Non-crossing geometric path
covering red and blue points in
the plane
Mikio Kano
Ibaraki University
Japan
October 2002
R=a set of red points in the plane={ }
B=a set of blue points in the plane={ }
We always assume that no three points in
R U B lie on the same line.
Theorem If |R|=|B|, then there exists a
perfect non-crossing geometric
alternating matching that covers R U B.
Proof of the previous theorem by using
Ham-sandwich and by induction
f(n)=2f(n/2)+O(n)
f(n)=O(n log n)
Problem For given RUB, can it be
covered by geometric alternating paths Pn
of order n without crossing ?
Path P4=Pn
=
=8
When we consider paths of odd order,
the number of red points might not be
equal to the number of blue points.
Path P3=Pn
=2.3+1.2=10
=1.3+2.2=7
Theorem
(Kaneko, MK, Suzuki)
If |R|=|B|=km and 2m <=14, then
RUB can be covered by path P2m
without crossings.
If |R|=k(m+1)+hm, |B|=km+h(m+1)
and 2m+1 <=11, then RUB can be
covered by P2m+1 without crossings.
2,3, …,11,12,14 are OK.
13,15,16,… NO
Sketch of Proof (I)
• We show that there exist a balanced convex
subdivision of the plane such that each
convex polygon contains either 2m
red+blue points or 2m+1 red+blue points.
• If 2m<=14 or 2m+1<=11, then every
configuration of 2m red+blue points or
2m+1 red+blue points has P2m or P2m+1
covering without crossings.
• In other case, there exists a configuration
having no Pn coverings.
Sketch of Proof II
Step 1: Convex balanced subdivision of the plane
Step 2: For each subdivision, there exists
a non-crossing path
P6
|R|=|B|=18
P5
|R|=3*4+2*2=16, |B|=2*4+3*2=14
We show some configurations which
have no path covering.
2m+1=13
2m=14
2m+1=15
2m+1=16
2m+1=17
2m+1=18
Balanced convex subdivision of
the plane
2m=6
Theorem (Bespamyatnikh, Kirkpatrick,Snoeyink,
Sakai and Ito, Uehara,Yokoyama)
If |R|=ag and |B|=bg, then there exists a
subdivision X1 U X2 U … U Xg of the plane
into g disjoint convex polygons such that every
Xi contains exactly a red points and b blue
points.
An equitable subdivision of 2g red points
and 4g blue points.
Not convex
n^(4/3) (log n)^3 log g time algorithm
Applying the above theorem with a=b=m
to our RUB, we can obtain the desired
convex subdivision of the plane.
Namely, if |R|=|B|=km, then there exists
a subdivision X1U … UXk of the plane
into k disjoint convex polygons such that
every Xi contains exactly m red points and
m blue points.
Theorem (Kaneko, MK and K.Suzuki)
If |R|=(m+1)k+mh and |B|=mk+(m+1)h, then
there exists a subdivision
X1 U … U Xk U Y1 U … U Yh
of the plane into k+h disjoint convex polygons
such that every Xi contains m+1 red points and
m blue points, and every Yj contains m red
points and m+1 blue points.
m=2 and m+1=3
• We can prove the above theorem in the
same way as the proof by Bespamyatnikh,
Kirkpatrick, Snoeyink.
• However we generalize the key lemma as
follows. The proof is the same as the proof
given by the above people.
Three cutting Theorem
Let |R|=g1+g2+g3 and |R|=h1+h2+h3.
Suppose that for every line l with |left(l) R|=gi,
it follows that |left(l) B|<hi.
Then there exists three rays r1, r2 and r3
such that
g2 red points and
r1 h2 blue points r2
g1 red points and
h1 blue points
g3 red points and
h3 blue points
r3
Conditions of 3-cutting Theorem
g3 red points
less than h3 blue points
g2 red points
g1 red points
less than h1
blue points
less than h2 blue points
A balanced convex subdivision
h2 blue points
g2 red points
h1 blue points
g1 red points
g3 red points
h3 blue points
A vertical line
Not convex
Remark on the above theorem
• Let a and b be integers s.t. 1<=a, a+2<=b.
• Then there exist configurations of |R|=ak+bh
red points |B|=bk+ah blue points for which
there exist no convex balanced subdivisions
of the plane.
• Thus the above theorem cannot be
generalized.
a=2, b=4
Each polygon contains either 2 red points and 4 blue
points, or 4 red points and 2 blue points.
We finally show that if either
|R|=|B| and |R U B|<=14, or
|R|=|B|+1 and |R U B|<=11,
then R U B can be covered by Pn.
|R U B|=4
P4
|R U B|=5
P5
|R U B|=6
P6
Lemma |RUB|=5. If RUB is a
configuration given in the figure,
there exists a path P5 which covers
RUB and starts with x.
x
Bad case
|R U B|=9
A new line satisfies good
condition.
x
Lemma |RUB|=6. If a red point x can
see a blue convex point y, then
there exists a path P6 which covers
RUB and starts with x.
y
We can show that
we may assume that
there exits a line in
the figure.
|R U B|=11
Conjecture (A.K, M.K)
Let g=g1+g2+ … +gk such that each gi < g/3.
If |R|=ag and |B|=bg, then there exists a
subdivision X1 U X2 U … U Xk of the plane
into k disjoint convex polygons such that each
Xi contains exactly agi red points and bgi blue
points.
agi red points
bgi blue points
Xi
Find a nice problem from the
following figure, and solve it
Thank you