-I_-
_.--.--
-
~_._-11-
_ .-._-_--.-.__. .._--
Geometric graphs with few disjoint edges
Pave1 Vdtr+
G&a T&W
Abstract
tains b + 1 pairwise disjoint edges. By a result of
Hopf and Pannwitz [HP] and Erdiis [El, cl(n) ,I n.
Alon and ErdZs [AE] showed that ez(n) 5 6n which
was improved by Goddard, Katchalski and Kleitman
[GKK] to ez(n) 5 3n. The best known lower bound,
ez(n) 2 2.5n - 4, is due to Perles (see [PA]). In [GKK]
it was also shown that 3.5n - 6 < es(n) 5 10n and that
> 4). For any fixed h, Path
ek(n) 2 w(log n) k-3 (k _
and T&Gcsik [PT] were first to prove that ek(n) in linear in n; their bound was ek(n) < L”n. On the other
hand, eI;(n) I: bn for n 2 2X:follows from a result of
Kupitz [Kl]. In this paper we further improve both the
upper and lower bounds for general k.
A geonuh=ic graph. is a graph drawn in the plane so that
bhe vertices are represented by points in general posit.ion, the edgesare represented by straight line segments
connect,ing the corresponding points.
Improving a result of Path and TM&k,
we show
that a geometric graph on n vertices with no X:+ 1 pairwise dis-ioint edgeshas at most b3(n+ 1) edges. On the
other hand, we construct geometric graphs with n verbites and approximately $(X:- 1)n edges,containing no
X:+ 1 pairwise disjoint edges.
We also give an 0(m2) algorithm for finding at le&
(m/(n + 1))1/3 p airwise disjoint edges in a geometric
graph with n vertices and m edges (m > n).
1
Theorem
Introduction
1 For n 2 215,
$k - l)n - 2b2 < ek(n) 2 k3(n + 1).
A geometric graph G is a graph drawn in the plane by
(possibly crossing) straight line segments, i.e., it is defined as a pair G = (V, E), where V is a set of points
in general position in t-heplane and E is a set of closed
segments whose endpoints belong to V.
The following question was raised by Avital and
Hanani [AH], Kupitz [Kl], Erdiis and Perles. Determine the smallest number ex-(n) such that any geometric graph with n vertices and m > er;(n) edges con-
In any geometric graph G, one can find a set of diajoint edges,at least as large as guaranteed by the upper
bound in Theorem 1, in 0(m2) time, where m in the
number of edges of G.
Theorem 2 There is an O(m2) time algorithm for
pairwise disjoint edges in a
finding at least ($)lf3
geometric graph with n vertices and m edges (m 3 n).
‘Courant Inst,itute, NYU and DIMACS Center, Rutgers
University. E-m&
gezaQdimacs .rutgers. edu. Supported
by OTKA-T-020914,
OTKA-F-22234 and NSF grant CCR-
For k = 3, the above mentioned bounds can be
slightly improved.
94-24395.
‘Charles University, Prague and DIMACS Center, Rutgers University.
E-mail: valtrQkam .mff . cuni. cz. Supported by t,he Czech Republic grant GACR 0194 and Charles
University grants GAUK 193, 194.
Theorem
3 For any n L. 6,
4n - 9 2 es(n) < 8.5n.
bmision tomakedi&d or hard copiesof all or partof& MQ~for
Or C~~OOm
USC k. @~~tcd witbout fee providrd t& copies
a~ not madeor distributedfor profit or commercialadva&ge andad
coPic3bearthis noticeandthe full citation on the fi page.
TOcopy
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to lists,
rCipik?sprior sppecific
pcdion andfora fee.
323 98 hhneapolis Minnesota USA
Cqv-ight Khf 1998O-8979I-973498l6...%5.00
Px‘Jnal
Theorems 1, 2 and 3 are proved in Sections 2, 3 and
4, respectively. Throughout the paper, we do not make
any notational distinction between an edge and the segment representing it.
184
e 41 e’:
\ e’
e 42 e’:
//
01
e’
e
'
e -t3 e’:
/
,'
h
25
/
e 44 e’:
\
e
Figure 1: The relations +.
2
2.1
The general case
The upper
bound
Our proof, as the proof of Path and T&8&k
bsscd on Dilworth’s theorem [D].
Dilworth’s
[PT], is
Theorem.
Let P be a partially ordewd
set containing no chain (totally ordered subset) of size
k -I- 1. Then P can be covered by k antichains (subsets
of pairurioe incomparable elements).
z(v2) < z(vi),
and e lies below e’,
e -t3 e’, if
z(vl) c z(vi),
z(v2) > z(vi),
and e lies below e’,
e 4’4 e’, if
Z(Q) > z(vi),
z(v2) < z(vi),
and e lies below e’.
Obviously, each of the relations -+ is a partial ordering, and any pair of disjoint edges in G is comparable
by at least one of them.
Since G does not contain k+ 1 disjoint edges,(E, -Q)
does not contain a chain of length k + 1. Therefore, by
Dihvorth’s theorem, E can be covered by k antichains
with respect to 31. Let El be the largest of these antichains, thus ]I&] 1 [El/k. Applying Dilworth’s theorem on (El, <s), we similarly get an antichain (with
respect to +) EZ C El of size lEzl> /311/k 2 (E(/k2.
In the rest of the proof we estimate the size of E2 from
above.
Since E2 is an antichain with respect to -Q and
42, n(e) f~ x(e)) # 0 for any e, e’ E E2. Therefore,
eEE2n(e) # 0, so there is a vertical line e which intersects addges in23
Let Gs = (V, E2) be a directed geometric graph obtained from (V, E2) by replacing each edge e = vlve in
Let G = (V, E) be. a geometric graph on n vertices,
containing no k -t 1 pairwise disjoint edges. For a vertex v, let z(v) and y(v) denote its z- and y-coordinate,
rcspcctivcly. We can assume without loss of generality
that no two vertices have the same z-coordinate.
An edge e is said to lie below an edge e’, if no vertical
lint crossing both e and e’ crossese strictly above e’.
Finally, Iet n(e) denote the orthogonal projection of e
to the z-axis.
Define four binary relations -ii (i = 1,. . .,4) on
the edge set E as follows (see also [PT, PA]). Let
e = vrv2,e’ = vivi be two disjoint edges of G, where
Z(Q) < m(v2) and z(vi) < z(v$ Then (see Fig. 1)
e 41 e’, if
I
< z(vi),
e +, e’, if
z(vl) > z(v~), z(v2) > z(vi),
n
and e lies below e’,
185
all 5 maximal zigzag paths in &
Figure 2: h/Iaximal zigzag paths.
e+
W-1
ei
ei
%<
Vi+1
'vi+2
ei+2
Z
ei+2
2ei+2
t>
ei+2
Figure 3: ei+2 + ei or ei+2 + e;.
E2 by the two oriented edges K$ and va.
For two
edges el = .uo.ui,ez = VA forming a path in g, we
say that ez is a zag of el, if the following two conditions
hold:
Lemma
5 G has at most n+ 1 mazimal zigzag p&s.
Lemmas 4 and 5 immediately give Theorem 1: every edge in ?$ is contained in a maximal zigzag path;
therefore, by Lemmas 4 and 5,
(i) r(el) n r(ea) has positive length,
ISI 2 2qn + l),
(ii) for any z E (n(el) II x(e2)) \ (?r(vl)}, the vertical
line through z intersects e2 below el, and it intersecbsno other edge going from 211between el and
and, consequently,
IE[ 2 k2[E21 = k213/2
_< k3(n + 1).
e2.
It remains to prove Lemmas 4 and 5.
Observe that each edge in z has at most one zag. We
call an oriented path Ble2 . . .e, in s a zigzag path, if
ei+l is the zag of ei, for each i = 1,. . . , r-l (seeFig. 2).
Lemma
edges.
4 Every zigzag path in z
Proof of Lemma 4.
Let P = 6162. . .e, be a zigzag
pathin&,andletei=viifori=l,...,T.
We
need the following claim.
Claim One of the two sequences
has at most 2X:
186
04
e2ktl
Figure 4: el, es,. . . , e2k+l are disjoint.
is decreasing, and the other one is increasing.
Proof of Claim.
Suppose the claim is false. Then
there is an index i such that either z(v(-1) < z(vi+l)
and ~(210< m(vt+z), or Z(VU~-1)> z(v~+I) and z(v~) >
Z(Vur+z). Consequently, t?i+z + ei or ei+z -$ ej (see
Fig. 3 for all four possible cases), which is a contradiction with the definition of E2. The claim is proved. 0
The length of P is at most 2k, otherwise el, es,. . . , ezE+l
would bc k + 1 disjoint edges according to Claim (see
Fig. 4). Cl
Proof of Lemma 6.
Let v E V be a vertex lying
to the rig2 of 1, and let 9, P2 be two different zigzag
paths in Gz ending in v. If the slope of the last edge in
PI is, say, smaller than the slope of the last edge in P2,
then the last edge of PI has a zag and, consequently, PI
can bc extended to a longer zigzag path. It follows that
at most one maximal zigzag path ends in v, and this is
similarly true for any vertex not lying on 1. Similarly,
at most two zigzag paths end in any vertex v on 4 (one
coming to v from the left, the other one from the right).
The lemma now follows from the assumption that no
pair of vertices lies on a vertical line. 0
2.2
The segments of El intersect the line r in k + 1
. . . , Tk/z such that it always lies
f-k/2$
r-k/2+1,
on the segments piqj with i + j - (Z + 1) = t. Let
R be the set of the centers of the segments rtrt+l
(t = -k/2, -k/2 + 1,. . ., k/2 - 2), and let EZ be the
set of all edges (segments) joining vertices of R with
vertices of P U Q (see Fig. 5).
We now show that the geometric graph G = (P U Q U
R, B1 U E2) gives the lower bound in Theorem 1.
First, observe that G has z + z + (k - 1) = n vertices
and
Points
IElI +
=
lE2l
(n-k+l)(k+l)-4(1+2$...+k/2)
2
+(n -k
+ l)(k - 1)
edges. It remains to show that G contains no k-b1 pairwise disjoint edges. Suppose that D is a set of disjoint
edges in G. If ID n El1 5 1, then IDI 5 lRl+ 1 = k.
Otherwise, let the leftmost edge of D n El, el, intersects f in a point r8, and the rightmost one, ez, in a
point Tt. Since there are s + k/2 vertices of R to the
left of el, D contains at most s + k/2 edges to the left
of el. Similarly, D contains at most k/2 - t edges to
the right of es. Since there are t - s - 2 vertices of
P U Q between el and ez, D contains at most t - s - 2
edges between el and es. Altogether, D contains at
most (s+k/2)+(k/2-t)+(t-s-2)+2
= k edges. 0
The lower bound
l?or simplicity, suppose that k is even and n is odd.
Set z = (n - k $1)/2. Let P be a set of z points
~1, . . , , p8 placed equidistantly in thii order from left to
right on a horizontal line p. Let Q = (~1,. . . , qr} be a
translation of P such that qi always corresponds to pi
and plpzqzql is a square. Let q be the line containing Q,
and let r bc the line parallel top and q, halfway between
them, Let El be the set of all segments(edges)piqj with
-k/2 ,< i 3-j - (x + 1) < k/2 (see Fig. 5).
187
Pl
P2
43
44
45
q6 = !?a
P3
P4
P5
P6 =Pz
Pl
P2
P3
P4
P6
P6 = Pz
E2
El
Figure 5: The edges of 231 and ,332(for k = 4, n = 15, z = 6).
3
Algorithm
113
are done, in the latter casewe proceed as follows below.
This step can be done again in O(m2) + O(m) = O(mz)
time.
Following the proof of Theorem 1, we build the oriented geometric graph 6 = (V, 3) from .Ez in O(m)
time. Further, we order the edgesfrom each vertex in
clockwise order (say) and consequently determine the
zag of each edge in z. This can be done in O(mlogm)
time. Then, we partition 3 into (at most n + I) maximal zigzag paths, and take every other edge in the
largest maximal zigzag paths (O(m) time). This gives
us at least (m/k2)/(n + 1) = k pairwise disjoint edges.
Let G = (V,E) be a geometric
graph with n vertices and m edges. Consider the set
E partially ordered by 41. For each edge e E E, let
Rankr(e) be the length of the longest chain in (E, +)
with t-he largest element e. We can find the numbers
Rankr(e),e E E, in O(m2) time as follows. We denote
t,he edgesof E by er, e2,. . . , e, according to increasing
:c-coordinate of the leftmost point (vertex). Obviously,
Rankr(er) = 1 and, for i > 1,
Sd
k :=
(
&yi
>
.
R.ankr(ei) = 1 + max{Rankr(ej) 1j < i and ej -=~rei},
4
unless j ,+?I i for all j < i in which case we have
Rank1(ei) = 1. Using this, we can determine the number Rankr(e) one by one for the edges er, ez, . . . , e, in
t,his order. Obviously, this can be done in O(m2) time.
If one of the numbers Rankr(ei) is at least L, then we
have a chain with respect! to 41 consisting of at-least
b, and its members are at least L disjoint edges. Otherwise, in O(m) time we find an i, 1 5 i < k, for which
]{e E E : Rankr(e) = i}] is maximal. For this i, let
El = (e E E : Rankr(e) = i}. The set El has size at
leashm/X: and forms an antichain with respect to <r.
Doing t.he same procedure for El and 42, we get eit,her at, least k edges which form a chain with respect
to -:z and thus pairwise disjoint, or we get a subset
& c El of size at least,m/k2 which forms an antichain
with respect to -+ (and also in -XI). In the first casewe
4.1
The case k = 3
The upper
bound
We use the following result.
Lemma 6 ([GKK])
1, a geometric graph G of n vertices does not contain 4 pairwise disjoint edges and there
is a line which intersects every edge of G and containa
no vertez of G, then G has at most 7n edges.
Lemma 6 is not stated in [GKK] explicitly. However,
its proof (relatively long case-analysis) is readily contained in the proof of Theorem 2 in [GKK].
Let G = (V, E) be a geometric graph without 4 pairwise disjoint edges. Denote the vertices by ~1,. . . , wun
185
Figure 6: A geometric graph with no 4 pairwise disjoint edges.
For n even, we take the above geometric graph on
n + 1 vertices, and remove a vertex and the six edges
. incident to it. The resulting graph on n vertices has
4n - 9 edges. •I
from left to right, and assumethat no pair of them liea
on a vertical line. For any 1 5 i < n let Di be the
aubgraph of G which contains only those edgesvavp of
G where cy,< i < /3.
It follow from Lemma 6 and the assumption that for
any 1s i < n l~Y(zi)l < 7n. For any 15 i < a, let G;
and Gil. be the subgraph of G induced by the vertices
‘VI,, , ., it and by vi,. . ., vn, respectively. Let
5
By a little modification, it is possible to slightly improve
the upper bound of Theorem 1.
j = max {i 1Gr does not contain 2 disjoint edges}.
Since G& contains two disjoint edges, G&.2 does
not, Suppose without loss of generality that j < n/2.
Then
E(G) = E(G;)
Theorem
7 For any k 5 n/2,
em(n) 2 Ak3n + 0(k2n).
U E(ci+t2) U E(Gj+l)
Proof.
(Sketch) We use the same partial orderings,
41, ‘(2, 43,414, as in the proof of Theorem 1 (seeFig. 1).
For any edge e of a geometric graph G = (YE), let
Rank:(e) be the maximum number Q 2 0 such that
there exist er,e2 ,...e,,EE,e~3el~3e2~3e3---~3
e,, and let w(e)
be the maximum number b 3 0
such that there exist el,e2,. . .,eb E E, el 44 e2 44
U{vivjs.~ E E(G) I i 2 j}.
Thcrcforc,
lE(G)I < cl(j) + el(n - j - 1) + 7n + j < 8.5n.
cl
4.2
Remarks
The lower bound
Firat, let n be odd. Take the vertices of a regular (n2)-gon, and join each of them with the furthermost 4
vcrticca. Add two vertices near the center of the (n-2)gon, and join each of them by an edge to all the other
vertices. The resulting geometric graph on n vertices
baa 4(n - 2)/2 + 2(n - l)-1=4n-7edges.
Itiseasy
to act that no 4 edgesare pairwise disjoint, see Fig. 6.
e3--‘+4
eb 44
e-
Lemma
8 Let G = (V, E) be a geometsdc graph with
a vertices and with no k + 1 p&wise disjoint edges.
Syqose that there is a line which avoids all vetiices
and crosses all edgesof G. Then
IEI 5 ;k3n + O(k2n).
189
a edges
el
.
L
b edges
>
Figure 7: I’C+ 1 disjoint edges.
Proof
of Lemma
8. For any 0 5 a, b < L - 1 let
disjoint edgesof G. We get the recursion
Ea,a = (e E E 1Rank:(e) = a, Rank,(e) = b} .
ek(n) 5 ik3n + O(k2)n + qkj21 (b/21> + q.k/2J (1nP.l)
Clearly, Ea,b is an antichain with respect to 43 and 44.
We can define Ea,band its maximal zigzag paths just, as
in the proof Theorem 1. It is easy to seethat there are at
most n maximal zigzag paths and every edge of G is
contained in at least one of them (in fact, in exactly one
of them, but here we do not need that). Suppose that
t.hereis a zigzag path Z,G!,. . ., &! of x = 2(L-a-b)+1
edges. Thene,e,i$
,..., zareX:-a-b+lparrwise .
disjoint edges. Moreover, we can add a edges above 3
and h edgesbelow z to get h+l pairwise disjoint edges,
contradicting our assumption. (See Fig. 7.)
Therefore, every maximal zigzag path of za,r, has at
most 2(X:- a - b) edges,so
and Theorem 7 follows. 0
If the number of edgesin a geometric graph is at least
0(n2), then Theorem 1 guarantees Q(Tz~/~)pairwisc disjoint edges. This is improved by the following result of
Path [P].
Theorem 9 ([PI) For any c > 0 there is a c’ > 0 such
that every geometric graph of n vertices and at leaat en2
edges has at least c’fi pairwise disjoint edges.
References
[AE] N. Alon and P. Erdiis, Disjoint edges in geometric graphs, Discrete and Computational GeometnJ 4
(1989), 257-290.
IEI=
c
o<o,b
jEa,bl<n
a+b<k
c
O<a,b
(k-a-b)=
a-W.3
ki2
(
[AH] S. Avital and H. Hanani, Graphs, Gilyonot Lematematika 3 (1966), 2-S (in Hebrew).
n,.
>
[D] R. P. Dilworth, A decomposition theorem for partially ordered sets, Annals ofMathematics 51 (1050),
161-166.
cl
Return to the proof of Theorem 7. It is not hard to
see that there is a line 1 which avoids all vertices of G
and on one side of 4 there are [n/2] vertices and at most
[h/2] pairwise disjoint edges,while on the other side of
e t(here are [n/2J vertices and at most [k/2J pairwise
[GKK] W. Goddard, M. Katchalski, and D. J. Kleitman, Forcing disjoint segments in the plane, EUTOpean Journal of Combinatorics 17 (1996), 391-395.
190
[HP] H. Hopf and E. Pannwitz, Aufgabe No. 167,
Jah?sbcTicht
dcr Deutschen Mathematiker-velreinigung 43 (1934), 114.
[Kl] Y. Kupitz, Extremal problems in combinatorial
geometry, Au&s Univewity Lectwe Notes Series,
53, (1979), Aarhus University, Denmark.
[LMPT] D. Larman, J. Matouljek, J. Path, and J.
TBrBcsik, A Ramsey-type result for planar convex
acta, The Bulletin of the London Mathematical Society 20 (1994), 132-136.
[I)] J. Path, personal communication.
[PA] f, Path and P,K. Agarwal, Combinatorial
m&y, John Wiley, New York, 1995.
Geo-
[PT] J. Path and J. Tii&sik, Some geometric applications of Dilworth’s theorem, Discrete and Computational Geometry 12 (1994), l-7.
191