A Tale of Two Proofs Kristin Camenga Penn State Altoona, Altoona, - - PowerPoint PPT Presentation

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A Tale of Two Proofs Kristin Camenga Penn State Altoona, Altoona, PA January 19, 2016 Camenga (Penn State) 1/19/16 January 19, 2016 1 / 66 The Problem Can every map be colored with four colors? Camenga (Penn State) 1/19/16 January 19,

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A Tale of Two Proofs

Kristin Camenga

Penn State Altoona, Altoona, PA

January 19, 2016

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SLIDE 2

The Problem

Can every map be colored with four colors?

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Example

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Example

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Example

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Example

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Example

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SLIDE 8

Example

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SLIDE 9

Example

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Example

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Example

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Maps and Coloring more precisely

Guidelines for maps: Maps must be on a plane or sphere. Countries (sections) of the map must be connected. A good coloring is one in which two adjacent countries - countries which share a border, not just a point - are colored with different colors.

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Good, better, best

We can do better!

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SLIDE 14

Good, better, best

We can do better!

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SLIDE 15

Good, better, best

We can do better!

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SLIDE 16

Good, better, best

We can do better!

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SLIDE 17

Good, better, best

We can do better!

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SLIDE 18

Good, better, best

We can do better!

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SLIDE 19

Good, better, best

We can do better!

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SLIDE 20

Good, better, best

We can do better! This only needs three colors. So we say the chromatic number is 3.

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SLIDE 21

Timeline for the Four Color Theorem

1852 1879 & 1880 1890 & 1891 1976 The problem was posed to Augustus DeMorgan by

  • ne of his students, Frederick Guthrie, in 1852.

Frederick Guthrie Augustus DeMorgan

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SLIDE 22

A student of mine asked me today to give him a reason for a fact which I did not know was a fact - and do not yet. He says that if a figure be anyhow divided and the compartments differently coloured so that figures with any portion of common boundary line are differently coloured - four colours may be wanted, but not more - the following is the case in which four colours are wanted. Query cannot a necessity for five or more be

  • invented. ...... If you retort with some very simple case which makes me
  • ut a stupid animal, I think I must do as the Sphynx did....

DeMorgan writing to Hamilton

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Timeline for the Four Color Theorem

1852 1879 & 1880 1890 & 1891 1976 A proof was first given in 1879 by Alfred Bray Kempe and another in 1880 by Peter Guthrie Tait. Alfred Bray Kempe Peter Guthrie Tait

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SLIDE 24

Timeline for the Four Color Theorem

1852 1879 & 1880 1890 & 1891 1976 Gaps were found in Kempe and Tait’s proofs by Percy Heawood in 1890 and Julius Petersen in 1891, respec-

  • tively. Oops!

Percy Heawood Julius Petersen

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Timeline for the Four Color Theorem

1852 1879 & 1880 1890 & 1891 1976 After work by many mathematicians, the theorem was finally proved in 1976 by Kenneth Appel and Wolfgang Haken. It took 1200 hours of computer time.

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Modeling the problem

We will represent every country by a point, called a vertex:

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Modeling the problem

We connect two vertices with an edge if the two corresponding countries share a boundary.

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Modeling the problem

We connect two vertices with an edge if the two corresponding countries share a boundary.

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Modeling the problem

The set of vertices and edges is called a (vertex-edge) graph. Any graph coming from a map can be drawn so that the edges do not

  • cross. Any graph that can be drawn this way is called a planar graph.

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Modeling the problem

Now, instead of coloring the countries, we color the vertices. For a good coloring, any two vertices connected by an edge must be colored different colors. If there is a good coloring with k colors, we say the graph is k-colorable. The Four Color Theorem states that all planar graphs are 4-colorable.

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Five Color Theorem

We will prove the Five Color Theorem: Every planar graph is 5-colorable.

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Kempe & Heawood proof - set up

We define: V is the number of vertices E is the number of edges F is the number of faces (regions, including the outside region) the degree of a vertex is the number of edges at that vertex To prove this theorem, we are going to make our job harder: Adding edges can never reduce the number of colors needed, so we will add edges until we have a triangulation.

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Kempe & Heawood proof - a key fact

We will show that there is a vertex in our triangulated graph that has degree at most 5. We do this by looking at the number of edges.

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Kempe & Heawood proof - a key fact

We will show that there is a vertex in our triangulated graph that has degree at most 5. We do this by looking at the number of edges. Since each face is a triangle, every face is surrounded by 3 edges. Each of these edges is counted for 2 faces. This gives that 3F = 2E

  • r

F = 2 3E.

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Kempe & Heawood proof - a key fact

We will show that there is a vertex in our triangulated graph that has degree at most 5. We do this by looking at the number of edges. Since each face is a triangle, every face is surrounded by 3 edges. Each of these edges is counted for 2 faces. This gives that 3F = 2E

  • r

F = 2 3E. Euler’s relation tells us that V − E + F = 2. Substitute into Euler’s relation: V − E + 2

3E = 2 or V − 1 3E = 2.

Solving for E, we get that E = 3V − 6 .

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Kempe & Heawood proof - a key fact

Assumption: Suppose all the vertices have degree at least 6.

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Kempe & Heawood proof - a key fact

Assumption: Suppose all the vertices have degree at least 6. Then if we add up all the degrees at the vertices, we get at least 6V .

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Kempe & Heawood proof - a key fact

Assumption: Suppose all the vertices have degree at least 6. Then if we add up all the degrees at the vertices, we get at least 6V . But when we add up all the degrees, we count each edge twice, once from each end, to get exactly 2E.

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SLIDE 39

Kempe & Heawood proof - a key fact

Assumption: Suppose all the vertices have degree at least 6. Then if we add up all the degrees at the vertices, we get at least 6V . But when we add up all the degrees, we count each edge twice, once from each end, to get exactly 2E. Therefore 2E ≥ 6V

  • r

E ≥ 3V

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SLIDE 40

Kempe & Heawood proof - a key fact

Assumption: Suppose all the vertices have degree at least 6. Then if we add up all the degrees at the vertices, we get at least 6V . But when we add up all the degrees, we count each edge twice, once from each end, to get exactly 2E. Therefore 2E ≥ 6V

  • r

E ≥ 3V Compare our two results: E = 3V − 6 E ≥ 3V

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SLIDE 41

Kempe & Heawood proof - a key fact

Assumption: Suppose all the vertices have degree at least 6. Then if we add up all the degrees at the vertices, we get at least 6V . But when we add up all the degrees, we count each edge twice, once from each end, to get exactly 2E. Therefore 2E ≥ 6V

  • r

E ≥ 3V Compare our two results: E = 3V − 6 E ≥ 3V These can’t both happen! Therefore we have a contradication and our assumption was wrong: not all the vertices have degree at least 6. Therefore, there is at least one vertex of degree at most 5.

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Kempe & Heawood proof - base cases

Suppose we have 5 or fewer vertices in our graph. Is the graph 5-colorable?

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Kempe & Heawood proof - base cases

Suppose we have 5 or fewer vertices in our graph. Is the graph 5-colorable? OF COURSE!

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Kempe & Heawood proof - base cases

Suppose we have 5 or fewer vertices in our graph. Is the graph 5-colorable? OF COURSE! Suppose we have 6 vertices in our graph. We know we can color the first

  • five. What happens when we try to color the sixth?

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Kempe & Heawood proof - base cases

Suppose we have 5 or fewer vertices in our graph. Is the graph 5-colorable? OF COURSE! Suppose we have 6 vertices in our graph. We know we can color the first

  • five. What happens when we try to color the sixth?

The worst case scenario is that we have used all five colors on the first five vertices and the 6th vertex is connected to each of the first five: Number the colors of the first 5 vertices clockwise from 1 to 5.

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SLIDE 46

Kempe & Heawood proof - base cases

Suppose we have 5 or fewer vertices in our graph. Is the graph 5-colorable? OF COURSE! Suppose we have 6 vertices in our graph. We know we can color the first

  • five. What happens when we try to color the sixth?

The worst case scenario is that we have used all five colors on the first five vertices and the 6th vertex is connected to each of the first five: We can change color 2 to 4, leav- ing color 2 for the center...

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SLIDE 47

Kempe & Heawood proof - base cases

Suppose we have 5 or fewer vertices in our graph. Is the graph 5-colorable? OF COURSE! Suppose we have 6 vertices in our graph. We know we can color the first

  • five. What happens when we try to color the sixth?

The worst case scenario is that we have used all five colors on the first five vertices and the 6th vertex is connected to each of the first five: ...unless there is an edge between vertices 2 and 4 so they have to be different colors.

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SLIDE 48

Kempe & Heawood proof - base cases

Suppose we have 5 or fewer vertices in our graph. Is the graph 5-colorable? OF COURSE! Suppose we have 6 vertices in our graph. We know we can color the first

  • five. What happens when we try to color the sixth?

The worst case scenario is that we have used all five colors on the first five vertices and the 6th vertex is connected to each of the first five: Then try to do the same color switch with 3 and 5.

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SLIDE 49

Kempe & Heawood proof - base cases

Suppose we have 5 or fewer vertices in our graph. Is the graph 5-colorable? OF COURSE! Suppose we have 6 vertices in our graph. We know we can color the first

  • five. What happens when we try to color the sixth?

The worst case scenario is that we have used all five colors on the first five vertices and the 6th vertex is connected to each of the first five: Then try to do the same color switch with 3 and 5. But any edge from 3 to 5 will cross another edge, which is impossible since the graph is planar.

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Kempe & Heawood proof - base cases

Suppose we have 5 or fewer vertices in our graph. Is the graph 5-colorable? OF COURSE! Suppose we have 6 vertices in our graph. We know we can color the first

  • five. What happens when we try to color the sixth?

The worst case scenario is that we have used all five colors on the first five vertices and the 6th vertex is connected to each of the first five: In that case we can switch color 3 to color 5 and color the center vertex 3.

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Kempe & Heawood proof - Kempe chains

Suppose we have more than 5 vertices. We have shown that one of those vertices must have degree 5 or less. So color the rest of the graph and then consider how to color the vertex of degree 5 or less. Again, number the colors.

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Kempe & Heawood proof - Kempe chains

Suppose we have more than 5 vertices. We have shown that one of those vertices must have degree 5 or less. So color the rest of the graph and then consider how to color the vertex of degree 5 or less. Consider all the vertices that are col-

  • red 2 or 4.

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SLIDE 53

Kempe & Heawood proof - Kempe chains

Suppose we have more than 5 vertices. We have shown that one of those vertices must have degree 5 or less. So color the rest of the graph and then consider how to color the vertex of degree 5 or less. If there is no path between the original vertices colored 2 and 4, switch colors in the group connected to vertex 2.

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Kempe & Heawood proof - Kempe chains

Suppose we have more than 5 vertices. We have shown that one of those vertices must have degree 5 or less. So color the rest of the graph and then consider how to color the vertex of degree 5 or less. If there is a path between the original vertices colored 2 and 4, consider the colors 3 and 5.

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SLIDE 55

Kempe & Heawood proof - Kempe chains

Suppose we have more than 5 vertices. We have shown that one of those vertices must have degree 5 or less. So color the rest of the graph and then consider how to color the vertex of degree 5 or less. Consider all the vertices that are col-

  • red 3 or 5.

There cannot be a path using these vertices from vertex 3 to vertex 5 or else it would cross the path of colors 2 and 4.

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SLIDE 56

Kempe & Heawood proof - Kempe chains

Suppose we have more than 5 vertices. We have shown that one of those vertices must have degree 5 or less. So color the rest of the graph and then consider how to color the vertex of degree 5 or less. Therefore, we can switch colors 3 and 5 in one section and color the center vertex.

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SLIDE 57

Thomassen & Another Proof

In 1994, Carsten Thomassen gave a different proof of the Five Color Theorem using a different kind of coloring.

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SLIDE 58

List-coloring

Each vertex is given a list of possible colors. We have to choose a color each vertex from the corresponding list to get a good coloring.

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SLIDE 59

List-coloring

Each vertex is given a list of possible colors. We have to choose a color each vertex from the corresponding list to get a good coloring.

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SLIDE 60

List-coloring

Each vertex is given a list of possible colors. We have to choose a color each vertex from the corresponding list to get a good coloring.

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SLIDE 61

List-coloring

Each vertex is given a list of possible colors. We have to choose a color each vertex from the corresponding list to get a good coloring.

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SLIDE 62

List-coloring

Each vertex is given a list of possible colors. We have to choose a color each vertex from the corresponding list to get a good coloring.

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SLIDE 63

List-coloring

Each vertex is given a list of possible colors. We have to choose a color each vertex from the corresponding list to get a good coloring.

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SLIDE 64

List-coloring

Each vertex is given a list of possible colors. We have to choose a color each vertex from the corresponding list to get a good coloring. A graph is k-choosable if there is a good coloring for any lists of length at least k.

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Thomassen’s Proof - set up

Again, we restrict the theorem to make the proof more straightforward. Triangulate the planar graph (inner faces only). Pick two adjacent vertices

  • n the boundary and color them 1 and 2. Let the rest of the vertices on

the boundary have lists of size 3, and all the vertices in the interior have lists of size 5. Then the graph can be colored from its lists.

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Thomassen’s Proof - base case

Suppose that we have only 3 vertices. Then everything is on the boundary and it is obvious we can color the graph from its lists.

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Thomassen’s Proof - induction step

Suppose that we have more than 3 vertices. Case 1: There is a chord across the graph. Then split the graph in half. Color the first part by induction. Then with the two fixed colors, color the rest of the graph by induction.

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SLIDE 68

Thomassen’s Proof - induction step

Suppose that we have more than 3 vertices. Case 1: There is a chord across the graph. Then split the graph in half. Color the first part by induction. Then with the two fixed colors, color the rest of the graph by induction.

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SLIDE 69

Thomassen’s Proof - induction step

Suppose that we have more than 3 vertices. Case 1: There is a chord across the graph. Then split the graph in half. Color the first part by induction. Then with the two fixed colors, color the rest of the graph by induction.

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Thomassen’s Proof - induction step

Suppose that we have more than 3 vertices. Case 2: There is no chord across the graph. Look at the vertex v next to the one colored 1 and its neighbors. Remove v and delete the other two colors in v’s list in its interior neighbors.

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Thomassen’s Proof - induction step

Suppose that we have more than 3 vertices. Case 2: There is no chord across the graph. Look at the vertex v next to the one colored 1 and its neighbors. Remove v and delete the other two colors in v’s list in its interior neighbors.

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SLIDE 72

Thomassen’s Proof - induction step

Suppose that we have more than 3 vertices. Case 2: There is no chord across the graph. Color this graph by induction. Color v with one of the deleted colors. This gives a coloring of the whole graph.

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SLIDE 73

Thomassen’s Proof - induction step

Suppose that we have more than 3 vertices. Case 2: There is no chord across the graph. Color this graph by induction. Color v with one of the deleted colors. This gives a coloring of the whole graph.

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SLIDE 74

Thomassen’s Proof implies the Five Color Theorem

Start with a planar graph and triangulate the interior.

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SLIDE 75

Thomassen’s Proof implies the Five Color Theorem

Start with a planar graph and triangulate the interior. Put the list {1, 2, 3, 4, 5} on every vertex.

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SLIDE 76

Thomassen’s Proof implies the Five Color Theorem

Start with a planar graph and triangulate the interior. Put the list {1, 2, 3, 4, 5} on every vertex. Choose two adjacent boundary vertices to be colored 1 and 2.

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SLIDE 77

Thomassen’s Proof implies the Five Color Theorem

Start with a planar graph and triangulate the interior. Put the list {1, 2, 3, 4, 5} on every vertex. Choose two adjacent boundary vertices to be colored 1 and 2. Delete two random colors from every other list on a boundary vertex.

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SLIDE 78

Thomassen’s Proof implies the Five Color Theorem

Start with a planar graph and triangulate the interior. Put the list {1, 2, 3, 4, 5} on every vertex. Choose two adjacent boundary vertices to be colored 1 and 2. Delete two random colors from every other list on a boundary vertex. Color the graph using Thomassen’s theorem.

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SLIDE 79

Thomassen’s Proof implies the Five Color Theorem

Start with a planar graph and triangulate the interior. Put the list {1, 2, 3, 4, 5} on every vertex. Choose two adjacent boundary vertices to be colored 1 and 2. Delete two random colors from every other list on a boundary vertex. Color the graph using Thomassen’s theorem. We end up with a coloring of the graph using just 5 colors! This means the graph is 5-colorable.

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SLIDE 80

The moral of the story

Making a stronger statement or using more advanced concepts may make the theorem more straightforward to prove.

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SLIDE 81

The moral of the story

Making a stronger statement or using more advanced concepts may make the theorem more straightforward to prove. Perhaps someday, a new approach or concept will let us prove the Four Color Theorem without the assistance of a computers.

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SLIDE 82

Graphics credits

8county map color WNYGS http://www.wnygs.org/ Frederick Guthrie: http://www.wikiwand.com/en/Frederick Guthrie Augustus DeMorgan: http://www-groups.dcs.st-and.ac.uk/history/PictDisplay/De Morgan.html Alfred Bray Kempe: http://www-history.mcs.st-and.ac.uk/PictDisplay/Kempe.html Peter Guthrie Tait: http://www-history.mcs.st-and.ac.uk/PictDisplay/Tait.html Percy Heawood: http://www-history.mcs.st-and.ac.uk/PictDisplay/Heawood.html Julius Petersen: http://www-history.mcs.st-andrews.ac.uk/PictDisplay/Petersen.html Kenneth Appel and Wolfgang Haken: http://www.las.illinois.edu/100/excellence/ Gardner April Fool graph http://mathworld.wolfram.com/Four-ColorTheorem.html Carsten Thomassen http://www2.mat.dtu.dk/people/C.Thomassen/

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SLIDE 83

April Fool’s!

In 1975, Martin Gardner played an April Fool’s joke by claiming the Four Color Theorem was not true and giving the following map as a

  • counterexample. See if you can find how to color the map with four colors!

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