Asymptotic cost of cutting down random free trees

0.16 MB·English

by  Elahe Zohoorian Azad

Asymptotic cost of cutting down random free
trees
2
1
0
2 Elahe Zohoorian Azad
n Damghan university of Iran
a
J January 31, 2012

8
2

Abstract
]
R
In this work, we calculate the limit distribution of the total cost in-
P
curred by splitting a tree uniformly distributed on the set of all finite
.
h free trees, appears as an additive functional induced by a toll equal to
t the square of the size of tree. The main tools used are the recent results
a
connecting the asymptotics of generating functions with the asymptotics
m
of theirHadamard product, and themethod of moments.
[
keywords. Additivefunctional,Generatingfunctions,Limitlaw,Recurrence.
2
v
9 1 Introduction
4
0
3 Trees are structures suitable for data storage and supporting computer algo-
. rithms, two fundamental aspects of data processing, with applications in many
8
fields. The cost of “divide-and-conquer” algorithms can be represented as an
0
0 additivefunctionaloftrees. Whiletherearemanystudiesonadditivefunctional
1 (see, for example, [11, 7, 12]), not enough attention has been given to the dis-
: tributions of functional defined on trees under the uniform model. However, a
v
i main motivation for undertaking this investigation is that it is key to analyz-
X
ingaspecialtypeofaDrop-Pushmodelofpercolationandcoagulation(see[15]).
r
a
In this paper, we consider the additive functional defined on the trees uni-
formly selected from the set of all the free trees of size n, for n given (called
Cayley trees in [2]), induced by the toll sequence (n2) (see definition of the
n 0
≥
Section 2). Our main result, Theorem 1, provides the limit distribution for a
suitably normalized version of this functional.
Theorem 1. Let X betheadditive functional definedon theuniform free trees
n
of size n, induced by the toll (n2) . Then,
n 0
≥
n−5/2 Xn L √2 ξ,
−→
1
Cost of cutting down random trees 2
whereξ isarandomvariablewhosedistributionischaracterizedbyitsmoments.:
k!√π
E(ξk)= a¯ ,
2(7k 2)/2Γ(5k 1) k
− 2−
where
k 1
−
a¯ =2(5k 6)(5k 4)a¯ + a¯ a¯ k 2; a¯ =√2.
k k 1 j k j 1
− − − − ≥
j=1
X
Curiously, the moments of our limit distribution are proportional to the
moments of the distribution of the average of the minimum of a normalized
Brownian Excursion, obtained by [8, Theorem 3.3].
Inwhatfollows,e=(e(t)) indicatesanormalizedBrownianExcursion.
0 t 1
≤ ≤
Theorem 2. The moments of the random variable η, defined by
η =4 min e(u)dsdt,
Z Z0<s<t<1s≤u≤t
are given by the formula
k!√π
E(ηk)= ω ,
2(7k 4)/2Γ(5k 1) k
− 2−
where
k 1
−
ω =2(5k 6)(5k 4)ω + ω ω k 2; ω =1.
k k 1 j k j 1
− − − − ≥
j=1
X
It is not unusual, in this kind of problem to have more than one characteri-
zationofa limit distribution. For instance, the Wiener index ofcertaintrees is
given by its moments involving Airy functions, and is alternatively character-
ized in terms of a Brownian.
For the demonstration of Theorem 1, we apply the strategy used in [4] to
obtain the limiting distributions of the additive functionals defined on Catalan
trees, in particular the singularity analysis of the generating series [6]. Indeed,
the Hadamard products appear naturally when one analyzes the moments of
additive functionals of trees. Theorem 1 extends to the moments of all order,
although the analysis of asymptotic behavior of the first moment, was made
already in [5]. The steps taken here allow a rather mechanical calculation
of asymptotic moments of each order, thus facilitating the application of the
method of moments.
Cost of cutting down random trees 3
2 Generating functions
We firstestablishheresomenotation. LetT be abinarytree andlet T denote
| |
the number of nodes in T. Suppose moreover that L(T) and R(T) indicate,
respectively, the left and right subtrees rooted at the 2 children of the root of
T. When the tree is notbinary,one canstill havetwosubtreesL(T)andR(T),
by cutting an edge which can be considered as root.
Definition 1. A functional f defined on a binary tree is called an additive
functional if it satisfies the recurrence
f(T)=f(L(T))+f(R(T))+b ,
T
| |
for any tree T with T 1. Here (b ) is a given sequence, henceforth called
n n 1
the toll function. | |≥ ≥
We analyze here a special additive functional on the trees, uniformly dis-
tributed on T : T = n , for n given . By a result attributed to Cayley [2],
{ | | }
there are U = nn 2 free trees (U connected acyclic labelled graphs) on n
n − n
nodes and accordingly, there are T = nn 1 rooted trees (in which a labelled
n −
node,iscalledroot oftree). Considerthe modelinwhichinitiallyeachfreetree
of size n is taken uniformly at random. Choose an edge at random among the
n 1edgesofthetree,orientitinarandomway,thencutit. Thisseparatesthe
−
treeinto anorderedpairofsmallertrees,thatarenow rooted;wecallthem the
leftandrightsubtrees. Continue the processwitheachofthe resultingsubtree,
discarding the root. Assume1 that the cost incurred by selecting the edge and
splitting the tree in a tree of size n is n2. Then X , the total cost incurred for
n
splitting a random tree of size n, satisfies, for n 1, the recurrence
≥
X =X +X +n2, (1)
n Ln Rn
where the indexes L and R are, respectively, the sizes of left and right sub-
n n
trees, obtained by division of the initial tree of size n. So X appears as the
n
additive functional induced by the toll sequence (n2) .
n 1
≥
A motivation, coming from the analysis of algorithms, is as follows. If time
isreversed,thismodeldescribedtheevolutionofarandomgraph,fromagraph
completely disconnected to a tree and which was used to analyze of the union-
find algorithms [3, 13, 14]. Knuth and Scho¨nhage provided a first analysis of it
1Onecansee[15,Proposition1]forthemainmotivationofgivingthisassumption. Briefly,
[15] analyzes a Drop-Push model of coagulation in which particles are dropped onto a one
dimensional lattice and carry out a random walk until they encounter an empty site where
theybecome stuck. Insuch amodel, the movements of theparticles, onthelattice, forman
additive coalescence processes which gives the good algorithmic reasons for considering the
recurrence (1). In fact, in the Drop-Push model, the cost of coalescence of two clusters of
particles,atthedroppingmomentofaparticle,isgivenasthenumberofstepsoftheparticle
until it sticks in an empty site and it is proven, [15, relation (8)], that the expected cost of
coalescenceoftwoclustersisproportionaltothesquareofthelengthoftheclusteronwhich
aparticledrops.
Cost of cutting down random trees 4
in 1978 ([10]), for different tolls however.
Let p be the probability for a tree of size n to have the left and right
n,k
subtrees respectively of sizes k and n k. Then
−
n kk 1(n k)n k 1
− − −
p = − . (2)
n,k k 2(n 1)nn 2
(cid:18) (cid:19) − −
The binomial coefficient n takes into account the labelling of the left and
k
rightsubtrees,andthequantitykk 1(n k)n k 1 isthenumberofrootedtrees
− − −
(cid:0) (cid:1) −
of sizes k and n k. In the denominator, nn 2 is the number of free trees,
−
−
n 1 is the number of the edges of the initial tree, and finally the coefficient 2
−
corresponds to the random orientation of the selected edge. It is convenient to
write this probability in the form:
n c c
k n k
pn,k = − ,
2(n 1) c
n
−
where, k 1,
∀ ≥ kk 1
−
c = .
k
k!
Let us start with the average of the cost function, a := E(X ), n 1,
n n
≥
which is obtained recursively by conditioning on the size of L :
n
a = E E (X +X +n2)
n L L n L
−
= EL(cid:2)(aL+an L)+n2 (cid:3)
−
n 1
−
= p (a +a )+n2
n,j j n j
−
j=1
X
n 1
= − n cjcn−j(aj +an j)+n2.
2(n 1) cn −
j=1 −
X
This recurrence can be rewritten as
n 1
n 1 − n 1
− c a = c a c + − c b , (3)
n n j j n j n n
n − n
j=1
X
where b =n2.
n
Remark. We replaced n2 by b , distinguishing the general form of the gen-
n
erating function, so that one can always consider any toll function in the place
of n2.
Cost of cutting down random trees 5
Definition2. The Hadamard productof twoentireseries F(z)= f zn and
n n
G(z)= g zn, denoted F(z) G(z), is the entire series defined by
n n ⊙ P
P (F G)(z) F(z) G(z):= f g zn.
n n
⊙ ≡ ⊙
n
X
Multiplying the equality (3) by zn/en and summing over n 1, we get
≥
z ωndω
A(z) C(z/e) a c (4)
⊙ − n nen ω
Z0 n
X
n 1 zn
=(A(z) C(z/e))C(z/e)+ − c b , (5)
⊙ n n nen
n
X
where A(z) and C(z) denote the ordinary generating function of (a ) and
n n 1
≥
(c ) , respectively.
n n 1
≥
In view of a result of Knuth and Pittel, [9], we know the singular expansion
at the dominant singularity z =e 1 of C(z):
−
C(z)=1 √2(1 ez)1/2+O(1 ez ). (6)
− − | − |
Moreover C satisfies the functional relation C(z)=zeC(z).
By differentiation,the relation(5)transformsinto a lineardifferentialequa-
tionof the firstorder,whichcan be readilysolvedby the variation-of-constants
method. Briefly,putting f(z):=A(z) C(z/e)andt(z):= n 1c b e nzn,
⊙ n −n n n −
the relation (5) takes the form
P
z dω
f(ω) =f(z)(1 C(z/e)) t(z). (7)
ω − −
Z0
By taking derivatives, we obtain
df(z) 1/z dC(z/e) 1 dt(z)
+f(z) − − dz = .
dz 1 C(z/e) 1 C(z/e) dz
− ! (cid:18) − (cid:19)
On the other hand, the equality C(z/e)= zeC(z/e) implies
e
dC(z/e) 1 dC(z/e)
=C(z/e) + .
dz z dz
(cid:18) (cid:19)
Assuming now (without loss of generality) the initial condition a c = b = 0,
1 1 1
the solution found will be in the form
C(z/e) z n 1 ωn dω
A(z) C(z/e)= ∂ ( − c b ) . (8)
⊙ 1 C(z/e) ω n n nen C(ω/e)
− Z0 n
X
And finally as n−n1cn = jn=−11 12cjcn−j, we have
P1 C(z/e) z dω
A(z) C(z/e)= ∂ [B(ω) C(ω/e)2] , (9)
ω
⊙ 21 C(z/e) ⊙ C(ω/e)
− Z0
where B(ω) denote the ordinary generating function of (b ) .
n n 1
≥
Cost of cutting down random trees 6
3 Moments by singularity analysis
Thanks to the singularity analysis technique, we can derive the asymptotics of
momentsofeachorder. Thesingularityanalysisisasystematiccomplex-analytic
technique that relates the asymptotic behavior of sequences to the behavior of
their generating functions in the proximity of their singularities. The applica-
bility of singular analysis rests on a technical condition: the ∆-regularity. See
[5, 6] for more details.
Definition3. AfunctiondefinedbyaTaylor series abouttheorigin withradius
of convergence equal to 1 is ∆-regular if it can be analytically continued in a
domain of form
∆(φ,η):= z : z <1+η, arg(z 1) >φ ,
{ | | | − | }
for some η > 0 and 0 < φ < π/2. A function f is said to admit a singular
expansion at z = 1, if it is ∆-regular and if one can find a sequence of com-
plex numbers (c ) , and an increasing sequence of real numbers (α ) ,
j 0 j J j 0 j J
satisfying α <A,≤wh≤ere A is a real number, such that the relation ≤ ≤
j
J
f(z)= c (1 z)αj +O(1 z A)
j
− | − |
j=0
X
holds uniformly in z ∆(φ,η). It is said to satisfy a singular expansion with
∈
logarithmic terms if,
J
1
f(z)= c (L(z))(1 z)αj +O(1 z A), L(z):=log ,
j
− | − | 1 z
j=0 −
X
where each c (.) is a polynomial.
j
Recall the definition of the generalized polylogarithm:
Definition 4. For α an arbitrary complex number and r a nonnegative integer,
the generalized polylogarithm function Li is defined for z <1, by
α,r
| |
(logn)r
Li (z):= zn.
α,r nα
n 1
X≥
In particular, Li (z) = L(z). Moreover, a useful property of generalized
1,0
polylogarithm functions is
Li Li =Li .
α,r β,s α+β,r+s
⊙
ThesingularexpansionofthepolylogarithminvolvestheRiemannzetafunc-
tion (see for example [5, Theorem 4]).
Cost of cutting down random trees 7
Lemma1. ThefunctionLi (z)is∆-regular,andforα / 1,2,... itsatisfies
α,r
∈{ }
the singular expansion
( 1)j
Liα,0(z) Γ(1 α)tα−1+ − ζ(α j)tj, (10)
∼ − j! −
j 0
X≥
where
(1 z)l
t= logz = − .
− l
l 1
X≥
For r >0, the singular expansion of Li is obtained using formal derivations:
α,r
∂r
Li (z)=( 1)r Li (z).
α,r − ∂αr α,0
Anaturalconsequenceofthislemma(whichisaparticularcaseof[4,Lemme
2.6]), is that
Li (z)=Γ(1 α)(1 z)α 1+O(1 z α)+ζ(α)1I ; α<1. (11)
α,0 − α>0
− − | − |
Another result, whichis veryuseful inwhat follows,is the decompositionof
the Hadamard product of (1 z)a (1 z)b (cf. [5, Proposition 8]).
− ⊙ −
Lemma 2. For the real numbers a and b,
(1 z)k (1 z)a+b+1+k
(1 z)a (1 z)b ∼ λ (a,b) − + µ (a,b) − ,
k k
− ⊙ − k! k!
k 0 k 0
X≥ X≥
where the coefficients λ and µ are given by
Γ(1+a+b) ( a)k¯( b)k¯
λ (a,b) = − − ,
k Γ(1+a)Γ(1+b) ( a b)k¯
− −
Γ( 1 a b)(1+a)k¯(1+b)k¯
µ (a,b) = − − − ,
k Γ( a)Γ( b) (2+a+b)k¯
− −
where xk¯ is defined as x(x+1)...(x+k 1), for k nonnegative entire.
−
Now, equipped with the singularity analysis toolkit, we are in a position to
find the asymptotic average from the relation (9).
Lemma 3. The expected value of the total cost, induced by the toll n2 in the
model of random free trees defined in Section 2, is
a = π/8 n5/2+O(n3/2). (12)
n
p
Cost of cutting down random trees 8
Proof. Since b =n2, we have B(z)=Li (z) and the equality (11) implies
n 2,0
−
B(z)=2(1 z) 3+O(1 z 2). (13)
− −
− | − |
Considering the singular expansion (6) of the generating function of the tree,
Lemma 2 gives
B(z) C(z/e)2 =2−1/2(1 z)−3/2+O(1 z −1).
⊙ − | − |
Consequently,
z ∂ [B(ω) C(ω/e)2] z 3(1 ω) 5/2
ω ⊙ dω = − − +O(1 ω −2) dω
C(ω/e) 2√2 | − |
Z0 Z0 (cid:20) (cid:21)
1
= (1 z) 3/2+O(1 z 1).
− −
√2 − | − |
Finally by the relation (9) we have
1
A(z) C(z/e)= (1 z) 2+O(1 z 3/2). (14)
− −
⊙ 4 − | − |
Moreover,for α positive, we have (see [6], for example)
n+α 1 Γ(n+α)
[zn](1 z)−α = −
− n Γ(α)Γ(n+1)
(cid:18) (cid:19)
nα 1
−
= (1+O(1/n)), (15)
Γ(α)
where[zn](1 z) αdenotesthen-thcoefficientofznintheexpansionof(1 z) α
− −
− −
in entire series. The last equality is obtained applying the Stirling formula.
Then, by the expansion of (14) and singularity analysis, we obtain
n
a c e n = (1+O(1/n))+O(n1/2).
n n −
4Γ(2)
Finally with c = n−3/2en(1+O(1/n)), we obtain (12).
n √2π
Now estimating the moments of higher order, we return to the recurrence
(1). For k 0,n 1, put
≥ ≥ µ (k):=E(X k),
n n
and
µ˜ (k):=c e nµ (k).
n n − n
LetM (z)denote the ordinarygeneratingfunction ofµ˜ (k), with z markingn.
k n
For k =1,
µ˜ (1):=c e na and M (z)=A(z) C(z/e).
n n − n 1
⊙
Cost of cutting down random trees 9
For k 2, we have
≥
k
X k = Xk1Xk2 b k3,
n k1+kX2+k3=k(cid:18)k1,k2,k3(cid:19) Ln n−Ln n
or again
k
X k =Xk +Xk + Xk1Xk2 b k3.
n Ln n−Ln k1+k1kX,2k+2<k3k=k(cid:18)k1,k2,k3(cid:19) Ln n−Ln n
Conditioning on the size of L , we obtain
n
n
k n c c
µn(k) = n2k3 j n−jµj(k1)µn j(k2)
k1+k1kX,2k+2<k3k=k(cid:18)k1,k2,k3(cid:19) Xj=1 2(n−1) cn −
n
n c c
j n j
+ − (µj(k)+µn j(k)).
2(n 1) cn −
j=1 −
X
Multiplying the latter by n 1c , we obtain
n−en n
n 1
n 1 − cn j
−n µ˜n(k)= en−jµ˜j(k)+rn(k), (16)
j=1 −
X
where
n 1
k − 1
r (k)= b k3 µ˜ (k )µ˜ (k ).
n n j 1 n j 2
k1+k1kX,2k+2<k3k=k(cid:18)k1,k2,k3(cid:19) Xj=1 2 −
Let R (z) denote the ordinarygeneratingfunction ofr (k), with z marking
k n
n. Therefore
k
Rk(z)= k ,k ,k (B(z)⊙k3)⊙[1/2Mk1(z)Mk2(z)], (17)
k1+k1kX,2k+2<k3k=k(cid:18) 1 2 3(cid:19)
where
B(z)⊙k3 :=B(z) B(z).
⊙···⊙
k3 time
Multiplying (16) by zn and summing| over n{z 1, we}obtain
≥
z dω
M (z)= M (ω) M (z)C(z/e)+R (z),
k k k k
ω
Z0
which is identified in the equality (7) if there we choose f(z) = M (z) and
k
t(z)=R (z). Finally, the solution of this equation is
k
C(z/e) z dω
M (z)= ∂ R (ω) . (18)
k ω k
1 C(z/e) C(ω/e)
− Z0
Cost of cutting down random trees 10
Proposition 1. For k 1, the generating function M (z) of µ˜ (k) satisfies
k n
≥
√2
Mk(z)= Ak (1 z)−5k/2+21 +O(1 z −5k/2+1), (19)
2 − | − |
where the coefficients A are defined by the recurrence
k
k 1
Ak = − k AjAk−j +kAk 1Γ(5k/2−1), k 2; A1 =2−3/2. (20)
j 2 − Γ(5k/2 3) ≥
j=1(cid:18) (cid:19) −
X
Proof. The proof is carried out by induction. For k = 1, the proposition has
been established in view of (14). For k 2, we demonstrate that R (z) has a
k
≥
singular expansion in the form
Rk(z)=Ak(1 z)−5k/2+1+O(1 z −5k/2+23). (21)
− | − |
Analyzing the various terms on the right hand side of (17), we observe that A
k
are defined by the recurrence (20):
(I) By induction hypothesis, when k and k are both nonzero, and k =0,
1 2 3
the contribution to R (z) is
k
21Mk1(z)Mk2(z) = 12 Ak1(1−z)−52k1+12 +O(|1−z|−52k1+1)
× Ahk2(1−z)−52k2+12 +O(|1−z|−52k2+1) i
= h12Ak1Ak2(1−z)−25k+1+O(|1−z|−25k+i3/2).
(II) When k , k and k are all nonzero, by relation (11) and the relation
1 2 3
below
12Mk1(z)Mk2(z)= 2Γ(5A(kk11+Ak2k)2 1)Li−25k+5k23+2,0(z)+O(|1−z|−5(k12+k2)+3/2),
2 −
and since B(z)⊙k3 =Li−2k3,0(z), the contribution to Rk(z) is
1 A A
Li (z) [ M (z)M (z)] = k1 k2 Li (z)
−2k3,0 ⊙ 2 k1 k2 2Γ(5(k1+k2) 1) −25k+k23+2,0
2 −
+ Li−2k3,0(z)⊙O(|1−z|−5(k12+k2)+3/2)
= O(1 z −25k+3/2).
| − |