EECS 310: Discrete Math Lecture 6 Graph Theory Reading: MIT OpenCourseWare 6.042 Def: clique number !(G): largest k s.t. G Chapter 5.3-5.5 contains a complete graph on k vertices. Claim: χ(G) ≥ !(G) Graph Coloring Def: max degree ∆(G): degree of max-degree vertex. Matching: edges represent compatibility. Claim: χ(G) ≤ ∆(G) + 1 Coloring: edges represent conflict. Proof: By induction. 22exam scheduling, coloring a map, up-33 On what? Choices: max degree, nodes, edges. 66dates to servers where certain pairs can't77 44be taken down simultaneously since they55 cover similar critical functionality, etc. • By induction on number of vertices: Def: k-coloring: label nodes with one of k P (n) = for all k, a graph on n vertices colors so adjacent nodes get different colors. with max deg. k is (k + 1)-colorable. Question: How many colors do you need for an n-node graph of type: • Base case: 1 vertex needs 1 color, so (0+ 1) = 1-colorable. • empty graph • Inductive hypothesis: n vertices with • line graph max deg. k is (k + 1)-colorable. • cycle, n = 2k • Inductive step: • \star" graph { Let G be an (n + 1) vertex graph • bipartite graph with max deg at most k. • cycle, n = 2k + 1 { Remove v and incident edges to get 0 0 • complete graph G . Color G inductively. { Add back v and edges. Since v has Def: chromatic number χ(G): smallest k s.t. at most k neighbors, must be an G is k-colorable. available color. ""NP-hard to find smallest, if you can you## get clay prize of $1M. But can bound de- gree based on other properties. Ideas? 2 1 Paths and Walks { case 1: fi; jg 2 E then Aij = 1 and there's 1 walk Def: walk: sequence of vertices from i to j of length one. of G, v0; : : : ; vk, and edges of G, { case 2: fi; jg 62 E fv0; v1g;:::; fvk−1; vkg: then Aij = 0 and there's no walk • may repeat vertices or edges from i to j of length one. • Inductive Step. • starts at v0 { Let P k be number walks of length • ends at vk xy k from x to y. • length is k (number of times traverse an k k { By I.H., Ptj is tj'th entry of A , call edge) k it Atj. Def: path: walk where all vi's are distinct { Group walks from i to j by first edge fi; tg: Example: Draw graph and specify a walk and a path. k+1 X k Pij = Ptj Claim: If there's a walk, then there's a path. t:fi;tg2E Proof: well-ordering, see text { Since A = 1 iff fi; tg 2 E, Note: length of path at most length of walk it n k+1 X k Pij = AitPtj Numbers of walks t=1 [[how many ways to get from here to there?]] { Using I.H., Def: adjacency matrix A: Aij = 1 if fi; jg 2 n E, 0 otherwise k+1 X k Pij = AitAtj Example: adjacency matrix of square with t=1 one diagonal k+1 { Which is Aij by matrix multipli- Claim: Number of walks of length k from i cation. to j is ij'th entry of Ak. Example: square and cube adjacency ma- Question: Find length of shortest paths? trix of previous example k compute powers of A until Aij > 0 22proof gives insight to relationship be-33 66tween adjacency matrix multiplication77 44and numbers of walks, important ingre-55Connectivity dient in pagerank. Proof: By induction on k. ""can we get from here to there? can each## node in the network send packets to each • Base case, k = 1. other node? 2 Def: i; j connected if there is a path from i components merge when add to j back, so G has n − m − 1 = Note: i connected to itself by convention n − (m + 1) connected compo- nents. Def: G connected if all pairs of nodes con- nected ""induction on edges and on nodes very## Example: Draw a disconnected graph, two common, we've seen both this lecture, edges and a triangle. questions? Def: connected components: subgraph of G Note: In inductive step, take (m + 1)-edge consisting of a node and every node con- graph (or (n + 1)-node graph) and delete ele- nected to it. ment. Bounds on number connected components in graph G of n nodes? 22Bounds on number of edges in connected33Build-up Error 66graph G of n nodes? How about G with n77 44connected components? 2 connected com-55Claim: If every node has degree ≥ 1, G is ponents? connected. Claim: Every graph with n vertices and m Question: Counterexample? edges has at least n − m connected compo- [[Graph of two edges ]] nents. Example: draw graph on two edges Proof: Induction on m. Let P (m) = 8n 2 Proof: Use induction on n. N;G with m edges has n−m connected com- ponents. • Let P (n) = if every node in n-vertex • Base case: m = 0, each vertex is con- graph has degree ≥ 1, then G is con- nected component so there are n con- nected. nected components. • Base case: only one 1-node graph, degree • Inductive step: Assume for every m-edge zero, statement vacuously true. graph. Consider (m + 1)-edge graph. • Inductive step: { remove arbitrary edge fi; jg { by induction, remaining graph G0 { Consider n-node graph G0 where has n − m connected components each vertex has degree 1. { add back edge { By assumption G0 connected, i.e., ∗ if i; j in same connected compo- there's a path between each i; j in 0 nent of G0, then G has n−m > G n − (m + 1) connected compo- { Now add one more node k with de- nents. gree one to get (n + 1)-node graph ∗ if i; j in different connected G. This node must connected to a components of G0, then two node, say l, in G0. 3 { Then G is connected since for any i; j 6= k we can use path from G0 and for i = k, we can use edge fk; lg with path from l to j in G0. 22Where's error? We can't get every33 66(n + 1)-node graph with min-degree 1 by77 44adding to n-node graph with min-degree55 1! Example above is counter-example. MANTRA: SHRINK-DOWN, GROW-BACK. [[repeat out loud ]] Note: if we shrink-down grow-back in above, • Inductive step: { Consider (n + 1)-node graph G where each node has degree ≥ 1 { Delete a node to get G0 { Then in G0 each node has degree ≥ ::: uh oh! 4.