On the maximum size of an anti-chain of linearly separable sets and convex pseudo-discs

0.15 MB·English

by  Rom Pinchasi

On the maximum size of an anti-chain of linearly
separable sets and convex pseudo-discs1
Rom Pinchasi Gu¨nter Rote
∗ †
8
0 February 1, 2008
0
2

n
a
J Abstract

8
WeansweraquestionraisedbyWalterMorris,andindependentlyby

] AlonEfrat,aboutthemaximumcardinalityofananti-chaincomposedof
G
intersections of a givenset ofn points in the plane with half-planes. We
M approachthis problemby establishing the equivalence with the problem
ofthe maximummonotone pathin anarrangementofn lines. Arelated
.
h
problem on convex pseudo-discs is also discussed in the paper.
t
a
m
1 Introduction
[

2
Let P be a set of n points in the plane, no three of which are collinear. A
v
1 subset of P is called linearly separable if it is the intersection of P with a
1 closed half-plane. A k-set of P is a subsetof k points from P which is linearly
3
separable. LetA = A (P)denotethecollection ofallk-setsofP. Itisawell-
0 k k
. known open problem to determine f(k), the maximum possible cardinality of
7
0 Ak, where P varies over all possible sets of n points in general position in
7 the plane. The current records are f(k) = O(nk1/3) by Dey ([D98]) and
0 f(⌊n/2⌋) ≥ neΩ(√logn) by To´th ([T01]).
:
v Let A = A(P) = ∪n A be the family of all linearly separable subsets
i k=0 k
X of P. The family A is partially ordered by inclusion. Clearly, each A is an
k
r anti-chain in A. The following problem was raised by Walter Morris in 2003
a
in relation with the convex dimension of a point set (see [ES88]) and, as it
turns out, it was independently raised by Alon Efrat 10 years before, in 1993:
Problem 1. What is the maximum possible cardinality g(n) of an anti-chain
in the poset A, over all sets P with n points?
1ThisresearchwassupportedbyaGrantfromtheG.I.F.,theGerman-IsraeliFoundation
for Scientific Research and Development.
Mathematics Dept., Technion—Israel Instituteof Technology, Haifa 32000, Israel.
∗
[email protected].
†Institut fu¨r Informatik, Freie Universit¨at Berlin, Takustr. 9, 14195 Berlin, Germany.
[email protected]
1
In Section 2 we show that in fact g(n) can be very large, and in particular
much larger than f(n).
2 d
Theorem 1. g(n) = Ω(n −√logn), for some absolute constant d> 0.
Inanattempttoboundfromabovethefunctiong(n)onecanviewlinearly
separable sets as a special case of a slightly more general concept:
Definition 1. Let P be a set of n points in general position in the plane.
A Family F of subsets of P is called a family of convex pseudo-discs if the
following two conditions are satisfied:
1. Every set in F is the intersection of P with a convex set.
2. If A and B are two different sets in F, then both sets conv(A)\conv(B)
and conv(B)\conv(A) are connected (or empty).
One natural example for a family of convex pseudo-discs is the family
A(P), where P is a set of n points in general position in the plane. To
see this, observe that every linearly separable set is the intersection of P
with a convex set, namely, a half-plane. It is therefore left to verify that
if A = P ∩ H and B = P ∩ H , where H and H are two half-planes,
A B A B
then both conv(A)\conv(B) and conv(B)\conv(A) are connected. Let A =
′
A \ H = A \ B = A \ conv(B). Since conv(A) ∩ conv(B) = ∅, we have
B ′
conv(A)\conv(B)⊃ conv(A). For any x∈ conv(A)\conv(B), we claim that
′
there is a point a ∈ A such that the line segment [x,a] is fully contained in
′ ′ ′
conv(A)\conv(B). Thiswillclearlyshowthatconv(A)\conv(B)isconnected.
Let a ,a ,a be three points in A such that x is contained in the triangle
1 2 3
a a a . If each line segment [x,a ], for i = 1,2,3, contains a point of conv(B),
1 2 3 i
it follows that x∈ conv(B), contrary to our assumption. Thus there must be
a line segment [x,a ] that is contained in A = A\conv(B), and we are done.
i ′
In Section 3 we boundfrom above the maximum size of a family of convex
pseudo-discs of a set P of n points in the plane, assuming that this family of
subsets of P is by itself an anti-chain with respect to inclusion:
Theorem 2. Let F be a family of convex pseudo-discs of a set P of n points
in general position in the plane. If no member of F is contained in another,
then F consists of at most 4 n +1 members.
2
(cid:0) (cid:1)
Clearly, in view of Theorem 1, the result in Theorem 2 is nearly best
possible. We show by a simple construction that Theorem 2 is in fact tight,
apart from the constant multiplicative factor of n2.
2 Large anti-chains of linearly separable sets
Instead of considering Problem 1 directly, we will consider a related problem.
2
Definition 2. For a pair x,y of points and a pair ℓ ,ℓ of non-vertical lines,
1 2
we say that x,y strongly separate ℓ ,ℓ if x lies strictly above ℓ and strictly
1 2 1
below ℓ , and y lies strictly above ℓ and strictly below ℓ .
2 2 1
We will also take the dual viewpoint and say that ℓ ,ℓ strongly separate
1 2
x,y. (In fact, this relation is invariant under the standard point-line duality.)
If we have a set L of lines, we say that the point pair x,y is strongly
separated by L, if L contains two lines ℓ ,ℓ that strongly separate x,y.
1 2
A pair of lines ℓ ,ℓ is said to be strongly separated by a set P of points
1 2
if there are two points x,y ∈ P that strongly separate ℓ and ℓ .
1 2
Using the above terminology one can reduce Problem 1 to the following
problem:
Problem 2. Let P be a set of n points in the plane. What is the maximum
possible cardinality h(n) (taken over all possible sets P of n points) of a set
of lines L in the plane such that for every two lines ℓ ,ℓ ∈ L, P strongly
1 2
separates ℓ and ℓ .
1 2
Figure 1: Problem 2.
To see the equivalence of Problem 1 and Problem 2, let P be a set of n
points and L be a set of h(n) lines that answer Problem 2. We can assume
that none of the points lie on a line of L. Then with each of the lines ℓ ∈ L
we associate the subset of P which is the intersection of P with the half-plane
belowℓ. Wethusobtainh(n)subsetsofP each ofwhichisalinearlyseparable
subset of P. Because of the condition on L and P, none of these linearly
separable sets may contain another. Therefore we obtain h(n) elements from
A(P) that form an anti-chain, hence g(n) ≥ h(n).
Conversely,assumewehaveananti-chainofsizeg(n)inA(P)forasetP of
npoints. Each linearly separablesetis theintersection ofP withahalf-plane,
3
which is bounded by some line ℓ. We can assume without loss of generality
thatnoneof theselines isvertical, andatleast half ofthehalf-spaces lie below
their bounding lines. These lines form a set L of at least ⌈g(n)/2⌉ lines, and
each pair of lines is separated by two points from the n-point set P. Thus,
h(n) ≥ ⌈g(n)/2⌉.
Before reducing Problem 2 to another problem, we need the following
simple lemma.
Lemma 1. Let ℓ ,...,ℓ be n non-vertical lines arranged in increasing order
1 n
of slopes. Let P be a set of points. Assume that for every 1 ≤ i < n, P
strongly separates ℓ and ℓ . Then for every 1 ≤ i < j ≤ n, P strongly
i i+1
separates ℓ and ℓ .
i j
Proof. We prove the lemma by induction on j − i. For j = i + 1 there is
nothing to prove. Assume j −i ≥ 2. We first show the existence of a point
x∈ P that lies above ℓ and below ℓ . Let B denote the intersection point of
i j
ℓ and ℓ . Let r denote the ray whose apex is B, included in ℓ , and points
i j i i
to the right. Similarly, let r denote the ray whose apex is B, included in ℓ ,
j j
and points to the right.
Since the slope of ℓ is between the slope of ℓ and the slope of ℓ , ℓ
i+1 i j i+1
must intersect either r or r (or both, in case it goes through B).
i j
Case 1. ℓ intersects r . Then there is a point x∈ P that lies above ℓ and
i+1 i i
below ℓ . This point x must also lie below ℓ .
i+1 j
Case 2. ℓ intersects r . Then,bytheinductionhypothesis,thereisapoint
i+1 j
x∈ P that lies above ℓ and below ℓ . This point x must also lie above ℓ .
i+1 j i
Theexistenceofapointy thatlies aboveℓ andbelow ℓ issymmetric.
j i
By Lemma 1, Problem 2 is equivalent to following problem.
Problem 3. What is the maximum cardinality h(n) of a collection of lines
L = {ℓ ,...,ℓ } in the plane, indexed so that the slope of ℓ is smaller than
1 h(n) i
the slope of ℓ whenever i <j, such that there exists a set P of n points that
j
strongly separates ℓ and ℓ , for every 1 ≤ i< h(n)?
i i+1
We will consider the dual problem of Problem 3:
Problem 4. What is the maximum cardinality h(n) of a set of points P =
{p ,...,p } in the plane, indexed so that the x-coordinate of p is smaller
1 h(n) i
than the x-coordinate of p , whenever i< j, such that there exists a set L of
j
n lines that strongly separates p and p , for every 1 ≤ i< h(n)?
i+1 i
We will relate Problem 4 to another well-known problem: the question of
the longest monotone path in an arrangement of lines.
Consider an x-monotone path in a line arrangement in the plane. The
length of such a path is the number of different line segments that constitute
the path, assuming that consecutive line segments on the path belong to
different lines in the arrangement. (In other words, if the path passes through
a vertex of the arrangement without making a turn, this does not count as a
new edge.)
4
Problem 5. What is the maximum possible length λ(n) of an x-monotone
path in an arrangement of n lines?
A construction of [BRSSS04] gives a simple line arrangement in the plane
which consists of n lines and which contains an x-monotone path of length
2 d
Ω(n −√logn) for some absolute constant d > 0. No upper bound that is
asymptotically better than the trivial bound of O(n2) is known.
Problem 5 is closely related to Problem 4, and hence also to the other
problems:
Proposition 1.
λ(n)+1
h(n) ≥ , (1)
(cid:24) 2 (cid:25)
λ(n) ≥ h(n)−2 (2)
Proof. We first prove h(n) ≥ ⌈(λ(n)+1)/2⌉. Let L be a simple arrangement
of n lines that admits an x-monotone path of length m = λ(n). Denote by
x ,x ,...,x the vertices of a monotone path arranged in increasing order of
0 1 m
x-coordinates. In this notation x ,...,x are vertices of the line arrange-
1 m 1
−
ment L, while x and x are chosen arbitrarily on the correspondingtwo rays
0 m
which constitute the first and last edges, respectively, of the path. For each
1 ≤ i < m let s denote the line that contains the segment x x , and let r
i i 1 i i
−
denote the line through the segment x x .
i i+1
For 1 ≤ i < m, we say that the path bends downward at the vertex x
i
if the slope of s is greater than the slope of r , and it bends upward if the
i i
slope of s is smaller than the slope of r . Without loss of generality we may
i i
assume that at least half of the vertices x ,...,x of the monotone path
1 m 1
−
are downward bends.
p1 xi2 p2 xi4
xi3
xi1
p0 xik
pk
p3
Figure 2: Constructing a solution for Problem 4.
5
Let i < i < ··· < i be all indices such that x is a downward bend,
1 2 k ij
where k ≥ (m−1)/2. Observe that for every 1 ≤ j < k, the monotone path
between x and x is an upward-bending convex polygonal path.
ij ij+1
We will now definek+1 points p ,p ,...,p such that for every 0 ≤ j < k
0 1 k
the x-coordinate of p is smaller than the x-coordinate of p , and the line
j j+1
r lies above p and below p while the line s lies below p and above
ij j j+1 ij j
λ(n)+1
p . This construction will thus show that h(n) ≥ ⌈ ⌉.
j+1 2
For every 1 ≤ j ≤ k let U and W denote the left and respectively the
j j
right wedges delimited by r and s . That is, U is the set of all points that
ij ij j
lie below r and above s . Similarly, W is the set of all points that lie above
ij ij j
r and below s .
ij ij
Claim 1. For every 1 ≤ j < k, W and U have a nonempty intersection.
j j+1
Proof. We consider two possible cases:
Case 1. i = i +1. In this case r = s . Therefore any point above the
j+1 j ij ij+1
line segment [x x ] that is close enough to that segment lies both below
ij ij+1
s and below r and hence W ∩U 6= ∅.
ij ij+1 j j+1
Case 2. i − i > 1. In this case, as we observed earlier, the monotone
j+1 j
path between x and x is a convex polygonal path. Therefore, r and
ij ij+1 ij
s are different lines that meet at a point B whose x-coordinate is between
ij+1
the x-coordinates of x and x . Any point that lies vertically above B and
ij ij+1
close enough to B belongs to both W and U .
j j+1
Now it is very easy to construct p ,p ,...,p , see Figure 2. Simply take
0 1 k
p to be any point in U , and for every 1 ≤ j < k let p be any point in
0 1 j
W ∩U . Finally, let p be any point in W . It follows from the definition
j j+1 k k
of U ,...,U and W ,...,W that for every 0 ≤ j < k, r lies above p and
1 k 1 k ij+1 j
below p and the line s lies below p and above p .
j+1 ij+1 j j+1
We now prove the opposite direction: λ(n)≥ h(n)−2.
Assume we are given h(n) points p ,...,p sorted by x-coordinate and
1 h(n)
a set of n lines L such that every pair p ,p is strongly separated by L.
i i+1
By perturbing the lines if necessary, we can assume that none of the lines
goes through a point, and no three lines are concurrent. For 1 < i < h(n),
let f be the face of the arrangement that contains p , and let A and B be,
i i i i
respectively, the left-most and right-most vertex in this face. (The faces f
i
are bounded, and therefore A and B are well-defined.) The monotone path
i i
will follow the upper boundary of each face f from A to B .
i i i
We have to show that we can connect B to A by a monotone path.
i i+1
This follows from the separation property of L. Let s ,r be a pair of lines
i i
that strongly separates p and p in such a way that r lies above p and
i i+1 i i
below p and s lies below p and above p . Since B lies on the boundary
i+1 i i i+1 i
of the face f that contains p , B lies also between r and s , including the
i i i i i
possibility of lying on these lines. We can thus walk on the arrangement from
B to the right until we hit r or s , and from there we proceed straight to the
i i i
intersection pointQ ofr ands . Similarly, thereisapathinthearrangement
i i i
6
from A to the left that reaches Q . and these two paths together link B
i+1 i i
with A .
i+1
To count the number of edges of this path, we claim that there must be at
least one bend between B and A (including the boundary points B and
i i+1 i
A ). If there is no bend at Q , the path must go straight through Q , say,
i+1 i i
on r . But then the path must leave r at some point when going to the right:
i i
if the path has not left r by the time it reaches A and A lies on r ,
i i+1 i+1 i
then the path must bend upward at this point, since it proceeds on the upper
boundary of the face f that lies above r .
i+1 i
Thus, the path makes at least h(n)−3 bends (between B and A , for
i i+1
1 < i< h(n)−1) and contains at least h(n)−2 edges.
Now it is very easy to give a lower bound for g(n), and prove Theorem 1.
λ(n)+1 2 d
Indeed, this follows because g(n) ≥ h(n) and h(n) ≥ ⌈ ⌉ =Ω(n −√logn),
2
The close relation between Problems 1 and 5 comes probably as no big
surprise if one considers the close connection between k-sets and levels in
arrangements of lines (see [E87, Section 3.2]). For a given set of n points
P, the k-sets are in one-to-one correspondence with the faces of the dual
arrangements of lines which have k lines passing below them and n−k lines
passing above them (or vice versa). The lower boundaries of these cells form
thek-thlevelinthearrangement,andtheupperboundariesformthe(k+1)-st
level.
OurchainofequivalencefromProblem1toProblem5extendsthisrelation
between k-sets and levels in a way that is not entirely trivial: for example, es-
tablishingthatwegetsetsthatformanantichain requiressomework,whereas
for k-sets this property is fulfilled automatically.
3 Proof of Theorem 2
The heart of our argument uses a linear algebra approach first applied by
Tverberg[T82]inhiselegantproofforatheoremofGrahamandPollak[GP72]
on decomposition of the complete graph into bipartite graphs.
Let F be a collection of convex pseudo-discs of a set P of n points in
general position in the plane. We wish to bound from above the size of F
−→
assuming that no set in F contains another. For every directed line L = xy
passing through two points x and y in P we denote by L the collection of
x
all sets A ∈ F that lie in the closed half-plane to the left of L such that L
touches conv(A) at the point x only. Similarly, let L be the collection of all
y
sets A ∈ F that lie inthe closed half-planeto theleft of Lsuch that L touches
conv(A) at the point y only. Finally, let L be those sets A ∈ F that lie in
xy
the closed half-plane to the left of L such that L supportsconv(A) at the edge
xy.
Definition 3. LetAandB betwo setsinF. LetLbeadirected linethrough
two points x and y in P. We say that L is a common tangent of the first kind
7
L
B
A
B A
L
L
Figure 3: The two cases of common tangents in Lemma 2
with respect the pair (A,B) if A ∈ L and B ∈ L .
x y
We say that L is a common tangent of the second kind with respect to
(A,B) if A ∈ L and B ∈ L , or if A∈ L and B ∈ L .
xy y x xy
The crucial observation about any two sets A and B in F is stated in the
following lemma.
Lemma 2. Let A and B be two sets in F. Then exactly one of the following
conditions is true.
1. There is precisely one common tangent of the first kind with respect to
(A,B) and no common tangent of the second kind with respect to (A,B),
or
2. there is no common tangent of the first kind with respect to (A,B), and
there are precisely two common tangents of the second kind with respect
(A,B).
Proof. The idea is that because A and B are two pseudo-discs and none of
conv(A) and conv(B) contains the other, then as we roll a tangent around
C = conv(A∪B), there is precisely one transition between A and B, and this
is where the situation described in the lemma occurs (see Figure 3).
Formally, by our assumption on F, none of A and B contains the other.
Any directed line L that is a common tangent of the first or second kind with
respect to A and B must be a line supporting conv(A∪B) at an edge.
Letx ,...,x denotethevertices ofC = conv(A∪B)arrangedincoun-
0 k 1
−
terclockwise order on the boundary of C. In what follows, arithmetic on
indices is done modulo k.
There must be an index i such that x ∈ A \ B, for otherwise every x
i i
belongs to B and therefore conv(B) = conv(A∪B) ⊃ conv(A) and therefore
B ⊃ A (because both A and B are intersections of P with convex sets) in
contrast to our assumption. Similarly, there must be an index i such that
x ∈ B \A.
i
8
Let I be the set of all indices i such that x ∈ A\B, and let I be the
A i B
set of all indices i such that x ∈ B \A.
i
We claim that I (and similarly I ) is a set of consecutive indices. To
A B
see this, assume to the contrary that there are indices i,j,i,j arranged in a
′ ′
cyclic order modulo k such that x ,x ∈ A\B and x ,x ∈ B. Then it is
i i j j
′ ′
easy to see that conv(A)\conv(B) is not a connected set because x and x
i i
′
are in different connected components of this set.
We have therefore two disjoint intervals I = {i ,i + 1,...,j } and
A A A A
I = {i ,i +1,...,j }. It is possible that i = j or i = j .
B B B B A A B B
Observethat x ,x ,x ,x arearranged in this counterclockwise cyclic
iA jA iB jB
order on the boundary of C, and for every index i ∈/ I ∪ I , x ∈ A∩B.
A B i
The only candidates for common tangents of the first kind or of the second
−−−−→
kind with respect to A and B are of the form x x , that is, they must pass
i i+1
through two consecutive vertices of C.
We distinguish two possible cases:
1. i = j + 1. In this case the line through x and x is the only
B A jA iB
common tangent of the first kind with respect to (A,B) and there are
no common tangents of the second kind with respect to (A,B).
2. i 6= j +1. In this case, there is no common tangent of the first kind
B A
with respect to (A,B). The line through x and x and the line
iB 1 iB
−
through x and x are the only common tangents of the second
jA jA+1
kind with respect to (A,B).
This completes the proof of the lemma.
Let A ,...,A be all the sets in F, and for every 1 ≤ i ≤ m let z be an
1 m i
−→
indeterminate associated with A . For each directed line L = xy, define the
i
following polynomial P :
L
P (z ,...,z ) =
L 1 m
1 1
z z + z z + z z
i j i j i j
(cid:18) (cid:19)(cid:18) (cid:19) 2(cid:18) (cid:19)(cid:18) (cid:19) 2(cid:18) (cid:19)(cid:18) (cid:19)
AXi Lx AXj Ly AXi Lx AjXLxy AXi Ly AjXLxy
∈ ∈ ∈ ∈ ∈ ∈
This polynomial contains a term z z whenever L is a tangent line for the
u v
pair (A ,A ) or for the pair (A ,A ) (of the first or of the second kind, and
u v v u
with coefficient 1 or 1, accordingly). If we sum this equation over all directed
2
lines L, it follows by Lemma 2 that every term z z with u6= v appears with
u v
coefficient 2:
P (z ,...,z ) = 2z z = (z +···+z )2−(z2+···+z2 ) (3)
L 1 m u v 1 m 1 m
XL uX<v
Considerthesystemof linear equations z = 0and z = 0,
−→ Ai∈Lx i Ai∈Ly i
where L = xy varies over all directed linePs determined by PP. Add to this
9
system the equation z +···+z = 0. There are 4 n +1 equations in this
1 m 2
system andif m > 4 n +1, theremustbeanontrivia(cid:0)l s(cid:1)olution. However, it is
2
easily seen that a no(cid:0)nt(cid:1)rivial solution (z ,...,z ) will result in a contradiction
1 m
to (3). This is because the left-hand side of (3) vanishes, while the right-hand
side equals −(z2+···+z2 )6= 0. We conclude that |F| = m ≤ 4 n +1.
1 m 2
(cid:0) (cid:1)
We now show by a simple construction that Theorem 2 is tight apart
from the multiplicative constant factor of n2. Fix three rays r ,r , and r
1 2 3
emanatingfromtheoriginsuchthattheanglebetweentworaysis120degrees.
For each i = 1,2,3, let pi,...,pi be n points on r , indexed according to their
1 n i
increasing distance from the origin. Slightly perturb the points to get a set P
of 3n points in general position in the plane. For every 1≤ j,k,l ≤ n define
F = {p1,...,p1}∪{p2,...,p2}∪{p3,...,p3}.
jkl 1 j 1 k 1 l
It can easily be checked that the collection of all F such that 1 ≤ j,k,l ≤ n
jkl
and j + k + l = n + 2 is an anti-chain of convex pseudo-discs of P. This
collection consists of n+1 sets.
2
(cid:0) (cid:1)
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10