Transcript Document

Polynomial
Partitioning
By Or Yarnitzky
1
Introduction
• Based on “Simple Proofs of Classical Theorem in Discrete
Geometry” by Kaplan, Matousek and Sharir, arXiv:
1102.5391, 2011.
2
Introduction
Distinct Distances Problem:
• Given a set 𝑃 of 𝑛 points in the plane, how many distinct
distances are there between them?
• We have seen the bound Ω(𝑛6 7 )
• We will see next week Ω(𝑛 log(𝑛))
3
Introduction
• In 2010, Guth and Katz obtained a nearly complete
solution to this problem using the Polynomial
Partitioning Theorem, which allowed them to obtain a
partition of the set P with favorable properties.
• Today we will prove the theorem and use it to prove
various bounds.
4
Incidences
• Given a line 𝑙 and a point 𝑝 in the plane ℝ2 , we say they
are incident to each other if the point 𝑝 lies on 𝑙.
• Given a set of points 𝑃 and a set of lines (or more
generally any type of curves) 𝐿, denote the number of
incidences between them by 𝐼 𝑃, 𝐿 .
5
Theorem 1 - Szemerédi-Trotter
• For sets 𝑃 of 𝑚 points and 𝐿 of 𝑛 lines we proved the
following tight upper bound on 𝐼 𝑃, 𝐿 :
𝐼 𝑃, 𝐿 = 𝑂 𝑚2 3 𝑛2
3
+𝑚+𝑛
• First proof was based on the crossing lemma for simple
graphs.
• Second proof used cuttings and the KST (Kővari-SósTurán) bound on 𝐼 𝑚, 𝑛
6
Theorem 2
• Let 𝑏, 𝑘, 𝐶 be constants. Given a set 𝑃 of 𝑚 points and Γ
a set of 𝑛 algebraic curves (defined later) such that:
• Every 𝛾 ∈ Γ is of a degree at most 𝑏
• For every 𝑘 different points in 𝑃 there exist at most 𝐶 distinct
curves in Γ passing through all of them
• Then 𝐼 𝑃, Γ = 𝑂(𝑚𝑘 (2𝑘−1) 𝑛(2𝑘−2) (2𝑘−1) + 𝑚 + 𝑛)
with constant proportional to 𝑏, 𝑘, 𝐶.
• Special case of “𝑘 degrees of freedom” presented in
previous lectures.
7
Geometric Graphs
• Let 𝑃 be a finite set of points. A geometric graph on 𝑃 is
a graph 𝐺 whose vertex set is 𝑃 and edges are straight
segments in the plane.
• Crossing number of 𝐺 is the maximal number of edges a
single line not passing through 𝑃 and not coinciding with
edges of 𝐺 intersects.
• This is not the crossing number we have seen.
8
Theorem 3
• Every set of 𝑛 points in ℝ2 has a geometric spanning tree
with crossing number 𝑂( 𝑛).
𝑛
𝑛
9
Lemmas about Polynomials
• Bivariate polynomials: 𝑓 =
𝑖,𝑗 𝑎𝑖,𝑗 𝑥
𝑖𝑦𝑗
∈ ℝ[𝑥, 𝑦]
• Denote:
𝑍 𝑓 = 𝑥, 𝑦 ∈ ℝ2 𝑓 𝑥, 𝑦 = 0}
deg 𝑓 = max 𝑖 + 𝑗 𝑎𝑖,𝑗 ≠ 0}
• The analysis can be extended to 𝑑-variate polynomials.
10
Lemma 1
• A nonzero bivariate polynomial (with at least one
nonzero coefficient) does not vanish on all ℝ2
Proof:
• Let 𝑓 ∈ ℝ[𝑥, 𝑦] be a nonzero polynomial. 𝑓 is also a
polynomial in 𝑦 with coefficients from ℝ 𝑥 (rational
functions in 𝑥)
• ℝ 𝑥 is a infinite field, therefore there is a constant 𝑏
such that 𝑓 𝑥, 𝑏 ≢ 0 (here 0 is the zero function)
• 𝑓(𝑥, 𝑏) is a nonzero polynomial over 𝑥 and therefore
there is a constant 𝑎 such that 𝑓 𝑎, 𝑏 ≠ 0
11
Lemma 2
• If 𝑙 is a line and 𝑓 ∈ ℝ 𝑥, 𝑦 is of degree 𝐷, then:
• 𝑙 ⊆ 𝑍(𝑓)
or
• 𝑍 𝑓 ∩𝑙 ≤𝐷
Proof:
𝑙 = 𝑣 + 𝑡𝑢 𝑡 ∈ ℝ
𝑍(𝑓)
𝑢
𝑣
12
Lemma 2
• If 𝑙 is a line and 𝑓 ∈ ℝ 𝑥, 𝑦 is of degree 𝐷, then:
• 𝑙 ⊆ 𝑍(𝑓)
𝑢
or
𝑣
• 𝑍 𝑓 ∩𝑙 ≤𝐷
Proof:
• Write 𝑙 in a parametric form 𝑣 + 𝑡𝑢 𝑡 ∈ ℝ , 𝑣, 𝑢 ∈ ℝ2 .
• 𝑍 𝑓 ∩ 𝑙 is the set of roots of the univariate polynomial
𝑔 𝑡 ≔ 𝑓 𝑣 + 𝑡𝑢 . Hence:
• 𝑔 ≡ 0 ⇒ 𝑙 ⊆ 𝑍(𝑓)
or
• 𝑍 𝑓 ∩ 𝑙 = 𝑍 𝑔 ≤ deg 𝑔 ≤ deg 𝑓 = 𝐷
13
Lemma 3
• If 𝑓 ∈ ℝ 𝑥, 𝑦 is nonzero and of degree 𝐷, then 𝑍(𝑓)
contains at most 𝐷 different lines
Proof: By Lemma 1, there is a 𝑝 ∈ ℝ2 ∖ 𝑍(𝑓).
• Suppose 𝑍(𝑓) contains 𝑘 distinct lines 𝑙1 , … , 𝑙𝑘 .
• Let 𝑙 be a line passing through 𝑝 which is not parallel to
any 𝑙𝑖 and does not pass through any 𝑙𝑖 ∩ 𝑙𝑗
• There is such 𝑙 because only a finite number of directions needed
to be avoided
• 𝑙 has exactly 𝑘 intersections with
𝑘
𝑖=1 𝑙𝑖
⊆𝑍 𝑓 .
• 𝑙 ⊈ 𝑍 𝑓 . By Lemma 2 it has at most 𝐷 intersections with
𝑍(𝑓), therefore 𝑘 ≤ 𝐷
14
Harnack’s Curve Theorem
• For the proof of Theorem 3 (Geometric Spanning Trees
with 𝑂( 𝑛) crossing number), we will need the following
result:
• Let 𝑓 ∈ ℝ 𝑥, 𝑦 , deg 𝑓 = 𝐷.
• Then the number of path-connected components of
𝑍(𝑓) is 𝑂 𝐷2 .
• The tight bound is 1 +
Example:
𝑛
𝑖=1
𝑥 − 𝑥𝑖
2
+
𝑛
𝑖=1
𝐷−1
2
𝑦 − 𝑦𝑖
deg 𝑓 = 2𝑛,
components
𝑛2
2
(we will not prove it).
− 𝜀2 = 0
connected
15
(Weak) Bézout's Theorem
• To prove Harnack’s Theorem, we will use without proof
the following result:
Let 𝑓, 𝑔 ∈ ℝ 𝑥, 𝑦 be polynomials of degrees 𝐷𝑓 , 𝐷𝑔 . Then:
• If the system 𝑓 = 𝑔 = 0 has a finite number of solutions
𝑘, ⇒ 𝑘 ≤ 𝐷𝑓 𝐷𝑔
• If the system 𝑓 = 𝑔 = 0 has a infinitely many solutions
⇒ 𝑓 and 𝑔 have a nontrivial common factor (not a
constant).
• Bézout's Theorem is a generalization to Lemma 2
• Note that for two polynomials 𝑓|𝑔 ⇔ 𝑍 𝑓 ⊆ 𝑍(𝑔)
16
Lemma 4
• Let 𝑓 ∈ 𝑅[𝑥, 𝑦] be a square-free polynomial, then 𝑓, 𝑓𝑦 =
𝜕𝑓
𝜕𝑦
have no common nontrivial factor (assuming 𝑓𝑦 ≢ 0)
Proof:
• Proof by induction on the degree 𝐷 of 𝑓. Assume that
𝑓 = ℎ ⋅ 𝑔 and 𝑓𝑦 = ℎ ⋅ 𝑘 for polynomials ℎ, 𝑔, 𝑘 where
ℎ, 𝑔, 𝑘 are not trivial.
• 𝑓𝑦 = ℎ𝑦 ⋅ 𝑔 + ℎ ⋅ 𝑔𝑦 = ℎ ⋅ 𝑘, so ℎ divides ℎ𝑦 ⋅ 𝑔. By
induction ℎ, ℎ𝑦 have no common factors so ℎ divides g, a
contradiction because 𝑓 is square free
17
Harnack’s Theorem Proof
• We can assume that 𝑓 is square free, otherwise we
remove repeating factors without changing 𝑍(𝑓).
• For every bounded component assign the rightmost
point. We can assume it exists and unique.
• Such points satisfy the equation 𝑓 = 𝑓𝑦 = 0 (Implicit
Function Theorem), but because of Lemma 4 and
Bézout's Theorem this equation has at most 𝐷(𝐷 − 1)
different solutions.
• Therefore, there are at most 𝐷 𝐷 − 1 bounded
components of 𝑍(𝑓) 𝑦
18
𝑥
Harnack’s Theorem Proof Continued
• There are at most 2𝐷 unbounded components.
• Assume by contradiction there are more.
• A circle 𝐶 (which is 𝑍 𝑥 2 + 𝑦 2 − 𝑟 ) with arbitrarily large
radius will not be a subset of 𝑍(𝑓) but will intersect all
components.
• Therefore the system 𝑓 = 𝑥 2 + 𝑦 2 − 𝑟 = 0 will have
more than 2𝐷 solutions, a contradiction.
• In total we get a bound of 𝐷 𝐷 − 1 + 2𝐷 = 𝑂(𝐷2 ), as
required
19
Ham-Sandwich Theorem
(Stone-Tukey)
• We say a finite set 𝐴 is bisected by a hyperplane ℎ if the
halfspaces created by ℎ have at most
each.
𝐴
2
points of 𝐴
We assume without proof the following theorem:
• For every 𝑑 finite sets in ℝ𝑑 there is a hyperplane ℎ
simultaneously bisecting all them.
20
Polynomial Ham-Sandwich
Theorem
• Let 𝐴1 , … 𝐴𝑠 ⊂ ℝ2 be finite sets, and let 𝐷 ∈ ℕ such that
𝐷+2
− 1 ≥ 𝑠,
2
• Then there is a nonzero bivariate polynomial of degree at
most 𝐷 that simultaneously bisects each 𝐴𝑖
• Here “bisects 𝑨” means that 𝑓 > 0 on at most
and 𝑓 < 0 on at most
𝐴
2
𝐴
2
points
points.
• This definition is equivalent to the previous if 𝑓 is linear
(then 𝑍(𝑓) is a hyperplane)
• There is a generalized result
21
Polynomial Ham-Sandwich
Theorem Proof
• Note 𝐷+2
is the number of monomials in a bivariate
2
polynomial of degree 𝐷, which is the number of
solutions to 𝑖 + 𝑗 ≤ 𝐷 where 𝑖, 𝑗 ∈ ℕ
• Set 𝑘 ≔
𝐷+2
2
− 1, Number of non-constant monomials
• Denote Φ: ℝ2 → ℝ𝑘 the Veronese map:
Φ 𝑥, 𝑦 ≔ 𝑥 𝑖 𝑦 𝑗
𝑖,𝑗 |1≤𝑖+𝑗≤𝐷
∈ ℝ𝑘
• The coordinates in ℝ𝑘 are denoted by pairs (𝑖, 𝑗) where
1≤𝑖+𝑗 ≤𝐷
• Example 𝐷 = 2: Φ 𝑥, 𝑦 = (𝑥, 𝑦, 𝑥 2 , 𝑥𝑦, 𝑦 2 )
22
Polynomial Ham-Sandwich
Theorem Proof - Continued
• We assume that 𝑠 = 𝑘, and we denote 𝐴′𝑖 = Φ 𝐴𝑖 .
• 𝐴′𝑖
𝑘
𝑖=1
are 𝑘 sets in ℝ𝑘
• Let ℎ be a hyperplane in ℝ𝑘 simultaneously bisecting 𝐴′𝑖 .
• ℎ has equation of the form: ℎ 𝑧 = 𝑎00 + 𝑖𝑗 𝑎𝑖𝑗 𝑧𝑖𝑗 = 0
where 𝑧𝑖𝑗 1≤𝑖+𝑗≤𝐷 are the coordinates in ℝ𝑘 .
• By the definition of Φ, it is easy to see that the
polynomial 𝑓 𝑥, 𝑦 = 𝑖,𝑗 𝑎𝑖𝑗 𝑥 𝑖 𝑦 𝑗 (we include 𝑎00 ) is the
desired polynomial, and deg 𝑓 ≤ 𝐷
• Indeed, 𝑓 𝑥, 𝑦 = ℎ(Φ 𝑥, 𝑦 )
23
General Polynomial HamSandwich Theorem
• Let 𝐴1 , … 𝐴𝑠 ⊂ ℝ𝑑 be finite sets, and let 𝐷 ∈ ℕ such that
𝐷+𝑑
− 1 ≥ 𝑠,
𝑑
• Then there is a nonzero 𝑑-variate polynomial of degree
at most 𝐷 that simultaneously bisects each 𝐴𝑖
24
Definition
• Let 𝑃 ⊂ ℝ2 be a set of 𝑚 points, let 𝑟 ∈ ℕ, 1 ≤ 𝑟 ≤ 𝑚
be a parameter.
• We say that 𝑓 ∈ ℝ[𝑥, 𝑦] is an r-partitioning polynomial
for 𝑃 if every connected component of ℝ2 ∖ 𝑍(𝑓)
(“Cell”) contains at most 𝑚 𝑟 points of 𝑃. Note every cell
is open
• No bound on 𝑃 ∩ 𝑍 𝑓
25
Polynomial Partitioning
Theorem
• For every set of 𝑚 points 𝑃 ⊂ ℝ2 and 1 ≤ 𝑟 ≤ 𝑚 there is
an r-partitioning polynomial of degree at most 𝑂 𝑟
• The number of cells is 𝑂(𝑟) (we will not prove or use it)
Proof Structure:
26
Polynomial Partitioning
Theorem Proof
• Construct collections 𝒫0 , 𝒫1 , … inductively, each consists
disjoint subsets of 𝑃 such that 𝒫𝑗 ≤ 2𝑗 .
• Start with 𝒫0 ≔ {𝑃}. Assume 𝒫𝑗 was constructed.
• By the Polynomial Ham-Sandwich Theorem there is a
polynomial 𝑓𝑗 bisecting all sets in 𝒫𝑗 , deg 𝑓𝑗 ≤ 2 ⋅ 2𝑗 .
• We can use the theorem because
2⋅2𝑗 +2
2
− 1 ≥ 2𝑗 ≥ |𝒫𝑗 |.
• For every 𝑄 ∈ 𝒫𝑗 subset of 𝑃 denote
𝑄+ = {𝑝 ∈ 𝑄|𝑓𝑗 𝑝 > 0} and 𝑄− = {𝑝 ∈ 𝑄|𝑓𝑗 𝑝 < 0}
• Set 𝒫𝑗+1 ≔
𝑄∈𝒫𝑗 {𝑄
+, 𝑄−}
• By induction ∀𝑄 ∈ 𝒫𝑗 , 𝑄 ≤ |𝑃|
2𝑗 .
27
Polynomial Partitioning
Theorem Proof - Continued
• Let 𝑡 = log 2 𝑟 . Then ∀𝑄 ∈ 𝒫𝑡 𝑄 ≤ |𝑃| 𝑟.
• Set 𝑓 ≔ 𝑓1 𝑓2 … 𝑓𝑡
• No cell contains points of two different sets in 𝒫𝑡 . Otherwise,
for points 𝑝, 𝑞 in different sets there is 𝑓𝑗 such that 𝑓𝑗 𝑝 > 0,
𝑓𝑗 𝑞 < 0.
• There is a path between the two points, so somewhere on
that path 𝑓𝑗 𝑐 = 0, so it crosses 𝑍(𝑓) which is a contradiction
• Bounding the degree of 𝑓:
𝑡
deg 𝑓 =
𝑡
2𝑗 2
deg(𝑓𝑗 ) ≤ 2
𝑗=1
𝑗=1
≤
2
2−1
2𝑡
2
= 𝑂( 𝑟)
28
Cuttings - Reminder
• Given 𝑛 lines, 𝒕-cutting is dividing the plane into
(generalized) triangles such that each cell is intersected
at most by 𝑛 𝑡 lines
• Cutting Lemma: For every 𝑛 and a parameter 1 ≤ 𝑡 ≤ 𝑛
there is a 𝑡-cutting with 𝑂(𝑡 2 ) triangles
29
Credit to
Peleg
Partitioning compared to
Cuttings
Both partitioning and cuttings divide the plane into cells:
𝒓-Partitioning
Points in a cell
|𝑃| 𝑟
Cells a line intersects
𝑂( 𝑟)
Lines intersecting a
cell
Average of 𝑂(|𝐿|
𝒓-Cutting
Average of 𝑂(|𝑃| 𝑟)
𝑂
𝑟)
𝑟 (? )
|𝐿|
𝑟
30
Theorem 1 Reminder
• For sets 𝑃 of 𝑚 points and 𝐿 of 𝑛 lines:
𝐼 𝑃, 𝐿 = 𝑂 𝑚2 3 𝑛2
3
+𝑚+𝑛
31
Theorem 1 Lemma
• Lemma: 𝐼 𝑃, 𝐿 ≤ 𝑛 + 𝑚2
Proof:
• Denote by 𝐿′ the lines incident to at most 1 point of 𝑃,
and 𝐿′′ the lines incident to 2 points or more. 𝐼 𝑃, 𝐿 =
𝐼 𝑃, 𝐿′ + 𝐼(𝑃, 𝐿′′ )
• 𝐼 𝑃, 𝐿′ ≤ 𝐿′ ≤ 𝑛.
• For any 𝑝 ∈ 𝑃, there are at most 𝑚 − 1 lines passing
through 𝑝 and another point of 𝑃.
• Hence 𝐼 𝑃, 𝐿′′ ≤ 𝑚 𝑚 − 1 < 𝑚2
32
Theorem 1 Proof
• We can now assume 𝑛 ≤ 𝑚.
• Also assume 𝑚 ≤ 𝑛, by considering the dual problem:
𝑎, 𝑏 ⟼ 𝑦 = 𝑎𝑥 − 𝑏, 𝑦 = 𝑚𝑥 + 𝑛 ⟼ 𝑚, −𝑛 .
• 𝑏 = 𝑚𝑎 + 𝑛 ⟺ −𝑛 = 𝑎𝑚 − 𝑏
• Denote 𝑟 = 𝑚4
3
𝑛2 3 .
• By assumptions 1 ≤ 𝑟 ≤ 𝑚.
• Let 𝑓 be an r-partitioning polynomial of 𝑃.
𝐷 ≔ deg 𝑓 = 𝑂
𝑟 = 𝑂(𝑚2
3
𝑛1 3 )
33
Theorem 1 Proof - Continued
• Let 𝑍 = 𝑍(𝑓),
• 𝐶1 , … , 𝐶𝑠 - Cells of ℝ2 ∖ 𝑍
• 𝑃0 ≔ 𝑃 ∩ 𝑍 – Points in 𝑍
• 𝑃𝑖 ≔ 𝑃 ∩ 𝐶𝑖 - Points in the i-th cell
• 𝐿0 = 𝑙 ∈ 𝐿 𝑙 ⊂ 𝑍 - Lines in 𝑍
• 𝐿𝑖 = 𝑙 ∈ 𝐿 𝑙 ∩ 𝐶𝑖 ≠ 𝜙 - Lines intersecting the i-th cell
• We know ∀𝑖 ≥ 1 ,
𝑃𝑖 ≤ 𝑚 𝑟 = 𝑛2
3
𝑚1
• Now 𝐼 𝑃, 𝐿 = 𝐼 𝑃0 , 𝐿0 + 𝐼 𝑃0 , 𝐿 ∖ 𝐿0 +
3
𝑠
𝑖=1 𝐼(𝑃𝑖 , 𝐿𝑖 )
34
𝑟 = 𝑚4
3
𝑛2
3
Theorem 1 Proof - Continued
𝐷 = 𝑂(𝑚2
𝑚 𝑟 = 𝑛2
• By Lemma 3 𝐿0 ≤ 𝐷, hence:
3
3
𝑛1 3 )
𝑚1 3
𝐼 𝑃0 , 𝐿0 ≤ 𝑃0 𝐿0 ≤ 𝑚𝐷 ≤ 𝑛𝐷 = 𝑂(𝑚2 3 𝑛2 3 )
• Every 𝑙 ∈ 𝐿 ∖ 𝐿0 intersects 𝑍 at most 𝐷 times (Lemma 2):
𝐼 𝑃0 , 𝐿 ∖ 𝐿0 ≤ 𝐷 𝐿\L0 = 𝑂(𝑚2 3 𝑛2 3 )
• By the lemma: 𝑠𝑖=1 𝐼 𝑃𝑖 , 𝐿𝑖 ≤ 𝑠𝑖=1( 𝐿𝑖 + 𝑃𝑖 2 )
• Because every line passes through 𝑍 at most 𝐷 times, it
intersects at most 𝐷 + 1 cells. Therefore:
𝑠
𝑖=1 |𝐿𝑖 | = 𝑂 𝐷 +
𝑠
2
≤ max 𝑃𝑖
𝑖=1 𝑃𝑖
𝑖
1 𝑛 = 𝑂(𝑚2 3 𝑛2
⋅
𝑠
𝑖=1
𝑃𝑖 ≤
𝑚
𝑟
3
+ 𝑛)).
⋅ 𝑚 = 𝑂(𝑚2 3 𝑛2 3 ).
35
Theorem 2
• An algebraic curve 𝜸 of degree 𝒃 is defined to be 𝑍(𝑓)
for some 𝑓 ∈ ℝ 𝑥, 𝑦 , deg 𝑓 = 𝑏.
• Given a set 𝑃 of 𝑚 points and Γ a set of 𝑛 algebraic
curves such that:
• Every 𝛾 ∈ Γ is of a degree at most 𝑏
• For every 𝑘 different points in 𝑃 there exist at most 𝐶
distinct curves in Γ passing through all of them
• Then 𝐼 𝑃, Γ = 𝑂(𝑚𝑘 (2𝑘−1) 𝑛(2𝑘−2)
with constant proportional to 𝑏, 𝑘, 𝐶
(2𝑘−1)
+ 𝑚 + 𝑛)
• In the proof we will assume every 𝛾 ∈ Γ is irreducible
36
Theorem 2 Lemma
Lemma:
• 𝐼 𝑃, Γ = 𝑂 𝑛 + 𝑚𝑘
• 𝐼 𝑃, Γ = 𝑂 𝑚 + 𝑛2
Proof:
• Separate between curves with fewer than 𝑘 incidences,
which have 𝑂(𝑛) incidences, and the other curves.
𝑘−1
• For each 𝑝 ∈ 𝑃 there are at most 𝐶 𝑚−1
=
𝑂(𝑚
)
𝑘−1
curves incident to it and to 𝑘 − 1 other points. Indeed,
for each choice of 𝑘 − 1 points there are at most 𝐶
curves incident to them and 𝑝. In total 𝑂(𝑚𝑘 ) incidences
37
Theorem 2 Lemma-continued
Lemma:
• 𝐼 𝑃, Γ = 𝑂 𝑛 + 𝑚𝑘
• 𝐼 𝑃, Γ = 𝑂 𝑚 + 𝑛2
Proof:
• By Bézout the curves have at most 𝑏 2 intersections.
Points lying on at most 1 curve generate 𝑂 𝑚
incidences
• Remaining points lie on at least 2 curves. Each 𝛾
contributes at most 𝑏 2 𝑛 − 1 = 𝑂(𝑛) incidences
with such points
• In total 𝑂 𝑚 + 𝑛𝑂 𝑛 = 𝑂(𝑚
+ 𝑛2 )
38
Theorem 2 Proof
• Assume 𝑚 < 𝑛2 and 𝑛 ≤ 𝑚𝑘 , otherwise by the lemma
𝐼 𝑃, Γ = 𝑂(𝑚 + 𝑛)
• Set 𝑟 = 𝑚2𝑘 (2𝑘−1) 𝑛2 (2𝑘−1) , by assumptions 1 ≤ 𝑟 ≤
𝑚. Let 𝑓 be an 𝑟-partitioning polynomial for 𝑃 such that
deg 𝑓 = 𝑂 𝑟 = 𝑂(𝑚𝑘 2𝑘−1 𝑛1 2𝑘−1 )
• Similarly to Theorem 1, denote 𝑍, 𝐶𝑖 , 𝑃0 , 𝑃𝑖 , Γ0 , Γ𝑖 .
• Again: 𝐼 𝑃, Γ = 𝐼 𝑃0 , Γ0 + 𝐼 𝑃0 , Γ ∖ Γ0 +
𝑠
𝑖=1 𝐼(𝑃𝑖 , Γ𝑖 )
• Assume 𝛾 ∈ Γ0 is infinite, otherwise include it in Γ ∖ Γ0
39
Theorem 2 Proof-continued
• Note every 𝛾 ∈ Γ0 is irreducible and infinite.
• 𝛾 is 𝑍(𝑔) for some polynomial irreducible 𝑔. Therefore
by Bézout's Theorem 𝑔 is a factor of 𝑓, so Γ0 ≤ deg 𝑓
= 𝑂( 𝑟).
• By the lemma 𝐼 𝑃0 , Γ0 = 𝑂 𝑚 + Γ0
= 𝑂(𝑚)
2
=𝑂 𝑚+𝑟
40
Theorem 2 Proof - Continued
𝑟 = 𝑚𝑘
2𝑘−1
𝑛1
2𝑘−1
• Applying Bézout's Theorem to 𝛾 ∈ Γ ∖ Γ0 and 𝑓,
𝛾 ∩ 𝑍 ≤ 𝑏 ⋅ deg 𝑓 = 𝑂 𝑟 , hence:
𝐼 𝑃0 , Γ ∖ Γ0 = 𝑂 𝑛 𝑟 = 𝑂(𝑚𝑘
(2𝑘−1) (2𝑘−2) (2𝑘−1)
𝑛
)
• Because each 𝛾 ∈ Γ ∖ Γ0 crosses 𝑍 at most 𝑂(deg 𝑓 )
times, it passes through 𝑂(deg 𝑓 ) sets 𝐶𝑖 , therefore
𝑠
=𝑂 𝑛 𝑟
𝑖=1 |Γ𝑖 | = 𝑂 𝑛 ⋅ deg 𝑓
𝑠
𝑖=1 𝐼
• By lemma:
𝑂 𝑛 𝑟 + max 𝑃𝑖
𝑖
𝑠
𝑖=1
𝑃𝑖 , Γ𝑖 = 𝑂
𝑘−1
𝑂
𝑠
𝑖=1
𝑃𝑖
Γ𝑖 + 𝑃𝑖
𝑘
≤
=𝑂 𝑛 𝑟+
41
Theorem 2 Applications
• Lines: 𝐼 𝑚, 𝑛 = 𝑂 𝑚2 3 𝑛2
3
+𝑚+𝑛
• Unit circles: 𝐼 𝑚, 𝑛 = 𝑂 𝑚2 3 𝑛2
3
+𝑚+𝑛
• General circles: 𝐼 𝑚, 𝑛 = 𝑂 𝑚3 5 𝑛4
5
+𝑚+𝑛
42
Theorem 3 - Reminder
• Every set of 𝑛 points in ℝ2 has a geometric spanning tree
with crossing number 𝑂( 𝑛)
• We will actually prove there is a geometric spanning path
with crossing number 𝑂( 𝑛)
43
Theorem 3 – Motivation
• Range Searching: Given a set 𝑃 of 𝑛 points in ℝ2 and a
collection of subsets of ℝ2 (“Ranges”), create an efficient
algorithm that receives a range and outputs the points in
the range. We will focus on half-plane range searching
(the ranges are half-planes)
44
Theorem 3 – Motivation-continued
• Approach: Preprocess the points.
• Given a geometric spanning tree with a low crossing
number, there are algorithms which use it to store the
points efficiently and answer half-plane queries in a
relatively short time
45
Theorem 3 – Lower Bound
𝑛 points
𝑛 − 1 lines
𝑛 points
𝑛 − 1 lines
⇒ There is a line with
intersections
𝑛−1
2 𝑛−2
= Ω( 𝑛)
46
Definition
• Let 𝑋 ⊆ ℝ2 , we say 𝑋 has crossing number at most 𝑘 if
each line (with possibly finitely many exceptions)
intersects 𝑋 in at most 𝑘 points.
𝑝5
𝑝4
𝑝1
𝑝2
𝑝6
𝑝7
𝑝3
𝑝8
47
Lemma A
• Let 𝑃 be a set of 𝑛 points, let 𝑋 ⊆ ℝ2 be a pathconnected set containing 𝑃 with crossing number at
most 𝑘.
• Then exists a geometric spanning tree of 𝑃 with crossing
number at most 2𝑘
• Given this lemma, to prove Theorem 3 we only need to
construct a path-connected set 𝑋 containing 𝑃 with
𝑝5
crossing number at most 𝑂( 𝑛)
𝑝4
𝑝5
𝑝4
𝑝1
𝑝2
𝑝3
𝑝1
𝑝6
𝑝7
𝑝8
𝑝6
𝑝2
𝑝3
48
𝑝7
𝑝8
Lemma A Stage 1 Drawing
𝑝5
𝑝5
𝑝4
𝑝4
𝑞4
𝑝1 /𝑞1
𝑝1
𝑝6
𝑝2
𝑆
𝑝2 /𝑞2 /𝑞3
𝑞5
𝑝6 /𝑞6
𝑞7
𝑝7
𝑝7
𝑝3
𝑝3
𝑝8
𝑝8
𝑝1 /𝑞1
𝑝1
𝑝1 /𝑞1
𝑝2 /𝑞2
𝑝2
𝑝3
49
Lemma A Proof Stage 1
• Stage 1: We build a geometric tree S which spans 𝑃 =
{𝑝1 , … , 𝑝𝑛 } and some other points Q, whose edges are
arcs (not necessarily straight segments) contained in 𝑋.
• We set 𝑆1 = 𝑝1 . Assuming we built such a tree 𝑆𝑖 for
points {𝑝1 , … , 𝑝𝑖 }, we choose an arc 𝛼𝑖 that connects
𝑝𝑖+1 to some point 𝑞𝑖 ∈ 𝑆𝑖 such that 𝑆𝑖 ∩ 𝛼𝑖 = {𝑞𝑖 }.
• We set 𝑆𝑖+1 ≔ 𝑆𝑖 ∪ 𝛼𝑖 and 𝑆 ≔ 𝑆𝑛 . Because 𝑆 ⊆ 𝑋 and
𝑋 has crossing number at most 𝑘, 𝑆 has a crossing
number at most 𝑘 as well.
50
Lemma A Proof - Stage 2
• Stage 2: We replace the arcs of 𝑆 by straight segments.
• This yields a tree 𝑇 spanning 𝑃 and 𝑄 = {𝑞1 , … , 𝑞𝑛−1 },
which has a crossing number of at most 𝑘.
• Indeed, if a line crosses 𝑡 edges of 𝑇 it will cross at least 𝑡
arcs of 𝑆, but the crossing number of 𝑆 is at most 𝑘
51
Lemma A – Stage 3 Drawing
𝑝1 , 𝑝2 , 𝑝3 , 𝑞4 , 𝑝5 , 𝑞5 , 𝑝4 , 𝑝6 , 𝑞7 , 𝑝8 , 𝑝7
𝑝5
𝑝4
𝑞4
𝑞5
𝑝1
𝑝6
𝑝2
𝑞7
𝑝3
𝑝8
𝑝7
52
Lemma A – Stage 3 Drawing
𝑝1 , 𝑝2 , 𝑝3 , 𝑞4 , 𝑝5 , 𝑞5 , 𝑝4 , 𝑝6 , 𝑞7 , 𝑝8 , 𝑝7
𝑝5
𝑝4
𝑞4
𝑞5
𝑝1
𝑝6
𝑝2
𝑞7
𝑝3
𝑝8
𝑝7
53
Lemma A Proof - Stage 3
• Stage 3: We preform a DFS run on the tree starting from
an arbitrary root, creating a list of the points in 𝑃 and 𝑄.
We delete the points of 𝑄 from the list, and connect each
pair of consecutive points by a straight segment.
• This is a geometric spanning tree of 𝑃, which has a
crossing number at most 2𝑘
𝑝5
𝑝4
𝑞4
54
𝑞5
Lemma A –Stage 3 Explanation
• Assume a line crosses 𝑡 new edges. For each new edge
crossed, the same line crosses an edge from the tree 𝑇,
we match between new edges to their edges in 𝑇,
resulting 𝑡 edges of 𝑇.
• Each edge of 𝑇 appears in this list at most twice,
therefore the number of unique edges in the list is at
least 𝑡 2, and the number of unique edges from 𝑇 the
line crosses is at most 𝑘.
• Hence 𝑡 2 ≤ 𝑘, or 𝑡 ≤ 2𝑘 as required.
𝑝5
𝑝4
𝑞4
𝑞5
55
Lemma B
• Let 𝑃 be a set of 𝑛 points, then there exist a set 𝑋 ⊆ ℝ2 that
contains 𝑃 has at most 𝑛 2 connected components and
crossing number 𝑂( 𝑛)
Proof:
• Let 𝑓 be an 𝑟-partitioning polynomial, 𝑟 specified later.
• By Harnack’s Theorem the number of connected components
of 𝑍 ≔ 𝑍(𝑓) is at most 𝑘 ⋅ deg 𝑓 2 for some constant 𝑘, and
deg 𝑓 ≤ 𝑚 𝑟 for some constant 𝑚.
• We choose 𝑟 = 𝑛 𝑐 where 𝑐 = 2𝑚2 𝑘 is a constant. The
number of connected components of 𝑍 is at most 𝑛 2
• Note that in order to use polynomial partitioning we require
1 ≤ 𝑟 ≤ 𝑛 ⇔ 1 ≤ 𝑐 ≤ 𝑛, but we can choose such 𝑐 and
assume 𝑛 is arbitrarily large
56
Lemma B Proof
• For every 𝑝 ∈ 𝑃 ∖ Z, define 𝜎𝑝 a straight segment connecting
𝑝 to 𝑍, avoiding 𝑍 otherwise.
• Define 𝑋 ≔ 𝑍 ∪
components.
𝑝∈𝑃∖Z 𝜎𝑝 .
𝑋 also has at most 𝑛 2 connected
• Let 𝑙 be a line not contained in 𝑍 and not containing any 𝜎𝑝
(By Lemma 3 only finitely many lines are avoided).
• 𝑙 intersects 𝑍 in at most deg 𝑓 = 𝑂( 𝑛) points. 𝑙 crosses at
most deg 𝑓 + 1 components of ℝ2 ∖ 𝑍, each contains at
most 𝑐 points of 𝑃 (Because 𝑓 is 𝑛 𝑐-partitioning)
• Therefore 𝑙 crosses at most 𝑐 1 + deg 𝑓 = 𝑂 𝑛
segments 𝜎𝑝 , hence 𝑙 crosses 𝑋 at most 𝑂 𝑛 times
57
Lemma B Drawing
58
Theorem 3 Proof
• By Lemma A, we only need to construct a connected set
𝑋 containing 𝑃 with crossing number at most 𝑂( 𝑛)
• To create 𝑋, apply Lemma B iteratively. We construct a
sequence of sets 𝐵0 , 𝐵1 , … such that each 𝐵𝑖 contains 𝑃
and has at most 𝑛 2𝑖 connected components
• We begin with 𝐵0 ≔ 𝑃, and having constructed 𝐵𝑖 , we
choose a point in each of its connected components
resulting a set 𝑅𝑖 , 𝑅𝑖 ≤ 𝑛 2𝑖 .
• Lemma B provides us with a set 𝑋𝑖 , 𝑅𝑖 ⊆ 𝑋𝑖 with crossing
number 𝑂( 𝑛 2𝑖 ) and at most 𝑛 2𝑖+1 connected
components.
59
Theorem 3 Proof - Continued
• We define 𝐵𝑖+1 ≔ 𝐵𝑖 ∪ 𝑋𝑖
• 𝐵𝑖+1 has at most 𝑛 2𝑖+1 connected components
because each point is reachable from one of the
components of 𝑋𝑖
• For some 𝑖0 we reach a connected 𝐵𝑖0 , which will be 𝑋.
• By definition 𝑋 =
is at most
𝑂( 𝑛)
𝑖≤𝑖0 𝑂(
𝑖≤𝑖0 𝑋𝑖 , therefore its crossing number
−𝑖 2
𝑛 2𝑖 ) ≤ 𝑂 𝑛 ⋅ ∞
𝑂
2
=
𝑖=0
60
Summary
• We’ve proved the Polynomial Partitioning Theorem
• Used it to prove Szemerédi-Trotter theorem
• Used it to generalize the theorem to algebraic curves
• Proved the existence of geometric spanning trees with
low crossing number
• Next lecture: (We) will prove the lower bound of the
distinct distances problem
Questions?
61
Thank You!
62