1.6
The Clarkson-Shor Averaging Technique
The Crossing-Lemma technique can be simplified for certain cases. For example, let P be a
set of n points in R2 . Construct the graph Gk (P ) on P by adding the edge (pi , pj ) to Gk iff
there exists a disk containing both pi , pj , and at most k other points of P . Such a graph is
called a k-Delaunay graph.
Claim: For any P , G0 (P ) is a planar graph.
An edge (pi , pj ) is in G0 iff there exists a disk containing only the points pi and pj . Draw
these edges as straight-line edges in the plane. Now no two edges can cross in this embedding,
as then their two disks’s boundary circles would intersect four times, an unusual thing for
circles.
Theorem 13 (Edges in Gk ). For any P , the graph Gk (P ) has at most O(nk) edges.
The basic reason is the following: assume Gk has ‘lots’ of edges. Then one can carefully
select a subset P 0 ⊆ P such that the graph G0 (P 0 ) has ‘lots’ of edges, a contradiction to the
earlier claim that it is planar, and so has at most 3|P 0 | edges.
And we will ‘carefully select’ by randomly picking a subset P 0 . Add each point p ∈ P to
P 0 with probability p. The number of edges in G0 (P 0 ) is now a random variable, and lets
double-count its expectation. On one hand:
E[|G0 (P 0 )|] ≤ 3E[|P 0 |] = 3np
Now, lets look at the probability that an edge (pi , pj ) of Gk (P ) ends up as an edge of G0 (P 0 ).
For that to happen, we have to pick both pi , pj and not pick any of the at-most k vertices
contained in the disk of pi , pj . This is at least p2 · (1 − p)k , and so is the expected value of
whether (pi , pj ) is present in G0 (P 0 ). Summing up over all m edges, we have:
E[|G0 (P 0 )|] ≥
X
p2 · (1 − p)k = m · p2 · (1 − p)k
e∈Gk (P )
Putting the two bounds together, and setting p = 1/2k, one gets the required bound.
Two flavors. We can again implement the above idea both combinatorially as well as
inductively. Combinatorially, count, over all subsets P 0 of P of size t, the number of edges
in G0 (P 0 ). The upper-bound is trivial:
X
P 0 ⊂P,
|P 0 |=t
n
|G0 (P )| ≤
· 3t
t
0
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For the lower-bound, count, for each edge (pi , pj ), the number of subsets P 0 in which it
appears in the graph G0 (P 0 ):
X
P 0 ⊂P,
|P 0 |=t
|G0 (P )| =
0
X
(pi ,pj )
n−2−k
| {P s.t. (pi , pj ) ∈ G0 (P )} | ≥ m ·
t−2
0
0
Setting t = n/2k and solving the upper- and lower-bounds gives the required result.
Disks. Here’s another example. Let P be a set of n points in R2 . A disk D is spanned by
P iff it has three points of P on its boundary. There are at most O(n2 ) such disks, say the
set D. Our goal is to upper-bound the number of disks in D containing at most k points of
P in their interior. As usual, the claim about empty disks:
Claim: There are at most 3n disks in D that contain 0 points of P .
This again follows from a planarity-based argument.
Now, add each point of P with probability p to our sample S, and count the number of
disks of D that end up as empty disks in S. The upper-bound, following from the above
Claim, is 3np. The lower-bound is derived similar to before: each disk D ∈ D is present as
an empty disk in S iff the three points defining D are picked into S, and none of the (at
most k) points contained in D are picked. The probability of this happening for a fixed D
is at least p3 · (1 − p)k . Summing up over all D, and from the upper-bound, we have
|{D ∈ D s.t. D contains at most k points of P }| · p3 (1 − p)k ≤ 3np
Setting p = 1/2k gives the bound of O(nk 2 ).
Rectangles. Consider a set P of n points in R2 , all lying above the x-axis. Let R be the
set of rectangles which are ‘anchored’ on the x-axis (i.e., their bottom edge is lying on the
x-axis), and that contain a point of P on each of their three other edges. Our goal is to
prove that the number of rectangles in R that contain at most k points of P are O(nk 2 ).
First the claim about the ‘zero’-th level rectangles:
Claim: There are at most n rectangles in R that contain 0 points of P .
Each point p ∈ P can have at most one such rectangle containing p on its upper edge.
As each rectangle in R contains some point of P on its upper edge, there can be only n
rectangles.
The rest of the argument is word-for-word same as above; I cut-and-paste for your (and my)
comfort. Now, add each point of P with probability p to our sample S, and count the number
of rectangles of R that end up as empty rectangles in S. The upper-bound, following from
the above Claim, is np. The lower-bound is derived similar to before: each rectangle R ∈ R
is present as an empty rectangle in S iff the three points defining R are picked into S, and
none of the (at most k) points contained in R are picked. The probability of this happening
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for a fixed R is at most p3 · (1 − p)k . Summing up over all R, and from the upper-bound, we
have
|{R ∈ R s.t. R contains at most k points of P }| · p3 (1 − p)k ≤ np
Setting p = 1/2k gives the bound of O(nk 2 ).
Final Remarks.
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