1. Non-crossing geometric path covering red and blue points in the plane Mikio Kano Ibaraki University Japan October 2002

2. 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.

3. Theorem If |R|=|B|, then there exists a perfect non-crossing geometric alternating matching that covers  R U B.

4. 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)

5. Problem For given R U B, can it be covered by geometric alternating paths P n of order n without crossing ? ==8 Path P 4 = P n

6. 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=7 Path P 3= P n

7. Theorem ( Kaneko, MK, Suzuki) If |R|=|B|=km and 2m <=14, then R U B can be covered by path P 2m without crossings. If |R|=k(m+1)+hm, |B|=km+h(m+1) and 2m+1 <=11, then R U B can be covered by P 2m +1 without crossings. 2,3, …,11,12,14 are OK. 13,15,16,… NO

8. 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 P 2m or P 2m+1 covering without crossings. In other case, there exists a configuration having no Pn coverings.

9. Sketch of Proof II Step 1: Convex balanced subdivision of the plane Step 2: For each subdivision, there exists a non-crossing path P 6 |R|=|B|=18

10. P5P5 |R|=3*4+2*2=16, |B|=2*4+3*2=14

11. 2m+1=13 We show some configurations which have no path covering.

12. 2m=14

13. 2m+1=15

14. 2m+1=16

15. 2m+1=17

16. 2m+1=18

17. Balanced convex subdivision of the plane 2m=6

18. 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.

19. An equitable subdivision of 2g red points and 4g blue points. Not convex n^(4/3) (log n)^3 log g time algorithm

20. Applying the above theorem with a=b=m to our R U B, we can obtain the desired convex subdivision of the plane. Namely, if |R|=|B|=km, then there exists a subdivision X1 U … U Xk of the plane into k disjoint convex polygons such that every Xi contains exactly m red points and m blue points.

21. 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.

22. m=2 and m+1=3

23. 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.

24. 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 r 1, r 2 and r 3 such that g3 red points and h3 blue points g1 red points and h1 blue points g2 red points and h2 blue points r1r1 r2r2 r3r3

25. less than h 1 blue points g 2 red points g 3 red points less than h 3 blue points Conditions of 3-cutting Theorem g 1 red points less than h 2 blue points

26. 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

27. 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.

28. a=2, b=4 Each polygon contains either 2 red points and 4 blue points, or 4 red points and 2 blue points.

29. P4P4 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

30. P5P5 |R U B|=5

31. P6P6 |R U B|=6

32. Lemma |R U B|=5. If R U B is a configuration given in the figure, there exists a path P 5 which covers R U B and starts with x. x Bad case A new line satisfies good condition. |R U B|=9

33. Lemma |R U B|=6. If a red point x can see a blue convex point y, then there exists a path P 6 which covers R U B and starts with x. x y We can show that we may assume that there exits a line in the figure. |R U B|=11

34. Conjecture (A.K, M.K) Let g=g 1 +g 2 + … +g k such that each g i < g/3. If |R|=ag and |B|=bg, then there exists a subdivision X 1 U X 2 U … U X k of the plane into k disjoint convex polygons such that each Xi contains exactly ag i red points and bg i blue points. Xi ag i red points bg i blue points

35. Find a nice problem from the following figure, and solve it Thank you