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Ma/CS 6b Class 13: Ramsey Theory 2
Frank P. Ramsey
By Adam Sheffer
Ramsey Numbers (Special Case)
푝1, … , 푝푘 – positive integers. The Ramsey number 푅 푝1, … , 푝푘; 2 is the minimum integer 푛 such that for every coloring of the edges of 퐾푛 using 푘 colors, at least one of the following subgraphs exists:
◦ A 퐾푝1 colored only with color 1.
◦ A 퐾푝2 colored only with color 2. ◦ …
◦ A 퐾푝푘 colored only with color 푘.
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An Example
The expression 푛 = 푅 3,3,3,3; 2 means that every coloring of the edges of 퐾푛 using four colors contains a monochromatic triangle.
Recall: 푅(3,3)
We already proved that 푅 3,3; 2 = 6.
Any red-blue coloring 퐾6 contains a monochromatic triangle.
◦ This is not the case for 퐾5.
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Recall: 푅(2,4)
What is 푅(2,4)?
◦ Either there is a red 퐾2, which is just a red edge, or there is a blue 퐾4. ◦ This is the case for any coloring of 퐾4, but not for any coloring of 퐾3. ◦ Thus, 푅(2,4) = 4.
Homogenous Subsets
푆 – a set of elements (e.g., vertices). 푆 ◦ For an integer 1 ≤ 푟 ≤ 푆 , we let the set 푟 of all subsets of 푟 elements of 푆. 푆 ◦ We give every subset of a color. 푟 ◦ We say that 푆′ ⊂ 푆 is 푖-homogeneous if every 푆′ subset of is of color 푖. 푟
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Example: Subsets of Size 2
When the subsets are of size two, this is equivalent to coloring edges of a complete graph.
Subsets of Size 3
If 푟 = 3, we color triples of vertices. ◦ Can be thought of as coloring the triangular faces of 퐾푛.
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General Ramsey Numbers
푟, 푝1, … , 푝푘 – positive integers. The Ramsey number 푅 푝1, … , 푝푘; 푟 is the minimum integer 푛 such that every 1,2, … , 푛 coloring of using 푘 colors 푟 yields an 푖-homogeneous set of size 푝푖, for some 1 ≤ 푖 ≤ 푘.
A Bit of Intuition
If 푟 = 3, we color triples of vertices. ◦ Can be thought of as coloring the triangular faces of 퐾푛. ◦ We are looking for a large subset 푆 of the vertices, such that each triangle that is spanned by three vertices of 푆 has the same color.
Monochromatic 퐾4
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The Case of Larger 푟
We already know that 푅 푝1, 푝2; 2 is finite for any positive 푝1 and 푝2. ◦ We now generalize this to the case of larger 푟. Theorem. For any positive integers 푟, 푝1, 푝2, the Ramsey number 푅 푝1, 푝2; 푟 is finite.
What Do We Need to Prove?
For any positive integers 푟, 푝1, 푝2, we need to prove that there exists 푛 such that every coloring of the 푟-tuples of a set of size 푛 in red and blue contains:
◦ Either a subset of 푝1 elements such that every 푟-tuple in it is red,
◦ or a subset of 푝2 elements such that every 푟-tuple in it is blue.
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Proof
We use a double induction. ◦ We prove the theorem by induction on 푟. ◦ We prove the induction step using an induction on 푝1 + 푝2.
Induction basis (on 푟).
◦ When 푟 = 1, we can simply take 푝1 + 푝2 − 1 elements. ◦ We already proved the case of 푟 = 2.
Induction Step
We prove the induction step for a given value of 푟 by induction on 푝1 + 푝2. ◦ Induction basis. If 푝1 < 푟 or 푝2 < 푟 then the claim vacuously holds for sets of 푝푖 elements, since they have zero 푟-tuples.
◦ Induction step. We set 푞1 = 푅 푝1 − 1, 푝2; 푟 , 푞2 = 푅 푝1, 푝2 − 1; 푟 , and 푛 = 1 + 푅 푞1, 푞2; 푟 − 1 . ◦ By the hypotheses, 푞1, 푞2, and 푛 are finite.
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Proof (cont.)
푞1 = 푅 푝1 − 1, 푝2; 푟 , 푞2 = 푅 푝1, 푝2 − 1 , 푛 = 1 + 푅 푞1, 푞2, 푟 − 1 . Let 푆 be a set of 푛 elements, with every 푟-tuple colored either red or blue. ◦ Pick an element 푥 ∈ 푆 and let 푆′ = 푆 ∖ {푥}. ◦ We color an 푟 − 1 -tuple 푇 ⊂ 푆′ using the same color as the 푟-tuple 푇 ∪ 푥 . ′ ◦ Since 푆 = 푛 − 1 = 푅 푞1, 푞2, 푟 − 1 , there is either a red subset of 푞1 elements (with respect to the colors of the 푟 − 1 -tuples), or a blue subset of 푞2 elements.
Completing the Proof
푞1 = 푅 푝1 − 1, 푝2; 푟 , 푞2 = 푅 푝1, 푝2 − 1 , 푛 = 1 + 푅 푞1, 푞2, 푟 − 1 . 푆 – a set of 푛 elements. 푆′ = 푆 ∖ 푥 .
WLOG, we assume that there is a red subset 푆푟 of 푞1 elements (with respect to 푟 − 1 -tuples). We consider the 푟-tuples of 푆푟. Since 푆푟 = 푞1 = 푅 푝1 − 1, 푝2; 푟 , either there is a blue subset of size 푝2, or a red subset of size 푝1 − 1. In the latter case, by adding 푥 to the subset, we obtain a red subset of size 푝1.
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Ramsey’s Theorem
A straightforward extension of the proof yields a more general result. Theorem. For any positive integers 푟, 푝1, … , 푝푘, the Ramsey number 푅 푝1, … , 푝푘; 푟 is finite. ◦ Intuitively: In every sufficiently large arbitrary object (i.e., an arbitrary coloring) there must be some order (i.e., a monochromatic subset).
Some History
In the early 1930’s in Budapest, a few students used to regularly meet on Sundays in a specific city park bench. Among the participants were Paul Erdős, George Szekeres, and Esther Klein. Klein told the group that for any set of five points with no three on a line, four of the points are the vertices of a convex quadrilateral.
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The convex hull of a point set 푆 is the smallest convex polygon that contains 푆. Given a set of five points: ◦ If the convex hall of the point set has at least four points in it, then we are done.
◦ If the convex hull consists of three points, we consider the line ℓ that passes through the two interior points. ◦ We take the two interior points and the two points that are on the same side of ℓ.
The Story Continues
The Budapest students started to think about whether there exists an 푛 such that every set of 푛 points with no three on a line contains the vertices of a convex pentagon. ◦ More generally, does a similar condition hold for every convex 푘-gon?
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The Happy Ending Problem
Even though Erdős was part of the group, the first to prove the claim was Szekeres. A couple of years later, Esther Klein, who suggested the problem married George Szekeres who solved it. ◦ Since then, this problem is called “the happy ending problem”. ◦ They lived together up to their 90’s.
The Happy Ending Theorem
Theorem. For every integer 푚 ≥ 3, there exists an integer 푁(푚) such that any set of 푁(푚) points with no three on a line contains a subset of 푚 points that are the vertices of a convex 푚-gon.
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First Claims
Straightforward claim. Any four vertices of a convex 푛-gon span a convex quadrilateral. Less straightforward claim. If every four vertices of an 푛-gon 푃 form a convex quadrilateral, then 푃 is convex.
Proving the Claim
Claim. If every four vertices of an 푛-gon 푃 form a convex quadrilateral, then 푃 is convex. Proof. Assume for contradiction that 푃 is not convex. ◦ There is a vertex 푣 that is not in the convex hull of the vertices of 푃. ◦ Triangulate the convex hull of 푃. ◦ 푣 together with the three vertices 푣 of the triangle containing 푣 do not span a convex quadrelateral!
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Proving the Theorem
Set 푁 푚 = 푅 푚, 5; 4 . ◦ Given a set of 푁 푚 points with no three on a line, we color every 4-tuple of points. ◦ A 4-tuple is colored red if it spans a convex quadrilateral, and otherwise blue. ◦ By Ramsey’s theorem, either there is a red subset of size 푚 and or blue one of size 5. ◦ But we proved that for any 5 points there must be a convex (=red) quadruple! ◦ Thus, there is a red subset of size 푚, and by the previous claim it spans a convex 푚-gon.
The Erdős-Szekeres Conjecture
In the 1930’s, Erdős and Szekeres proved 2푚 − 4 2푚−2 + 1 ≤ 푁 푚 ≤ + 1. 푚 − 2 Conjecture (Erdős and Szekeres). 푁 푚 = 2푚−2 + 1.
Hardly any progress until 2016. 2/3 ◦ Suk proved 푁 푚 ≤ 2푚+6푚 log 푚.
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Frank P. Ramsey
Died in 1930 at the age of 26. By then, he: ◦ Wrote mathematical papers, including getting a whole subfield named after him. ◦ Wrote several philosophical works. Wittgenstein mentions him in the introduction to his Philosophical Investigations as an influence. ◦ Wrote several economics papers, as a student to John Maynard Keynes. ◦ Had a wife, kids, etc.
Graph Ramsey Theory
So far we looked for monochromatic copies of some 퐾푚 in colorings of 퐾푛. ◦ We now move to searching monochromatic copies of other types of graphs.
◦ Given two graphs 퐺1, 퐺2, we denote by 푅 퐺1, 퐺2 the minimum number 푛 such that every coloring of 퐾푛 contains either a blue copy of 퐺1 or a red copy of 퐺2. Example.
◦ 푃푚 – a graph that is a path of length 푚. ◦ Then 푅 푃2, 푃2 = 3.
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Copies of Triangles
We denote as 푚퐾3 a set of 푚 disjoint copies of 퐾3. Theorem. 푅 푚퐾3, 푚퐾3 = 5푚, for every 푚 ≥ 2. Proof. We begin with the lower bound.
◦ Consider a red 퐾3푚−1 and another red 퐾1,2푚−1. Every other edge in the graph is blue. ◦ There are 5푚 − 1 vertices, with 푚 − 1 disjoint red triangles and 푚 − 1 disjoint blue triangles.
◦ Thus, 푅 푚퐾3, 푚퐾3 > 5푚 − 1.
Illustration
퐾3푚−1 퐾1,2푚−1
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Proof: Upper Bound
Theorem. 푅 푚퐾3, 푚퐾3 = 5푚, for 푚 ≥ 2. Proof. We prove an upper bound by induction on 푚. Induction basis. The case of 푚 = 2 is not trivial, but we will skip it.
Induction step. Consider a coloring of 퐾5푚. We repeatedly look for a monochromatic triangle and remove its vertices from the graph. Since 푅 3,3 = 6, this process continues as long as at least six vertices remain. Since 5푚 − 3푚 ≥ 6 for 푚 ≥ 3, we have at least 푚 disjoint monochromatic triangles.
Proof: Upper Bound (cont.)
We consider a coloring of 퐾5푚 and find 푚 monochromatic triangles in it. If all triangles are of the same color, we are done. Thus, assume that we have a red triangle Δ푎푏푐 and a blue triangle Δ푑푒푓. WLOG, assume that at least 5 of the 9 edges between the two triangles are red. WLOG, assume that two of these 5 edges meet in 푑. We thus have a red triangle and a blue 푑 푒 푏 triangle with a common vertex . 푑 푎
푓 푐
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Proof: Upper Bound (cont.)
We consider a coloring of 퐾5푚. It remains to consider the case of a red triangle and a blue triangle have a common vertex. By the induction hypothesis, the remaining 5푚 − 5 vertices contain 푚 − 1 disjoint triangles of the same color. By adding one of the two triangles, we obtain 푚 triangles of the same color. Bowtie
The Case of a Tree
Theorem. Let 푇 be a tree with 푚 vertices. Then 푅 푇, 퐾푛 = 푚 − 1 푛 − 1 + 1. Proof. We begin with a lower bound.
◦ Consider 푛 − 1 copies of 퐾푚−1 colored completely in blue. Edges between vertices of different copies are colored red. ◦ This is a set of (푚 − 1)(푛 − 1) vertices containing no red 퐾푛 and no blue 푇. ◦ Thus, 푅 푇, 퐾푛 > 푚 − 1 푛 − 1 .
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Proof: Upper Bound
Claim. Let 푇 be a tree with 푚 vertices. Then 푅 푇, 퐾푛 = 푚 − 1 푛 − 1 + 1. Proof. We prove the upper bound by induction on 푚 + 푛. ◦ Induction basis: if 푚 = 1 or 푛 = 1, the claim obviously holds. ◦ Induction step: Set 푁 = 푚 − 1 푛 − 1 + 1. Consider a coloring of 퐾푁 and a vertex 푣. ◦ If 푣 is connected to > 푚 − 1 푛 − 2 red edges, by the hypothesis, either the neighbors span a blue 푇 or 푣 and its neighbors span a red 퐾푛.
Proof: Upper Bound (cont.)
We set 푁 = 푚 − 1 푛 − 1 + 1. If any vertex is incident to more than (푚 − 1)(푛 − 2) red edges, we are done. Assume that every vertex is incident to at most (푚 − 1)(푛 − 2) red edges. That is, every vertex is adjacent to at least 푚 − 1 blue edges. Easy to prove: Any graph with minimum degree at least 푚 − 1 contains every tree with 푚 vertices.
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The End
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