)
Proof. It is enough to show the following formula
1 n+1
|T | = k! |T |··· T . (4)
n,k (k +1)! n +1,...,n +1 n1,0 nk+1,0 
n1+···+Xnk+1=n−k(cid:18) 1 k+1 (cid:19) (cid:12) (cid:12)
 (cid:12) (cid:12)
First of all, we will make a tuple (T ,...,T ) of local minimum trees satisfying two
1 k+1
conditions:
(i) The tuple (L(T ),...,L(T )) is an ordered partition of the set [0,n] and
1 k+1
(ii) root(T ) < root(T ) < ··· < root(T ),
1 1 k+1
where L(T) means a set of labels of vertices in T and root(T) a label of the root of T.
Consider a tuple (S ,...,S ) of local minimum trees with the only condition (i). For
1 k+1
a given sequence n ,...,n of nonnegative integers with n + ···+ n = n − k, the
1 k+1 1 k+1
number of tuples (S ,...,S ) with |L(S )| = n +1 is equal to
1 k+1 i i
n+1
|T |··· T .
n +1,...,n +1 n1,0 nk+1,0
(cid:18) 1 k+1 (cid:19)
(cid:12) (cid:12)
So the number of all tuples (S ,...,S ) with the condi(cid:12)tion (i)(cid:12)is equal to
1 k+1
n+1
|T |··· T .
n +1,...,n +1 n1,0 nk+1,0
n1+···+Xnk+1=n−k(cid:18) 1 k+1 (cid:19) (cid:12) (cid:12)
(cid:12) (cid:12)
From the condition (ii), the number of all tuples (T ,...,T ) is equal to
1 k+1
1 n+1
|T |··· T .
(k +1)! n +1,...,n +1 n1,0 nk+1,0
n1+···+Xnk+1=n−k(cid:18) 1 k+1 (cid:19) (cid:12) (cid:12)
(cid:12) (cid:12)
Since the number of decreasing subtrees on {root(T ),...,root(T )} is k!, we get the
1 k+1
(cid:3)
formula (4).
From Lemma 4, we can deduce the following result.
ON THE ENUMERATION OF ROOTED TREES WITH FIXED SIZE OF MAXIMAL DECREASING TREES5
Theorem 5. We have three exponential generating functions:
n tk+1 xn+1 xn+1
1+ |T | = exp t nn , (5)
n,k
k! (n+1)! (n+1)!
!
n≥0 k=0 n≥0
XX X
n xn+1 xn+1 −1
1+ |T |(k +1)tk+1 = 1−t nn , (6)
n,k
(n+1)! (n+1)!
!
n≥0 k=0 n≥0
XX X
n xn+1 xn+1
|T |tk+1 = −ln 1−t nn . (7)
n,k
(n+1)! (n+1)!
!
n≥0 k=0 n≥0
XX X
Proof. From Lemma 4, left-hand side of three formulas become
n n+1 tk+1 xn+1
1+ n n1···n nk+1 , (8)
n +1,...,n +1 1 k+1 (k +1)!(n+1)!
Xn≥0Xk=0Xni (cid:18) 1 k+1 (cid:19)
n n+1 xn+1
1+ n n1···n nk+1tk+1 , (9)
1 k+1
n +1,...,n +1 (n+1)!
n≥0 k=0 ni (cid:18) 1 k+1 (cid:19)
XXX
n n+1 tk+1 xn+1
n n1···n nk+1 , (10)
n +1,...,n +1 1 k+1 k +1(n+1)!
n≥0 k=0 ni (cid:18) 1 k+1 (cid:19)
XXX
where n means n +···+n = n−k. Using the compositional formula for exponential
i 1 k+1
structures [Sta99, Theorem 5.5.4], three formulas are of form F(tG(x)) where
xn+1
G(x) = nn .
(n+1)!
n≥0
X
In case (8), the corresponding F(x) is given by
(i+1)! xi+1
F(x) = 1+ = exp(x).
(i+1)! (i+1)!
i≥0
X
In case (9), the corresponding F(x) is given by
(i+1)! xi+1 1
F(x) = 1+ = .
1 (i+1)! 1−x
i≥0
X
In case (10), the corresponding F(x) is given by
(i+1)! xi+1 1
F(x) = = ln .
(i+1) (i+1)! 1−x
i≥0
X
(cid:3)
These complete the proof.
6 SEUNGHYUNSEO AND HEESUNGSHIN
By definition of T , we have
n,k
n
|T | = (n+1)n, (11)
n,k
k=0
X
which can be also induced from t = 1 in (7). Similarly, putting t = 1 in (6) and applying
the equation (5.67) in [Sta99], we get
n xn+1 xn+1 −1
1+ |T |(k +1) = 1− nn
n,k
(n+1)! (n+1)!
!
n≥0 k=0 n≥0
XX X
xn+1
= 1+ (n+2)n .
(n+1)!
n≥0
X
Thus we have
n
(k +1)|T | = (n+2)n. (12)
n,k
k=0
X
From (11) and (12), we are able to deduce the followings.
Theorem 6. For a nonnegative integers n, k, and α, we have
n
k +α
|T | = (n+1+α)n. (13)
n,k
α
k=0(cid:18) (cid:19)
X
Proof. Since we have proved |T | = |F | in Theorem 3, it is enough to show
n,k n,k
n
k +α
|F | = (n+1+α)n. (14)
n,k
α
k=0(cid:18) (cid:19)
X
For α = 0, let G be the set of functions g from [n] to [0,n] with [0,k−1] ⊂ g([n]) but
n,k
k 6∈ g([n]). By definition of G ,
n,k
n
|G | = (n+1)n.
n,k
k=0
X
There is a simple bijection ϕ from F to G as follows: Given a f ∈ F , consider a
n,k n,k n,k
function g from [n] to [0,n] defined by
f(i)−1 if f(i) ≤ k
g(i) =
f(i) otherwise.
(
Since the images of g includes 0,...,k −1 but does not include k, the function g belongs
to G and ϕ(f) = g is well-defined. Since ϕ is reversible, it is a bijection. So it holds that
n,k
|F | = |G |
n,k n,k
for all 0 ≤ k ≤ n.
ON THE ENUMERATION OF ROOTED TREES WITH FIXED SIZE OF MAXIMAL DECREASING TREES7
For α > 0, let H be the set of functions g from [n] to [−α,n] with
n,k,α
|[−α,k −1]\g([n])| = α
and k 6∈ g([n]), where [a,b] := {a,...,b}. For every function h from [n] to [−α,n], since
the cardinality of the domain is less than the cardinality of the codomain by α+1, there
exists a unique k satisfying above conditions. Note that k is the (α + 1)-st element in
[−α,n]\g([n]). Thus
n
|H | = (n+1+α)n.
n,k,α
k=0
X
Let A be the set of all α-elements subsets of [−α,k −1]. Clearly, |A| = k+α . There is
α
a bijection from A×F to H as follows: For a given (A,f) ∈ A×F , we make a
n,k n,k,α n,k
(cid:0) (cid:1)
(A,ϕ(f)) ∈ A×G . Consider the order-preserving bijection σ from [0,n] to [−α,n]\A.
n,k
Then we can define the function h from [n] to [−α,n] by
h = σ ◦(ϕ(f))
and this function h is contained in H . Hence,
n,k,α
k +α
|H | = |A×F | = |A|·|F | = |F |
n,k,α n,k n,k n,k
α
(cid:18) (cid:19)
for all 0 ≤ k ≤ n. (cid:3)
For example, let n = 5, k = 2, and α = 3. Take A = {−2,−1,1} ∈ [−3,1] and f =
3
(f(1),...,f(5)) = (5,2,1,3,2) ∈ F . Then g = ϕ(f) ∈ G and h = σ ◦(ϕ(f)) ∈ H
5,2 5,2 5,2,3
(cid:0) (cid:1)
are given by
(g(1),...,g(5)) = (5,1,0,3,1),
(h(1),...,h(5)) = (5,0,−3,3,0).
Let us consider the equation (13) or (14) for a negative integer α. In fact, the left hand
sides of these equations are not well-defined even for α = −1, nevertheless the right hand
sides are. Here we find, however, the coefficients of |T | that can replace the term k−1
n,k −1
as follows.
(cid:0) (cid:1)
Theorem 7. For a positive integer n, we have
n n
1 1
|T | = |F | = nn. (15)
n,k n,k
k k
k=1 k=1
X X
Proof. From Proposition 2 and Theorem 3, expanding (n−i)n by the binomial theorem,
we have
k
|T | = |F | = (−1)i (n−i)n
n,k n,k
i
i≥0 (cid:18) (cid:19)
X
k n
= (−1)i nn−j(−i)j.
i j
i≥0 (cid:18) (cid:19) j≥0 (cid:18) (cid:19)
X X
8 SEUNGHYUNSEO AND HEESUNGSHIN
Hence the left hand side of (15) is
n n
1 1 k n
|T | = (−1)i nn−j(−i)j.
n,k
k k i j
k=1 j≥0 k=1 i≥0 (cid:18) (cid:19)(cid:18) (cid:19)
X XXX
We divide it into three cases; j = 0, j = 1, and j > 1.
In case of j = 0,
n n
1 k 1
(−1)i nn = nn (1−1)k = 0. (16)
k i k
k=1 i≥0 (cid:18) (cid:19) k=1
XX X
In case of j = 1,
n n
1 k k −1
(−1)i nn(−i) = nn (−1)i−1
k i i−1
k=1 i≥0 (cid:18) (cid:19) k=1 i≥0 (cid:18) (cid:19)
XX XX
n
= nn (1−1)k−1 = nn. (17)
k=1
X
In case of j > 1,
n
1 k n
(−1)i nn−j(−i)j
k i j
j>1 k=1 i≥0 (cid:18) (cid:19)(cid:18) (cid:19)
XXX
n
n k −1
= (−1)i nn−j(−1)jij−1
j i−1
j>1 i≥0 (cid:18) (cid:19) k=1(cid:18) (cid:19)
XX X
n n
= (−1)i+j nn−jij−1
j i
j>1 i≥0 (cid:18) (cid:19) (cid:18) (cid:19)
XX
n n
= (−1)j nn−j (−1)i ij−1 = 0. (18)
j i
" #
j>1 (cid:18) (cid:19) i≥0 (cid:18) (cid:19)
X X
Note that, by the principle of inclusion and exclusion, the expression (−1)i n ij−1
i≥0 i
in (18) is the number of surjections from [j −1] to [n]. Since j −1 < n, it is zero.
From (16), (17), and (18), we finally obtain n 1 |T | = 0+nn +0P= nn. (cid:0) (cid:1) (cid:3)
k=1 k n,k
P
4. Inverse Relation
For nonnegative integer n, let A(n) be the square matrix of size n+1 defined by
i+j
A(n) := .
i
(cid:20)(cid:18) (cid:19)(cid:21)0≤i,j≤n
ON THE ENUMERATION OF ROOTED TREES WITH FIXED SIZE OF MAXIMAL DECREASING TREES9
Define two column vectors t(n) and p(n) by
|T | (n+1)n
n,0
|T | (n+2)n
n,1
t(n) :=  .  and p(n) :=  . .
. .
. .
 |T |   (2n+1)n 
 n,n   
   
Then the equation (13) can be interpreted as
A(n)t(n) = p(n).
The matrix A(n) is nonsingular. Moreover, we can compute its inverse directly. Let B(n)
be the square matrix of size n+1 defined by
n
m m
B(n) := (−1)i+j .
i j
" #
mX=0(cid:18) (cid:19)(cid:18) (cid:19) 0≤i,j≤n
Theorem 8. For a nonnegative integer n, the two matrices A(n) and B(n) are inverse
matrices of each other.
Proof. Here is the calculation:
n n n
i+l m m
A(n) B(n) = (−1)l+j
i,l l,j
i l j
l=0 l=0 (cid:18) (cid:19) m=0(cid:18) (cid:19)(cid:18) (cid:19)
X X X
n n
m i+l m
= (−1)m−j (−1)m−l
j l m−l
" #
m=0 (cid:18) (cid:19) l=0 (cid:18) (cid:19)(cid:18) (cid:19)
X X
Comparing the coefficients of qm for (1−q)−(i+1)(1−q)m = (1−q)−(i+1−m), we obtain
n
i+l m i
(−1)m−l = .
l m−l m
l=0 (cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
X
Thus, we have
n n
m i
A(n) B(n) = (−1)m−j
i,l l,j
j m
l=0 m=0 (cid:18) (cid:19)(cid:18) (cid:19)
X X
n
i i−j i
= (−1)m−j (1−1)i−j = δ
i,j
j m−j j
(cid:18) (cid:19)m=0 (cid:18) (cid:19)(cid:18) (cid:19)
X
(cid:3)
which completes the proof.
From the matrix identity B(n)p(n) = t(n), we obtain another expression for |T |.
n,k
Corollary 9. For nonnegative integers n and k, we have
m m
|T | = (−1)k+l (n+1+l)n. (19)
n,k
k l
0≤l≤m≤n (cid:18) (cid:19)(cid:18) (cid:19)
X
10 SEUNGHYUNSEO AND HEESUNGSHIN
5. Remarks
Since |T | = |F |, it is desired to construct a bijection between T and F for all
n,k n,k n,k n,k
0 ≤ k ≤ n. Also, it is natural to ask a bijective proof of (15). Recently, Jang Soo Kim
[Kim11] constructed bijections for the above questions. It would be interesting to give a
combinatorial explanation of (19).
Acknowleagement
The authors thank to Fr´ed´eric Chapoton who first asked this question while the second
authorinLyon, andto theanonymousreferees fortheir valuablecomments andsuggestions
to improve this paper. This study was supported by 2008 Research Grant from Kangwon
National University to the first author. For the second author, this work was supported
by INHA UNIVERSITY Research Grant (INHA-42830).
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power series and algebraic combinatorics (Moscow, 2000), pages146–157.Springer,Berlin, 2000.
[Kim11] J. S. Kim. Bijections on rooted trees with fixed size of maximal decreasing subtrees. ArXiv e-
prints, August 2011. arXiv:1108.6038.
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(Seunghyun Seo) Kangwon National University
E-mail address: [email protected]
(Heesung Shin) Inha University
E-mail address: [email protected]