PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS

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1 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS PETER BALLEN Abstract. The theory of perfect graphs relates the concept of graph colorings to the concept of cliques. In this paper, we introduce the concept of a perfect graph as well as a number of graph classes that are always perfect. We next introduce both the Weak Perfect Graph Theorem and the Strong Perfect Graph Theorem and provide a proof of the Weak Perfect Graph Theorem. We also demonstrate an application of perfect graphs, using perfect graphs to prove both Mirsky's Theorem and Dilworth's Theorem. 1. Introduction The theory of perfect graphs relates the concept of graph colorings to the concept of cliques. Aside from having an interesting structure, perfect graphs are considered important for three reasons. First, several common classes of graphs are known to always be perfect. Second, a number of important algorithms only work on perfect graphs. Finally, perfect graphs can be used in a wide variety of applications, ranging from scheduling to order theory to communication theory. Ÿ2 introduces several basic graph theory denitions as well as a few technical results that will be used in later sections. Ÿ3 introduces the concept of a perfect graph and proves a few classes of graphs are perfect. Ÿ3 also introduces the Weak Perfect Graph Theorem and the Strong Perfect Graph Theorem and provides a proof of the Weak Theorem. Ÿ4 discusses applications of perfect graphs both within and outside of graph theory. Our primary application will be using perfect graphs to prove two order theory theorems: Mirsky's Theorem and Dilworth's Theorem. 2. Graph Theory Concepts Ÿ2 is broken into three subsections. Ÿ2.1 introduces the concept of a graph. Ÿ2.2 introduces cliques and independence sets. Ÿ2.3 introduces graph colorings Graphs. Denition 2.1. A graph G = (V, E) consists of a set of vertices V and a set of edges E. Elements of V are distinct. Elements of E are 2-sets of the form {x, y}, where x and y are both vertices in V. The order of a graph G is equal to the cardinality of V and is denoted G. V is called the vertex set of G and E is the edge set of G. In this paper, we'll use the phrase let G = (V, E) to mean let G be a graph with vertex set V and edge set E. Additionally, the phrase let x G will mean let x be a vertex in G. There are two common methods used to describe a graph. The rst method is to mathematically describe V and E. The second method is to use a picture where 1

2 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 2 vertices are represented as points and edges are represented as lines that connect two vertices. We will use both methods in this paper. Example 2.2. We will use a graph throughout the paper to illustrate several graph theory concepts. The house graph is a graph with vertex set V = {a, b, c, d, e} and edge set E = {{a, b}, {b, c}, {c, d}, {d, e}, {e, a}, {c, e}}. The house graph has order 5. Figure 2.1 depicts a visual representation of the house graph. Figure 2.1. The house graph Remark. We draw attention to three special classes of graphs. Let G = (V, E). If V is empty, we refer to G as an order-zero graph. Order-zero graphs are generally regarded as uninteresting, but can be annoying to work with; many graph theorems and denitons do not make sense when discussed in the context of order-zero graphs. If V is innite, G is called an innite graph. Innite graphs are quite interesting, but are beyond the scope of this paper: see [5] for more information on the subject. Finally, if E contains no duplicates, and {x, x} / E for any x G, then G is called a simple graph. In a simple graph, no two vertices are connected by more than one edge, and no vertex is connected to itself. This paper only discusses simple graphs that are not innite and not order-zero. Thus, when we use the word graph, we mean a simple graph with a nite non-zero number of vertices. Denition 2.3. Let G = (V, E). G = (V, E ) is called a subgraph of G if V V and E E. G is a proper subgraph of G if G G. G is an induced subgraph of G if for every x, y V, {x, y} E if and only if {x, y} E. This paper is primarily interested in induced subgraphs. An induced subgraph can be uniquely identied by its vertices; the edge set can be determined from V and E. Example 2.4. Figure 2.2 and 2.3 both depict subgraphs of the house graph. The graph in Figure 2.2 is not an induced subgraph because it is missing the edge {d, e}. The graph in Figure 2.3 is an induced subgraph, and could be described as the induced subgraph with vertex set {b, c, d, e}.

3 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 3 Figure 2.2. A noninduced subgraph Figure 2.3. An induced subgraph Denition 2.5. Let G = (V, E) be a graph. G = (V, E ) is called the converse of G if E = {{x, y} : x, y V, {x, y} / E}, i.e. every vertex that was connected by an edge in G is not connected in G, and every vertex that was not connected by an edge in G is connected in G. It is useful to note that G = G. Example 2.6. Let G be the house graph. Then G = (V, E ), where V = {a, b, c, d, e} and E = {{e, b}, {b, d}, {d, a}, {a, c}}. G is depicted in Figure 2.4. Figure 2.4. The converse of the house graph

4 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 4 Denition 2.7. Let G = (V, E) and G = (V, E ). We say G is isomorphic to G if there is a bijective function ψ : V V such that {x, y} E if and only if {ψ(x), ψ(y)} E. ψ is called a graph isomorphism. Graph isomorphism are often discussed in the context of vertex relabeling. For example, we could relabel vertices of the house graph {v 1, v 2, v 3, v 4, v 5 } instead of {a, b, c, d, e}. Relabeling vertices does not change the fundamental structure of the graph. For example, if G 1 is isomorphic to G 2, then G 1 is isomorphic to G Neighbors and Independence. Given two vertices x and y, we are often interested in discussing whether x and y are connected by an edge. Denition 2.8. Two distinct vertices x, y G are independent if {x, y} is not an edge in G. Vertices x and y are neighbors if {x, y} is an edge in G. Example 2.9. Returning to the house graph in Figure 2.1, a is independent from c and d, a is neighbors with b and e, and a is neither neighbors nor independent with itself. If x and y are independent vertices in G, then x and y are neighbors in G. Similarly, if x and y are neighbors in G, then they are independent in G. Denition Let G = (V, E). An independence set of G is a set A V where every pair of vertices in A is independent. The independence number of a graph is the size of the largest independence set of that graph and is denoted α(g). Example Let G be the house graph. The house graph has nine independence sets, described below. The largest independence set has two vertices, so α(g) = 2 Five independence sets with a single vertex {a}.{b}, {c}, {d}, {e} Four independence sets with two vertices {a, c}, {a, d}, {b, d}, {b, e} Denition Let G = (V, E). A clique of G is a set K V where every pair of vertices in K are neighbors. The clique number of a graph is the size of the largest clique of that graph and is denoted ω(g). Example Let G be the house graph. The house graph contains 12 cliques, described below. The largest clique has three vertices, so ω(g) = 3. Five cliques with a single vertex: {a}, {b}, {c}, {d}, {e} Six cliques with two vertices: {a, b}, {b, c}, {c, d}, {d, e}, {e, a}, {b, d} One clique with three vertices: {c, d, e} There are two important relationships between cliques and independence sets. Just as independent vertices in G are neighboring vertices in G, independence sets of G are cliques of G. Similarly, cliques of G are independence sets of G. Thus, for any graph G, α(g) = ω(g) and ω(g) = α(g). Proposition Let G be a graph. Let A be an independence set of G and let K be a clique of G. Then A K 1. Proof. Suppose A K > 1. Let x, y A K where x y. Then x, y A, so x and y are independent and not neighbors. But x, y K, so x and y are neighbors.

5 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 5 Relationship between vertices are preserved under graph isomorphisms. Let G and G be graphs with graph isomorphism ψ. Then x and y are independent in G if and only if ψ(x) and ψ(y) are independent in G. The same is true for neighboring vertices. Similarly, A is an independence set of G if and only if ψ(a) is an independence set of G, and by extension α(g) = α(g ). The same is true for cliques and clique number. We'll end the subsection by denining a special class of graphs: complete graphs. Denition A graph G = (V, E) is a complete graph if every vertex of G is neighbors with every other vertex. A complete graph with n vertices is often denoted K n. If K n is a complete graph of order n, α(k n ) = 1 and ω(k n ) = n. Figure 2.5 depicts complete graphs of order 1, 2, 3, 4, and 5. Figure 2.5. From left to right, complete graphs of order 1, 2, 3, 4, and Graph Colorings. Denition A coloring of a graph G = (V, E) is a surjective function c : V S such that if x and y are neighboring vertices in G then c(x) c(y). If S has k elements, we call c a k-coloring of G. S is often thought of as either a set of colors {red, green, blue,...} or as a set of integers {1, 2, 3,...}. If G is a graph with order n, there always exists an n-coloring of G by assigning every vertex a unique color. However, because c is surjective, S V, and G cannot have an (n + 1)-coloring. Remark. In this paper, we will never use black as part of a coloring. Vertices drawn in black (such as the previous graphs) will be used when we are not interested in discussing colorings. Example Let G be the house graph. We dene a 3-coloring c of G by the following function, which is visually represented in Figure 2.6. c(a) = c(c) = green, c(b) = c(e) = blue, c(d) = red There are often multiple k-colorings for a given graph; we could create a dierent 3-coloring for the house graph by setting c(a) = red. We can also create a 4- coloring for the house graph by setting c(a) = purple or a 5-coloring by assigning every vertex a unique color. Is there a 2-coloring of the house graph? No, as the following proposition will demonstrate.

6 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 6 Figure 2.6. The house graph along with a 3-coloring. Proposition Let G be a graph and K be a clique of G. If c is a coloring of G, c assigns every vertex in K a dierent color. Proof. If x and y are distinct vertices in K, they are neighbors so by denition c(x) c(y). The set {c, d, e} is a clique of the house graph, so each of these three vertices must be assigned a dierent color. We are often interested in nding the smallest possible coloring of a graph. Denition Let G be a graph such that there exists a k-coloring of G but there does not exist a (k 1)-coloring of G. The chromatic index of G is equal to k is denoted χ(g). The house graph has chromatic index 3, since a 3-coloring of the house graph exists, but there is no 2-coloring of the house graph. Finding the chromatic index of an arbitrary graph can be dicult. It is often easier to nd lower and upper bounds on the chromatic index. This paper only takes advantage of two easily proven lower bounds, but a great deal of research has gone into nding much more precise bounds. Proposition For any graph G, ω(g) χ(g), i.e. graph is always less than or equal to its chromatic index. the clique number of a Proof. There exists a clique K of G with cardinality ω(g). By Proposition 2.18, every vertex in K must be assigned a dierent color. Thus, ω(g) χ(g). One immediate consequence of this proposition is that the chromatic index of a complete graph is equal to its order, since n = ω(k n ) χ(k n ) n. While this proposition is useful, we will occasionally apply it in a slightly modied form. Corollary Let G be a graph along with a k-coloring c. If ω(g) = k, then ω(g) = χ(g). Proof. If there exists a k-coloring of G, then χ(g) k = ω(g) by denition. But by Proposition 2.20, ω(g) χ(g). Thus, ω(g) = χ(g).

7 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 7 Proposition For any graph G, G α(g) χ(g). Proof. Let G = (V, E) be a graph and let c : V S be a χ(g)-coloring of G. For any r S, let V r = {v V : c(v) = r}. No two neighboring vertices can be assigned the same color, so no two vertices in V r are neighbors and V r is an independence set. Since α(g) is the size of the largest independence set of G, V r α(g). Furthermore, since every vertex is colored by exactly one color, and there are exactly χ(g) colors, the following equation holds: G = r S V r r S α(g) = α(g) S = α(g) χ(g) G An immediate consequence of this theorem is that for any graph G, α(g) χ(g). We end the subsection by stating that chromatic index is preserved under graph isomorphism; if G and G are isomorphic then χ(g) = χ(g ). 3. Perfect Graphs and the Perfect Graph Theorems Ÿ3 is broken into three subsections. Ÿ3.1 introduces a few classes of graphs that are always perfect. Ÿ3.2 introduces and proves the Weak Perfect Graph Theorem. Ÿ3.3 introduces the Strong Perfect Graph Theorem. Without further delay, we are ready to dene a perfect graph. Denition 3.1. A graph G is perfect if for every induced subgraph H of G, ω(h) = χ(h). This includes the improper subgraph where H = G. Note that if H is an induced subgraph of G, every induced subgraph of H is also an induced subgraph of G. Thus, if G is a perfect graph, then every induced subgraph of G is also perfect. Perfection is preserved under graph isomorphism; if G and G are isomorphic then G is perfect if and only if G is perfect Examples of Perfect Graphs. One reason perfect graphs are considerd important is the wide range of graph classes that end up being perfect. Both graphs we introduced in Ÿ2, complete graphs and the house graphs, are examples of perfect graphs. Theorem 3.2. All complete graphs are perfect. Proof. We will prove this theorem using induction on the order of the graph. The complete graph of order 1 (K 1 ) is perfect, since ω(k 1 ) = χ(k 1 ) = 1 and K 1 has no proper induced subgraphs. Assume that complete graphs of order n are perfect. Let K n+1 be a complete graph of order n + 1 and let H be an induced subgraph of K n+1. All induced subgraphs of complete graphs are complete graphs. If H is a proper subgraph of K n+1, it is a complete graph of order n or less and is perfect by the induction hypothesis, thus ω(h) = χ(h). Finally, if H = K n+1, ω(k n+1 ) = χ(k n+1 ) = n + 1. Every order 1 graph is a complete graph, so every order 1 graph is perfect. Theorem. The house graph is perfect.

8 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 8 We could certainly consider every induced subgraph H of the house graph and show ω(h) = χ(h). However, there are 51 induced subgraphs to check. Rather than prove the theorem using the denition of perfect graphs, we defer this proof until Ÿ4, where we will use the Weak Perfect Graph Theorem to easily show the house graph is perfect. We now introduce a new class of graphs: open chains. Denition 3.3. Let G = (V, E). G is an open chain if V = {v 1, v 2,... v n } and E = {{v 1, v 2 }, {v 2, v 3 },... {v n 1, v n }}. Figure 3.1 depicts an open chain of order 7 along with a 2-coloring: v i is colored red if i is odd and is colored blue if i is even. Figure 3.1. An open chain of order 7 Theorem 3.4. All open chains are perfect. Proof. Let G = (V, E) be an open chain, using the same notation as Denition 3.3. Let H be an induced subgraph of G. There are two potential cases. Case I: H contains no neighboring vertices, for example Figure 3.2. If H contains no neighboring vertices, ω(h) = 1. The function c(v i ) = red for all v H is a 1-coloring of H, so by Corollary 2.21, ω(h) = χ(h). Case II: H contains neighboring vertices, for example Figure 3.3. There are no cliques of size three or greater, so ω(h) = 2. The function c (dened below) is a 2-coloring of H, so χ(h) = ω(h). { red i is odd c(v i ) = blue i is even Figure 3.2. An induced subgraph with no neighbors Figure 3.3. An induced subgraph with neighbors There are several other classes of perfect graphs. We will introduce two more classes of perfect graphs in Ÿ4. A short selection of interesting graphs that are perfect includes: bipartite graphs, interval graphs, permutation graphs, rigid-circuit graphs, split graphs, and threshold graphs. We do not discuss all of these graphs in this paper; interested readers are directed to [7].

9 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS The Weak Perfect Graph Theorem. Checking every induced subgraph is one way to show that a graph is perfect. However, for complicated graphs with large order, this quickly becomes unfeasable. Even the relatively simple house graph has 51 induced subgraphs to check. Fortunately, the Weak and Strong Perfect Graph Theorems provide a way to prove that a graph or class of graphs is perfect without checking every subgraph. Both theorems were conjectured by Claude Berge in 1961[1]. The Weak Perfect Graph Theorem was proved by László Lovász in 1972 [8]. The proof we present is based o of Lovász's work, as well as the work done by Reinhard Diestel in [3]. Before we can present the proof, we require a bit of setup. Denition 3.5. Let G = (V, E) be a graph and let x be a vertex in G. Let G = (V, E ) where V = V {x } and E is the union of E, the edge {x, x}, and the edges {x, n} for each vertex n G that is neighbors with x. We refer to G as the graph obtained from G by expanding x to an edge {x, x }. Example 3.6. Let G be the graph obtained from the house graph by expanding c to an edge {c, c }. G is drawn in Figure 3.4, along with the 4-coloring c(a) = c(c) = green, c(b) = c(e) = blue, c(d) = red, c(c ) = yellow. Figure 3.4. The graph obtained from the house graph by expanding c to an edge {c, c } Lemma 3.7. Let G be a perfect graph and x G. Let G be the graph obtained by expanding x to an edge {x, x }. Then α(g) = α(g ). Proof. Every independence set of G is also an independence set of G, so α(g) α(g ). Let A be an independence set of G. If x / A, A is also an independence set of G. If x A, then neither x nor any vertex neighboring x is in A, since all of these vertices neighbor x. Then A = (A {x }) {x} is an independence set of G, since no vertex that neighbors x is in A. Note that A = A. Therefore, for every independence set of G, there exists an independence set of G with the same size. Thus, α(g) α(g ). Lemma 3.8. Let G be a graph with x G and let G be the graph obtained by expanding x to an edge {x, x }. If G is perfect, then G is perfect.

10 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 10 Proof. We will use induction on the order of G. If G has order 1, G is a complete graph of order 1 and G is a complete graph of order 2, thus G is perfect. Assume the lemma holds for all graphs of order n or less. Let G be a perfect graph of order n + 1 with x G, and let G be obtained by expanding x to an edge {x, x }. Set ω = ω(g), χ = χ(g), and let c be a χ-coloring of G. Note that since G is perfect, ω = χ. Let H be an induced subgraph of G. There are ve potential cases. Once we show ω(h ) = χ(h ) for each case, the lemma will be proven. Case I: H G and x / H. Then H is an induced subgraph of G, so H is perfect and ω(h ) = χ(h ). Case II: H G and x H, but x / H. Then H is isomorphic to an induced subgraph of G (relabel x to x) and is perfect, so ω(h ) = χ(h ) Case III: H G and x, x H. Let V H = {v G : v x } and let H be the induced subgraph of G with vertex set V H. Then H is obtained by expanding x H to an edge {x, x }. But H has less than n vertices, so H is perfect by the induction hypothesis and ω(h ) = χ(h ). Case IV: H = G and there exists a clique K of size ω in G such that x K. This is the case represented in Figure 3.4 where G is the house graph and x = c. When x is expanded, K is expanded to a clique of size ω + 1 since x K, so ω(g ) = ω + 1. Furthermore, a (χ + 1)-coloring of G exists by taking c and setting c(x ) = u, where u is a unique color not assigned to any other vertex. Thus, ω(g ) = ω + 1 = χ + 1 = χ(g ) and ω(h ) = χ(h ). Case V: H = G and for every clique K of size ω in G, x / K. First, observe that every clique of G is also a clique of G, so ω ω(g ). In a moment, we will show χ(g ) χ. Since G is perfect, χ = ω and χ(g ) χ = ω ω(g ). By Proposition 2.20, ω(g ) χ(g ). Therefore, ω(g ) = χ(g ) and ω(h ) = χ(h ). Now to show χ(g ) χ. Let c be a χ-coloring of G. Without loss of generalty, say c(x) = red. This implies every vertex neighboring x is not colored red by c. We will construct a k-coloring c 2 of G where k χ and c 2 (v) = red if and only if c(v) = red and x v. We can extend c 2 to be a k-coloring of G by setting c 2 (x ) = red and thus prove χ(g ) = k χ. Now to construct c 2. Let V notred = {v G : c(v) red} and let G be the induced subgraph of G with vertex set {x} V notred. Suppose there exists a clique K of size ω in G. By Proposition 2.18, c assigns every vertex in K a dierent color. There are ω vertices in K and χ = ω colors of c, so one vertex in K must be colored red. The only vertex c colors red in G is x, so x K. This is a contradiction: x is in no clique of size ω. Thus, ω(g ) < ω. Set k = ω(g ). Since G is an induced subgraph of perfect graph G, G is perfect and χ(g ) = ω(g ) = k. Thus, there exists a k -coloring c of G. Do not let red be a color in this coloring. { Set k = k + 1. We can construct a k-coloring of G by dening c (v) v G c 2 (v) = red v / G. Since k < ω, k + 1 = k χ and we have found our coloring. This entire process is illustrated in Figure 3.5. Denition 3.9. Let G = (V, E) where V = {v 1,..., v n }. Let f : V N be a function and let G i = (V i, E i ) be a complete graph of order f(v i ) for all i {1,... n}. Let G = (V, E ) be a graph where V = V 1 V 2... V n and E = E 1 E 2 E n {{x, y} : x V i, y V j, {v i, v j } E}. Then G is the graph obtained by replacing every vertex of G with a complete graph of order f.

11 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 11 Figure 3.5. Demonstration of Case V. On left, the original graph G with 4-coloring c. In center, the induced subgraph G with 3-coloring c. On right, the graph G along with 4- coloring c 2. Example Let f(a) = f(c) = 3, f(b) = 4, f(d) = 2, f(e) = 1 and let G = (V, E ) be the graph obtained by replacing every vertex of the house graph with a complete graph of order f. G is depicted in Figure 3.6 along with a 7-coloring (we omit an explicit denition of this coloring for space). Observe that V = V a V b V c V d V e. While it might not be obvious from the picture, V b V c is a clique, so ω(g ) = 7 and χ(g ) = ω(g ). Figure 3.6. The graph obtained by replacing every vertex of the house graph with a perfect graph Lemma Let G = (V, E) be a graph along with function f : V N. Let G be the graph obtained by replacing every vertex of G with a complete graph of order f. Then α(g) = α(g ). Furthermore, if G is perfect, then G is perfect. Proof. Let G = (V, E) where V = {v 1,..., v n }. Consider the algorithm described below. Every loop of j will construct a clique V i = {v i, v i1, v i2,..., v if(vi }. This ) 1 clique will have cardinality f(v i ) and will be connected to clique V j if v i and v j are neighbors in G. Thus, when the algorithm completes, G cur is isomorphic to G. Note that if G cur and G are isomorphic, α(g cur ) = α(g ) and G cur is perfect

12 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 12 if and only if G is perfect. α(g cur ) is initialized to α(g) and is unchanged by the algorithm, so α(g ) = α(g). Additionally, if G is perfect, then G cur is initialized to be perfect and running the algorithm maintains the perfection of G cur, so G is perfect. G cur = G for i between 1 and n: for j between 1 and f(v i ) 1: G cur = The graph obtained from G by expanding v i to an edge {v i, v ij } We are now ready to introduce and prove the Weak Perfect Graph Theorem. Theorem 3.12 (Weak Perfect Graph Theorem). A graph G is perfect if and only if its converse G is perfect. Proof. We need only prove that if G is perfect, then G is perfect. Once we do so, it follows that if G is perfect then G = G is perfect. We will use induction on the order of G. When G has order 1, G = G and the theorem holds. Assume the theorem holds for graphs of order n or less, and let G = (V, E) be a perfect graph of order n + 1. Let H be an induced subgraph of G. If H G, then let H be the induced subgraph of G where H = H. H has order n or less since H = H < n + 1. H is perfect since it is an induced subgraph of perfect graph G. By the induction hypothesis, H is perfect and ω(h) = χ(h). Thus, we only need to prove the case where H = G, that is show ω(g) = χ(g). Let K be the set of all cliques of G, and let A be the set of all independence sets of G with exactly α(g) elements. We will show that there exists a nonempty clique K K such that K A for all A A, i.e. K is a clique that intersects every independence set of size α(g). Let M denote the induced subgraph of G with vertex set {v G : v / K} and let M denote its converse. Every independence set of size α(g) intersects K and is thus not an independence set of M, so α(m) < α(g). This implies that ω(m) = α(m) < α(g) = ω(g), which can be rewritten as ω(m) + 1 ω(g). Furthermore, recall that if K is a clique of G, K is an independence set of G. Let c be a χ(m) coloring of M that does{ not use red as a color. We can construct a χ(m) + 1 coloring c of G: c c(v) v M (v) = red v / M i.e. v K. Since K is an independence set, this is a valid coloring. Thus, χ(g) χ(m) + 1. M is an induced subgraph of G with order n or less, so by the induction hypothesis M is perfect and χ(m) = ω(m). Therefore, χ(g) χ(m) + 1 = ω(m) + 1 ω(g). This implies χ(g) = ω(g) by Proposition 2.20 and G is perfect. We now will show that such a K exists. For the house graph, K = {c, d, e} is a clique that intersects every independence set of cardinality two. For the open chain, K = {v 1, v 2 } intersects every independence set with cardinality equal to the independence number. But what about an arbitrary graph? Suppose towards a contradiction that no such K exists, i.e. for every K K, there exists an A K A such that K A K =. We don't know enough about arbitrary graph G to do anything meaningful, so we will create a new graph S with a better dened structure. Let f(v) = {K K : v A K } and let S be the induced subgraph of G with vertex set {v G : f(v) > 0}. Since G is perfect, S is perfect. Let S be the graph obtained by replacing every vertex in S with a complete graph of order f, that is to say replace every vertex v i G with a complete graph of order f(v i ) and

13 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 13 vertex set V i. By Lemma 3.11, S is perfect and α(s) = α(s ). Since S is perfect, ω(s ) = χ(s ). We will now work towards a contradiction showing ω(s ) < χ(s ). The largest clique of S has the form V i where V i is the vertex set of the complete graph v i X expanded from v i and X S is a clique of G (and thus X K). Then, ω(s ) = V i = v i X v i Xf(v i ) = {(v i, K} : v i X, K K, v i A k } = K K X A K. X is a clique and A K is an independence set, so X A K 1 by Proposition Furthermore, X A X = 0 since K A K = for any K K. Thus, ω(s ) = X A K K 1. K K Additionally, observe that S = V i = f(v i ) = {(v i, K} : v i V, K v i S v i V K, v i A k } = A K = K α. K K By Proposition 2.22,χ(S ) S α(s ) = S α(s). Since S is an induced subgraph of G, every independence set of S is also an independence set of G. Thus, α(s) α(g) and χ(s ) S α(g) = K α α = K. Therefore, ω(s ) K 1 < K < χ(s ) and ω(s ) < χ(s ). Therefore, there must exist a nonempty clique K K such that K A for all A A The Strong Perfect Graph Theorem. As mentioned earlier, Berge conjectured both the Weak and Strong Perfect Graph Theorems. The Strong Perfect Graph Theorm was proven in 2002 by the team of Maria Chudnovsky, Neil Robertson, Paul Seymour, and Robin Thomas [2]. Denition Let G = (V, E). G is a closed chain if V = {v 1,... v n } and E = {{v 1, v 2 }, {v 2, v 3 },... {v n 1, v n }, {v n, v 0 }}. While closed chains are similar to open chains there is one key dierence: all open chains are perfect but not all closed chains are perfect. Example Let G = (V, E), where V = {a, b, c, d, e} and E = {{a, b}, {b, c}, {c, d}, {d, e}, {e, a}}. G is a closed chain depicted in Figure 3.7. G is not perfect because ω(g) = 2 but χ(g) = 3. Figure 3.7. A closed chain of order 5

14 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 14 Any closed chain with an odd number of vertices will not be perfect. Furthermore, if H is a closed chain with odd order and G is a graph such that H is an induced subgraph of G, then G will not be perfect. This is the idea that inspires the Strong Perfect Graph Theorem. Theorem 3.15 (Strong Perfect Graph Theorem). Let G be a graph and G be its converse. G is perfect if and only if: (1) There is no induced subgraph H of G such that H is a closed chain of odd order; and (2) There is no induced subgraph H of G such that H is a closed chain of odd order. The proof of this theorem is 150 pages long and beyond the scope of this paper. Interested readers are directed to [2]. However, we can easily see that the Strong Perfect Graph Theorem implies the Weak Perfect Graph Theorem. The two conditions hold for G if and only if they hold for G. If G is perfect, the conditions hold for G, so the conditions hold for G, and G is perfect. 4. Applications of Perfect Graphs and the Perfect Graph Theorems Ÿ4 discusses a few applications of perfect graphs and the Perfect Graph Theorems. Ÿ4.1 introduces and proves Mirsky's Theorem. Ÿ4.2 introduces and proves Dilworth's Theorem. We mentioned earlier that the Weak Perfect Graph Theorem could be used to prove that the house graph is perfect. We now present that proof. Theorem 4.1. The house graph is perfect. Proof. Let G be the house graph. G = (V, E), where E = {{e, b}, {b, d}, {d, a}, {a, c}}. G is an open chain, so G is perfect. By the Weak Perfect Graph Theorem, G is perfect. The Perfect Graph theorems are quite useful when trying to prove a graph is perfect or that a class of graphs are perfect. There are a number of occasions when it is convenient to know a graph is perfect. In the eld of graph theory, several additional results are built on knowing a given graph is perfect. In the eld of algorithms, computing the clique number, independence number, and chromatic index of an arbitrary graph is known to be computationally expensive. However, there exist algorithms that can compute the clique number, independence number, and chromatic index of a perfect graph in polynomial time [6]. In the eld of order theory, a number of theorems can be proven using perfect graphs. We will discuss two of these theorems in greater depth Mirsky's Theorem. The rst theorem we will discuss is Mirsky's Theorem. Mirsky's Theorem was proven by Leon Mirsky using order theory [9]. We will provide an alternative proof using perfect graphs. However, before we can do so, we must introduce a few order theory denitions. Denition 4.2. Let X be a set. A relation is a partial order if the following properties hold for all a, b, c X. Reexive: a a. Antisymmetric: If a b and b a then a = b. Transitive: If a b and b c then a c.

15 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 15 Two example partial orders are the standard less-than-or-equals relation when X R and the standard subset relation when X is a set of sets. Denition 4.3. Let X be a set with partial order and a, b X. If either a b or b a, then we say a and b are comparable and a b. If a b and b a, then we say a and b are incomparable and a /b. Denition 4.4. Let P be a set with partial order and let L = (l 1, l 2,..., l n ) be a list where l i P and l i l j if i j. L is called a chain if l i l j for all l i, l j L. L is called an antichain if l i /l j for all l i, l j L where l l l j. L is called an increasing chain if l i l i+1 for all l i L except l n. Example 4.5. Let P = {A, B, C, D, E, F }, where A = {1, 2, 3, 4}, B = {1, 2, 3}, C = {2, 3, 4}, D = {2, 3}, E = {3, 4}, F = {4}, along with partial order, the standard subset relation. Then (F, E, C, A) is an increasing chain since F E C A. (E, F, C, A) is a chain but not an increasing chain, since E F C A. (B, C) is an antichain since B C and C B so B /C. (B, E, F ) is neither a chain nor an antichain, since B /E but E F. Theorem 4.6 (Mirsky's Theorem). Let P be a nite set with partial order. P can be partitioned into k antichains, where k is the length of the longest chain. To prove Mirsky's Theorem, we will introduce a new class of graphs: comparability graphs. Denition 4.7. Let P be a set with partial order and let G = (P, E). G is a comparability graph if E = {{x, y} : x, y P, x y, x y}. Example 4.8. Let P be the set with partial order described in Example 4.5. The comparability graph associated with P is drawn below along with the coloring c(a) = blue = 1; c(b) = c(c) = green = 2; c(d) = c(e) = yellow = 3; c(f ) = red = 4. Figure 4.1. A comparability graph It is useful to note that all cliques of a comparability graph are chains, since if K is a clique and x, y K, then x and y are neighbors so x y. Similarly, independence sets of comparability graphs are antichains.

16 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 16 Theorem 4.9. Comparability graphs are perfect. Proof. We will prove this theorem using induction on the order of the graph. Comparability graphs of order 1 are perfect. Assume that comparability graphs of order n are perfect. Let G = (P, E) be a comparability graph of order n + 1 and let H be an induced subgraph of G. All induced subgraphs of comparability graphs are comparability graphs. Thus, if H is a proper subgraph of G, it is a comparabilty graph of order n or less and is perfect by the induction hypothesis, so ω(h) = χ(h). Thus, to prove the theorem, we need only prove ω(g) = χ(g). Let c : P N be the function c(v) = the longest increasing chain that starts at v. Since all cliques are chains, the size of the longest clique is equal to the size of the longest increasing chain, so ω(g) = max v P c(v). Furthermore, if x and y are neighbors, then either x y or y x. If x y, c(x) c(y) + 1, since you can construct an increasing chain by appending x to the front of the longest increasing chain that starts at y. Similarly, if y x, c(y) c(x) + 1. Thus, if x and y are neighbors, c(x) c(y). Therefore, c is a χ(g)-coloring of G, where χ(g) max v P Thus, ω(g) = χ(g). We are now ready to prove Mirsky's Theorem. c(v) = ω(g). Proof. Let P be a partial set with partial order. Let G = (P, E) be the associated comparability graph. All cliques of G are chains, so the length of the largest chain is equal to ω(g). Let c be a χ(g) coloring of G. If A i is the set of all vertices colored i, then A i is an independence set since no two vertices with the same color can be neighbors. All independence sets of G are antichains, so c partitions P into χ(g) antichains. Finally, since G is perfect, ω(g) = χ(g) Dilworth's Theorem. A second order theory result is Dilworth's Theorem. Robert Dilworth originally proved it using order theory results[4], but we will use perfect graphs. Theorem 4.10 (Dilworth's Theorem). Let P be a nite set with partial order. P can be partitioned into k chains, where k is the length of the longest antichain. To prove this theorem, we will introduce a new class of graphs: incomparability graphs. Denition Let P be a set with partial order and let G = (P, E). G is a incomparability graph if E = {{x, y} : x, y P, x /y}. It is useful to note that all cliques of an incomparability graph are antichains, since if K is a clique and x, y K, then x and y are neighbors so x /y. Similarly, independence sets of incomparability graphs are chains. Theorem Incomparability graphs are perfect. Proof. Proving incomparability graphs are perfect by showing ω(h) = χ(h) is dicult, since there is no analogue to a longest increasing chain to use as the basis of the coloring. Instead, we note that the converse of any incomparability graph is a comparability graph. Since all comparability graphs are perfect, by the Weak Perfect Graph Theorem, all incomparability graphs are perfect. Our proof of Dilworth's Theorem is similar to our proof of Mirsky's Theorem.

17 PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 17 Proof. Let P be a set with partial order. Let G = (P, E) be the associated incomparability graph. All cliques of G are antichains, so the length of the largest antichain is equal to ω(g). Let c be a χ(g) coloring of G. If A i is the set of all vertices colored i, then A i is an independence set. All independence sets of G are chains, so c partitions P into χ(g) chains. Since G is perfect, ω(g) = χ(g). 5. Conclusion Perfect graphs are one of the deepest and most fascinating graph theory topics to emerge in the late 20th century. In this paper, we introduced the concept of perfect graphs. We proved that the house graph, complete graphs, open chains, comparability graphs, and incomparability graphs are all perfect. We also proved the Weak Perfect Graph Theorem, which states that the converse of a perfect graph is perfect. Finally, we demonstrated an application of perfect graphs, using them to prove Mirsky's and Dilworth's Theorems. We have only touched the surface of what is possible with perfect graphs. There are several additional classes of perfect graphs that can be used in countless applications; research on perfect graphs and their applications is ongoing. References [1] C Berge. Färbung von graphen deren sämtliche bzw., ungerade kreise starr sind (zusammenfassung) math. Nat. ReiheWiss. Z. Martin Luther Univ. Halle Wittenberg, page 114, [2] Maria Chudnovsky, Neil Robertson, Paul Seymour, and Robin Thomas. The strong perfect graph theorem. Annals of Mathematics, 164(1):pp , [3] R. Diestel. Graph Theory: Springer Graduate Text GTM 173. Springer Graduate Texts in Mathematics (GTM) [4] R. P. Dilworth. A decomposition theorem for partially ordered sets. Annals of Mathematics, 51(1):pp , [5] Martin Charles Golumbic. Algorithmic graph theory and perfect graphs, volume 57. Elsevier, [6] Martin Grötschel, Laszlo Lovász, and Alexander Schrijver. Polynomial algorithms for perfect graphs. Ann. Discrete Math, 21:325356, [7] László Lovász. Perfect graphs. Selected topics in graph theory, 2:5587, [8] L. Lovász. Normal hypergraphs and the perfect graph conjecture. Discrete Mathematics, 2(3): , [9] L. Mirsky. A dual of dilworth's decomposition theorem. The American Mathematical Monthly, 78(8):pp , 1971.