Graph Coloring Algorithms

Transcription

1 Graph Coloring Algorithms Course Notes Extension COMP5408 Advanced Algorithms Lucas Rioux-Maldague November 10 th Introduction Note: in this text, unless otherwise noted, a graph is always simple. A coloring of a graph is an assignment of labels to certain elements of a graph. More commonly, elements are either vertices (vertex coloring), edges (edge coloring), or both edges and vertices (total colorings). The most common form asks to color the vertices of a graph such that no two adjacent vertices share the same color (label). This is called a proper vertex coloring. For a graph G, the minimum number of colors such that G has a proper vertex coloring is denoted χ(g), the chromatic number of G. For example, for any path P n on n vertices, χ(p n ) = 2. Observe that this bound also applies to any bipartite graph: we can assign color 1 to vertices of the first partition, and color 2 to vertices of the second, there will be no two adjacent vertices that share the same color since vertices of the same class are never adjacent to each other by definition of a bipartite graph. The lower bound on χ is obviously the clique number ω, but what about an upper bound? One would think about V, but other than the complete graph, this bound is never tight. With further thought, we can see that χ(g) (G) + 1 (the proof is left to the reader). A result by Brooks states the following: Theorem 1.1 (Brooks Theorem [3]). If G is a connected graph that is not a complete graph or an odd cycle, then χ(g) (G). 2 Problem Complexity One could ask if we can find an optimal algorithm to find an optimal proper coloring of any graph. However, computing the chromatic number for a general graph is NP-hard [9]. The problem to decide if, given any integer k 3, a graph admits a k-coloring, is also NP-complete. Note that for k = 2, the problem is equivalent to checking if the graph is bipartite, this can be done in linear time using an algorithm such as breadth first search. As we will see later, there also exist polynomial time algorithms for specific families of graphs, such as planar or perfect graphs. We will also look at some heuristics that can speed up the process; however, their oputput will not be optimal in most cases. Even if the problem is NP-hard for the general problem, there are still interesting exponential time solutions to solve the problem. We will now look at such an algorithm, which is based on the idea of contracting edges. Edge contraction For a given graph G and two vertices u, v V (G), a contraction G/uv consists of removing u and v from the graph, and replacing them by a vertex w that is adjacent to all the vertices previously adjacent to at least one of u, v. If u and v are non adjacent vertices, then we can see that in a coloring of G, u and v will either share the same color or have distinct colors. Thus, we can analyse both cases, which are respectively equivalent to finding a coloring of G in which u and v are contracted (forces the same color), and a coloring in which there is an edge (u, v) (forces different colors). This gives 1

2 rise to the following recurrence relation, which is due to Zykov [30] (G + (u, v) denotes the graph G with the edge (u, v) added): χ(g) = min{χ(g + (u, v)), χ(g/uv)} for (u, v) / E(G) (1) Using this recurrence relation, we can build this algorithm which recursively computes each case, building a Zykov tree. Algorithm 1 Coloring number of a graph using Equation 1. Input: A graph G in adjacency list form. Output: The exact value of χ(g). 1: procedure FindColoring(Graph G) 2: if G is a complete graph then 3: return V (G) 4: else 5: Let u, v be a pair of vertices such that (u, v) / E(G) 6: return min{findcoloring(g + (u, v)), FindColoring(G/uv)} 7: end if 8: end procedure 9: return FindColoring(G) This algorithm runs in exponential time, the exact time will vary depending on the heuristic that is used to select u and v. We can see that every leaf of the Zykov tree will be a complete graph. Thus, we want every branch to converge into a complete graph in the fewest number of steps. Thus, we want G + (u, v) to have many edges, and G/uv to have as few as possible. To do this, we could select u and v that have a high number of common neighbors, which will eliminate many edges. Another application of edge contractions that also result in a recurrence relation is for the chromatic polynomial. This polynomial P (G, t) counts the number of different proper colorings on t colors of G. Note that since finding the chromatic number of a graph is NP-complete, finding the chromatic polynomial is also NP-complete, as given P (G, t) we can find the exact value of χ(g) by trying every value of k starting at 1, and stopping when P (G, t) > 0. This verification can be done in polynomial time. For this polynomial, suppose we want to find the number of possible colorings in a graph with no edge (u, v). We can distinguish two cases: if u and v have the same color, or if u and v have different colors. Hence, the total number of colorings in the original problem is the sum of the number of colorings in each of those cases. This gives the following recurrence relation: P (G (u, v), k) = P (G/uv, k) + P (G, k) for (u, v) E(G) P (G, k) = P (G (u, v), k) P (G/uv, k) (2) This gives rise to a procedure called the deletion-contraction algorithm, which can be used not only to compute the chromatic polynomial and number, but also various other graph invariants such as the number of spanning trees, the number of acyclic colorings and others. In the next section, we will look at various heuristics and near-optimal solutions. 3 Heuristics and Greedy Algorithms We now present a greedy type of algorithm that can color any graph in linear time. The downside is that this algorithm is sometimes far from the optimal. This algorithm is based on an ordering of the vertices. The ordering can be chosen in a specific way, or it can be random. We will see how different orderings can produce different results. Claim 3.1. Algorithm 2 executes in O( V + E ) time. 2

3 Algorithm 2 Greedy algorithm for coloring a graph G Input: An ordering v 1 < v 2 < < v n of the vertices of a graph G. Output: A coloring c of the vertices of G 1: c(v 1 ) 1 2: c(v i ) 0 for 2 i n 3: m 1 4: for i = 2 to n do 5: if k, 1 k m such that c(v) k for all neighbors v of v i and k is the smallest such color then 6: c(v i ) k 7: else 8: c(v i ) m + 1 9: m m : end if 11: end for Proof. The initialization part takes O( V ) time. The loop is executed O( V ) time, each iteration taking deg(v) time where v is the vertex considered during the iteration. Since v V (G) deg(v) = 2 E, the running time must be O( V + E ). The coloring produced by Algorithm 2 is clearly valid, but it may not be optimal. In fact, it may be arbitrarily far from the optimal chromatic number, one such example is with a bad ordering of a crown graph, which is a complete bipartite graph with the edges of a perfect matching removed [29]. Figure 1 shows the result of two different orderings of the same graph; one is optimal, while the other uses n/2 colors. This shows how much the output of the algorithm depends on the ordering. Thus, it would be interesting to find a method that gives an optimal coloring; that is, an ordering for which the greedy algorithm produces an optimal coloring. It is known that an optimal optimal exists for every graph [7]. However, it follows from the NP completeness of the optimal coloring problem that finding an optimal coloring is also NP-complete. Figure 1: Two different orderings of the same graphs for the greedy ordering. The number inside each vertex represents its position in the ordering. Figure from [29] Since finding an optimal ordering is usually infeasible, many heuristics that attempt to give better results have been proposed. One very common strategy for selecting the next vertex to be colored is to select the vertex of smallest degree that is not already colored. This way, the number of colors used by the algorithm will be at most + 1, where is the maximum degree of the graph. An improvement over this heuristic can improve the upper by one for all graphs that have chromatic index at most : select as the two first vertices of the ordering two vertices not adjacent to each other, and adjacent to the last vertex, and each subsequent vertex must be adjacent to at least one earlier vertex. This condition is the same as in a part of Brooks s Theorem, which is where the bound of comes from. 3

4 Similar to the idea of selecting a vertex of low degree, there is also the idea of selecting a vertex of high degree: at each step, we select the vertex of highest degree. Again, the upper bound on the number of colors required by this algorithm is + 1, but this ordering can result in a better coloring for some graphs. Other degree based heuristics include the saturation degree ordering, where the saturation degree is the number of distinct colors on adjacent neighbors of a vertex, and incidence degree ordering, where the incidence degree is the number of colored neighbors adjacent to a vertex. Other heuristics also exist that are based on the independent set of a graph. 3.1 Outerplanar Graphs For some classes of graphs, it is fast to find a near optimal ordering. One easy example is outerplanar graphs, which are graphs that can be embedded in the plane such that no two edges cross and all the vertices are adjacent to the unbounded region (see Figure 2). The key for this ordering is in the following theorem: Theorem 3.2. Let G be an outerplanar graph. Then, G contains at least one vertex of degree 2. We find the algorithm as follows: Let G be an outerplanar graph and S = s i,, s V (G) be a sequence of vertices such that each s 1 has degree 2 or less in G and s i has degree 2 in V {s 1,, s i 1 } for 1 < i V (G). The ordering is the reverse of S. This produces a three coloring: each time a vertex is colored, it is adjacent to at most two vertices, so at least one color is available. Note that this coloring is optimal for any outerplanar graph that has at least one face of odd degree (the degree of a face is the number of vertices it is adjacent to), such as a triangulation of an outerplanar graph, as any odd cycle requires three colors. This result is the base of the short proof of the Art Gallery theorem by Fisk [8]. (a) A planar graph (b) An outerplanar graph Figure 2: Planar and outerplanar graphs 3.2 Chordal graphs A chordal graph is a graph that contains a chord on each cycle of four or more vertices, where a chord is an edge disjoint from the cycle that connect two vertices of the cycle (see Figure 3). Chordal graphs are a subset of the perfect graph, for which perfect orderings can be found in polynomial time. For chordal graphs, the perfect ordering can be found efficiently in linear time, using the lexicographic breadth-first search algorithm of Rose, Lueker and Tarjan [23]. The algorithm finds a perfect elimination ordering, which is an ordering of the vertices of the graph such that for each vertex v, v and its neighbors that occur after v in the ordering form a clique in the graph. For chordal graphs, the reverse of the perfect elimination ordering is an optimal ordering for the greedy coloring algorithm [16]. We will not further discuss the topic of lexicographic breadth-first search here, as it has so many variants and applications that it warrants a chapter of its own. 4 Planar Graphs Planar graphs, graphs that can be embedded in the plane such that no two edges cross (see Figure 2), are one of the families of graphs for which a near-optimal coloring (at most one from the optimal), can 4

5 Figure 3: A section of a chordal graph. Chords are drawn in green. Note that if any of the chords is removed, the resulting graph will not be chordal, as there will be a chordless cycle on four vertices. Figure from [28]. be found in polynomial time. We previously saw that this was also true for outerplanar graphs, a family of planar graphs. The very famous Four Color Theorem [1] states that any graph can be colored with four colors, and its proof leads to a polynomial time algorithm. The quest for a proof of this theorem also leads to many other results in various areas of Graph theory, one of them is a linear time algorithm to five color planar graphs. A very famous incorrect proof by Kempe in 1879 [15] put the problem back into active research, and provided many constructions which were used in the first correct proof in Kempe s mistake was discovered in 1890 by Percy J. Heawood, who found that Kempe s proof, while incorrect for four-coloring, did prove that five colors were sufficient, and due to its constructive nature, lead to a five coloring algorithm. In the next section, we will look at a variation of this algorithm, and present a quadratic time algorithm for four coloring. 4.1 Five Coloring In his 1879 paper, as a building block for his proof, Kempe showed that every planar graph contains a vertex of degree five or less. Many refinements of this results have been published, we make use a version by Wernicke. This result is the basis of a simple linear time five coloring algorithm that was proposed by Robertson, Sanders, Seymour and Thomas [22]. Theorem 4.1 (Wernicke s Theorem [27]). Let G be a planar simple graph with minimum degree 5. Then, G has a vertex of degree 5 that is adjacent to another vertex of degree at most 6. The algorithm is recursive, in each step, a vertex of degree five or less is found and the procedure recurses on the smaller graph. 5

6 Algorithm 3 Algorithm to 5-color a planar graph in O( V ) time. Input: A planar graph G in adjacency list form. Output: A coloring c of G on five colors. 1: procedure Color(Graph G) 2: if V (G) = 1 then 3: Let x be the only vertex of G 4: Set c(x) any color in {1, 2, 3, 4, 5} 5: else 6: Let x be a vertex of degree 5 or less. 7: S = 8: if x has degree 5 then 9: Let v 1, v 2,, v 5 be vertices around x in clockwise order, with deg(v 1 ) 6 10: if v 1 is not adjacent to v 3 then 11: S = {v 1, v 3 } 12: else v 1 is adjacent to v 3 so, by planarity, v 2 is not adjacent to v 4 13: S = {v 2, v 4 } 14: end if 15: end if 16: Merge all vertices of S in G and let w be the resulting vertex. 17: c = Color(G x) 18: v V (S {x}), set c(v) = c (v). 19: v S, set c(v) = c (w). 20: c(x) any color in {1, 2, 3, 4, 5} not used by any of its neighbors. 21: end if 22: return c 23: end procedure 24: return Color(G) 6

7 Claim 4.2. Algorithm 3 returns a valid five coloring of any planar graph G. Proof. We can use induction on the number of vertices in the graph. Let n = V (G). Induction base: n = 1. In this case, the algorithm assigns any color in {1, 2, 3, 4, 5} to the only vertex of the graph. Induction step: suppose this is true for any graph on less than n vertices. The algorithm selects a vertex x with degree five or less. If the vertex has degree five, then it selects two non-adjacent vertices that are adjacent to x, and merges them together. We can see that this always exists: if v 1, v 2, v 3, v 4, v 5 are the vertices in order around x, then if v 1 and v 3 are adjacent together, then by planarity, v 2 and v 4 must not be, otherwise the (v 2, v 4 ) edge would cross the (v 1, v 3 ) edge. Then, the graph G resulting from the merger of vertices in S (if any), and the removal of x is colored with five vertices. By induction hypothesis, this is a valid five coloring of G. Back to the original graph G, all vertices get the same coloring as in G, with some new colors added for vertices in S and x. Note that since the vertices in S are non adjacent, having the same color is valid. Furthermore, since they were merged together in G, their color is valid with regard to their other neighbors. For x, if its degree is less than five, then at most four of {1, 2, 3, 4, 5} are used by its neighbors, so affecting one of the remaining colors is valid. Otherwise, if it has degree five, then at least two of its neighbors, vertices in S, have the same color. Thus, at most four colors are used by its neighbors so a valid color is available for x. We only used colors in {1, 2, 3, 4, 5}, so this completes the proof. Claim 4.3. Algorithm 3 runs in O( V ) time. Proof. All the operations in each iteration of the procedure Color can be done in constant time (S has size at most 2, and the coloring does not have to be recopied at line 14, it can be the same data structure to which we remove w and add every vertex in S), with the exception of finding x, but this can be implemented in amortized linear time by maintaining a stack of vertices of degree at most five. Each time the algorithm recurses, there is one less vertex in the input. Therefore the total running time is O( V ). 4.2 Four Coloring The proof of the Four Color Theorem by Appel and Haken also provides an algorithm that runs in O(n 4 ). The proof was simplified by Robertson, Sanders, Seymour and Thomas [21], and the authors also gave a new algorithm that uses the improvements of the new proof. The algorithm is simpler and runs in O(n 2 ) time [22]. Note that some parts of the algorithm color a triangulation of the graph (see below), but such a triangulation can always be created by adding edges to the graph, and the coloring of this triangulation will clearly be valid for the non-triangulated graph. We now look at some key definitions that are crucial in the understanding of the algorithm. Definition 4.4. A plane triangulation is a triangulation where each face has degree 3. Definition 4.5. A plane near-triangulation is a triangulation where each finite face has degree 3. Definition 4.6. A k-ring is a cycle on k edges with at least one vertex in each of its interior and exterior. Definition 4.7. A non-trivial k-ring is a cycle on k edges with at least two vertices in each of its interior and exterior. Definition 4.8. A bending k-ring is a cycle C on k edges with at least four vertices in each of its interior and exterior and a vertex not on C that is adjacent to three consecutive vertices of C. Definition 4.9. A plane triangulation G is nearly 7-connected if 1. G has no k-ring for k {2, 3, 4}, 2. G has no non-trivial 5-ring, 7

8 (a) A plane triangulation (b) A plane near-triangulation Figure 4: Triangulations u (a) A 5-ring. The cycle consists of the vertices in red, the exterior is blue and the interior is orange. (b) A non-trivial 5-ring. There is at least two vertices in each of the outside and inside of the cycle. Figure 5: Rings (c) A bending 5-ring. u is adjacent to three consecutive vertices of the cycle. 3. G has no bending 6-ring A graph is 5-connected if it satisfies condition 1, and is internally 6-connected if it satisfies conditions 1 and 2. Definition K is a configuration if K is a near-triangulation G(K) together with a function γ K : V (G(K)) N such that the three following conditions hold: 1. For every v V (G(K)), G(K) v has at most two components, and if there are exactly two, γ K (v) = deg(v) + 2, 2. For every v V (G(K)), if v is not incident with the outside face, then γ K (v) = deg(v) and otherwise, γ K (v) > deg(v). In either case, γ K (v) The ring size of K is at least 2. When drawing configurations (see Figure 6), a common convention is to use specific shapes for the vertices depending on the value of γ K : 1. γ K (v) = 5 Solid dot 2. γ K (v) = 6 No shape 3. γ K (v) = 7 Circle 4. γ K (v) = 8 Square 8

9 5. γ K (v) = 9 Triangle 6. γ K (v) = 10 Pentagon Definition The ring-size of a configuration K is v V γ K(v) deg(v) 1 where V = {v V (G(K)) v is adjacent to the outer face }. (a) A configuration of ring size 14 (b) A configuration of ring size 9 Figure 6: Configurations We need two more definitions that describe how configurations are embedded into a graph. Definition A configuration K is weakly contained in a plane graph G if there is a function f from V (G(K)) to V (G) such that: 1. For every v V (G(K)), deg(f(v)) = γ K (v) 2. For every v V (G(K)) such that G(K) v is connected, if the cyclic neighborhood of v is (u 1,, u j ) such that for every 1 i < j, u i u i+1 v is a triangle of G(K), then the cyclic neighborhood of f(v) is (f(u 1 ),, f(u j )) if deg(f(v)) = γ K (v), or else, (f(u 1 ),, f(u j ), w 1,, w m ) for some m {1, 2, 3}. 3. For every v V (G(K)) such that G(K) v is not connected, if the cyclic neighborhood of v is (u 1,, u j, v 1,, v k ) such that for every 1 i < j, u i u i+1 v is a triangle of G(K) and for every 1 i < k, v i v i+1 v is a triangle of G(K), then the cyclic neighborhood of f(v) is (f(u 1 ),, f(u j ), w, f(v 1 ),, f(v k ), x). Definition A configuration K is strongly contained in a plane graph G if K is weakly contained in G by a function f that also satisfies the following: 1. For every v, w V (G(K)), if f(v) = f(w), then v = w. 2. For every v, w V (G(K)), if f(v) is adjacent to f(w) in G, then v is adjacent to w in G(K). Parts of the incorrect proof of the Four Color Theorem given by Kempe [15] are still found in the proof of the Four Color Theorem by Appel and Haken. One such part is to find a vertex of low degree, then to analyse the various cases. The case where the degree is four is correct, but the degree five case is incorrect, which is where his proof failed. The degree three or two are trivial, but looking at the degree four case will be a helpful introduction to the actual algorithm. First, we need to define the concept of a Kempe chain (see Figure 8 for an illustration). Definition Given a planar graph G, two colors a, b and a vertex v V (G) with c(v) = a, a Kempe chain is a maximal connected subset of G that contains v and whose vertices all have color a or b. Kempe s proof finds a vertex u of low degree, in this case, u has degree 4. We remove u from the graph, which can be colored with four colors by induction hypothesis (Kempe uses strong induction on the number of vertices, note that this leads to a recursive algorithm). If the four vertices adjacent to u are colored with less than four different colors, then there is one color available for u. Otherwise, all 9

10 Figure 7: The configuration in Figure 6a is strongly contained in the left graph and weakly contained in the right graph. Why? Here is a hint: look at the bottom two vertices of the configuration in the right graph: they are adjacent in the right graph, but not in the configuration, which contradicts the second condition of a strongly contained configuration. Figure from [22]. Figure 8: {a, b, c, d, e} is a Kempe chain. No other adjacent vertices can be added to the chain without contradicting the definition. four colors are used. Let a, b, c, d be the vertices in order around u. Without loss of generality, suppose a is colored with red and c with green. Find a Kempe chain from a on red and green. If the chain does not contain c (Figure 9a), then we can recolor each vertex of the chain that has color green to red, and red to green (Figure 9b). The coloring will stay valid and this will free the color red which can be used for u. Otherwise, the chain contains c (Figure 9c). Then we can simply use a Kempe chain on the two other colors, blue and orange, from b, and apply the same reasoning. The existence of a cycle from a to c prevents the Kempe chain from containing b and d by planarity of the graph. Along with Kempe chains, it is also crucial to understand how the proof of the Four Color Theorem works. The proof uses the concept of reducible configurations. To understand this, we can look at a simple example. As we mentioned earlier, every graph must contain a vertex of degree five or less. If we have a set A that contains all the vertices of degree less than six, then A is an unavoidable set in any planar graph, as any planar graph cannot avoid A. Now suppose the Four Color Theorem is false. Thus, there exist graphs that require five colors. Let G be the smallest of all such graphs. Since G is the smallest, then any graph on less than V (G) vertices must be colorable with four colors. A reducible configuration is any arrangement of vertices that cannot be contained in G. Such an example is vertices with degree four or less: if G contains such a vertex u, then let G = G u. V (G ) < V (G), so we can color G with four colors as G is the smallest graph that requires five colors. Then, if the four vertices adjacent to u have four different colors, the coloring can be rearranged using a Kempe chain so that only three are used. There is then a color available for u and G can be four-colored. Thus, G cannot contain 10

11 vertices of degree four or less. The proof of the Four Color Theorem shows that any planar graph contains one of 633 reducible configurations. Since this covers all planar graphs, then five colors are never required. This set of 633 reducible configurations is denoted U. The exact composition of this set can be found in the paper by Robertson et al. [21]. Note that the original proof of the theorem by Appel and Haken [1] used 1936 configurations, but this was reduced a few times after redundancies were discovered: to 1834, then 1482, then to the current 633. In their improved proof, Robertson et al. prove the following theorem: Theorem Every plane triangulation of minimum degree 5 weakly contains a configuration from U. (a) The Kempe chain does not contains c (b) We switch the color of every vertex in the chain, this frees a color. (c) The Kempe chain contains c, we switch to the other chain b. Figure 9: Degree 4 analysis in Kempe s proof With this, we can now look at the algorithm. Note that to Kempe the vertices refers to flipping colors in a Kempe chain. 11

12 Algorithm 4 Algorithm to color a planar graph using four colors Input: A plane graph G in adjacency list form on n vertices. Output: A coloring c of G on four colors. 1: procedure RingAnalysis(Ring R, Graph G) 2: Let E, I be the subgraphs of G with E on the exterior of R and I on the interior 3: Let H 1 = G E 4: Triangulate H 1 5: c 2 Color(H 1 ) V (H 1 ) < V (G) 6: Let H 2 = G I 7: Triangulate H 2 8: c 2 Color(H 2 ) V (H 1 ) < V (G) 9: Kempe c 1 and c 2 to match on R and let c be the resulting coloring 10: V (H 1 ) + V (H 1 ) V (G) : return c 12: end procedure 13: procedure Color(Graph G) 14: if n 4 then 15: Assign each vertex a different color, and return c. 16: end if 17: if G has a face F of degree at least 4 then 18: Let x, y be two non-adjacent vertices on F. They exist by planarity 19: Let H be a graph formed by merging x and y into a vertex z in G 20: c = Color(H) 21: c(x), c(y) c(z) 22: return c 23: else if G has a vertex x with deg(x) 4 then 24: Then, the cycle C surrounding x is a k-ring. 25: return RingAnalysis(C, G) 26: else G has minimum degree 5, so by Theorem 4.15, it contains a configuration in U 27: Let K be a configuration of U weakly contained in G. 28: if K is not strongly contained in G then 29: Let C be a k-ring for k 4 or a non-trivial 5-ring on G. 30: return RingAnalysis(C, G) 31: else K is strongly contained in G 32: Let r be the ring size of K r 14 since this is true of any configuration of U 33: Let H = G V (G) 34: Let T E(H) be a set of edges according to the definition of reducibility 35: Let J be the graph obtained by contracting T in H. 36: c Color(J) c is a coloring of J 37: Kempe the vertices of V (H) until we find a coloring c that extends into K. 38: end if 39: end if 40: end procedure 41: return Color(G) 12

13 Claim Let G be any subgraph of a graph G. If Color(G ) returns a valid coloring of G, then the procedure RingAnalysis(R, G) for a ring R of G returns a valid coloring of G. Proof. By our hypothesis, c 1 and c 2 are valid colorings of H 1 and H 2. Birkhoff [2], to whom this procedure is due, proved that it is possible to Kempe the vertices so that the colorings match on R. Thus, the resulting coloring c is valid. Claim Algorithm 4 computes a valid four-coloring of any planar graph G. Proof. We can prove this using strong induction on the number of vertices n in the graph. Case n 4: Clearly, the coloring returned is proper as each vertex gets its own color. Induction step: suppose this is true for any graph on less than n vertices. We can enumerate all cases: 1. If G contains a face of degree at least 4, then by planarity, there exists two non adjacent vertices x, y, so we can force them to have the same color. By induction, the coloring of the smaller graph is valid, and assigning the same color to x, y will result in a valid coloring. 2. If G has a vertex of degree less than five, then by Claim 4.16, the coloring produced by RingAnalysis(C, G) is valid. 3. Otherwise, by Theorem 4.15, G contains a configuration K in U. If the configuration is not strongly contained in G, then we can find a k-ring C, and Claim 4.16 completes the proof. Otherwise, since K is reducible, a coloring of H can be found [22]. This completes the proof. Claim Algorithm 4 runs in O( V 2 ) time. Proof. Each of the operations done in Color can be done in O( V ) time. Linear time algorithms can be used to find a configuration in U, since each vertex in such a configuration has degree at most 11. For RingAnalysis, all the operations can also be done in linear time: finding E and I takes linear time, so does triangulating them and kemping the colorings. Also, notice that each time the algorithm recurses, the number of vertices is reduced at least by 1. Thus, at most O( V ) recursive calls are made, which implies the algorithm takes O( V 2 ) time. By Claims 4.18 and 4.17, we get the following result: Theorem Algorithm 4 computes a valid four-coloring of any planar graph G in O( V 2 ) time. 4.3 Three Coloring What about three coloring planar graphs, as not all planar graphs require three colors? It turns out this problem is NP-hard, even if we restrict the problem to graphs where the maximum degree is 4 [9]. For two coloring, it is, as mentioned earlier, equivalent to testing if a graph is bipartite, which can be done in linear time. Thus, we can always find a coloring that uses at most one extra color than the optimal in polynomial time. 5 Edge Coloring The edge coloring problem is a variant on graph colorings where we want to assign a color to each edge such that no two adjacent (incident to the same vertex) edges share the same color. This parameter is denoted χ (G), the coloring index, which is the minimum number of colors required such that there exists a proper edge coloring of a graph G. While the coloring number can vary from a small constant to + 1, the coloring index is quite stable. We have the following: Theorem 5.1 (Vizing s Theorem [25]). Let G be a graph with maximum degree. Then, χ (G)

14 v u Figure 10: The color of (u, v), red, is free on u. v x 1 x2 x 3 Figure 11: A fan F = [x 1, x 2, x 3] of v (dashed edges are uncolored), (v, x 1), (v, x 2), (v, x 3) are the fan edges. Note that F = [x 1, x 2] is also a fan of v, but it is not maximal. Similarely to vertex coloring, deciding if χ(g) is (G) or (G)+1 for a graph G is also NP-complete. However, there are polynomial algorithms that can find a + 1 coloring in polynomial time. In the following section, a relatively simple such algorithm that is due to Misra and Gries [20] is presented. As will be proved later, this algorithm computes the coloring in O(mn) time where m = E, n = V. Before presenting the algorithm, we need to define the concept of a fan. Definition 5.2. The color x of an edge (u, v) is free on u if c(u, z) x for all (u, z) E(G) : z v. Definition 5.3. A fan of a vertex u is a sequence of vertices F [1 : k] that satisfies the following conditions: 1. F [1 : k] is a non-empty sequence of distinct neighbors of u 2. (F [1], u) E(G) is uncolored 3. The color of F [i + 1] is free on F [i] for 1 i < k Definition 5.4. Given a fan F, any edge (F [i], X) for 1 i k is a fan edge. Let c, d be colors. A cd X -path is an edge path that goes through vertex X, only contains edges colored c and d and is maximal (we cannot add any other edge as it would include edges with a color not in c e a g f b d Figure 12: Examples of cd x paths: ac, cg, gd is a red-green c path, bd, dg is a red-orange d path, ac is a red-orange a path. 14

15 {c, d}). Note that only one such path exists for a vertex X, as at most one edge of each color can be adjacent to a given vertex. The operation invert the cd-path switches every edge on the path colored c to d and every edge colored d to c. Inverting a path can be useful to free a color on X if X is one of the endpoints of the path: if X was only adjacent to c, it will now be adjacent to d, freeing c for another edge adjacent to X. Also, the flipping operation will not alter the validity of the coloring since for the endpoints, only one of {c, d} can be adjacent to the vertex, and for other members of the path, the operation only switches the color of edges, no new color is added. Given a fan F [1 : k] of a vertex X, the rotate fan operation does the following (in parallel): 1. c(f [i], X) = c(f [i + 1], X) 2. Uncolor F [k] This operation leaves the coloring valid (why? this is left to the reader as an exercice). c e c e a g f a g f b d b d Figure 13: Inverting a cd x path: in this example, inverting the red-green a path from the graph on the left results in the graph on the right. v v x 1 x2 x 3 x 1 x2 x 3 Figure 14: Rotating a fan: in this example, rotating the fan F = [x 1, x 2, x 3] on the left results in the fan on the right. Algorithm 5 Edge-coloring algorithm Input: A graph G. Output: A coloring c of the edges of G 1: Let U E(G) 2: while U do 3: Let (u, v) be any edge in U. 4: Let F [1 : k] be a maximal fan of u starting at F [1] = v. 5: Let c be a color that is free on u and d be a color that is free on F [k]. 6: Invert the cd u path 7: Let w V (G) be such that w F, F = [F [1]...w] is a fan and d is free on w. 8: Rotate F and set c(u, w) = d. 9: U U {(u, v)} 10: end while 15

16 Claim 5.5. The inversion of the cd path guarantees a vertex w such that w F, F = [F [1]...w] is a fan and d is free on w. Proof. Prior to the inversion, there are two cases: 1. The fan has no edge colored d. Since F is a maximal fan and d is free on F [k], then there is no edge with color d adjacent to u, otherwise if there was, it would be after F [k], as d is free on f[k], but F was maximal, which is a contradiction. Thus, d is free on u, and since c is also free on u, the cd u path is empty and the inversion has no effect on the graph. We can set w = F [k]. 2. The fan has one edge colored d. Let F [x + 1] be this edge. Note that x since F [1] is uncolored. Thus, d is free on F [x]. Also, x k since the fan has length k but there exists a F [x + 1]. We can now show that after the inversion, for each y {1,, x 1, x + 1,, k}, the color of (F [y + 1], u) is free on y (1). Note that prior to the inversion, the color of (u, F [y + 1]) is not c or d, since c is free on u and (u, F [x + 1]) has color d and the coloring is valid. The inversion only affects edges that are colored c or d, so (1) holds. F [x] can either be in the cd u path or not. If it is not, then the inversion will not affect the set of free colors on F [x], and d will remain free on it. We can set w = F [x]. Otherwise, we can show that F is still a fan and d remains free on F [k]. Since d was free on F [x] before the inversion and F [x] is on the path, F [x] is an endpoint of the cd u path and c will be free on F [x] after the inversion. The inversion will change the color of (u, F [x + 1]) from d to c, so since c is now free on F [x] and (1) holds, F remains a fan. We also have that d remains free on F [k], since F [k] is not on the cd u path (suppose that it is; since d is free on F [k], then it would have to be an endpoint of the path, but u and F [x] are the endpoints). Select w = F [k]. Note that in any case, the fan F is a prefix of F, which implies F is also a fan. This completes the proof. Claim 5.6. The coloring produced by Algorithm 5 is proper. Proof. By induction on the number of colored edges. Base case: no edge is colored, this is valid. Induction step: suppose this was true at the end of the previous iteration. In the current iteration, after inverting the path, d will be free on u, and by Claim 5.5, it will also be free on w. Rotating F does not compromise the validity of the coloring. Thus, after setting c(u, w) = d, the coloring is still valid. Claim 5.7. Algorithm 5 requires at most + 1 colors. Proof. In a given step, we need to find colors c and d. Since u is adjacent to at least one uncolored edge and its degree is bounded by, at least one color in {1,, } is available for c. For d, F [k] may have degree and no uncolored adjacent edge. Thus, a color + 1 may be required. Theorem 5.8. Algorithm 5 computes a proper edge coloring on using + 1 colors in O( E V ) time. Proof. At each step, the rotation uncolors (u, w) and colors (u, v) which was previously uncolored. Thus, one additional edge gets colored. Hence, the loop will run O( E ) times. Finding the maximal fan, the colors c and d and invert the cd u path can be done in O( V ) time. Finding w and rotating F takes O(deg(u)) O( V ) time. Finding and removing the edge (u, v) can be done using a stack in constant time (pop the last element) and this stack can be populated in O( E ) time. Thus, each iteration of the loop takes O( V ) time, and the total running time is O( E + E V ) = O( E V ). The rest follows by Claims 5.6 and Exercices 1. Prove that for a graph G, χ(g) (G)

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