infinite problems and finite applications

Infinite Problemsand Finite Applications

Nathaniel Dean

Chair & Professor, Department of MathematicsTexas State University, San Marcos, TX 78666

nd17@txstate.edu

The 28th Clemson Mini-Conference onDiscrete Mathematics and Algorithms

October 3, 2013

Some Infinite Problems

• Chromatic Number of the plane– Integer & Rational Distances– Konig’s Infinity Lemma– Unit Distance Graphs– Single-walled Carbon Nanotubes (SWNTs)

• Ulam’s Lattice Point Problem– Konig’s Infinity Lemma– Ulam Number– Incomprehensible Graphs

Graph G = (V, E)a set V of objects and

a set E of (unordered) pairs of objects

E = {{1,2}, {1,3}, {1,7}, {1,3}, {2,6}, {3,4}, {3,6}, {4,6}, {4,7}, {4,8}, {5,6}, {5,7}, {5,9}, {7,8}, {7,9}, {8,9}}

V = {1, 2, 3, 4, 5, 6, 7, 8, 9}

vertexnodepoint

edgelinkline

1

2

3

4 5

6

7

8

9

Edge List12, 13,17, 13,26, 34,36, 46,47, 48,56, 57,59, 78,79, 89

Graph G = (V, E)a set V of objects and

a set E of (unordered) pairs of objects

Points at Rational Distance

What is the largest possible subset of points in the plane with no three of the points collinear and no four of the points concyclic, and having all interpoint distances rational?

Jessica Poole – NREUP, Summer 2003Tawfig Haj – M.S. Thesis, 2005

Charles Gabi – Undergrad Course, Fall 2006

Rational Distancesfor Points on a Circle

(7/10,-12/5)(-7/10,-12/5)

(-7/10,12/5)

(5/2,0)(-5/2,0)

(7/10,12/5)

Circles

There exists an infinite set of points on the circle x2 + y2 = r2 where the distance between any pair of them is rational. Hence, there are arbitrarily many points, no 3 collinear, with integer distances.

Parabola

How many points can be found on the (half) parabola y = x2, x ≥ 0, so that the distance between any pair of them is rational? > 4?

Points on a Parabola

(r,r²)

d2

d1

d3

y = x 2y

x

Infinitely Many Rational Distance Sets of Size 3

Proposition: There are infinitely many rational distance sets of 3 points on the parabola y = x2.

Proof: Pick two rational numbers, say 1 and 3, and let those be the lengths of the segments AB and BC. Fix these distances and slide the points up the parabola. The distance AC will increase, bounded above by 4. Since the rationals are dense in the reals, there are many placements of the points where AC has rational length. Hence, there are infinitely many such triples.

More Recent Results

• Avion Braveman (2001) and others produced three point solutions.

• Campbell (1997) used elliptic curves to compute a four point solution and proved that there are infinitely many such sets.

• Tawfig Haj (2005 Master’s thesis) by clever use of Pythagorean triples proved there are infinitely many 4-point solutions.

Ellipses?

Hyperbolas?

An infinite tree which is locally finite must contain an infinite path.

N

Konig’s Infinity LemmaHierarchical Structure

Konig’s Infinity Lemma

Proof Idea:Since there are finitely many branches, at least one of them must have an infinite subtreeGo in that direction.

Konig’s Infinity Lemma

Proof Idea:

Find an infinite branch of the tree.

Go in that direction.

Proof Idea:

Find an infinite branch of the tree.

Go in that direction.

Konig’s Infinity Lemma

Proof Idea:

Find an infinite branch of the tree.

Go in that direction.

Konig’s Infinity Lemma

Konig’s Infinity Lemma

Proof Idea:

Find an infinite branch of the tree.

Go in that direction.

Application of KL to Coloring(applied later to Ulam Number)

Corollary: Suppose G is a locally finite graph and any finite subgraph of G is k-colorable. Then G is k-colorable.

Proof. Create a tree as follows. Enumerate the vertices of G as v1, v2, … , and let the nth level of the tree be given by the collection Ln of all k-colorings of v1, v2, … vn. Draw an edge from any k-coloring of v1, v2, … vn to a k-coloring of v1, v2, … vn+1 iff the coloring of the n + 1 vertices extends our coloring of n vertices. This is a tree! It has infinitely many vertices, and by KL there's an infinite path. This path made of colorings that all agree with each other provides a k-coloring of all of G.

Chromatic Number of the Plane P

Let G be the graph in the plane where V = R2 and two points are joined by an edge when their Euclidean distance is 1.

Define P = (G)

Hadwiger-Nelson (~1950): P = ?

Any subgraph of G is called a unit distance graph.

Unit Distance Graphs

Some Known UDGs

Petersen Graph

Cycle Cn

Grid Graph

4 P (The Moser Spindle)

P 7 (Hexagonal Tiling)

P 6 ?(Pritikin -1998)

Every UDG of order 6197 is 6-colorable.

Every UDG of order 12 is 4-colorable.

MaximalNot Maximal

A UDG is maximal if the addition of any edge results in a graph which is not a UDG.

Number of Edges in a UDG

• U(n) is the maximum size of a maximal UDG of order n.

• u(n) is the minimum size of a maximal UDG of order n.

n 2 3 4 5 6 7u(n) 1 3 5 7 8 10U(n) 1 3 5 7 9 12

Purdy and Purdy (1988, CRAY computer)Chilakamarri and Mahoney (1995, by hand)

Maximal UDGs on 7 vertices

The Minimal Forbidden Graphs on ≤ 7 Vertices

Separable UDGs

All maximal UDGs are connected.

There are no separable UDGs with < 8 vertices.

There are only 3 separable maximal UDGs on 8 vertices.

ApproachGoal: Enumerate all maximal UDGs on 8 vertices

1. Find u(n) for 8 vertices– Show U(n) = 14

2. Use maximal UDGs on 7 vertices to create a list of possible UDGs on 8 vertices.

3. Reduce the list of possibilities using minimal forbidden subgraphs.

4. Use mathematical algorithms to help identify UDGs from the remaining list.

5. Draw UDGs.

6. Find exact coordinates to verify UDGs.

Creating a List of Possible Maximal UDGs

Starting with each of maximal UDGs on 7 vertices.

Process of Elimination

Created list of possible non-isomorphic graphs from maximal UDGs on 7 vertices

Eliminated graphs not 2-connected

Eliminated graphs containing forbidden graphs

Eliminated graphs without a triangle

Identified rigid subgraphs by hand

Eliminated non-maximal UDGs

Proved or disproved each using Geometer’s Sketchpad

Verified most results with algebraic geometry formulation

Found exact coordinates from drawings and systems of equations

1216

118

94

61 Maximal UDGs on 8 Vertices

Ongoing Work

• Find all forbidden graphs on 8 vertices.

• Enumerate all maximal UDGs on 9 vertices.

• Proofs for general properties

• UDGs in higher dimensions• Heawood graph

Carbon Nanotubes

Mathematical Programming Model of Bond Length and Angular Resolution of

Carbon Nanotubes

Homomorphism

• If xy E(G), then f(x)f(y) E(H) or f(x) = f(y) and

• If ab E(H), then there exists x,y V(G) such that f(x)=a, f(y)=b, and xy E(G).

A surjective map f: V(G) V(H) of G onto H where

Homomorphism

Homomorphism

Another Homomorphism

A homomorph H of G is a uniformly bounded homomorph if for some integer m every vertex x of H satisfies

.|)(| 1 mxf

Ulam number u(G) = min {(H): H is a uniformly bounded homomorph of G}.

• u(G) (G).• H is a homomorph of G u(H) u(G).• F G u(F) u(G).

(K. Chilakamarri and N. Dean, 1995)

Corollary: Every finite homomorph of contains as a subgraph.

Konig: Every infinite graph contains• infinitely many components,• a vertex of infinite degree, or• an infinite path.

NN

Corollary: 2)( Nu

.L

(K. Chilakamarri and N. Dean, 1995)

A graph has a good drawing in k if it can be drawn in k so that

• a function h:++ every subset of k of diameter ≤ d contains ≤ h(d) verticesAND

• L + every edge has length ≤ L.

If G has a good drawing in a plane stripof fixed width, then u(G) 2.

L

Shrinking each cell to a vertex yields a homomorph isomorphic to a collection of paths.

If G has a good drawing in the plane, then u(G) 6.

L

Shrinking each cell to a vertex yields a homomorph isomorphic to a subgraph of the triangular grid.

(not a good drawing )

u(G) 3 every drawing of G in any strip [0,N] x R is incomprehensible.

u(G) 7 every drawing of G in the plane is incomprehensible.

• d + such that, for any integer N, there is a region of diameter ≤ d containing ≥ N verticesOR

• Edges are arbitrarily long.

Final Remarks

• Key IdeaKonig’s Infinity Lemma

• Goals– Generalize it (infinite).– Apply it (finite).