PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS PETER BALLEN 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. Abstract. 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. 2 introduces several basic graph theory denitions as well as a few technical results that will be used in later sections. 3 introduces the concept of a perfect graph and proves a few classes of graphs are perfect. 3 also introduces the Weak Perfect Graph Theorem and the Strong Perfect Graph Theorem and provides a proof of the Weak Theorem. 4 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 2 is broken into three subsections. 2.1 introduces the concept of a graph. 2.2 introduces cliques and independence sets. 2.3 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 2.1. cardinality of V V and is denoted is called the vertex set of use the phrase let edge set E . G = (V, E) |G|. G and E G. In this paper, we'll G be a graph with vertex set V and x ∈ G will mean let x be a vertex in is the edge set of to mean let Additionally, the phrase let 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 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 edge set house graph is a graph with vertex set E = {{a, b} , {b, c} , {c, d} , {d, e} , {e, a} , {c, e}}. V = {a, b, c, d, e} and The house graph has order 5. Figure 2.1 depicts a visual representation of the house graph. Figure 2.1. The house graph Remark. V We draw attention to three special classes of graphs. Let is empty, we refer to G as an order-zero graph. G = (V, E). If 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 G is innite, 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 a E contains no duplicates, and simple graph. {x, x} ∈ /E for any x ∈ G, then G is called 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. G = (V, E). G0 = (V 0 , E 0 ) is called a subgraph of G if V 0 ⊂ V 0 0 and E ⊂ E . G is a proper subgraph of G if G 6= G . G is an induced subgraph of 0 0 G if for every x, y ∈ V , {x, y} ∈ E if and only if {x, y} ∈ E . 0 Let 0 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 and V0 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}. PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 2.2. A Figure non- Figure induced subgraph Denition 2.5. 2.3. An 3 induced subgraph G = (V, E) be a graph. G = (V, E 0 ) 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. Let 0 It is useful to note that Example 2.6. and G = G. G be the house graph. E = {{e, b}, {b, d}, {d, a}, {a, c}}. G 0 Let Then G = (V, E 0 ), where V = {a, b, c, d, e} is depicted in Figure 2.4. Figure 2.4. The converse of the house graph PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS Denition 2.7. G = (V, E) Let 0 4 G0 = (V 0 , E 0 ). We say G is isomorphic to ψ : V → V 0 such that {x, y} ∈ E if and only if and G if there is a bijective function {ψ(x), ψ(y)} ∈ E 0 . 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 {a, b, c, d, e}. graph. For example, if 2.2. G1 is isomorphic to Neighbors and Independence. interested in discussing whether Denition 2.8. edge in G. d, a and x and Two distinct vertices Vertices Example 2.9. c {v1 , v2 , v3, v4, v5 } instead of Relabeling vertices does not change the fundamental structure of the x and y are G2 , then G1 Given two vertices y b x and y, we are often are connected by an edge. x, y ∈ G are independent if {x, y} if {x, y} is an edge in G. neighbors Returning to the house graph in Figure 2.1, is neighbors with G2 . is isomorphic to and e, and a a is not an is independent from is neither neighbors nor independent with itself. If x and Similarly, if y x are independent vertices in and Denition 2.10. y are neighbors in Let G, G, then x and y are neighbors in then they are independent in G = (V, E). An independence set of G A is independent. The independence every pair of vertices in is a set number the size of the largest independence set of that graph and is denoted Example 2.11. Let G be the house graph. A⊆V α(G). The house graph has nine independence α(G) = 2 {a}.{b}, {c}, {d}, {e} {a, c}, {a, d}, {b, d}, {b, e} Five independence sets with a single vertex Four independence sets with two vertices Denition 2.12. of vertices in K Let G = (V, E). are neighbors. A The clique of G is a set K ⊆ V where every pair clique number of a graph is the size of the largest clique of that graph and is denoted Example 2.13. Let G ω(G). be the house graph. The house graph contains 12 cliques, described below. The largest clique has three vertices, so • • • where of a graph is sets, described below. The largest independence set has two vertices, so • • G. G. ω(G) = 3. {a}, {b}, {c}, {d}, {e} {a, b}, {b, c} , {c, d} , {d, e} , {e, a} , {b, d} vertices: {c, d, e} Five cliques with a single vertex: Six cliques with two vertices: One clique with three There are two important relationships between cliques and independence sets. G are neighboring vertices in G, independence sets G. Similarly, cliques of G are independence sets of G. Thus, for G, α(G) = ω(G) and ω(G) = α(G). Just as independent vertices in of G are cliques of any graph Proposition 2.14. Let G be a graph. Let A be an independence set of G and let be a clique of G. Then |A ∩ K| ≤ 1. K Proof. y Suppose |A∩K| > 1. Let x, y ∈ A∩K where x 6= y . Then x, y ∈ A, so x and x, y ∈ K , so x and y are neighbors. are independent and not neighbors. But PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 5 G x and y are independent 0 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) 0 0 is an independence set of G , and by extension α(G) = α(G ). The same is true for Relationship between vertices are preserved under graph isomorphisms. Let and G0 be graphs with graph isomorphism ψ. Then cliques and clique number. We'll end the subsection by denining a special class of graphs: complete graphs. Denition 2.15. A graph G = (V, E) is a complete graph if every vertex of G is neighbors with every other vertex. n vertices is often denoted K n . If K n is a complete graph n of order n, α(K ) = 1 and ω(K ) = n. Figure 2.5 depicts complete graphs of order A complete graph with n 1, 2, 3, 4, and 5. Figure 2.5. From left to right, complete graphs of order 1, 2, 3, 4, and 5 Graph Colorings . Denition 2.16. A coloring of a graph G = (V, E) is a surjective function c : V 2.3. S such that if x elements, we call S y are neighboring k -coloring of G. and c a vertices in G then c(x) 6= c(y). If S → k has {red, green, blue, ...} or as a set of n, there always exists an n-coloring color. However, because c is surjective, is often thought of as either a set of colors integers {1, 2, 3, ...}. If G is a graph with order G by assigning every vertex a unique |S| ≤ |V |, and G cannot have an (n + 1)-coloring. of 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 2.17. 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 c(a) = red. c(a) = purple or 3-coloring for the house graph by setting We can also create a 4- coloring for the house graph by setting 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. PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 6 Figure 2.6. The house graph along with a 3-coloring. Proposition 2.18. 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) 6= 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 2.19. Let there does not exist a k is denoted G be a graph such that there exists a k -coloring of G but (k − 1)-coloring of G. The chromatic index of G is equal to 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. bounds on the chromatic index. It is often easier to nd lower and upper 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 2.20. For any graph G, ω(G) ≤ χ(G), i.e. the clique number of a graph is always less than or equal to its chromatic index. 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 2.21. 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). PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 7 Proposition 2.22. For any graph G, |G| ≤ α(G) ∗ χ(G). Proof. G = (V, E) be a graph and let c : V → S be a χ(G)-coloring of G. r ∈ S , let Vr = {v ∈ V : c(v) = r}. No two neighboring vertices can be assigned the same color, so no two vertices in Vr are neighbors and Vr is an independence set. Since α(G) is the size of the largest independence set of G, |Vr | ≤ α(G). Furthermore, since every vertex is colored by exactly one color, and there are exactly χ(G) colors, the following equation holds: X X |G| = |Vr | ≤ α(G) = α(G) ∗ |S| = α(G) ∗ χ(G) Let For any r∈S r∈S |G| α(G) ≤ χ(G). We end the subsection by stating that chromatic index is preserved under graph 0 0 isomorphism; if G and G are isomorphic then χ(G) = χ(G ). An immediate consequence of this theorem is that for any graph G, 3. Perfect Graphs and the Perfect Graph Theorems 3 is broken into three subsections. 3.1 introduces a few classes of graphs that are always perfect. 3.2 introduces and proves the Weak Perfect Graph Theorem. 3.3 introduces the Strong Perfect Graph Theorem. Without further delay, we are ready to dene a perfect graph. Denition 3.1. χ(H). A graph G is perfect Note that if H is an induced subgraph of also an induced subgraph of subgraph of G and 3.1. if for every induced subgraph This includes the improper subgraph where G0 G G. Thus, if G G, H of G, ω(H) = H = G. H every induced subgraph of is is a perfect graph, then every induced is also perfect. Perfection is preserved under graph isomorphism; if are isomorphic then G is perfect if and only if Examples of Perfect Graphs. G0 is perfect. One reason perfect graphs are considerd important is the wide range of graph classes that end up being perfect. Both graphs we introduced in 2, 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 1 ω(K 1 ) = χ(K 1 ) = 1 and K 1 has no proper induced subgraphs. Assume that complete graphs of order n are perfect. n+1 Let K be a complete graph of order n + 1 and let H be an induced subgraph n+1 of K . All induced subgraphs of complete graphs are complete graphs. If H n+1 is a proper subgraph of K , it is a complete graph of order n or less and is n+1 perfect by the induction hypothesis, thus ω(H) = χ(H). Finally, if H = K , ω(K n+1 ) = χ(K n+1 ) = n + 1. complete graph of order 1 (K ) is perfect, since Every order 1 graph is a complete graph, so every order 1 graph is perfect. Theorem. The house graph is perfect. PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS We could certainly consider every induced subgraph show ω(H) = χ(H). H 8 of the house graph and However, there are 51 induced subgraphs to check. Rather than prove the theorem using the denition of perfect graphs, we defer this proof until 4, 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 E = {{v1, v2 }, {v2 , v3 }, . . . {vn−1 , vn }}. an open chain if V = {v1 , v2, . . . vn } vi Figure 3.1 depicts an open chain of order 7 along with a 2-coloring: red if i is odd and is colored blue if i and is colored is even. Figure 3.1. An open chain of order 7 Theorem 3.4. All open chains are perfect. Proof. Let H Let G = (V, E) be an open chain, using the same notation as Denition 3.3. be an induced subgraph of Case I: H G. There are two potential cases. contains no neighboring vertices, for example Figure 3.2. If no neighboring vertices, ω(H) = 1. c(vi ) = red ω(H) = χ(H). The function for all H contains v ∈ H is a H , so by Corollary 2.21, 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(vi ) = blue i is even 1-coloring of Case II: Figure 3.2. An in- 3.3. An Figure duced subgraph with no duced subgraph neighbors neighbors inwith There are several other classes of perfect graphs. We will introduce two more classes of perfect graphs in 4. 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]. PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 3.2. The Weak Perfect Graph Theorem. 9 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 G0 = (V 0 , E 0 ) where V 0 = V ∪ {x0 } and E 0 is the union of E , the edge {x0 , x}, and 0 0 the edges {x , n} for each vertex n ∈ G that is neighbors with x. We refer to G as 0 the graph obtained from G by expanding x to an edge {x, x }. Example 3.6. G0 be the graph obtained from the house graph by expanding c to an edge {c, c }. G0 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(c0 ) = yellow. Let 0 Figure 3.4. The graph obtained from the house graph by expanding c to an edge {c, c0 } Lemma 3.7. Let G be a perfect graph and x ∈ G. Let G0 be the graph obtained by expanding x to an edge {x, x0 }. Then α(G) = α(G0 ). Proof. Every independence set of G is also an independence set of G0 , so α(G) ≤ α(G0 ). Let A0 be an independence set of G0 . If x0 ∈ / A0 , A0 is also an independence 0 0 0 set of G. If x ∈ A , then neither x nor any vertex neighboring x is in A , since all 0 0 0 of these vertices neighbor x . Then A = (A − {x }) ∪ {x} is an independence set 0 of G, since no vertex that neighbors x is in A. Note that |A| = |A |. Therefore, for 0 every independence set of G , there exists an independence set of G with the same 0 size. Thus, α(G) ≥ α(G ). Lemma 3.8. Let G be a graph with x ∈ G and let G0 be the graph obtained by expanding x to an edge {x, x0 }. If G is perfect, then G0 is perfect. PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS Proof. We will use induction on the order of graph of order 1 and G0 G. If G has order 1, is a complete graph of order 2, thus G0 G 10 is a complete 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 G0 be obtained by expanding x to an edge {x, x0 }. 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 0 be an induced subgraph of G0 . There are ve potential cases. Once 0 0 we show ω(H ) = χ(H ) for each case, the lemma will be proven. 0 0 0 Case I: H 6= G and x ∈ / H 0 . Then H 0 is an induced subgraph of G, so H 0 is 0 0 perfect and ω(H ) = χ(H ). 0 0 0 0 Case II: H 6= G and x ∈ H , but x ∈ / H 0 . Then H 0 is isomorphic to an induced 0 0 0 subgraph of G (relabel x to x) and is perfect, so ω(H ) = χ(H ) 0 0 0 0 0 0 Case III: H 6= G and x, x ∈ H . Let VH = {v ∈ G : v 6= x } and let H be 0 the induced subgraph of G with vertex set VH . Then H is obtained by expanding x ∈ H to an edge {x, x0 }. But H has less than n vertices, so H 0 is perfect by the 0 0 induction hypothesis and ω(H ) = χ(H ). 0 0 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. 0 When x is expanded, K is expanded to a clique of size ω + 1 since x ∈ K , so 0 0 ω(G ) = ω + 1. Furthermore, a (χ + 1)-coloring of G exists by taking c and setting c(x0 ) = u, where u is a unique color not assigned to any other vertex. Thus, ω(G0 ) = ω + 1 = χ + 1 = χ(G0 ) and ω(H 0 ) = χ(H 0 ). 0 0 Case V: H = G and for every clique K of size ω in G, x ∈ / K . First, observe 0 0 that every clique of G is also a clique of G , so ω ≤ ω(G ). In a moment, we will 0 0 0 show χ(G ) ≤ χ. Since G is perfect, χ = ω and χ(G ) ≤ χ = ω ≤ ω(G ). By 0 0 0 0 0 0 Proposition 2.20, ω(G ) ≤ χ(G ). Therefore, ω(G ) = χ(G ) and ω(H ) = χ(H ). 0 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 c2 of G where k ≤ χ and c2 (v) = red if and only 0 if c(v) = red and x 6= v . We can extend c2 to be a k -coloring of G by setting 0 0 c2 (x ) = red and thus prove χ(G ) = k ≤ χ. ∗ Now to construct c2 . Let Vnotred = {v ∈ G : c(v) 6= red} and let G be the induced subgraph of G with vertex set {x} ∪ Vnotred . 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 ∗ ∗ c2 (v) = . Since k < ω , k + 1 = k ≤ χ and we have found our red v∈ / G∗ coloring. This entire process is illustrated in Figure 3.5. Denition 3.9. G = (V, E) where V = {v1 , . . . , vn }. Let f : V → N be Gi = (Vi , Ei ) be a complete graph of order f (vi ) for all i ∈ {1, . . . n}. Let G0 = (V 0 , E 0 ) be a graph where V 0 = V1 ∪ V2 ∪ . . . Vn and E 0 = E1 ∪ E2 ∪ · · · ∪ En ∪ {{x, y} : x ∈ Vi , y ∈ Vj , {vi , vj } ∈ E}. Then G0 is the graph obtained by replacing every vertex of G with a complete graph of order f . Let a function and let PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 11 Figure 3.5. Demonstration of Case V. On left, the original graph G∗ G coloring Example 3.10. 0 0 (V , E ) with 4-coloring with 3-coloring c∗ . c. In center, the induced subgraph On right, the graph G0 along with 4- c2 . Let f (a) = f (c) = 3, f (b) = 4, f (d) = 2, f (e) = 1 and let G0 = be the graph obtained by replacing every vertex of the house graph with a complete graph of order f . G0 is depicted in Figure 3.6 along with a 7-coloring (we omit an explicit denition of this coloring for space). Va ∪ Vb ∪ Vc ∪ Vd ∪ Ve . While it might not 0 0 0 clique, so ω(G ) = 7 and χ(G ) = ω(G ). V0 = Vb ∪ Vc is a Observe that be obvious from the picture, Figure 3.6. The graph obtained by replacing every vertex of the house graph with a perfect graph Lemma 3.11. Let G = (V, E) be a graph along with function f : V → N. Let G0 be the graph obtained by replacing every vertex of G with a complete graph of order f . Then α(G) = α(G0 ). Furthermore, if G is perfect, then G0 is perfect. Proof. Let G = (V, E) where V = {v1 , . . . , vn }. Consider the algorithm described Vi = {vi , vi1 , vi2 , . . . , vif (vi )−1 }. This Vj if vi and vj are neighbors in G. Thus, when the algorithm completes, Gcur is isomorphic to G0 . Note that if Gcur and G0 are isomorphic, α(Gcur ) = α(G0 ) and Gcur is perfect below. Every loop of j will construct a clique clique will have cardinality f (vi ) and will be connected to clique PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 12 G0 is perfect. α(Gcur ) is initialized to α(G) and is unchanged by the 0 algorithm, so α(G ) = α(G). Additionally, if G is perfect, then Gcur is initialized 0 to be perfect and running the algorithm maintains the perfection of Gcur , so G is perfect. if and only if Gcur = G for i between 1 and n: for j between 1 and f (vi ) − 1: Gcur = The graph obtained from G by expanding vi to an edge {vi , vij } 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 G is perfect then G = G is perfect. We will use induction on the order 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 6= 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 6= ∅ 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 ( c(v) v ∈ M 0 0 construct a χ(M ) + 1 coloring c of G: c (v) = . Since red v ∈ / M i.e. v ∈ K 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 follows that if of G. When a clique that intersects every independence set of cardinality two. chain, K = {v1, v2 } independence number. But what about an arbitrary graph? contradiction that no such A such that For the open intersects every independence set with cardinality equal to the K ∩ AK = ∅. K exists, i.e. for every K ∈ K, Suppose towards a there exists an AK ∈ G to We don't know enough about arbitrary graph do anything meaningful, so we will create a new graph S0 with a better dened f (v) = |{K ∈ K : v ∈ AK }| and let S be the induced subgraph of G 0 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 vi ∈ G with a complete graph of order f (vi ) and structure. Let PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS Vi . By ω(S ) = χ(S 0 ). vertex set Lemma 3.11, S0 is perfect and α(S) = α(S 0 ). Since 13 S0 is perfect, 0 We will now work towards a contradiction showing clique of S 0 S has the form Vi where Vi ω(S 0 ) < χ(S 0 ). The largest is the vertex set of the complete graph vi ∈X X ⊂ S is a clique of G (and thus X ∈ K) . Then, X X X 0 ω(S ) = |Vi | = f (vi ) = |{(vi , K} : vi ∈ X, K ∈ K, vi ∈ Ak }| = |X∩AK |. expanded from vi and vi ∈X vi ∈X K∈K AK is an independence set, so |X ∩ AK | ≤ 1 by Proposition |X ∩ AX | = 0 since K ∩ AK = ∅ for any K ∈ K. Thus, P ω(S 0 ) = |X ∩ AK | ≤ |K| − 1. K∈K P P 0 Additionally, observe that |S | = |Vi | = f (vi ) = |{(vi , K} : vi ∈ V, K ∈ vi ∈S vi ∈V P K, vi ∈ Ak }| = |AK | = |K| ∗ α. X is a clique and 2.14. Furthermore, K∈K 0 0 |S | |S | ) ≥ α(S 0 ) = 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) |S 0 | |K|∗α 0 and χ(S ) ≥ = |K|. α(G) = α 0 0 0 0 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 6= ∅ for all A ∈ A. By Proposition 2.22,χ(S 3.3. 0 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 3.13. Let G = (V, E). G is a closed E = {{v1, v2 }, {v2 , v3 }, . . . {vn−1 , vn }, {vn , v0 }}. chain if V = {v1 , . . . vn } and 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 3.14. 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 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 induced subgraph of G, then G G is a graph such that H is an 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 for G, G if and only if they hold for so the conditions hold for G, and G G. If G is perfect, the conditions hold is perfect. 4. Applications of Perfect Graphs and the Perfect Graph Theorems 4 discusses a few applications of perfect graphs and the Perfect Graph Theorems. 4.1 introduces and proves Mirsky's Theorem. 4.2 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. 4.1. 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 properties hold for all • • • Reexive: X be a set. a, b, c ∈ X . A relation ≤ is a partial order if the following a ≤ a. a ≤ b and b ≤ a then a = b. a ≤ b and b ≤ c then a ≤ c. Antisymmetric: If Transitive: If PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS 15 Two example partial orders are the standard less-than-or-equals relation when R and the standard subset relation when X 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 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. or Denition 4.4. Let P be a set with partial order ≤ and let L = (l1 , l2 , . . . , ln ) li ∈ P and li 6= lj if i 6= j . L is called a chain if li ⊥ lj for all li , lj ∈ L. L is called an antichain if li //lj for all li , lj ∈ L where ll 6= lj . L is called an increasing chain if li ≤ li+1 for all li ∈ L except ln . be a list where 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 ≤. can be partitioned into k antichains, where k is the length of the longest chain. P 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 6= 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 independence sets of comparability graphs are antichains. x ⊥ y. Similarly, 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. Com- parability graphs of order 1 are perfect. Assume that comparability graphs of order n are perfect. Let G = (P, E) G. be a comparability graph of order be an induced subgraph of n+1 and let H All induced subgraphs of comparability graphs are H is a proper subgraph of G, it is a comparabilty 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 . comparability graphs. Thus, if graph of order Since all cliques are chains, the size of the longest clique is equal to the size of the longest increasing chain, so v∈P or y ≤ x. Furthermore, if x and y are neigh- x ≤ y , c(x) ≥ c(y) + 1, since you can construct 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) 6= c(y). Therefore, c is a χ(G)-coloring of G, where χ(G) ≤ max c(v) = ω(G). bors, then either x≤y ω(G) = max c(v). If an increasing chain by appending v∈P ω(G) = χ(G). Thus, We are now ready to prove Mirsky's Theorem. Proof. Let P ≤. Let G = (P, E) be the associated G are chains, so the length of the largest chain χ(G)−coloring of G. If Ai is the set of all vertices be a partial set with partial order comparability graph. All cliques of is equal to ω(G). Ai colored i, then Let c be a is an independence set since no two vertices with the same color can be neighbors. All independence sets of χ(G) 4.2. antichains. Finally, since Dilworth's Theorem. G G are antichains, ω(G) = χ(G). so c partitions P into is perfect, 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 4.11. 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 4.12. 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. PERFECT GRAPHS AND THE PERFECT GRAPH THEOREMS Proof. Let P be a set with partial order incomparability graph. All cliques of G ≤. Let G = (P, E) 17 be the associated are antichains, so the length of the largest ω(G). Let c be a χ(G)−coloring of G. If Ai is the set of all Ai is an independence set. All independence sets of G are c partitions P into χ(G) chains. Since G is perfect, ω(G) = χ(G). antichain is equal to vertices colored chains, so i, then 5. Conclusion Perfect graphs are one of the deepest and most fascinating graph theory topics to emerge in the late 20th century. perfect graphs. In this paper, we introduced the concept of 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, 1961. [2] Maria Chudnovsky, Neil Robertson, Paul Seymour, and Robin Thomas. The strong perfect graph theorem. Annals of Mathematics, 164(1):pp. 51229, 2006. [3] R. Diestel. Graph Theory: Springer Graduate Text GTM 173. Springer Graduate Texts in Mathematics (GTM). 2012. [4] R. P. Dilworth. A decomposition theorem for partially ordered sets. Annals of Mathematics, 51(1):pp. 161166, 1950. [5] Martin Charles Golumbic. Algorithmic graph theory and perfect graphs, volume 57. Elsevier, 2004. [6] Martin Grötschel, Laszlo Lovász, and Alexander Schrijver. Polynomial algorithms for perfect graphs. Ann. Discrete Math, 21:325356, 1984. [7] László Lovász. Perfect graphs. Selected topics in graph theory, 2:5587, 1983. [8] L. Lovász. Normal hypergraphs and the perfect graph conjecture. Discrete Mathematics, 2(3):253 267, 1972. [9] L. Mirsky. A dual of dilworth's decomposition theorem. The American Mathematical Monthly, 78(8):pp. 876877, 1971.