Paradigm 5. Data Structure. Suffix trees. What is a suffix tree? Suffix tree. Simple applications. Simple applications. Algorithms
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1 Prdigm. Dt Struture Known exmples: link tble, hep, Our leture: suffix tree Will involve mortize method tht will be stressed shortly in this ourse Suffix trees Wht is suffix tree? Simple pplitions History Algorithms More pplitions 0//00 CS, OUC 0//00 CS, OUC Wht is suffix tree? A tree to represent ll possible suffixes. E.g. S=g. eh edge is lbeled with string. every suffix is spelled out by pth from the root to lef, nd vie vers. Why? To void some suffix tht is prefix of some other suffix The root g Pth-lbel of the node is g A lef edge Lef, =the strting position of the suffix Suffix tree A suffix tree hs extly n leves nd t most n- nodes. The lbel of eh edge n be represented using indies Thus, suffix tree n be represented using O(n log n) bits,,,,,,,,,, S= g 0//00 CS, OUC 0//00 CS, OUC Simple pplitions Simple pplitions Find ll ourrenes of give Q in S Strt from the root, go down long the unique pth Q to reh point x. All the leves below x re the ourrenes of Q. Time: O( Q +ourrenes) g g E.g., S= Q= Ourrenes:, Find the longest repeted substring in S Find the deepest internl node. Word deep is in terms of the length the string the node represents Time: O(n) g g E.g., S= This node is of depth The longest repet is 0//00 CS, OUC 0//00 CS, OUC
2 Simple pplitions Find the longest ommon substrings of two or more strings About 0 yrs go, Knuth onjeture tht liner time lgorithm For this problem is impossible. Suffix tree llows liner time solution E.g. onsider two strings S nd S Build generlized suffix tree for both S nd S Mrk every internl node tht embres leves representing suffixes of S nd S Report the deepest mrked node g g g E.g., S =, S =g Report this node-- Strightforwrd onstrution of suffix trees Consider S[..n], where S[n]= Algorithm: Initilize the tree with only root For i=n to do Inludes S[i..n] into the tree Time: O(n ), qudrti time 0//00 CS, OUC 0//00 CS, OUC 8 Less thn qudrti time? Yes, we n do it in O(n) time Short history of suffix trees. 9, Weiner s lgorithm, lled position tree Liner time for onstnt size lphbet, but muh spe Knuth hs lled it the lgorithm of 9 9(JACM), MGreight s Liner time for onstnt size lphbet, qudrti spe 99(Algorithmi), Ukkonen s lgorithm On line, liner time for onstnt size lphbet, liner spe 99(FOCS), Frh s lgorithm Liner time for generl lphbet Tody, we disuss Ukkonen s lgorithm, whih is the oneptully esiest liner-time onstrution lg. The ide of Ukkonen s lg. Construt sequene of impliit suffix trees; The lst of whih is onverted to true suffix tree. 0//00 CS, OUC 9 0//00 CS, OUC 0 Impliit suffix trees Ukkonen s lg. t high level A impliit suffix tree is obtined by Removing from edge lbels Removing edges with no lbel Removing nodes with only one hild E.g. S= Denote T i be the impliit suffix tree for S[..i]. Construt T. For i= to n- /*phse i */ Construt T i+ from T i. Convert T n to true suffix tree T. 0//00 CS, OUC 0//00 CS, OUC
3 Illustrtion S= Step Step Step Step 0//00 CS, OUC T Phse T Phse T Phse T Phse i: Construting T i+ from T i For j= to i+ /*extension of the suffix S[j..i] */ Strting from the root, find the endpoint of the pth lbeled β= S[j..i] the pth with hrter S[i+] Rule : if β ends t lef, S[i+] is ppended to the lbel of the lst (lef) edge Rule : if every pth from β strts with hrter S[i+], rete new lef (nd my lso new internl node) nd lef edge lbeled with S[i+], number the reted lef i. Rule : if some pth from β strts with hrter S[i+], do nothing 0//00 CS, OUC Exmple: From T to T S=, I= J= suffix Rule J= suffix Rule J= suffix Rule J= suffix Rule Algorithm: Summry. Construt T. For i= to n- /*phse I: T i T i+ */ For j= to i+ /* extension of the suffix S[j..i] */ Lote the endpoint O( β ) of the pth β= S[j..i] the pth with S[i+] by the O() rules Time: Eh extension O(n) time, O(n ) extensions, Totl time=o(n ). The lgorithm my seem foolish sine we lredy know strightforwrd lgorithm in O(n ) time. We will redue O(n ) to O(n) with two observtions, nd n implement trik. 0//00 CS, OUC 0//00 CS, OUC Observtion Consider phse i, one we pply rule to extend S[j..i], then rule will be pplied for extending S[k..i] for k=j+,,i Thus, nothing to do for k=j+,,i Proof: Sine rule is pplied to extend S[j..i], pth lbeled S[j..i] in the tree T i must ontinue with hrter S[i+]. Thus, there is lso pth for S[k..i], followed by S[i+]. Remrk Bsed on the previous observtion, in phse i, one we hve pplied rule, we n stop. This sves lot of work. 0//00 CS, OUC 0//00 CS, OUC 8
4 Observtion One we dd lef for suffix in T i, tht lef remins in T i+, T i+, Proof: We never remove lef. From the bove we n infer Ft : If in Phse i we hve used rule or to extend A[j..i], then pth A[j..i+] will be in T i+ nd end t lef, nd onsequently, in Phse i+, the extension for A[j..i+] will use rule. Remrk In phse i, let j i be the lst extension involving lef. In other words, for extension due to k j i, we do not perform ny rule (I.e., ll by rule or ). In phse i+, when we perform n extension due to k j i,we lwys enounter lef t the end of S[k..i+], thus, only rule is pplied (ording to Ft in Slide 9). 0//00 CS, OUC 9 0//00 CS, OUC 0 Algorithm for phse i /* for j= j i, extension of j is bsed on rule, so we do nothing */ For j=j i + i+, Find the endpoint of the pth from the root lbeled with S[j..i] the pth with hrter S[i+] bsed on rule,, or If we extend the pth with rule /* extension j for j =j+ i+ re bsed on rule. So we need to do nothing */ Set j i+ =j-; Brek Whole proess Summry Phse : ompute extension..j +. Phse : ompute extension j +..j +. Phse i: ompute extension j i +..j i+ +. Phse n-: ompute extension j n- +..j n +. In totl we will do t most n extensions. For n extension due to j, it tkes O(n) time beuse we need to find the endpoint of S[j..i]. The totl time is O(n ). The proess n be elerted using suffix link. 0//00 CS, OUC 0//00 CS, OUC Suffix link For n internl node v with pth-lbel x, if there is nother node s(v) with pth-lbel, then we rete suffix link from v to s(v) g g Is suffix link well defined? For (impliit) suffix tree, every internl node (exept the root) hs suffix link. Proof: Consider ny internl node v with pth-lbel x. x is the ommon prefix of S[i..n] nd S[j..n] The two leves lbeled i nd j under v is the ommon prefix of S[i+..n] nd S[j+..n] Thus, there is n internl node u with pth-lbel. Suffix link of v=u. 0//00 CS, OUC 0//00 CS, OUC
5 How to use suffix link? Time omplexity Assume before extension due to j, we mintin suffix links for ll internl nodes. In the extension for j, we hve loted w whih is the end of S[j..i]. To strt extension for j+, we need to go to the end of S[j+..i].. From w, go up on edge to v. Through suffix link, go to s(v). Go down until we find the end of S[j+..i], sy, w. If w hs led to newly reted internl node, rete suffix link from w to w v w s(v) w Find the end of S[j+..i]: Step,, nd tke O() time. Step tkes mortized O() time. (?) So, eh extension n be done in mortized O() time. As there re n extensions, the totl time is O(n). 0//00 CS, OUC 0//00 CS, OUC Why Step tkes mortized O() time? The skip/ount trik Step is to wlk down from node s(v) long pth lbeled. There surely must be suh pth from s(v). Diret implemented, this wlk tkes O( ) time. A simple trik, lled skip/ount trik, will redue the trversl time to O( of edges on the pth). So, define node-depth of u to be the of edges on the pth from the root to u. Our tsk is then to justify the bove lim bout skip/ount nd tht By mortiztion, eh step goes down O() edge. Let g =, u=s(v) Repet Find the edge e=(u, u ) whose first hrter= []. Let l= lbel(e) If l<g then Else = [l+,g]; g=g-l; u=u Skip to lbel(e)[g]; exit v w s(v) w 0//00 CS, OUC 0//00 CS, OUC 8 Step go down mortized O() edges Note tht for eh extension, Step redues the node-depth by Step redues the node-depth by t most Step inreses the node-depth Sine there re n extensions, All steps nd n redue the node-depth by t most n Sine the mximum node-depth is n-, All steps n t most inrese the node-depth by n- By mortiztion, eh step goes down O() nodes. Creting the true suffix tree Convert the impliit suffix tree T n to true suffix tree in O(n) time Append S with the terminl hrter Independently perform phse n+ on T n with S. Impliit T S= phse Expliit T 0//00 CS, OUC 9 0//00 CS, OUC 0
6 More pplitions Mximum unique mth. O(n) Given two strings S nd S. Find ll substrings w suh tht w pper extly one in both strings, nd w is mximl (I.e., ny substring x inluding w nnot pper extly one in both strings) Longest ommon prefix. O(n) Given string S[..n], for i,j, the problem is to find the length of the longest ommon prefix of S[i,..n] nd S[j..n] Mximum plindrome ( 最大回文 ) Plindrome is string X s.t. X=X R. e.g., level The problem is to find the longest substring of S tht is plindrome. 0//00 CS, OUC Additionl pplitions Ziv-Lempel dt ompression Minimum length enoding of DNA All-pirs suffix-prefix mthing For Reover DNA Dt ompression 0//00 CS, OUC
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