Faster exact computation of rspr distance

Transcription

1 DOI /s Faster exact comptation of rspr distance Zhi-Zhong Chen Ying Fan Lsheng Wang Springer Science+Bsiness Media New Yk 2013 Abstract De to hybridiation eents in eoltion, stdying two different genes of a set of species may yield two related bt different phylogenetic trees f the set of species. In this case, we want to measre the dissimilarity of the two trees. The rooted sbtree prne and regraft (rspr) distance of the two trees has been sed f this prpose, and many algithms and software tools hae been deeloped f compting the rspr distance of two gien phylogenetic trees. The preiosly fastest exact algithm f this problem rns in O ( d n ) time, where n and d are the nmber of leaes and the rspr distance of the inpt trees, respectiely. In this paper, we present a faster exact algithm which rns in O ( d n ) time. Or experiments show that the new algithm is significantly faster than the newest ersion (namely, 1.1.1) of the preiosly best software (namely, rspr) f RSPR distance. Keywds Phylogenetic tree rspr distance Fixed-parameter algithm Z.-Z. Chen (B) Diision of Infmation System Design, Tokyo Denki Uniersity, Ishiaka, Hatoyama, Hiki, Saitama , Japan chen@mail.dendai.ac.jp Y. Fan L. Wang Department of Compter Science, City Uniersity of Hong Kong, Tat Chee Aene, Kowloon, Hong Kong SAR yingying1988@gmail.com L. Wang cswangl@city.ed.hk

2 1 Introdction When stdying the eoltionary histy of a set of existing species, one can obtain a phylogenetic tree of the set of species with high confidence by looking at a segment of seqences a set of genes (Ma et al. 1999; Ma and Zhang 2011). When looking at another segment of seqences, a different phylogenetic tree can be obtained with high confidence, too. In this case, we want to measre the dissimilarity of the two trees. The rooted sbtree prne and regraft (rspr) distance of the two trees has been sed f this prpose (Hein et al. 1996). Unftnately, it is NP-hard to compte the rspr distance of two gien phylogenetic trees (Bdewich and Semple 2005; Hein et al. 1996). So, it is challenging to deelop programs that can compte the rspr distance of two gien trees of large rspr distance. Indeed, this has motiated researchers to design approximation algithms (Bonet et al. 2006; Bdewich et al. 2008; Hein et al. 1996; Rodriges et al. 2007) exact algithms (Bdewich and Semple 2005; W 2009; Whidden and Zeh 2009; Whidden et al. 2010), as well as heristic algithms (Beiko and Hamilton 2006; Goloboff 2007; Hill et al. 2010; MacLeod et al. 2005; Than et al. 2008), f compting the rspr distance of two gien trees. In particlar, Gasieniec et al. (1999) show that rspr distance is hard to approximate. The preiosly fastest exact algithm is de to Whidden et al. (2010) and rns in O ( d n ) time, where n and d are the nmber of leaes and the rspr distance of the inpt trees, respectiely. Recently, we hae shown that a fast exact algithm f rspr distance can be sed to speed p the comptation of hybridiation nmber and the constrction of minimm hybridiation netwks (Chen and Wang 2012). In this paper, we present a faster exact algithm which rns in O ( d n ) time. Or algithm bilds on Whidden et al. s bt reqires a nmber of new ideas. The main idea is to repeatedly find a pair of sibling leaes in one of the inpt trees me careflly. We hae implemented the algithm and also applied it to other related problems sch as the problem of compting the hybridiation nmber of two me phylogenetic trees. The experimental reslts are presented in Chen and Wang (2013), where a bg in the ersion of Whidden et al. s RSPR is also pointed ot. Sbseqently, Whidden et al. fixed the bg and released the newest ersion (namely, 1.1.1) of RSPR. Inthe release of the bggy ersion 1.1.0, they also conjectre that their algithm rns in O ( 2 d n ) time. Unftnately, the crected ersion is slower than and we dobt that their conjectre is tre. Indeed, the rnning time of the preiosly fastest algithm f rspr distance remained to be O ( d n ). Althogh the wst-case time bond O ( d n ) of new algithm may look marginally better than the wst-case time bond O ( d n ) of Whidden et al. s, experiments show that een when d is not large (say, d 50), new algithm is significantly faster than the newest ersion of Whidden et al. s RSPR f randomly generated instances. A possible explanation f this phenomenon is that the best case f algithm is not considered in Whidden em et al. s algithm and it occrs qite often when the inpt is randomly generated. The remainder of this paper is ganied as follows. In Sect. 2, we gie the basic definitions that will be sed throghot the paper. In Sect. 3, we sketch Whidden

3 et al. s algithm f rspr distance, becase new algithm f rspr distance will bild on theirs. In Sect. 4, we otline algithm f rspr distance. In Sect. 5, we detail a main step in algithm f rspr distance. In Sect. 6, we analye the time complexity of algithm f rspr distance. Finally, we compare the perfmance of new algithm with that of Whidden et al. s RSPR in Sect Preliminaries Throghot this paper, a rooted fest always means a directed acyclic graph in which eery node has in-degree at most 1 and ot-degree at most 2. Let F be a rooted fest. The roots (respectiely, leaes) of F are those nodes whose in-degrees (respectiely, ot-degrees) are 0. A leaf of F is a root leaf if it is also a root of F. Thesie of F, denoted by F, is the nmber of roots in F mins 1. A node of F is nifrcate if it has only one child in F. If a root of F is nifrcate, then contracting in F is the operation that modifies F by deleting. If a non-root node of F is nifrcate, then contracting in F is the operation that modifies F by first adding an edge from the parent of to the child of and then deleting. F conenience, we iew each node of F as an ancest and descendant of itself. F a node of F, thesbtree F of F rooted at is the sbgraph of F whose nodes are the descendants of in F and whose edges are those edges connecting two descendants of in F.If is a non-root node of F, then F is called a pendant sbtree of F. On the other hand, if is a root of F, then F is a component tree of F. F is a rooted tree if it has only one root. If and are two leaes in the same component tree of F, then the pendant sbtrees of F between and are the pendant sbtrees whose roots w satisfy that the (ndirected) path between and in F does not contain w bt contains the parent of w. IfU is a set of leaes in a component tree of F, then LCA F U denotes the lowest common ancest (LCA) of the leaes in U. A rooted binary fest is a rooted fest in which the ot-degree of eery non-leaf node is 2. Let F be a rooted binary fest. F is a rooted binary tree if it has only one root. If is a non-root node of F with parent p and sibling, then detaching the pendant sbtree with root is the operation that modifies F by first deleting the edge ( p, ) and then contracting p. A detaching operation on F is the operation of detaching a pendant sbtree of F. If is a root leaf of F, then eliminating from F is the operation that modifies F by simply deleting. On the other hand, if is a non-root leaf of F, then eliminating from F is the operation that modifies F by first detaching the sbtree rooted at and then deleting. A phylogenetic tree on a set X of species is a rooted binary tree whose leaf set is X. LetT 1 and T 2 be two phylogenetic trees on the same set X of species. If we can apply a seqence of detaching operations on each of T 1 and T 2 so that they become the same fest F, then we refer to F as an agreement fest (AF) of T 1 and T 2.A maximm agreement fest (MAF) of T 1 and T 2 is an agreement fest of T 1 and T 2 whose sie is minimied oer all agreement fests of T 1 and T 2. The sie of an MAF of T 1 and T 2 is called the rspr distance between T 1 and T 2.

4 3 Sketch of Whidden et al. s algithm f rspr distance In this section, we sketch the preiosly fastest algithm [de to Whidden et al. (2010)] f compting the rspr distance of two gien phylogenetic trees. Whidden et al. s algithm indeed soles the following problem (denoted by rsprdc, f conenience): Inpt (k, T 1, ), where k is a nonnegatie integer, T 1 is a phylogenetic tree on a set X of species, and is a rooted fest obtained from some phylogenetic tree T 2 on X by perfming ero me detaching operations. Otpt Yes if perfming k me detaching operations on leads to an AF of T 1 and T 2 ; no otherwise. Obiosly, to compte the rspr distance between two gien phylogenetic trees T 1 and T 2, it sffices to sole rsprdc on inpt (k, T 1, T 2 ) f k = 0, 1, 2, (in this der), ntil a yes is otptted. Note that the inpt integer k to rsprdc mst be nonnegatie. So, eery time befe we call an algithm A f rsprdc on an inpt (k, T 1, ), we need to check if k 0. If k 0, we proceed to call A on inpt (k, T 1, ); otherwise, we do not make the call. Howeer, in der to keep the description of A simpler, we do not explicitly mention this checking process when we describe A. Whidden et al. s algithm f rsprdc is recrsie and proceeds as follows. In the base case, k = 0 and it sffices to check if each component tree of is a pendant sbtree of T 1. If each component tree of is a pendant sbtree of T 1, then we otpt yes and stop. Otherwise, we otpt no and retrn. So, sppose that k > 0. Then, wheneer T 1 has two sibling leaes and sch that and are also sibling leaes in, we modify T 1 and by merging and into a single leaf (say, ). Meoer, wheneer has a root that is also a leaf, we modify T 1 and by eliminating from them. We repeat these two types of modifications of T 1 and ntil none of them is possible. After that, if becomes empty, then we can otpt yes and stop. Otherwise, we select two arbitrary sibling leaes and in T 1 and se them to distingish three cases as follows. Case 1: and are in different component trees of. In this case, in der to transfm T 1 and into identical fests, it sffices to try two choices to modify them, namely, by either eliminating eliminating. F each choice, we frther recrsiely sole rsprdc on inpt (k 1, T 1, ). So, we make two recrsie calls here. Case 2: and are in the same component tree of and either (I) and the parent of are siblings in (II) and the parent of are siblings in. See Fig. 1. In this case, if (I) [(respectiely, (II)] holds, then in der to transfm T 1 and into identical fests with the minimm nmber of detaching operations, it sffices to modify by detaching the sbtree rooted at the sibling w of (respectiely, ), and frther recrsiely sole rsprdc on inpt (k 1, T 1, ) Whidden et al. (2010). So, we make only one recrsie call here. Case 3: and are in the same component tree of and neither (I) n (II) in Case 2 holds. In this case, in der to transfm T 1 and into identical fests, it sffices to try three choices to modify them. The first two choices are the same as those in Case 1. In the third choice, we cont the nmber b of pendant sbtrees of

5 T 1 (1) w Fig. 1 The best case in Whidden et al. s algithm: 1 The sbtree of T 1 rooted at the parent of and, 2 The sbtree of rooted at the LCA of and, where each black triangle indicates a pendant sbtree of (2) w T 1 Fig. 2 The wst case in Whidden et al. s algithm, where each black triangle indicates a pendant sbtree of between and, modify by detaching the pendant sbtrees between and, and recrsiely sole rsprdc on inpt (k b, T 1, ). Note that b 2 and we make three recrsie calls here. Let t(k) be the time needed by the algithm to sole rsprdc on inpt (k, T 1, ). Whidden et al. (2010) show that t(k) = x k n, where n is the nmber of leaes in T 1 and x is the smallest real nmber satisfying the ineqality x 2 2x + 1. Intitiely speaking, this ineqality iginates from Case 3 aboe where b = 2 (see Fig. 2). Since x = 1 + 2, their algithm rns in O ( k n ) time. 4 Ideas f improing Whidden et al. s algithm Like Whidden et al. s Algithm, algithm f rsprdc will be recrsie. So, its exection on inpt (k, T 1, ) can be modeled by a tree Ɣ in which the root cresponds to (k, T 1, ), each other node cresponds to a recrsie call, and a recrsie call A is a child of another call B if and only if B calls A directly. We call Ɣ the search tree on inpt (k, T 1, ). F conenience, we define the sie of Ɣ to be the nmber of its nodes. The depth of a node in Ɣ is the distance between the root and in Ɣ. In particlar, the depth of the root is 0. The depth of Ɣ is the maximm depth of a node in Ɣ. In the remainder of this paper, s(k) denotes the maximm sie of a search tree of algithm on inpt (k, T 1, ), where the maximm is taken oer all T 1 and. Clearly, we hae s(0) = 1 and s(k) = 0 f k < 0. (4.1) To improe Whidden et al. s algithm, idea is to find two sibling leaes and in T 1 me careflly as follows. Step 1. Case 2 in Sect. 3 isthe best case in Whidden et al. s algithm becase we make only one recrsie call in this case. So, we first try to find two sibling leaes and (in T 1 ) satisfying the condition in the best case (cf. Fig. 1). If sch two sibling leaes and exist in T 1, then we se them to modify and frther recrsiely sole

6 Fig. 3 The best case in algithm T 1 p q q q rsprdc as in Case 2 of Whidden et al. s algithm. Clearly, the following ineqality holds in this case: s(k) s(k 1) + 1. (4.2) We hereafter assme that this case does not occr. Step 2. We then try to find two sibling leaes and (in T 1 ) sch that the sibling q of the parent of and in T 1 is also a leaf of T 1 and either (1) and q are sibling leaes in (2) and q are sibling leaes in (see Fig. 3). If sch two sibling leaes and exist in T 1, then we say that the best case in algithm occrs becase this case is essentially symmetric to the best case in Whidden et al. s algithm 1. So, if sch two sibling leaes and exist in T 1 and (1) [(respectiely, (2)] holds, then we can modify T 1 and by eliminating (respectiely, ) from them, and frther recrsiely sole rsprdc on inpt (k 1, T 1, ). Clearly, Ineqality 4.2 holds in this case. We hereafter assme that this case does not occr. Step 3. We try to find two sibling leaes and (in T 1 ) satisfying the condition in Case 1 of Whidden et al. s algithm. If sch two sibling leaes and exist in T 1, then we se them to modify T 1 and and frther recrsiely sole rsprdc as in Case 1 of Whidden et al. s algithm. Clearly, the following ineqality holds in this case: s(k) 2s(k 1) + 1. (4.3) We hereafter assme that this case does not occr. Step 4. We try to find two sibling leaes and (in T 1 ) sch that there are at least three pendant sbtrees of between and. If sch two sibling leaes and exist in T 1, then we se them to modify and frther recrsiely sole rsprdc as in Case 3 of Whidden et al. s algithm. Clearly, the following ineqality holds in this case: s(k) 2s(k 1) + s(k 3) + 1. (4.4) We hereafter assme that this case does not occr. Step 5. We select two sibling leaes and (in T 1 ) whose distance from the root of T 1 is maximied. By Steps 1, 3, and 4, has exactly two pendant sbtrees between and. Letq be the sibling of the parent of and in T 1. By choice of and, it follows that either q is a leaf both children and of q in T 1 are leaes (see Fig. 4). So, we hae two cases to consider. Since both cases are complicated, we detail them in the next section. 1 In other wds, we can se a similar argment of Whidden et al. (2010) to proe the crectness of the processing of T 1 and in this case.

7 Fig. 4 The two possible cases f two sibling leaes and farthest from the root T 1 p q T 1 p q Fig. 5 The two possible cases of f two sibling leaes in T 1, where each black triangle indicates a pendant sbtree of (1) (2) 5 Details of step 5 in Sect. 4 In this section, we detail Step 5 in Sect. 4. By the assmptions in Steps 1 throgh 4 in Sect. 4, we know that f two arbitrary sibling leaes and in T 1, = LCA F2 {,} satisfies exactly one of the following conditions: C1. is the grandparent of both and in [see Fig. 5(1)]. C2. is the parent of one of and and is the great-grandparent of the other in [see Fig. 5(2)]. Recall that in Step 5, we first select two sibling leaes and (in T 1 ) whose distance from the root of T 1 is maximied. Let p be the parent of and in T 1 and q be the sibling of p in T 1. By choice of and, it follows that either q is a leaf both children of q in T 1 are leaes (see Fig. 4). So, we hae two cases to consider. The difficlty is how to se and to modify T 1 and/ and frther recrsiely sole rsprdc. In Sect. 5.1, we detail how to do this in the case where q is a leaf. The other case is detailed in Sect When analying a case in the two sections, we always make the following assmption: Assmption 1 None of the preios cases occrs. 5.1 The case where q is a leaf Since and are sibling leaes in T 1, the assmptions in Steps 1 throgh 4 in Sect. 4 imply that and q are not sibling leaes in and neither are and q in. Since (respectiely, ) and q are not sibling leaes in, the (ndirected) path between (respectiely, ) and q contains at least three edges if q belongs to the component tree of that contains and. We can also assme that the distance from to is not shter than the distance from to, where = LCA F2 {,}. So, one of the following cases occrs. Case 1.1: q does not belong to the same component tree of as and (see Fig. 6). In this case, idea f speeding p Whidden et al. s algithm is as follows. Basically, we emlate Whidden et al. s algithm by sing the sibling pair (,) to try three different choices of modifying T 1 and. The first choice is to eliminate from T 1 and. A crcial point is that after the elimination of, we know that and

8 Fig. 6 The sbtrees of rooted at LCA F2 {,} and q in Case 1.1, where each black triangle indicates a pendant sbtree q q q become a new sibling pair of T 1 and they belong to different component trees of, implying that we need to se the pair (, q) to frther try two different choices of modifying T 1 and (namely, eliminating either q from T 1 and ). So, the first choice leads to two ways of modifying T 1 and one of which is to eliminate and from T 1 and and the other is to eliminate and q from T 1 and. Similarly, the second choice is to eliminate from T 1 and and it leads to two ways of modifying T 1 and one of which is to eliminate and from T 1 and and the other is to eliminate and q from T 1 and. Note that the first and the second choices lead to f ways of modifying T 1 and and hence lead to f recrsie calls. The really crcial point here is that the f recrsie calls can be redced to three, becase two of them are identical. Finally, the third choice is to modify by detaching the pendant sbtrees between and. In smmary, in der to transfm T 1 and into identical fests, it sffices to try the following f different choices to modify them and frther make recrsie calls: 1. We eliminate and from T 1 and, and then recrsiely sole rsprdc on inpt (k 2, T 1, ). 2. We eliminate and q from T 1 and, and then recrsiely sole rsprdc on inpt (k 2, T 1, ). 3. We eliminate and q from T 1 and, and then recrsiely sole rsprdc on inpt (k 2, T 1, ). 4. We detach the two pendant sbtrees of between and, and then recrsiely sole rsprdc on inpt (k 2, T 1, ). Clearly, the following ineqality holds in this case: s(k) 4s(k 2) + 1. (5.1) Case 1.2: LCA F2 {,} is not an ancest of q in (see Fig. 7). In this case, in der to transfm T 1 and into identical fests, it sffices to try f different choices to modify them and frther make recrsie calls: 1. We eliminate from T 1 and, and then recrsiely sole rsprdc on inpt (k 1, T 1, ). Fig. 7 A ption of in Case 1.2, where each ig-ag line indicates a path containing at least one edge and each black triangle indicates a pendant sbtree q q

9 q q q (1) (2) (3) Fig. 8 (1) The sbtree of rooted at LCA F2 {,} in Case 1.3, (2) the sitation in Case 1.3 immediately after eliminating from, and (3) the sitation in Case 1.3 immediately after eliminating from, where each black triangle indicates a pendant sbtree 2. We eliminate and q from T 1 and, and then recrsiely sole rsprdc on inpt (k 2, T 1, ). 3. We first eliminate from T 1 and.letb be the nmber of pendant sbtrees of between and q. Note that b 2. We then detach the b pendant sbtrees of between and q. After that, we recrsiely sole rsprdc on inpt (k 1 b, T 1, ). 4. We detach the two pendant sbtrees of between and, and then recrsiely sole rsprdc on inpt (k 2, T 1, ). Clearly, the following ineqality holds in this case: s(k) s(k 1) + 2s(k 2) + s(k 3) + 1. (5.2) Case 1.3: The (ndirected) path between and q in contains exactly three edges and so does the (ndirected) path between and q in [(see Fig. 8(1)]. In der to transfm T 1 and into identical fests, it sffices to try three difference choices to modify them and frther make recrsie calls: 1. We first eliminate from T 1 and [see Fig. 8(2)]. Then, the best case in Whidden et al. s algithm occrs. Me precisely, we can frther detach the sbtree of rooted at the sibling of q. After that, we recrsiely sole rsprdc on inpt (k 2, T 1, ). 2. We first eliminate from T 1 and (see Fig. 8(3)]. Then, the best case in Whidden et al. s algithm occrs. Me precisely, we can frther detach the sbtree of rooted at the sibling of. After that, we recrsiely sole rsprdc on inpt (k 2, T 1, ). 3. We detach the two pendant sbtrees of between and, and frther recrsiely sole rsprdc on inpt (k 2, T 1, ). Clearly, the following ineqality holds in this case: s(k) 3s(k 2) + 1. (5.3) Case 1.4: The (ndirected) path P,q between and q in is not shter than the (ndirected) path between and q in (see Fig. 9). In this case, Assmption 1 implies that P, contains at least f edges, and we proceed as in Case 1.2. Case 1.5: The (ndirected) path between and q in is shter than the (ndirected) path P,q between and q in (see Fig. 10). In this case, Assmption 1

10 Fig. 9 A ption of in Case 1.4, where each ig-ag line indicates a path containing at least one edge and each black triangle indicates a pendant sbtree q q Fig. 10 A ption of in Case 1.5, where each ig-ag line indicates a path containing at least one edge and each black triangle indicates a pendant sbtree q q implies that P,q contains at least f edges. In der to transfm T 1 and into identical fests, it sffices to try f different choices to modify them and frther make recrsie calls: 1. We eliminate from T 1 and, and then recrsiely sole rsprdc on inpt (k 1, T 1, ). 2. We eliminate and q from T 1 and, and then recrsiely sole rsprdc on inpt (k 2, T 1, ). 3. We first eliminate from T 1 and.letb be the nmber of pendant sbtrees of between and q. Note that b 2. We then detach the b pendant sbtrees of between and q. After that, we recrsiely sole rsprdc on inpt (k 1 b, T 1, ). 4. We detach the two pendant sbtrees of between and, and then recrsiely soling rsprdc on inpt (k 2, T 1, ). Clearly, Ineqality 5.2 holds in this case. 5.2 The case where q is a non-leaf Let and be the children of q in T 1.Let = LCA F2 {,} and = LCA F2 {, }. By the assmptions in Steps 1 throgh 4 in Sect. 4, we know that Condition C1 C2 holds f and and also f and. We can also assme that the distance from to in is not shter than the distance from to in. Similarly, we can frther assme that the distance from to in is not shter than the distance from to in. So, one of the following cases occrs. Case 2.1: and belong to different component trees of (see Fig. 11). In this case, in der to transfm T 1 and into identical fests, it sffices to try eleen different choices to modify them and make recrsie calls as follows. 1. We modify T 1 and as follows. First, we eliminate and from T 1 and. Then, and become two sibling leaes in T 1. The crcial point is that and belong to different component trees in. We proceed to frther modify T 1 and in two different ways and accdingly make two recrsie calls as in Case 1 in Whidden et al. s algithm. In me detail, the two recrsie calls sole rsprdc on inpt

11 Fig. 11 The sbtrees of rooted at and in Case 2.1, where each black triangle indicates a pendant sbtree (k 3, T 1, ) and (k 3, T 1, ), respectiely. Intitiely speaking, we are able to aoid the wst case in Whidden et al. s algithm when processing the sibling leaes and. 2. We modify T 1 and similarly as in Item 1. The only difference is that we first eliminate and from T 1 and and then proceed to se the sibling leaes and to frther modify T 1 and in two different ways and accdingly make two recrsie calls as in Case 1 in Whidden et al. s algithm. 3. We modify T 1 and as follows. First, we eliminate from T 1 and. Then, we detach the two pendant sbtrees of between and. After these modifications, we recrsiely sole rsprdc on inpt (k 3, T 1, ). 4. We modify T 1 and as in Item 1. The only difference is that we first eliminate and from T 1 and and then proceed to se the sibling leaes and to frther modify T 1 and in two different ways and accdingly make two recrsie calls as in Case 1 in Whidden et al. s algithm. 5. We modify T 1 and similarly as in Item 1. The only difference is that we first eliminate and from T 1 and and then proceed to se the sibling leaes and to frther modify T 1 and in two different ways and accdingly make two recrsie calls as in Case 1 in Whidden et al. s algithm. 6. We modify T 1 and as follows. First, we eliminate from T 1 and. Then, we detach the two pendant sbtrees of between and. After these modifications, we recrsiely sole rsprdc on inpt (k 3, T 1, ). 7. We modify by detaching the two pendant sbtrees of between and.after these modifications, we recrsiely sole rsprdc on inpt (k 2, T 1, ). Clearly, the following ineqality holds in this case: s(k) s(k 2) + 10s(k 3) + 1. (5.4) Case 2.2: is neither an ancest n a descendant of in. In this case, we can frther assme that wheneer Condition C2 holds f and, it holds f and as well. See Fig. 12. We next distingish two sbcases as follows. Case 2.2.1: and are not siblings in Condition C1 holds f and. In this case, in der to transfm T 1 and into identical fests, it sffices to try fifteen different choices to modify them and make recrsie calls as follows.

12 (1) (2) Fig. 12 A ption of T 1 and in Case 2.2, where each black triangle indicates a pendant sbtree and each ig-ag line indicates a path containing at least one edge T 1 (1) (2) (3) Fig. 13 A ption of T 1 and in Case 2.2 immediately after eliminating and from T 1 and,where each black triangle indicates a pendant sbtree and each ig-ag line indicates a path containing at least one edge 1. We modify T 1 and as follows. First, we eliminate and from T 1 and (see Fig. 13). Then, and become two sibling leaes in T 1.Letb be the nmber of pendant sbtrees of between and. The crcial point is that b 3. We proceed to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. In me detail, the three recrsie calls sole rsprdc on inpt (k 3, T 1, ), (k 3, T 1, ), and (k 2 b, T 1, ), respectiely. Intitiely speaking, we are able to aoid the wst case in Whidden et al. s algithm when processing the sibling leaes and. 2. We modify T 1 and similarly as in Item 1. The only difference is that we first eliminate and from T 1 and (so that and become siblings in T 1 and the nmber b of pendant sbtrees of between and is at least 4) and then proceed to se the sibling leaes and to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden em et al. s algithm. 3. We modify T 1 and as follows. First, we eliminate from T 1 and. Then, we detach the two pendant sbtrees of between and. After these modifications, we recrsiely sole rsprdc on inpt (k 3, T 1, ). 4. We modify T 1 and similarly as in Item 1. The only difference is that we first eliminate and from T 1 and (so that and become siblings in T 1 and the nmber b of pendant sbtrees of between and is at least 4) and then proceed to se the sibling leaes and to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. 5. We modify T 1 and similarly as in Item 1. The only difference is that we first eliminate and from T 1 and (so that and become siblings in T 1 and the

13 Fig. 14 A ption of T 1 and in Case 2.2.2, where each black triangle indicates a pendant sbtree T 1 nmber b of pendant sbtrees of between and is at least 4) and then proceed to se the sibling leaes and to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. 6. We modify T 1 and as follows. First, we eliminate from T 1 and. Then, we detach the two pendant sbtrees of between and. After these modifications, we recrsiely sole rsprdc on inpt (k 3, T 1, ). 7. We modify by detaching the two pendant sbtrees of between and.after these modifications, we recrsiely sole rsprdc on inpt (k 2, T 1, ). Clearly, the following ineqality holds in this case: s(k) s(k 2) + 10s(k 3) + s(k 5) + 3s(k 6) + 1. (5.5) Case 2.2.2: and are siblings in and Condition C2 holds f and. In this case, in der to transfm T 1 and into identical fests, it sffices to try fifteen different choices to modify them and make recrsie calls as follows. 1. We modify T 1 and similarly as in Item 1 in Case The only difference is that after eliminating and from T 1 and, the nmber b of pendant sbtrees of between and is We modify T 1 and similarly as in Item 2 in Case The only difference is that after eliminating and from T 1 and, the nmber b of pendant sbtrees of between and is We modify T 1 and as follows. First, we eliminate from T 1 and. Then, we detach the two pendant sbtrees of between and.now, and are siblings in both T 1 and. So, we modify T 1 and by merging and into a single leaf (say, ). As can be seen from Fig. 14, the best case in Whidden et al. s algithm occrs. Me precisely, we can frther detach the sbtree of rooted at the sibling of. After that, we recrsiely sole rsprdc on inpt (k 4, T 1, ). 4. We modify T 1 and similarly as in Item 4 in Case The only difference is that after eliminating and from T 1 and, the nmber b of pendant sbtrees of between and is Same as Item 5 in Case Same as Item 6 in Case Same as Item 7 in Case Clearly, the following ineqality holds in this case: s(k) s(k 2) + 9s(k 3) + 2s(k 4) + 2s(k 5) + s(k 6) + 1. (5.6)

14 Fig. 15 Thesbtreeof rooted at = in Case 2.3 = = Fig. 16 Thesbtreeof rooted at LCA F2 {, } in Case 2.3. immediately after eliminating Fig. 17 Thesbtreeof rooted at in Case 2.4 Fig. 18 (1) The sbtree of rooted at LCA F2 {, } in Case 2.4 immediately after eliminating from T 1 and. (2) The sbtree of rooted at LCA F2 {, } in Case 2.4 immediately after eliminating and (1) (2) Case 2.3: = (see Fig. 15). In this case, in der to transfm T 1 and into identical fests, it sffices to try two different choices to modify them and make recrsie calls as follows. 1. We modify T 1 and as follows. First, we eliminate from T 1 and (see Fig. 16). Then, the best case in Whidden et al. s algithm occrs. Me precisely, we can frther modify T 1 and by eliminating. After that, we recrsiely sole rsprdc on inpt (k 2, T 1, ). Note that eliminating first and next is eqialent to eliminating first and next. 2. We modify by detaching the two pendant sbtrees of between and.after these modifications, we recrsiely sole rsprdc on inpt (k 2, T 1, ). Clearly, the following ineqality holds in this case: s(k) 2s(k 2) + 1. (5.7) Case 2.4: is a child of in and the sibling of in is (see Fig. 17). In this case, in der to transfm T 1 and into identical fests, it sffices to try fie different choices to modify them and make recrsie calls as follows. 1. We modify T 1 and as follows. First, we eliminate from T 1 and [(see Fig. 18(1)]. Then, the best case in Whidden et al. s algithm occrs. Me precisely, we can frther modify by detaching the sbtree rooted at the parent of.after that, we recrsiely sole rsprdc on inpt (k 2, T 1, ).

15 Fig. 19 A ption of in Case 2.6 (1) (2) Fig. 20 (1) The sbtree of rooted at LCA F2 {, } in Case 2.6 immediately after eliminating from T 1 and, and (2) the sbtree of rooted at LCA F2 {, } in Case 2.6 immediately after eliminating and 2. We modify T 1 and as follows. First, we eliminate and from T 1 and [see Fig. 18(2)]. Then, the best case in Whidden et al. s algithm occrs. Me precisely, we can frther modify by detaching the sbtree rooted at the sibling of. After that, we recrsiely sole rsprdc on inpt (k 3, T 1, ). 3. We modify T 1 and by eliminating and from T 1 and. We then recrsiely sole rsprdc on inpt (k 2, T 1, ). 4. We modify T 1 and by eliminating from T 1 and. We then modify by detaching the two pendant sbtrees of between and. After the modification, we recrsiely sole rsprdc on inpt (k 3, T 1, ). 5. We modify by detaching the two pendant sbtrees of between and.after the modification, we recrsiely sole rsprdc on inpt (k 2, T 1, ). Clearly, the following ineqality holds in this case: s(k) 3s(k 2) + 2s(k 3) + 1. (5.8) Case 2.5: is a child of in and the sibling of in is. This case is similar to Case 2.4. Case 2.6: is a child of in and the sibling of in is not (see Fig. 19). In this case, in der to transfm T 1 and into identical fests, it sffices to try fie different choices to modify them and make recrsie calls as follows. 1. We modify T 1 and as follows. First, we eliminate from T 1 and [see Fig. 20(1)]. Then, the best case in Whidden et al. s algithm occrs. Me precisely, we can frther modify by detaching the sbtree rooted at the parent of. After that, we recrsiely sole rsprdc on inpt (k 2, T 1, ). 2. We modify T 1 and as follows. First, we eliminate and from T 1 and [(see Fig. 20(2)]. Then, the best case in Whidden et al. s algithm occrs. Me precisely, we can frther modify by detaching the sbtree rooted at the sibling of. After that, we recrsiely sole rsprdc on inpt (k 3, T 1, ).

16 Fig. 21 All possibilities of the relatie locations of,,,,, and in in Case 2.8, where each black triangle indicates a pendant sbtree and each ig-ag line indicates a path containing at least one edge 3. We modify T 1 and by eliminating and from T 1 and. We then recrsiely sole rsprdc on inpt (k 2, T 1, ). 4. We modify T 1 and by eliminating from T 1 and. We then modify by detaching the two pendant sbtrees of between and. After the modification, we recrsiely sole rsprdc on inpt (k 3, T 1, ). 5. We modify by detaching the two pendant sbtrees of between and.after the modification, we recrsiely sole rsprdc on inpt (k 2, T 1, ). Clearly, Ineqality 5.8 holds in this case. Case 2.7: is a child of in and the sibling of in is not. This case is similar to Case 2.6. Case 2.8: There is a directed path P from to in and P contains at least two edges. Then, Fig. 21 shows all possibilities of the relatie locations of,,,,, and in. Note that each ig-ag line in the figre indicates a path containing at least one edge. Depending on whether each sch path contains one, two, at least three edges, we distingish seeral sbcases as follows. Case 2.8.1: The sibling of in is and Condition C1 holds f and (see Fig. 22). Obiosly, if the sitation in Fig. 22(2) occrs, then by switching and, we come to the sitation in Fig. 22(1). So, withot loss of generality, we may assme that the sibling of in is always. Now, in der to transfm T 1 and into identical fests, it sffices to try eleen different choices to modify them and make recrsie calls as follows. 1. We modify T 1 and as follows. First, we eliminate and from T 1 and. Then, and become two sibling leaes in T 1. The crcial point is that there are three pendant sbtrees of between and. We proceed to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. In me detail, the three recrsie calls sole rsprdc on inpt (k 3, T 1, ), (k 3, T 1, ), and (k 5, T 1, ), respectiely. Intitiely

17 (1) (2) (3) Fig. 22 Thesbtreeof rooted at LCA F2 {, } in Case T 1 (1) (2) Fig. 23 (1) The sbtree of T 1 rooted at {, } in Case immediately after merging the sbtree rooted at the paren of and into a single leaf, and (2) the sbtree of rooted at LCA F2 {, } in Case immediately after detaching the pendant sbtrees of between and and frther merging the sbtree rooted at the parent of and into a single leaf speaking, we are able to aoid the wst case in Whidden et al. s algithm when processing the sibling leaes and. 2. We modify T 1 and by eliminating and from them, and then recrsiely sole rsprdc on inpt (k 2, T 1, ). 3. We modify by first eliminating and then detaching the two pendant sbtrees of between and. We frther recrsiely sole rsprdc on inpt (k 3, T 1, ). 4. We modify T 1 and similarly as in Item 1. The only difference is that we first eliminate and from T 1 and and then proceed to se the sibling leaes and to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. 5. We modify T 1 and by eliminating and from them, and then recrsiely sole rsprdc on inpt (k 2, T 1, ). 6. We modify T 1 and by first eliminating from them and then detaching the two pendant sbtrees of between and. We frther recrsiely sole rsprdc on inpt (k 3, T 1, ). 7. We modify T 1 and as follows. First, we detach the two pendant sbtrees of between and. Then, the sbtree of T 1 rooted at the parent of and is identical to the sbtree of rooted at the parent of and. So, we can modify T 1 and by merging the sbtree into a single leaf (say, ). Now, the sitation is as shown in Fig. 23. As can be seen from Fig. 23, the best case in algithm occrs and so we can frther modify T 1 and by eliminating. After that, we recrsiely sole rsprdc on inpt (k 3, T 1, ).

18 (1) Fig. 24 Thesbtreeof rooted at LCA F2 {, } in Case (2) (3) Fig. 25 The sitation in Case immediately after eliminating and from T 1 and T 1 Clearly, the following ineqality holds in this case: s(k) 2s(k 2) + 7s(k 3) + 2s(k 5) + 1. (5.9) Case 2.8.2: The sibling of in is, and Condition C2 holds f and (see Fig. 24). Obiosly, if the sitation in Fig. 24(2) occrs, then by switching and, we come to the sitation in Fig. 24(1). So, withot loss of generality, we may assme that the sibling of in is always. Now, in der to transfm T 1 and into identical fests, it sffices to try nine different choices to modify them and make recrsie calls as follows. 1. We modify T 1 and by eliminating and from them, and then recrsiely sole rsprdc on inpt (k 2, T 1, ). 2. We modify T 1 and as follows. First, we eliminate and from them. Then, the sitation is as shown in Fig. 25. As can be seen from Fig. 25, the best case in Whidden et al. s algithm occrs and so we can frther modify by detaching the sbtree rooted at the sibling of. After that, we recrsiely sole rsprdc on inpt (k 3, T 1, ). 3. We modify by first eliminating and then detaching the two pendant sbtrees of between and. We frther recrsiely sole rsprdc on inpt (k 3, T 1, ). 4. We modify T 1 and as follows. First, we eliminate and from T 1 and. Then, and become two sibling leaes in T 1. The crcial point is that there are three pendant sbtrees of between and. We proceed to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. In me detail, the three recrsie calls sole rsprdc on inpt (k 3, T 1, ), (k 3, T 1, ), and (k 5, T 1, ), respectiely. Intitiely speaking, we are able to aoid the wst case in Whidden et al. s algithm when processing the sibling leaes and.

19 Fig. 26 Thesbtreeof rooted at LCA F2 {, } in Case We modify T 1 and by eliminating and from them, and then recrsiely sole rsprdc on inpt (k 2, T 1, ). 6. We modify by first eliminating and then detaching the two pendant sbtrees of between and. We frther recrsiely sole rsprdc on inpt (k 3, T 1, ). 7. We modify T 1 and as follows. First, we detach the two pendant sbtrees of between and. Then, the sbtree of T 1 rooted at the parent of and is identical to the sbtree of rooted at the parent of and. So, we can modify T 1 and by merging the sbtree into a single leaf (say, ). Now, the sitation is as shown in Fig. 23. As can be seen from Fig. 23, the best case in algithm occrs and so we can frther modify T 1 and by eliminating. After that, we recrsiely sole rsprdc on inpt (k 3, T 1, ). Clearly, the following ineqality holds in this case: s(k) 2s(k 2) + 6s(k 3) + s(k 5) + 1. (5.10) Case 2.8.3: is a grandchild of in, the sibling of in is neither n, and Condition C1 holds f and (see Fig. 26). In this case, in der to transfm T 1 and into identical fests, it sffices to try seenteen different choices to modify them and make recrsie calls as follows. 1. We modify T 1 and as follows. First, we eliminate and from T 1 and. Then, and become two sibling leaes in T 1. The crcial point is that there are three pendant sbtrees of between and. We proceed to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. In me detail, the three recrsie calls sole rsprdc on inpt (k 3, T 1, ), (k 3, T 1, ), and (k 5, T 1, ), respectiely. Intitiely speaking, we are able to aoid the wst case in Whidden et al. s algithm when processing the sibling leaes and. 2. We modify T 1 and similarly as in Item 1. The only difference is that we first eliminate and from T 1 and and then proceed to se the sibling leaes and to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. 3. We modify by first eliminating and then detaching the two pendant sbtrees of between and. We frther recrsiely sole rsprdc on inpt (k 3, T 1, ). 4. We modify T 1 and similarly as in Item 1. The only difference is that we first eliminate and from T 1 and and then proceed to se the sibling leaes and to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm.

20 T 1 Fig. 27 The sitation in Case immediately after detaching the pendant sbtrees of between and and frther merging the identical sbtree of T 1 and rooted at the parent of and into a single leaf T 1 T 1 (1) Fig. 28 (1) The sitation in Item 7a of Case immediately after eliminating from T 1 and.(2)the sitation in Item 7b of Case immediately after eliminating from T 1 and (2) 5. We modify T 1 and similarly as in Item 1. The only difference is that we first eliminate and from T 1 and and then proceed to se the sibling leaes and to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. 6. We modify by first eliminating and then detaching the two pendant sbtrees of between and. We frther recrsiely sole rsprdc on inpt (k 3, T 1, ). 7. We modify T 1 and as follows. First, we detach the two pendant sbtrees of between and. Then, the sbtree of T 1 rooted at the parent of and is identical to the sbtree of rooted at the parent of and. So, we can modify T 1 and by merging the sbtree into a single leaf (say, ). Now, the sitation is as shown in Fig. 27. We frther try three different choices to modify T 1 and as follows. (a) We eliminate from T 1 and. Then, the sitation is as shown in Fig. 28(1). As can be seen from Fig. 28(1), the best case in Whidden et al. s algithm occrs and so we can frther modify by detaching the sbtree rooted at the sibling of. After that, we recrsiely sole rsprdc on inpt (k 4, T 1, ). (b) We eliminate from T 1 and. Then, the sitation is as shown in Fig. 28(2). As can be seen from Fig. 28(2), the best case in Whidden et al. s algithm occrs and so we can frther modify by detaching the sbtree rooted at the sibling of. After that, we recrsiely sole rsprdc on inpt (k 4, T 1, ). (c) We modify by detaching the two pendant sbtrees between and.after that, we recrsiely sole rsprdc on inpt (k 4, T 1, ). Clearly, the following ineqality holds in this case: s(k) 10s(k 3) + 3s(k 4) + 4s(k 5) + 1. (5.11) Case 2.8.4: is a grandchild of in, the sibling of in is neither n, and Condition C2 holds f and (see Fig. 29). In this case, in der to transfm T 1 and into identical fests, it sffices to try thirteen different choices to modify them and make recrsie calls as follows.

21 Fig. 29 Thesbtreeof rooted at LCA F2 {, } in Case We modify T 1 and by eliminating and from them, and then recrsiely sole rsprdc on inpt (k 2, T 1, ). 2. We modify T 1 and by eliminating and from them, and then recrsiely sole rsprdc on inpt (k 2, T 1, ). 3. We modify by first eliminating and then detaching the two pendant sbtrees of between and. We frther recrsiely sole rsprdc on inpt (k 3, T 1, ). 4. We modify T 1 and as follows. First, we eliminate and from T 1 and. Then, and become two sibling leaes in T 1. The crcial point is that there are three pendant sbtrees of between and. We proceed to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. In me detail, the three recrsie calls sole rsprdc on inpt (k 3, T 1, ), (k 3, T 1, ), and (k 5, T 1, ), respectiely. Intitiely speaking, we are able to aoid the wst case in Whidden et al. s algithm when processing the sibling leaes and. 5. We modify T 1 and similarly as in Item 4. The only difference is that we first eliminate and from T 1 and and then proceed to se the sibling leaes and to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. 6. We modify by first eliminating and then detaching the two pendant sbtrees of between and. We frther recrsiely sole rsprdc on inpt (k 3, T 1, ). 7. We modify T 1 and as follows. First, we detach the two pendant sbtrees of between and. Then, the sbtree of T 1 rooted at the parent of and is identical to the sbtree of rooted at the parent of and. So, we can modify T 1 and by merging the sbtree into a single leaf (say, ). Now, the sitation is as shown in Fig. 27. Ths, we can proceed to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Items 7a throgh 7c in Case Clearly, the following ineqality holds in this case: s(k) 2s(k 2) + 6s(k 3) + 3s(k 4) + 2s(k 5) + 1. (5.12) Case 2.8.5: The sibling of the parent y of in is, neither n is a descendant of y, and Condition C1 holds f and (see Fig. 30). Obiosly, if the sitation in Fig. 30(2) occrs, then by switching and, we come to the sitation in Fig. 30(1). So, withot loss of generality, we may assme that the sibling of the parent of in is always. Now, in der to transfm T 1 and into identical fests, it sffices to try seenteen different choices to modify them and make recrsie calls as follows.

22 y y y (1) Fig. 30 Thesbtreeof rooted at LCA F2 {, } in Case (2) (3) 1. We modify T 1 and as follows. First, we eliminate and from T 1 and. Then, and become two sibling leaes in T 1. The crcial point is that there are f pendant sbtrees of between and. We proceed to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. In me detail, the three recrsie calls sole rsprdc on inpt (k 3, T 1, ), (k 3, T 1, ), and (k 6, T 1, ), respectiely. Intitiely speaking, we are able to aoid the wst case in Whidden et al. s algithm when processing the sibling leaes and. 2. We modify T 1 and as follows. First, we eliminate and from T 1 and. Then, and become two sibling leaes in T 1. The crcial point is that there are three pendant sbtrees of between and. We proceed to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. In me detail, the three recrsie calls sole rsprdc on inpt (k 3, T 1, ), (k 3, T 1, ), and (k 5, T 1, ), respectiely. Intitiely speaking, we are able to aoid the wst case in Whidden et al. s algithm when processing the sibling leaes and. 3. We modify by first eliminating and then detaching the two pendant sbtrees of between and. We frther recrsiely sole rsprdc on inpt (k 3, T 1, ). 4. We modify T 1 and similarly as in Item 1. The only difference is that we first eliminate and from T 1 and and then proceed to se the sibling leaes and to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. 5. We modify T 1 and similarly as in Item 2. The only difference is that we first eliminate and from T 1 and and then proceed to se the sibling leaes and to frther modify T 1 and in three different ways and accdingly make three recrsie calls as in Case 3 in Whidden et al. s algithm. 6. We modify T 1 and by first eliminating from them and then detaching the two pendant sbtrees of between and. We frther recrsiely sole rsprdc on inpt (k 3, T 1, ). 7. We modify T 1 and as follows. First, we detach the two pendant sbtrees of between and. Then, the sbtree of T 1 rooted at the parent of and is identical to the sbtree of rooted at the parent of and. So, we can modify T 1 and by merging the sbtree into a single leaf (say, ). Now, the sitation is as shown in Fig. 31. We frther try three different choices to modify T 1 and as follows.

23 T 1 Fig. 31 The sitation in Case immediately after detaching the pendant sbtrees of between and and frther merging the identical sbtree of T 1 and rooted at the parent of and into a single leaf Fig. 32 The sitation in Item 7b of Case immediately after eliminating from T 1 and T 1 y y y (1) Fig. 33 Thesbtreeof rooted at LCA F2 {, } in Case (2) (3) (a) We eliminate from T 1 and. After that, we recrsiely sole rsprdc on inpt (k 3, T 1, ). (b) We eliminate from T 1 and. Then, the sitation is as shown in Fig. 32.As can be seen from Fig. 32, the best case in Whidden et al. s algithm occrs and so we can frther modify by detaching the sbtree rooted at the sibling of. After that, we recrsiely sole rsprdc on inpt (k 4, T 1, ). (c) We modify by detaching the two pendant sbtrees between and.after that, we recrsiely sole rsprdc on inpt (k 4, T 1, ). Clearly, the following ineqality holds in this case: s(k) 11s(k 3) + 2s(k 4) + 2s(k 5) + 2s(k 6) + 1. (5.13) Case 2.8.6: The sibling of the parent y of in is, neither n is a descendant of y, and Condition C2 holds f and (see Fig. 33). Obiosly, if the sitation in Fig. 33(2) occrs, then by switching and, we come to the sitation in Fig. 33(1). So, withot loss of generality, we may assme that the sibling of the parent of in is always. Now, in der to transfm T 1 and into identical fests, it sffices to try fifteen different choices to modify them and make recrsie calls as follows. 1. We modify T 1 and as follows. First, we eliminate and from T 1 and. Then, and become two sibling leaes in T 1. The crcial point is that there are three