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MC302 GRAPH THEORY Thursday, 9/5/13 Today: Graph degrees and edges The Havel-Hakimi Theorem Reading: [CH] 1.4 [HR] Rest of 1.1, 1.2 Hand-in HW #1 will be on our website by this evening Exercises: Due Friday 9/13/13 by 3 PM [CH] 1.4.2, 1.4.9 [HR] 1.1.8, 1.1.9 Thursday, 9/5/13, Slide #1 Exercise: Counting edges and degrees 1. How many edges are there in the complete graph Kn, as a function of n? 2. How many edges are there in the complete bipartite graph Kp,q, as a function of p and q? 3. For each of Kn and Kp,q, what are all the vertex degrees? 4. If G is any graph, and we add up the vertex degrees, how are the answers related to the numbers of edges in G? Thursday, 9/5/13, Slide #2 The Degree-Sum Theorem First Theorem of Graph Theory (Degree-Sum Theorem): The sum of all the vertex degrees equals twice the number of edges. Symbolically, n ∑ d ( v ) = 2e k k =1 Is this also true for non-simple graphs, i.e., graphs with parallel edges and loops? Corollary: In any graph there is an even number of vertices with odd degree. Proof? Thursday, 9/5/13, Slide #3 Regular Graphs A regular graph is one in which all vertices have the same degree. Exercises: 1. Is Kn regular? Is Kp, q regular? 2. If a graph with n vertices is d-regular, how many edges does it have? 3. If two graphs on n vertices are both d-regular for some d, must they be isomorphic? 4. For each k = 0, ..., 5, draw a k-regular graph on 5 vertices, or explain why one cannot exist. Thursday, 9/5/13, Slide #4 Degree Sequences The degree sequence of a graph is the sequence of its vertex degrees, usually written in decreasing order. Exercises: 1. What’s the degree sequence of the graph below? 2. A graphic sequence is a sequence that is the degree sequence of some graph. Which of the following are graphic sequences? 5, 4, 3, 2, 2 5, 4, 3, 2, 2, 1 5, 4, 3, 3, 2, 2, 1 Thursday, 9/5/13, Slide #5 Havel-Hakimi Theorem Theorem. Let S =(d1, d2, …, dn) be a sequence of n integers written in decreasing order: d1 ≥ d2 ≥ … ≥ dn-1 ≥ dn. Then S is a graphic sequence if and only if the following sequence S’ of n-1 integers is graphic, where k = d1: S’ = (d2-1 , d3-1 , … , dk+1-1 , dk+2 ,…, dn-1 , dn) Notes: S’ is obtained from S by removing the first number d1, and then subtracting 1 from each of the next d1 numbers of S. To apply the theorem, S must be in decreasing order. However, the set S’ may not be in decreasing order. Exercise: For the graph on the right, compute S and S’. Thursday, 9/5/13, Slide #6 Applying Havel-Hakimi Given sequence S with n = 9: S = (6, 5, 5, 4, 3, 3, 2, 2, 2) Apply H-H to S to get S’ with n = 8: S’ = (5-1, 5-1, 4-1, 3-1, 3-1, 2-1, 2, 2) = (4, 4, 3, 2, 2, 1, 2, 2) Put S’ in decreasing order : S’ = (4, 4, 3, 2, 2, 2, 2, 1) Apply H-H again, getting set with n= 7. Keep repeating until answer is clearly yes or clearly no! Thursday, 9/5/13, Slide #7 Generating a graph with a given degree sequence If we reach a sequence S’ that is graphic, we can reverse the process of H-H, adding vertices as we go: When we add back a vertex corresponding to a number we’ve removed, we must remember to attach it to the corresponding vertices whose degrees were reduced. 4, 4, 3, 2, 2, 2, 2, 1- - > 3, 2, 1, 1, 2, 2, 1 = 3, 2, 2, 2, 1, 1, 1- - > 1, 1, 1, 1, 1, 1 1, 1, 1, 1, 1, 1 Thursday, 9/5/13, Slide #8 “If and only if” theorems If a theorem says “A if and only if B,” where A and B are statements, this means: “If A then B” and “If B then A” “If A then B” is called “necessity:” “If B then A” is called “sufficiency:” If A is true, then B must (necessarily) be true. If B is true that is sufficient evidence that A is also true. Havel-Hakimi says “S is graphic if and only if S’ is graphic” The construction on previous slide give idea of one proof direction – which one? Thursday, 9/5/13, Slide #9 Proof of Necessity: If S is graphic, then S’ is graphic It would be nice if we could say: If S is graphic, just delete the highest degree vertex, leaving a graph with degree sequence S’. Thus S’ is graphic This might be true, but it doesn’t have to be. For the graph below, compute S and S’, and then delete the highest degree vertex and compute the degree sequence of the new graph. Thursday, 9/5/13, Slide #10 Proof of Necessity (If S is graphic, then S’ is graphic) Assume S = (d1 ≥ d2 ≥ … ≥ dn) is graphic. Prove that S’ = (d2-1 , d3-1 , … , dk+1-1 , …, dn-1 , dn) is graphic. By assumption, there is some graph G with V(G) = {v1, v2, …, vn} such that deg(vi) = di for each i = 1, …, n. Let k = d1 = deg(v1). If N(v1) = {v2, …, vk+1}, we’re done. Otherwise, repeatedly “adjust” G to get new graph with N(v1) = {v2, …, vk+1} Thursday, 9/5/13, Slide #11 Details of “adjusting” G 1. 2. 3. 4. 5. 6. Assume: v1 is not adjacent to at least one vi in {v2, ..., vk}. Then: There must be some vj in {vk+1, ..., vn} that is adjacent to v1 – why? If deg(vi) = deg(vj), we just switch their labels to make v1 adjacent to vi. Otherwise, deg(vi) > deg(vj) – why? So there is another vertex w, adjacent to vi, but not to vj. Now we remove and add edges to make v1 adjacent to vi, without changing the degree of any vertex. Keep repeating this process as long as v1 is not adjacent to some vertex in S. See file HavelHakimi.pdf for write-up of this proof. Thursday, 9/5/13, Slide #12

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