The Computational Complexity of the Unrooted Subtree Prune and Regraft Distance. Technical Report CS

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1 The Computational Complexit of the Unroote ubtree rune an egraft Distane Glenn Hike Frank Dehne Anrew au-chaplin Christian Blouin Tehnial eport C Jul, 006 Fault of Computer iene 6050 Universit Ave., Halifax, Nova otia, B3H W5, Canaa

2 The Computational Complexit of the Unroote ubtree rune an egraft Distane Glenn Hike Frank Dehne Anrew au-chaplin Christian Blouin Abstrat We show that omputation of the subtree prune an regraft () istane between unroote binar phlogeneti trees is N-Har an fixe parameter tratable. imilar results exist for the relate tree bisetion an reonnetion (TB) istane, as well as the istane between roote trees but the omplexit of the unroote ase has heretofore remaine unknown. Introution Binar phlogeneti trees are use to esribe evolutionar relationships between organisms. Tpiall, speies represente b DNA or protein sequene information are assoiate with the leaves of the tree an the internal noes orrespon to speiation events. In orer to moel anestor-esenant relationships on the tree, a iretion must be assoiate with its eges. This is ahieve b rooting the tree with a vertex of egree, representing a ommon anestor to all speies in the tree. Often, insuffiient information exists to etermine the roots of trees an, as suh, the are left unroote. Unroote trees still provie a notion of evolutionar similarit between organisms even if the iretion of esent remains unknown. The phlogeneti tree representation has reentl ome uner srutin with ritis laiming that it is too simple to properl represent mirobial evolution, partiularl in the presene of Lateral Gene Transfer (LGT) events [4]. An LGT is the transfer of geneti material between speies b other means than inheritane. Thus, LGT events annot be represente in a tree as the woul form les. The prevalene of LGT events in mirobial evolution an, however, be stuie using phlogeneti trees. Given a pair of trees of the same speies, eah onstrute using ifferent sets of genes, an LGT event will orrespon to a isplaement of a ommon subtree, referre to a subtree prune an regraft () operation. Determining the minimum number of operations that explain the topologial ifferenes between a pair of trees thus iels the most parsimonious LGT senario []. Computing this number of operations, known as the istane, is thus ke to the stu of the prevalene of LGT in baterial evolution. In this paper, we investigate the omputational omplexit of this problem for unroote trees. The outline is as follows. Basi efinitions are provie in the remainer of this setion. In etion we show that istane omputation is N-Har for unroote trees an in etion 3 we show that it is fixe parameter tratable. esearh partiall supporte b the Natural ienes an Engineering esearh Counil of Canaa an Genome Atlanti. Fault of Computer iene, Dalhousie Universit, Halifax, Canaa, hike@s.al.a hool of Computer iene, Carleton Universit, Ottawa, Canaa, frank@ehne.net Fault of Computer iene, Dalhousie Universit, Halifax, Canaa, ar@s.al.a Fault of Computer iene, Dalhousie Universit, Halifax, Canaa, blouin@s.al.a

3 Definition.. () An unroote binar phlogeneti tree T (or more briefl a phlogeneti tree) is a tree whose leaves (egree verties) are labele bijetivel b a (speies) set, an suh that eah non-leaf vertex is unlabele an has egree three. () An ege of a tree T inient with a leaf is a penant ege, otherwise we sa it is an internal ege. Let L(T ) enote the leaf set of a tree T ; the other verties are sai to be internal. (3) A fore ontration is an operation on a tree T in whih we elete a vertex v of egree two an replae the two eges inient to v b a single ege. Definition.. A subtree prune an regraft () operation on a phlogeneti tree T is efine as utting an ege an thereb pruning a subtree, t, an then regrafting the subtree t b the same ut ege to a new vertex obtaine b subiviing a preexisting-existing ege in T. All verties of egree are then elete using fore ontrations. Definition.3. A tree bisetion an reonnetion (TB) operation on a phlogeneti tree T is efine as removing an ege, giving two new subtrees, t an t, whih are then reonnete b reating a new ege between the mipoints of an ege in t an an ege in t. Again fore ontrations are applie to ensure the resulting tree is binar. In the ase that one of the subtrees is a single leaf, then the ege onneting t an t is inient to the leaf. Definition.4. An path or TB path between two trees T an T is a sequene of operations or TB operations, respetivel, that onverts T into T. Definition.5. The istane ( ) an TB istane ( T B) between two trees T an T is the minimum number of operations an TB operations, respetivel, require to onvert T into T. Distane Computation is N-Har for Unroote Trees Hein et al. [6] showe that etermining if (T, T ) is equal to some onstant k is N- Complete b proviing two polnomial time reutions. The first transforms an instane of known N-Complete problem Exat Cover b 3-ets (X3C) into an instane of Maximum Agreement Forest (MAF) size of two roote phlogeneti trees. The seon transforms MAF size into istane. This reution from MAF to is speifie in Lemma 7 from [6] whih states that (T, T ) = MAF (T, T ) for an pair of roote (or unroote) phlogeneti trees T an T. Allen an teel [], however, have sine provie a ounterexample to Lemma 7 in [6], showing that the equation = MAF oes not alwas hol for roote an unroote trees, invaliating the N-Harness proof in Hein et al. [6]. Allen an teel [] were able to show that the relationship hols for the TB istane ( T B) of unroote trees. Borewih an emple use a revise efinition of the MAF to show that istane omputation is N-har for roote trees [3]. However, the omplexit of omputing the istane between unroote trees has, to the best of our knowlege, remaine an open problem. The remainer of this setion is evote to proving that unroote is inee N- Har. We first review the reution from X3C to MAF in [6] an verif that the trees use in this reution an be unroote without altering the result. We then show that the istane of the tree instanes use in this reution is equal to their TB istane, thus showing that even though Lemma 7 from [6] is inorret in the general ase, it is vali for the trees in the reution from X3C.

4 Definition.. An agreement forest for two trees is an ommon forest that an be obtaine from both trees b utting the same number of eges from eah tree, appling fore ontrations after eah ut. A maximum agreement forest (MAF) for two trees is an agreement forest with a minimum number of omponents. [6] Definition.. The exat over b 3-sets (X3C) problem is efine as follows [5]: Given a set X with X = n = 3q an a olletion C of m 3-element subsets of X. Does C ontain an exat over for X, i.e., a sub-olletion C C suh that ever element of X ours in exatl one member of C? NOTE: emains N-Complete if no element ours in more than three subsets. Also note that this problem remains N-Complete if eah element ours in exatl three subsets. This seon propert is implie b [6] though never expliitl state. A supplemental proof is provie in Lemma A. of Appenix A.. eution from X3C to oote MAF We now review the polnomial-time reution from X3C to MAF for roote trees provie b Hein et. al. [6], larifing their notation to reflet that eah element of X belongs to exatl three subsets in C, i.e. X = C = 3q = m = n, a fat implie but not learl state in their paper. An instane of X3C is transforme into two roote phlogeneti trees shown in Figure. Eah element of X is represente b a triplet of the form {a, u, v} an eah triplet appears 3 times in eah tree, one for eah ourrene in a subset in C. Tree T is illustrate in Figure (a). Eah subtree A i T, shown in Figure (b) orrespons to a subset i C. Eah subtree of A i inue b the triple {a i,j, u i,j, v i,j } where j {,, 3} orrespons to a single element of X. Tree T, shown in Figure (), has the same leaf set as T but a ifferent topolog. Eah D i subtree of T, as seen in Figure (e), orrespons to a subset in C exept onl the a-part of eah triplet is present. Eah B i subtree of T, as seen in Figure (), orrespons to an element in X. Eah suh element x = {a, u, v} in the set X appears in three ifferent subsets of C: j, k, an l. Without loss of generalit, assume it onsists of the first element of j, seon element of k, an thir element of l. The orresponing B tree woul have leaves {u j,j, u k,k, u l,l, v j,j, v k,k, v l,l } where j =, k =, l = 3. Hein et. al. show that MAF (T, T ) = 0q + if an onl if C ontains an exat over of X.. eution from X3C to Unroote We begin b verifing that the reution from X3C to roote MAF from [6] esribe above an be triviall applie to the unroote ase. Lemma.. Given an instane of X3C where X = C = 3q, MAF (T, T ) = MAF (U, U ) where T an T are the trees obtaine b the reution in [6] (Figure ) an U an U are unroote versions of T an T isplae in Figure. Aoringl, the instane of X3C has a solution if an onl if MAF (U, U ) = 0q + as this equalit was shown for MAF (T, T ) in [6]. 3

5 A x x A x 3 v i, x 4 u i, a i, A n x n x n v i, v i,3 u i, a i, n n u i,3 a i,3 (a) Tree T (b) ubtree A i x x x n B B n n n u j,j u k,k a i, D u l,l D n D n v j,j v k,k v l,l a i, a i,3 () Tree T () ubtree B i (e) ubtree D i Figure : eution of an instane of X3C to MAF (T, T ) from [6]. Eah element of X orrespons to an {a, u, v} triplet. The instane of X3C has a solution if an onl if MAF (T, T ) = 0q + (where n = 3q). 4

6 z z x x x n A x x B A x 3 x 4 B n n n A n x n x n D n n D n D n (a) Tree U (b) Tree U Figure : Unroote version of T an T from Figure. The unrooteness oes not affet the number of omponents in the MAF. roof. Trees T an T an be unroote b aing a leaf z as a penant to the root, reating trees U an U shown in Figure. eall from [6] that MAF (T, T ) ontains a omponent onsisting of the hain x,...x n,,..., n. Observe that MAF (U, U ) MAF (T, T ) as an agreement forest for U an U an be reate from MAF (T, T ) b aing z to the x hain. Furthermore, MAF (U, U ) MAF (T, T ) as, for the same reasons outline in the proof of Lemma 3., aing a leaf to both trees annot erease their TB istane. It follows that MAF (T, T ) = MAF (U, U ). We now provie a transformation from X3C to unroote. Lemma.. For an instane of X3C where X = C = 3q, (U, U ) = MAF (U, U ) Where U an U are unroote versions of the trees obtaine using the reution in [6] as esribe above. Note that this is a ver restrite version of Lemma 7 from [6]. roof. It is suffiient to show that the inequalit (U, U ) MAF (U, U ) is true as (U, U ) MAF (U, U ) follows from Lemma.7(b) an Theorem.3 from []. MAF (U, U ) is forme b the utting eges from A i subtrees (an the orresponing subtrees in U ) in either of two possible was [6]:. Cut leaves u i,, v i,, u i,, v i,, u i,3, v i,3 an then prune the remaining subtree forme b leaves {a i,, a i,, a i,3 }. uh a proeure ontributes 7 omponents to the MAF.. Cut the leaves a i,, a i,, a i,3 then ut eah of the remaining two-leaf subtrees: {u i,, v i, }, {u i,, v i, }, an {u i,3, v i,3 }. These operations ontribute 6 omponents to the MAF 5

7 We now show that given two trees U an U an their MAF, whih was reate using the above ut operations, there exists MAF operations that an transform U to U. In partiular, for eah set of ut operations, there exists an equivalent set of operations.. rune leaves u i,, v i,, u i,, v i,, u i,3, v i,3 from A i an regraft them onto the hain, forming B i subtrees in the require positions. rune the subtree {a i,, a i,, a i,3 } an regraft into the position of D i. In this ase, 7 operations are performe.. rune the leaves a i,, a i,, a i,3 an regraft them onto the hain, forming a D i subtree in the proper position. rune the remaining two-leaf subtrees: {u i,, v i, }, {u i,, v i, }, an {u i,3, v i,3 } an regraft them onto the hain, forming B i subtree omponents in the require position. 6 operations are use. There is a one-to-one orresponene between uts forme when reating the MAF an operations that an transform U to U. Thus (U, U ) MAF (U, U ). Theorem.3. istane is N-Har for unroote phlogeneti trees. roof. B Lemma., an instane of X3C with X = C = 3q an be reue to a pair of unroote trees U an U suh that X3C has a over if an onl if MAF (U, U ) = 0q +. Lemma. shows that (U, U ) = MAF (U, U ), ompleting the reution from X3C to istane. Note that, in fat, we have atuall showe that eiing if the istane between unroote trees equals a given onstant is N-Complete. ine atuall fining the istane woul solve this eision problem, istane omputation is N- Har for unroote trees. 3 Distane Computation is Fixe-arameter Tratable for Unroote Trees We will now show that istane omputation for unroote trees is fixe-parameter tratable, where the parameter is the istane. This was onjeture in [] but oul not be proven so far. We will re-use the kernelization for TB istane alulation introue b Allen an teel []. As the show, the repeate appliation of the following ule an ule operations to a set of two trees reues their sizes to a linear funtion of their TB istane while preserving the TB istane. Definition 3.. of ULE : eplae an penant subtree that ours in both trees b a single leaf with a new label. ee Figure 3(a). Definition 3.. of ULE : eplae an hain of penant subtrees that our ientiall in both trees b three new leaves with new labels orretl oriente to preserve the iretion. ee Figure 3(b). 6

8 T T C C C n C C C n ule T T ule T T s s 3 3 T T (a) (b) Figure 3: ules an from Allen an teel []. Allen an teel [] showe that ule also leaves the istane unhange. The onjeture the same to be true for ule. Lemma 3.. ule preserves -Distane []. In the remainer of this setion, we prove Allen an teel s onjeture that ule preserves -Distane. Lemma 3.. If an single ommon leaf is prune from both T an T, their istane is either unaffete or reue b. roof. If an single ommon leaf is prune from both T an T, the same path still onverts T into T. Hene, the istane an not inrease. uppose that the istane ereases b (or more). Take the path for the reue trees an a a single operation to hanle the elete leaf. This results in a path between T an T that is shorter than their istane, a ontraition. We now introue a new ule 3. Its properties will be use to show that ule preserves istane. Definition 3.3. of ULE 3: eplae an (non-empt) hain of penant leaves of length n that our ientiall in both trees b a hain of n new leaves with new labels orretl oriente to preserve the iretion of the hain as shown in Figure 4(a). Lemma 3.3. uessive appliations of ule 3 to ommon hains of length > annot reue the -Distane b more than. roof. Let T an T be two phlogeneti trees sharing a ommon hain C = {,,..., n } of length n > whose istane is k as shown in Figure 4(b). epeatel appling ule 3 to this hain n times iels T an T whose ommon hain ontains a single element an whose istane is k. We now show that k k +. Let be the penant subtree of T roote at the vertex ajaent to suh that C =. imilarl, let be the penant subtree of T roote at the vertex ajaent to n suh that C =. Let an be the subtrees of T analogous to an respetivel. runing an regrafting to the stem of an pruning an regrafting to the stem of iels T an T respetivel as shown in Figure 4(b). Observe that C is now a ommon subtree of both T an T an 7

9 thus, b Lemma 3., (T, T ) = (T, T ) = k as C an be reue via ule to a hain with a single leaf without affeting the istane. Thus, T an be transforme to T b performing a single to iel T, then k s to give T an then a single to transform T into T. Therefore k k +. The above steps have shown that reuing a hain of length n to a hain of length using ule 3 annot erease the istane b more than. It follows from Lemma 3. that using ule 3 to reue a hain of length n to a hain of an length less than n annot erease the istane b more than. n =k n T T n n = T =k T = T ule 3 T n =k n n n T T T" T" (a) ule 3. (b) Illustration of Lemma 3.3: Appliation of ule 3 annot reue (T, T ) b more than sine T an alwas be transforme into T with k + operations regarless of the length of the ommon hain. Figure 4: A New ule 3 an Illustration of Lemma 3.3. Lemma 3.4. If ule 3 preserves istane on ommon hains of length n, then it preserves istane for ommon hains of an length > n. roof. uppose the two trees share the hain C = {,,..., n+ }. Note that these trees also share a ommon subhain C = {,,..., n }. If ule 3 is istane preserving for hains of length n, then we an appl it to C ieling a new ommon hain: D = {,,..., n }. The onl element left from C is n+ whih follows n in both trees. C has thus been transforme into the ommon hain C = {,,..., n, n+ }. There are n elements in C. C has been reue from n + to n elements. We have shown that if ule 3 is istane preserving for hains of length n, then it is also istane preserving for hains of length n +. Thus, if ule 3 is istane preserving for hains of length n, then b inution it is istane preserving for hains of an length > n. Lemma 3.5. ule preserves -istane. (This was onjeture in [].) roof. ule an be eompose into appliations of ule an ule 3, the latter onl on hain lengths greater than 3. Thus, proving that ule 3 is istane preserving on suh hain lengths is suffiient to prove that ule is alwas istane preserving. Let T an T be two trees sharing a ommon hain C = {,,..., n }. We will use inution on hain length 8

10 n to show that ule 3 an be applie for an n > 3 without altering the istane. Lemma 3.4 has alrea prove the inution step so all that remains is to show that ule is istane preserving in the base ase. The base ase will be when n = 4. Let = k in this ase. Now suppose that ule 3 Does Not preserve the istane when n = 4. roof that this is a ontraition follows. If ule 3 is not istane preserving for n = 4 then it is not istane preserving for < n < 4. Otherwise it woul ontrait Lemma 3.4: uppose that ule 3 preserves istane for some i, < i < 4. Then ule 3 must preserve istane for an n > i whih inlues n = 4. This is a ontraition. eall that if ule 3 is not istane preserving then b Lemma 3., its appliation will reue the istane b whenever it is applie to hains of length n >. Now bak to our hain of length 4. We have shown that eah time we reue the length of this hain with ule 3, we erease the istane. That is, for n = 4 with = k b appling ule 3 we obtain n = 3 with = k, an b appling ule 3 again we obtain n = with = k, an b appling ule 3 a thir time we obtain n = with = k 3. However, Lemma 3.3 states that a ifferene in hain length an hange b at most! Assuming that ule 3 Does Not preserve istane for n = 4 is therefore a ontraition beause it implies that the istane an be reue b 3 using hain reutions. Thus, there must be some hain length 4 suh that appliation of ule 3 oes not affet the istane. As a onsequene of Lemma 3.4, ule 3 oes not affet istane when n = 4 an the base ase is vali. Theorem 3.6. istane omputation for unroote trees is fixe-parameter tratable. roof. Lemmas 3. an 3.5 show that ule an ule preserve the istane. What remains to be shown is that the tree size after kernelization is boune b a funtion of the istane onl. This follows b appling two results from [] for two unroote phlogeneti trees T an T : T B(T, T ) (T, T ) ([], Lemma.4) an n 4( T B(T, T ) ) ([], Theorem 3.8) where n is the size of the trees after kernelization an 7 is a onstant. Combining these we obtain n 8( (T, T ) ) whih ompletes the proof. 4 Conlusion We have shown that omputing the istane between two unroote phlogeneti trees is N-Har an FT. These results are not altogether surprising given that the same properties have been shown for the relate TB an roote metris [, 3], but solve a long-staning open problem none the less. resentl, we are eveloping an algorithm for unroote istane base on the FT kernelization presente in etion 3 with the intent of testing its performane on real ata. eferenes [] Benjamin L. Allen an Mike teel. ubtree transfer operations an their inue metris on evolutionar trees. Annals of Combinatoris, 5(): 5, 00. [] obert G. Beiko an Niholas Hamilton. hlogeneti ientifiation of lateral geneti transfer events. BMC Evolutionar Biolog, 5(6),

11 [3] Magnus Borewih an Charles emple. On the ompuational omplexit of the roote subtree prune an regraft istane. Annals of Combinatoris, 8(4):409 43, 004. [4] W. For Doolittle. hlogeneti lassifiation an the universal tree. iene, 84:4 8, 999. [5] Mihael. Gare an Davi. Johnson. Computers an Intratabilit: A Guie to the Theor of N-Completeness. W.H. Freeman an Compan, 979. [6] Jotun Hein, Tao Jiang, Lusheng Wang, an Kaizhong Zhang. On the omplexit of omparing evolutionar trees. Disrete Applie Mathematis, 7:53 69, 996. Appenix A X3C Clarifiation Lemma A.. X3C is N-Complete if eah element ours in exatl three subsets. roof. Consier an instane of X3C in whih no element ours in more than three subsets. We provie a polnomial time reution from suh an instane, known to be N-Complete, into an instane in whih eah element ours in exatl three subsets. Let: Y X : Elements of X that appear in exatl one subset Y X : Elements of X that appear in exatl two subsets Y 3 X : Elements of X that appear in exatl three subsets o Y + Y + 3 Y 3 = X = 3q For eah element to appear in exatl three subsets, we must a Y + Y elements to subsets in C. Let multiset Z = {z 0, z,..., z 3p } = Y + Y + Y be these elements we have to a. Note that Z = 3p where p = (q Y 3 ) Y. Let X = {x 0, x,..., x 3p } be a set of new elements suh that X = 3p an X X =. We now reate a olletion C of new subsets out of Z an X so that eah element in X X appears in a subset in C + C exatl three times. For eah i = 0, 3,..., 3p, we a four subsets to C : 4i = {x i, x i+, x x+ } 4i+ = {z i, x i, x i+ } 4i+ = {z i+, x i+, x i+ } 4i+3 = {z i+, x i+, x i } We now show that X X an C + C form an instane of X3C suh that ever element of X X appears in 3 subsets in C + C an X has a over in C if an onl if X X has a over in C + C. 0

12 (if): If X has a over in C, then X X has a over in C + C : Let C be the over of X. Then p is a over X X. (onl if): If X X has a over in C + C, then X has a over in C: imilar to above, the onl wa to over X is with p an no other elements of C an be part of an exat over. This means that X is overe entirel b subsets in C so X is exatl overe b C.