Noncrossing geometric path covering red and blue points in the plane

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Non-crossing geometric path covering red and
blue points in the plane

• Mikio Kano
• Ibaraki University
• Japan
• October 2002

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Ra set of red points in the plane Ba set
of blue points in the plane We always
assume that no three points in R U B lie on the
same line.
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Theorem If RB, then there exists a perfect
non-crossing geometric alternating matching that
covers R U B.
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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)
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Problem For given RUB, can it be covered by
geometric alternating paths Pn of order n
without crossing ?

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Path P4Pn
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When we consider paths of odd order, the number
of red points might not be equal to the number of
blue points.
2.31.210
Path P3Pn
1.32.27
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Theorem (Kaneko, MK, Suzuki)

• If RBkm and 2m can be covered by path P2m without crossings.
• If Rk(m1)hm, Bkmh(m1) and 2m1
  crossings.
• 2,3, ,11,12,14 are OK. 13,15,16, NO

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Sketch of Proof (I)

• We show that there exist a balanced convex
  subdivision of the plane such that each convex
  polygon contains either 2m redblue points or
  2m1 redblue points.
• If 2mof 2m redblue points or 2m1 redblue points has
  P2m or P2m1 covering without crossings.
• In other case, there exists a configuration
  having no Pn coverings.

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Sketch of Proof II
Step 1 Convex balanced subdivision of the
plane Step 2 For each subdivision, there exists
a non-crossing path
P6 RB18
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P5
R342216, B243214
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We show some configurations which have no path
covering.
2m113
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2m14
13
2m115
14
2m116
15
2m117
16
2m118
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Balanced convex subdivision of the plane
2m6
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• Theorem (Bespamyatnikh, Kirkpatrick,Snoeyink,
• Sakai and Ito, Uehara,Yokoyama)
• If Rag and Bbg, 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.

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An equitable subdivision of 2g red points and 4g
blue points.
Not convex
n(4/3) (log n)3 log g time algorithm
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• Applying the above theorem with abm
• to our RUB, we can obtain the desired
• convex subdivision of the plane.
• Namely, if RBkm, 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.

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• Theorem (Kaneko, MK and K.Suzuki)
• If R(m1)kmh and Bmk(m1)h, then
• there exists a subdivision
• X1 U U Xk U Y1 U U Yh
• of the plane into kh disjoint convex polygons
• such that every Xi contains m1 red points
  and m blue points, and every Yj contains m red
  points and m1 blue points.

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m2 and m13
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• 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.

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Three cutting Theorem Let Rg1g2g3 and
Rh1h2h3. Suppose that for every line l with
left(l) Rgi, it follows that left(l)
B such that
g2 red points and h2 blue points
r1
r2
g1 red points and h1 blue points
g3 red points and h3 blue points
r3
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Conditions of 3-cutting Theorem
g3 red points less than h3 blue points
g2 red points
less than h2 blue points
g1 red points
less than h1 blue points
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A balanced convex subdivision
h2 blue points
g2 red points
Not convex
h1 blue points
g3 red points
g1 red points
h3 blue points
A vertical line
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Remark on the above theorem

• Let a and b be integers s.t. 1
• Then there exist configurations of Rakbh red
  points Bbkah blue points for which there
  exist no convex balanced subdivisions of the
  plane.
• Thus the above theorem cannot be generalized.

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a2, b4
Each polygon contains either 2 red points and 4
blue points, or 4 red points and 2 blue points.
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We finally show that if either RB and R U
BB can be covered by Pn.
R U B4
P4
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R U B5
P5
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R U B6
P6
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x
Lemma RUB5. If RUB is a configuration given
in the figure, there exists a path P5 which
covers RUB and starts with x.
Bad case
R U B9
A new line satisfies good condition.
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x
Lemma RUB6. 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 B11
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Conjecture (A.K, M.K)

• Let gg1g2 gk such that each gi
• If Rag and Bbg, 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
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Find a nice problem from the following figure,
and solve it
Thank you