March 2014 Graph Coloring - PowerPoint Presentation

March 2014 Graph Coloring 1 Graph Coloring prepared and Instructed by Shmuel WimerEng. Faculty, Bar-Ilan University

Vertex Coloring March 2014 Graph Coloring 2 A -coloring of a graph is a labeling . A coloring is proper if no two vertices and connected with an edge have same color, i.e. .   is -colorable if it has proper -coloring.   The chromatic number is the smallest such that has proper -coloring. is called -chromatic.   If , but for every proper subgraph , then is -critical.  

March 2014 Graph Coloring 3 The vertices having same color in a proper -coloring must be independent. Therefore, is the minimum number of independent sets covering . Hence, is -colorable iff is -partite.   Examples. Every bipartite graph is 2-colorable.Every even cycle graph is 2-colorable (it is bipartite).Every odd cycle graph is 3-colorable and 3-critical. 2-colorability can be tested with BFS. (how?) We compute the distance from a vertex . A connected graph is bipartite iff and are independent sets, where and are vertices of even and odd distance from , respectively.  

March 2014 Graph Coloring 4 The largest clique size satisfies .The largest independent set size satisfies , since every color class is an independent set, therefore having at most vertices.   Is possible?     Yes! Proper coloring of requites colors . , but hence .   Could it be constructed with rather than ?              

March 2014 Graph Coloring 5 Example. Minimizing exam period in school . How to schedule exams in minimum parallel sessions, where no two concurrent exams have a common student?Define , where corresponds to course, and iff courses and have a common student.An independent set of vertices implies a parallel exam session. is the smallest number of parallel sessions.  Example. Chemical storage . Store different chemicals. The interaction between some pairs is explosive. What is the smallest required number of compartments in the storage? .  

Upper Bounds of Chromatic Number March 2014 Graph Coloring 6 Easy bounds are , , and , all hold with equality for cliques.  Better than upper bounds can be obtained by coloring algorithms.   A greedy algorithm w.r.t assigns to the smallest color index not incident so far to .  Proposition. There is . ( is the largest vertex degree.)  Proof. By construction. A vertex has no more than neighbors. Upon coloring there must be at least one of colors unused. ■  

March 2014 Graph Coloring 7 Different orderings may yield smaller upper bounds. Finding the best ordering is hard. Is there an ordering yielding ? It can be shown that such exists.  Example. Register allocation and interval graphs.Consider the registers used by a compiler, each has start and end time. What is the smallest number of physical registers that can be used? Assign the symbols , , , … to the registers in the code, and draw their usage time intervals.                 Proposition . If is an interval graph then .  

March 2014 Graph Coloring 8 Proof. By left-to-right traversal of the time intervals, pre sorted by their starting time. Initializing . Increasing to at starting point and decreasing to at ending point. ■ The bound may still be very poor.   For -vertex star , whereas .   For -vertex wheel , whereas .  

March 2014 Graph Coloring 9 The bound can be further improved by considering the vertices with high degree first.  Proposition. (Welsh-Powell 1967) If the vertices are ordered in non increasing degree, , then .   Proof . When vertex is colored, its already colored neighbors have at most distinct colors.   Its (proper) color is therefore . Maximization over yields the upper bound. ■  

March 2014 Graph Coloring 10 Lemma. If is -critical graph, then . Proof. Assume in contrary that . Let be a vertex for which .   We colored properly with colors, which contradicts with being -critical graph ( . ■  Since is -critical, is by definition -colorable. Use any colors to color properly .Since , consume colors at most. Let us color by one not consumed by .   The minimum degree in can also be used to deduce upper bounds.  

March 2014 Graph Coloring 11 Corollary. (Szekeres-Wilf 1968)   Proof. Let () and be a -critical subgraph of .By the above lemma   There is also , yielding the desired bound. ■   Show a graph where . (homework)  

  Coloring of Directed Graphs March 2014 Graph Coloring 12 Theorem . Let a graph be directed with longest path , then . Furthermore, there are orientations of ’s edges such that equality holds.   Proof . Let be a maximal acyclic sub digraph of (not necessarily a tree).   must have some vertices with outgoing arcs only.    

March 2014 Graph Coloring 13 Define to be a coloring function assigning color to vertex (longest path from an outgoing vertex).  strictly increases along a path in using the colorson .   For each edge there exists a path in between and , since either there was or is closing a cycle of .       1 1 2 3 4 5 6

March 2014 Graph Coloring 14 That implies since increases along paths of . Consequently, is a proper coloring and .  To prove the existence of an orientations of ’s edges satisfying , an orientation satisfying is shown.   Each edge is oriented iff . Since is a proper coloring, this defines an orientation.  Let be an optimal coloring satisfying . We derive a digraph as follows.  

March 2014 Graph Coloring 15 Since the color labels along paths in strictly increase, and there are only labels, there is , hence . ■   1 2 3 4    

Brooks’ Theorem March 2014 Graph Coloring 16 The bound holds for any graph.Brook showed that cliques and odd cycles are essentially the only graphs where holds.  Theorem. (Brooks 1941) If is connected and other than a clique or an odd cycle, then .   Proof . Let have nodes and be connected, neither a clique, nor an odd cycle. Let and assume , as otherwise for it is single edge, and cycle for. 

March 2014 Graph Coloring 17 Consider first the case where is not k-regular.Choose a vertex for which and grow a spanning tree rooted at (by any search, e.g. BSF).  Index the vertices in decreasing order as they are being reached by the search, yielding the order .   Each vertex other than has a higher-indexed neighbor along its path to root, hence it has at most lower-indexed neighbors.   Using the greedy coloring with the above vertex ordering (starting at leaves) obtains proper -coloring .  

March 2014 Graph Coloring 18 In the remaining cases is -regular. 3 cases possible.  1st case: is 1-connected. Let be a cut-vertex.        Let be a component of together with . The degree of in is less than and a proper -coloring is possible.  

March 2014 Graph Coloring 19 That can repeat for every components of , yielding -proper coloring for each ( included).  By permuting colors of the subgraphs, we can make the colorings agree on , yielding -proper coloring of . Find a vertex with two non adjacent neighbors and (why such exist?) whose deletion leaves a connected subgraph (otherwise G was 2-conncted).         2 nd case: G is not 2-connected .

March 2014 Graph Coloring 20 is connected and a spanning tree rooted at can be constructed (e.g. BFS).The labels are assigned to the vertices in decreasing order as they are reached.  Starting coloring from and , they use same color.   Each vertex other than has at most lower-indexed neighbors so colors can be used for those.   All in all, proper coloring of has been obtained.   has neighbors, of which and already used the same color. The rest neighbors used at most other colors, and can therefore be properly colored.  

  March 2014 Graph Coloring 21 3 rd case: is 2-connected . Choose a vertex such that vertex connectivity . T hat is possible by choosing to be one of the two disconnecting vertices.   has a neighbor in every block of obtained by deleting the 2nd vertex in a cut-set, otherwise was 1-connected rather than 2-connected.      

March 2014 Graph Coloring 22 There is no edge connecting and since they reside in different blocks.   is connected since blocks have no cut-vertices and are not such.   implies is also connected.   All in all this is the same situation as the case of not 2-connected. ■   Brooks’ Theorem implies that the cliques and the odd cycles are the only -regular -critical graphs. (homework) 

March 2014 Graph Coloring 23 Example. Prove that for any graph , there is a partition , ( ) , , such that .   Proof . Consider any coloring of with colors. Pick color classes and denote by the vertices of these color classes. Let . The above construction yields a proper coloring of by colors and proper coloring of by colors.Consequently, and . 

March 2014 Graph Coloring 24 On the other hand, cannot be colored with less than colors.Otherwise, together with the coloration of by colors, could be colored with less than colors, which is impossible. Thus.  Similarly and symmetrically . In conclusion .  

Chromatic Polynomials March 2014 Graph Coloring 25We shall associate with any graph a function telling whether or not it is 4-colorable. This study was motivated by the hope to prove the Four-Color Theorem, which by that time was a conjecture.Let denote the number of proper colorings of a graph with colors. is called the chromatic function of .  Example. . The first vertex can be colored in ways, while each of the other two in ways.  

March 2014 Graph Coloring 26 For there is and for there is .   If then . For there is .   The F our-Color Theorem for planar graph states that .  It is difficult to compute by inspection, but it can be systematically obtained as a sum of chromatic functions of complete graphs. For a tree of vertices there is .  

March 2014 Graph Coloring 27 Theorem. Let be not adjacent, and let and be obtained from by adding the edge , and by identifying and , respectively. Then .                            

March 2014 Graph Coloring 28 28 Proof . In a proper coloring of , and may have either the same color or different colors.  The number of proper colorings where and have different colors does not change if an edge would exist, yielding .   Similarly, the number of proper colorings where and have same color does not change if and are merged, yielding . ■   Corollary. The chromatic function is a polynomial.Proof. The procedure of the theorem results in two graphs. In the number of edges is increased. In the number of vertices is decreased. 

March 2014 Graph Coloring 29 The process is finite. It ends with producing complete graphs, whose chromatic functions are polynomial.The chromatic function is therefore a finite sum of polynomials, which must be polynomial too. ■For -vertex graph the degree of is , the coefficient of is 1 and that of is , the sign of the coefficients is alternating, and the free coefficient is zero. (homework)  Example . Scheduling feasibility . Lectures scheduling is in order, for which some time slots are given (e.g. campus is open). There is no limit on available rooms.

March 2014 Graph Coloring 30 It is known that some lectures cannot take place in parallel (e.g. some students are registered to both).Is scheduling feasible? How many schedules there are? Solution. Define a graph where corresponds to a lecture, and corresponds to lectures that cannot be scheduled simultaneously. Derive the chromatic polynomial , where is the number of time slots.Given , evaluation of yields the number of distinct schedules. indicates non existence of feasible scheduling. ■  

March 2014 Graph Coloring 31   Example. = + = + + +

March 2014 Graph Coloring 32 = + + + 2  

March 2014Graph Coloring 33 Show that if and are disjoint (no common vertices) then . (homework)Show that if is complete then . (homework)  

Edge Coloring March 2014 Graph Coloring 34 Edge coloring partitions into sets (some possibly empty) . A -edge-coloring of a graph is a labeling .   An edge coloring is proper if adjacent edges have different colors. All coloring henceforth are assumed proper.

March 2014 Graph Coloring 35 Edge coloring thus partitions into sets of matchings. (Only loopless graphs admit proper edge coloring).   is -edge-colorable if it has -edge-coloring.   The edge chromatic number is the smallest such that has -edge-coloring. is called -edge-chromatic.Clearly, .  Clearly, is -edge-colorable, where .   Not 3-edge-colorable, hence .  

March 2014 Graph Coloring 36 Example. Timetabling. teachers …, and classes …, are given. Teacher is required to teach class a lessen of period . Schedule a complete timetable having minimum total duration.   Solution . The scheduling is represented by a bipartite graph , , vertices and are connected with parallel edges. The minimum number of colors required for edge-coloring ensures minimum duration.No schedule overlap for a teacher.No lesson overlap for a class. ■ 

March 2014 Graph Coloring 37 Homework. Show that a -regular graph is -edge colorable iff can be partitioned into perfect matchings.Show by an appropriate edge coloring that . Given graph , , , show that . Eight schoolgirls go for a walk in pairs every day. Can they arrange their outgoing so that every girl has a different companion at every day of the week?  

Edge Coloring of Bipartite Graphs March 2014 Graph Coloring38 Let the subgraph span ( ), and be a -edge-coloring of . A color is available for an edge if it is available in its two end vertices. If is uncolored, any of its available colors can be assigned to extend to a -edge-coloring of .   For , each component of is either an even cycle or a path (called -path). (why?)   Theorem. If is bipartite then .  

March 2014 Graph Coloring 39 Proof. By induction on . Let . Assume that has a .   If a color is available for we are done. Otherwise, each of the colors is represented either at or at .   Since the degrees of and in are at most, there are colors , where is available at and exists in , and is available at and exists in .  Consider the subgraph . Because has a degree one in , the component containing is an -path .  

March 2014 Graph Coloring 40 cannot terminate at . If it did, it would started from with color and end at with color , hence comprising even number of edges.  would then be an odd cycle in , impossible for a bipartite graph.  Interchanging the colors of , a new -edge-colorable is obtained, where color is available at both and .   Assigning color to obtains a -edge-coloring of . ■                 

March 2014 Graph Coloring 41 Clearly, , and for bipartite graphs there is .What can be said about an upper bound? Surprisingly, it is very tight. Theorem. (Vizing 1964, Gupta 1966). Let be a simple graph (no parallel edges, loopless). Then   Proof . Let be a proper subgraph of , edge-colored with colors, but could not be colored. We present a recoloring procedure to include .   Upper Bound of  

March 2014 Graph Coloring 42 Since more than colors are used, every vertex has a missing color.  Let be missing at and be missing at . ( must be presented at and at .)             Let be a neighbor of such that is colored . Some color must be missing at since colors are used.  

March 2014 Graph Coloring 43 Suppose does not appear on . We could recolor with , free from , and then color with .         So we suppose that appears on . The process continues for .         Finding a new color that appears at , let be the neighbor of such that the edge is colored .  

March 2014 Graph Coloring 44 At some color must be missing.                                       If is missing at , we downshift color from to for .   We are finished, unless appears at , in which case the process continues to and a color .   There are only colors, hence the repetitive selection of must eventually repeat a color.  

March 2014 Graph Coloring 45 Let be the shortest non repetitive color list such that is missing at and repeats one of .               Let for some . This color was missing at and appeared on .       If is missing , we use on and downshift colors from to complete the augmentation.  

March 2014 Graph Coloring 46 Hence we assume that appears at and does not. Let be the longest alternating path of edges colored and that begins at (with ). is unique. (why ?)     Depending on the opposite end of , recoloring can take place to complete the augmentation. There are three possibilities of end: at , at , and elsewhere.                  

March 2014 Graph Coloring 47 If ends at , it is with since is colored with .  Interchange colors along , color with , and downshift colors from . This completes the edge coloring augmentation.                                                

March 2014 Graph Coloring 48 If ends at , it is with since is missing from .  Interchange colors along , color with , and downshift colors from . This completes the edge coloring augmentation.                                              

March 2014 Graph Coloring 49 Finally, suppose that neither ends at nor , so it ends at some vertex outside                           ends with edge colored either by or , so either of or is missing from the far end of , as otherwise would not be longest.  

March 2014 Graph Coloring 50 Notice that the vertices along can also touch any of , since has been assumed to present at these (otherwise coloring downshift had been possible). We downshift from , assign color to , and interchange colors along . ■                                                  

Line Graphs March 2014 Graph Coloring 51Many questions about vertices have natural analogues involving edges. Definition. The line graph is defined by and if and , where .   Independent sets have no pairs of adjacent vertices; matchings have no adjacent edges. Vertex coloring partitions the vertices into independent sets; edges can be partitioned into matching.                                      

Line Graphs Characterization March 2014 Graph Coloring 52 Theorem. (Krausz 1943) A simple graph is the line graph of some simple graph iff has a partition into cliques using each vertex of at most twice. Proof. Necessity follows from the fact that the edges adjacent at a vertex of are represented in by vertices connected in a clique.  Since an edge connects two vertices, those vertices imply two cliques at most. For sufficiency , suppose has such a partition, using cliques . We shall construct H satisfying . Assume that has no isolated varices.  

March 2014 Graph Coloring 53 Let be the vertices of that appear in a single clique of (if such one exist). We define a vertex in for each set of .   Edges of are defined such that vertices are adjacent if the corresponding sets (cliques) of intersect.   By its construction, each vertex of G appears in exactly two sets of .   Also, two vertices cannot both appear in two sets of , as otherwise a clique was split among the sets. 

March 2014 Graph Coloring 54 Hence there are no parallel edges in , so it is simple, and there is one edge in for each vertex of .   clique clique impossible     => Adjacent vertices in appear together in some and the corresponding edges of share the vertex corresponding to . Hence . ■   Krausz’s theorem does not directly yield an efficient test for line graph, which the following does.

March 2014 Graph Coloring 55 Theorem. (Bieneke 1968) A simple graph is the line graph of some simple graph iff does not contain any of the following subgraphs as an induced subgraph.