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 inR U B lie on the same line.

Theorem If |R|=|B|, then there exists a perfect non-crossing geometric alternating matching that coversR 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 ?

= =8Path P4=Pn

When we consider paths of odd order,the number of red points might not be equal to the number of blue points.

=2.3+1.2=10

=1.3+2.2=7Path P3=Pn

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 planeStep 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

2m+1=13

We show some configurations which have no path covering.

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

g3 red points and h3 blue points

g1 red points and h1 blue points

g2 red points and h2 blue pointsr1 r2

r3

less than h1

blue points

g2 red points

g3 red points less than h3 blue points

Conditions of 3-cutting Theorem

g1 red points less than h2 blue points

h1 blue points

g2 red points

g3 red points

A balanced convex subdivision

A vertical line

g1 red points

h2 blue points

h3 blue points

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.

P4

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

P5|R U B|=5

P6|R U B|=6

Lemma |RUB|=5. If RUB is aconfiguration given in the figure,there exists a path P5 which covers RUB and starts with x.

x

Bad case

A new line satisfies goodcondition.

|R U B|=9

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.

x

y

We can show thatwe may assume thatthere exits a line inthe 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.

Xi

agi red points

bgi blue points

Find a nice problem from the following figure, and solve it

Thank you