Edge colorings Edge coloring problems often arise when objects being scheduled are pairs of underlying elements, e.g., a pair of teams that play each other, a pair consisting of a teacher and a class, etc. (See slides for an example.) Definition A k-edge-coloring of a graph G is a function f : E(G)!{1, 2,...,k}. Itisproper if edges that share an endpoint have di erent colors, i.e., e 1 \ e 2 6= ; =) f(e 1 ) 6= f(e 2 ). G is said to be k-edge-colorable if it has a proper k-edge-coloring. Definition The minimum k such that G has a proper k-edge-coloring is called the edge chromatic number of G and is denoted 0 (G). Remark Graphs with loops cannot be properly colored, so we consider loopless graphs only. Question What are easy upper and lower bounds on 0 (G)? Proposition. For any graph G, (G) apple 0 (G) apple E(G). The above lower bound may seem trivial. However, this bound holds with equality for large classes of graphs in particular, for bipartite graphs, which we will prove. Examples For a cycle C n, (C n )=2and 0 (C n )= ( 2 if n even 3 if n odd. Math 104 - Prof. Kindred - Lecture 15 Page 1
The complete graph K 2n has (K 2n )=2n 1. Then we can color the edges of K 8 as illustrated below: Exercise outside of class Show that 0 (K 2n+1 )=2n +1. Remark Note that in vertex coloring, each color class was an independent set. In the context of edge coloring, each color class is a matching. n Recall that (G) (G) since color classes are independent sets for vertex coloring. Can we obtain an analogous result for 0 (G)? Proposition. For any graph G, 0 (G) E(G) 0 (G). Fundamental theorem coming up =) 0 (G) isalwaysverycloseto (G) for simple graphs G. Theorem (Vizing). Let G be a simple graph (no loops, no multiple edges). Then (G) apple 0 (G) apple (G)+1. We omit the proof of the upper bound, but you can find it in the textbook (algorithmic proof). Remark The simple graphs G for which 0 (G) = (G) arecalledclass 1 graphs, and those for which 0 (G) = (G)+1 are called Class 2 graphs. The problem of deciding to which class a given graph belongs is very hard (NP-complete). Math 104 - Prof. Kindred - Lecture 15 Page 2
Question Does this result hold for graphs with multiple edges (loopless graphs)? Counterexample: fat triangle Given 3 vtcs, say there are l edges between each pair of vtcs. (So the graph has 3l edges total.) Then (G) =2l and 0 (G) =3l. The generalization of Vizing s theorem for loopless graphs (or multigraphs without loops) is the following. Theorem. For any (loopless) graph G, (G) apple 0 (G) apple (G)+µ(G) where µ(g) =max uv2e(g) µ(u, v) and µ(u, v) is the # of edges with endpoints u and v. So µ(g) isthemaximumoftheedgemultiplicitiesing. Our next goal is to show that bipartite graphs are Class 1 graphs that is, (G)-edge-colorable. We first need a lemma. (G)-regular bipartite super- Lemma. Every bipartite graph G has a graph (a graph containing G). Proof. Let G be an X, Y -bipartite graph with k = supergraph G 0 of G can be constructed as follows. (G). A large k-regular Clone G. If G is not regular, add a vertex to X for each vertex of Y and a vertex to Y for each vertex of X. Onthenewvertices,construct another copy of G. Math 104 - Prof. Kindred - Lecture 15 Page 3
Add edges between vertex and its clone, as needed. For each v 2 V (G) withd G (v) <k,joinitstwocopiesinthenewgraph with k d G (v) edgestogetg 0. Now G 0 is a k-regular bipartite supergraph of G. (G 0 is connected if G was connected.) See slides for an example of above construction. Remark If G is a simple bipartite graph, then we can use a variation of the construction above to obtain a simple (G)-regular bipartite supergraph of G. Replace 2nd step above with the following: For each v 2 V (G) with d G (v) <k,joinitstwocopieswithasingleedgeinthenew graph to get G 0. Then (G 0 )= (G) =k, (G 0 )= (G)+1, and G 0 is a supergraph of G. Iteratingthe2-stepprocessk times yields the desired simple bipartite supergraph H. (G) Recall Hall s Marriage Thm: Theorem. For k 2 N, every k-regular bipartite graph has a perfect matching. Math 104 - Prof. Kindred - Lecture 15 Page 4
Theorem (König). If G is a bipartite graph, then 0 (G) = (G). Proof. Since 0 (G) (G) foranygraphg, itissu cienttoshowthat 0 (G) apple (G) wheng is bipartite. We give a proof by induction on (G). Base case: Assume (G) =1. Thennotwoedgesshareanendpoint (since such a shared vertex would have degree at least 2), and so every edge of G can be assigned the same color. Thus, 0 (G) apple 1= (G). Induction hypothesis: Suppose any bipartite graph G with (G) =k has 0 (G) apple (G). Suppose G is a bipartite graph with (G) =k +1. Bypreviouslemma, there exists a (G)-regular bipartite graph H that contains G. ByHall s Marriage Thm, H has a perfect matching M. ThenH M is a k-regular bipartite graph and so by the IH, H M satisfies 0 (H M) apple (H M) = (H) 1=k. Consider any k-edge-coloring of H M. Extendittoanedge-coloringof H by assigning all edges of M color k +1. Thisisaproper(k +1)-edgecoloring of H since no two edges of M share an endpoint. Since any subgraph of H must then be (k +1)-edge-colorable, we have that 0 (G) apple k +1. Hmwk problem 7.1.21 asks you to give an algorithmic proof of König s Thm. The proof yields a polynomial-time algorithm for constructing a (G)-edge-coloring of a bipartite graph G. Math 104 - Prof. Kindred - Lecture 15 Page 5
Line graphs Definition Given a graph G, theline graph of G, denotedl(g), is the graph whose vertices are the edges of G and ef 2 E(L(G)) () e and f are edges in G with a common endpoint. Alinegraphistheintersection graph of the edges of G. Example G L(G) Observations If e = uv is an edge in G, thend L(G) (e) =d G (u)+d G (v) 2. The numbers of vertices and edges in L(G) are V (L(G)) = E(G) and E(L(G)) = X v2v (G) dg (v) 2 The line graph of a connected graph G is connected. (Note that the converse is not necessarily true: a disconnected graph may have a connected line graph.) For a simple graph G, verticesformacliqueinl(g) ifandonlyifthe corresponding edges in G have one common endpoint (a star) or form atriangle.(thus,!(l(g)) = (G) unless (G) =2andG contains atriangle.) Math 104 - Prof. Kindred - Lecture 15 Page 6
The only connected graph that is isomorphic to its line graph is a cycle graph C n for n 3. For a graph G, 0 (G) = (L(G)). So edge coloring is really a special case of vertex coloring. We can characterize the graphs that exist as line graphs. A graph H is the line graph of some other graph if and only if it is possible to find a collection of cliques in H, partitioningtheedgesofh, suchthateach vertex of H belongs to at most two of the cliques. Aforbiddensubgraphcharacterizationexistsforlinegraphs,namelya set S of nine graphs exists such that a simple graph G is a line graph of some simple graph if and only if G does not have any graph in S as an induced subgraph. This characterization yields an algorithm to test whether or not G is alinegraphinpolynomialtime(inn = V (G) ). Math 104 - Prof. Kindred - Lecture 15 Page 7
Edge coloring and line graphs Math 104, Graph Theory March 14, 2013 Scheduling problem Problem In a school, there are a teachers,,...,x a and b classes,,...,y b. Given that teacher x i is required to teach class y j for p ij periods, schedule a complete timetable in the minimum # of possible periods.
Edge-coloring a complete graph Building a bipartite, regular supergraph
Forbidden subgraph characterization for line graphs