Pigeonhole Principle and Ramsey’s Theorem Scott Fletcher

Let’s play a game

Let’s play a game

Let’s play a game

Let’s play a game

Let’s play a game

Let’s play a game

Let’s play a game

Let’s play a game My hypothesis: given this shape, there will always be a winner. Can we prove this?

The Pigeonhole Principle If one tries to fit n pigeons into m pigeonholes, and n > m, there will always be at least one pigeonhole which holds more than one pigeon.

Pigeonhole Proof We assume that the Pigeonhole Principle is invalid. We choose natural numbers n and m, such that n > m. We put n pigeons in the m pigeonholes. Since the Pigeonhole Principle is invalid, each of the m pigeonholes will contain at most 1 pigeon. Let m’ represent the number of non-empty pigeonholes. m’ ≤ m Since all pigeons were assigned to the pigeonholes, n ≤ m’ ≤ m (Contradicts n > m)

Haircount The Pigeonhole Principle can be used to prove seemingly nonsensical statements. One of the most famous of these this that: In New York, there are at least two people with the exact same number of hairs on their head.

Haircount The average person has around 150,000 hairs on their head, we assume however, due to the size of a hair follicle, that no one can have more than 6 million hairs on their head. Smallest recorded hair follicle 1/1500 inch. 1/1500 * 6000000 = 4000 inch = 333 feet = 111 yards = 101.5 m

Haircount The average person has around 150,000 hairs on their head, we assume however, due to the size of a hair follicle, that no one can have more than 6 million hairs on their head.

Mathematical Notation Let N and M be sets with |N| = n and |M| = m Let F : N → M be a function. There exists then a set S with |S| = s and s ≥ ceiling(n / m) And an element c0 ∈ M with: S := { x | x ∈ N, F(x) = c0 }

Proving s = ceiling(n / m) We need to show that for c*m < n there will be a pigeonhole with c+1 pigeons. Using the same method as before, we assume that this isn’t valid. ⇒ each pigeonhole contains at most c pigeons. We define n’ as the number of pigeons in the pigeonholes. ⇒ n’ ≤ c*m < n (Contradiction, n’ < n).

Infinite Pigeonhole Principle Let’s assign a pigeon to every natural number and put them into 10 pigeonholes. Following from our mathematical notation, we can assume that at least one of the pigeonholes contains infinite pigeons.

Is the Pigeonhole Principle useful? Doesn’t just apply to pigeons and pigeonholes; the principle can be extended onto mathematical Relations and Functions, and can be a useful crutch for proofs. Is there a way we could use the Pigeonhole Principle to prove my hypothesis from before?

Ramsey’s Theorem Ramsey’s Theorem uses the Pigeonhole Principle to assert that certain patterns are guaranteed to appear in any large enough data set. Specifically, it looks at complete graphs with arbitrary edge labelings. F. P. Ramsey 1903 - 1930

Complete Graph A set of vertices and edges (V, E), such that each vertex is connect to all other vertices by an edge. http://mathworld.wolfram.com/images/eps-gif/CompleteGraphs_801.gif

Edge Labeling Each edge in the graph is assigned a color out of a finite color set. https://en.wikipedia.org/wiki/Ramsey%27s_theorem#/media/File:RamseyTheory_K5_no_mono_K3.svg

Arbitrary Edge Labeling ‘Arbitrary’ simply denotes the lack of rules in the Edge Labeling; all possible labelings are valid.

Ramsey Number (generalization) Ramsey’s Theorem asserts that: for every natural number n and every set set of natural numbers {x1, x2, … , xn} There exists a minimum natural number r such that Any arbitrarily edge-labeled complete graph with r vertices and n colors contains a monochrome clique with xi vertices. i ∈ {1, 2, … , n} We call r the Ramsey Number for set {x0, x1, … , xn} and indicate it with R(x1, x2, … , xn)

Subgraph Given any graph (V, E), a subgraph is a set of vertices and edges (V’, E’) such that V’ ⊆ V E’ ⊆ E

Clique Subgraph which is itself a complete graph

Monochrome Subgraph Subgraph of an edge-labeled graph, where all of its corresponding edges in the parent graph are of the same color.

Review Once again, the Ramsey Number r for set {x1, x2, … , xn} is defined as the minimum natural number, such that: Any arbitrarily edge-labeled complete graph with r vertices and n colors contains a monochrome clique with xi vertices. i ∈ {1, 2, … , n}

R(x1, x2, … , xn) format When using this format to denote a Ramsey Number, |{x1, x2, … , xn}| = n is the number of colors in the Edge Labeling and Each element xi i ∈ {1, 2, …, n} is a unique color in the Labeling

Example R(3, 2) We assign the two colors in the notation as red and blue R(3, 2) Now we try to find a complete graph with r vertices, such that its coloring contains A monochrome red clique with 3 vertices OR A monochrome blue clique with 2 vertices Regardless of how it is colored.

Looking for: R(3, 2) > 2 All permutations:

R(3, 2) = 3 Using Pigeonhole Principle Looking for: R(3, 2) = 3 Using Pigeonhole Principle Three edges, two colors. ⇒ at least two edges are the same color

R(3, 2) = 3 Using Pigeonhole Principle Looking for: R(3, 2) = 3 Using Pigeonhole Principle Already fulfills requirements

R(3, 2) = 3 Using Pigeonhole Principle Looking for: R(3, 2) = 3 Using Pigeonhole Principle Both fulfill the requirements

R(s, 2) = s A complete graph with s-1 vertices completely colored red does not pass A complete graph with s vertices either Is completely red Contains at least one blue edge And thus meets all requirements.

R(3, 3) = 5? Can we find a complete graph with 5 vertices that contains neither nor

R(3, 3) > 5

R(3, 3) = 6?

R(3, 3) = 6? Pick a starting Vertex

R(3, 3) = 6? Consider its edges. The Pigeonhole Principle tells us that at least 3 of the 5 edges will be the same color. We can assume that those edges are blue (if not, we just switch out blue for red)

R(3, 3) = 6? Now, look at the vertices connected to the starting vertex by blue edges. Their three shared edges build a clique. These three vertices were chosen for clarity. This works just as well when any other three are chosen.

R(3, 3) = 6? If any of the edges in the Clique are blue, the requirements are fulfilled

R(3, 3) = 6? If none of the edges of the clique are blue, the requirements are fulfilled.

R(3, 3) = 6? In other words, we have reduced the problem down to the one we just solved. R(3, 2) = 3 ⇒ R(3, 3) = 6

Sim Our game from before, but it works. Guaranteed to have exactly one winner with one problem: The first player is extremely likely to win.

Sim Winner Problem It has been proven that the second player can win when playing perfectly. Still an open mathematical problem to develop a practical strategy.

Ramsey Number Computation Formerly approached by calculating upper and lower bounds, in the hopes of finding an exact formula. Lower bounds are found by producing Ramsey Graphs, which are examples that show the Ramsey Number must be greater than a certain number, like our 5 vertex example from before.

Ramsey Number Computation Upper bounds much more difficult to determine, and rely mostly on the following inequality: R(s, r) ≤ R(s-1, r) + R(s, r-1) -1

Brute Force cn(n-1)/2 possible colorings for c colors O(cn^2) The best known way to determine a Ramsey Number is brute force. Complete Graphs become exponentially more complex: n(n-1)/2 edges for n vertices The amount of possible edge labelings becomes almost unmanageable: cn(n-1)/2 possible colorings for c colors O(cn^2)

R(3, 3, 3) = 17

Final Thoughts “Erdős asks us to imagine an alien force, vastly more powerful than us, landing on Earth and demanding the value of R(5, 5) or they will destroy our planet. In that case, he claims, we should marshal all our computers and all our mathematicians and attempt to find the value. But suppose, instead, that they ask for R(6, 6). In that case, he believes, we should attempt to destroy the aliens.” -Joel Spencer, mathematician and professor at the Courant Institute of Mathematical Sciences in NYC. Specialist in Ramsey Theory.