Tolerance intersection graphs of degree bounded subtrees of a tree with constant tolerance 2

Discrete Mathematics 340 (2017) 209–222

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Tolerance intersection graphs of degree bounded subtrees of a tree with constant tolerance 2 Elad Cohen a, *, Martin Charles Golumbic b,a , Marina Lipshteyn b , Michal Stern c,b a b c

Department of Computer Science, University of Haifa, Mount Carmel, 31905, Israel Caesarea Rothschild Institute, University of Haifa, Mount Carmel, 31905, Israel The Academic College of Tel-Aviv - Yaffo, Rabeno Yeruham 2, 68182, Tel-Aviv, Israel

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Article history: Received 29 May 2011 Received in revised form 12 August 2016 Accepted 14 August 2016

Keywords: Chordal graph Weakly chordal graph Intersection graph of subtrees of a tree Complete bipartite graph

a b s t r a c t An (h, s, t)-representation of a graph G consists of a collection of subtrees {Sv : v ∈ V (G)} of a tree T , such that (i) the maximum degree of T is at most h, (ii) every subtree has maximum degree at most s, and (iii) there is an edge between two vertices in the graph if and only if the corresponding subtrees in T have at least t vertices in common. Jamison and Mulder denote the family of graphs that admit such a representation as [h, s, t ]. Our main theorem shows that the class of weakly chordal graphs is incomparable with the class [h, s, t ]. We introduce new characterizations of the graph K2,n in terms of the families [h, s, 2] and [h, s, 3]. We then present our second main result characterizing the graphs in [4, 3, 2] as being the graphs in [4, 4, 2] avoiding a particular family of substructures, and we give a recognition algorithm for the family [4, 3, 2]. © 2016 Elsevier B.V. All rights reserved.

1. Introduction and motivation A graph is chordal if it contains no chordless cycle of length greater than 3. The class of chordal graphs is widely investigated and plays a fundamental role in graph theory. One of the reasons is that chordal graphs can be modeled as intersection graphs of subtrees of a tree (c.f. [2,5] and [18]). One can construct in linear time such an intersection representation, called a ‘‘clique tree’’, such that each node of the host tree corresponds to a maximal clique in the chordal graph. In many real world applications, the intersection representation of a graph is more important than the graph itself. In [15] and [16], the intersection representation of a graph on a tree is defined as follows. An (h, s, t)-representation ⟨S , T ⟩ of G consists of a collection of subtrees S = {Sv : v ∈ V (G)} of a tree T , such that (i) the maximum degree of T is at most h, (ii) every subtree has maximum degree at most s, and (iii) there is an edge between two vertices in G if and only if the corresponding subtrees in T have at least t vertices in common. The class of graphs that have an (h, s, t)-representation is denoted by [h, s, t ]. Throughout the paper we will use the following definition. Definition 1.1. We say that G is sharply contained in [h, s, t ] if G ∈ [h, s, t ], G ̸ ∈ [h′ , ∞, t ] and G ̸ ∈ [∞, s′ , t ], where h′ < h and s′ < s, respectively. Thus, the family of chordal graphs is the same as [∞, ∞, 1]. This result was strengthened in [17] and [16], respectively, showing that the family of chordal graphs is also equivalent to [3, 3, 1] and [3, 3, 2]. The notation ∞ here means that no

*

Corresponding author. Fax: +972 4 8288181. E-mail addresses: [email protected] (E. Cohen), [email protected] (M.C. Golumbic), [email protected] (M. Lipshteyn), [email protected] (M. Stern). http://dx.doi.org/10.1016/j.disc.2016.08.015 0012-365X/© 2016 Elsevier B.V. All rights reserved.

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Fig. 1. The forbidden subgraphs from Theorem 1.2.

restriction is imposed. The family of interval graphs, by definition, is the family [2, 2, 1] and is equivalent to [2, 2, 2]. There are other papers that study [h, s, t ], for specific values of h, s and t, although without using this notion. For example, the family of edge intersection graphs of paths in a tree [7] (EPT graphs) is the class [∞, 2, 2], and the family of vertex intersection graphs of paths in a tree [6,13] (VPT graphs or path graphs) is [∞, 2, 1]. A graph is weakly chordal if neither the graph nor its complement contains a chordless cycle of length greater than 4. The class of weakly chordal graphs is also well studied and has a number of known applications [1,4,12,14]. It was shown in [10] that [4, 2, 2] is equivalent to the intersection of the family of weakly chordal graphs and [∞, 2, 2], and in [8,12] the following theorem characterizing the graphs in [4, 4, 2] was given. Theorem 1.2 ([8,12]). A graph G is a weakly chordal (K2,3 , 4P2 , P2 ∪ P4 , P6 , H1 , H2 , H3 )-free graph (see Fig. 1) if and only if the graph G has a (4, 4, 2)-representation. Our main motivation in this paper has been to determine if there is an [h, s, t ] class of graphs that corresponds to the class of weakly chordal graphs. From our results on the complete bipartite graphs, which are a family of weakly chordal graphs, we are able to confirm that weakly chordal graphs cannot be characterized within the [h, s, t ] framework. Our second important goal has been to characterize the class [4, 3, 2] which is the last open case for the families [h, s, t ] with 4 ≥ h ≥ s and t ≥ 2. In Section 2, we investigate the complete bipartite graph K2,n , where one part of the bipartition has two vertices and the other part has n vertices. In [15] and [16], a function f (h, s, t) is given, such that for any n > f (h, s, t), the graph K2,n has no (h, s, t)-representation. We strengthen their results and prove new theorems characterizing n such that K2,n has an (h, s, 2)-representation and those such that K2,n has an (h, s, 3)-representation. In Section 3, we characterize the family [4, 3, 2]. In particular, we introduce a new family of graphs called ‘‘4-flowers’’, we prove that a 4-flower graph is sharply contained in [4, 4, 2] and that [4, 3, 2] is the family of 4-flower-free graphs belonging to [4, 4, 2]. We also provide a polynomial-time recognition algorithm for the class [4, 3, 2] in Section 5. The recognition algorithm is used to prove the characterization theorem. In Section 4, we show a hierarchy of families of graphs between chordal and weakly chordal within the [h, s, t ] framework. The hierarchy also includes all cases of [∞, s, 2]. 2. The (h, s, t)-representations of K2,n Jamison and Mulder (c.f. [15] and [16]) investigated the intersection graph of subtrees of a tree by studying the representations of the complete bipartite graph K2,n . They found an upper bound for the size of n as a function of s and t, but they observed that this bound is far from being optimal. Let R(s, t) denote the complete balanced rooted tree whose root has s children, internal nodes have s − 1 children and all leaves are at distance t − 1 from the root. Let γ (s, t) be the number of subtrees of R(s, t) which have exactly t nodes and which contain the root. Jamison and Mulder prove in [16] the following: Theorem 2.1 ([16]). If h, s and t are positive integers with h ≥ s, and n is a positive integer with n > γ (s, t)(t + 1), then K2,n ̸ ∈ [h, s, t ]. In this section, we will improve the bound of Jamison and Mulder for t = 2 and t = 3. The case of t = 2 will be used in Section 4 for particular separation examples in the hierarchy shown in Fig. 5. The case of t = 3 continues the search of Jamison and Mulder for (h, s, t)-representations for the graph K2,n . We introduce the following notation for an (h, s, t)-representation of K2,n . Let {a, b} be the vertices of the 2-side of K2,n , and let A and B be the subtrees of the host tree T that correspond to a and b. Similarly, let {si } be the vertices of the n-side of K2,n , and S ′ = {Si } the subtrees on the host tree T that correspond to {si } respectively, for 1 ≤ i ≤ n.

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2.1. The case t = 2 We first prove the following lemma that is needed for Theorem 2.3. Lemma 2.2. For n > 1, in any (∞, ∞, 2)-representation of K2,n on a host tree T , the subtrees A and B must share exactly one vertex r ∈ T . Moreover, choose r as the root of T , and let ra , rb , {ri } denote the roots of A, B and {Si } respectively. Then, the vertex r, ra , rb , and all ri are all the same. Proof. Suppose that A and B do not intersect. There exists a unique path p in T from A to B, and every Si must contain this path. Thus, the set of the vertices {si : 1 ≤ i ≤ n} is not an independent set, a contradiction. Therefore, A and B intersect in a unique vertex, denoted by r. Since r is the root of T and is the unique common vertex of A and B, ra = rb = r. Each si is a neighbor of a and b, therefore Si intersects in at least one edge with both A and B. Hence, Si must contain r and ri = r, for 1 ≤ i ≤ n. □ Theorem 2.3. K2,s is sharply contained in [2s, s, 2]. Proof. We first show K2,s ∈ [2s, s, 2]. Consider a star of degree 2s with half the edges belonging to A and half to B. Assign each Si an A-edge and a B-edge, such that no new intersection is introduced. Clearly, this is a (2s, s, 2)-representation of K2,s . Suppose that K2,s has an (h′ , s∗ , 2)-representation for some h′ < 2s and s∗ ≥ s. We consider the construction of Lemma 2.2. Since the degree of the root r is at most h′ < 2s, one of A or B will share an edge with at most s − 1 subtrees in S ′ , a contradiction. Therefore, K2,s has an (h′ , s∗ , 2)-representation if and only if h′ ≥ 2s. Suppose that K2,s has an (h∗ , s′ , 2)-representation for some s′ and h∗ with s′ < s and h∗ ≥ h. Consider the construction of Lemma 2.2. The degree of the subtree A at r is at most s′ . Each subtree Si is rooted at r and shares at least one edge with A. Since no two subtrees in S ′ share an edge, this is a contradiction. Therefore, K2,s has an (h∗ , s′ , 2)-representation if and only if s′ ≥ s. □ Corollary 2.4. K2,n ∈ [h, s, 2] if and only if h ≥ 2s and s ≥ n. 2.2. The case t = 3 We first prove two lemmas. Lemma 2.5. For n > s2 − 1, in any (h, s, 3)-representation of K2,n on a host tree T , the subtrees A and B share exactly one edge. Proof. Suppose that A and B do not share a vertex. The nonempty path from A to B contains exactly one edge e, which is shared by all subtrees in S ′ . Since the degree of A and B is at most s, the number of subtrees in S ′ is at most s, a contradiction to the number of subtrees n > s in S ′ . Thus, A and B share a vertex. Suppose that A and B share exactly one vertex; denote it by r. Choose r to be the root of T , and let ra , rb , {ri } be the roots of A, B, {Si }, respectively. Since A and B share only r, we have ra = rb = r. Furthermore, ri = r for 1 ≤ i ≤ n. Let the edges of A containing r be called A-links, and similarly let the edges of B containing r be called B-links. Note that each of A and B has at most s edges containing r, and each Si must include r and at least one A-link and one B-link. Any subtrees Si and Sj that contain the same A-link must contain different B-links, otherwise at least two Si will share two edges, a contradiction. Therefore, an A-link can be contained in at most s subtrees. Let l be an A-link that is contained in s subtrees, and is the only common edge of these subtrees. Each one of these subtrees must contain an additional edge of A. Since the degree of A is at most s, at least one of these subtrees contains another A-link. Let S ′ be a subtree that contains two A-links l and l′ . Since there is at least one such subtree S ′ , there are at most s2 − 1 subtrees, which contradicts the number of subtrees n > s2 − 1 in S ′ . Hence, A and B do not share exactly one vertex. Since A and B cannot share two edges or more, they must share exactly one edge. □ Lemma 2.6. Consider K2,n ∈ [h, s, 3] such that n > s2 − 1. If h < 2s − 1, then n ≤ 2(s − 1)2 + 2(s − 1) − 1 = 2s(s − 1) − 1. Otherwise, n ≤ 2(s − 1)2 + 2(s − 1) = 2s(s − 1). Proof. By Lemma 2.5, A and B share exactly one edge e. Let e be called the bridge. Edges incident to e are called the ramps. A ramp that is contained in A or B is called an A-ramp or a B-ramp, respectively. There are at most s − 1 A-ramps and s − 1 B-ramps on each side of the bridge. There are two kinds of subtrees in S ′ : Bridge crossing: Subtrees in S ′ that share the bridge together with A and B. Each bridge crossing subtree must contain an A-ramp on one side of the bridge and a B-ramp on the other side. If h < 2s − 1, then without loss of generality there are at most s − 2 A-ramps. Since all bridge crossing subtrees share the bridge, no two such subtrees share a ramp. Therefore, if h < 2s − 1, there can be at most 2(s − 1) − 1 bridge-crossing subtrees in S ′ . Otherwise, if h ≥ 2s − 1, then there can be at most 2(s − 1) bridge-crossing subtrees in S ′ .

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Fig. 2. K2,n representation for t = 3 with examples of red (bold) and blue (mid) paths. (For interpretation of the references to colour in this figure legend, the reader is referred to the web version of this article.)

Bridge non-crossing: Subtrees in S ′ that contain ramps from only one side. Let x and y be the endpoints of e. Each bridge non-crossing subtree must contain an A-ramp, a B-ramp and either x or y. Any two subtrees containing x and the same A-ramp must contain different B-ramps. Since there are at most s − 1 A-ramps and s − 1 B-ramps containing x, there can be at most (s − 1)2 bridge non-crossing subtrees containing x in S ′ . The same argument holds for all bridge non-crossing subtrees containing y in S ′ . Thus, if h < 2s − 1, there are at most 2(s − 1)2 + 2(s − 1) − 1 subtrees in S ′ . Otherwise, there are at most 2(s − 1)2 + 2(s − 1) subtrees in S ′ . □ Theorem 2.7. K2,f (s) is sharply contained in [2s − 1, s, 3], where f (s) = 2(s − 1)2 + 2(s − 1) = 2s(s − 1). Proof. We first show K2,f (s) ∈ [2s − 1, s, 3]. We describe a (2s − 1, s, 3)-representation of K2,f (s) . See Fig. 2. Consider two stars with (2s − 1) edges sharing one edge e = (x, y). Let e be called the bridge, and the rest of the star edges called ramps, where the ramps incident to x form the x-side and the ramps incident to y form the y-side. Also, add s − 1 edges at the endpoint of each ramp not incident to e to form a star with degree s. Construct subtrees A and B that each contain the bridge, so that on each side of the bridge half of the ramps (and their incident edges) are assigned to A and half to B. Label the ramps belonging to A and B on each side of the bridge as A-rampk or B-rampk , respectively, for 1 ≤ k ≤ s − 1, and their star’s edges as A-legk,j and B-legk,j , for 1 ≤ j ≤ s − 1. We define three types of subtrees Si : subtrees that are totally in the x-side, subtrees that are totally in the y-side, and subtrees that cross the bridge. We construct each Si to be a path. On the x-side of the bridge, assign for each {Sl }, writing l = j + (k − 1)(s − 1) (uniquely, in base s − 1 arithmetic), a unique red path [A-legk,j , A-rampk , B-rampj , B-legj,k ], for 1 ≤ k, j ≤ s − 1. Notice that no two red paths share a leg nor the same two ramps. Similarly, we construct (s − 1)2 red paths on the y-side. The rest of the subtrees of {Si } intersect on the bridge, and contain only ramps (no legs). Half are being assigned with blue paths [x-side-A-rampk , bridge, y-side-B-rampk ], and the other half with blue paths [y-side-B-rampk , bridge, x-side-A-rampk ], for 1 ≤ k ≤ s − 1. Note that no two blue paths share a ramp. Thus, in all cases, no two Si ’s share more than one edge. Therefore, in K2,f (s) the set {si } is an independent set, as expected. Thus, K2,f (s) ∈ [2s − 1, s, 3]. Since f (s) > s2 − 1, if h < 2s − 1 then by Lemma 2.6, there are at most 2(s − 1)2 + 2(s − 1) − 1 vertices in the larger side of the bipartition. Since f (s) = 2(s − 1)2 + 2(s − 1), this is a contradiction. Therefore, h ≥ 2s − 1. Suppose K2,f (s) has a (2s − 1, s′ , 3)-representation for s′ < s. By Lemma 2.6, there are at most 2(s′ − 1)2 + 2(s′ − 1) vertices. Since f (s) > 2(s′ − 1)2 + 2(s′ − 1), this is a contradiction. □ Corollary 2.8. K2,n ∈ [h, s, 3] if and only if h ≥ 2s − 1 and n ≤ 2(s − 1)2 + 2(s − 1). 3. Characterization of [4, 3, 2] In this section, we characterize the class [4, 3, 2]. These results are based in part on the previous results of [8,12] for the class [4, 4, 2], as stated in Theorem 1.2. We define a new family of graphs, namely, the 4-flower graphs. Definition 3.1. Let GQ be a graph with vertex set Q = {qij : 1 ≤ i < j ≤ 4} and edge set defined as follows: Let Qi = {qjk : j = i or k = i, 1 ≤ j < k ≤ 4}. Each vertex set Qi is a clique in GQ . A 4-flower graph consists of: 1. An induced subgraph GQ ′ of GQ , such that |Qi′ | ≥ 2, for every 1 ≤ i ≤ 4, where Qi′ = Qi ∩ Q ′ . 2. A vertex r, which is adjacent to all vertices in Q ′ .

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(a) An example of a 4-flower graph.

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(b) The unique 4-flower graph of order 9.

Fig. 3. Examples of 4-flower graphs.

3. Four distinct sets of vertices P1 , . . . , P4 , where Pi is either a single vertex that is adjacent to at least two elements of Qi′ or an edge that is adjacent to at least two elements of Qi′ . The vertex r is adjacent to at least one vertex in each Pi . Figs. 3(a) and 3(b) show examples of 4-flowers and Figs. 4(a) and 4(b) show their representations, respectively. A 4-flower is an example of a graph which is in [4, 4, 2], but is not in [4, 3, 2], as we will show in Lemma 3.4. The chordless cycle Cn ∈ [∞, 2, 2], and Golumbic and Jamison [7] show that there is a unique (∞, 2, 2)-representation of Cn called a pie. Definition 3.2. A pie is a star subgraph of a tree T with n edges (a1 , q), (a2 , q), . . . , (an , q), such that each ‘‘slice’’ (ai , q) ∪ (ai+1 , q), for i = 1, . . . , n (arithmetic modulo n), is contained in a different member of S . The vertex q is called the center of the pie. In [8,12], the authors generalize the result of [7] to the case of [∞, ∞, 2]: Theorem 3.3 ([8,12]). If a graph G ∈ [∞, ∞, 2] contains a chordless cycle C = (x1 , x2 , . . . , xn , x1 ) (n ≥ 4), then the representation T contains a pie on these n vertices. We now prove that the 4-flower graphs are sharply contained in [4, 4, 2]. Lemma 3.4. 4-flower is sharply contained in [4, 4, 2]. Proof. Suppose there exists a (3, s, 2)-representation ⟨S , T ⟩ of a 4-flower graph G = (V , E). By definition a 4-flower graph contains an induced C4 . Then by Theorem 3.3 we have a contradiction. Hence, a 4-flower graph does not have an (3, s, 2)representation for any s. Suppose there exists an (h, 3, 2)-representation ⟨S , T ⟩ of a 4-flower graph G = (V , E). If all six vertices of ∪Qi , for 1 ≤ i ≤ 4, are in V then the cycle C = (q12 , q23 , q34 , q14 , q12 ) is an induced C4 in G. Without loss of generality, if q24 ̸ ∈ V then by Definition 3.1 the vertices q12 , q23 , q34 , q14 are in V . So, we have C as an induced C4 in G. By Theorem 3.3, the subtrees corresponding to the vertices of C form a pie in ⟨S , T ⟩ . Let q be the center vertex of a substar with neighbors a1 , a2 , a3 , a4 in ⟨S , T ⟩ , such that the edges (q, a1 ), (q, a2 ), (q, a3 ), (q, a4 ) are the edges of the pie and (q, a1 ) ∪ (q, a2 ), (q, a2 ) ∪ (q, a3 ), (q, a3 ) ∪ (q, a4 ), (q, a4 ) ∪ (q, a1 ) are the ‘‘slices’’ that are contained in the subtrees which correspond to q12 , q23 , q34 , q14 , respectively. Let Ti be the host subtree containing ai , obtained from T by deleting (q, ai ). Since r is adjacent to each vertex in C , the subtree Sr must contain the center vertex q. Moreover, by assumption, Sr has degree at most 3, and therefore Sr has an edge in at most three of the {Ti }. The vertex r is adjacent to at least one vertex in each element of {Pl }, 1 ≤ l ≤ 4. Therefore, there is one host subtree Tk , 1 ≤ k ≤ 4, and a pair of indices i, j, where i ̸ = j, such that the subtrees corresponding to vertices in Pi and Pj have edges only in Tk . By definition, each of Pi and Pj has at least two neighbors on C . Moreover, Pi and Pj have at most one common neighbor on C . Hence, at least three subtrees which correspond to the vertices of C share the edge (q, ak ), a contradiction. Thus, the 4-flower graph does not have an (h, 3, 2)-representation for any h. However, the 4-flower graph has a (4, 4, 2)path of length two, and when |Pi | = 2, Ti is isomorphic to a representation as follows: when |Pi | = 1, Ti is isomorphic to a ⋃ K1,4 centered at ai . Each subtree corresponding to a vertex p ∈ Pi consists of a subpath Sp . When |Pi | = 1, Sp contains one edge ep , and when |Pi | = 2, Sp contains a distinct subpath of length two of Ti . The subtree corresponding to qij consists of the

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(a) The (4, 4, 2)-representation of the 4-flower graph shown in Fig. 3(a).

(b) The (4, 4, 2)-representation of the unique 4-flower graph of order 9 shown in Fig. 3(b).

Fig. 4. The (4, 4, 2)-representations of the 4-flower graph examples.

edges (q, ai ), (q, aj ) and the set of edges {ep }, for⋃each p ∈ Pi that is adjacent to qij . Similarly, the subtree Sr consists of the entire pie, as well as the edges ep , for each p ∈ Pi adjacent to r. □

⋃

Corollary 3.5. The class [4, 3, 2] is strictly contained in the class [4, 4, 2]. In Section 5, we present a new algorithm to recognize [4, 3, 2]. A direct consequence of the correctness of the algorithm, which is stated in Theorem 5.7, is the following corollary: Corollary 3.6. If G ∈ [4, 4, 2] and is a 4-flower-free graph, then G ∈ [4, 3, 2]. Lemma 3.4 and Corollary 3.6 yield a characterization of the class [4, 3, 2]. Theorem 3.7. A graph G ∈ [4, 3, 2] if and only if G is a 4-flower-free graph and G ∈ [4, 4, 2]. Proof. If G ∈ [4, 3, 2], then it is certainly in [4, 4, 2]. Moreover, G does not contain a 4-flower graph by Lemma 3.4. Conversely, if G is a 4-flower-free graph and G ∈ [4, 4, 2], then by Corollary 3.6, G ∈ [4, 3, 2]. □ 4. The hierarchy In this section, we investigate the relationship between various classes in the framework of [h, s, t ] and the well-known families of chordal and weakly chordal graphs. Specifically, we demonstrate the results illustrated in the hierarchy shown in Fig. 5. We also prove that the class of weakly chordal graphs is incomparable with [h, s, t ] for t > 2 or t = 2 and h ≥ 5. We say that a hierarchy is complete, when all containment relationships are given. That is, (1) classes that appear in the same box are equivalent, (2) a downward edge from class A to class B indicates that class A contains class B, (3) an example appearing along the edge between two classes is a separating example for those classes, (4) the lack of a hierarchical (containment) relation indicates that the classes are incomparable. The following lemma is needed for the completeness of the hierarchy presented in Fig. 5: Lemma 4.1. The n-wheel Wn is sharply contained in [n, ⌈n/2⌉, 2]. Proof. Let ⟨S , T ⟩ be an (h, s, 2)-representation of the n-wheel Wn . By Theorem 3.3, the n-cycle forms a pie in any (∞, ∞, 2)representation. Therefore, h ≥ n. Let q be the center vertex of a star with neighbors a1 , . . . , ah , where h ≥ n, such that a1 , . . . , an form the pie. Since a maximal clique in a cycle is 2, every edge (ai , q) is contained in at most two subtrees corresponding to v1 , . . . , vn . Therefore, the subtree S corresponding to the vertex that is adjacent to all of v1 , . . . , vn , must contain at least n/2 of the edges {(ai , q)} for even n and (n + 1)/2 of the edges {(ai , q)} for odd n. Therefore, the degree of S at q is at least ⌈n/2⌉. The n-wheel Wn , for even n, has an (n, n/2, 2)-representation consisting of the pie and the subtree S containing the edges (a1 , q), (a3 , q), . . . , (an−1 , q). The n-wheel Wn , for odd n, has an (n, (n + 1)/2, 2)-representation consisting of the pie and the subtree S containing the edges (a1 , q), (a3 , q), . . . , (an−2 , q), (an−1 , q). □

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Fig. 5. The hierarchy.

Theorem 4.2. The hierarchy in Fig. 5 is complete. Proof. Equivalences. The equivalence of the class [4, 4, 2] and the class of weakly chordal (K2,3 , P6 , 4P2 , P2 ∪ P4 , H1 , H2 , H3 )-free graphs is proved in [8,12]. The equivalence of [4, 3, 2] and 4-flower-free [4, 4, 2] is proved in Theorem 3.7. All other equivalences are proved in [7,11,15] and [16]. Containments. Every graph in [h, s, t ] is also in [h′ , s′ , t ] for h ≤ h′ and s ≤ s′ . All the containments immediately follow by definition. Separation examples. The following are the separation examples as shown in Fig. 5. The graph K2,n , for n ≥ 3, is not in [4, 4, 2] by [8,12], but is a weakly chordal graph. The graph K2,3 ∈ [∞, 3, 2]. The graphs K2,s+1 and W2s+1 are not in [∞, s, 2], but are in [∞, s + 1, 2], by Corollary 2.4 and Lemma 4.1, respectively. The graph C5 is not weakly chordal by definition, but does belong to [∞, 2, 2] by [7]. The graph C4 is not chordal, but does belong to [4, 2, 2] by [7]. The 4-flower graph is in [4, 4, 2] by Lemma 3.4. The graph A4 is not an EPT graph by [11], but is

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Fig. 6. A (3, 2, 3)-representation of the graph C5 .

chordal. Finally, as proved in [3], the graph T2 is not an interval graph, but can be easily verified to be a chordal graph and an EPT graph. Incomparabilities. The family of weakly chordal graphs is incomparable with [∞, s, 2], for every fixed finite s ≥ 2: the graph K2,s+1 and the graph C5 are the separation examples. The incomparability of the class of chordal graphs and [∞, 2, 2] is proved in [11], with C4 and A4 as separation examples. □ We now state the main result of this section: Theorem 4.3. The class [h, s, t ], for any fixed h, t, is incomparable with the class of weakly chordal graphs if and only if either (i) t > 2 and h ≥ 3, or (ii) t = 2 and h ≥ 5. Proof. (⇐) Let [h, s, t ] be a class of graphs with fixed t ≥ 2. Then, by Theorem 2.1, we have K2,n ̸ ∈ [h, s, t ], for n > γ (s, t)(t + 1), but K2,n is weakly chordal. The graph C5 is not weakly chordal, but has a (5, 2, 2)-representation consisting of a degree 5 pie (see Theorem 3.3), and a (3, 2, 3)-representation as shown in Fig. 6. Hence, by containment C5 ∈ [h, s, t ]. Therefore, [h, s, t ] is incomparable with the class of weakly chordal graphs. (⇒) Let [h, s, t ] be a class of graphs which is incomparable with the class of weakly chordal graphs. Then t ≥ 2, since the graphs in [h, s, 1] are chordal by [5], and therefore are contained in the class of weakly chordal graphs. Suppose t = 2. We consider all possibilities for h < 5. The graph classes [4, 4, 2], [4, 3, 2], [4, 2, 2], [3, 3, 2], [3, 2, 2] are contained in the class of weakly chordal graphs by Theorem 4.2. The class [2, 2, 2] is the family of interval graphs, which are weakly chordal graphs. Therefore, h ≥ 5. In fact, the class [2, 2, t ], for t ≥ 1, is the family of interval graphs. Hence, if t > 2 then h ≥ 3. □ 5. Recognition algorithm of [4, 3, 2] graphs We now present the recognition algorithm Recognize [4, 3, 2]: After verifying that G is weakly chordal (Step 1) and that G contains none of the forbidden subgraphs K2,3 , 4P2 , P2 ∪ P4 , P6 , H1 , H2 , H3 , 4-flow er (Step 2), the algorithm then initializes a (4, 4, 2)-representation (Step 3), using the algorithm of [8,12]. The heart of our algorithm is the iterative loop (Step 4), which refines the representation at each vertex u of the host tree, by reducing the degree of those subtrees having degree 4 at u (and proved correct at Lemma 5.4). The algorithm answers either that the input graph is not in [4, 3, 2], when stopping in either Step 1 or Step 2, or finds a (4, 3, 2)-representation of the input graph.

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Recognize [4,3,2] input: A graph G output: ‘no’ (if G ∈ / [4, 3, 2]), otherwise, a (4, 3, 2)-representation ⟨S , T ⟩ of G begin if G is not weakly chordal (verify using a known algorithm such as [1]) then return ‘no’ ; if G has an induced subgraph isomorphic to one of {K2,3 , 4P2 , P2 ∪ P4 , P6 , H1 , H2 , H3 , 4-flower} then return ‘no’;

⟨S , T ⟩ ← Construct (4, 4, 2)-representation(G) (given in [8, 12]); while ∃ a vertex u ∈ T , such that ∃ subtree S ∈ ⟨S , T ⟩ with degree 4 at u do ⟨S , T ⟩ ← Subtree-degree-reduce4(⟨S ,T ⟩,u); end

Definition 5.1. Let ⟨S , T ⟩ be an (h, s, t)-representation of G. Let u be a vertex in T and let GU be the induced subgraph of G, such that every vertex in GU corresponds to a subtree in ⟨S , T ⟩ that contains the vertex u. The collection of subtrees SU

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corresponds to the vertex set of GU . Every subtree in SU is called a core subtree, and the other subtrees in ⟨S , T ⟩ are called non-core subtrees, with respect to the vertex u. A core edge in T is one that is contained only in core subtrees, and the other edges in T are called non-core edges. Definition 5.2. A two-pair in a graph G is a pair of vertices {x, y} such that every chordless path between x and y contains exactly two edges. Clearly, the common neighborhood of a two-pair {x, y} is a minimal (x, y)-separator, which we denote by Sep(x, y). Theorem 5.3 ([14]). A graph G is weakly chordal if and only if every induced subgraph of G either has a two-pair or is a clique.

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Subtree-degree-reduce4 procedure input: a (4, 4, 2)-representation ⟨S , T ⟩ of G and a vertex u ∈ T at which there is a subtree of degree 4 output: a (4, 4, 2)-representation ⟨S ′ , T ′ ⟩ of G with fewer vertices with a subtree of degree 4, such that u has no subtree with degree 4 begin ⟨S , T ⟩ ← PreprocessingA(⟨S , T ⟩, u); find the induced subgraph GU ; if GU is a clique then ⟨S ′ , T ′ ⟩ ← TransformationB(⟨S , T ⟩, u); end procedure; //GU is not a clique find a two-pair {x, y} in GU and Sep(x, y); if Sep(x, y) is a clique then ⟨S ′ , T ′ ⟩ ← TransformationC-subtree(⟨S , T ⟩, u, Sx ); end procedure; //Sep(x, y) is not a clique find {C ij }, 1 ≤ i < j ≤ 4 and C 5 in GU by Definition 5.5; MultiColoringD ←MultiColoringD-subtree(⟨S , T ⟩, u, {C ij } for 1 ≤ i < j ≤ 4, C 5 ); ⟨S ′ , T ′ ⟩ ← TransformationD-subtree(⟨S , T ⟩, u, MultiColoringD); end

The following is the main lemma needed to prove the correctness of the recognition algorithm. Lemma 5.4. Let ⟨S , T ⟩ be a (4, 4, 2)-representation of a 4-flower-free graph G. If ⟨S , T ⟩ is an input to Subtree-degreereduce4(⟨S , T ⟩, u), then the output is a (4, 4, 2)-representation of G with fewer vertices at which there is a subtree of degree 4, and such that u has no subtree of degree 4. Moreover, no subtree in ⟨S , T ⟩ has increased its maximal degree beyond 3 in the output representation. Proof. By Theorem 1.2, G is a weakly chordal (K2,3 , 4P2 , P2 ∪ P4 , P6 , H1 , H2 , H3 )-free graph. The Steps of the procedure Subtreedegree-reduce4 are justified with correspondingly numbered Claims. We first outline the flow of the proof: Step 1. The procedure PreprocessingA(⟨S , T ⟩, u) finds a (4, 4, 2)-representation of G, such that every subtree in S uses either no edges with the endpoint u or two edges or four edges with the endpoint u according to Claim for Step 1. Moreover, the number of vertices in T contained in a subtree with degree 4 remains the same. Step 2. We find the induced subgraph GU , where each vertex in GU corresponds to a subtree that contains the vertex u in T . This can be easily done. By the hereditary property, the induced subgraph GU is also a weakly chordal graph. Step 3. According to Theorem 5.3, the subgraph GU is either a clique or has a two-pair. If GU is a clique, then at Step 3 we call TransformationB(⟨S , T ⟩, u) and Subtree-degree-reduce4 ends. The output is a (4, 4, 2)-representation of G with fewer vertices with subtree of degree 4, such that u has no subtree with degree 4, according to Claim for Step 3. Moreover, no subtree has increased its maximal degree beyond 3 in the output representation. Otherwise, at Steps 4–8 we assume that GU is not a clique and therefore has a two-pair. Step 4. We find a two-pair {x, y} and the set Sep(x, y). See [14] for further details. Step 5. If Sep(x, y) is a clique, then we perform TransformationC-subtree(⟨S , T ⟩, u, Sx ), and Subtree-degree-reduce4 ends. By Claim for Step 5, this finds a (4, 4, 2)-representation of G with fewer vertices with subtree of degree 4, such that u has no subtree of degree 4. Moreover, no subtree has increased its maximal degree beyond 3 in the output representation. Otherwise, at Steps 6–8 we assume that Sep(x, y) is not a clique. Step 6. We find the sets {Cij }, 1 ≤ i < j ≤ 4, and C C5 in GU by Definition 5.5. Step 7. We call MultiColoringD-subtree(⟨S , T ⟩, u, {Cij } for 1 ≤ i < j ≤ 4, C5 ) to obtain MultiColoringD, which has the properties proved in Claim for Step 7.

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Step 8. We call TransformationD-subtree(⟨S , T ⟩, u, MultiColoringD) and according to Claim for Step 8 obtain a (4, 4, 2)representation with fewer vertices with subtree of degree 4, such that u has no subtree with degree 4. Moreover, no subtree has increased its maximal degree beyond 3 in the output representation.

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PreprocessingA procedure input: ⟨S , T ⟩, u output: ⟨S , T ⟩ //let v1 , . . . , v4 be the neighbors of u in T ; foreach star edge (vi , u) do find S (vi , u), which is the collection of subtrees in S that contains only the edge (vi , u) among (v1 , u), . . . , (v4 , u); split the edge (vi , u) into two edges, by adding a dummy vertex wi such that: foreach subtree S ∈ S (vi , u) do replace (vi , u) by the edge (vi , wi ) in S (thus making wi the endpoint of S); foreach subtree S ∈ / S (vi , u) and (vi , u) is contained in S do replace (vi , u) by the two edges (vi , wi ) and (wi , u) in S.

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find S ∗ ⊂ S the collection of subtrees such that each contains exactly three edges among (w1 , u), . . . , (w4 , u); if S ∗ ̸ = ∅ then foreach subtree Sv ∈ S ∗ do add the non-existing edge (wi , u) to Sv ;

TransformationB procedure input: ⟨S , T ⟩, u output: ⟨S ′ , T ′ ⟩ add the edge (w1 , w2 ); foreach subtree S ∈ S U that contains the edge (u, w2 ) do replace (u, w2 ) by the two edges (u, w1 ) and (w1 , w2 ) in S; remove the edge (u, w2 );

TransformationC-Subtree procedure input: ⟨S , T ⟩, u, Sx output: < S ′ ,T ′ > //Suppose Sx contains the star edges (w1 , u) and (w2 , u) ; add the edge (w1 , w2 ); foreach subtree S ∈ S U that contains the edge (u, w2 ) do replace (u, w2 ) by the two edges (u, w1 ) and (w1 , w2 ) in S; remove the edge (u, w2 );

MultiColoringD-Subtree procedure input: ⟨S , T ⟩, u, {C ij } for 1 ≤ i < j ≤ 4, C 5 output: MultiColoringD (C (e) for each e ∈ T and C (S) for each non-core subtree S) define seven colors {cλ }, λ ∈ {12, 13, 14, 23, 24, 34, 5}, such that each cλ corresponds to C λ ; multicolor each edge e ∈ T with all colors cλ , such that e is contained in a core subtree in C λ ; repeat if an edge e ∈ T is contained in a non-core subtree S with an edge colored cλ then color the edge e with color cλ ; until no further coloring is possible; for all edges e ∈ T , C (e) ← the set of colors of e; for all non-core subtrees S, C (S) ← ∪{C (e) : e is contained in S }; Proofs of the Claims: In all Claims that we are about to prove, we assume that ⟨S , T ⟩ is a (4, 4, 2)-representation of a 4-flower-free graph G. Claim for Step 1. By performing PreprocessingA we obtain a (4, 4, 2)-representation of G, such that every subtree in S either uses no edges with the endpoint u or uses two or four edges with the endpoint u. Moreover, the number of vertices in T contained in a subtree with degree 4 remains the same.

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TransformationD-subtree procedure input: ⟨S , T ⟩, u, MultiColoringD output: ⟨S ′ , T ′ ⟩ find 1 ≤ piv ot ≤ 4, such that S vpivot is empty; ⟨S , T ⟩ ← MovePivot(⟨S , T ⟩, u, MultiColoringD, Tvpivot ); foreach subtree Sv , v ∈ C 5 do Remove the edge (u, vpiv ot ) and its edges in Tvpivot from Sv ;

MovePivot procedure input: ⟨S , T ⟩, u, MultiColoringD, Tv output: ⟨S , T ⟩ foreach core edge e = (v, a) ∈ Tv , c5 ∈ C (e) do ⟨S , T ⟩ ← MovePivot(⟨S , T ⟩, u, MultiColoringD, Ta ); while ∃ a non-core edge e = (v, a) ∈ Tv , c5 ∈ C (e) do E ← {e} ∪ {e′ = (v, a′ ) ∈ Tv : e′ is a non-core edge and C (e′ ) = C (e)}; find star edge e′′ ̸ = (u, vpiv ot ) and C (e′′ ) ⊇ C (e); split the star edge e′′ into two edges, by adding a dummy vertex w ; replace the star edge e′′ by the two edges in every core subtree containing it; add a dummy edge (w ′ , w ); foreach edge (v, v ′ ) ∈ E do replace (v, v ′ ) by (w, w ′ ), (w ′ , v ′ ) in every core subtree containing it; replace (v, v ′ ) by (w ′ , v ′ ) in every non-core subtree containing it; delete the edge (v, v ′ ) from T ;

Proof. At Step 1, any two subtrees that share a common edge (vi , u), 1 ≤ i ≤ 4, in the input correspond to adjacent vertices in G, and these subtrees share the common edge (vi , wi ) in the output. Moreover, any two subtrees that do not share a common edge (vi , u), 1 ≤ i ≤ 4, in the input correspond to non-adjacent vertices in G, and these subtrees do not share a common edge after Step 1 is performed. Thus, each subtree either uses no edges from (w1 , u), . . . , (w4 , u) or uses at least two edges among (w1 , u), . . . , (w4 , u). After Step 1 is performed, if a subtree Sv in the input contains exactly three edges among (vi , u), 1 ≤ i ≤ 4, then v is adjacent to all vertices corresponding to subtrees containing u ∈ T . Therefore, by adding a non-existing edge (wi , u) to Sv , any two subtrees that do not share a common edge (wi , u), 1 ≤ i ≤ 4, after Step 1 is performed, do not share a common edge after Step 3 is performed. Thus each subtree uses two or four edges among (w1 , u), . . . , (w4 , u). Therefore, the output is a (4, 4, 2)-representation of G. Since there exists a subtree with degree 4 at u ∈ T in the input, the number of vertices in T contained in a subtree with degree 4 remains the same. □ Claim for Step 3. If GU is a clique and there exists a subtree of degree 4 at u, then the output of TransformationB is a (4, 4, 2)representation of G with fewer vertices with subtree of degree 4, such that u has no subtree with degree 4. Moreover, no subtree has increased its maximal degree beyond 3 in the output representation. Proof. We arbitrarily choose to change the edge (w2 , u). Since SU is a clique, the subtrees that contain (w1 , u) or (w2 , u) in ⟨S , T ⟩ correspond to adjacent vertices in G, and these subtrees contain the common edge (w1 , u) in ⟨S ′ , T ′ ⟩ . All the other subtrees remain unchanged. Therefore, ⟨S ′ , T ′ ⟩ is a (4, 4, 2)-representation of G. In ⟨S ′ , T ′ ⟩ the degree of u is 3 and clearly there is no subtree with degree 4 at u. The degree of w is 3. The degree of all the other vertices in ⟨S ′ , T ′ ⟩ is the same as in ⟨S , T ⟩. Hence, ⟨S ′ , T ′ ⟩ is a (4, 4, 2)-representation of G with fewer vertices with subtree of degree 4, such that u has no subtree with degree 4. Moreover, no subtree has increased its maximal degree beyond 3 in the output representation. □ Claim for Step 5. If Sep(x, y) is a clique, then the output of TransformationC-subtree is a (4, 4, 2)-representation of G with fewer vertices with subtree of degree 4, such that u has no subtree of degree 4. Moreover, no subtree has increased its maximal degree beyond 3 in the output representation. Proof. Since Sx contains the edges (w1 , u) and (w2 , u), Sy contains the edges (w3 , u) and (w4 , u), and the subtrees that contain either (w1 , u) or (w2 , u) correspond to vertices adjacent to x in GU . The subtrees that contain (w1 , u) and (w2 , u) correspond to a clique in GU .

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Suppose there exist two subtrees, such that each contains only one from (w1 , u) and (w2 , u) and the subtrees do not share an edge. Then, each of the two subtrees contains one from (w3 , u) and (w4 , u) and their corresponding vertices are adjacent to y. Therefore, these subtrees correspond to non-adjacent vertices in Sep(x, y). Contradiction! Therefore, all the subtrees that contain (w1 , u) or (w2 , u) correspond to a clique in GU . Only the subtrees that contain (w1 , u) or (w2 , u) are changed to contain the common edge (w1 , u). Since these subtrees correspond to a clique in GU , the output is a (4, 4, 2)-representation of G. In ⟨S ′ , T ′ ⟩ the degree of u is 3 and clearly there is no subtree with degree 4 at u. The degree of w is 3. The degree of all the other vertices in ⟨S ′ , T ′ ⟩ is the same as in ⟨S , T ⟩. Hence, ⟨S ′ , T ′ ⟩ is a (4, 4, 2)-representation of G with fewer vertices with subtree of degree 4, such that u has no subtree with degree 4. Moreover, no subtree has increased its maximal degree beyond 3 in the output representation. □ Definition 5.5. Let v1 , . . . , v4 be the neighbors of u ∈ T . For 1 ≤ i < j ≤ 4, we define the cliques Cij = {v ∈ GU : Sv contains exactly the star edges (vi , u) and (vj , u) among the star edges}. In addition, C5 = {v : v ∈ GU − {Cij }, 1 ≤ i < j ≤ 4}. Note that these seven cliques partition SU and, by Claim for Step 1, each subtree Sv , v ∈ C5 , contains (v1 , u), . . . , (v4 , u). Definition 5.6. In MultiColoringD, a non-core subtree S touches a color c ∈ C (S) if S shares an edge with at least one core subtree corresponding to color c. Let Tvi be the subtree rooted at vi ∈ T , obtained by removing the vertex u from T . Let Svi , 1 ≤ i ≤ 4, be the collection of non-core subtrees S that are contained in Tvi , such that c5 ∈ C (S) and |C (S)| > 2. For any vertex a ∈ Tvi , we further define the subtree Ta of Tvi to be rooted at a. Claim for Step 7. In the output of MultiColoringD-subtree the following hold: I. Suppose there exists i, such that Svi ̸ = ∅. (i) ∃S ∈ Svi that touches at least two colors. (ii) At least one of the following holds: (a) ∃S ∈ Svi that touches c5 and touches at least two other colors. (b) ∃S , S ′ ∈ Svi such that S and S ′ share an edge in T , and S ∪ S ′ touches c5 and at least two other colors. II. ∃i, such that Svi = ∅. III. Let e be a non-core edge, such that c5 ∈ C (e) and e is contained in a subtree Tvi , where Svi = ∅. Then there exists a star edge (u, vj ), such that C (e) ⊆ C ((u, vj )) where i ̸ = j. Proof. I. (i) Suppose all subtrees in Svi touch at most one color. Since Svi ̸ = ∅, there exist subtrees Sva , Svb , Svc ∈ Svi , such that each one touches a different color ca , cb , cc . We choose, from all possible such triplets, subtrees such that the path from va to vb in G is minimal and the path from vb to vc in G is minimal. Then (va , vb ) ∈ E(G), since otherwise the path from va to vb together with the neighbor of va in the clique corresponding to ca and the neighbor of vb in the clique corresponding to cb form a chordless cycle of size greater than 4. If (vb , vc ) ∈ E(G), then (va , vc ) ∈ E(G), since otherwise the vertices va , vb , vc together with the neighbor of va in the clique Ca and the neighbor of vc in the clique corresponding to Cc form a chordless cycle of size greater than 4. Thus, the vertices va , vb , vc together with one of their neighbors in the cliques Ca , Cb , Cc form the forbidden subgraph C6 in G. Otherwise, let P be the minimal path from vb to vc in G. Let vb′ be the last vertex on P, such that Sv ′ touches the color cb (possibly vb′ = vb ). b Then (vb′ , vc ) ∈ E(G), since otherwise vb′ , vc together with a neighbor of vb′ in the clique Cb and a neighbor of vc in the clique ′ Cc form a chordless cycle of size greater than 4. Thus, the vertices va , vb , vc together with their neighbors in the cliques Ca , Cb , Cc form the forbidden subgraph C6 in G. (ii) Suppose (a) does not hold. By (i), there exists a subtree Sv ∈ Svi that touches colors ca , cb ∈ C (Sv ). Since Svi ̸ = ∅, there exists a subtree Sw ∈ Svi that touches cc ∈ C (Sw ) where one of ca , cb , cc is c5 . We choose Sv and Sw such that the path P = [v, . . . , w] is minimal in G and P does not contain a vertex corresponding to a core subtree. If Sv and Sw share an edge, then (b) holds. Otherwise, Sv and Sw do not share an edge. The subtree Sw possibly touches ca or cb , but not both. Without loss of generality, suppose Sw does not touch cb . Then we show a contradiction in each one of the following cases: Case 1: None of the subtrees corresponding to the vertices on the path P, besides Sv , touches cb . Then the path P together with a neighbor of v in Cb and a neighbor of w in Cc form a chordless cycle greater than 4. Contradiction! Case 2: There exists a subtree corresponding to a vertex on P that touches cb . Let z be the last vertex on P, such that its corresponding subtree Sz touches cb . Then (z , w ) must be an edge on P, since otherwise [z , . . . , w] together with a neighbor of z in Cb and a neighbor of w in Cc form a chordless cycle greater than 4. Similarly, if z ′ is the last vertex on the path P, such that its corresponding subtree Sz ′ touches ca , then (z ′ , w ) must be an edge on P. Thus, z = z ′ and Sz touches both ca and cb . This is a contradiction to the minimality of P, since the path from w to z is shorter than P. II. For a contradiction, suppose that for all i, 1 ≤ i ≤ 4, Svi ̸ = ∅. We now construct a 4-flower graph as defined in Definition 3.1.

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If there exists S ∈ Svi that satisfies I.(ii).(a), then let Pi = {pi } be the vertex corresponding to S. Otherwise, there exist subtrees S , S ′ ∈ Svi that satisfy I.(ii).(b). Then let Pi = {pi , p′i } be the vertex set corresponding to {S , S ′ }. Since for all i, 1 ≤ i ≤ 4, Svi ̸ = ∅, then Pi ̸ = ∅. Let qij correspond to a vertex from Cij for all i < j and let r correspond to a vertex from C5 ̸ = ∅. This graph is a 4-flower, according to Definition 3.1. Contradiction! III. Since Svi = ∅, every non-core subtree in Tvi colored c5 has at most two colors. Therefore, every non-core edge in Tvi is colored c5 and possibly another color cλ , λ = ij. The star edge (u, vj ) is colored c5 and cλ , λ = ij. Therefore, C (e) ⊆ C ((u, vj )). □ Consider TransformationD-subtree procedure. In Step 1, a pivot is chosen according to Claim for Step 7(II). In Step 2, the recursive procedure MovePivot is used to modify the representation by identifying and moving a rooted subtree of Tvpivot to some other Tvj . In Step 3, we remove edges of Tvpivot from each core subtree corresponding to C5 . Claim 1. In the output representation ⟨S , T ⟩ of MovePivot the following hold: (i) There is no non-core edge e, C5 ∈ C (e) in Tvpivot . (ii) Every dummy edge (w, w ′ ) ∈ T is contained in core subtrees with edges colored with c5 and at most one color of {cij }. (iii) The output representation is a (4, 4, 2)-representation of G. Moreover, no subtree has increased its maximal degree beyond 3 in the output representation. Proof. Only edges that are contained in Tvpivot in ⟨S , T ⟩ are transformed. Therefore, each transformed edge e is colored with c5 and at most one of {Cij } colors. (i) Follows from the procedure. (ii) By Claim for Step 7(III) and the choice of piv ot, all edges in every E are colored c5 and at most one other color. Every edge (w, w ′ ) is a replacement of some edge (v, a) ∈ E . Therefore, (w, w ′ ) is colored c5 and at most one other color. (iii) Let us prove that the output representation is a (4, 4, 2)-representation of G. We first prove that every subtree in S is connected as follows: whenever a non-core edge (v, v ′ ) is replaced in the procedure by (w ′ , v ′ ), the host tree T remains a tree. Each one of the edges e ∈ E is replaced by (v ′ , w ′ ), (w ′ , w ) in every core subtree, where w is a dummy vertex on the path [u, vj ] of T . By Claim for Step 7(III), C ((u, vj )) ⊇ C (e). Therefore, each core subtree that contains the edge e in the input is a connected subtree in the output. Since all edges of a non-core subtree are colored with the same set of colors, either all the edges (v, v ′ ) of a non-core subtree are in E or it has no edge in E . Each edge (v, v ′ ) in E is replaced by (w ′ , v ′ ) in every non-core subtree. Therefore, every non-core subtree remains connected. We now prove that the adjacencies remain unchanged. By the procedure, two subtrees contain a transformed edge in the input if and only if the two subtrees contain the transformed edge in the output. Only the edges {(w, w ′ )} are added by the procedure. Only the core subtrees contain the edges {(w, w ′ )}. From connectivity, if two core subtrees contain the edge (w, w ′ ), then these subtrees contain the path [u, vj ] and therefore share the edge (u, vj ) in the input. Thus, the output of MovePivot is a (4, 4, 2)-representation of G. It remains to prove that the maximal degree of each subtree is at most 3 in the output. The only added vertices are {w} and {w ′ }. Each dummy vertex w is of degree 3. The transformed edge (v, v ′ ) is substituted by (w ′ , v ′ ) in all the subtrees containing it. Therefore, the degree of each vertex w ′ in the output equals the degree of the corresponding vertex v in the input. □ Claim for Step 8. The output of TransformationD-subtree is a (4, 4, 2)-representation of G with fewer vertices with subtree of degree 4, such that u has no subtree of degree 4. Moreover, no subtree has increased its maximal degree beyond 3 in the output representation. Proof. By Claim for Step 7(II), there exists i, 1 ≤ i ≤ 4, such that Svi = ∅. Therefore, there exists a 1 ≤ piv ot ≤ 4 that is found at Step 1 of the procedure. After MovePivot is performed at Step 2, by Claim 1(iii), we obtain a (4, 4, 2)-representation of G, where the maximal degree of each subtree is at most 4. Moreover, after MovePivot is performed, there is no non-core subtree in Tvpivot that shares an edge with {Sv : v ∈ C5 } by Claim 1(i). Also, every subtree in {Sv : v ∈ C5 } shares an edge with every core subtree, since every core subtree contains at least one star edge (u, vi ), i ̸ = piv ot. Every non-core subtree that shares an edge with Sv is contained in Tvi , i ̸ = piv ot. Thus, we have a (4, 4, 2)-representation of G. After Step 3 is performed, every subtree in {Sv : v ∈ C5 } contains only the three star edges {(u, vi )}, i ̸ = piv ot. Therefore, there is no subtree that has degree 4 at u in ⟨S ′ , T ′ ⟩. Thus, the output representation of TransformationD-subtree has fewer vertices with subtree of degree 4, such that u has no subtree of degree 4. Moreover, no subtree has increased its maximal degree beyond 3 in the output representation. □ All Claims are proven as required, thus, the proof of the Lemma is complete. □

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We are finally ready to prove the correctness of the algorithm Recognize [4, 3, 2]. Theorem 5.7. Let G be a 4-flower-free graph that has a (4, 4, 2)-representation. Then, the algorithm Recognize [4, 3, 2] finds a (4, 3, 2)-representation of G. Proof. By Theorem 1.2 and since G is 4-flower-free, G will successfully pass Steps 1 and 2 in the algorithm Recognize [4, 3, 2]. In Step 3, the algorithm will produce a (4, 4, 2)-representation of G using the algorithm in [8,12]. By Lemma 5.4, in each iteration of Step 4, the number of subtrees in the representation with degree 4 is reduced. Since G is a finite graph, the loop will eventually finish and the algorithm will output a (4, 3, 2)-representation of G. □ 6. Conclusions Following the introduction of the [h, s, t ] paradigm by Jamison and Mulder in Refs. [15,16], we have undertaken a long term study to characterize each of these classes for all values of the parameters h ≤ 4, s ≤ 4, t ≤ 2 and to provide recognition algorithms for them. This is the fourth (and final) journal paper in our series, the previous ones being [10–12]. The complete hierarchy has been illustrated in Fig. 5. We have shown that (unlike the chordal graphs) the weakly chordal graphs cannot be characterized within the [h, s, t ] paradigm. In [9], a question inspired by Jamison and Mulder was proposed: Does [∞, 2, t ] ⊊ [∞, 2, t + 1] hold, for any positive integer t? Theorems 2.3 and 2.7 give a positive answer for the cases of t = 2 and t = 3. In fact, these results prove that the answer is positive for subtrees in general and not just subpaths, for t = 2 and t = 3. Hence, we propose to extend the question to the following form. Conjecture 6.1. Do [∞, s, t ] ⊊ [∞, s, t ′ ] and [∞, s, t ] ⊊ [∞, s′ , t ] hold, whenever t ′ > t and s′ > s, respectively? Acknowledgment Support for the first author during the final stages of this research was provided in part by ISF grant 347/09. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18]

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