Discrete Math II Howon Kim 2017. 12

Agenda 11.1 Definitions 11.2 Subgraphs, Complements, & Graph Isomorphism 11.3 Vertex Degree: Euler Trails & Circuits 11.4 Planar Graphs 11.5 Hamilton Paths and Cycles 11.6 Graph Coloring and Chromatic Polynomials

Hamilton Path & Cycle Definition If G=(V,E) is a graph or multigraph with|V| 3, we say that G has a Hamilton cycle if there is a cycle in G that contains every vertex in V. (cycle that passes just once through each vertex of a given graph) A Hamilton path is a path (and not a cycle ) in G that contains each vertex. K4 Q3

Hamilton Path & Cycle Definition That is, Hamilton cycle: a spanning cycle, that is, a cycle including all the vertices of a graph. Hamiltonian graph: a graph that contains a Hamiltonian cycle. Hamilton path: a path that includes all the vertices of a graph.

Example Hamilton Cycle Non-Hamilton Cycle 정12면체 Dodecahedron

Theorem 11.7 Existence of a Hamilton path in Kn* Let Kn* be a complete directed graph. That is, Kn* has n vertices and for each distinct pair x,y of vertices, exactly one of the edges (x,y) or (y,x) is in Kn*. Such a graph (called a tournament ) always contains a (directed) Hamilton path. K5* K4* 리그전(league 戰, 문화어: 련맹전)은 스포츠 경기에서 각 팀이 다른 팀과 모두 최소 한 번씩 경기를 치르는 경기 방식이다. 영어권에서 리그라는 표현은 일반적으로 대회라는 뜻으로 사용되므로 이 방식을 지칭할 경우 라운드 로빈 토너먼트(round-robin tournament) 또는 올 플레이 올 토너먼트(all-play-all tournament)라는 표현을 사용한다. 한국어에서 일반적으로 토너먼트가 싱글 엘리미네이션 토너먼트를 가리키는 것과는 상반된다. Tournament is a directed graph obtained by choosing a direction for each edge in an undirected complete graph. The name tournament originates from such a graph's interpretation as the outcome of a round-robin tournament in which every player encounters every other player exactly once. [Wikipedia] 참고: single elimination tournament

Proof of Theorem 11.7 Complete directed graph에서의 vertex 수: n Let Pm (m  2) a path containing the m-1 edges (v1,v2), (v2,v3),…,(vm-1, vm). If m=n, we are finished. … If not, let v be a vertex that doesn’t appear in Pm. If (v,v1) is an edge in Kn*, we can extend Pm by adjoining this edge. … Exist

Proof of Theorem 11.7 (v,v1)아니면 (v1,v).둘중 하나. Complete graph 이므로. If not, then (v1,v) must be an edge. Now suppose that (v,v2) is in the graph. Then we have the larger path: (v1,v),(v,v2),(v2,v3),…,(vm-1,vm). … … Ex)

Proof of Theorem 11.7 If (v,v2) is not an edge in Kn*, then (v2,v) must be. So, we can construct the path the way that we have described just before. 9 9

Proof of Theorem 11.7 As we continue this process there are only two possibilities: (a) For some the edges (vk,v),(v,vk+1) are in Kn* and we replace (vk,vk+1) with this pair of edges: or (b) (vm,v) is in Kn* and we add this edge to pm. Either case results in a path pm+1 that includes m+1 vertices and has m edges. … (a) (b)

Proof of Theorem 11.7 This process can be repeated until a Hamilton path is found.

Traveling Salesman Problem The problem is: given a number of cities and the costs of travelling from any city to any other city, what is the least-cost round-trip route that visits each city exactly once and then returns to the starting city? It is related to the search for Hamilton cycles in a graph. It is to find a minimum-cost Hamilton cycle in a complete graph whose edges are labeled with costs

Traveling Salesman Problem Greedy Algorithm (Nearest-Neighbor Algorithm) Starting at a given vertex, chooses the edge with the least weight to the next possible vertex, that is, to the “closest” vertex. This strategy is continued at each successive vertex until a Hamiltonian cycle is completed. It cannot provide us the optimal solution. K5 14 12 10 13 8 7 11 9 6 5 e d a b c Greedy solution (40) ; a  d  b  c  e  a 7 6 8 5 14 Another solution (37) ; a  d  b  e  c  a 7 6 9 5 10

Traveling Salesman Problem Greedy Algorithm (Nearest-Neighbor Algorithm)

Agenda 11.1 Definitions 11.2 Subgraphs, Complements, & Graph Isomorphism 11.3 Vertex Degree: Euler Trails & Circuits 11.4 Planar Graphs 11.5 Hamilton Paths and Cycles 11.6 Graph Coloring and Chromatic Polynomials

Graph Coloring Definition chromatic number If G = (V,E) is an undirected graph, a proper coloring of G occurs when adjacent vertices have different colors. chromatic number The minimum number of colors needed to properly color G, which is written (G). (G) = 3

Graph Coloring There is no simple way to actually determine whether an arbitrary graph is n-colorable. However, the following statement gives a simple characterization of 2-colorable graphs. G is 2-colorable is equivalent to G is bipartite.

Coloring Complete Graphs The chromatic number of a complete graph Kn is n. (Kn) = n K3,3 ? (K5 ) = 5 K5

Coloring Planar Graph Any planar graph without loops is 4-colorable. Four Color Conjecture If the regions of any map M are colored so that adjacent regions have different colors, then no more than four colors are required. Posed in 1850s, proved by K.Appel & W.Haken in 1976 through analyzing 2,000 different cases. The proof of the above theorem uses computers in an essential way The examination of each different type of graph seems to be beyond the grasp of human beings without the use of a computer. Thus the proof, unlike most proofs in mathematics, is technology dependent.

Chromatic Polynomial Let G be an undirected graph G=(V,E), and let λ be the number of colors that we have available for properly coloring the vertices of G. Chromatic polynomial P(G,λ), λN, of graph G=(V,E) is the function whose value at λ(λ=1,2,3…) that is the number of proper colorings f:V {1,2,3,…, λ} of G with at most λ colors The chromatic polynomial function P(G, λ), in the variable λ, will tell us in how many different ways we can properly color the vertices of G, using at most λ colors.

Chromatic Polynomial If G=(V,E) with |V|=n and E=Ø, then G consists of n isolated points P(G,λ)= λ* λ*…* λ= λn If G=Kn, then at least n colors must be available for us to color G properly. P(G,λ)= λ (λ-1) (λ-2) …(λ-(n-1)) If G is a path on n vertices, then P(G,λ)= λ (λ-1)n-1

Chromatic Polynomial If G is a path on n vertices, then P(G,λ)= λ (λ-1)n-1 P(G1,λ)= λ (λ-1)3 P(G2,λ)= λ (λ-1)4 λ=1, P(G1, λ)= P(G2, λ)=0 λ=2, P(G1, λ)= P(G2, λ)=2 That is, (G1) = (G2)=2 If five colors are available, then we can properly color G1 in 5(4)3=320 ways, G2 in 5(4)4=1280 ways.

Chromatic Polynomial If G is made up of components G1,G2,…,Gk, then; P(G,λ)=P(G1,λ) * P(G2,λ)*…*P(Gk,λ)

Chromatic Polynomial Decomposition Thm. For Chromatic Polynomials If G=(V,E) is connected graph and eE, then P(Ge,λ)=P(G,λ) + P(Ge’,λ)

Chromatic Polynomial Let G=(V,E) be an undirected graph. For e={a,b} E, let Ge denote the subgraph of G obtained by deleting e from G, without removing vertices a and b; That is, Ge=G-e From Ge a subgraph of G is obtained by coalescing the vertices a and b. It is denoted by Ge’

Chromatic Polynomial P(Ge,λ)=P(G,λ) + P(Ge’,λ) P(G,λ) =P(Ge,λ)-P(Ge’,λ) = λ(λ-1)3- λ(λ-1) (λ-2) = λ(λ-1)(λ2-3 λ+3) = λ4-4 λ3+6λ2-3 λ Since P(G,1) =0 while P(G,2)=2>0, we know that (G)=2

Chromatic Polynomial Chromatic numbers for some graphs Graph G (G) Path graph Pn, n≥3 2 Cycle graph Cn, n even, n≥2 Cycle graph Cn, n odd, n≥3 3 Complete graph Kn, n even, n≥2 n Complete graph Kn, n odd n≥3 Complete bipartite graph Km,n, m, n≥1 Petersen graph Cycle graph