Self-Improving Algorithms

Self-Improving Algorithms
c 2011 Society for Industrial and Applied Mathematics
Vol. 40, No. 2, pp. 350–375
Abstract. We investigate ways in which an algorithm can improve its expected performance
by fine-tuning itself automatically with respect to an unknown input distribution D. We assume
here that D is of product type. More precisely, suppose that we need to process a sequence I1 , I2 , . . .
of inputs I = (x1 , x2 , . . . , xn ) of some fixed length n, where each xi is drawn independently from
some arbitrary, unknown distribution Di . The goal is to design an algorithm for these inputs
that eventually the expected running time will be optimal for the input distribution D = i Di .
We give such self-improving algorithms for two problems: (i) sorting a sequence of numbers and
(ii) computing the Delaunay triangulation of a planar point set. Both algorithms achieve optimal
expected limiting complexity. The algorithms begin with a training phase during which they collect
information about the input distribution, followed by a stationary regime in which the algorithms
settle to their optimized incarnations.
Key words. average case analysis, Delaunay triangulation, low entropy, sorting
AMS subject classifications. 68Q25, 68W20, 68W40
DOI. 10.1137/090766437
1. Introduction. The classical approach to analyzing algorithms draws a familiar litany of complaints: worst-case bounds are too pessimistic in practice, say the
critics, while average-case complexity too often rests on unrealistic assumptions. The
charges are not without merit. Hard as it is to argue that the only permutations
we ever want to sort are random, it is a different level of implausibility altogether to
pretend that the sites of a Voronoi diagram should always follow a Poisson process
or that ray tracing in a binary space partitioning (BSP) tree should be spawned by a
Gaussian. Efforts have been made to analyze algorithms under more complex models
(e.g., Gaussian mixtures and Markov model outputs) but with limited success and
lingering doubts about the choice of priors.
Suppose we wish to compute a function f that takes I as input. We get a sequence
of inputs I1 , I2 , . . . , and wish to compute f (I1 ), f (I2 ), . . . . It is quite plausible to
assume that all these inputs are somehow related to each other. This relationship,
though exploitable, may be very difficult to express concisely. One way of modeling
this situation is to postulate a fixed (but complicated) unknown distribution D of
inputs. Each input Ij is chosen independently at random from D. (Refer to Figure
1.1.) Is it possible to quickly learn something about D so that we can compute f (I) (I
chosen from D) faster? (Naturally, this is by no means the only possible input model.
For example, we could have a memoryless Markov source, where each Ij depends only
∗ Received
by the editors July 28, 2009; accepted for publication (in revised form) December 28,
2010; published electronically April 5, 2011. Preliminary versions appeared as N. Ailon, B. Chazelle,
S. Comandur, and D. Liu, Self-improving algorithms, in Proceedings of the 17th SODA, 2006, pp.
261–270; and K. L. Clarkson and C. Seshadhri, Self-improving algorithms for Delaunay triangulations, in Proceedings of the 24th SoCG, 2008, pp. 148–155. This work was supported in part by NSF
grants CCR-998817 and 0306283 and ARO grant DAAH04-96-1-0181.
† Computer Science Faculty, Technion, Haifa, Israel (
‡ Department of Computer Science, Princeton University, Princeton, NJ (chazelle@cs.princeton.
§ IBM Almaden Research Center, San Jose, CA (,
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I1 , I 2 , I 3 , . . .
f (I1 ), f (I2 ), f (I3 ), . . .
Fig. 1.1. A self-improving algorithm A processes a sequence I1 , I2 , . . . of inputs drawn independently from a random source D.
on Ij−1 . However, for simplicity we will here focus on a fixed source that generates
the inputs independently.)
That is what a self-improving algorithm attempts to do. Initially, since nothing
is known about D, our self-improving algorithm can provide only a worst-case guarantee. As the algorithm sees more and more inputs, it can learn something about
the structure of D. We call this the training phase of the self-improving algorithm.
During this phase, the algorithm collects and organizes information about the inputs
in the hope that it can be used to improve the running time (with respect to inputs from D). The algorithm then moves to the limiting phase. Having decided that
enough has been learned about D, the algorithm uses this information to compute
f (I) faster. Note that this behavior is tuned to the distribution D.
Obviously, there is no reason why we should get a faster running time for all D.
Indeed, if f is the sorting function and D is the uniform distribution over permutations, then we require an expected Ω(n log n) time to sort. On the other hand, if D was
a low-entropy source of inputs, it is quite reasonable to hope for a faster algorithm.
So when can we improve our running time? An elegant way of expressing this is to
associate (using information theory) an “optimal” running time to each distribution.
This is a sort of estimate of the best expected running time we can hope for, given
inputs chosen from a fixed distribution D. Naturally, the lower the entropy of D, the
lower this running time will be. In the limiting phase, our self-improving algorithm
should achieve this optimal running time.
To expect a good self-improving algorithm that can handle all distributions D
seems a bit ambitious, and indeed we show that even for the sorting problem there
can be no such space-efficient algorithm (even when the entropy is low). Hence, it
seems necessary to impose some kind of restriction on D. However, if we required D
to be, say, uniform or a Gaussian, we would again be stuck with the drawbacks of
traditional average-case analysis. Hence, for self-improvement to be of any interest,
the restricted class of distributions should still be fairly general. One such class is
given by product distributions.
1.1. Model and results. We will focus our attention on distributions D of
product type. Think of each input as an n-dimensional vector (x1 , . . . , xn ) over some
appropriate domain. This could be a list of numbers (in the case of sorting) or a list of
points (for Delaunay triangulations). Each
xi is generated independently at random
from an arbitrary distribution Di , so D = i Di . All the Di ’s are independent of each
other. It is fairly natural to think of various portions of the input as being generated
by independent sources. For example, in computational geometry, the convex hull
of uniformly independently distributed points in the unit square is a well-studied
Note that all our inputs are of the same size n. This might appear to be a rather
unnatural requirement for (say) a sorting algorithm. Why must the 10th number in
our input come from the same distribution? We argue that this is not a major issue
(for concreteness, let us focus on sorting). The right way to think of the input is as a
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set of sources D1 , D2 , . . . , each independently generating a single number. The actual
“order” in which we get these numbers is not important. What is important is that
for each number, we know its source. For a given input, it is realistic to suppose that
some sources may be active, and some may not (so the input may have less than n
numbers). Our self-improving sorters essentially perform an independent processing
on each input number, after which O(n) time is enough to sort.1 The algorithm is
completely unaffected by the inactive sources. To complete the training phase, we
need to get only enough information about each source. What if new sources are
introduced during the stationary phase? Note that as long as O(n/ log n) new sources
(and hence new numbers) are added, we can always include these extra numbers in
the sorted list in O(n) time. Once the number of new sources becomes too large, we
will have to go back to the training phase. This is, of course, quite acceptable: if
the underlying distribution of inputs changes significantly, we have to recalibrate the
algorithm. For these reasons, we feel that it is no loss of generality to deal with a
fixed input length, especially for product distributions.
Our first result is a self-improving sorter. Given a source D = i Di of realnumber sequences I = (x1 , . . . , xn ), let π(I) denote the permutation induced by the
ranks of the xi ’s, using the indices i to break ties. Observe that since I is a random
variable, so is π(I). We can define the entropy H(π(I)), over the randomness of D,
and the limiting complexity of our algorithm will depend on H(π(I)). Note this
quantity may be much smaller than the entropy of the source itself but can never
exceed it.
As we mentioned earlier, the self-improving algorithm initially undergoes a training phase. At the end of this phase, some data structures storing information about
the distributions are constructed. In the limiting phase, the self-improving algorithm
is fixed, and these data structures do not change. In the context of sorting, the
self-improving sorter becomes some fixed comparison tree.
Theorem 1.1. There exists a self-improving
sorter of O(n + H(π(I))) limiting
complexity for any input distribution D = i Di . Its worst-case running time is
O(n log n). No comparison-based algorithm can sort an input from D in less than
H(π(I)) time. For any constant ε > 0, the storage can be made O(n1+ε ) for an
expected running time of O(ε−1 (n+H(π(I)))). The training phase lasts O(nε ) rounds,
and the probability that it fails is at most 1/n.
Why do we need a restriction on the input distribution? In section 3.3, we show
that a self-improving sorter that can handle any distribution requires an exponentially large data structure. Fredman [31] gave an algorithm that could optimally sort
permutations from any distribution D. His algorithm needs to know D explicitly, and
it constructs lookup tables of exponential size. Our bound shows that Fredman’s algorithm cannot be improved. Furthermore, we show that even for product distributions
any self-improving sorter needs superlinear space. Hence, our time-space tradeoffs
are essentially optimal. We remind the reader that we focus on comparison-based
Theorem 1.2. Consider a self-improving algorithm that, given any fixed distribution D, can sort a random input from D in an expected O(n + H(π(I))) time. Such
an algorithm requires 2Ω(n log n) bits of storage.
Let ε ∈ (0,
1). Consider a self-improving algorithm that, given any product distribution D = i Di , can sort a random input from D in an expected ε−1 (n + H(π(I)))
time. Such an algorithm requires a data structure of bit size n1+Ω(ε) .
1 The
self-improving Delaunay triangulation algorithms behave similarly.
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For our second result, we take the notion of self-improving algorithms to the
geometric realm and address the classical problem of computing the Delaunay
triangulation of a set of points in the Euclidean plane. Given a source D = i Di of
sequences I = (x1 , . . . , xn ) of points in R2 , let T (I) denote the Delaunay triangulation
of I. If we interpret T (I) as a random variable on the set of all undirected graphs with
vertex set {1, . . . , n}, then T (I) has an entropy H(T (I)), and the limiting complexity
of our algorithm depends on this entropy.
Theorem 1.3. There exists a self-improving algorithm for planar Delaunay triangulations of O(n + H(T (I))) limiting complexity for any input distribution D = Di .
Its worst-case running time is O(n log n). For any constant ε > 0, the storage can be
made O(n1+ε ) for an expected running time of O(ε−1 (n + H(T (I)))). The training
phase lasts O(nε ) rounds, and the probability that it fails is at most 1/n.
From the linear time reduction from sorting to computing Delaunay triangulations [14, Theorems 8.2.2 and 12.1.1], the lower bounds of Theorem 1.2 carry over to
Delaunay triangulations.
Both our algorithms follow the same basic strategy. During the training phase,
we collect data about the inputs in order to obtain a typical input instance V for D
with |V | = O(n), and we compute the desired structure S (a sorted list or a Delaunay
triangulation) on V . Then for each distribution Di , we construct an entropy optimal
search structure Di for S (i.e., an entropy optimal binary search tree or a distribution
sensitive planar point location structure). In the limiting phase, we use the Di ’s in
order to locate the components of a given input I in S. The fact that V is a typical
input ensures that I will be broken into individual subproblems of expected constant
size that can be solved separately, so we can obtain the desired structure for the input
V ∪ I in expected linear time (plus the time for the Di -searches). Finally, for both
sorting and Delaunay triangulation it suffices to know the solution for V ∪ I in order
to derive the solution for I in linear expected time [21, 22]. Thus, the running time
of our algorithms is dominated by the Di -searches, and the heart of the analysis lies
in relating this search time to the entropies H(π(I)) and H(T (I)), respectively.
1.2. Previous work. Concepts related to self-improving algorithms have been
studied before. List accessing algorithms and splay trees are textbook examples of
how simple updating rules can speed up searching with respect to an adversarial
request sequence [5, 15, 35, 45, 46]. It is interesting to note that self-organizing data
structures were investigated over stochastic input models first [4, 6, 13, 32, 40, 44].
It was the observation [11] that memoryless sources for list accessing are not terribly
realistic that partly motivated work on the adversarial models. It is highly plausible
that both approaches are superseded by more sophisticated stochastic models such as
hidden Markov models for gene finding or speech recognition, or time-coherent models
for self-customized BSP trees [8] or for randomized incremental constructions [23].
Recently, Afshani, Barbay, and Chan [1] introduced the notion of instance optimality,
which can be seen as a generalization of output-sensitivity. They consider the inputs
as sets and try to exploit the structure within each input for faster algorithms.
Much research has been done on adaptive sorting [30], especially on algorithms
that exploit near-sortedness. Our approach is conceptually different: we seek to
exploit properties, not of individual inputs, but of their distribution. Algorithmic
self-improvement differs from past work on self-organizing data structures and online
computation in two fundamental ways. First, there is no notion of an adversary: the
inputs are generated by a fixed, oblivious, random source D, and we compare ourselves
against an optimal comparison-based algorithm for D. In particular, there is no
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concept of competitiveness. Second, self-improving algorithms do not exploit structure
within any given input but, rather, within the ensemble of input distributions.
A simple example highlights this difference between previous sorters and the selfimproving versions. For 1 ≤ i ≤ n, fix two random integers ai , bi from {1, . . . , n2 }.
The distribution Di is such that Pr[xi = ai ] = Pr[xi = bi ] = 1/2, and we take D =
i=1 Di . Observe that every permutation generated by D is a random permutation,
since the ai ’s and bi ’s are chosen randomly. Hence, any solution in the adaptive, selforganizing/adjusting framework requires Ω(n log n) time, because no input Ij exhibits
any special structure to be exploited. On the other hand, our self-improving sorter
will sort a permutation from D in expected linear time during the limiting phase:
since D generates at most 2n different permutations, we have H(π(I)) = O(n).
2. Entropy and comparison-based algorithms. Before we consider sorting
and Delaunay triangulations, let us first recall some useful properties of information
theoretic entropy [28] and explain how it relates to our notion of comparison-based
Let X be a random
variable with a finite range X . The entropy of X, H(X), is
defined as H(X) := x∈X Pr[X = x] log(1/ Pr[X = x]). Intuitively, the event that
X = x gives us log(1/ Pr[X = x]) bits of information about the underlying elementary
event, and H(X) represents the expected amount of information that can be obtained
from observing X. We recall the following well-known property of the entropy of the
Cartesian product of independent random variables [28, Theorem 2.5.1].
Claim 2.1. Let H(X1 , . . . , Xn ) be the joint entropy of independent random variables X1 , . . . , Xn . Then
H(X1 , . . . , Xn ) =
H(Xi ).
We now define our notion of a comparison-based algorithm. Let U be an arbitrary
universe, and let X be a finite set. A comparison-based algorithm to compute a
function X : U → X is a rooted binary tree A such that (i) every internal node of
A represents a comparison of the form f (I) ≤ g(I), where f, g : U → R are arbitrary
functions on the input universe U; and (ii) the leaves of A are labeled with outputs
from X such that for every input I ∈ U, following the appropriate path for I leads
to the correct output X(I). If A has maximum depth d, we say that A needs d
comparisons (in the worst case). For a distribution D on U, the expected number of
comparisons (with respect to D) is the expected length of a path from the root to a
leaf in A, where the leaves are sampled according to the distribution that D induces
on X via X.
Note that our comparison-based algorithms generalize both the traditional notion of comparison-based algorithms [27, Chapter 8.1], where the functions f and g
are required to be projections, as well as the notion of algebraic computation trees
[9, Chapter 16.2]. Here the functions f and g must be composed of elementary
functions (addition, multiplication, square root) such that the complexity of the
composition is proportional to the depth of the node. Naturally, our comparisonbased algorithms can be much stronger. For example, deciding whether a sequence
x1 , x2 , . . . , xn of real numbers consists of n distinct elements needs one comparison
in our model, whereas every algebraic computation tree for the problem has depth
Ω(n log n) [9, Chapter 16.2]. However, for our problems of interest, we can still derive
meaningful lower bounds.
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Claim 2.2. Let D be a distribution on a universe U, and let X : U → X be a
random variable. Then any comparison-based algorithm to compute X needs at least
H(X) expected comparisons.
Proof. This is an immediate consequence of Shannon’s noiseless coding theorem [28, Theorem 5.4.1], which states that any binary encoding of an information
source such as X(I) must have an expected code length of at least H(X). Any
comparison-based algorithm A represents a coding scheme: the encoder sends the
sequence of comparison outcomes, and the decoder descends along the tree A, using
the transmitted sequence to determine comparison outcomes. Thus, any comparisonbased algorithm must perform at least H(X) comparisons in expectation.
Note that our comparison-based algorithms include all the traditional sorting
algorithms [27] (selection sort, insertion sort, quicksort, etc.) as well as classic algorithms for Delaunay triangulations [12] (randomized incremental construction, divide
and conquer, plane sweep). A notable exception are sorting algorithms that rely
on table lookup or the special structure of the input values (such as bucket sort or
radix sort) as well as transdichotomous algorithms for sorting [33, 34] or Delaunay
triangulations [16, 17, 18].
The following lemma shows how we can use the running times of comparisonbased algorithms to relate the entropy of different random variables. This is a very
important tool that will be used to prove the optimality of our algorithms.
Lemma 2.3. Let D be a distribution on a universe U, and let X : U → X and
Y : U → Y be two random variables. Suppose that the function f defined by f :
(I, X(I)) → Y (I) can be computed by a comparison-based algorithm with C expected
comparisons (where the expectation is over D). Then H(Y ) = C + O(H(X)), where
all the entropies are with respect to D.
Proof. Let s : X(U) → {0, 1}∗ be a unique binary encoding of X(U). By unique
encoding, we mean that the encoding is 1 − 1. We denote the expected code length of s
with respect to D, ED [|s(X(I))|], by Es . By another application of Shannon’s noiseless
coding theorem [28, Theorem 5.4.1]), we have Es ≥ H(X) for any unique encoding s
of X(U), and there exists a unique encoding s∗ of X(U) with Es∗ = O(H(X)).
Using f , we can convert s∗ into a unique encoding t of Y (U). Indeed, for every
I ∈ U, Y (I) can be uniquely identified by a string t(I) that is the concatenation of
s∗ (X(I)) and additional bits that represent the outcomes of the comparisons for the
algorithm to compute f (I, X(I)). Thus, for every element y ∈ Y (U), we can define
t(y) as the lexicographically smallest string t(I) for which Y (I) = y, and we obtain a
unique encoding t for Y (U). For the expected code length Et of t, we get
Et = ED [|t(Y (I))|] ≤ ED [C + |s∗ (X(I))|] = C + Es∗ = C + O(H(X)).
Since Shannon’s theorem implies Et ≥ H(Y ), the claim follows.
3. A self-improving sorter. We are now ready to describe our self-improving
sorter. The algorithm takes
an input I = (x1 , x2 , . . . , xn ) of real numbers drawn
from a distribution D = i Di (i.e., each xi is chosen independently from Di ). Let
π(I) denote the permutation induced by the ranks of the xi ’s, using the indices i
to break ties. By applying Claim 2.2 with U = Rn , X the set of all permutations
on {1, . . . , n}, and X(I) = π(I), we see that any sorter must make at least H(π(I))
expected comparisons. Since it takes Ω(n) steps to write the output, any sorter needs
Ω(H(π(I)) + n) steps. This is, indeed, the bound that our self-improving sorter
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For simplicity, we begin with the steady-state algorithm and discuss the training
phase later. We also assume that the distribution D is known ahead of time and that
we are allowed some amount of preprocessing before having to deal with the first input
instance (section 3.1). Both assumptions are unrealistic, so we show how to remove
them to produce a bona fide self-improving sorter (section 3.2). The surprise is how
strikingly little of the distribution needs to be learned for effective self-improvement.
3.1. Sorting with full knowledge. We consider the problem of sorting I =
(x1 , . . . , xn ), where each xi is a real number drawn from a distribution Di . We can
assume without loss of generality that all the xi ’s are distinct. (If not, simply replace
xi by xi + iδ for an infinitesimally small δ > 0, so that ties are broken according to
the index i.)
The first step of the self-improving sorter is to sample D a few times (the training
phase) and create a “typical” instance to divide the real line into a set of disjoint,
sorted intervals. Next, given some input I, the algorithm sorts I by using the typical
instance, placing each input number in its respective interval. All numbers falling
into the same intervals are then sorted in a standard fashion. The algorithm needs a
few supporting data structures.
• The V -list. Fix
an integer parameter λ = log n, and sample λ input instances from Di . Form their union, and sort the resulting λn-element multiset into a single list u1 ≤ · · · ≤ uλn . Next, extract from it every λth item
and form the list V = (v0 , . . . , vn+1 ), where v0 = 0, vn+1 = ∞, and vi = uiλ
for 1 ≤ i ≤ n. Keep the resulting V -list in a sorted table as a snapshot of a
“typical” input instance. We will prove the remarkable fact that, with high
probability, locating each xi in the V -list is linearly equivalent to sorting I.
We cannot afford to search the V -list directly, however. To do that, we need
auxiliary search structures.
• The Di -trees. For any i ≥ 1, let BiV be the predecessor2 of a random y
from Di in the V -list, and let HiV be the entropy of BiV . The Di -tree is
defined to be an optimum binary search tree
[41] over the keys of the V -list,
where the access probability of vk is PrDi xi ∈ [vk , vk+1 ) = Pr BiV = k for
any 0 ≤ k ≤ n. This allows us to compute BiV using O(HiV + 1) expected
The self-improving sorter. The input I is sorted by a two-phase procedure.
First we locate each xi in the V -list using the Di -trees. This allows us to partition
I into groups Z0 < Z1 < · · · of xi ’s sharing
the same predecessor in the V -list. The
first phase of the algorithm takes an O(n + i HiV ) expected time.3 The next phase
involves going througheach Zk and sorting their elements naively, say, using insertion
sort, in total time O( k |Zk |2 ). See Figure 3.1.
The expected running time is O(n + ED [ i HiV + k |Zk |2 ]), and the total space
used is O(n2 ). This can be decreased to O(n1+ε ) for any constant ε > 0; we describe
how at the end of this section. First, we show how to bound the running time of the
first phase. This is where we really show the optimality of our sorter.
Lemma 3.1.
HiV = O(n + H(π(I))).
2 Throughout this paper, the predecessor of y in a list refers to the index of the largest list element
≤ y; it does not refer to the element itself.
3 The H V ’s themselves are random variables depending on the choice of the V -list. Therefore,
this is a conditional expectation.
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Training phase
I1 , I 2 , I 3 , . . .
Limiting phase
I = (x1 , x2 , . . . , xn )
V ∪I
Fig. 3.1. The self-improving sorter: during the training phase, the algorithm constructs a
typical sorted list, the V -list, and a sequence D1 , D2 , . . . of optimal search trees for V with respect
to D1 , D2 , . . . . In the limiting phase, the algorithm uses the Di ’s to locate the xi ’s in the V -list,
sorts the individual buckets, and removes the elements from V .
Proof. Our proof actually applies to any linear-sized sorted list V . Let B V :=
, . . . , BnV ) bethe sequence of predecessors for all elements in I. By Claim 2.1, we
have H(B V ) = i HiV , so it suffices to bound the entropy of H(B V ). By Lemma 2.3
applied with U = Rn , X(I) = π(I), and Y (I) = B V , it suffices to give a comparisonbased algorithm that can determine B V (I) from (I, π(I)) with O(n) comparisons. But
this is easy: just use π(I) to sort I (which needs no further comparisons), and then
merge the sorted list I with V . Now the lemma follows from Claim 2.1.
Next we deal with the running time of the second phase. As long as the groups
Zk are small, the time to sort each group will be small. The properties of the V -list
ensure that this is the case.
Lemma 3.2. For 0 ≤ k ≤ n, let Zk = {xi | vk ≤ xi < vk+1 } be the elements with
predecessor k. With
probability atleast2 1 − n over the construction of the V -list,
we have ED |Zk | = O(1) and ED |Zk | = O(1) for all 0 ≤ k ≤ n.
Proof. Remember that the V -list was formed by taking certain elements from a
sequence Iˆ = s1 , s2 , . . . , sλn that was obtained by concatenating λ = log n inputs
ˆ and let t = [si , sj ). Note that all
I1 , I2 , . . . . Let si ≤ sj be any two elements from I,
the other λn − 2 numbers are independent of si and sj . Suppose we fix the values
of si and sj (in other words, we condition on the values of si and sj ). For every
∈ {1, . . . , λn} \ {i, j}, let Y be the indicator random variable for the event that
s ∈ t, and let Y (t) := Y . Since all the Y ’s are independent, by Chernoff’s
bound [42, Theorem 4.2], for any β ∈ [0, 1],
Pr[Y (t) ≤ (1 − β)E[Y (t) ]] ≤ exp −β 2 E[Y (t) ]/2 .
Setting β = 10/11, we see that if E[Y (t) ] > 11log n, then Y (t) > log n with
probability at least 1 − 1/(λ2 n4 ). Note that this is true for every fixing of si and sj .
Therefore, we get the above statement even with the unconditioned random variable
Y (t) . Now, by applying the same argument to any pair si , sj with i = j and taking
such pairs, we get that with probability at least 1 − n−2
a union bound over all λn
over the construction of I the following holds for all half-open intervals t defined by
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pairs si , sj with i = j: if E[Y (t) ] > 11log n, then Y (t) > log n. From now on we
assume that this implication holds.
The V -list is constructed such that for tk = [vk , vk+1 ), Y (tk ) ≤ log n, and hence
(t )
E[Y (tk ) ] = O(log n). Let Xi k be the indicator random variable for the event that
(t )
xi ∈R Di lies in tk , and let X (tk ) := i Xi k = |Zk |. Note that (where a and b
denote the indices of vk and vk+1 in I)
E[Y (tk ) ] =
(tk )
(tk )
] − 2 = log nE[X (tk ) ] − 2,
and therefore E[X (tk ) ] = O(1). Now, since the expectation of X (tk ) is constant, and
since X (tk ) is a sum of independent indicator random variables, we can apply the
following standard claim in order to show that the second moment of X (tk ) is also
Claim 3.3. Let X = i Xi be a sum of independent positive random variables
with Xi = O(1) for all i and E[X] = O(1). Then E[X 2 ] = O(1).
Proof. By linearity of expectation,
2 ⎤
Xi ⎦ =
E Xi2 + 2
E[Xi ]E[Xj ]
E X = E⎣
O (E[Xi ]) +
E[Xi ]
= O(1).
This concludes the proof of Lemma 3.2.
Combining Lemmas 3.1 and 3.2, we get the running time of our self-improving
sorter to be O(n + H(π(I))). This proves the optimality of time taken by the sorter.
We now show that the storage can be reduced to O(n1+ε ) for any constant ε > 0.
The main idea is to prune each Di -tree to depth ε log n. This ensures that each tree
has size O(nε ), so the total storage used is O(n1+ε ). We also construct a completely
balanced binary tree T for searching in the V -list. Now, when we wish to search for xi
in the V -list, we first search using the pruned Di -tree. At the end, if we reach a leaf
of the unpruned Di -tree, we stop since we have found the right interval of the V -list
which contains xi . On the other hand, if the search in the Di -tree was unsuccessful,
then we use T for searching.
In the first case, the time taken for searching is simply the same as with unpruned
Di -trees. In the second case, the time taken is O((1+ε) log n). But note that the time
taken with unpruned Di -trees is at least ε log n (since the search on the pruned Di -tree
failed, we must have reached some internal node of the unpruned tree). Therefore,
the extra time taken is only a O(ε−1 ) factor of the original time. As a result, the
space can be reduced to O(n1+ε ) with only a constant factor increase in running time
(for any fixed ε > 0).
3.2. Learning the distribution. In the last section we showed how to obtain a
self-improving sorter if D is known. We now explain how to remove this assumption.
The V -list is built in the first log n rounds, as before. The Di -trees will be built
after O(nε ) additional rounds, which will complete the training phase. During that
phase, sorting is handled via, say, mergesort, to guarantee O(n log n) complexity. The
training part consists of learning basic information about BiV for each i. For notational
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simplicity, fix i, and let pk = Pr[BiV = k] = Pr Di [ vk ≤ xi < vk+1 ]. Let M = cnε for
a large enough constant c. For any k, let χk be the number of times, over the first M
rounds, that vk is found to be the V -list predecessor of xi . (We use a standard binary
search to compute predecessors in the training phase.) Finally, define the Di -tree to
be a weighted binary search tree defined over all the vk ’s such that χk > 0. Recall
that the crucial property of such a tree is that the node associated with a key of
weight χk is at depth O(log(M/χk )). We apply this procedure for each i = 1, . . . , n.
This Di -tree is essentially the pruned version of the tree we used in section 4.1.
Like before, its size is O(M ) = O(nε ), and it is used in a similar way as in section 4.1,
with a few minor differences. For completeness, we go over it again: given xi , we
perform a search down the Di -tree. If we encounter a node whose associated key vk is
such that xi ∈ [vk , vk+1 ), we have determined BiV , and we stop the search. If we reach
a leaf of the Di -tree without success, we simply perform a standard binary search in
the V -list.
Lemma 3.4. Fix i. With probability at least 1 − 1/n3 , for any k, if pk > n−ε/3 ,
then M pk /2 < χk < 3M pk /2.
Proof. The expected value of χk is M pk . If pk > n−ε/3 , then, by Chernoff’s
bound [7, Corollary A.17], the count χk deviates from its expectation by more than
a = M pk /2 with probability less than (recall that M = cnε )
2 exp(−2a2 /M ) = 2 exp(−M p2k /2) < 2 exp(−(c/2)n2ε/3 ) ≤ n−4
for c large enough. A union bound over all k completes the proof.
Suppose now the implication of Lemma 3.4 holds for all k (and fixed i). We show
now that the expected
search time for xi is O(ε−1 HiV + 1). Consider each element in
the sum Hi = k pk log(1/pk ). We distinguish two cases.
• Case 1: pk > n−ε/3 . In this case, vk must be in Di , as otherwise we would
have χk = 0 by the definition of Di , which is a contradiction (for n large
enough) to Lemma 3.4, which states that χk > M n−ε/3 /2 . Hence, the cost
of the search is O(log(M/χk )), and its contribution to the expected search
time is O(pk log(M/χk )). By Lemma 3.4, this is also O(pk (1 + log p−1
k )), as
• Case 2: pk ≤ n−ε . The search time is always O(log n); hence the contribution
to the expected search time is O(ε−1 pk log p−1
k ).
By summing up over all k, we find that the expected search time is O(ε−1 HiV +1).
This assumes the implication of Lemma 3.4 for all i. By a union bound, this holds
with probability at least 1 − 1/n2 . The training phase fails when either this does not
hold, or if the V -list does not have the desired properties (Lemma 3.2). The total
probability of this is at most 1/n.
3.3. Lower bounds. Can we hope for a result similar to Theorem 1.1 if we
drop the independence assumption? The short answer is no. As we mentioned earlier, Fredman [31] gave a comparison-based algorithm that can optimally sort any
distribution of permutations. This uses an exponentially large data structure to decide which comparisons to perform. Our lower bound shows that the storage used by
Fredman’s algorithm is essentially optimal.
To understand the lower bound, let us examine the behavior of a self-improving
sorter. Given inputs from a distribution D, at each round, the self-improving sorter is
just a comparison tree for sorting. After any round, the self-improving sorter may wish
to update the comparison tree. At some round (eventually), the self-improving sorter
must be able to sort with expected O(n + H(π(I))) comparisons: the algorithm has
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“converged” to the optimal comparison tree. The algorithm uses some data structure
to represent (implicitly) this comparison tree.
We can think of a more general situation. The algorithm is explicitly given an
input distribution D. It is allowed some space where it stores information about D (we
do not care about the time spent to do this). Then, (using this stored information) it
must be able to sort a permutation from D in expected O(n + H(π(I))) comparisons.
So the information encodes some fixed comparison-based procedure. As shorthand for
the above, we will say that the algorithm, on input distribution D, optimally sorts D.
How much space is required to deal with all possible D’s? Or just to deal with product
distributions? These are the questions that we shall answer.
Lemma 3.5. Let h = (n log n)/α for some sufficiently large constant α < 0,
and let A be an algorithm that can optimally sort any input distribution D with
H(π(I)) ≤ h. Then A requires 2Ω(n log n) bits of storage.
Proof. Consider the set of all n! permutations of {1, . . . , n}. Every subset Π of
2h permutations induces a distribution DΠ defined by picking every permutation in
Π with equal probability and none other. Note that the total number of such dis h
) = h, where D<
is the distribution
tributions DΠ is 2n!h > (n!/2h )2 and H(D<
on the output π(I) induced by D . Suppose there exists a comparison-based procedure AΠ that sorts a random input from DΠ in expected time at most c(n + h) for
some constant c > 0. By Markov’s inequality this implies that at least half of the
permutations in Π are sorted by AΠ in at most 2c(n + h) comparisons. But, within
2c(n + h) comparisons, the procedure AΠ can only sort a set P of at most 22c(n+h)
permutations. Therefore, any other Π such that AΠ = AΠ will have to draw at least
half of its elements from P . This limits the number of such Π to
2c(n+h) h−1
< (n!)2 2c(n+h)2 .
2 /2
2 /2
This means that the number of distinct procedures needed exceeds
(n!/2h )2 /((n!)2
2c(n+h)2 ) > (n!)2
2−(c+1)(n+h)2 = 2Ω(2
n log n)
assuming that h/(n log n) is small enough. A procedure is entirely specified by a string
of bits; therefore, at least one such procedure must require storage logarithmic in the
previous bound.
We now show that a self-improving sorter dealing with product distributions
requires superlinear size. In fact, the achieved tradeoff between the O(n1+ε ) storage
bound and an expected running time off the optimal by a factor of O(1/ε) is optimal.
Lemma 3.6. Let c > 0 be a large enough parameter, and let A be an algorithm
that, given a product distribution D, can sort a random permutation from D in expected
time c(n + H(π(I))). Then A requires a data structure of bit size n1+Ω(1/c) .
Proof. The proof is a specialization of the argument used for proving Lemma 3.5.
Let h = (n log n)/(3c) and κ = 2h/n . We define Di by choosing κ distinct integers
in {1, . . . , n} and making them equally likely to be picked as xi . (For convenience, we
use the tie-breakingrule that maps xi → nxi +i−1. This ensures that π(I) is unique.)
We then
n set D := i Di . By Claim 2.1, D has entropy n · h/n = Θ(h). This leads
to nκ > (n/κ)κn choices of distinct distributions D. Suppose that A uses s bits of
storage and can sort each such distribution in c(n + h) expected comparisons. Some
fixing S of the bits must be able to accommodate this running time for a set G of at
least (n/κ)κn 2−s distributions D. In other words, some comparison-based procedure
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can deal with (n/κ)κn 2−s distributions D. Any input instance that is sorted in at
most 2c(h + n) time by S is called easy: the set of easy instances is denoted by E.
Because S has to deal with many distributions, there must be many instances
that are easy for S. This gives a lower bound for |E|. On the other hand, since easy
instances are those that are sorted extremely quickly by S, there cannot be too many
of them. This gives an upper bound for |E|. Combining these two bounds, we get a
lower bound for s. We will begin with the easier part: the upper bound for |E|.
Claim 3.7. |E| ≤ 22c(h+n)+2 .
Proof. In the comparison-based algorithm represented by S, each instance I ∈ E
is associated with a leaf of a binary decision tree of depth at most 2c(h + n), i.e.,
with one of at most 22c(h+n) leaves. This would give us an upper bound on s if each
I ∈ E were assigned a distinct leaf. However, it may well be that two distinct inputs
I, I ∈ E have π(I) = π(I ) and lead to the same leaf. Nonetheless, we have a collision
bound, saying that for any permutation π, there are at most 4n instances I ∈ E with
π(I) = π. This implies that |E| ≤ 4n 22c(h+n) .
To prove the collision bound, first fix a permutation π. How many instances can
map to this permutation? We argue that knowing that π(I) = π for an instance
I ∈ E, we need only 2n − 1 additional bits to encode I. This immediately shows that
there must be fewer than 4n such instances I. Write I = (x1 , . . . , xn ), and let I be
sorted to give the vector I = (y1 , . . . , yn ). Represent the ground set of I as an n-bit
vector α (αi = 1 if some xj = i, or else αi = 0). For i = 2, . . . , n, let βi = 1 if
yi = yi−1 ; else βi = 0. Now, given α and β, we can immediately deduce the vector I,
and by applying π −1 to I, we get I. This proves the collision bound.
Claim 3.8. |E| ≥ nn κ−2n 2−2s/κ .
Proof. Each Di is characterized by a vector vi = (ai,1 , . . . , ai,κ ), so that D itself is
specified by v = (v1 , . . . , vn ) ∈ Rnκ . (From now on, we view v both as a vector and as a
distribution of input instances.) Define the jth projection of v as v j = (a1,j , . . . , an,j ).
Even if v ∈ G, it could well be that none of the projections of v are easy. However,
if we consider the projections obtained by permuting the coordinates of each vector
vi = (ai,1 , . . . , ai,κ ) in all possible ways, we enumerate each input instance from v
the same number of times. Note that applying these permutations gives us different
vectors which also represent D. Since the expected time to sort an input chosen
from D ∈ G is at most c(h + n), by Markov’s inequality, there exists a choice of
permutations (one for each 1 ≤ i ≤ n) for which at least half of the projections of the
vector obtained by applying these permutations are easy.
Let us count how many distributions have a vector representation with a choice of
permutations placing half its projections in E. There are fewer than |E|κ/2 choices of
such instances, and, for any such choice, each vi = (ai,1 , . . . , ai,κ ) has half its entries
already specified, so the remaining choices are fewer than nκn/2 . This gives an upper
bound of nκn/2 |E|κ/2 on the number of such distributions. This number cannot be
smaller than |G| ≥ (n/κ)κn 2−s ; therefore |E| ≥ nn κ−2n 2−2s/κ , as desired.
It now just remains to put the bounds together:
nn κ−2n 2−2s/κ
=⇒ n log n − 2n log κ − 2s/κ
2ch + 2cn + 2
=⇒ κn(log n − 2 log κ) − 2cκh − 2cκn − 2κ
We have κ = nΘ(1/c) and h = (n log n)/(3c). Since c is sufficiently large, we get
s = n1+Ω(1/c) .
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4. Delaunay triangulations. We now consider self-improving algorithms for
Delaunay triangulations. The aim of this section is to prove Theorem 1.3. Let I =
(x1 , . . . , xn ) denote an input instance, where each xi is a point in the plane, generated
by a point distribution Di . The distributions Di are arbitrary and may be continuous,
although we never explicitly use such a condition. Each xi is independent of
the others,
so in each round the input I is drawn from the product distribution D = i Di , and
we wish to compute the Delaunay triangulation of I, T (I). To keep our arguments
simple, we will assume that the points of I are in general position (i.e., no four points
in I lie on a common circle). This is no loss of generality and does not restrict the
distribution D, because the general position assumption can always be enforced by
standard symbolic perturbation techniques [29]. Also we will assume that there is
a bounding triangle that always contains all the points in I. Again, this does not
restrict the distribution D in any way, because we can always simulate the bounding
triangle symbolically by adding virtual points at infinity.
The distribution D induces a (discrete) distribution on the set of Delaunay triangulations, viewed as undirected graphs with vertex set {1, . . . , n}. Consider the
entropy of this distribution: for each graph G on {1, . . . , n}, let pG be the probability
that it represents theDelaunay triangulation of I ∈R D. We have the output entropy H(T (I)) := − G pG log pG . By Claim 2.2, any comparison-based algorithm
to compute the Delaunay triangulation of I ∈R D needs at least H(T (I)) expected
comparisons. Hence, an optimal algorithm will be one that has an expected running
time of O(n + H(T (I))) (since it takes O(n) steps to write the output).
We begin by describing the basic self-improving algorithm. (As before, we shall
first assume that some aspects of the distribution D are known.) Then, we shall
analyze the running time using our information theory tools to argue that the expected
running time is optimal. Finally, we remove the assumption that D is known and give
the time-space tradeoff in Theorem 1.3.
4.1. The algorithm. We describe the algorithm in two parts. The first part explains the learning phase and the data structures that are constructed (section 4.1.1).
Then, we explain how these data structures are used to speed up the computation
in the limiting phase (section 4.1.2). As before, the expected running time will be
expressed in terms of certain parameters of the data structures obtained in the learning phase. In the next section (section 4.2), we will prove that these parameters are
comparable to the output entropy H(T (I)). First, we will assume that the distributions Di are known to us and the data structures described will use O(n2 ) space.
Section 4.3 repeats the arguments of section 3.2 to remove this assumption and to
give the space-time tradeoff bounds of Theorem 1.3.
As outlined in Figure 4.1, our algorithm for Delaunay triangulation is roughly a
generalization of our algorithm for sorting. This is not surprising, but note that while
the steps of the two algorithms, and their analyses, are analogous, in several cases a
step for sorting is trivial, but the corresponding step for Delaunay triangulation uses
some relatively recent and sophisticated prior work.
4.1.1. Learning phase. For each round in the learning phase, we use a standard
algorithm to compute the output Delaunay triangulation. We also perform some extra
computation to build some data structures that will allow speedup in the limiting
The learning phase is as follows. Take the first λ := log n input lists I1 , I2 , . . . , Iλ .
Merge them into one list Iˆ of λn = nlog n points. Setting ε := 1/n, find an ε-net
V ⊆ Iˆ for the set of all open disks. In other words, find a set V such that for
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Intervals (xi , xi ) containing no values
of I
Typical set V
log n training instance points with the
same BV value
Expect O(1) values of I within each
bucket (of the same B V index)
Optimal weighted binary trees Di
Sorting within buckets
Sorted list of V ∪ I
Build sorted V from sorted V ∪ I (trivial)
(analysis) merge sorted V and I
(analysis) recover the indices BiV from
the sorted I (trivial)
Delaunay triangulation
Delaunay disks
Range space ε-net V [26, 39], ranges are
disks, ε = 1/n
log n training instance points in each
Delaunay disk
Expect O(1) points of I in each Delaunay disk of V
Entropy-optimal planar point location
data structures Di [10]
Triangulation within V(Zs ) ∩ s (Claim
T (V ∪ I)
Build T (I) from T (V ∪ I) [21, 22]
(analysis) merge T (V ) and T (I) [19]
(analysis) recover the triangles BiV in
T (V ) from T (I) (Lemma 4.8)
Fig. 4.1. Delaunay triangulation algorithm as a generalization of the sorting algorithm.
ˆ C contains at
any open disk C that contains more than ελn = log n points of I,
least one point of V . It is well known that that there exist ε-nets of size O(1/ε) for
disks [26, 38, 39, 43], which here is O(n). Furthermore, it is folklore that our desired
ε-net V can be constructed in time n(log n)O(1) , but there seems to be no explicit description of such an algorithm for our precise setting. Thus, we present an algorithm
based on a construction by Pyrga and Ray [43] in the appendix.
Having obtained V , we construct the Delaunay triangulation of V , which we
denote by T (V ). This is the analogue of the V -list for the self-improving sorter. We
also build an optimal planar point location structure (called D) for T (V ): given a
point, we can find in O(log n) time the triangle of T (V ) that it lies in [12, Chapter 6].
Define the random variable BiV to be the triangle of T (V ) that xi falls into.4 Now let
the entropy of
BiV be HiV . If the probability that xi falls into triangle t of T (V ) is pti ,
then Hi = − t pti log pti . For each i, we construct a search structure Di of size O(n)
that finds BiV in an expected O(HiV ) time. These Di ’s can be constructed using the
results of Arya et al. [10], for which the expected number of primitive comparisons is
HiV + o(HiV ). These correspond to the Di -trees used for sorting.
We will now prove an analogue to Lemma 3.2 which shows that the triangles of
T (V ) do not contain many points of a new input I ∈R D on the average. Consider a
triangle t of T (V ), and let Ct be its circumscribed disk; Ct is a Delaunay disk of V .
If a point xi ∈ I lies in Ct , we say that xi is in conflict with t and call t a conflict
triangle for xi . Refer to Figure 4.2. (The “conflict” terminology arises from the fact
that if xi were added to V , triangles with which it conflicts would no longer be in the
Delaunay triangulation.) Let Zt := I ∩ Ct , the random variable that represents the
points of I ∈R D that fall inside Ct , the conflict set of t. Furthermore, let Xt := |Zt |.
4 Assume that we add the vertices of the bounding triangle to V . This will ensure that x will
always fall within some triangle BiV .
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T (V )
Fig. 4.2. Conflicts between T (V ) and the inputs: the input point x conflicts with triangles t1
and t2 , y conflicts with t1 , t2 , and t3 , and z conflicts only with t3 .
Note that the randomness comes from the random distribution of Iˆ (on which V and
T (V ) depend) as well as the randomness of I. We are interested in the expectation
E[Xt ] over I of Xt . All expectations are taken over a random input I chosen from D.
Lemma 4.1. For any triangle t of T (V ), let Zt = {xi | xi ∈ Ct } be the conflict
set of t, and define Xt := |Zt |. With probability at least 1 − n−2 over the construction
of T (V ), we have E[Xt ] = O(1) and E[Xt2 ] = O(1) for all triangles t of T (V ).
Proof. This is similar to the argument given in Lemma 3.2 with a geometric twist.
Let the list of points Iˆ be s1 , . . . , sλn , the concatenation of I1 through Iλ . Fix three
distinct indices i, j, k and the triangle t with vertices si , sj , sk (so we are effectively
conditioning on si , sj , sk ). Note that all the remaining λn − 3 points are chosen
independently of si , sj , sk from some distribution D . For each ∈ {1, . . . , λn} \
{i, j, k}, let Y be the indicator variable for the event that s is inside Ct . Let
Y (t) = Y . Setting β = 11/12 in (3.1), we get that if E[Y (t) ] > 12log n, then
> log n with probability at least 1 − 1/(λ3 n5 ). This is true for every fixing
of si , sj , sk , so it is also true unconditionally. By applying the same argument to
any triple i, j, k of distinct indices, and taking a union bound over all λn
we obtain that with probability at least 1 − n−2 , for any triangle t generated by the
ˆ if E[Y (t) ] > 12log n, then Y (t) > log n. We henceforth assume that
points of I,
this event happens.
Consider a triangle t of T (V ) and its circumcircle Ct . Since T (V ) is Delaunay,
Ct contains no point of V in its interior. Since V is a (1/n)-net for all disks with respect
ˆ Ct contains at most log n points of I,
ˆ that is, Y (t) ≤ log n. This implies that
to I,
E[Y ] = O(log n), as in the previous paragraph. Since E[Y (t) ] > log nE[Xt ] − 3, we
obtain E[Xt ] = O(1), as claimed. Furthermore, since Xt can be written as a sum of
independent indicator random variables, Claim 3.3 shows that E[Xt2 ] = O(1).
4.1.2. Limiting phase. We assume that we are done with the learning phase
and have T (V ) with the property given in Lemma 4.1: for every triangle t ∈ T (V ),
E[Xt ] = O(1) and E[Xt2 ] = O(1). We have reached the limiting phase, where the
algorithm is expected to compute the Delaunay triangulation with the optimal running
time. We will prove the following lemma in this section.
Lemma 4.2. Using the data structures from the learning phase, and the properties
the Delaunay
of them that hold with probability at least 1 − 1/n2, in the limiting
triangulation of input I can be generated in an expected O(n + i=1 HiV ) time.
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T (V )
Fig. 4.3. Determining the conflict set for xi : the triangle BiV containing xi is found via Di .
Then we perform a breadth-first search from BiV until we encounter triangles that no longer conflict
with xi . The dark gray triangles form the conflict set of xi , and the light gray triangles mark the
end of the breadth-first search. Since the conflict set Si is connected, and since the dual graph has
bounded degree, this takes O(|Si |) steps.
The algorithm, and the proof of this lemma, has two steps. In the first step,
T (V ) is used to quickly compute T (V ∪ I), with the time bounds of the lemma.
In the second step, T (I) is computed from T (V ∪ I), using a randomized splitting
algorithm proposed by Chazelle et al. [21], who provide the following theorem.
Theorem 4.3 (see [21, Theorem 3]). Given a set of n points P and its Delaunay
triangulation, for any partition of P into two disjoint subsets P1 and P2 , the Delaunay triangulations T (P1 ) and T (P2 ) can be computed in O(n) expected time, using a
randomized algorithm.
The remainder of this section
n is devoted to showing that T (V ∪I) can be computed
in an expected time O(n + i=1 HiV ). The algorithm is as follows. For each xi ∈ I,
we use Di to find the triangle BiV of T (V ) that contains
n it. By the properties of
the Di ’s as described in section 4.1.1, this takes O( i=1 HiV ) expected time. We
now need to argue that given the BiV ’s, the Delaunay triangulation T (V ∪ I) can be
computed in expected linear time. For each xi , we walk through T (V ) and find all the
Delaunay disks of T (V ) that contain xi , as in incremental constructions of Delaunay
triangulations [12, Chapter 9]. This is done by breadth-first search of the dual graph
of T (V ), starting from BiV . Refer to Figure 4.3. Let Si denote the set of triangles
whose circumcircles contain xi . We remind the reader that Zt is the conflict set of
triangle t.
Claim 4.4. Given all BiV ’s, all Si and Zt sets can be found in expected linear
Proof. To find all Delaunay disks containing xi , do a breadth-first search from
BiV . For any triangle t encountered, check if Ct contains xi . If it does not, then we
do not look at the neighbors of t. Otherwise, add t to Si and xi to Zt and continue.
Since Si is connected in the dual graph of T (V ),5 we will visit all Ct ’s that contain xi .
The time taken to find Si
is O(|Si |). The total time taken to find all Si ’s (once all
the BiV ’s are found) is O( i=1 |Si |). Define the indicator function χ(t, i) that takes
value 1 if xi ∈ Ct and zero otherwise. We have
5 Since
|Si | =
i=1 t∈T (V )
χ(t, i) =
t∈T (V ) i=1
χ(t, i) =
Xt .
the triangles in Si cover exactly the planar region of triangles incident to xi in T (V ∪{xi }).
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V(V )
GI (V )
Fig. 4.4. (a) V(V ) is dual to T (V ). Each vertex t of V(V ) corresponds to the center of
the circumcircle of a triangle t of T (V ), and it has the same conflict set Zt of size Xt . (b) The
geode triangulation GI (V ) is obtained by connecting the vertices of each region of V(V ) to the
lexicographically smallest incident vertex with the smallest Xt . The conflict set of a triangle s is
the union of the conflict sets of its vertices and point v defining the region.
Therefore, by Lemma 4.1,
|Si | = E
Xt =
E[Xt ] = O(n).
This implies that all Si ’s and Zt ’s can be found in expected linear time.
Our aim is to build the Delaunay triangulation T (V ∪ I) in linear time using
the conflict sets Zt . To that end, we will use divide-and-conquer to compute the
Voronoi diagram V(V ∪ I), using a scheme that has been used for nearest neighbor
searching [24] and for randomized convex hull constructions [20, 25]. It is well known
that the Voronoi diagram of a point set is dual to the Delaunay triangulation, and
that we can go from one to the other in linear time [12, Chapter 9]. Refer to Figure
4.4(a). Consider the Voronoi diagram of V , V(V ). By duality, the vertices of V(V )
correspond to the triangles in T (V ), and we identify the two. In particular, each
vertex t of V(V ) has a conflict set Zt , the conflict set for the corresponding triangle in
T (V ), and |Zt | = Xt , by our definition of Xt (see Figure 4.4(a)). We triangulate the
Voronoi diagram as follows: for each region r of V(V ), determine the lexicographically
smallest Voronoi vertex tr in r with minimum Xt . Add edges from all the Voronoi
vertices in r to tr . Since each region of V(V ) is convex, this yields a triangulation6 of
V(V ). We call it the geode triangulation of V(V ) with respect to I, GI (V ) [20, 24].
Refer to Figure 4.4(b). Clearly, GI (V ) can be computed in linear time. We extend
the notion of conflict set to the triangles in GI (V ): Let s be a triangle in GI (V ),
and let t1 , t2 , t3 be its incident Voronoi vertices. Then the conflict set of s, Zs , is
defined as Zs := Zt1 ∪ Zt2 ∪ Zt3 ∪ {v}, where v ∈ V is the point whose Voronoi region
contains the triangle s. In the following, for any two points x and y, |x − y| denotes
the Euclidean distance between them.
6 We need to be a bit careful when handling unbounded Voronoi regions: we pretend that there is
a Voronoi vertex p∞ at infinity which is the endpoint of all unbounded Voronoi edges, and when we
triangulate the unbounded region, we also add edges to p∞ . By our bounding triangle assumption,
there is no point in I outside the convex hull of V , and hence the conflict set of p∞ is empty.
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s v
B(v, y)
Fig. 4.5. The nearest neighbor of a point y ∈ s is either v or needs to be in the conflict set of
one of its vertices.
Claim 4.5. Let s be a triangle of GI (V ), and let Zs be its conflict set. Then the
Voronoi diagram of V ∪ I restricted to s, V(V ∪ I) ∩ s, is the same as the Voronoi
diagram of Zs restricted to s, V(Zs ) ∩ s.
Proof. Consider a point p in the triangle s, and let y be the nearest neighbor
of p in V ∪ I. If y ∈ V , then y has to be v, since s lies in the Voronoi region of v
with respect to V . Now suppose that y ∈ I. Let B(v, y) be the perpendicular bisector
of the line segment (v, y) (i.e., the line containing all points in the plane that have
equal distance from v and y). Refer to Figure 4.5. Let B + be the halfplane defined
by B(v, y) that contains y. Since B + intersects s, by convexity it also contains a
vertex of s, say, t1 . Because t1 and y are on the same side (B + ), |y − t1 | < |v − t1 |.
Note that Ct1 has center t1 and radius |v − t1 |, because t1 is a vertex of the Voronoi
region corresponding to v (in V(V )). Hence, y ∈ Zt1 . It follows that y ∈ Zs , so
V(V ∪ I) ∩ s = V(Zs ) ∩ s, as claimed.
Claim 4.5 implies that V(V ∪ I) can be found as follows: for each triangle s of
GI (V ), compute V(Zs ) ∩ s, the Voronoi diagram of Zs restricted to s. Then, traverse
the edges of GI (V ) and fuse the bisectors of the adjacent diagrams, yielding V(V ∪ I).
Lemma 4.6. Given V(V ), the Voronoi diagram V(V ∪ I) can be computed in
expected O(n) time.
Proof. The time to find V(Zs ) ∩ s for a triangle s in GI (V ) is O(|Zs | log |Zs |) =
O(|Zs |2 ) [12, Chapter 7]. For a region r of V(V ), let S(r) denote the set of triangles
of GI (V ) contained in r, and let E(r) denote the set of edges in V(V ) incident to r.
Recall that t
r denotes the common vertex of all triangles in S(r). The total running
time is O(E[ s∈GI (V ) |Zs |2 ]), which is proportional to
r∈V(V ) s∈S(r)
|Zs |2 ⎦ ≤ E ⎣
(1 + Xtr + Xt1 + Xt2 )2 ⎦
r∈V(V ) (t1 ,t2 )∈E(r)
≤ E⎣
(1 + 2Xt1 + Xt2 )2 ⎦ ,
r∈V(V ) (t1 ,t2 )∈E(r)
since Xtr ≤ min(Xt1 , Xt2 ). For e = (t1 , t2 ), letYe = 1 + 2Xt1 + Xt2 . Note that
E[Ye ] = O(1), by Lemma 4.1. We can write Ye = i (1/n + 2χ(t1, i) + χ(t2 , i)), where
χ(t, i) was the indicator random variable for the event that xi ∈ Ct . Hence, since
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1/n + 2χ(t1 , i) + χ(t2 , i) < 4, Claim 3.3 implies that E[Ye2 ] = O(1). Thus,
|Zs |2 ⎦ ≤
E[(Ye )2 ] =
s∈GI (V )
r∈V(V ) e∈E(r)
e=(t1 ,t2 )
r∈V(V ) e∈E(r)
e=(t1 ,t2 )
The number of edges in V(V )
is linear, and each edge e is incident to exactly two
Voronoi regions r. Therefore, E[ s∈GI (V ) |Zs |2 ] = O(n). Furthermore, assembling
the restricted diagrams takes time O E s∈GI (V ) |Zs | , and as |Zs | ≤ |Zs |2 , this is
also linear.
4.2. Running time analysis. In this section, we prove that the running time
bound in Lemma 4.2 is indeed optimal. As discussed at the beginning of section 4,
Claim 2.2 implies that any comparison-based algorithm for computing the Delaunay
triangulation of input I ∈R D needs at least H(T (I)) expected comparisons.
that by Lemma 4.2, the expected running time of our algorithm is O(n + i HiV ).
The following is the main theorem of this section.
Theorem 4.7. For HiV , the entropy of the triangle BiV of T (V ) containing xi ,
and H(T (I)), the entropy of the Delaunay triangulation of I, considered as a labeled
HiV = O(n + H(T (I))).
Proof. Let B V := (B1V , . . . , BnV ) be thevector of all the triangles that contain
the xi’s. By Claim 2.1, we have H(B V ) = i HiY . Now we apply Lemma 2.3 with
U = R2 , X = T (I), and Y . In Lemma 4.8 we will show that the function f :
(I, T (I)) → (B1V , . . . , BnV ) can be computed in linear time, so H(BiV ) = O(n+H(T (I)),
by Lemma 2.3. This proves the theorem.
We first define some notation. For a point set P ⊆ V ∪ I and p ∈ P , let ΓP (p)
denote the neighbors of p in T (P ). It remains to prove the following lemma.7
Lemma 4.8. Given I and T (I), for every xi in I we can compute the triangle
BiV in T (V ) that contains xi in total expected time O(n).
Proof. First, we compute T (V ∪ I) from T (V ) and T (I) in linear time [19, 36].
Thus, we now know T (V ∪I) and T (V ), and we want to find for every point xi ∈ I the
triangle BiV of T (V ) that contains it. For the moment, let us be a little less ambitious
and try to determine, for each xi ∈ I, a conflict triangle CiV in T (V ); i.e., CiV is a
triangle t with xi ∈ Zt . If x ∈ I and v ∈ V such that xv is an edge of T (V ∪ I), we
can find a conflict triangle for x in T (V ) in time O(n) by inspecting all the incident
triangles of v in T (V ). Actually, we can find conflict triangles for all neighbors of v
in T (V ∪ I) that lie in I by merging the two neighbor lists (see below). Noting that
on average the size of these lists will be constant, we could almost determine all the
CiV , except for one problem: there might be inputs x ∈ I that are not adjacent to any
v ∈ V in T (V ∪ I). Thus, we need to dynamically modify T (V ) to ensure that there
is always a neighbor present. Details follow.
Claim 4.9. Let p ∈ V ∪ I and write Vp := V ∪ {p}. Suppose that T (V ∪ I)
and T (Vp ) are known. Then, in total time O(|ΓV ∪I (p)| + |ΓVp (p)|), for every xi ∈
ΓV ∪I (p) \ Vp , we can compute a conflict triangle Ci p of xi in T (Vp ).
7 A similar lemma is used in [22] in the context of hereditary algorithms for three-dimensional
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Proof. Let xi ∈ ΓV ∪I (p) \ Vp , and let Ci p be the triangle of T (Vp ) incident to
p that is intersected by line segment pxi . We claim that Ci p is a conflict triangle
for xi . Indeed, since pxi is an edge of T (V ∪ I), by the characterization of Delaunay
edges (e.g., [12, Theorem 9.6(ii)]), there exists a circle C through p and xi which
does not contain any other points from V ∪ I. In particular, C does not contain any
other points from Vp ∪ {xi }. Hence pxi is also an edge of T (Vp ∪ {xi }), again by the
characterization of Delaunay edges applied in the other direction. Therefore, triangle
Ci p is destroyed when xi is inserted into T (V ∪ J) and is a conflict triangle for xi
in T (Vp ). It follows that the conflict triangles for ΓV ∪I (p) \ Vp can be computed by
merging the cyclically ordered lists ΓV ∪I (p) and ΓVp (p). This requires a number of
steps that is linear of the size of the two lists, as claimed.
For certain pairs of points p, xi , the previous claim provides a conflict triangle
Ci p . The next claim allows us to get CiV from this, which is what we wanted in the
first place.
Claim 4.10. Let xi ∈ I, and let p ∈ V ∪ I. Let Ci p be the conflict triangle for
xi in T (Vp ) incident to p, as determined in Step 2(c). Then we can find a conflict
triangle CiV for xi in T (V ) in constant time.
Proof. If p ∈ V , there is nothing to prove, so assume that p ∈ I. If Ci p has all
vertices in V , then it is also a triangle in T (V ), and we are trivially done. So assume
that one vertex of Ci p is p. Let e be the edge of Ci p not incident to p, and let v, w
be the endpoints of e. We will show that xi is in conflict with at least one of the two
triangles in T (V ) that are incident to e. Given e, such a triangle can clearly be found
in constant time. Refer to Figure 4.6 for a depiction of the following arguments.
Since v, w ∈ V , by the characterization of Delaunay edges, it follows that e is also
an edge of T (V ). If xi does not lie in Ci p , then xi must also be in conflict with the
other triangle t that is incident to e (since t is intersected by the Delaunay edge pxi ).
Note that t cannot have p as a vertex and is a triangle of T (V ).
Suppose xi lies in Ci p . Since Ci p is a triangle in T (Vp ), the interior has no points
other than xi . Thus, the segments vxi and wxi are edges of T (Vp ∪ {xi }). These must
also be edges of T (V ∪ {xi }). But this means that xi must conflict with the triangle
in T (V ) incident to e at the same side as Ci p .
Algorithm 1 Determining the conflict triangles.
1. Let Q be a queue containing the elements in V .
2. While Q = ∅:
(a) Let p be the next point in Q.
(b) If p = xi ∈ I, then insert p into T (V ) using the conflict triangle CiV for
xi , to obtain T (Vp ). If p ∈ V , then T (Vp ) = T (V ).
(c) Using Claim 4.9, for each unvisited neighbor xj ∈ ΓV ∪I (p) ∩ I, compute
a conflict triangle Cj p in T (Vp ).
(d) For each unvisited neighbor xj ∈ ΓV ∪I (p) ∩ I, using Cj p , compute a
conflict triangle CjV of xj in T (V ). Then insert xj into Q, and mark it
as visited.
The conflict triangles for all points in I can now be computed using breadth-first
search (see Algorithm 1). The loop in Step 2 maintains the invariant that for each
point xi ∈ Q ∩ I, a conflict triangle CiV in T (V ) is known. Step 2(b) is performed as in
the traditional randomized incremental construction of Delaunay triangulations [12,
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Ci p
Ci p
Fig. 4.6. (a) If xi is outside Ci p , it conflicts with the triangle t of T (V ) on the other side of e.
Ci p ,
it conflicts with the triangle t of T (V ) at the same side of e, since vxi
(b) If xi lies inside
and wxi are both edges of T (V ).
Chapter 9]: walk from CiV through the dual graph of T (V ) to determine the conflict
set Si of xi (as in the proof of Claim 4.4), insert new edges from all points incident
to the triangles in Si to xi , and remove all the old edges that are intersected by these
new edges. The properties of the conflict set ensure that this yields a valid Delaunay
triangulation. By Claim 4.10, Step 2(d) can be performed in constant time.
The loop in Step 2 is executed at most once for each p ∈ V ∪ I. It is also
executed at least once for each point, since T (V ∪ I) is connected and in Step 2(d) we
perform a breadth-first search. The insertion in Step 2(b) takes O(|ΓVxi (xi )|) time.
Furthermore, by Claim 4.9, the conflict triangles of p’s neighbors in T (V ∪ I) can be
computed in O(|ΓVp (p)| + |ΓV ∪I (p)|) time. Finally, as we argued above, Step 2(d)
can be carried out in total O(|ΓV ∪I (p)|) time. Now note that for xi ∈ I, |ΓVxi (xi )| is
proportional to |Si |, the number of triangles in T (V ) in conflict with xi . Hence, the
total expected running time is proportional to
|ΓVp (p)| + |ΓV ∪I (p)| ⎦
p∈V ∪I
= E⎣
|ΓV (v)| +
|Si | +
|ΓV ∪I (p)|⎦ = O(n).
p∈V ∪I
Finally, using breadth-first search as in the proof of Claim 4.4, given the conflict
triangles CiV , the triangles BiV that contain the xi ’s can be found in an O(n) expected
time, and the result follows.
4.3. The time-space tradeoff. We show how to remove the assumption that
we have prior knowledge of the Di ’s (to build the search structures Di ) and prove
the time-space tradeoff given in Theorem 1.3. These techniques are identical to those
used in section 3.2. For the sake of clarity, we give a detailed explanation for this
setting. Let ε ∈ (0, 1) be any constant. The first log n rounds of the learning phase
are used as in section 4.1.1 to construct the Delaunay triangulation T (V ). We first
build a standard search structure D over the triangles of T (V ) [12, Chapter 6]. Given
a point x, we can find the triangle of T (V ) that contains x in O(log n) time.
The learning phase takes M = cnε rounds for some large enough constant c. The
main trick is to observe that (up to constant factors), the only probabilities that are
relevant are those that are at least n−ε/3 . In each round, for each xi , we record
the triangle of T (V ) that xi falls into. Fix i, and for any triangle t of T (V ), let χt be
the number of times over the first M rounds that BiV = t. At the end of M rounds,
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we take the set Ri of triangles t with χt > 0. We remind the reader that p(t, i) is the
probability that xi lies in triangle t. The proof of the following lemma is identical to
the proof of Lemma 3.4.
Lemma 4.11. Fix i. With probability at least 1 − 1/n3 , for every triangle t of
T (V ), if p(t, i) > n−ε/3 , then M p(t, i)/2 < χt < 3M p(t, i)/2.
For every triangle t in Ri , we estimate p(t, i) as p̂(t, i) = χt /M , and we use p̂(t, i)
to build the approximate search structure Di . For this, we take the planar subdivision
Gi induced by the triangles in Ri , compute the convex hull of Gi , and triangulate the
remaining polygonal facets. Then we use the construction of Arya et al. [10] to build
an optimal planar point location structure Di for Gi according to the distribution
p̂i (the triangles of Gi not in Ri are assigned probability 0). This structure Gi has
the property that a point in a triangle t with probability p̂(t, i) can be located in
O(log(1/p̂(t, i))) steps [10, Theorems 1.1 and 1.2].
The limiting phase uses these structures to find BiV for every xi : given xi , we
use Di to search for it. If the search does not terminate in log n steps or Di fails to
/ Ri ), then we use the standard search structure, D, to find BiV .
find BiV (i.e., BiV ∈
Therefore, we are guaranteed to find BiV in O(log n) time. Clearly, each Di stores
O(M ) = O(nε ) triangles, so by the bounds given in [10], each Di can be constructed
with size O(nε ) in O(nε log n) time. Hence, the total space is bounded by n1+ε and
the time required to build all the Di ’s is O(n1+ε log n).
Now we just repeat the argument given in section 3.2. Instead of doing it through
words, we write down the expressions (for some variety). Let s(t, i) denote the time
to search for xi given that p(i, t) > n−ε/3 . By Lemma 4.11, we have χt > M n−ε/3 /2,
so t ∈ Ri , for c large enough, and thus s(t, i) = O(log(1/p̂(t, i))) = O(1 − log p(t, i)).
p(t, i)s(t, i) = O ⎝
p(t, i)(1 − log p(t, i))⎠
= O ⎝1 −
p(t, i) log p(t, i)⎠ .
We now bound the expected search time for xi :
p(t, i)s(t, i) =
p(t, i)s(t, i) +
p(t, i)s(t, i)
= O ⎝1 +
p(t, i) log n −
p(t, i) log p(t, i)⎠ .
Noting that for p(t, i) ≤ n−ε/3 , we have O(log n) = O(ε−1 log(1/p(t, i))), we get
p(t, i)s(t, i)
= O ⎝1 − ε−1
=O 1−ε
p(t, i) log p(t, i) −
p(t, i) log p(t, i)
p(t, i) log p(t, i)⎠
= O(1 + ε−1 HiV ).
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If follows that the total expected search time is O(n + ε−1 i HiV ). By the analysis of
section 4.1 and Theorem 4.7, we have that the expected running time in the limiting
phase is O(ε−1 (n + H(T (I)))). If the conditions in Lemmas 4.1 and 4.11 do not hold,
then the training phase fails. But this happens with probability at most 1/n. This
completes the proof of Theorem 1.3.
5. Conclusions and future work. Our overall approach has been to deduce
a “typical” instance for the distribution and then use the solution for the typical
instance to solve the current problem. This is a very appealing paradigm; even though
the actual distribution D could be extremely complicated, it suffices to learn just
one instance. It is very surprising that such a single instance exists for product
distributions. One possible way of dealing with more general distributions is to have
a small set of typical instances. It seems plausible that even with two typical instances,
we might be able to deal with some dependencies in the input.
We could imagine distributions that are very far from being generated by independent sources. Maybe we have a graph labeled with numbers, and the input is
generated by a random walk. Here, there is a large dependency between various components of the input. This might require a completely different approach than the
current one.
Currently, the problems we have focused upon already have O(n log n) time algorithms. So the best improvement in the running time we can hope for is a factor
of O(log n). The entropy optimality of our algorithms is extremely pleasing, but our
running times are always between O(n) and O(n log n). It would be very interesting
to get self-improving algorithms for problems where there is a much larger scope for
improvement. Ideally, we want a problem where the optimal (or even best known)
algorithms are far from linear. Geometric range searching seems to a good source of
such problems. We are given some set of points, and we want to build data structures
that answer various geometric queries about these points [2]. Suppose the points came
from some distribution. Can we speed up the construction of these structures?
A different approach to self-improving algorithms would be to change the input
model. We currently have a memoryless model, where each input is independently
drawn from a fixed distribution. We could have a Markov model, where the input
Ik depends (probabilistically) only on Ik−1 , or maybe on a small number of previous
Appendix. Constructing the ε-net V . Recall that λ = log n. Given a
set Iˆ of m := nλ points in the plane, we would like to construct a set V ⊆ Iˆ of size
ˆ > λ intersects V . (This is a (1/n)-net
O(n) such that any open disk C with |C ∩ I|
for disks.) We describe how to construct V in deterministic time n(log n)O(1) , using
a technique by Pyrga and Ray [43]. This is by no means the only way to obtain V .
Indeed, it is possible to use the older techniques of Clarkson and Varadarajan [26] to
get a another, randomized, construction with a better running time.
We set some notation. For a set of points S, a k-set of S is a subset of S of size k
obtained by intersecting S with an open disk. A (> k)-set is is such a subset with
size more than k. We give a small sketch of the construction. We take the collection
ˆ We need to obtain a small hitting set for Iˆ=λ . To do this, we
Iˆ=λ of all λ-sets of I.
trim Iˆ=λ to a collection of λ-sets that have small pairwise intersection. Within each
such set, we will choose an ε-net (for some ε). The union of these ε-nets will be our
final (1/n)-net. We now give the algorithmic construction of this set and argue that
it is a (1/n)-net. Then, we will show that it has size O(n).
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It is well known that the collection Iˆ=λ has O(mλ) sets [25, 37] and that an
explicit description of Iˆ=λ can be found in time O(mλ2 ) [3, 37], since Iˆ=λ corresponds
ˆ each of whose cells represents some λ-set
to the λth-order Voronoi diagram of I,
of I [37]. Let I ⊆ I=λ be a maximal subset of Iˆ=λ such that for any J1 , J2 ∈ I,
|J1 ∩ J2 | ≤ λ/100. We will show in Claim A.1 how to construct I inO(mλ5 ) time.
To construct V , take a (1/200)-net VJ for each J ∈ I, and set V := J∈I VJ .8 It is
well known that each VJ has constant size and can be found in time O(|J|) = O(λ)
ˆ if an open disk C intersects Iˆ
[20, p. 180, Proof I]. The set V is an (1/n)-net for I:
in more than λ-points, by the maximality of I, it must intersect a set J ∈ I in more
than λ/100 points. Now V contains a (1/200)-net for J (recall that |J| = λ), so V
must meet the disk C. We will argue in Claim A.2 that |V | = O(n). This completes
the proof.
Claim A.1. The set I can be constructed in time O(mλ5 ).
Proof. We use a simple greedy algorithm. For each J ∈ Iˆ=λ , construct the
collection J>λ/100 of all (> λ/100)-sets of J. The set J has size λ, and the total
number of disks defined by the points in J is at most λ3 . Thus, there are at most
λ3 sets in J>λ/100 , and they can all be found in O(λ4 ) time. Since there are at most
O(mλ) sets J (as we argued earlier), the total number of (> λ/100)-sets is O(mλ4 ),
a radix sort on the multiset
and they
can be obtained in O(mλ ) time. Next, perform
J := J∈Iˆ=λ J>λ/100 . This again takes time O(mλ5 ). Note that for any J1 , J2 ∈ Iˆ=λ ,
|J1 ∩ J2 | > λ/100 precisely if J1 and J2 share some (> λ/100)-set. Now I is obtained
as follows: pick a set J ∈ Iˆ=λ , put J into I, and use the sorted multiset J to find all
J ∈ Iˆ=λ that share a (> λ/100)-set with J. Discard those J from Iˆ=λ . Iterate until
Iˆ=λ is empty. The resulting set I has the desired properties.
Claim A.2. |V | = O(n).
Proof. The set V is the union of (1/200)-nets for each set J ∈ I. Since each
net has constant size, it suffices to prove that I has O(n) sets. This follows from a
charging argument due to Pyrga and Ray [43, Theorem 12]. They show [43, Lemma 7]
how to construct a graph GI = (I, EI ) on vertex set I with at most |EI | ≤ 24|I|
ˆ let Ip be the set of all J ∈ I that
edges with the following property: for p ∈ I,
contain p, and let Gp = (Ip , Ep ) be the induced subgraph on vertex set Ip . Then, for
all p, |Ep | ≥ |Ip |/4 − 1. Thus,
ˆ = m.
(|Ip |/4 − |Ep |) ≤ |I|
Consider the sum
exactly λ points, so each set
p∈Iˆ |Ip |. All sets in I contain
contributes λ to the sum. By double counting, p∈Iˆ |Ip |/4 = λ|I|/4. Furthermore,
an edge (J1 , J2 ) ∈ EI can appear in Ep only if p ∈ J1 ∩J2 , so again by double-counting,
|Ep | ≤ λ|EI |/100 ≤ 24λ|I|/100.
Hence, m ≥
p∈Iˆ(|Ip |/4
− |Ep |) ≥ λ|I|/100, and |I| = O(m/λ) = O(n).
8 That is, V is a subset of J such that any open disk that contains more than |J|/200 points
from J intersects VJ .
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