Improved Master Theorems for Divide-and-Conquer
Recurrences
SALVADOR ROURA
Universitat Politècnica de Catalunya, Barcelona, Catalonia, Spain
Abstract. This paper presents new theorems to analyze divide-and-conquer recurrences, which
improve other similar ones in several aspects. In particular, these theorems provide more information, free us almost completely from technicalities like floors and ceilings, and cover a wider set of
toll functions and weight distributions, stochastic recurrences included.
Categories and Subject Descriptors: F.2 [Analysis of Algorithms and Problem Complexity]; G.2.1
[Discrete Mathematics]: Combinatorics—recurrences and difference equations
General Terms: Algorithms, Measurement, Performance
Additional Key Words and Phrases: Asymptotic analysis, divide-and-conquer, master theorem
1. Introduction
This work presents new theorems to solve many divide-and-conquer recurrences
that arise in practice. Recall that a recurrence is a definition of a function F n in
terms of the values of F at indices smaller than n; a recurrence is divide-andconquer—DAC, for short—if the average size of these indices is a fraction of n.
Our reference source here is the Master Theorem—MT, for short—as it can be
found in Sedgewick and Flajolet [1996]. Other references in this subject include
the (classic) Master Theorem [Aho et al. 1974; Bentley et al. 1980; Cormen et al.
1990], several improvements [Kao 1997; Verma 1994; 1997; Wang and Fu 1996]
as well as other related results [Karp 1994].
Assume that we have the recurrence F n ⫽ t n ⫹ W 䡠 F S n , with t n ⬎ 0 and S n ⫽
Z 䡠 n ⫹ ᏻ(1) for some 0 ⬍ Z ⬍ 1. If F n describes the cost to solve with a
certain algorithm a problem of size n, then t n — customarily called toll function—is the cost of the divide and combine steps, W is the fixed number of
This research was supported by the ESPRIT BRA Project ALCOM II, contract #7141, by the
ESPRIT LTR Project ALCOM-IT, contract #20244 and by a grant from CIRIT (Comissió Interdepartamental de Recerca i Innovació Tecnològica). A previous version of this work was partially
written while the author was visiting Princeton University.
Author’s address: Departament de Llenguatges i Sistemes Informàtics, Universitat Politècnica de
Catalunya, E-08028 Barcelona, Catalonia, Spain. E-mail: [email protected].
Permission to make digital / hard copy of part or all of this work for personal or classroom use is
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© 2001 ACM 0004-5411/01/0300-0170 $05.00
Journal of the ACM, Vol. 48, No. 2, March 2001, pp. 170 –205.
Improved Master Theorems for Divide-and-Conquer Recurrences
171
recursive calls at each step, and S n is the size of the subproblems to be
recursively solved. Let ␣ ⫽ ⫺logZ W. Then, the MT states that the solution to
the recurrence is
Fn ⫽
冦
⌰ 共 n ␣兲 ,
if
⌰共tn log n兲, if
⌰ 共 t n兲 ,
if
tn ⫽ ᏻ共na兲 and a ⬍ ␣;
tn ⫽ ⌰共n␣ logc n兲 and c ⱖ 0;1
tn ⫽ ⍀共na兲 and a ⬎ ␣.
(To be completely rigorous, the last case above requires additional conditions of
smoothness for the toll function t n .)
Usually, a MT does not provide an explicit solution but partial information
about the function under study, like bounds on its growing order. Nevertheless, a
MT has two interesting properties: First, it typically provides results which rely
exclusively upon the asymptotic behaviour of the toll function and of the
distribution of weights; for instance, in the MT above we only use the main term
of t n and the value ␣, which depends on W and Z. In particular, this means that
the values of F n at small indices are irrelevant to most MTs. Second, a MT is
easy and fast to use. As an example, we can solve the DAC recurrence
M n ⫽ n ⫺ 1 ⫹ M n/ 2 ⫹ M n/ 2
(1)
for n ⱖ 2, with M 0 ⫽ M 1 ⫽ 0, which defines the number of comparisons to sort
an array of n keys with mergesort in the worst case [Flajolet and Golin 1994].
The term ᏻ(1) in the definition of S n covers expressions with floors and ceilings
in the argument of the recursive call. Hence, a direct application of the rules
above yields M n ⫽ ⌰(n log n).
In this paper, we will make the MT more flexible and more informative. For
instance, consider the recurrence
Bn ⫽ 1 ⫹
共 n ⫺ 1 兲 / 2 
n
䡠 B (n⫺1)/ 2 ⫹
共 n ⫺ 1 兲 / 2 
n
䡠 B (n⫺1)/ 2
(2)
for n ⱖ 2, with B 0 ⫽ 0 and B 1 ⫽ 1, which defines the expected number of steps
of a binary search for a random key in an array of size n. This recurrence does
not follow the MT pattern utterly, since we have 1 ⫺ 1/n expected recursive calls
at each step. Therefore, to use the MT we must assume that the solution to the
recurrence F n ⫽ 1 ⫹ F n/ 2 must be close to B n —which is true—, and conclude
B n ⫽ ⌰(log n). We need a posterior reasoning to rigorously prove that this
approximation does not lead to a wrong answer.
By contrast, the MT presented in Section 2 directly deals with recurrences
where the number of recursive calls is not constant but tends to a constant. As a
consequence, we will get rid of the annoying technicalities related to factors with
floors and ceilings. And we will see that in some cases it is possible to get the
multiplying factor of the dominating term of the function under study. This
improvements will yield the result B n ⫽ log2 n ⫹ o(log n) for the example
above.
1
Through all this work, logc n means (log n) c .
172
SALVADOR ROURA
Yet we will extend the MT further, to deal with recurrences where the
asymptotic sizes of the subproblems to be recursively solved consist in several
fixed fractions of the original problem,2 like
冉 冑冊
Fn ⫽ tn ⫹ 2 ⫺
1
n
F (n⫹1)/3 ⫹ 4F (n/ 2)⫺ 冑7 n ⫹
冉 冊
1
2
⫹
1
n
F (4n/5)⫹ln2 n
(3)
for n large enough. (Since all the theorems in this work rely only upon the
asymptotic properties of the toll function and the distribution of weights, we can
avoid explicitly stating the initial values of the recurrence.)
It is frequent that the analysis of the cost of a given algorithm or data structure
results in a stochastic recurrence, which does not follow any of the patterns
mentioned above. As an example of stochastic recurrence, consider
Qn ⫽ n ⫹ 1 ⫹
2
n
冘
Qk
(4)
0ⱕk⬍n
for n ⱖ 1, with Q 0 ⫽ 0. This recurrence defines the expected number of
comparisons to sort an n-key random array with quicksort. Clearly, the MTs we
know so far cannot solve it, since the size of the subarrays to be recursively
sorted is not fixed; it can be either an insignificant part of the whole input or
almost it all. In this case, though, other techniques lead us to a closed solution
[Hoare 1962].
Much more difficulties presents the analysis of the stochastic recurrence
Qn ⫽ 1 ⫹
4
n共n ⫹ 1兲
冘 共n ⫺ k兲Q
k
(5)
0ⱕk⬍n
for n ⱖ 1, with Q 0 ⫽ 0, where Q n denotes the expected number of comparisons
during a half-defined search in a random quad-tree of size n. An accurate
asymptotic expression for Q n can be obtained after a thorough analysis by means
of generating functions [Flajolet et al. 1993], but this technique requires a deep
knowledge and expertise.
Section 3 presents a new MT for stochastic recurrences, which shares the good
properties of the MT in Section 2: it is simple and fast to use, and only the main
term of t n and information on the asymptotic distribution of weights will be
relevant. In addition, we will see how a simple iteration of the former MT
sometimes yields several of the main terms of the function under study with their
corresponding multiplicative factors.
The rest of the paper is organized as follows: Sections 4 and 5 present
general-purpose theorems that solve many DAC recurrences that do not have to
follow any particular pattern. Those theorems will allow us to prove in Sections 6
and 7 the MTs given in Sections 2 and 3, but also have other applications. For
2
To the best of the author’s knowledge, Kao [1997] was the first to develop a basic MT for this kind
of recurrences in 1986 (personal communications). Some additional results about these recurrences
may be found in Verma [1997].
Improved Master Theorems for Divide-and-Conquer Recurrences
173
instance, they are useful for the analysis and asymptotic improvement of
quicksort and quickselect [Martı́nez and Roura 1998].
In Section 8, we will see that it is possible to directly compute the main term of
the variance of the cost of some algorithms, if the nonrecursive cost is large
enough and the algorithm follows one of two typical patterns (quickselect and
quicksort are prototypical examples of those patterns).
Finally, Section 9 presents some open problems.
Preliminary versions of this work appeared in Roura [1997a; 1997b].
2. The Discrete Master Theorem
In this section, we cast in the form of a Master Theorem some of the results that
will be proved in latter sections. This MT deals with recurrences like (1), (2), and
(3), where the problem is broken into pieces such that the asymptotic size of each
one is a fixed fraction of the size of the whole problem. We call them discrete
recursive definitions.
Definition 2.1.
We say that
Fn ⫽
再
bn ,
tn ⫹
冘
1ⱕdⱕD
if
Rd,nFSd, n , if
0ⱕn⬍N
nⱖN
is a discrete recursive definition of F n iff D ⱖ 1; R d, n ⫽ w d ⫹ r d, n ⱖ 0, where
w d ⬎ 0 and 兺 1ⱕdⱕD 兩r d, n 兩 ⫽ ᏻ(n ⫺ ␳ ) for some ␳ ⬎ 0; and S d, n ⫽ z d 䡠 n ⫹ s d, n ,
where 0 ⬍ z d ⬍ 1 and 兺 1ⱕdⱕD 兩s d, n 兩/n ⫽ ᏻ(n ⫺ ␴ ) for some ␴ ⬎ 0.
Here, D is the finite number of subproblems to be recursively solved; R d, n is
the number of recursive calls to deal with the dth subproblem, where w d is the
asymptotic number of calls to it; and S d, n is the (integer) size of the dth
subproblem, where z d is the asymptotic fraction of the original problem to be
solved by the dth recursive call.
For example, (1) is a discrete recursive definition. There we have two
subproblems to recursively deal with: D ⫽ 2; whose size is asymptotically 1/2 of
the size of the original problem: z 1 ⫽ z 2 ⫽ 1/ 2, ⫺1 ⱕ s 1, n ⱕ 0, 0 ⱕ s 2, n ⱕ 1;
and there is exactly one call to each one: w 1 ⫽ w 2 ⫽ 1, r 1, n ⫽ r 2, n ⫽ 0. For the
bounds of s n, k , we have used the fact that r ⫺ 1 ⱕ r ⱕ r and r ⱕ r ⱕ r ⫹
1 for every real r. Notice that ␳ ⫽ ␴ ⫽ 1 is a possible choice in this example.
The Discrete MT (Theorem 2.3) covers the presence of sublogarithmical
factors in the toll function, that is, factors whose growing order is smaller than
log␧ n for any ␧ ⬎ 0. This allows us to deal with toll functions like t n ⫽ n 2 ln3 n 䡠
ln ln n, for instance. We define precisely this concept.
Definition 2.2. Let ␮ n be a strictly positive nondecreasing function for large
n. Moreover, assume that for every ␧ ⬎ 0 the function log␧ n/ ␮ n is increasing as
long as n is large enough. Then we say that ␮ n is a sublogarithmical function.
Now we are ready to state the Discrete MT (examples of its use can be found
in Appendix A).
THEOREM 2.3 (DISCRETE MASTER THEOREM). Let Fn be a function defined by
a discrete recursive definition, and let B na lnc n 䡠 ␰n be the main term of tn, where
B ⬎ 0, a and c are arbitrary constants, and ␰n ⫽ ␮n or ␰n ⫽ 1/␮n for some
174
SALVADOR ROURA
sublogarithmical function ␮n. Let ⌽( x) ⫽ 兺1ⱕdⱕD wd 䡠 ( zd)x, and Ᏼ ⫽ 1 ⫺ ⌽(a).
Then,
(1) if Ᏼ ⬎ 0, then F n ⬃ t n /Ᏼ;
(2) if Ᏼ ⫽ 0, then
(2.1) if c ⬎ ⫺1, then F n ⬃ t n ln n/Ᏼ⬘, where Ᏼ⬘ ⫽ ⫺(c ⫹ 1)兺 1ⱕdⱕD w d 䡠
( z d ) a ln z d ;
(2.2) if c ⫽ ⫺1, then F n ⫽ ᏻ(n a log␧ n) for any ␧ ⬎ 0, and F n ⫽ ⍀(n a ) if
F n ⱖ 0 for every n ⱖ 0;
(2.2⬘) if c ⫽ ⫺1 and ␮ n ⫽ 1, then F n ⬃ t n ln n 䡠 ln ln n/Ᏼ⬙, where Ᏼ⬙ ⫽
⫺兺 1ⱕdⱕD w d 䡠 ( z d ) a ln z d ;
(2.3) if c ⬍ ⫺1, then F n ⫽ ᏻ(n a ) (F n ⫽ ⌰(n a ), if F n ⱖ 0 for every n ⱖ
0);
(3) if Ᏼ ⬍ 0, then F n ⫽ ᏻ(n ␣ ) (F n ⫽ ⌰(n ␣ ), if F n ⱖ 0 for every n ⱖ 0), where
␣ is the unique solution of ⌽( ␣ ) ⫽ 1.
PROOF.
Section 6 is devoted to prove this theorem.
e
The Discrete MT could be trivially adapted to deal with negative toll functions
by changing the conditions “B ⬎ 0” and “F n ⱖ 0 for every n ⱖ 0” to “B ⬍ 0”
and “F n ⱕ 0 for every n ⱖ 0”, respectively.
Also, though we will not prove it in this work, we could relax Definition 2.1 to
permit the value z d ⫽ 1 for some indices, as long as (a) there is, at least, one
index d such that 0 ⬍ z d ⬍ 1; and (b) the total sum of weights w d of the indices
d with z d ⫽ 1 is strictly smaller than 1. Notice that neither
A n ⫽ t n ⫹ A n⫺3
(6)
B n ⫽ t n ⫹ 2B n/ 2 ⫹ 2B n⫺1
(7)
nor
is, even under the relaxation above, a discrete recursive definition. The first
recurrence does not satisfy any of the two conditions, while the second recurrence fails to fulfill the condition (b). They are out of the scope of this paper. By
contrast, the recurrence
F n ⫽ t n ⫹ F n/ 2 ⫹
1
2
䡠 F n⫺2
is a discrete recursive definition, and so we could analyze it exactly as stated by
the Discrete MT.
The case z d ⫽ 0 for some indices 1 ⱕ d ⱕ D is especially awkward. We avoid
dealing with it.
3. The Continuous Master Theorem
This section covers the analysis of recurrences like (4) and (5), which we will call
continuous. But first we need to define the concept of shape function (the reason
for this name will be clear after Definition 3.2).
Improved Master Theorems for Divide-and-Conquer Recurrences
175
Definition 3.1. Let ␻ ( z) ⱖ 0 be a function over [0, 1] such that 兰 10 ␻ ( z) dz
exists and is at least 1. Furthermore, assume that there is some ␮ ⬍ 0 such that
兰 10 ␻ ( z) z ␮ dz also converges.3 Then we say that ␻ ( z) is a shape function.
Definition 3.2.
We say that
Fn ⫽
再
bn ,
tn ⫹
0ⱕn⬍N
if
冘
0ⱕk⬍n
␻n,kFk , if
nⱖN
is a continuous recursive definition of F n iff there exist some shape function ␻ ( z),
some constant 0 ⬍ q ⱕ 1 and some function M n ⫽ ⌰(n q ) with integer values
such that, if we define z n, j ⫽ j/M n for every 0 ⱕ j ⱕ M n , I n, j ⫽ [ z n, j 䡠 n,
z n, j⫹1 䡠 n) for every 0 ⱕ j ⬍ M n , and
␧ n, j ⫽
冏冘
冕
␻ n, k ⫺
k僆I n, j
z n, j⫹1
␻ 共 z 兲 dz
z n, j
冏
(8)
for every 0 ⱕ j ⬍ M n , then 兺 0ⱕj⬍M n ␧ n, j ⫽ ᏻ(n ⫺ ␳ ) for some ␳ ⬎ 0.
Usually, M n ⫽ n is a possible choice, and in this case (8) reduces to
冏
␧ n, k ⫽ ␻ n, k ⫺
冕
(k⫹1)/n
冏
␻ 共 z 兲 dz .
k/n
Loosely speaking, we can use the integral in the expression above to find a good
approximation to ␻ n, k . Notice that there can only be one shape function ␻ ( z)
related to a given continuous recursive definition, except for bizarre shape
functions obtained from ␻ ( z) by changing its value at a finite number of points,
or by similar minor perturbations.
For instance, consider the recurrence
Sn ⫽ n ⫹ 1 ⫹
冘
2
n2
kS k
(9)
0ⱕk⬍n
for the expected number of comparisons to select the ith out of n keys using
quickselect—also known as FIND [Hoare 1961]—when i is chosen at random. Its
shape function is ␻ ( z) ⫽ 2z (see Figure 1), since
冕
(k⫹1)/n
k/n
␻ 共 z 兲 dz ⫽
冕
(k⫹1)/n
k/n
2z dz ⫽ z
2
冏
(k⫹1)/n
⫽
k/n
2k
n
2
⫹
1
n
2
⫽ ␻ n, k ⫹
1
n2
,
and hence ␧ n, k ⫽ 1/n 2 and 兺 0ⱕk⬍n ␧ n, k ⫽ 1/n ⫽ ᏻ(n ⫺ ␳ ) with ␳ ⫽ 1.
Therefore, ␻ ( z) is nothing except the asymptotic shape of the distribution of
weights, which now is very similar to a continuous probability distribution, where
the area beneath the function is the asymptotic number of recursive calls. Since,
by definition, 兰 10 ␻ ( z) dz ⱖ 1, we are assuming that there is at least one
3
The author’s conjecture is that this technical property is always true if 兰10 ␻ ( z) dz exists and is at
least 1. Section 7 provides more details.
176
SALVADOR ROURA
FIG. 1.
Shape function for FIND.
recursive call for large n. This condition (very likely to hold in practice)
simplifies the study of these recurrences.
On the other hand, we have allowed M n to be ⌰(n q ) for some q ⱕ 1, as long
as q ⬎ 0. So it is not necessary that ␻ ( z) fits the n weights individually, but a
polynomial number of groups of weights. This relaxation could be useful when
dealing with particularly difficult recurrences.
THEOREM 3.3 (CONTINUOUS MASTER THEOREM). Let Fn be a function defined
by a continuous recursive definition, and let B na lnc n 䡠 ␰n be the main term of tn,
where B ⬎ 0, a and c are arbitrary constants, and ␰n ⫽ ␮n or ␰n ⫽ 1/␮n for some
sublogarithmical function ␮n. Let ␸( x) ⫽ 兰10 ␻( z) zx dz, and Ᏼ ⫽ 1 ⫺ ␸(a). Then,
(1) if Ᏼ ⬎ 0, then F n ⬃ t n /Ᏼ;
(2) if Ᏼ ⫽ 0, then
(2.1) if c ⬎ ⫺1, then F n ⬃ t n ln n/Ᏼ⬘, where Ᏼ⬘ ⫽ ⫺(c ⫹ 1) 兰 10 ␻ ( z) z a
ln z dz;
(2.2) if c ⫽ ⫺1, then F n ⫽ ᏻ(n a log␧ n) for any ␧ ⬎ 0, and F n ⫽ ⍀(n a ) if
F n ⱖ 0 for every n ⱖ 0;
(2.2⬘) if c ⫽ ⫺1 and ␮ n ⫽ 1, then F n ⬃ t n ln n 䡠 ln ln n/Ᏼ⬙, where Ᏼ⬙ ⫽
⫺兰 10 ␻ ( z) z a ln z dz;
(2.3) if c ⬍ ⫺1, then F n ⫽ ᏻ(n a ) (F n ⫽ ⌰(n a ), if F n ⱖ 0 for every n ⱖ
0);
(3) if Ᏼ ⬍ 0 (including the case Ᏼ ⫽ ⫺⬁) then F n ⫽ ᏻ(n ␣ ) (F n ⫽ ⌰(n ␣ ), if
F n ⱖ 0 for every n ⱖ 0), where ␣ is the unique solution of ␸ ( ␣ ) ⫽ 1.
PROOF.
Section 7 is devoted to prove this theorem.
e
To use the Continuous Master Theorem, we must first identify the shape
function ␻ ( z) for the distribution of weights. This is equivalent to identifying the
values {w d } 1ⱕdⱕD and { z d } 1ⱕdⱕD for the discrete case. One possibility is to
conjecture that ␻ ( z) ⫽ n 䡠 ␻ n, zn . For instance, for (9) this technique yields
␻ ( z) ⫽ n 䡠 2/n 2 䡠 zn ⫽ 2z, which we have already proved to be right. But for (5)
we have additional problems, since the expression n 䡠 ␻ n, zn ⫽ 4(1 ⫺ z)/(1 ⫹
Improved Master Theorems for Divide-and-Conquer Recurrences
177
1/n) includes n. Lemma 7.2 in Section 7 provides a way to find ␻ ( z) that works
in many practical situations.
Examples of how to use the Continuous Master Theorem can be found in
Appendix B.
4. The Core Theorems
In this section and Section 5, we will prove some general theorems that apply to
many DAC recurrences. In Sections 6 and 7, we will see how to use these results
to prove the statements of both the Discrete and the Continuous MTs.
We begin introducing the concept of divide-and-conquer recursive definition
formally.
Definition 4.1.
Let F n be a function defined for every n ⱖ 0, and let
0ⱕk⬍n
Ᏺ ⫽ 关 N, 兵 b n 其 0ⱕn⬍N , 兵 t n 其 nⱖN , 兵 w n, k 其 nⱖN
兴
be a recursive definition, with N ⱖ 1 and w n, k ⱖ 0. For every n ⱖ N, let W n ⫽
兺 0ⱕk⬍n w n, k and Z n ⫽ 兺 0ⱕk⬍n (w n, k /W n )(k/n). Then we say that Ᏺ is a DAC
recursive definition of F n if and only if (a) F n ⫽ b n for every 0 ⱕ n ⬍ N; (b) for
every n ⱖ N,
Fn ⫽ tn ⫹
冘
w n, k F k ;
(10)
0ⱕk⬍n
and (c) it exists some upper bound U ⬍ 1 such that Z n ⱕ U for all n large
enough. We also define W ⫽ limn3⬁ W n and Z ⫽ limn3⬁ Z n , if they exist.
This definition can be easily interpreted: W n is the total number of recursive
calls to solve a problem of size n; Z n is the average fraction of the original
problem that is solved by a recursive call (it turns out that 0 ⱕ Z n ⬍ 1); and for
large n, the average size of the recursive calls is, at most, a fraction of the whole
problem. For example, for (9), we have W n ⫽ 1 ⫺ 1/n, W ⫽ 1, Z n ⫽ 2/3 ⫺
1/3n and Z ⫽ 2/3, while for (1) we have W n ⫽ 2, W ⫽ 2, Z n ⫽ 1/ 2 and Z ⫽
1/ 2. By contrast, (6) is not a DAC recursive definition.
Most of the results from which both MTs are derived refer to canonical
recursive definitions, which are defined as follows:
Definition 4.2. Let Ᏺ be a DAC recursive definition of a function F n . We say
that Ᏺ is a canonical recursive definition if and only if both these properties hold:
(a) The constant W exists and is equal to 1. (b) If we define m n ⫽ W n ⫺ W, then
兩m n 兩 ⫽ ᏻ(n ⫺ ␳ ) for some ␳ ⬎ 0.
Hence, a DAC recursive definition is canonical if the number of recursive calls
tends to 1, with a minimum convergence speed. For instance, for (9), we have
W ⫽ 1, and 兩m n 兩 ⫽ 1/n ⫽ ᏻ(n ⫺ ␳ ) with ␳ ⫽ 1.
Now, assume that F n ⱖ 0 is defined by (10). Then we have F n ⫽ ⍀(t n ),
because F n cannot grow more slowly than t n does. So the question is identifying
under which conditions F n can grow faster than t n . For the recursive definitions
we deal with and roughly speaking, we will prove that there is a growing order
⌰(n ␣ ) associated to every distribution of weights, irrespectively of how small t n
178
SALVADOR ROURA
is. This would be similar to state that F n ⫽ ⌰(max{t n , n ␣ }), which is almost
true.
For instance, consider the recurrence
F n ⫽ t n ⫹ F n/4 .
(11)
We will see in this section that ␣ ⫽ 0 for any canonical recurrence, such as this
one. Therefore, for “large” values of t n (case 1 of the MTs) like n, n 3 or even 2 n ,
we should get F n ⫽ ⌰(t n ), which is true. For “small” values of t n (case 3 of the
MTs) like 1/n, 1/n 3 or 2 ⫺n , F n should be ⌰(1), which is also true. However,
things are not so easy for values of t n close to ⌰(1) (case 2 of the MTs). For
example, for t n ⫽ 1 the growing order of F n turns out to be ⌰(log n) instead of
⌰(1). We will see in Section 5 why this additional factor appears.
At this point, it is worth commenting on the word “Core” in the title of this
section. It tries to suggest the idea that recursive definitions with a toll function
small enough—and thus inside the zone dominated by the term ⌰(n ␣ ) associated
to the distribution of weights—are the most difficult to analyze. Indeed, there is
no way to find the lower order terms of F n nor even the multiplicative factor of
the main term n ␣ , but to consider all the values of F n , the values at small indices
included.
For example, consider (11), and assume that it holds for n ⱖ 1. Let t n ⬎ 0 be
small enough. Then F n ⫽ ⌰(1), but modifying F 0 yields a significant change in
F n for every n. In fact, changing the value of the function at any index
significantly affects F at an infinite number of indices. Set in terms of a recursion
tree (see Cormen et al. [1990], for example), the solution to the recurrence is
dominated by the values at the leaves.
Moreover, in some cases, the multiplicative factor of the main term n ␣ is not
even asymptotically constant. Consider the following recurrence,
Fn ⫽ tn ⫹
再
F 2n/4 ,
if
F2(n⫺1)/4⫹1 , if
n is even
n is odd
for every n ⱖ 2. Let t n ⬎ 0 be small enough and such that t 2n ⫽ t 2n⫹1 , and let
F 0 ⫽ 0 and F 1 ⫽ 1. Then it is easy to see that F n ⫽ ⌰(1), but F n cannot tend
to a constant, since it fluctuates periodically.
We end this section with two theorems which formalize one of the claims
above, namely that any canonical recurrence is on the one hand ⍀(1), and on the
other hand ᏻ(1) for t n small enough. The three technical conditions in the
statement of Theorem 4.3 typically hold and, loosely speaking, avoid cases like
“everything is zero” and “the negative and the positive contributions cancel out.”
THEOREM 4.3. Let Fn be defined by a canonical recursive definition, and let
bn ⱖ 0, tn ⬎ 0 and 兺0ⱕk⬍N wn,k ⫽ ᏻ(n⫺␴) for some ␴ ⬎ 0. Then Fn ⫽ ⍀(1).
PROOF. By hypothesis, 兩m n 兩 ⫽ 兩W n ⫺ 1兩 ⫽ ᏻ(n ⫺ ␳ ), where ␳ ⬎ 0. Choose
some ␯ such that 0 ⬍ ␯ ⬍ max{␳, ␴}, and define W⬘n ⫽ 兺 Nⱕk⬍n w n, k . Taking
into account that W n ⱖ 1 ⫺ 兩m n 兩, we have W⬘n ⫽ W n ⫺ 兺 0ⱕk⬍N w n, k ⱖ 1 ⫹
ᏻ(n ⫺ ␳ ) ⫺ ᏻ(n ⫺ ␴ ) ⱖ 1 ⫺ n ⫺ ␯ , as long as n is large enough.
Improved Master Theorems for Divide-and-Conquer Recurrences
179
Now define
Z⬘n ⫽
冘
Nⱕk⬍n
w n, k
1⫺n
⫺␯
䡠
k
n
ⱕ
冘
wn,k k
Wn
Wn
䡠 ⫽
䡠 Zn ⱕ
䡠 U,
⫺␯
1⫺n
1 ⫺ n ⫺␯
0ⱕk⬍n Wn n
Wn
1⫺n
⫺␯
where 0 ⬍ U ⬍ 1 is the upper bound for the Z n ’s that exists by definition.
Choose any V such that U ⬍ V ⬍ 1. Then it is clear that Z⬘n ⱕ V for large n.
For the next step of the proof, choose some N⬘ ⬎ N large enough to get W⬘n ⱖ
1 ⫺ n ⫺ ␯ , Z⬘n ⱕ V, and also n ⫺ ␯ ⱕ 1/ 2 for every n ⱖ N⬘. Let h( x) ⫽
exp(2V ␯ /(1 ⫺ V ␯ ) x ␯ ). In what follows, we will use the easy-to-prove fact that
h( x) ⱖ 1 is a decreasing convex function in the interval [1, ⫹⬁).
Set M ⫽ min{F n /h(n)} Nⱕn⬍N⬘ . Notice that b n ⱖ 0 and t n ⬎ 0 imply M ⬎ 0.
We can prove by induction that F n ⱖ M 䡠 h(n) holds not only for all N ⱕ n ⬍
N⬘ but also for all n ⱖ N⬘: Assuming the induction hypothesis, we have
Fn ⫽ tn ⫹
冘
i.h.
w n, k F k ⱖ
冘
wn,kM 䡠 h共k兲 ⫽ M 䡠 W⬘n
Nⱕk⬍n
0ⱕk⬍n
冘
Nⱕk⬍n
wn,k
W⬘n
䡠 h共k兲.
The convexity of h( x), the bound for W⬘n and the fact that h( x) is decreasing
yields
F n ⱖ M 䡠 W⬘n 䡠 h
冉
冘
Nⱕk⬍n
w n, k
W⬘n
冊
䡠 k ⱖ M 共 1 ⫺ n ⫺␯兲 䡠 h
冉
冘
Nⱕk⬍n
wn,k
1 ⫺ n ⫺␯
冊
䡠k .
From the definition of Z⬘n , F n ⱖ M(1 ⫺ n ⫺ ␯ ) 䡠 h(nZ⬘n ). Taking into account
that 1 ⫺ x ⱖ exp(⫺2x) for all 0 ⱕ x ⱕ 1/ 2, and replacing x by n ⫺ ␯ , we get 1 ⫺
n ⫺ ␯ ⱖ exp(⫺2n ⫺ ␯ ). Furthermore, Z⬘n ⱕ V and h( x) is decreasing, and thus we
have F n ⱖ M exp(⫺2n ⫺ ␯ ) 䡠 h(nV) ⫽ M 䡠 h(n), the last step obtained through
simple manipulations. This ends the inductive proof.
Finally, h(n) ⱖ 1 yields F n ⱖ M ⬎ 0 for every n ⱖ N, and the statement of
the theorem follows. e
THEOREM 4.4. Let Fn be defined by a canonical recursive definition, and let
tn ⫽ ᏻ(logc n) for some c ⬍ ⫺1. Then Fn ⫽ ᏻ(1).
PROOF. Let Ᏺ be the recursive definition for F n . First of all, we introduce a
recursive definition Ᏻ for a new function G n , by modifying Ᏺ as follows: Let ␴ n, k
denote the weights of Ᏺ, Y n denote 兺 0ⱕk⬍n ␴ n, k , and assume without loss of
generality that Y n ⬎ 0—recall that limn3⫹⬁ Y n ⫽ 1. Define w n, k , the weights of
Ᏻ, to be equal to ␴ n, k if Y n ⱖ 1, and equal to ␴ n, k /Y n if Y n ⬍ 1 (this is just a
normalization of the weights so that they sum at least 1). Let u n be the toll
function of Ᏺ; we define the toll function of Ᏻ as t n ⫽ 兩u n 兩. Altogether, ⫺G n ⱕ
F n ⱕ G n trivially holds for every n ⱖ 0, so it is enough to bound G n by above to
prove the theorem.
Let W n , W, etc. be as in Definitions 4.1 and 4.2 but for the set of weights w n, k
of Ᏻ. Then, on the one hand, we have that W is still 1, but now W n ⱖ 1 and
m n ⱖ 0. On the other hand Z n remains the same and hence Z n ⱕ U for some
0 ⬍ U ⬍ 1. Choose some V such that U ⬍ V ⬍ 1, and some ␯ such that 0 ⬍
␯ ⬍ ␳, where ␳ ⬎ 0 is the constant in Definition 4.2. Choose some a ⱖ 1 large
180
SALVADOR ROURA
enough to get V ␯ /(1 ⫺ V ␯ ) ⱕ a ␯ . We introduce two auxiliar functions, f( x) ⫽
exp(⫺V ␯ /(1 ⫺ V ␯ )( x ⫹ a) ␯ ), and g( x) ⫽ 兺 iⱖ1 g i ( x) c , where g i ( x) ⫽ ln( x ⫹
a) ⫺ i ln V. It is not difficult to prove that, for every x ⱖ 0, 0 ⬍ f( x) ⱕ 1 is an
increasing concave function, and g( x) ⱖ 0 is a well-defined decreasing convex
function.
There are three bounds for the quantities introduced so far that we will find
helpful through this proof (they all hold for n large enough, say for n ⱖ N⬘ for
some constant N⬘). First, taking into account that m n ⫽ ᏻ(n ⫺ ␳ ), we have 0 ⱕ
m n ⱕ 1/((n ⫹ a) ␯ ⫹ 1). Second, we define Z⬘n ⫽ (nZ n ⫹ a)/(n ⫹ a) ⫽ Z n ⫹
a(1 ⫺ Z n )/(n ⫹ a) ⱕ U ⫹ a/n ⱕ V. Finally, by the initial hypothesis about t n
we know that there exists some constant K ⬎ 0 such that t n ⱕ K lnc (n ⫹ a).
Assume n ⱖ N⬘, and introduce
Sn ⫽
冘
冘
w n, k f 共 k 兲 ⫽
0ⱕk⬍n
w n, k
0ⱕk⬍n
Wn
䡠 f共k兲 ⫹
mn
Wn
冘
w n, k f 共 k 兲 ,
0ⱕk⬍n
where for the second equality above we have used the fact that W n ⫽ 1 ⫹ m n .
Then, since f( x) is positive, increasing and concave, and m n is positive, we can
bound S n to get
Sn ⱕ f
冉
冘
w n, k
0ⱕk⬍n
Wn
冊
䡠k ⫹
mn
Wn
䡠 W n f 共 n 兲 ⫽ f 共 nZ n 兲 ⫹ m n f 共 n 兲 .
By the definition of Z⬘n , we know that nZ n ⫹ a ⫽ (n ⫹ a) Z⬘n ⱕ (n ⫹ a)V.
Therefore,
f 共 nZ n 兲 ⫽ exp
冉
⫺V␯
␯
共1 ⫺ V 兲共nZn ⫹ a兲
␯
冊 冉
ⱕ exp
⫺1
共1 ⫺ V␯兲共n ⫹ a兲␯
冊
.
(12)
For the next step we use an auxiliar function A n ⫽ 1 ⫺ 1/((n ⫹ a) ␯ ⫹ 1) ⫽
(n ⫹ a) ␯ /((n ⫹ a) ␯ ⫹ 1). Since 1/A n ⫽ 1 ⫹ 1/(n ⫹ a) ␯ , and 1 ⫹ x ⱕ exp( x)
for every x ⱖ 0, it follows that 1/A n ⱕ exp(1/(n ⫹ a) ␯ ), and thus A n ⱖ
exp(⫺1/(n ⫹ a) ␯ ). Hence, A n f(n) ⱖ exp(⫺1/(n ⫹ a) ␯ ) f(n) ⫽ exp(⫺(1 ⫺
V ␯ )/(n ⫹ a) ␯ ), and from (12), m n ⱕ 1/((n ⫹ a) ␯ ⫹ 1), and the definition of
A n we conclude S n ⱕ f(n). We will use this inequality in a moment.
Now introduce R n ⫽ 兺 0ⱕk⬍n w n, k g(k), and use the convexity of g( x) to get
Rn ⱖ
冘
0ⱕk⬍n
w n, k
Wn
䡠 g共k兲 ⱖ g
冉冘
0ⱕk⬍n
w n, k
Wn
冊
䡠 k ⫽ g 共 nZ n 兲 ⫽
冘 g 共 nZ 兲 .
i
n
c
iⱖ1
By definition, g i (nZ n ) ⫽ ln(nZ n ⫹ a) ⫺ i ln V ⫽ ln((n ⫹ a) Z⬘n ) ⫺ i ln V.
Assume n ⱖ N⬘; the bound for Z⬘n yields g i (nZ n ) ⱕ ln(n ⫹ a) ⫹ ln V ⫺ i ln
V ⫽ g i⫺1 (n). Therefore, R n ⱖ 兺 iⱖ1 g i⫺1 (n) c ⫽ lnc (n ⫹ a) ⫹ g(n).
We are ready to prove the theorem. Set M ⫽ max{(G n /K ⫹ g(n))/
f(n)} 0ⱕn⬍N⬘ . Then we have G n /K ⱕ M 䡠 f(n) ⫺ g(n) for any n ⬍ N⬘. We can
Improved Master Theorems for Divide-and-Conquer Recurrences
181
prove this bound by induction when n ⱖ N⬘:
Gn
K
⫽
tn
K
⫹
冘
0ⱕk⬍n
w n, k 䡠
Gk
K
i.h.
ⱖ lnc共n ⫹ a兲 ⫹
冘
wn,k共M 䡠 f共k兲 ⫺ g共k兲兲
0ⱕk⬍n
⫽M 䡠 Sn ⫹ lnc共n ⫹ a兲 ⫺ Rn ⱕ M 䡠 f共n兲 ⫺ g共n兲.
Finally, we have F n ⱕ G n ⱕ K(M 䡠 f(n) ⫺ g(n)) ⱕ K 䡠 M for every n ⱖ 0
(and F n ⱖ ⫺K 䡠 M), and the theorem is proved. e
5. The Flesh Theorems
In this section, we deal with the canonical recursive definitions whose toll function is
large enough to dominate the solution. These recurrences are typically easier to
analyze than those in Section 4 (and thus the “Flesh” in the title), and in most cases
we can get the multiplicative factor of the main term of the function under study.
Moreover, sometimes it is possible to extract one by one several main terms with
their multiplicative factors, until the core is reached (see Appendix B).
The reason for the name chosen in the following definition will be clear after
Theorem 5.3.
Definition 5.1. Let un be a function over the integers. We say that ␥n is a
bounding function of un if and only if there exist some constant N␥ ⱖ 1 and some
strictly positive nonincreasing function ␤( z) defined over (0, 1) such that (a) un␥n ⫺
uk␥k ⱖ ␤(k/n)un for every n ⬎ N␥ and every N␥ ⱕ k ⬍ n; (b) un, ␥n ⬎ 0 for every
n ⱖ N␥; (c) ␥n is a subpolynomial function; and (d) un␥n ⫽ ␻(1).
For instance, ␥ n ⫽ 1 is a bounding function of u n ⫽ n 2 . Take ␤ ( z) ⫽ 1 ⫺ z 2 .
Then we have u n ␥ n ⫺ u k ␥ k ⫽ n 2 ⫺ k 2 ⫽ (1 ⫺ (k/n) 2 )n 2 ⫽ ␤ (k/n)u n . The
other conditions trivially hold.
The word “entropy” is defined below with a meaning wider than usually. Our
definition reduces to the traditional one when t n ⫽ 1 and ␥ n ⫽ log2 n.
Definition 5.2. Let Ᏺ be a canonical recursive definition of a function F n , and
let ␥ n be a bounding function of t n . We define the entropy of Ᏺ with respect to ␥ n
at every n ⱖ max{N, N ␥ } as
Ᏼ n( ␥ ) ⫽ ␥ n ⫺
冘
w n, k 䡠
N ␥ ⱕk⬍n
tk
tn
䡠 ␥k .
(13)
We also define Ᏼ(␥) ⫽ limn3⬁ Ᏼ n( ␥ ) , if it exists.
The next theorem, together with Lemma 5.4, will allow us to bound the
solution of many canonical recursive definitions.
THEOREM 5.3. Let Fn be defined by a canonical recursive definition, and let ␥n
be a bounding function of un. If tn ⫽ ᏻ(un), then Fn ⫽ ᏻ(un␥n). If tn ⫽ o(un), then
Fn ⫽ o(un␥n).
PROOF. Let Ᏼ n( ␥ ) be defined as in (13) but using u n instead of t n . We first
prove that Ᏼ n( ␥ ) ⫽ ⍀(1), as follows. We know that Z n ⱕ U for large n, where
0 ⬍ U ⬍ 1. Choose some V such that U ⬍ V ⬍ 1, and let n be large enough to
182
SALVADOR ROURA
get nV ⬎ N ␥ . Then t n ␥ n ⬎ t n ␥ n ⫺ t k ␥ k ⱖ ␤ (k/n)t n ⱖ ␤ (V)t n for every N ␥ ⱕ
k ⬍ nV. Alternatively, we can write ␥ n ⬎ ␥ n ⫺ t k /t n 䡠 ␥ k ⱖ ␤ (V). We now use
these bounds in the definition of Ᏼ n( ␥ ) :
Ᏼ n( ␥ ) ⫽
冉冘
冊
w n, k ⫺ m n ␥ n ⫺
0ⱕk⬍n
冘
冘
ⱖ
ⱖ
冉
tk
wn,k ␥n ⫺
N␥ⱕk⬍nV
0ⱕk⬍N␥
w n, k 䡠
N ␥ ⱕk⬍n
冘
wn,k␥n ⫹
冘
tn
tk
tn
䡠 ␥k
冊
䡠 ␥ k ⫺ m n␥ n
wn,k␤共V兲 ⫺ mn␥n .
0ⱕk⬍nV
On the other hand, from
V
冘
冘
w n, k ⫽
nVⱕk⬍n
w n, k 䡠
nV
nVⱕk⬍n
n
冘
ⱕ
w n, k 䡠
0ⱕk⬍n
k
n
⫽ Z nW n ⱕ U 䡠 W n ,
we have 兺0ⱕk⬍nV wn,k ⫽ Wn ⫺ 兺nVⱕk⬍n wn,k ⱖ (1 ⫺ U/V)Wn. Altogether, we
deduce Ᏼn(␥) ⱖ (1 ⫺ U/V)␤(V)Wn ⫺ mn␥n. But mn␥n vanishes for large n, because
兩mn兩 ⫽ ᏻ(n⫺␳) for some ␳ ⬎ 0 and ␥n is a subpolynomial function. Therefore, since
1 ⫺ U/V ⬎ 0, ␤(V) ⬎ 0 and Wn tends to W ⫽ 1, we have Ᏼn ⱖ Q for some constant
Q ⬎ 0 and n large enough (say, larger than some constant NᏴ).
Now we can prove the ᏻ⵹ case. Since t n ⫽ ᏻ(u n ), there exist constants K ⬎ 0
and M such that 兩t n 兩 ⱕ Ku n for every n ⱖ M. Choose M to be at least N Ᏼ , and
introduce a new function,
Gn ⫽
再
Fn ,
Kun ⫹
0ⱕn⬍M
if
w
G
,
if
n,k k
0ⱕk⬍n
冘
n ⱖ M.
A simple proof by induction yields ⫺G n ⱕ F n ⱕ G n for every n. Now define
In ⫽
冦
Gn ,
if
Gn ⫺
K
Q
䡠 un␥n , if
0 ⱕ n ⬍ N␥
n ⱖ N␥ .
Let n ⱖ M. Using the definitions of I n and G n in the first step, and the
definition of I n back in the second step, we get
I n ⫽ Ku n ⫹
冘
K
w n, k G k ⫺
Q
0ⱕk⬍n
⫽ Kun ⫺
⫽ Kun ⫺
K
Q
K
Q
冉
u n␥ n ⫺
䡠 u nᏴ n ⫹
䡠 u n␥ n
冊
冘
wn,kuk␥k ⫹
冘
wn,kIk .
N␥ⱕk⬍n
0ⱕk⬍n
冘
0ⱕk⬍n
wn,kIk
Improved Master Theorems for Divide-and-Conquer Recurrences
183
From the bound for Ᏼ n , the toll function above is ⱕ Ku n ⫺ K/Q 䡠 u n Q ⫽ 0, as
long as n ⱖ M. Therefore, if we define
Jn ⫽
再冘
兩I n 兩,
0ⱕk⬍n
if
wn,k Jk , if
0ⱕn⬍M
n ⱖ M,
then we have I n ⱕ J n . But according to Theorem 4.4, J n ⫽ ᏻ(1), that is, J n ⱕ S
for some constant S ⬎ 0 and n large enough. This implies I n ⱕ S as well, and
hence F n ⱕ G n ⫽ I n ⫹ K/Q 䡠 u n ␥ n ⱕ S ⫹ K/Q 䡠 u n ␥ n or alternatively,
F n /u n ␥ n ⱕ S/u n ␥ n ⫹ K/Q. Taking into account that u n ␥ n ⫽ ␻ (1), the term
S/u n ␥ n will vanish for large n, and thus we can conclude F n /u n ␥ n ⱕ K⬘ (and
F n /u n ␥ n ⱖ ⫺K⬘) for any constant K⬘ ⬎ K/Q as long as n is large enough. The
case ᏻ⵹ is proved.
The proof of the case o⵹ follows the pattern above. Therefore, we only point
out the differences. We have to show that for every ␧ ⬎ 0 there is some N ␧ large
enough such that 兩F n 兩/u n ␥ n ⱕ ␧ as long as n ⱖ N ␧ . Set ␯ ⫽ Q 䡠 ␧/ 2. Then, by
hypothesis, it exists some M ␯ such that 兩t n 兩 ⱕ ␯ u n for every n ⱖ M ␯ . Define G n
using ␯ u n as toll function for every n ⱖ M ␯ , and I n as G n ⫺ ␯ /Q 䡠 u n ␥ n for
every n ⱖ N ␥ . Then we have I n ⱕ S ␯ for some constant S ␯ ⬎ 0 and every n
large enough, and thus F n /u n ␥ n ⱕ S ␯ /u n ␥ n ⫹ ␯ /Q. Since the first term vanishes
for large n, we can always pick some N ␧ large enough to get F n /u n ␥ n ⱕ ␯ /Q ⫹
␯ /Q ⫽ ␧ for every n ⱖ N ␧ (and also F n /u n ␥ n ⱖ ⫺␧), thus proving the case o⵹
and the theorem. e
The following lemma provides bounding functions for the most usual toll
functions.
LEMMA 5.4
—Let u n ⫽ n a ␦ n for every n ⱖ N u , where N u ⱖ 1, a ⬎ 0, and ␦ n is a strictly
positive nondecreasing function at every n ⱖ N u . Then ␥ n ⫽ 1 is a bounding
function of u n .
—Let v n ⫽ lnc n 䡠 ⑀ n for every n ⱖ N v , where N v ⱖ 2, c ⬎ ⫺1, and ⑀ n is a
strictly positive nondecreasing function at every n ⱖ N v . Then ␥ n ⫽ ln n is a
bounding function of v n .
—Let w n ⫽ ln⫺1 n for every n ⱖ N w , for some N w ⱖ 3. Then ␥ n ⫽ ln n 䡠 ln ln
n is a bounding function of w n .
PROOF. For the first case, let N ␥ ⫽ N u and take ␤ ( z) ⫽ 1 ⫺ z a . Thus,
␤ ( z) ⬎ 0 is nonincreasing in (0, 1). Take any n ⬎ N ␥ and any N ␥ ⱕ k ⬍ n.
Then u n ␥ n ⫺ u k ␥ k ⫽ n a ␦ n ⫺ k a ␦ k ⱖ n a ␦ n ⫺ k a ␦ n ⫽ (1 ⫺ (k/n) a )n a ␦ n ⫽
␤ (k/n)u n , and the condition (a) in Definition 5.1 is satisfied. The conditions (b),
(c), and (d) are trivial to prove.
For the second case, take ␤ ( z) ⫽ ⫺(c ⫹ 1)ln z, if ⫺1 ⬍ c ⱕ 0, or ␤ ( z) ⫽
⫺ln z, if c ⬎ 0. Let N ␥ ⫽ N v , and take any k and n such that N ␥ ⱕ k ⬍ n. For
the moment, we have v n ␥ n ⫺ v k ␥ k ⫽ lnc⫹1 n 䡠 ⑀ n ⫺ lnc⫹1 k 䡠 ⑀ k ⱖ lnc⫹1 n 䡠 ⑀ n
⫺ lnc⫹1 k 䡠 ⑀ n ⫽ (lnc⫹1 n ⫺ lnc⫹1 k) ⑀ n .
To check that the condition (a) in Definition 5.1 holds, we consider the last
expression separately for positive and negative c. For c ⬎ 0, it is not difficult to
prove that lnc⫹1 n ⫺ lnc⫹1 k ⱖ ⫺ln(k/n)lnc n when n ⱖ 1 and 1 ⱕ k ⱕ n.
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SALVADOR ROURA
Therefore, v n ␥ n ⫺ v k ␥ k ⱖ ⫺ln(k/n)lnc n 䡠 ⑀ n ⫽ ␤ (k/n)v n . Similarly, when
⫺1 ⬍ c ⱕ 0 it suffices to prove that lnc⫹1 n ⫺ lnc⫹1 k ⱖ ⫺(c ⫹ 1)ln(k/n)
lnc n when n ⱖ 1 and 1 ⱕ k ⱕ n, and then v n ␥ n ⫺ v k ␥ k ⱖ ⫺(c ⫹ 1)ln(k/n)lnc
n 䡠 ⑀ n ⫽ ␤ (k/n)v n . The rest of conditions in Definition 5.1 trivially hold.
For the third case, let N ␥ ⫽ N w , and take any n ⬎ N ␥ and any N ␥ ⱕ k ⬍ n.
Define ␤ ( z) ⫽ ⫺ln z, and let y ⫽ ln n and z ⫽ k/n. It can be shown that ln y ⫺
ln( y ⫹ ln z) ⱖ ⫺ln z/y for all z in (exp(⫺y), 1]. Therefore, w n ␥ n ⫺ w k ␥ k ⫽ ln
ln n ⫺ ln ln k ⫽ ln y ⫺ ln( y ⫹ ln z) ⱖ ⫺ln z/y ⫽ ␤ (k/n)w n . The rest of
conditions for a bounding function can be easily proved. e
The following corollary is an immediate consequence of Theorem 5.3 and
Lemma 5.4.
COROLLARY 5.5. Let the functions Fn, Gn, and Hn be defined by canonical
recursive definitions with toll functions tn ⫽ ᏻ(na␦n), t⬘n ⫽ ᏻ(logc n 䡠 ⑀n) and
ᏻ(log⫺1 n), respectively, as stated in Lemma 5.4. Then Fn ⫽ ᏻ(tn), Gn ⫽ ᏻ(t⬘n log n)
and Hn ⫽ ᏻ(log log n). Moreover, Fn ⫽ ⌰(tn) if Fn ⱖ 0 and tn ⫽ ᏻ(na␦n).
The last theorem of this section allows us to compute the constant factor of the
leading term of the solution of some recurrences.
THEOREM 5.6. Let Fn be defined by a canonical recursive definition, and let ␥n
be a bounding function of tn. Furthermore, assume that Ᏼ(␥) exists. Then Fn ⫽
tn␥n/Ᏼ(␥) ⫹ o(tn␥n).
PROOF. From the proof of Theorem 5.3 we know that Ᏼ n( ␥ ) ⫽ ⍀(1). So, if
Ᏼ exists, it is strictly positive. Define
(␥)
Gn ⫽
冦
Fn ,
Fn ⫺
0 ⱕ n ⬍ N␥
if
t n␥ n
Ᏼ (␥)
n ⱖ N␥ .
, if
For n ⱖ N ␥ , use the definition of G n in both directions, like in the proof of
Theorem 5.3. This yields
Gn ⫽ tn ⫺
Ᏼ n( ␥ )
Ᏼ (␥)
䡠 tn ⫹
冘
w n, k G k .
0ⱕk⬍n
Let t⬘n denote the toll function above, and let h n ⫽ Ᏼ n( ␥ ) ⫺ Ᏼ ( ␥ ) . Then t⬘n ⫽
t n ⫺ (Ᏼ ( ␥ ) ⫹ h n )/Ᏼ ( ␥ ) 䡠 t n ⫽ ⫺h n t n /Ᏼ ( ␥ ) ⫽ o(t n ), because h n ⫽ o(1).
Therefore, it suffices to apply Theorem 5.3 to get G n ⫽ o(t n ␥ n ), which implies
F n ⫽ t n ␥ n /Ᏼ ( ␥ ) ⫹ G n ⫽ t n ␥ n /Ᏼ ( ␥ ) ⫹ o(t n ␥ n ). e
As an example of application of Theorem 5.6, we can easily compute the
expected number of comparisons to find a key chosen at random in a trie with n
keys, when the digits are independent and equally likely to be 0 or 1. The
recurrence is
Tn ⫽ 1 ⫹
1
2
n
冘
0ⱕkⱕn
冉冊
n
T .
k k
Improved Master Theorems for Divide-and-Conquer Recurrences
185
(It would be a simple matter to write another recurrence with range 0 . . . n ⫺ 1
for k.) Note that this recurrence is not covered by any of the MTs presented in
this work; however, we can solve it anyway. The entropy with respect to ln n is
Ᏼ n(ln) ⫽ ln n ⫺
1
2
n
冘
2ⱕkⱕn
冉冊
n
ln k.
k
The distribution of weights gets closer to n/ 2 as long as n grows. This implies
Ᏼ n(ln) ⬃ ln n ⫺ ln(n/ 2), Ᏼ(ln) ⫽ ln 2, and T n ⬃ ln n/ln 2 ⫽ log2 n.
More formally, we can use the fact that
冉冊 冉 冊
冘
1
2n
␧ 2n
n
ⱕ exp ⫺
k
6
(1⫹␧)n/ 2⬍kⱕn
for any 0 ⱕ ␧ ⱕ 1 (see Hagerup and Rub [1990, page 306]), and set ␧ ⫽ ␧ n ⫽
n ⫺1/3 to get
Ᏼ n(ln) ⫽ ln n ⫺
1
2
ⱖln n ⫺ ln
n
冉
冘
2ⱕkⱕ(1⫹␧n)n/ 2
共 1 ⫹ ␧ n兲 n
2
冉冊
冘
冉冊
1
n
n
ln k ⫺ n
ln k
k
2 (1⫹␧n)n/ 2⬍kⱕn k
冊 冉 冊
⫺ exp ⫺
␧n2 n
6
ln n ⫽ ln 2 ⫹ o共1兲.
On the other hand, it is easy to prove an upper bound ln 2 for Ᏼ(ln). Therefore,
Ᏼ(ln) ⫽ ln 2, and finally T n ⬃ log2 n.
6. Proof of the Discrete Master Theorem
In this section, we will derive the Discrete MT. While doing this, we will
repeatedly make use of the quantities introduced in Definitions 2.1, 4.1 and 4.2.
LEMMA 6.1. ⌽( x) is a strictly decreasing continuous function such that limx3⫹⬁
⌽( x) ⫽ 0 and limx3⫺⬁ ⌽( x) ⫽ ⫹⬁.
PROOF.
By elementary calculus.
LEMMA 6.2.
e
A discrete recursive definition is canonical if and only if ⌽(0) ⫽ 1.
PROOF. For the “only if” proof, we observe that W ⫽ limn3⫹⬁ W n ⫽
兺 1ⱕdⱕD w d ⫽ ⌽(0).
For the “if” proof, the definition of discrete recursive definition tells us that
兩m n 兩 ⫽ 兩W n ⫺ ⌽(0)兩 ⱕ 兺 1ⱕdⱕD 兩r d, n 兩 ⫽ ᏻ(n ⫺ ␳ ). Furthermore, Z ⫽ limn3⬁
Z n ⫽ 兺 1ⱕdⱕD w d 䡠 z d ⫽ ⌽(1), which by Lemma 6.1 is strictly smaller than 1
when W ⫽ ⌽(0) ⫽ 1. e
Theorem 6.3 justifies the cases 2.3 and 3 of the Discrete MT when the
recurrence is canonical, that is, when ␣ ⫽ 0. Note that the case 2.3 reduces to
a ⫽ 0 and c ⬍ ⫺1, and the condition Ᏼ ⬍ 0 in the case 3 is equivalent to a ⬍ 0.
THEOREM 6.3. Let Cn be defined by a canonical discrete recursive definition,
and let tn ⫽ ᏻ(logc n), where c ⬍ ⫺1. Then Cn ⫽ ᏻ(1). Moreover, Cn ⫽ ⌰(1) if
Cn ⱖ 0 for every n ⱖ 0 and tn ⬎ 0 for n large enough.
186
SALVADOR ROURA
PROOF. It is enough to realize that 兺 0ⱕk⬍N w n, k ⫽ 0 for large n, and use
Theorems 4.3 and 4.4. e
For the next step, we need to define the concept of subpolynomial function
formally.
Definition 6.4. Let ␭ n be a strictly positive nondecreasing function for large
n. Moreover, assume that for every ␧ ⬎ 0 the function n ␧ / ␭ n is increasing as long
as n is large enough. Then we say that ␭ n is a subpolynomial function.
Lemma 6.5 is used to prove Theorems 6.6 and 7.10.
LEMMA 6.5. Let tn ⫽ na␰n, where a ⬎ 0 and ␰n ⫽ ␭n or ␰n ⫽ 1/␭n for some
subpolynomial function ␭n. Then 1 is a bounding function of tn.
PROOF. We only need to prove that n a ␰ n is as stated by Lemma 5.4, which is
obvious when ␰ n ⫽ ␭ n . When ␰ n ⫽ 1/ ␭ n , it is enough to express t n as n a/ 2 ␦ n ,
with ␦ n ⫽ n a/ 2 / ␭ n . e
Theorem 6.6 corresponds to the case 1 of the Discrete MT when ␣ ⫽ 0 and
a ⬎ 0. The statement of the MT applies to toll functions with subpolynomial
factors that are ᏻ(logc n) for some c, which is more restrictive than the functions
considered by Theorem 6.6.
THEOREM 6.6. Let a function be defined by a canonical discrete recursive
definition, and let tn ⫽ na␰n, where a ⬎ 0 and ␰n ⫽ ␭n or ␰n ⫽ 1/␭n for some
subpolynomial function ␭n. Then Ᏼ(1) ⫽ 1 ⫺ ⌽(a).
PROOF.
Ᏼ
(1)
n
Let us assume ␰ n ⫽ ␭ n . Then, for large n,
⫽1⫺
冘
R d, n 䡠
1ⱕdⱕD
t S d, n
tn
⫽1⫺
冘
冉
共 w d ⫹ r d, n 兲 z d ⫹
1ⱕdⱕD
s d, n
n
冊
a
␭ S d, n
␭n
.
But 1 ⱖ ␭ S d, n / ␭ n ⱖ (S d, n /n) ⑀ ⫽ ( z d ⫹ s d, n /n) ␧ for any 1 ⱕ d ⱕ D and any ␧ ⬎
0, as long as n is large enough. Moreover, the contribution to Ᏼ n(1) of r d, n and
s d, n vanishes for large n. Altogether,
1 ⫺ ⌽共a兲 ⫽ 1 ⫺
冘
1ⱕdⱕD
w d 䡠 z da ⱕ Ᏼ (1) ⱕ 1 ⫺
冘
w d 䡠 z da⫹␧ ⫽ 1 ⫺ ⌽ 共 a ⫹ ␧ 兲 .
1ⱕdⱕD
This is true no matter how small ␧ is, and ⌽( x) is a continuous function. We can
thus conclude Ᏼ (1) ⫽ 1 ⫺ ⌽(a).
A similar argument proves the case ␰ n ⫽ 1/ ␭ n . e
Note that the MTs do not deal with toll functions whose growing rate is larger
than polynomial, like 2 n . Nevertheless, several of the results in this work also
apply to superpolynomial toll functions, which are formally defined as follows:
Definition 6.7. Let u n be a positive function for large n. For every a ⬎ 0,
assume that u n /n a is increasing as long as n is large enough. Then we say that u n
is a superpolynomial function.
Notice that the conditions in the first case of Lemma 5.4 trivially hold for any
superpolynomial function. As a consequence, the statement of Corollary 5.5 for
Improved Master Theorems for Divide-and-Conquer Recurrences
187
toll functions like t n ⫽ ᏻ(n a ␦ n ) is true not only for polynomial toll functions but
for superpolynomial toll functions as well.
The next theorem deals with discrete recurrences when the toll function is
superpolynomial.
THEOREM 6.8. Let a function be defined by a canonical discrete recursive
definition, and let tn be a superpolynomial function. Then Ᏼ(1) ⫽ 1.
PROOF. The proof is very similar to that of Theorem 6.6. It is enough to
realize that Ᏼ (1) ⱖ 1 ⫺ ⌽(a) for every a ⬎ 0, no matter how large a is. e
We require Lemma 6.9 for the statements of Theorems 6.10 and 7.12.
LEMMA 6.9. Let tn ⫽ lnc n 䡠 ␰n, where c ⬎ ⫺1 and ␰n ⫽ ␮n or ␰n ⫽ 1/␮n for
some sublogarithmical function ␮n. Then ln n is a bounding function of tn.
PROOF. If ␰ n ⫽ ␮ n , then t n is clearly as stated by Lemma 5.4. If ␰ n ⫽ 1/ ␮ n ,
we only need to write t n as ln(c⫺1)/ 2 n 䡠 ␦ n , with ␦ n ⫽ ln(c⫹1)/ 2 n/ ␭ n . e
Theorem 6.10 corresponds to the case 2.1 of the Discrete MT when ␣ ⫽
a ⫽ 0.
THEOREM 6.10. Let a function be defined by a canonical discrete recursive
definition, and let tn ⫽ lnc n 䡠 ␰n, where c ⬎ ⫺1 and ␰n ⫽ ␮n or ␰n ⫽ 1/␮n for some
sublogarithmical function ␮n. Then Ᏼ(ln) ⫽ ⫺(c ⫹ 1)兺1ⱕdⱕD wd 䡠 ln zd.
PROOF.
Let us assume ␰ n ⫽ 1/ ␮ n (the case ␰ n ⫽ ␮ n is similar). Then
Ᏼ
(ln)
n
冘 共w ⫹ r
⫽ ln n ⫺
d
d,n
兲
lnc⫹1Sd,n
1ⱕdⱕD
lnc n
䡠
␮n
␮Sd, n
for large n. Besides,
lnc⫹1Sd,n
lnc n
ⱕ
lnc⫹1Sd,n
lnc n
䡠
␮n
␮Sd, n
ⱕ
lnc⫹1⫺␧Sd,n
lnc⫺␧n
for any 1 ⱕ d ⱕ D and any ␧ ⬎ 0, as long as n is large enough. Furthermore,
ln( z d ⫹ s d, n /n) ⫽ ln z d ⫹ ᏻ(n ⫺ ␴ ) for some ␴ ⬎ 0. Thus,
lnc⫹1Sd,n ⫽ 共ln n ⫹ ln zd ⫹ ᏻ共n⫺␴兲兲c⫹1
⫽ lnc⫹1n ⫹ 共c ⫹ 1兲lnc n ln zd ⫹ ᏻ共logc⫺1n兲 ⫹ ᏻ共n⫺␴ logc n兲.
Taking into account that ᏻ(logc⫺1 n) ⫹ ᏻ(n ⫺ ␴ logc n) ⫽ ᏻ(logc⫺1 n), and that
冘
兩r d, n 兩 䡠
1ⱕdⱕD
lnc⫹1Sd,n
lnc n
ⱕ
冘
兩rd,n兩ln n ⫽ ᏻ共n⫺␳log n兲,
1ⱕdⱕD
we can bound Ᏼ n(ln) as follows:
Ᏼ
(ln)
n
ⱕ ln n ⫺
冘 共w ⫹ r
d
1ⱕdⱕD
d,n
兲
lnc⫹1Sd,n
lnc n
188
SALVADOR ROURA
冘 w 共ln n ⫹ 共c ⫹ 1兲ln z ⫹ ᏻ共log
⫽ ⫺共c ⫹ 1兲 冘 w 䡠 ln z ⫹ ᏻ共log n),
⫽ ln n ⫺
d
d
⫺1
n))⫹ᏻ(n⫺␳log n兲
1ⱕdⱕD
d
⫺1
d
1ⱕdⱕD
where, for the last step, we have used the fact that W ⫽ 1. A similar reasoning
yields the bound
Ᏼ n(ln) ⱖ ⫺共c ⫹ 1 ⫺ ␧兲
冘
wd 䡠 ln zd ⫹ ᏻ共log⫺1n).
1ⱕdⱕD
From these last two bounds, and since the one above holds for any ␧ ⬎ 0, the
theorem is proved. e
Finally, Theorem 6.11 corresponds to the case 2.2 of the Discrete MT when
␣ ⫽ a ⫽ 0.
THEOREM 6.11. Let a function be defined by a canonical discrete recursive
definition, and let tn ⫽ ln⫺1 n. Then Ᏼ(ln 䡠 ln ln) ⫽ ⫺兺1ⱕdⱕD wd 䡠 ln zd.
PROOF.
By the definitions of entropy and discrete recursive definition,
Ᏼ n(ln䡠ln ln)
ln n
⫽ ln ln n ⫺
冘 共w ⫹ r
d
d,n
1ⱕdⱕD
兲ln ln
冉冉
zd ⫹
sd,n
n
冊冊
n ,
as long as n is large enough. Let y ⫽ ln n. Recall that 兩r d, n 兩 ⫽ ᏻ(n ⫺ ␳ ) and
兩s d, n 兩/n ⫽ ᏻ(n ⫺ ␴ ). Then,
ln)
Ᏼ (ln䡠ln
y
y
⫽ ln y ⫺
冘 共w ⫹ ᏻ共exp共⫺␳y兲兲兲ln共 y ⫹ ln共 z ⫹ ᏻ共exp共⫺␴y兲兲兲兲.
d
d
1ⱕdⱕD
The contributions of the terms ᏻ(exp(⫺ ␳ y)) and ᏻ(exp(⫺ ␴ y)) vanish for large
y. Thus,
ln)
Ᏼ (ln䡠ln ln) ⫽ lim Ᏼ(ln䡠ln
⫽ lim
y
y3⫹⬁
y3⫹⬁
冘
1ⱕdⱕD
wd共ln y ⫺ ln共 y ⫹ ln zd兲兲
1/y
.
Now it suffices to use l’Hôpital’s rule to get the statement of the theorem.
e
To derive both the Discrete and the Continuous MTs, we will use the
fundamental property that every discrete or continuous recursive definition is
either canonical or can be reduced to a canonical one by means of a definition
like C n ⫽ F n /n ␣ , for some appropriate constant ␣. This concept is formalized as
follows:
Definition 6.12. Let Ᏺ be a DAC recursive definition of a function F n . We
say that Ᏺ is a proper DAC recursive definition if and only if there exists some
Improved Master Theorems for Divide-and-Conquer Recurrences
constant ␣ such that, if we define C n ⫽ F n /n ␣ , then
Cn ⫽
tn
n␣
⫹
冘
0ⱕk⬍n
189
冉冊
冘
k
k␣ Fk
tn
w n, k 䡠 ␣ 䡠 ␣ ⫽ ␣ ⫹
w n, k
n k
n
n
0ⱕk⬍n
␣
Ck
is a canonical recursive definition.
For instance, any canonical recursive definition is a proper DAC recursive
definition with ␣ ⫽ 0, that is, with C n ⫽ F n . Typically, W ⱖ 1, and then ␣ is
positive. For W smaller than 1 (which is not usual in practice) ␣ is negative,
which is problematic for small indices. This is why we required W to be at least 1
for continuous recursive definitions; note that the term (k/n) ␣ is unbounded for
small k and negative ␣. By contrast, for discrete recursive definitions we allowed
W to be smaller than 1; this is a reason to avoid the case z d ⫽ 0.
As an example of nonproper DAC recursive definition, we have (7); it is easy
to see that it is not canonical nor can be converted into canonical through a
change like C n ⫽ F n /n ␣ . Theorems 6.13 and 7.14 state that this is never the case
with discrete and continuous recursive definitions.
THEOREM 6.13. All discrete recursive definitions are proper DAC recursive
definitions whose ␣ is the unique solution of ⌽(␣) ⫽ 1.
PROOF. From Lemma 6.1, ⌽(␣) ⫽ 1 has always a unique solution. Define
C n ⫽ F n /n ␣ . Then, for large n,
Cn ⫽
tn
n␣
⫹
冘
R⬘d , n C S d, n ,
1ⱕdⱕD
where R⬘d, n ⫽ (w d ⫹ r d, n )( z d ⫹ s d, n /n) ␣ . Set w⬘d ⫽ w d 䡠 z ␣d and r⬘d, n ⫽ R⬘d, n ⫺
w⬘d . Then we have 兺 1ⱕdⱕD 兩r⬘d, n 兩 ⫽ ᏻ(n ⫺ ␳ ⫹ n ⫺ ␴ ).
Let W n , W, etc. be defined over the new weights. On the one hand, W n ⫽
兺 1ⱕdⱕD R⬘d, n ⫽ 兺 1ⱕdⱕD (w⬘d ⫹ r⬘d, n ), and W ⫽ 兺 1ⱕdⱕD w⬘d ⫽ ⌽( ␣ ) ⫽ 1.
Furthermore, 兩m n 兩 ⫽ 兩W n ⫺ W兩 ⱕ 兺 1ⱕdⱕD 兩r⬘d, n 兩 ⫽ ᏻ(n ⫺ ␳ ⫹ n ⫺ ␴ ). On the
other hand,
Zn ⫽
冘
1ⱕdⱕD
1
R⬘d , n S d, n
䡠
⫽
Wn
n
Wn
冘
1ⱕdⱕD
冉
共 w⬘d ⫹ r⬘d , n 兲 z d ⫹
s d, n
n
冊
,
and hence Z ⫽ 兺 1ⱕdⱕD w⬘d 䡠 z d ⫽ ⌽( ␣ ⫹ 1) ⬍ 1. The theorem follows.
e
We now make use of Theorems 5.6 and 6.13 to summarize Theorems 6.3, 6.6,
6.8, 6.10, and 6.11 into Corollary 6.14. This corollary also includes the corresponding results for continuous recurrences (Theorems 7.9, 7.10, 7.11, 7.12, and
7.13), which will be presented and proved in Section 7.
COROLLARY 6.14. Let Fn be defined by a discrete (respectively, continuous)
recursive definition, and let ␣ be the unique solution to ⌽(␣) ⫽ 1 (respectively,
␸(␣) ⫽ 1).
—If t n ⫽ ᏻ(n ␣ logc n), where c ⬍ ⫺1, then F n ⫽ ᏻ(n ␣ ). Moreover, F n ⫽ ⌰(n ␣ )
if F n ⱖ 0 for every n ⱖ 0 and t n ⬎ 0 for large n.
190
SALVADOR ROURA
—If t n ⫽ n a ␰ n , where a ⬎ ␣ and ␰ n ⫽ ␭ n or ␰ n ⫽ 1/ ␭ n for some subpolynomial
function ␭ n , then F n ⬃ t n /(1 ⫺ ⌽(a)) (respectively, F n ⬃ t n /(1 ⫺ ␸ (a))).
—If t n is a superpolynomial function, then F n ⬃ t n .
—If t n ⫽ n ␣ lnc n 䡠 ␰ n , where c ⬎ ⫺1 and ␰ n ⫽ ␮ n or ␰ n ⫽ 1/ ␮ n for some
sublogarithmical function ␮ n , then F n ⬃ t n ln n/Ᏼ, where Ᏼ ⫽ ⫺(c ⫹
1)兺 1ⱕdⱕD w d 䡠 ( z d ) ␣ ln z d (respectively, Ᏼ ⫽ ⫺(c ⫹ 1) 兰 10 ␻ ( z) z ␣ ln z dz).
—If t n ⫽ n ␣ ln⫺1 n, then F n ⬃ t n ln n 䡠 ln ln n/Ᏼ, where Ᏼ ⫽ ⫺兺 1ⱕdⱕD w d 䡠
( z d ) ␣ ln z d (respectively, Ᏼ ⫽ ⫺兰 10 ␻ ( z) z ␣ ln z dz).
7. Proof of the Continuous Master Theorem
We begin with the conjecture included in Definition 3.1.
CONJECTURE 7.1. For every positive function ␻( z) such that 兰10 ␻( z) dz converges, there exists some ␮ ⬍ 0 such that 兰10 ␻( z) z␮ dz also converges.
The author of this paper could not prove that, irrespective of ␻ ( z), the
integral ␸ ( x) converges for every x 僆 [ ␮ , 0] for some ␮ ⬍ 0, though no
counterexample of this statement was found. However, this property clearly
holds for the shape functions we usually deal with. For instance, if ␻ ( z) is
bounded near 0, then any ␮ ⬎ ⫺1 fulfills the condition above. To the best of the
author’s knowledge, it is an open problem to determine if the statement of
Conjecture 7.1 is always true.
The following lemma provides an easy way to find the shape function of many
continuous recurrences.
LEMMA 7.2. Let ␻n,k ⫽ A 䡠 f1 . . . fm/( g1 . . . gm⫹1) be the weights of a given
recurrence, where A ⬎ 0 is an arbitrary constant, m ⱖ 0 is any integer constant, fi ⫽
(ain ⫹ bik ⫹ ci) for all 1 ⱕ i ⱕ m for some constants ai, bi, and ci such that at least
one of ai or bi is not 0, and gi ⫽ (n ⫹ di) for all 1 ⱕ i ⱕ m ⫹ 1, where the di’s are
arbitrary constants. Set ␴n,k ⫽ A 䡠 (a1n ⫹ b1k) . . . (amn ⫹ bmk)/nm⫹1 and ␻( z) ⫽
n 䡠 ␴n,zn, and suppose that 兰10 ␻( z) dz is at least 1. Then the given recurrence is
continuous and ␻( z) is its shape function.
PROOF. First, we prove that there exists some ␮ ⬍ 0 such that 兰 10 ␻ ( z) z ␮ dz
converges. Since ␻ ⬘( z) exists and is bounded in [0, 1], ␻ ( z) is also bounded in
[0, 1]. Set h ⫽ max{ ␻ ( z)} 0ⱕzⱕ1 , and choose any ⫺1 ⬍ ␮ ⬍ 0. Then 兰 10 ␻ ( z) z ␮
dz ⱕ h 兰 10 z ␮ dz ⬍ ⫹⬁.
To prove the conditions in Definition 3.2, let B ⱖ 0 be such that 兩 ␻ ⬘( z)兩 ⱕ B
for every 0 ⱕ z ⱕ 1. It is easy to see that this bound produces
冕
(k⫹1)/n
k/n
␻ 共 z 兲 dz ⱕ
冕 冉冉冊 冉 冊冊
(k⫹1)/n
k/n
␻
k
n
⫹ z⫺
k
n
B dz ⫽
␻ 共 k/n 兲
n
⫹
B
2n 2
,
(k⫹1)/n
and its symmetric 兰 k/n
␻ ( z) dz ⱖ ␻ (k/n)/n ⫺ B/ 2n 2 . Now, we set M n ⫽ n
2
to get ␧ n, k ⱕ B/ 2n and 兺 0ⱕk⬍n ␧ n, k ⫽ ᏻ(n ⫺1 ). Finally, it is enough to notice
Improved Master Theorems for Divide-and-Conquer Recurrences
that 兩 ␻ n, k ⫺ ␴ n, k 兩 ⫽ ᏻ(n ⫺2 ) to obtain
冘
0ⱕk⬍n
冘
␧ n, k ⱕ
0ⱕk⬍n
冉冏
冏 冏
␻ n, k ⫺ ␴ n, k ⫹ ␴ n, k ⫺
冕
(k⫹1)/n
␻ 共 z 兲 dz
k/n
冏冊
191
⫽ ᏻ 共 n ⫺1兲.
e
Hereafter, we will present results for continuous recurrences that are equivalent to those for discrete recurrences in Section 6, together with some technical
propositions.
LEMMA 7.3. ␸( x) is a strictly decreasing continuous function in the interval [␮,
⫹⬁), such that limx3⫹⬁ ␸( x) ⫽ 0.
PROOF. By hypothesis, ␸(0) ⱖ 1. This implies that there are z 1 and z 2 such
that 0 ⬍ z 1 ⬍ z 2 ⬍ 1 and 兰 zz 12 ␻ ( z) dz ⬎ 0. For every x ⬍ y such that ␸ ( x) and
␸ ( y) exist,
␸共 x兲 ⫺ ␸共 y兲 ⱖ
冕
z2
␻ 共 z 兲共 z x ⫺ z y 兲 dz ⱖ
z1
冕
z2
␻ 共 z 兲 dz 䡠 min兵 zx ⫺ zy其z1ⱕzⱕz2 ⬎ 0.
z1
Hence, ␸ ( x) is a strictly decreasing function.
Proving that ␸ ( x) tends to 0 as long as x grows is not more difficult. For every
␧ ⬎ 0, let 0 ⬍ z ␧ ⬍ 1 be large enough to get 兰 z1␧ ␻ ( z) dz ⱕ ␧/ 2. Assume x ⱖ 0.
Then,
␸共 x兲 ⫽
冕
z␧
0
ⱕ 共 z ␧兲 x
␻ 共 z 兲 z x dz ⫹
冕
冕
冕
1
␻ 共 z 兲 z x dz
z␧
z␧
␻共 z兲 dz ⫹
1
z␧
0
␧
␻共 z兲 dz ⱕ 共 z␧兲x␸共0兲 ⫹ ⱕ ␧
2
for x large enough.
It is left to prove that ␸ ( x) is a continuous function. Recall that ␮ ⬍ 0 is the
constant such that, according to Conjecture 7.1, 兰 10 ␻ ( z) z ␮ dz converges. Choose
any a such that a ⬎ ␮ . Then,
lim 兩␸共 x兲 ⫺ ␸共a兲兩 ⫽ lim
x3a
ⱕ
x3a
冕
1
0
冏冕
1
0
␻共 z兲 z␮共 zx⫺␮ ⫺ za⫺␮兲 dz
冏
␻共 z兲 z␮ dz 䡠 lim共max兵兩zx⫺␮ ⫺ za⫺␮兩其0ⱕzⱕ1兲 ⫽ ␸共␮兲 䡠 0 ⫽ 0.
e
x3a
The following two technical propositions will be used in several proofs in this
section.
PROPOSITION 7.4. Let a function be defined by a continuous recursive definition. For every x ⬎ 0, define
192
SALVADOR ROURA
␧ n, j 共 x 兲 ⫽
冏
冘
␻ n, k
k僆I n, j
冉冊 冕
x
k
⫺
n
z n, j⫹1
␻ 共 z 兲 z x dz
z n, j
冏
for all 0 ⱕ j ⬍ M n . Then for every x ⬎ 0, we have 兺 0ⱕj⬍M n ␧ n, j ( x) ⫽ ᏻ(n ⫺ ␴ ) for
some ␴ ⬎ 0.
PROOF. First of all, notice that ␧ n, j (0) equals ␧ n, j . Take any 0 ⱕ j ⬍ M n , and
assume that the first term in the definition of ␧ n, j ( x) is larger than the second
term. Then
冉 冊 冕
冉 冊 冘 冉 冊冕
冘
冉 冊冉 冘 冕
␧ n, j 共 x 兲 ⱕ
冘
␻ n, k
z n, j⫹1 䡠 n
⫽
⫽
␻n,k ⫺
k僆In, j
共 j ⫹ 1兲x ⫺ jx
共 M n兲 x
␻ 共 z 兲共 z n, j 兲 x dz
z n, j
x
Mn
z n, j⫹1
⫺
n
k僆I n, j
j⫹1
x
x
j
Mn
␻n,k ⫹
k僆In, j
zn, j⫹1
␻共 z兲 dz
zn, j
x
j
Mn
␻n,k ⫺
k僆In, j
zn, j⫹1
zn, j
冊
␻共 z兲 dz .
Now, if we set
␺n ⫽
再
1,
if
x
x
共Mn兲 ⫺ 共Mn ⫺ 1兲 , if
0⬍xⱕ1
,
x⬎1
then it is easy to see that ( j ⫹ 1) x ⫺ j x ⱕ ␺ n . Moreover, j ⬍ M n . Hence,
␧ n, j ( x) ⱕ ␺ n /(M n ) x 兺 k僆I n, j ␻ n, k ⫹ ␧ n, j . If the second term in the definition of
␧ n, j ( x) is larger than the first one, a similar reasoning yields ␧ n, j ( x) ⱕ ␺ n /(M n ) x
兰 zz n,n, jj⫹1 ␻ ( z) dz ⫹ ␧ n, j . Altogether,
冘
␧ n, j 共 x 兲 ⱕ
0ⱕj⬍M n
⫽
冘
0ⱕj⬍M n
␺n
共 M n兲 x
冉
冘
0ⱕk⬍n
冉 冉
␻n,k ⫹
␺n
共 M n兲 x
冕
1
0
冘
␻ n, k ⫹
冊
冘
k僆I n, j
␻共 z兲 dz ⫹
冕
z n, j⫹1
冊 冊
␻ 共 z 兲 dz ⫹ ␧ n, j
z n, j
0ⱕj⬍Mn
␧n,j ⫽
␺n
共 M n兲 x
⌰ 共 1 兲 ⫹ ᏻ 共 n ⫺␳兲 .
Finally, taking into account that ␺ n /(M n ) x ⫽ ᏻ(n ⫺qx ) when 0 ⬍ x ⱕ 1 and ␺ n /
(M n ) x ⫽ ᏻ(n ⫺q ) when x ⬎ 1, we can bound the expression above by ᏻ(n ⫺ ␴ ) for
some ␴ ⬎ 0 small enough. e
PROPOSITION 7.5. Let a function be defined by a continuous recursive definition. Then, for every x ⬎ 0, 兺0ⱕk⬍n ␻n,k(k/n)x ⫽ ␸( x) ⫹ ᏻ(n⫺␴) for some ␴ ⬎ 0.
Improved Master Theorems for Divide-and-Conquer Recurrences
PROOF.
冏
冘
0ⱕk⬍n
It is enough to observe that
␻ n, k
冉冊 冏 冏 冉
x
k
冘
⫺ ␸共 x兲 ⫽
n
0ⱕj⬍M n
冘
␻ n, k
k僆I n, j
冉冊
k
n
x
⫺
冕
z n, j⫹1
193
␻ 共 z 兲 z x dz
z n, j
冊冏
,
which is not larger than 兺 0ⱕj⬍M n ␧ n, j ( x). But, according to Proposition 7.4, this
sum is ᏻ(n ⫺ ␴ ) for some ␴ ⬎ 0. e
A continuous recursive definition is canonical if and only if
LEMMA 7.6.
␸(0) ⫽ 1.
PROOF. For the “only if” proof, a reasoning similar to the one in Proposition
7.5 yields 兩W n ⫺ ␸ (0)兩 ⫽ ᏻ(n ⫺ ␳ ). Hence, ␸(0) has to equal 1 for the recurrence
to be canonical.
For the “if” proof, when ␸ (0) ⫽ W ⫽ 1, we have 兩m n 兩 ⫽ ᏻ(n ⫺ ␳ ). Moreover,
W n Z n ⫽ 兺 0ⱕk⬍n ␻ n, k k/n ⫽ ␸ (1) ⫹ ᏻ(n ⫺ ␴ ) for some ␴ ⬎ 0 small enough.
Now, taking into account that W n ⫽ 1 ⫹ ᏻ(n ⫺ ␳ ), we obtain Z n ⫽ ␸ (1) ⫹
ᏻ(n ⫺ ␴ ⫹ n ⫺ ␳ ), and conclude that Z ⫽ ␸ (1) ⬍ 1. e
We require just two more technical propositions before proving the main
results for continuous recurrences.
PROPOSITION 7.7. Let ␻n,k be the weights of a continuous recursive definition, fn
be any function such that fn ⫽ ⍀(1) and fn ⫽ o(n), x ⱖ 0, and ␳ ⬎ 0 and 0 ⬍ q ⱕ
1 be the constants in the definition of continuous recursive definition. Define hn ⫽
max{ fn, n1⫺q}. Then 兺0ⱕk⬍fn ␻n,k(k/n)x ⫽ ᏻ(max{n⫺␳, (hn/n)⫺␮}), where ␮ ⬍ 0 is
such that ␸(␮) converges.
PROOF. Assume x ⫽ 0. Let g n ⫽ ⌰(h n 䡠 n q⫺1 ) be a function with integer
values such that g n 䡠 n/M n ⱖ f n for large n (notice that this implies g n ⱖ 1).
Then 兺 0ⱕk⬍f n ␻ n, k ⱕ 兺 0ⱕk⬍g n 䡠 n/M n ␻ n, k ⫽ 兺 0ⱕj⬍g n 兺 k僆I n, j ␻ n, k for large n. On
the other hand,
冏冘
0ⱕj⬍g n
冉
冘
␻ n, k ⫺
k僆I n, j
冕
z n, j⫹1
z n, j
␻ 共 z 兲 dz
冊冏
ⱕ
冘
␧ n, j ⫽ ᏻ 共 n ⫺␳兲.
0ⱕj⬍g n
Furthermore, 兺 0ⱕj⬍g n 兰 zz n,n, jj⫹1 ␻ ( z) dz ⫽ 兰 g0n /M n ␻ ( z) z ␮ z ⫺ ␮ dz, which can be
bounded by
冉 冊冕
gn
Mn
⫺␮
gn/Mn
␻共 z兲 z␮ dz ⱕ ᏻ共共hn/n兲⫺␮兲␸共␮兲 ⫽ ᏻ共共hn/n兲⫺␮兲.
0
Altogether, the proposition is proved for the case x ⫽ 0.
The proof for x ⬎ 0 is now trivial, since 兺 0ⱕk⬍f n ␻ n, k (k/n) x ⱕ 兺 0ⱕk⬍f n ␻ n, k .
e
PROPOSITION 7.8. Let ␻n,k be the weights of a continuous recursive definition,
x ⱖ 0, and N ⱖ 1 be any constant. Then 兺0ⱕk⬍N ␻n,k(k/n)x ⫽ ᏻ(n⫺u) for some
u ⬎ 0.
194
SALVADOR ROURA
PROOF. It is enough to use Proposition 7.7 with f n ⫽ N. Then h n ⫽ ⌰(n 1⫺q )
and the sum above is bounded by ᏻ(max{n ⫺ ␳ , n ␮ q }). e
THEOREM 7.9. Let Cn be defined by a canonical continuous recursive definition,
and let tn ⫽ ᏻ(logc n), where c ⬍ ⫺1. Then Cn ⫽ ᏻ(1). Moreover, Cn ⫽ ⌰(1) if
Cn ⱖ 0 for every n ⱖ 0 and tn ⬎ 0 for n large enough.
PROOF. By Proposition 7.8, we have 兺 0ⱕk⬍N w n, k ⫽ ᏻ(n ⫺u ) for some u ⬎ 0.
Hence, we only need to use Theorems 4.3 and 4.4 to complete this proof. e
THEOREM 7.10. Let a function be defined by a canonical continuous recursive
definition, and let tn ⫽ na␰n, where a ⬎ 0 and ␰n ⫽ ␭n or ␰n ⫽ 1/␭n for some
subpolynomial function ␭n. Then Ᏼ(1) ⫽ 1 ⫺ ␸(a).
PROOF.
Let us assume ␰ n ⫽ 1/ ␭ n . Then
Ᏼ
(1)
n
冘
⫽1⫺
␻ n, k 䡠
N ␥ ⱕk⬍n
tk
tn
⫽1⫺
冘
␻ n, k
N ␥ ⱕk⬍n
冉冊
k
n
a
␭n
␭k
.
By hypothesis, for every ␧ ⬎ 0 there exists some N ␧ such that k ␧ / ␭ k ⱕ (k ⫹
1) ␧ / ␭ k⫹1 and ␭ k ⱕ ␭ k⫹1 for all k ⱖ N ␧ . Choose any 0 ⬍ ␧ ⬍ a, and let k and
n be such that N ␧ ⱕ k ⬍ n. Then we have k ␧ / ␭ k ⱕ n ␧ / ␭ n and ␭ k ⱕ ␭ n .
Therefore,
冘
N ␧ ⱕk⬍n
␻ n, k
冉冊
k
n
a
冘
ⱕ
␻ n, k
N ␧ ⱕk⬍n
冉冊
k
n
a
␭n
␭k
冘
ⱕ
␻ n, k
N ␧ ⱕk⬍n
Now we can set N ␥ ⫽ N ␧ to get
1⫺
冘
N ␥ ⱕk⬍n
␻ n, k
冉冊
k
n
a
ⱖᏴ
(1)
n
ⱖ1⫺
冘
N ␥ ⱕk⬍n
␻ n, k
冉冊
冉冊
k
k
a⫺␧
.
n
a⫺␧
.
n
From Propositions 7.5 and 7.8, we deduce 1 ⫺ ␸ (a) ⱖ Ᏼ (1) ⱖ 1 ⫺ ␸ (a ⫺ ␧),
which is true no matter how small ␧ is. Since ␸ ( x) is a continuous function, we
conclude Ᏼ (1) ⫽ 1 ⫺ ␸ (a).
A similar argument proves the case ␰ n ⫽ ␭ n . e
THEOREM 7.11. Let a function be defined by a canonical continuous recursive
definition, and let tn be a superpolynomial function. Then Ᏼ(1) ⫽ 1.
PROOF.
The proof is very similar to that of Theorem 7.10.
e
THEOREM 7.12. Let a function be defined by a canonical continuous recursive
definition, and let tn ⫽ lnc n 䡠 ␰n, where c ⬎ ⫺1 and ␰n ⫽ ␮n or ␰n ⫽ 1/␮n for some
sublogarithmical function ␮n. Then Ᏼ(ln) ⫽ ⫺(c ⫹ 1) 兰10 ␻( z)ln z dz.
PROOF. First of all, we must make sure that Ᏼ(ln) exists and is strictly positive.
It is easy to prove that ⫺兰 10 ␻ ( z) ln z dz, if exists, is strictly greater than 0.
Proving that it does not diverge is not more difficult. Let ␮ ⬍ 0 be such that ␸(␮)
converges, and let 0 ⬍ u ⬍ 1 be such that ⫺ln z ⱕ z ␮ in the interval (0, u].
Then,
Improved Master Theorems for Divide-and-Conquer Recurrences
⫺
冕
1
冕
␻共 z兲ln z dz ⱕ
0
u
␻共 z兲 z␮ dz ⫺
冕
1
195
␻共 z兲ln z dz ⱕ ␸共␮兲 ⫺ ␸共0兲ln u ⬍ ⫹⬁.
u
0
Now, assume ␰ n ⫽ 1, that is, t n ⫽ lnc n. Then we have
冘
Ᏼ n(ln) ⫽ 共1 ⫺ Wn兲ln n ⫹ Wnln n ⫺
␻n,k 䡠
lnc⫹1k
N␥ⱕk⬍n
lnc n
.
Let g n ⫽ ⌰(n d ) be a function with integer values, where d ⫽ q/ 2. So, we have
0 ⬍ d ⬍ q and 0 ⬍ d ⫹ 1 ⫺ q ⬍ 1. For large n,
冘
Ᏼ n(ln) ⫽ ⫺mnln n ⫹
␻n,k 䡠
lnc⫹1n ⫺ lnc⫹1k
lnc n
gn䡠n/Mnⱕk⬍n
冘
⫹
冘
␻n,kln n ⫹
冉
␻n,k ln n ⫺
N␥ⱕk⬍gn䡠n/Mn
0ⱕk⬍N␥
lnc⫹1k
lnc n
冊
.
The term ⫺m n lnn above vanishes because m n ⫽ ᏻ(n ⫺ ␳ ). Moreover, the second
line above is positive, and can be bounded by (兺 0ⱕk⬍g n 䡠 n/M n ␻ n, k ) ln n. Setting
f n ⫽ ⌰(n 1⫺q/ 2 ) in Proposition 7.7 proves that this sum also vanishes for large n.
The rest of the proof is devoted to find the limit of the term that is left.
For every g n ⱕ j ⬍ M n , define
⌬ n, j ⫽
冏
冘
␻ n, k 䡠
lnc⫹1n ⫺ lnc⫹1k
lncn
k僆I n, j
冉 冕
⫺ ⫺共c ⫹ 1兲
zn, j⫹1
␻共 z兲ln z dz
zn, j
冊冏
Assume that the first term above is larger than the second one. Then
⌬ n, j
c⫹1
冘
ⱕ
␻ n, k 䡠
lnc⫹1n ⫺ lnc⫹1共 zn,j 䡠 n兲
k僆I n, j
冉
⫽ ⫺ln zn,j ⫹ ᏻ
共c ⫹ 1兲lncn
冉 冊冊 冘
1
log n
⫹
冕
zn, j⫹1
zn, j
␻n,k ⫹ ln zn,j⫹1
k僆In, j
␻共 z兲ln zn,j⫹1 dz
冕
zn, j⫹1
␻共 z兲 dz.
zn, j
But
⫺ln zn,j
冘␻
k僆In, j
⫽ ln zn,j 䡠
ⱕ ⫺ln
n,k
冉冕
冉冊
gn
Mn
⫹ ln zn,j⫹1
冕
zn, j⫹1
␻共 z兲 dz
zn, j
zn, j⫹1
␻共 z兲 dz ⫺
zn, j
䡠 ␧n,j ⫹ ln
冉
冘␻
n,k
k僆In, j
gn ⫹ 1
gn
冊冕
冊 冉
zn, j⫹1
zn, j
⫹ ln
zn,j⫹1
zn,j
␻共 z兲 dz.
冊冕
zn, j⫹1
zn, j
␻共 z兲 dz
.
196
SALVADOR ROURA
Taking into account the facts that ⫺ln( g n /M n ) ⫽ ⌰(log n q⫺d ) ⫽ ᏻ(log n) and
ln(1 ⫹ 1/g n ) ⫽ ⌰(1/g n ) ⫽ ᏻ(n ⫺d ), we have
⌬ n, j
c⫹1
⫽ ᏻ 共 log n兲␧n,j ⫹ ᏻ共n⫺d兲
冕
zn, j⫹1
␻共 z兲 dz ⫹ ᏻ
zn, j
冉 冊冘
1
log n
␻n,k .
k僆In, j
If the second term in the definition of ⌬ n, j above is larger than the first one, by
means of a similar reasoning, we get
⌬ n, j
c⫹1
⫽ ᏻ 共 log n兲␧n,j ⫹ ᏻ共n⫺d兲
冘␻
n,k
⫹ᏻ
k僆In, j
Thus, we can conclude
冘
冘
⌬ n, j ⫽ ᏻ 共 log n兲
g n ⱕj⬍M n
␧n,j ⫹ ᏻ
gnⱕj⬍Mn
冉 冊冕
1
log n
zn, j⫹1
␻共 z兲 dz.
zn, j
冉 冊 冉 冊
1
log n
Wn ⫹ ᏻ
1
log n
␸共0兲,
which vanishes for large n, because the sum of ␧ n, j ’s is ᏻ(n ⫺ ␳ ) for some ␳ ⬎ 0.
Now we are ready to finish the proof, since
冏
冘
␻ n, k 䡠
lnc⫹1n ⫺ lnc⫹1k
lncn
g n 䡠n/M n ⱕk⬍n
冉 冕
⫺ ⫺共c ⫹ 1兲
1
␻共 z兲 ln z dz
gn/Mn
冊冏
is bounded by above by 兺 g n ⱕj⬍M n ⌬ n, j ⫽ o(1). Therefore,
Ᏼ (ln) ⫽ lim ⫺共c ⫹ 1兲
n3⫹⬁
冕
1
␻共 z兲 ln z dz ⫽ ⫺共c ⫹ 1兲
gn/Mn
冕
1
␻共 z兲 ln z dz.
0
This ends the proof of the case ␰ n ⫽ 1.
The cases ␰ n ⫽ ␮ n and ␰ n ⫽ 1/ ␮ n for sublogarithmical ␮ n ⫽ ␻ (1) can be
proved by similar arguments (see the proof of Theorem 6.10). e
THEOREM 7.13. Let a function be defined by a canonical continuous recursive
definition, and let tn ⫽ ln⫺1 n. Then Ᏼ(ln 䡠 ln ln) ⫽ ⫺兰10 ␻( z)ln z dz.
PROOF. This proof follows the same pattern as the one of Theorem 7.12.
Therefore, we skip some steps, for the sake of brevity.
Let ␯ n ⫽ exp( 公ln n), and g n ⫽ ⌰(n q / ␯ n ) be a function with integer values. By
the definitions of entropy and continuous recursive definition,
Ᏼ n(ln䡠ln ln) ⫽ ⫺mn ln n 䡠 ln ln n ⫹ ln n 䡠
⫹ ln n 䡠 ln ln n
冘
0ⱕk⬍N␥
冘
␻n,k共ln ln n ⫺ ln ln k兲
gn䡠n/Mnⱕk⬍n
␻n,k ⫹ ln n 䡠
冘
␻n,k共ln ln n ⫺ ln ln k兲,
N␥ⱕk⬍gn䡠n/Mn
as long as n is large enough. The term ⫺m n ln n 䡠 ln ln n clearly vanishes
for large n. Moreover, the second line above is positive, and bounded by
(兺 0ⱕk⬍g n 䡠 n/M n ␻ n, k ) ln n ln ln n. By Proposition 7.7, this term is also o(1), since
Improved Master Theorems for Divide-and-Conquer Recurrences
197
f n ⫽ h n ⫽ n/ ␯ n and (h n /n) ⫺ ␮ ⫽ ␯ n␮ . Below, we prove that the limit of the
remaining term above is as stated by the theorem.
For every g n ⱕ j ⬍ M n , define
冏
⌬ n, j ⫽ ln n 䡠
冘␻
n,k
冉
共ln ln n ⫺ ln ln k兲 ⫺ ⫺
k僆In, j
冕
zn, j⫹1
␻共 z兲ln z dz
zn, j
冊冏
.
Assume that the first term above is larger than the second one (the opposite
produces a similar bound). Then
⌬ n, j ⱕ ln n 䡠
冘␻
n,k
共ln ln n ⫺ ln ln共 zn,j 䡠 n兲兲 ⫹
k僆In, j
冕
zn, j⫹1
␻共 z兲ln zn,j⫹1 dz.
zn, j
Now it is not difficult to get
冉
ln共1/zn,j兲
⌬ n, j ⱕ ⫺ln n 䡠 ln 1 ⫺
ln n
冉 冊冕
1
⫹ln 1 ⫹
gn
zn, j⫹1
冊冘
␻n,k ⫹ ln zn,j
k僆In, j
冕
zn, j⫹1
␻共 z兲 dz
zn, j
␻共 z兲 dz.
zn, j
The contribution to the entropy of the second line is asymptotically irrelevant,
because ln(1 ⫹ 1/g n ) ⫽ ⌰(1/g n ) ⫽ o(1). For the next step, we show that the
factor ln n 䡠 ln (1 ⫺ ln(1/z n, j )/ln n) equals ln z n, j ⫹ o(1), which is easy if we
take into account the fact that 1/z n, j ⫽ ᏻ(M n /g n ) ⫽ ᏻ( ␯ n ). Hence, ln(1/z n, j ) ⫽
ᏻ(ln ␯ n ) ⫽ ᏻ共 3 ln n兲, which implies ln(1/z n, j )/ln n ⫽ ᏻ(ln⫺2/3 n) ⫽ o(1). Finally,
we observe that ln(1 ⫺ x) ⫽ ⫺x ⫹ ᏻ( x 2 ) when x ⫽ o(1) to conclude
冑
冉
ln n 䡠 ln 1 ⫺
ln共1/zn,j兲
ln n
冊
⫽ ln n 䡠
冉
ln zn,j
ln n
冊
⫹ ᏻ共ln⫺4/3n兲 ⫽ ln zn,j ⫹ o共1兲.
Now it is enough to follow the last steps in the proof of Theorem 7.12 to finish
this proof. e
THEOREM 7.14. All continuous recursive definitions are proper DAC recursive
definitions whose ␣ is the unique solution of ␸(␣) ⫽ 1.
PROOF. From Lemma 7.3 and ␸(0) ⱖ 1, the equation above has a unique
solution. Define C n ⫽ F n /n ␣ . Then, for every n ⱖ N,
Cn ⫽
tn
1
n
n␣
⫹
␣
冘
␻ n, k F k ⫹
冘
␻ n, k 䡠
Nⱕk⬍n
0ⱕk⬍N
冘
k␣ Fk
䡠
⫽ t⬘n ⫹
␻ ⬘n , k C k ,
n␣ k␣
0ⱕk⬍n
where t⬘n ⫽ t n /n ␣ ⫹ (1/n ␣ )兺 0ⱕk⬍N ␻ n, k F k , and
␻ ⬘n , k ⫽
再
0,
if
␣
␻n,k共k/n兲 , if
0ⱕk⬍N
N ⱕ k ⬍ n.
198
SALVADOR ROURA
From Propositions 7.4 and 7.8, it is a simple matter to prove that ␻ ( z) z ␣ is the
shape function of Ꮿ (just set x ⫽ ␣ ). Moreover, since ␸(␣) ⫽ 1, we have
兰 10 ␻ ( z) z ␣ dz ⫽ 1. This proves that the recurrence for C n is a canonical
continuous recursive definition.
As a final remark, notice that the term (1/n ␣ ) 兺 0ⱕk⬍N ␻ n, k F k in t⬘n is
ᏻ(n ⫺ ␣ ⫺u ) for some u ⬎ 0 (use Proposition 7.8), and hence we are allowed to
forget about the contribution of this term to the main term of C n . e
8. Computing the Variance
We have devoted previous sections to analyze the (expected) cost of algorithms.
However, if the executions are random (in the different possible inputs of the
same size; or in the different possible executions with the same input, if the
algorithm is randomized), then the variance of the cost is also a valuable
measure. In this section, we will analyze the variance of two types of algorithms.
8.1. ONE-BRANCH ALGORITHMS. The first class includes the algorithms that
perform one recursive call with probability 1 ⫺ S n , and otherwise stop the
recursion chain with probability S n , where S n tends to zero as long as n grows.
FIND is a prototypical example of this class. It divides the original array into two
parts, and afterwards recursively continues in the subarray to the left or in the
subarray to the right of the current pivot, or either stops with probability 1/n. In
this setting, ␻ n, k is the probability that the algorithm performs the recursive call
over a subproblem with size k when the original problem has size n.
Let ᐀ n be the random variable that describes the nonrecursive cost, including
both the cost to split the problem and the cost to combine the recursive
solutions. We assume that these two costs do not depend on the particular input
nor the recursive solutions, but eventually depend on which was the “pivot”
(whatever it means depending on the algorithm). We also assume that the
following symmetry holds:
(k)
T n,
j ⫽ Pr兵᐀n ⫽ j兩the 共k ⫹ 1兲-th element is the pivot其
⫽ Pr兵᐀n ⫽ j兩the 共n ⫺ k兲-th element is the pivot其.
Under this hypothesis, the size of the recursive call implicitly gives us which was
the pivot (or its symmetric).
Let the total cost of the algorithm be described by the random variable Ᏺ n ,
and let P n, j ⫽ Pr{Ᏺ n ⫽ j}. Of course, we have 兺 jⱖ0 P n, j ⫽ 1. The P n, j ’s satisfy
the following recurrence:
P n, j ⯝
冘
0ⱕk⬍n
␻ n, k
冘P
k, i
(k)
T n,
j⫺i .
iⱖ0
The reason for the “⯝” above is that we dismiss the contribution to the cost of
the executions that stop the chain of recursive calls, since, by hypothesis, S n tends
to zero, and in most practical situations this cost is asymptotically irrelevant.
Let
E 共 n, k 兲 ⫽ E关᐀n兩the 共k ⫹ 1兲-th element is the pivot兴 ⫽
冘T
jⱖ0
(k)
n,j
䡠 j,
(14)
Improved Master Theorems for Divide-and-Conquer Recurrences
and
Q 共 n, k 兲 ⫽ E关共᐀n兲2兩the 共k ⫹ 1兲-th element is the pivot兴 ⫽
199
冘T
(k)
n,j
䡠 j 2.
(15)
jⱖ0
Define F n ⫽ E[Ᏺ n ] ⫽ 兺 jⱖ0 P n, j 䡠 j and G n ⫽ E[(Ᏺ n ) 2 ] ⫽ 兺 jⱖ0 P n, j 䡠 j 2 . On the
one hand, by means of simple rewriting we obtain
Fn ⯝ tn ⫹
冘
␻ n, k F k ,
(16)
0ⱕk⬍n
as expected. Recall that, despite the approximation above, the main term of the
solution of the recurrence is indeed the main term of F n . On the other hand,
from the equality j 2 ⫽ ( j ⫺ i) 2 ⫹ 2( j ⫺ i)i ⫹ i 2 , it is straightforward to get
Gn ⯝
冘
␻ n, k
冘
␻n,k共Q共n, k兲 ⫹ 2E共n, k兲 Fk兲 ⫹
k, i
iⱖ0
0ⱕk⬍n
⫽
冘P 冘T
(k)
n, j⫺i
䡠 j2
jⱖ0
0ⱕk⬍n
冘
␻n,kGk .
(17)
0ⱕk⬍n
To solve this recurrence we need to estimate the sums 兺 k ␻ n, k Q(n, k) and
兺 k ␻ n, k E(n, k) F k .
Assume that the set of ␻ n, k ’s corresponds to a canonical continuous recursive
definition, and that ᐀ n is not random, that is, ᐀ n ⫽ t n . Then Q(n, k) ⫽ (t n ) 2 ,
and from E(n, k) ⫽ t n and (16) we have 兺 0ⱕk⬍n ␻ n, k 2E(n, k) F k ⯝ 2t n F n ⫺
2(t n ) 2 . Thus, (17) reduces to
G n ⯝ 2t n F n ⫺ 共 t n 兲 2 ⫹
冘
␻ n, k G k .
(18)
0ⱕk⬍n
Notice that this recurrence is a continuous recursive definition with the same
shape function as (16).
Suppose first t n ⬃ n a ␰ n , where a ⬎ 0 and ␰ n includes subpolynomial terms.
According to the Continuous MT, F n ⬃ t n /(1 ⫺ ␸ (a)). Therefore, the toll
function in (18) is ⬃ 2n 2a ( ␰ n ) 2 /(1 ⫺ ␸ (a)) ⫺ n 2a ( ␰ n ) 2 ⫽ n 2a ( ␰ n ) 2 (1 ⫹
␸ (a))/(1 ⫺ ␸ (a)), and
Gn ⬃
1 ⫹ ␸共a兲
共 1 ⫺ ␸ 共 2a 兲兲共 1 ⫺ ␸ 共 a 兲兲
䡠 共 t n兲 2 .
Hence, the variance of Ᏺ n , V[Ᏺ n ] ⫽ E[(Ᏺ n ) 2 ] ⫺ E[Ᏺ n ] 2 , can be computed as
V 关 Ᏺ n兴 ⬃
␸共2a兲 ⫺ ␸共a兲2
共1 ⫺ ␸共2a兲兲共1 ⫺ ␸共a兲兲2
䡠 共 t n兲 2 .
Suppose now t n ⬃ lnc n 䡠 ␰ n , where c ⬎ ⫺1 and ␰ n includes sublogarithmical
terms. Then F n ⬃ t n ln n/(c ⫹ 1)I, where I ⫽ ⫺兰 10 ␻ ( z) ln z dz. Replacing this
result in (18) yields Gn ⬃ (tn)2 ln2 n/(c ⫹ 1)2I2, and we can only conclude V[Ᏺn] ⫽
o(Fn)2, but no more precise statement can be made in general.
200
SALVADOR ROURA
8.2. TWO-BRANCH ALGORITHMS. These algorithms perform two recursive
calls to subproblems with size k and n ⫺ 1 ⫺ k, respectively, with probability
1 ⫺ S n . Otherwise, the algorithm makes no recursive call. Again, S n tends to
zero as long as n grows. Quicksort is a prototypical example of this class. It
divides the array into two parts, and afterwards it recursively sorts the left and
the right subarrays.
Let ␺ n, k be the probability that the original problem of size n is broken into
two subproblems of sizes k and n ⫺ 1 ⫺ k, respectively. Like for one-branch
algorithms, assume that the algorithm’s cost to break and recombine does not
depend on the particular input nor the recursive solutions, but does depend on k,
the “pivot.” Again, let P n, j ⫽ Pr{Ᏺ n ⫽ j}. Then
P n, j ⯝
冘
␺ n, k
冘
P k, i P n⫺1⫺k, y T n, k, j⫺i⫺y .
i, yⱖ0
0ⱕk⬍n
Let E(n, k) be defined as in (14), and let F n ⫽ E[Ᏺ n ]. It is easy to see that F n
follows the recurrence
Fn ⯝ tn ⫹
冘
␺ n, k 共 F k ⫹ F n⫺1⫺k 兲 .
0ⱕk⬍n
Let Q(n, k) be defined as in (15), and let G n ⫽ E[(Ᏺ n ) 2 ]. Then
冘
Gn ⯝
␺ n, k
冘
P k, i P n⫺1⫺k, y
i, yⱖ0
0ⱕk⬍n
冘T
n, k, j⫺i⫺y
䡠 j 2.
jⱖ0
Using j 2 ⫽ ( j ⫺ i ⫺ y) 2 ⫹ 2( j ⫺ i ⫺ y)i ⫹ 2( j ⫺ i ⫺ y) y ⫹ 2iy ⫹ i 2 ⫹ y 2
yields
Gn ⯝
冘
␺ n, k 共 Q 共 n, k 兲 ⫹ 2E 共 n, k 兲共 F k ⫹ F n⫺1⫺k 兲 ⫹ 2F k F n⫺1⫺k 兲
0ⱕk⬍n
⫹
冘
␺n,k共Gk ⫹ Gn⫺1⫺k兲.
0ⱕk⬍n
We can solve the recurrences for F n and G n under some particular hypotheses.
Assume that E(n, k) ⫽ t n , Q(n, k) ⫽ (t n ) 2 , and that there exists some function
␺ ( z) such that
冘
0ⱕj⬍M n
冏冘
␺ n, k ⫺
k僆I n, j
冕
z n, j⫹1
冏
␺ 共 z 兲 dz ⫽ ᏻ 共 n ⫺␳兲,
z n, j
where ␺ ( z), M n , I n, j , z n, j and ␳ are like in Definition 3.2. Then F n follows the
continuous recurrence
Fn ⯝ tn ⫹
冘
␻ n, k F k ,
0ⱕk⬍n
where ␻ n, k ⫽ ␺ n, k ⫹ ␺ n, n⫺1⫺k , and thus ␻ ( z) ⫽ ␺ ( z) ⫹ ␺ (1 ⫺ z). Moreover,
␣ ⫽ 1, because
Improved Master Theorems for Divide-and-Conquer Recurrences
冕
1
␻ 共 z 兲 z 1 dz ⫽
0
冕
1
冕
共 ␺ 共 z 兲 z ⫹ ␺ 共 z 兲共 1 ⫺ z 兲兲 dz ⫽
0
1
201
␺ 共 z 兲 dz ⫽ 1.
0
Also, under the assumptions above, we have
G n ⯝ 2t n F n ⫺ 共 t n 兲 2 ⫹ 2
冘
冘
␺ n, k F k F n⫺1⫺k ⫹
0ⱕk⬍n
␻ n, k G k .
0ⱕk⬍n
Suppose first t n ⬃ n a ␰ n , where a ⬎ 1 and ␰ n includes subpolynomial terms.
Then F n ⬃ t n /(1 ⫺ ␸ (a)), and
Gn ⬃
冉
1 ⫹ ␸共a兲
共 1 ⫺ ␸ 共 2a 兲兲共 1 ⫺ ␸ 共 a 兲兲
⫹
2兰 10 ␺ 共 z 兲 z a 共 1 ⫺ z 兲 a dz
共 1 ⫺ ␸ 共 2a 兲兲共 1 ⫺ ␸ 共 a 兲兲 2
冊
共 t n兲 2 ,
where the integral above results from a reasoning similar to those used to derive
the Continuous MT. We can conclude
V 关 Ᏺ n兴 ⬃
␸共2a兲 ⫺ ␸共a兲2 ⫹ 2兰10 ␺共 z兲 za共1 ⫺ z兲a dz
共1 ⫺ ␸共2a兲兲共1 ⫺ ␸共a兲兲2
䡠 共 t n兲 2 .
Suppose now t n ⬃ n lnc n 䡠 ␰ n , where c ⬎ ⫺1 and ␰ n includes sublogarithmical terms. Then F n ⬃ t n ln n/Ᏼ, where Ᏼ ⫽ ⫺(c ⫹ 1) 兰 10 ␻ ( z) z ln z dz. Like
for one-branch algorithms, we can only conclude V[Ᏺ n ] ⫽ o(F n ) 2 , since
Gn ⬃
2 兰 10 ␺ 共 z 兲 z 共 1 ⫺ z 兲 dz
共 1 ⫺ ␸ 共 2 兲兲 Ᏼ 2
䡠 共 t n 兲 2 ln2n ⫽
共tn兲2 ln2n
Ᏼ2
.
For this last step, it is enough to notice that 2 兰 10 ␺ ( z) z(1 ⫺ z) dz ⫽ 1 ⫺ ␸ (2).
9. Open Problems
The results of this work could be extended in other directions. For instance,
consider the system of equations
F n(1)
F n(2)
⫽
⫽
t n(1) ⫹
⫽
(1, 1) (1)
␻ n,
⫹···⫹
i Fi
0ⱕi⬍n
t
(2)
n
⫹
t n(k) ⫹
冘
0ⱕi⬍n
冘
冘
(1, k) (k)
␻ n,
i Fi ,
0ⱕi⬍n
␻
(2, 1)
n, i
F
(1)
i
⫹···⫹
0ⱕi⬍n
···
F n(k)
冘
冘
(2, k) (k)
␻ n,
i Fi ,
0ⱕi⬍n
(k, 1) (1)
␻ n,
⫹···⫹
i Fi
冘
(k, k) (k)
␻ n,
i Fi ,
0ⱕi⬍n
where the weights above fulfill the discrete and/or the continuous pattern. The
problem is: Can we systematically solve such a system, for a wide set of toll
functions and weight distributions?
On the other hand, it would be interesting to dispose of theorems to
systematically solve recurrences on more than one variable, like
202
SALVADOR ROURA
F n, m ⫽ t n, m ⫹
冘 冘
w n, m, i, j F i, j ,
0ⱕi⬍n 0ⱕi⬍n
where the weights above follow a discrete or continuous pattern.
Finally, and mostly of theoretical interest, we could deal with toll functions
t n ⫽ n ␣ ␦ n for some ␦ n subpolynomial and superlogarithmical (like ␦ n ⫽
exp(公ln n)), or ␦ n ⫽ ␻ (log⫺1 n) and ␦ n ⫽ o(log⫺1⫹␧ n) for every ␧ ⬎ 0, or
␦ n ⫽ o(log⫺1 n) and ␦ n ⫽ ␻ (log⫺1⫺␧ n) for every ␧ ⬎ 0. The main interest of
the later case would be finding the maximum growing order for ␦ n such that
F n ⫽ ᏻ(n ␣ ).
Appendixes
A. Examples of the Use of the Discrete Master Theorem
We start solving (1). First, we identify the main term of toll function, and the sets
of values {w d } 1ⱕdⱕD and { z d } 1ⱕdⱕD . This yields t n ⬃ n 1 ln0 n, w 1 ⫽ w 2 ⫽ 1
and z 1 ⫽ z 2 ⫽ 1/ 2. Note that the bounds for r d, n and s d, n trivially hold here,
because Definition 2.1 always covers floors and ceilings. We now define ⌽( x) ⫽
2(1/ 2) x , and evaluate Ᏼ ⫽ 1 ⫺ ⌽(1) ⫽ 0. Since Ᏼ ⫽ 0, we compute Ᏼ⬘ ⫽ ⫺(0 ⫹
1)(2(1/2)1 ln(1/2)) ⫽ ln 2. Therefore, M n ⬃ n ln n/ln 2 ⫽ n log2 n.
Let us analyze (2). The main term in its toll function is n 0 ln0 n. Moreover,
w 1 ⫽ w 2 ⫽ 1/ 2 and z 1 ⫽ z 2 ⫽ 1/ 2. So, we define ⌽( x) ⫽ (1/ 2) x . Since Ᏼ ⫽
1 ⫺ ⌽(0) ⫽ 0, we compute Ᏼ⬘ ⫽ ⫺(0 ⫹ 1)((1/2)0 ln(1/2)) ⫽ ln 2, and conclude
B n ⬃ ln n/ln 2 ⫽ log2 n.
We now consider (3). Suppose t n ⫽ 6n 2 /ln5 n. The weights are w 1 ⫽ 2, w 2 ⫽
4 and w 3 ⫽ 1/ 2; the fractions are z 1 ⫽ 1/3, z 2 ⫽ 1/ 2 and z 3 ⫽ 4/5. It is a
simple matter to check that this recurrence follows Definition 2.1. So, we define
⌽( x) ⫽ 2(1/3) x ⫹ 4(1/ 2) x ⫹ 1/ 2 䡠 (4/5) x , and Ᏼ ⫽ 1 ⫺ ⌽(2) ⫽ ⫺122/225.
Since Ᏼ ⬍ 0, we have F n ⫽ ⌰(n ␣ ), where ␣ is the unique solution of the
equation ⌽(␣) ⫽ 1, which numerically is ␣ ⯝ 2.68723.
Solving
f 共 n 兲 ⫽ ⌰ 共 n 兲 ⫹ 2f
冉冊 冉冊 冉冊
n
2
⫹ 2f
n
3
⫹ 2f
n
6
is very similar. It yields f(n) ⫽ ⌰(n ␣ ), where ␣ ⯝ 1.72121 is the unique solution
of 2(1/2)␣ ⫹ 2(1/3)␣ ⫹ 2(1/6)␣ ⫽ 1. This is an upper bound to the system of
equations presented in Kuo and Chang [1994, pp. 79 – 80], where a very good
approximation, ␣ ⯝ 1.722, is given.4
Finally, set t n ⫽ n 2 for (11). Solving it is very easy: w 1 ⫽ 1 and z 1 ⫽ 1/4;
⌽( x) ⫽ (1/4) x and Ᏼ ⫽ 1 ⫺ ⌽(2) ⫽ 15/16; since Ᏼ ⬎ 0, we deduce F n ⬃
n 2 /(15/16) ⫽ 16n 2 /15.
In this example, we can get more information by means of a simple trick. Let
G n ⫽ F n ⫺ 16n 2 /15. Then,
4
Thanks are due to Fatos Xhafa for suggesting this example.
Improved Master Theorems for Divide-and-Conquer Recurrences
2
G n ⫽ n ⫹ F n/4 ⫺
16n 2
15
⫽⫺
n2
15
⫹

16 n
203
2
15 4
⫹ Gn/4 .
We can now find bounds for G n . Taking into account that n/4 ⫺ 3/4 ⱕ n/4 ⱕ
n/4, we have n 2 /16 ⫺ 3n/8 ⫹ 9/16 ⱕ n/4 2 ⱕ n 2 /16, and hence
⫺
2n
5
⫹
9
15
⫹ Gn/4 ⱕ Gn ⱕ Gn/4 .
We can use the Discrete MT again to solve G⬘n ⫽ ⫺2n/5 ⫹ 9/15 ⫹ G⬘n/4 . This
is even simpler, since the distribution of weights remains unchanged, and so does
⌽( x). The main term in the toll function is negative (⫺2n/5); recall that this is
not a problem. Computing Ᏼ yields Ᏼ ⫽ 1 ⫺ ⌽(1) ⫽ 3/4, and therefore G⬘n ⬃
⫺2/5 䡠 n/(3/4) ⫽ ⫺8n/15. We now solve G ⬙n ⫽ G ⬙n/4 , and get G ⬙n ⫽ ᏻ(n ␣ ) ⫽
ᏻ(1); not ⌰(1), since every term could be zero. Finally, we conclude F n ⫽
16n 2 /15 ⫹ G n , where ⫺8n/15 ⫹ o(n) ⱕ G n ⱕ ᏻ(1), thus achieving tighter
bounds for F n . We will see that for stochastic recurrences this technique provides
more accurate results.
B. Examples of the Use of the Continuous Master Theorem
We first analyze (4). It is trivial to show that ␻ ( z) ⫽ 2. Therefore, ␸ ( x) ⫽
兰 10 ␻ ( z) z x dz ⫽ 2/( x ⫹ 1) if x ⬎ ⫺1 (and ␸ ( x) ⫽ ⫹⬁ otherwise). Since Ᏼ ⫽
1 ⫺ ␸(1) ⫽ 0, we define Ᏼ⬘ ⫽ ⫺兰 10 ␻ ( z) z 1 ln z dz ⫽ 1/ 2, and finally Q n ⬃ n
ln n/(1/ 2) ⫽ (2 ln 2)n log2 n ⯝ 1.38n log2 n.
To solve (5) we start identifying the shape function for the weights, using
Lemma 7.2. First, we replace the term n ⫹ 1 in the denominator of ␻ n, k by n.
This yields ␴ n, k ⫽ 4(n ⫺ k)/n 2 . Second, we compute ␻ ( z) ⫽ n 䡠 ␴ n, zn ⫽ n 䡠
4(n ⫺ zn)/n 2 ⫽ 4(1 ⫺ z). Third and last, we check that 兰 10 ␻ ( z) dz ⫽ 2 ⱖ 1
indeed. Now we define ␸ ( x) ⫽ 4 兰 10 (1 ⫺ z) z x dz ⫽ 4(1/( x ⫹ 1) ⫺ 1/( x ⫹ 2))
if x ⬎ ⫺1 (and ␸ ( x) ⫽ ⫹⬁ otherwise). Since Ᏼ ⫽ 1 ⫺ ␸(0) ⫽ ⫺1 ⬍ 0, we
conclude F n ⫽ ⌰(n ␣ ), where ␣ is the unique solution of ␸(␣) ⫽ 1. This yields
␣ ⫽ (公17 ⫺ 3)/2 ⯝ 0.56155 (the only solution to ␣2 ⫹ 3␣ ⫺ 2 ⫽ 0 that is larger
than ⫺1).
We now solve (9), for which we already know that ␻ ( z) ⫽ 2z. In this case
␸ ( x) ⫽ 2/( x ⫹ 2) if x ⬎ ⫺2 (and ␸ ( x) ⫽ ⫹⬁ otherwise). Since Ᏼ ⫽ 1 ⫺
␸(1) ⫽ 1/3 ⬎ 0, it follows that S n ⬃ n/(1/3) ⫽ 3n.
In this example, we can get more information, using the technique that was
presented in Appendix A. Define G n ⫽ S n ⫺ 3n. Then,
Gn ⫽ n ⫹ 1 ⫹
2
n
2
冘
kS k ⫺ 3n ⫽ ⫺2n ⫹ 1 ⫹
0ⱕk⬍n
6
n
2
0ⱕk⬍n
But 兺 0ⱕk⬍n k 2 ⫽ n 3 /3 ⫺ n 2 / 2 ⫹ n/6. Thus,
G n ⫽ ⫺2 ⫹
1
n
⫹
2
冘
n2 0ⱕk⬍n
冘
kGk .
k2 ⫹
2
冘
n2 0ⱕk⬍n
kGk .
204
SALVADOR ROURA
Notice that, in contrast to what happens with discrete recurrences, the recurrence for G n above does not contain noise terms. Now we use the Continuous
MT again. The first step is already done, since the distribution of weights
remains the same, and ␸ ( x) too. Since Ᏼ ⫽ 1 ⫺ ␸(0) ⫽ 0, we compute Ᏼ⬘ ⫽
⫺兰 10 ␻ ( z) z 0 ln z dz ⫽ 1/ 2, and get G n ⬃ ⫺2 ln n/(1/ 2) ⫽ ⫺4 ln n.
We can go one step further, defining I n ⫽ G n ⫹ 4H n , which is slightly simpler
than defining I n ⫽ G n ⫹ 4 ln n. Notice that we are extracting the main terms of
S n one by one. This step yields
In ⫽
4H n
n
⫺
1
n
⫹
冘
2
n2
kI k ,
0ⱕk⬍n
where we have used the fact that 兺 0ⱕk⬍n kH k ⫽ n 2 H n / 2 ⫺ n 2 /4 ⫺ nH n / 2 ⫹
n/4. Now we get Ᏼ ⫽ 1 ⫺ ␸(⫺1) ⫽ ⫺1 ⬍ 0, and hence I n ⫽ ᏻ(n ␣ ) ⫽ ᏻ(1); we
cannot deduce I n ⫽ ⌰(1), since the toll function in the definition of I n includes
positive and negative terms together. At this point we cannot go any further,
because we have reached the core of the recurrence. As a conclusion, we have
S n ⫽ 3n ⫺ 4 ln n ⫹ ᏻ(1).
As a final remark, hybrid recurrences like
F n ⫽ n 2 ⫹ F n/ 2 ⫹
1
n
冘
Fk
0ⱕk⬍n
can be analyzed by combining both MTs. So, to solve this recurrence is enough to
find out that w 1 ⫽ 1, z 1 ⫽ 1/ 2 and ␻ ( z) ⫽ 1; compute Ᏼ ⫽ 1 ⫺ (1/ 2) 2 ⫺
兰 10 z 2 dz ⫽ 5/12 ⬎ 0; and conclude F n ⬃ n 2 /(5/12) ⫽ 12n 2 /5.
ACKNOWLEDGMENTS.
I thank my advisor Conrado Martı́nez for his support
while working on this problem, and several referees for their helpful comments.
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RECEIVED JULY
1998;
REVISED JULY
2000;
ACCEPTED JULY
2000
Journal of the ACM, Vol. 48, No. 2, March 2001.