On the expected number of ksets
Discrete Comput Geom
On the Expected Number of kSets 0 1
0 2Department of Computer Science,Rutgers University , Piscataway; NJ 08903 , USA
1 1Mathematical Institute, The Hungarian Academyof Sciences , Pf. 127, H1364 Budapest , Hungary
Given a set S of n points in Ra, a subset X of size d is called a ksimplex if the hyperplane aft(X) has exactly k points on one side. We study Ed(k, n), the expected number of ksimplices when S is a random sample of n points from a probability distribution P on Rd. When P is spherically symmetric we prove that Ea(k, n) < cn a ~. When P is uniform on a convex body K c R z we prove that E2(k , n) is asymptotically linear in the range cn < k < n/2 and when k is constant it is asymptotically the expected number of vertices on the convex hull of S. Finally, we construct a distribution P on Rz for which E2((n  2)/2, n) is cn log n.
1. Introduction and Summary
I. Bar~inyand W. Steiger
possess? It is easy to translate an upper bound for ed(k, n) into an upper bound
on ksets.
Clearly, O(nd) provides a trivial upper bound for ea(k, n). When d = 2, nontrivial
bounds were obtained by Lov~sz [15] for halving sets (n even, k = n/2), and later,
for general k < n/2, by E r d r s et al. [12]. A simple construction gives a set S with
n log k ksets, while a counting argument shows that e2(k, n ) = O(nx/k ). These
bounds were rediscovered several times, for example by Edelsbrunner and Welzl
[11], but had not been improved until Pach et al. [17] reduced the bound to
nx/k/log* k. Papers [1], [13], and [22] contain results related to the study of
e2(k, n).
Raimund Seidel (see [10]) extended the Lov~tsz lower bound construction to
d = 3 and showed that e3(k, n) = ~(nk log(k + 1)). The argument m a y be applied
inductively giving ed(k, n) = f~(nk d 2 log(k)).
A nontrivial upper bound for d = 3 was recently obtained by Bgr~my et al. [5].
They showed that e3(n/2, n) = n 3*, where e > 0 is some small constant. This, in
turn, was improved by Aronov et al. [2] to O(n 8/3 log 5/a n). Dey and Edelsbrunner
[9] have been able to remove the logarithmic factors from this bound. Recently, a
nontrivial upper bound for d > 3 was established via a result of Zivaljevi6 and
Vrerica [23]. They proved a colored version of Tverberg's theorem which now
implies that O(nde") is an upper bound for halving sets in R d, ed > 0 being a small
constant depending on d.
It appears likely that the truth is near the lower bound. Support comes from
the fact that in "typical" cases there are relatively few ksets. In this paper we
study Ed(k, n), the expected number of ksimplices when X is a sample of n r a n d o m
points from a probability measure P on R d. When there is no confusion we write
E(k, n). The following derivation gives an expression for Ed(k, n) that we use
throughout. Pick d points x 1. . . . . xd independently, according to P. Write I for the
hyperplane aft(x1. . . . . xd). We assume throughout that P vanishes on every
hyperplane so I is well defined with probability one. (In particular, P is nonatomic.)
Write l § and l  for the open halfspaces on the right and left of l, respectively,
and set F(/) = min(P(l§ P(l)), the probability content cut off by l. The random
variable F(/) has a distribution function
G(t) = P(F(I) < t)
(
1
)
(
2
)
which determines Ed(k, n) in the following way. Given a sample X = {x 1. . . . . xn}
from P, the expected number of ksimplices is
Ed(k, n) =
~
l <ij<"'<id<_n
Prob[there are k points on one side of aff(xi,, .... xid)]
=
k
[tk(1  t)~ak + (1  t)kt"ak] dG(t).
O u r first result is a simple one a b o u t spherically symmetric distributions (the
definition is given in Section 2).
Theorem 1. For a spherically symmetric distribution we have Ed(k, n) < c l n d  1, Cl
bein9 a constant depending only on d.
Next we deal with the case where P is the uniform distribution on a compact,
convex b o d y K ~ R 2. We assume that Area(K) = 1 so that P coincides with the
restriction of Lebesgue measure to K. Define v: K ~ R by
v(x) = inf{Area(K c~ H): x E H, H is a halfplane}
(
3
)
and A(t) = Ar(t ) = Area{x e K : v(x) < t}. The properties of v and A(t) have been
studied in [3], [6], and [21]. Here we prove that G is differentiable and that G'(t)
is essentially equal to A(t). This helps establish the following theorem.
Theorem 2. There are absolute constants c2 and ca such that, for the uniform
distribution over any convex set in the plane,
for every sufficiently large n and every k = O, 1. . . . . I(n  2)f2].
Sometimes we express the relation in (4) as
(
4
)
(
5
)
Remark 1. Since t <_A(t), we have c4 < A(t) < 1 when t > c4 > 0. T h e o r e m 2
then shows that when 89_> k/n > c,,
The behavior of A(t) (for small t) is given by Theorem 7 of [6]
/ k + l'x
E 2 ( k , n ) ~ n A ~ ) "
E2(k, n) ~ n.
1
cst log  <_ A(t) < c6 t2/3.
t
Schiitt and Werner [21] show that for a function f ( t ) with c s t l o g ( 1 / t ) <
f(t) < c6t 2/3 (and some additional properties) there is a convex set K = K f ~ R 2
of area 1 such that At(t) ~ f(t). This shows that not only does
n f k + 1~ 2/3
cs(k + l) lOg k + l _( E2(k, n) _( c6nk%)
I. B/trhny and W. Steiger
hold for P uniform on a convex body, but also, for (almost) any function
between these bounds, there is a convex body K with E2(k, n) behaving like that
function.
The special case k = 0 is interesting. Then Ez(k, n) equals the expected number
of edges of conv(X), which was known to behave like A(1/n) (see [6]). So Theorem
2 says that Ez(k, n) behaves like the expected number of edges of conv(X) when
k is a constant, and like n when k/n > t o.
Finally, we give an example of a distribution for which E2(k, n) is large. We
consider the case k = (n  2)/2 (n even), that is, the expected number of halving
segments. We give a distribution Pn such that
)
whenever the sample size m is within a constant factor of n. Then, using Pn, we
describe a distribution P for which
E / n  2
z \ 2
)
, n > c8n log n.
Finally, we point out the abstract of [7], where one of the present results
was announced, but with an erroneous proof. This is one of the reasons we
take some care in establishing the simple statements about Ea. The methods
are familiar in geometric probability and integral geometry (see [4], [161
and [19]). Nevertheless, the results seem to be the first ones concerning Ea
and in view of the fact that ksets have applications in computational
geometry and machine learning [14], [181, we feel that these theorems are useful and
interesting.
2. Spherically Symmetric Continuous Distributions
Suppose that P has a density function g : R d ~ R that only depends on Ixl,
the distance from x e R d to the origin. We say that such a P is spherically
symmetric. This defines another function f : R§ ~ R by f ( r ) = g(Ixl) when
r = l x l .
P r o o f o f Theorem 1. Set xd_ 1 = vold x(Sd 1). Clearly,
1 =
g(x) dx =
f ( r ) r ~ 1 dr du = tc~_ 1
f(r)r d 1 dr.
N o w let H(t) be an open halfspace with probability content t, 0 < t _< 89 and write
p = p(t) for the distance (from the origin) to H, the bounding hyperplane of H(t).
Then
t =
where r is the length of the c o m p o n e n t of x parallel to u and y = x  ur, u denoting
the unit n o r m a l to H.
Claim 1. G(t + At)  G(t) < caAt.
T h e o r e m 1 follows immediately because, from (2),
Ea(k, n) <
k
[tk(1  t)nak + (1  t)ktndk]c 9 dt
~ Ctond 1;
the last inequality is a consequence of the wellknown fact that
(m + l ) (
7
) f ~ tJ(l  t)~J dt = l.
Proof o f Claim 1.
which says that
We use the B l a s c h k e  P e t k a n t s c h i n formula (see p. 201 of [20])
d x l " " d x a = d! vold_ t(conv{yl . . . . . Yd}) dYl""dYa du dr,
where the points x 1. . . . . x d lie in the hyperplane ux = r with u ~ Sd 1, the unit
sphere in R d and r > 0, and Yi = xi  ru. With this fact,
G ( t + A t )  G ( t ) = f . . . f ~
g ( x t ) . . . g ( x d ) d x l . . . d x a
<F(I)<t+At
= f f [ i ( , + , o ; . ~ s d  ~ ; , ' " ; y ~
f ( x / r 2 + . y ' j 2 ) ' ' ' f ( x / r 2 + ' y a ] 2 ,
x d! vola t(conv{yl . . . . . Ya}) d y l " " d y a du dr
= d! x._ 1 c ' " ' f ."f" .
dp(t+At) di
dI
x vola_ l(conv{yl . . . . . Ya}) d Y t ' " d Y a dr.
f ( x / r 2 + l y l l E ) ' " f ( x / r 2 + ,yal z)
(
7
)
(
8
)
(
9
)
but
Notice that the innermost d integrals here denote the expectation of the
volume of conv{yl . . . . . Yd} when the points Yl . . . . . Yd are distributed on the
hyperplane H = {x: u x = r} according to density f ( x / ~ + [y[2). This is, again,
a spherically symmetric distribution in the hyperplane H with center ru which we
take for the origin of H and denote by 13. The signed volume of conv{yl . . . . . y~}
is
1 d o t C
( d  1)!
,o
...
I. Bfir/myand W. Steiger
det ( ; 1  . .
1 ) = det (10
"'" Ye
1 ... 1 ) + "" + det ( ~
Y2 "'" Ye
~) .
Consequently, with unsigned volumes,
Since every term on the righthand side has the same expectation,
d
Moreover,
by Hadamard's inequality. This way we get
1 1 d  I
vol(conv{O, Yl . . . . . Yd1})  (d  1)! Idet(yl . . . . . Ydl)l < (d  1)! i1=1I lYJ
G(t + At)  G(t) <_ dBrd_l
f ( x / r z + ly, IB)lY~l dy,
Jr=p(t+At) L i = I
x fy~ f ( x / r 2 + ty~l2) dy~ dr.
dI
f ( x / r 2 + y2)ly, I d y i = f
ES d2
We need s o m e further n o t a t i o n . G i v e n tp e [0, 2n] a n d t e (
0, 1
) there is a unique
directed line l(q~, t) with direction q~ t h a t has F(/) = t. l(q~, t) is clearly c o n t i n u o u s in
By (7)
and therefore
f r "u)
= p(t + At)
f R
d
f ( x / r 2 + lyal2) dyd dr = A t
G(t + A t )  G(t) < dEra_ 21ax a/\~2}
\ x a  x /
'At.
U n i f o r m Distribution on a Convex Set
Let K c R 2 be a c o n v e x set with Area(K) = 1. W e are interested in E2(k, n) w h e n
P is the Lebesgue m e a s u r e restricted to K. Since E 2 is i n v a r i a n t under
(nondegenerate) affine t r a n s f o r m a t i o n s o f K we m a y a s s u m e t h a t K is in " n o r m a l
position," i.e., t h a t
r B 2 ~ K c 2rB 2,
w h e r e B 2 is the unit disk, centered at the origin, a n d r is a universal c o n s t a n t (in
fact r = 33/4, b u t we do not need this precision). T h e existence of the " n o r m a l
p o s i t i o n " follows f r o m t h a t of the L 6 w n e r  J o h n ellipsoid [8].
It is m o r e c o n v e n i e n t to w o r k with the directed version of (
2
). So let l = x~
d e n o t e the line directed f r o m x to y. Write F(/) for the p r o b a b i l i t y content o f
the halfplane l + on the right o f l; this is equal to the a r e a of K c~ l § Set G(t) =
P r o b [ F ( / ) < t]. T h e n (
2
) b e c o m e s
= l s
f;
[ ]
(10)
I. B~rfiny and W. Steiger
Theorem 3.
G(t) is differentiable w h e n t ~ (
0, 1
) and
Theorem 4.
A s t ~ O,
W e m e n t i o n h e r e t h a t G ( t ) ~ tA(t) is p r o v e d in [3]. T h e o r e m s 3 a n d 4
e s t a b l i s h a different a n d a p p a r e n t l y m o r e s u b t l e p r o p e r t y o f t h e f u n c t i o n G. W e
n e e d :
L e m m a 1.
F o r each t ~ (
0, 1
) there is a constant Ct such that
]~0(tp, t)  ~(r
u)] < Ctlt  u]
f o r all tp e [0, 2 h i and u E [0, 1]. I n fact, Ct = 8r/min(t, 1  t).
T h e p r o o f is s t r a i g h t f o r w a r d u s i n g t h e n o r m a l p o s i t i o n a n d t h e f o l l o w i n g e a s y
facts (refer t o Fig. 1):
1. T h e c h o r d f u n c t i o n p ~ q,(tp, t(~o, p)) is c o n c a v e in p.
2. F o r all s ~ (
0, 1
), 4r(p(tp, 0)  p(tp, s)) > s a n d 4r(p(tp, s)  p(tp, 1)) > 1  s.
3. q4~o, sXp(~0, t)  p(~o, u)) < 2(u  t) if s = u o r s = t.
W e o m i t t h e details.
G(t + At)  G(t) = P r o b [ F ( / ) 6 [t, t + At)]
<F(~yI<t+At
=
Jp(cp,t+At)
IX  f~l dye d~ dp dq~.
<~eKnl
A n e l e m e n t a r y c o m p u t a t i o n reveals that
ff~
<y~Knl
I~  .~1 d~ d)~ = ~Z3(/),
where X(/) = ~k(~o,t(tp, p)) is the length o f the c h o r d K n I. S o
G(t + At)  G(t) = ~ f ~
Cp(q~~
Jp(~,t+At)
~/3((p, t((~, p)) dp d e
= ~
f ~lt
02(0, t)
f p(q~t.)
d p(r t + At)
+ ~ So~nfp(r t)
dp(tp,t+At)
[ 0 2 ( 0 , t(tp, p))  ~b2(0, t)]O(cp, t(r p)) dp dq~.
Therefore The first term here equals ~ S2" ~02(~0,t) d~o At since trivially (again, see Fig. 1)
At =
f p(O,t)
q/(q~, t(tp, p)) dp.
(
11
)
G(t + At)  G(t)
At

< 6At
1 ~'2,

6 J o
J pt~o,t+At)
1 ; ~ " f"~"')
< 6At Jp(~.,+ao
8rTz
3
C, At,
J~Oz(q~, t(q~, p))  ~2(q~, t)l~(q~, t(q2, p)) dp dq2
8rC, At@(~o, t(~o, p))dp dq~
where we used L e m m a 1 in the last inequality and (
11
) in the last equality.
[]
R e m a r k 2.
We point out that, for 89> t > t o > 0,
cll < G'(t) < c12.
(
12
)
The u p p e r b o u n d is trivial from T h e o r e m 3 because ~Ois bounded. F o r the lower
b o u n d it is enough to see that ~k(q~,t) > c13t. This follows easily from the normal
position of K.
Before the p r o o f of T h e o r e m 4 we need some preparation. The b o d y
K(v ~_ t) = { x ~ K : v(x) ~_ t}
is clearly convex. We assume t < to < 0.01, say, and then K(v > t) is n o n e m p t y as
well. T h u s the b o u n d a r y of K(v > t) is a convex curve V(t) with left and right
tangents at every z e V(t). These tangents coincide at all but countably m a n y
z ~ v(t).
Fix t ~ (0, to]. Given q~~ [0, 2n) let ).(q~,t) be the unique directed line (with
direction ~0)that is a supporting line to K(v > t) and has K(v > t) on its left. 2(q~, t)
has exactly one point (to be denoted by z(q~, t)) in c o m m o n with K(v > t) since, as is
proved in [3], V(t) contains no line segment. Call the angle ~0 reoular if ).(~0,t)
is tangent (left, right, or both) to the curve V(t) at z(q~, t). Write R for the
set of regular angles in [0, 2r0 a n d N R for its complement. It is not difficult
to see that R is a closed set. Therefore N R is a countable union of open intervals;
the point in the p r o o f o f T h e o r e m 4 is that the total length of these intervals
is O(t).
Recall t h a t l(tp, t) is a directed line t h a t cuts off a r e a t f r o m K. It follows f r o m
the p r o o f of L e m m a G in [3] that if ~o is regular, then 2(q~, t) a n d l(q~, t) coincide
and z(tp, t) is the m i d p o i n t of the c h o r d K n l(tp, t). Finally, let L(to, t) be the length
of the segment connecting z(q~, t) to the last p o i n t o n 2(q~,t) in K. O b s e r v e that,
for a regular angle, L(tp, t) = 89 t).
W e omit the simple p r o o f of the following.
C l a i m 2.
Area(K(v > t)) = 8912~ L2(cp, t) d~o.
L e m m a 2.
The total length o f the intervals in N R is O(t).
P r o o f A s s u m e t h a t (p is n o n r e g u l a r and let tp + a n d ~o be the direction of the
left a n d right t a n g e n t s l + and l  to V(t) at z(qo, t). Since q~+(q~) are regular, z(tp, t)
is the m i d p o i n t of the c o r r e s p o n d i n g c h o r d s which we d e n o t e b y u§ + a n d u v ,
as is s h o w n in Fig. 2. T h e n u  u + a n d v  v § s p a n parallel lines. Let S be the strip
between them. Clearly,
Area(rB2\S) <_ A r e a ( K \ S ) _< 2t.
An e l e m e n t a r y c o m p u t a t i o n reveals t h a t the width of S is at least 2 r
(t2/(2r)) 1/3 > 0.8, so
m i n ( l u   v  I , lu +  v+l)_> 0.8.
(
13
)
M o r e o v e r ,
1 = Area(K) < 2t + 89diam2(K)(n  (~0+  q~)),
V(t)
K
tl't"
UJ
u
Z
u
Fig. 2. Tangents at a nonregular point z = z(~0,t) on V(t).
b e c a u s e b o t h lines 1+ a n d 1 cut off a c a p f r o m K of a r e a t, a n d the r e m a i n d e r is
c o n t a i n e d in a c i r c u l a r s e c t o r with c e n t e r z(~p, t), r a d i u s e q u a l to d i a m ( K ) < 4r,
a n d a n g l e n  (tp §  tp). C o n s e q u e n t l y (see Fig. 2),
P r o o f ( s e e Fig. 3). If w I = w 2, t h e n this follows f r o m (
14
). O t h e r w i s e , let w be the
i n t e r s e c t i o n of the t a n g e n t s at w I a n d w 2. O b s e r v e t h a t the a n g l e w l w w 2 is at least
n/2 (since ~b2  ~'1 < rr/2), so
Iw  w2[ < Iwl  w21 < Izi  zi+ l[ < 0.20.
On the Expected N u m b e r of kSets
V(t)
[ ]
t >__A r e a ( c o n v { u l , u 2 , v l } ) = 89  u2l lul  vl[ sin(ff 2  ~kl)
w h e r e w e u s e d (
13
) as well.
P r o o f o f Theorem 4.
F o r a r e g u l a r direction, L(~o, t) = 89
t). T h e n
I f all d i r e c t i o n s a r e r e g u l a r , w e a r e finished. O t h e r w i s e
If:
= ~ 4L2(q~, t) dfp
"~R
1
(~/2((p, t)  4L2(qh t)) dq~
= ~A(t) + ~ 1
JN R
(~,2(tp, t)  4LZ(~p, t)) dcp.
IG'(t) 1 ~A(t)l < ~ f
JN R
I~J2(cp, t)  4LZ(cp, t)l dcp
m e a s ( N R ) < Cl4t.
I. Bfirfinyand W. Steiger
According to [6], A(t) > ct log(I/t) for s o m e absolute c o n s t a n t c, so we get
=
G'(t)
~ a ( t ) ( l +
O
T h e o r e m 2 follows easily f r o m T h e o r e m s 3 and 4 using s o m e properties of A(t),
namely:
It follows easily f r o m T h e o r e m 4 a n d the properties of G'(t) and A(t) that
f ~ tk(1 t)'~kG'(t) dt ~ f ~ tk(1 t)"kA(t) dt.
Therefore it is e n o u g h to show that, for all k = 1. . . . . [.m/2],
Write I(m, k) for the expression on the right, l(m, k) would decrease if we only
integrated on the interval [k/m, 1], and, since A(t) is increasing, it would decrease
further if we replaced A(t) by A(k/m). This shows that
I(m, k) > A (:)(m + 1)
/ra
tk(1  t)mk dt > c2A (:) 9
F o r the last inequality it should be p r o v e d that (m + 1)
tk(1  t)mk dt >_ c2 > 0
/m
for all k = 1. . . . . / m / 2 J a n d for all large e n o u g h m. This can be d o n e as follows.
The integrand is maximal at t = k/m and decreases on [k/m, 1]. So, for any
T e [k/m, 1],
f: tk(1 i t ) ~ I ~ dt > (T: )
Ira
Tk(1  T)mk.
Choosing T = (k + x/%)/m gives a good lower bound for the integral. We omit
the technical computations.
For the other inequality we observe that A(t) <_A(k/ra) when t < k/m. From
property 2 above,
ti tm'~ 2 li k \
when t > k/m. This gives
l(m,k) <_ (m +
and this is less than c3A(k/m) by (
9
).
3.1.
Higher Dimensions
; /,,
_< A ( k ) (m + 1)( m k ) [ ; tk(1  t)~k dt + c~5 ~mS
2 t~f+2(1/  ]t)"* dr
We mention a possible generalization to the case d > 2. In this case define
G(t) = P[V(x 1. . . . . xa) < t],
where F ( x l , . . . , xa) is the probability content of the halfspace on the righthand
side of aft(x1 . . . . . xa). Here x~...... xa are independent random points from P (on
Ra). Formula (2) is replaced by its directed version:
Let P be the uniform distribution on a convex body K c R d. Define v and A(t)
as in (
3
). It is proved in [3] that G(t) ~ ta ~A(t) for any convex body K ~ R a but
what we need here is the behavior of the derivative of G. This does not seem to
be easy to establish and we could only settle the case when K is smooth (say c~3)
[]
(
15
)
and so
with the G a u s s  K r o n e c k e r curvature b o u n d e d away from zero and infinity. In
this case we can prove
G'(t) ~ ta 2A(t)
E d ( k , n ) ~ ( : ) ( n k d ) f 2 t k + a  2 ( 1   t ) n  a  k A ( t ) d t .
It is k n o w n that, for a ~3 convex b o d y K , A(t) ,,. t 2/ta+~ which gives
Ea(k, n) .,~ k a 2 +2/(d + 1)n 1  2/(d + 1)
in view of (
9
). This shows, again, that Ea(k, n) behaves like the expected n u m b e r
of facets (or vertices, edges, etc.) of the r a n d o m p o l y t o p e inscribed in K when k is
constant and like n a  ~ when k > cn. This is p r o b a b l y true for all convex bodies
K c R a, not only for the cr ones.
4. A Distribution with Many Halving Lines
Erd6s et al. [12] exhibited a set T~ of nl = 3" 2i points which has at least cni log nl
halving segments. We use this example to construct distributions P for which
E2((n  2)/2, n) > c s n log n. First we review the example of [12] and point out
some new features that are needed for the analysis.
The example is sequential. At step i = 1 there are nl = 6 points; three are
vertices of an equilateral triangle and three are on rays from the center through
these vertices, as in Fig. 4(a). Clearly, there are h 1 = 6 halving segments. To
form T2,
each point u e T1 splits into two close points I/1, u2 which are positioned so they
define a halving line, as in Fig. 4(b). In addition each pair u, v that defined a
halving line in T1 now defines two halving lines, as shown in Fig. 4(b) (see also
Fig. 5). This gives nz = 3" 2 2 = 12 points with h2 = 18.
In general, T / h a s ni = 3" 2i points. It is shown in [12] that each point u ~ Ti
may be replaced by two close points ux, u2 which can be positioned so that:
1. u l u 2 is a halving segment in Ti+ 1.
2. If uv was a halving segment in T~, two new halving lines are formed from
ul, u2, vl, v2 (see Fig. 5).
If hi denotes the n u m b e r of halving segments in T~, properties 1 and 2, respectively,
show that
hi+ 1 = rli q 2hl,
hi = 6,
a recurrence with solution h i = 3" T  1(i + 1).
To describe o u r construction we need to k n o w f~(j), the n u m b e r of jsegments
in T, j = 0, 1. . . . . n J 2  1. We have used hi for f { n J 2  1) and we write h f =
f~(ni/2  2) for the n u m b e r of segments that are onelessthanhalving. F r o m Fig. 5,
if uv was a jsegment in T~, then the four segments u l v l , u~v2, u 2 v l , u2v2 form two
(2j + 1)segments and a 2jsegment and a (2j + 2)segment in Ti+ 1. However, when
j = nJ 2  1, the 2jsegment and the (2j + 2)segment are both
onelessthanhalving.,Tberefore fi+ 1(
0
) = f,{0) and
f~+ ~(2j) = f~(J) + f,(j  1),
ni
j = 1 , . . . , 2
2,
f~+ 1(2j + 1) = 2f,(j),
j = O, 1. . . . . n2l
I. Bhrhny and W. Steiger
f~+,(2j) = 2f~(j) + L ( J  1),
fi + ,(2j + 1) = 2f~(j) + n i,
E q u a t i o n (
19
) is the recurrence for hl, while (
18
) gives
ni
j = ~  I,
2
a n d linear b e t w e e n the points j/ni. E v a l u a t i n g (
19
) for j = nd2  1 shows t h a t
9~(89 = hJnl = (i + 1)/2 a n d for j = nl/2  2, t h a t g~(89 1/n3 = hi/nl > i  1.
F r o m (
17
) a n d (
18
), f o r j < nJ2  2, gi+l((J + 1)/n3 = 9i((J + 1)/nl) a n d this implies
91+,((J + 1)/n3 = 91((J + 1)/n3. Therefore, for t < ti = 89 1/nl, gZ+k(t) = 9i(t), by
the linearity o f gl. T h e s e relations allow the c o m p u t a t i o n o f all values o f f~(j).
W e n o w m a k e T~ into a set Si of positive a r e a b y replacing each p o i n t x e T~
b y the disk centered at x with r a d i u s ei, which m a y be chosen small e n o u g h so
t h a t the disks are in general position (no three s t a b b e d b y a line). It is n o t surprising
t h a t :
L e m m a 3. I f P i is the uniform distribution on Si, then EE((n  2)/2, n) = fl(n l o g n)
as long as an < nl < bn, f o r f i x e d 0 < a < b < oo.
P r o o f
W e have, a c c o r d i n g to (
2
),
E(n22~2 'n)=(2n)(n/n2 21)fo/22[t(1t)]"/21dG(t)
> 2 c16
_>
2
~
j ,/5,/,/~, 2"
nlJ
<__c l a n 3 / 2 e 2("In') f
d 1/2  I/x/'~
l/2
dG(t)
= c l s n ' / 2 e  2 t " / ' ) [ G ( 8 9
Now let x and y be two points chosen independently and randomly according
to P~. Write D for the event that x, y are not in the same disk; clearly, Prob[Dl =
1  1/nl. We have
1
x / ~ ) = P r o b [ F ( x y ) e [ ~
'
;]1
>_ Prob[F(xy) e [12
~fni'
D Prob[D]
1
.,/~ 1 f/(j)
> (ni  1) 2
j=n,(1/21lye) gi\ n l/ "
Note that ti/2 = )  1~rail2 > 89 1 / ~ / a n d that there are 2i  k  1 values o f j such
that tk < (j + 1)/ni < tk+ l. Therefore,
a( 89 a(89 1
2 ~ , ~ k~+'*
) >   Z
ni k=i/2 j=[k
( j + l ]
g, 
\ n i /I

2 ~~
Z g,( O
~i k=i/2
>
  ni k=i/2
( k  1)2ik1 >

Combined with (21),
and so E 2 >_ cTn log n.
n  2
E2 ~  ,
)
n >_q8n3/2e Z("/"')cl9
lognl
[]
On the other hand, it is straightforward to show that, as n * o% the expected
number of halving segments for n points chosen from P~ is O(n). The argument is
a simple calculation like the one in (
8
) using the fact that dG is bounded as n
increases.
Next we show that there is a single distribution for which E 2 grows at a
superlinear rate. Assume that a sequence w,, ~ 0 is given. We construct an
absolutely continuous distribution P for which Ez((n  2)/2, n) > C7Wnrtlog n for
any n. We use the same sequence of sets T~and system of disks St as before with the
nesting condition that Si = Si+ 1. This can be achieved if, in each step, the radii
of the disks are small enough.
Define P by requiring that P(Si) = ms with every disk in St having probability
I. Bfir/my and W. Steiger
c o n t e n t mJn~, i = 1, 2. . . . ; mi is specified later. Clearly, m 1 = 1 must h o l d a n d as
S~ ~ S~+1 we have ms > m~+~. If m~ > m~+l we define P, restricted to SI\Si+I, to
be u n i f o r m on S~\S~§ P is a p r o b a b i l i t y m e a s u r e for every sequence 1 =
m 1 > m 2 > .' of positive numbers.
A r g u i n g as in (21) we see that
E (n   2
) >_
EG( 89  G (11  ~)n ]n
As in the p r o o f of L e m m a 3 we let x, y d e n o t e a r a n d o m p a i r of p o i n t s d i s t r i b u t e d
a c c o r d i n g to P. Define i by requiring nl < n < ni + r Let Di d e n o t e the event that
b o t h x a n d y are in Si b u t b e l o n g to different disks of Si. Clearly,
P r o b [ D / ] = m
.
T h e previous c o m p u t a t i o n applies n o w in the following way:
If we c h o o s e ml = 1 for all i, then P is a p r o b a b i l i t y d i s t r i b u t i o n , with s u p p o r t
N Si a n d having E 2 ,~ n log n. This d i s t r i b u t i o n is c o n c e n t r a t e d in a small set. If
we c h o o s e a d e c r e a s i n g sequence m~ slowly t e n d i n g to zero, then P is an a b s o l u t e l y
c o n t i n u o u s m e a s u r e a n d E 2 > m2n log n.
Acknowledgments
T h e a u t h o r s are grateful to G i i n t e r R o t e who p o i n t e d out an e r r o r in an earlier
version, a n d m a d e m a n y useful suggestions for the c u r r e n t one. A referee m a d e
suggestions for i m p r o v i n g the p r e s e n t a t i o n .
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