Computational Biology Lecture 18: Genome rearrangements, finding maximal matches Saad Mneimneh

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1 Computtionl Biology Leture 8: Genome rerrngements, finding miml mthes Sd Mneimneh We hve seen how to rerrnge genome to otin nother one sed on reversls nd the knowledge of the preserved loks or genes. Now we re onerned with determining those preserved genes. More generlly, given two strings nd y we would like to find ll the miml mthes etween them. Glol lignments One possiility is to perform glol lignment of the two strings nd y with speil soring sheme; for instne, + for mth, 0 for mismth, nd 0 for gp. Then we ould identify ll the miml positively soring hunks of the lignment. The disdvntges of this pproh is tht it requires O(mn) running time, might not otin ll ndidte mthes, nd otins mthes tht re not neessrily miml. Here s n emple: = y = Then n optiml lignment might e: The ove lignment find the mthes ([..], y[..]) = nd ([7..9], y[..]) =. While the first mth is miml (it nnot e etended to the left or to the right), the seond mth is not sine it n e etended to. Moreover, the lignment missed mny miml mthes mong whih the miml mth ([..], y[..]) =. k-mers Another possiility for identifying possile miml mthes etween nd y is to find their ommon k-mers. Rell tht k-mer is simply k long sustring. Here s n lgorithm: Denote k-mer of y (w, 0, i) where w = i... i+k Denote k-mer of y y (z,, j) where z = y j...y j+k Sort them leiogrphilly (i.e. ditionry order) Dedue ll k long mthes etween nd y Emple (k = ): = y = s -mers: (, 0, ), (, 0, ), (, 0, ) y s -mers: (,, ), (,, ), (,, ) sort them: (, 0, ), (, 0, ), (,, ), (,, ), (, 0, ), (,, ) Identify mthes: (, 0, ), (,, ) nd (, 0, ), (,, )

2 The worst se running time of this lgorithm is still O(mn) e.g. we n hve O(m) k-mers in nd O(n) k-mers in y tht re identil. However, it is etter thn ove euse it is tully O((m + n) log(m + n) + L), where L is the numer of mthes; wheres the running time of glol lignment is O(mn) regrdless of the numer of mthes. The disdvntge of this pproh is tht mthes re not neessrily miml nd, therefore, we spend O(L) time on finding short mthes tht we re not relly looking for. A possile solution would e to inrese k, whih redues L nd hene the runnig time, nd inreses our hnes of otining miml mthes. However, this would men tht we will miss the short (less thn k, ut signifint) miml mthes! So y setting vlue for k, we re mking ompromise mong the long nd short miml mthes nd the running time of the lgorithm. An et solution Here s solution tht will give ll miml mthes etween two strings nd y. Construt the mtri P suh tht P (i, j)= if i = y j, nd P (i, j) = 0 otherwise. Clerly mth of length k strting t i nd y j will orrespond to digonl in the mtri P (i, j), P (i +, j + ),..., P (i + k, j + k ). The mth is miml if P (i, j ) = 0 (or i = or j = ) nd P (i + k, j + k) = 0 (or i + k = m or j + k = n). Therefore, ll miml mthes n e otined in O(mn) time in the following wy: First onstrut the mtri P. Then in rowwise mnner, hek if P (i, j) is the strt of miml mth, i.e. if P (i, j ) = 0 (or does not eist). If P (i, j) is the strt of miml mth, follow the mth long the digonl P (i, j), P (i+, j +),..., P (i+k, j +k ) suh tht P (i + k, j + k) = 0 (or does not eist). Construting the mtri tkes O(mn) time. Cheking whether P (i, j) is the strt of mth tkes O() time for totl of O(mn) time for ll P (i, j) s. Eh digonl is investigted one nd ll digonls re disjoint (miml); therefore, finding the mthes tkes n dditionl O(mn) time. Below we desrie dt struture tht will llow us to otin ll miml mthes in O(m + n + k), where k is their numer. Suffi trees To find miml mthes etween nd y we will use n effiient dt struture lled suffi tree. A suffi tree for string s sores ll suffies of s nd supports fst lookup, e.g. it requires O(l) time to find whether string of length l is sustring of s. Definition: A suffi tree T for string s of length m is rooted tree suh tht: It hs etly m leves numered to m Eh internl node other thn the root hs t lest two hildren Eh edge is leled y sustring of s No two edge lels on outgoing edges of node strt with the sme hrter For ny lef i, the ontention of the edge lels on the pth from the root to i spells out the suffi i... m Below is n emple of suffi tree for s = : 6 Figure : Suffi tree for s = The definition of suffi tree ove does not gurntee the eistene of suffi tree for every string s. In ft, suffi tree stisfying ll of the ove onditions is not gurnteed to eist.

3 Consider for instne the string s =, i.e. remove tht lst hrter from the ove emple. There must e pth from the root to lef numer tht spells. Also, there must e pth from the root to lef numer tht spells. Sine two edges t node nnot strt with the sme hrter, the portion of the pth tht spells must e shred y the two pths. Therefore, lef hs to e n internl node s well, whih nnot hppen! This prolem rises when suffi of s is prefi of nother suffi of s. Therefore, we n eliminte the prolem y terminting the string s with speil end mrker tht is gurnteed not to our nywhere in s. As result, suffi of s (now lwys ending in ) n never e prefi of nother suffi, nd this gurntees tht the suffi tree esits. Let us look t the properties of suffi tree T with set of nodes V nd set of edges E: E = V ( tree) Numer of leves is m (ssume s = m inluding the speil end mrker) Sine every internl node, other thn the root, hs t lest two hildren, E = O(leves) = O(m) More generlly, ny sutree S of T with k leves stisfies E S = O(k) Here s simple lgorithm for uilding suffi tree: Algorithm given s = s...s m insert speil end mrker t the end of s initilize the suffi tree T to one root for j = to m + do find the longest mth of j... m in T strting from the root nd following unique pth split the edge where the mth stops, dd new node w dd n edge (w, j) (j is the new lef) nd lel it with the remining unmthed hrters of j... m It is esy to prove tht the ove lgorithm produes suffi tree for s y heking ll the onditions tht suffi tree hs to stisfy. We will leve the proof s simple eerise. Here s the sme emple of s = repeted, showing the ove lgorithm step y step. () () () 6 () () (6) 7 6 (7)

4 Let us now nlyze the time nd spe requirement of the ove lgorithm. The running time is O(m ) sine eh suffi requires n O(m) time (dominted y the mthing of suessive hrters) to updte the suffi tree. The spe required y the ove lgorithm is O(m ) s well euse eh edge stores n O(m) size lel nd we hve O(m) edges s desried erlier (look t the properties of suffi tree ove). This spe requirement n, however, e redued to only O(m) y not epliitly storing lel i... j, ut impliitly y storing [i, j], i.e. the indies mrking the strt nd end of the lel in s. Therefore, we hve n O(m ) time nd O(m) spe lgorithm for uilding suffi tree T for string s of length m. There eists n O(m) time lgorithm (hene O(m) spe s well) for uilding suffi tree. We re not going to desrie it here euse it is too omplited nd not so importnt y itself. So we will ssume the eistene of liner time nd spe lgorithm for uilding suffi tree. But now tht we hve suffi tree dt struture, how n we use it effiiently to find ll miml mthed etween two strings nd y. Well, let us look t simpler prolem first whih is very fmous prolem in omputer siene: the string mthing prolem. String Mthing: Given string nd string y, find ll lotions in where y ours. Here s n lgorithm for string mthing: Algorithm uild suffi tree T for mth the hrters of y long unique pth in T until (se ) either y is ehusted, or (se ) no more mthes re possile O(m) O(n) if (se ), y does not our in else the k leves in the su-tree elow the point of the lst mth give the k lotion of y in (trverse the tree in liner time) O(k) Why is the ove lgorithm orret? Simply euse if y ours in t position i, then the i th suffi of must strt with y. As result, lef i must e rehed in T from the root y pth strting with y; therefore, ll suh leves re in the sutree indentified y the lgorithm. The ove lgorithm runs in O(m + n + k) where m is the length of, n is the legnth of y, nd k is the numer of lotions where y ours in. Sine the input hs size O(m + n) nd the output hs size O(k), this is optiml! For otining the miml mthes of nd y, we re going to follow the sme spirit of string mthing y finding lotions of sustrings of y in. But if we do it for every sustring of y, then we spend O(n (m + n)) = O(n m) time, nd for eh mth we find we hve to further hek whether it is miml or not. We will e more intelligent nd hek only for the suffies of y hene spending O(n(m + n)) = O(mn) time. Here s wht we re going to do: we re interested in mthes of the form i... i+l = y j...y j+l suh tht i... i+l y j...y j+l (nnot e etended to the right) nd i... i+l y j...y j+l (nnot e etended to the left). For suffi of y, y j...y n, we will find ll mthes strting t y j tht nnot e etended to the right. Then we will find wy to drop ll those mthes mong them tht n e etended to the left. Finlly, doing this for ll suffies of y results in ll miml mthes of nd y. Here s how to find ll mthes strting t y j tht nnot e etended to the right: Build suffi tree for (we will do this only one) Find the pth in T determined y the longest possile prefi of the suffi y j...y n (it ould stop in the middle of n edge e in tht se e is prt of the pth) Let v k, k =..p e n internl node on this pth nd T k e the sutree rooted t v k tht eludes v k+ Identify the leves in eh sutree, eh lef orresponds to lotion of mth tht nnot e etended to the right The following figure illustrtes the ide:

5 root L v L Lst point of mth v T v p- L p v p m leves T T p- T p Figure : Finding mthes strting t y j tht nnot e etended to the right In the figure ove, every lef i in sutree T k gives the lotion i in of mth etween i... i+ L L k nd y j...y j+ L L k tht nnot e etended to the right. Given the suffi tree for, these n e found in O( L k + m k ) = O(n + m) time. A mth found ove is not neessrily miml sine it is possile tht we n etend it to the left. Therefore, for every lef i we find, we hve to hek whether it orresponds to miml mth or not. Given lef i, let left(i) e the hrter i (define 0 nd y 0 to e distint speil hrters). When uilding the suffi tree for, left(i) for every lef i n e stored t the lef. Now it is ovious tht if left(i) y j, then i represents miml mth, nd this is simple hek. Therefore, we otin ll miml mthes etween nd y y repeting the ove proess for every suffi of y, resulting in n O(mn) time lgorithm. Algorithm uild suffi tree T for O(m) for j = to n O(n) do O(m + n) find the pth in T determined y the longest possile prefi of the suffi y j...y n (it ould stop in the middle of n edge e in tht se e is prt of the pth) let v k, k =..p e n internl node on this pth nd T k e the sutree rooted t v k tht eludes v k+ let l(v k ) = length of mth up to node v k identify ll leves i in eh sutree T k, k =..p, suh tht left(i) y j Suh lef i in sutree T k represents miml mth of length l(v k ) strting t position i in nd position j in y The running time of the lgorithm ove is O(mn), hene we find ll miml mthes of nd y in O(mn) time. We did not fulfill our promise for liner time lgoritm. As we mentioned efore, it is possile to find ll miml mthes of nd y in O(m + n + k) time, where k is the numer of those mthes. The following setion eplins how. Generlized sufi tree The min deffiieny using the O(mn) time ound for the lgorithm ove summrizes in tht we hve to repet the min mthing proess n times, one for every suffi of y. Wht if we ould inorporte ll suffies of s well s those of y in the sme single suffi tree?, nd do it one?, then we ould possily void the need to hek for every suffi of y; sine ll the informtion will e omptly ptured in one suffi tree. This is the min ide of this setion. A generlized suffi tree for strings s, s,..., s k is suffi tree tht stores ll suffies of s, ll sufies of s,..., ll suffies of s k.

6 Here s simple lgorithm to uild suh generlized suffi tree for set of strings sed on suffi tree lgorithm for single string. ppend different end mrker to eh string in the set {s,..., s k } ontente ll the strings together to form single string s = s...s k uild suffi tree for the ontented string s The resulting suffi tree will hve lef for eh suffi of the ontented string nd is uilt in O( s s k ) time, i.e. liner time in the size of ll k strings. The lef numers n e esily onverted to two numers, one identifying string s i nd the other strting position in s i. Figure shows the generlized suffi tree for s = nd s =. s =, s = s =,6,7,,,,,8,,,,,, Figure : Generlized suffi tree for s nd s As it might e ovious, defet of this lgorithm is tht the tree now represents suffies (of the ontented string) tht spn more thn one originl string. These syntheti suffies re not generlly of interest. However, euse end of string mrker is distint nd is not in ny of the originl strings, the lel on ny pth from the root to n internl node must e sustring of one of the originl strings. Hene y reduing the seond inde of the lel on lef edges only, ll unwnted syntheti suffies re removed. Figure illustrtes the ide.,7,6,,,,,,8,,,,, Figure : Fiing lels on lef edges only 6

7 Therefore, given two strings nd y of length m nd n respetively, we n uild suffi tree for oth in O(m + n) time. Eh lef in the tree represents either suffi for or suffi for y. A lef (, i) represents the i th suffi of nd lef (, j) represents the j th suffi of y. Now mrk eh internl node v with (y) if there is lef in the sutree of v representing suffi of (y). This n e done in liner time y ottom up trversl of the tree from leves to the root. Note tht if v is mrked (y), ll nestors of v re mrked (y). Lemm: If αp is sustring of nd αq is sustring of y for p q, then α orresponds to n internl node v mrked with oth nd y nd vie-vers. Proof: α ours in oth nd y suh tht the hrter to the right of α in differs from the hrter to the right of α in y. Therefore, there must e node in the suffi tree tht orresponds to pth leled α from whih there is n edge whose lel strts with p leding to lef for, nd n edge whose lel strts with q leding to lef for y. Hene, the node is n internl node mrked with oth nd y. Conversely, if we hve n internl node v leled oth nd y then let α e the ontention of lels on the pth from the root to v. There is pth from v to lef i for nd pth from v to lef j for y. If the two pths strt with distint edges e nd e, then we re done: let p e the first hrter of e s lel nd q e the first hrter of e lel, p q. Then αp is sustring of nd αq is sustring of y. So ssume e = e = e nd hene p = q. But sine v is rning node, there must e n edge f from v whose lel strts with w p. The edge f must led to either lef for or lef for y. If f leds to lef for, then e leds to lef j for y nd we re done. If f leds to lef for y, then f leds to lef i for nd we re done. Definition: An internl node v is left diverse iff it hs two hildren v nd v with lef i for in v s sutree nd lef j for y in v s sutree, suh tht left(i) left(j). Lemm: If uαp is sustring of nd wαq is sustring of y for u w nd p q, then α orresponds to left diverse node v nd vie-vers. Proof: similr to the previous proof. We ll suh n α miml ommon sustring. Therefore, we hve only O(m + n) miml ommon sustrings for nd y represented y the left diverse nodes of the suffi tree (ut eh miml ommon sustring might pper in multiple lotions). If we identify left diverse nodes in liner time, we need only O(m + n) time nd spe to ome up with this ompt representtion of ll miml ommon sustrings. A miml mth n e represented s (p, p, l) where p nd p re the lotions of miml ommon sustring of length l in nd y respetively. We n otin ll miml mthes in O(m + n + k) where k is their numer. We will not desrie the lgorithm in detil here ut you n refer to Gusfield D, Algorithms on Strings, Trees nd Sequenes pges -, 7, Prolem 7 pge 7. First we present n lgorithm for identifying left diverse nodes in liner time, then desrie riefly how to modify this lgorithm to otin ll miml mthes in O(m + n + k). The lgorithm for identifying left diverse nodes is s follows. For eh node v, the lgorithm reords two hrters (v) nd (v) suh tht the hrter (v) is: the left hrter of every lef for in v s sutree, or speil hrter ɛ if no lef for eists in v s sutree, or speil dinstint hrter The hrter (v) is defined similrly y repling ove with y. Computing (v) nd (v) n e done in uttom up pproh in liner time. Then it is ovious tht v is left diverse iff it hs two hildren v nd v suh tht: (v ) (v ), (v ) ɛ, (v ) ɛ, or (v ) (v ), (v ) ɛ, (v ) ɛ It tkes O( ) time (onstnt) to find two suh hildren or none, where is the lphet (eh node hs t most hildren). 7

8 To otin ll miml mthes we n mintin t eh node v nd for eh hrter of the lphet: list L v,, of leves for in v s sutree suh tht left(i) = for ll i L v,, list L v,y, of leves for y in v s sutree suh tht left(j) = for ll i L v,y, Clerly node v is left diverse iff if hs two hildren v nd v suh tht L v,, nd L v,y, nd. Then for every (left diverse) node v nd ll hrters, the rtesin produt of L v,, nd L v,y, for ll hildren v nd v of v gives the strting positions of ll miml mthes orresponding to node v. Building the lists t every node v n e done y merging ll lists of its hildren (opying is no good sine it will tke O(m + n) time t eh node). This of ourse will destroy the hildren s lists, ut lukily those lists re not needed one ll miml mthes orresponding to v hve een identified. Therefore, uilding the lists n e done in ottom up pproh nd tkes liner time (ssuming onstnt lphet size). Moreover, the time for performing the rtesin produt of lists is proportionl to the numer of miml mthes. Hene we hve n O(m + n + k) time lgorithm. Referenes Gusfield D., Algorithms on Strings, Trees, nd Sequenes, Chpters, 6, 7. 8