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

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, 2016 2 / 66

Example Camenga (Penn State) 1/19/16 January 19, 2016 3 / 66

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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. Camenga (Penn State) 1/19/16 January 19, 2016 12 / 66

Good, better, best We can do better! Camenga (Penn State) 1/19/16 January 19, 2016 13 / 66

Good, better, best We can do better! Camenga (Penn State) 1/19/16 January 19, 2016 14 / 66

Good, better, best We can do better! Camenga (Penn State) 1/19/16 January 19, 2016 15 / 66

Good, better, best We can do better! Camenga (Penn State) 1/19/16 January 19, 2016 16 / 66

Good, better, best We can do better! Camenga (Penn State) 1/19/16 January 19, 2016 17 / 66

Good, better, best We can do better! Camenga (Penn State) 1/19/16 January 19, 2016 18 / 66

Good, better, best We can do better! Camenga (Penn State) 1/19/16 January 19, 2016 19 / 66

Good, better, best We can do better! This only needs three colors. So we say the chromatic number is 3. Camenga (Penn State) 1/19/16 January 19, 2016 20 / 66

Timeline for the Four Color Theorem The problem was posed to Augustus DeMorgan by 1852 one of his students, Frederick Guthrie, in 1852. 1879 & 1880 1890 & 1891 1976 Frederick Guthrie Augustus DeMorgan Camenga (Penn State) 1/19/16 January 19, 2016 21 / 66

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 out a stupid animal, I think I must do as the Sphynx did.... DeMorgan writing to Hamilton Camenga (Penn State) 1/19/16 January 19, 2016 22 / 66

Timeline for the Four Color Theorem A proof was first given in 1879 by Alfred Bray Kempe 1852 and another in 1880 by Peter Guthrie Tait. 1879 & 1880 1890 & 1891 1976 Alfred Bray Kempe Peter Guthrie Tait Camenga (Penn State) 1/19/16 January 19, 2016 23 / 66

Timeline for the Four Color Theorem Gaps were found in Kempe and Tait’s proofs by Percy 1852 Heawood in 1890 and Julius Petersen in 1891, respec- tively. Oops! 1879 & 1880 1890 & 1891 1976 Percy Heawood Julius Petersen Camenga (Penn State) 1/19/16 January 19, 2016 24 / 66

Timeline for the Four Color Theorem After work by many mathematicians, the theorem 1852 was finally proved in 1976 by Kenneth Appel and Wolfgang Haken. It took 1200 hours of computer time. 1879 & 1880 1890 & 1891 1976 Camenga (Penn State) 1/19/16 January 19, 2016 25 / 66

Modeling the problem We will represent every country by a point, called a vertex : Camenga (Penn State) 1/19/16 January 19, 2016 26 / 66

Modeling the problem We connect two vertices with an edge if the two corresponding countries share a boundary. Camenga (Penn State) 1/19/16 January 19, 2016 27 / 66

Modeling the problem We connect two vertices with an edge if the two corresponding countries share a boundary. Camenga (Penn State) 1/19/16 January 19, 2016 28 / 66

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. Camenga (Penn State) 1/19/16 January 19, 2016 29 / 66

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 . Camenga (Penn State) 1/19/16 January 19, 2016 30 / 66

Five Color Theorem We will prove the Five Color Theorem : Every planar graph is 5-colorable. Camenga (Penn State) 1/19/16 January 19, 2016 31 / 66

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. Camenga (Penn State) 1/19/16 January 19, 2016 32 / 66

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. Camenga (Penn State) 1/19/16 January 19, 2016 33 / 66

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 F = 2 3 F = 2 E or 3 E . Camenga (Penn State) 1/19/16 January 19, 2016 33 / 66

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 F = 2 3 F = 2 E or 3 E . Euler’s relation tells us that V − E + F = 2. Substitute into Euler’s relation: V − E + 2 3 E = 2 or V − 1 3 E = 2. Solving for E , we get that E = 3 V − 6 . Camenga (Penn State) 1/19/16 January 19, 2016 33 / 66

Kempe & Heawood proof - a key fact Assumption: Suppose all the vertices have degree at least 6. Camenga (Penn State) 1/19/16 January 19, 2016 34 / 66

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 6 V . Camenga (Penn State) 1/19/16 January 19, 2016 34 / 66

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 6 V . But when we add up all the degrees, we count each edge twice, once from each end, to get exactly 2 E . Camenga (Penn State) 1/19/16 January 19, 2016 34 / 66

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 6 V . But when we add up all the degrees, we count each edge twice, once from each end, to get exactly 2 E . Therefore 2 E ≥ 6 V or E ≥ 3 V Camenga (Penn State) 1/19/16 January 19, 2016 34 / 66

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 6 V . But when we add up all the degrees, we count each edge twice, once from each end, to get exactly 2 E . Therefore 2 E ≥ 6 V or E ≥ 3 V Compare our two results: E = 3 V − 6 E ≥ 3 V Camenga (Penn State) 1/19/16 January 19, 2016 34 / 66

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 6 V . But when we add up all the degrees, we count each edge twice, once from each end, to get exactly 2 E . Therefore 2 E ≥ 6 V or E ≥ 3 V Compare our two results: E = 3 V − 6 E ≥ 3 V 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. Camenga (Penn State) 1/19/16 January 19, 2016 34 / 66

Kempe & Heawood proof - base cases Suppose we have 5 or fewer vertices in our graph. Is the graph 5-colorable? Camenga (Penn State) 1/19/16 January 19, 2016 35 / 66

Kempe & Heawood proof - base cases Suppose we have 5 or fewer vertices in our graph. Is the graph 5-colorable? OF COURSE! Camenga (Penn State) 1/19/16 January 19, 2016 35 / 66