Lecture 9: Suffix Trees

Transcription

1 Computtionl Genomics Prof. Ron Shmir & Prof. Roded Shrn School of Computer Science, Tel Aviv University גנומיקה חישובית פרופ' רון שמיר ופרופ' רודד שרן ביה"ס למדעי המחשב,אוניברסיטת תל אביב Lecture 9: Suffix Trees Decemer

2 Suffix Trees Description follows Dn Gusfield s ook Algorithms on Strings, Trees nd Sequences Slides sources: Pvel Shviko, (University of Trento), Him Kpln (Tel Aviv University) CG 12 Ron Shmir

3 Outline Introduction Suffix Trees (ST) Building STs in liner time: Ukkonen s lgorithm Applictions of ST CG Ron Shmir 3

4 Introduction CG Ron Shmir 4

5 Exct String/Pttern Mtching S = m, n different ptterns p 1 p n Text S eginning end Pttern occurrences tht overlp CG Ron Shmir 5

6 String/Pttern Mtching - I Given text S, nswer queries of the form: is the pttern p i sustring of S? Knuth-Morris-Prtt 1977 (KMP) string mtching lg: O( S + p i ) time per query. O(n S + Σ i p i ) time for n queries. Suffix tree solution: O( S + Σ i p i ) time for n queries. CG Ron Shmir 6

7 String/Pttern Mtching - II KMP preprocesses the ptterns p i ; The suffix tree lgorithm: preprocess S in O( S ): uilds suffix tree for S when pttern of length n is input, the lgorithm serches it in O(n) time using tht suffix tree. CG Ron Shmir 7

8 Donld Knuth CG Ron Shmir 8

9 Prefixes & Suffixes Prefix of S: sustring of S eginning t the first position of S Suffix of S: sustring tht ends t its lst position S=AACTAG Prefixes: AACTAG,AACTA,AACT,AAC,AA,A Suffixes: AACTAG,ACTAG,CTAG,TAG,AG,G P is sustring of S iff P is prefix of some suffix of S. Nottion: S[i,j] =S(i), S(i+1),, S(j) CG Ron Shmir 9

10 Suffix Trees CG Ron Shmir 10

11 Trie A tree representing set of strings. { } eef d fe fg c f e e c d f e c g CG Ron Shmir 11

12 Trie (Cont) Assume no string is prefix of nother Ech edge is leled y letter, no two edges outgoing from the sme node e re leled the sme. e d f c Ech string corresponds to lef. CG Ron Shmir f e g 12

13 Compressed Trie Compress unry nodes, lel edges y strings c c e e d f eef d f f c e g e g CG Ron Shmir 13

14 Def: Suffix Tree for S S = m 1. A rooted tree T with m leves numered 1,,m. 2. Ech internl node of T, except perhps the root, hs 2 children. 3. Ech edge of T is leled with nonempty sustring of S. 4. All edges out of node must hve edge-lels strting with different chrcters. 5. For ny lef i, the conctention of the edge-lels on the pth from the root to lef i exctly spells out S[i,m], the suffix of S tht strts t position i. S=xxc CG Ron Shmir this lel is red downwrds! 14

15 Existence of suffix tree S If one suffix S j of S mtches prefix of nother suffix S i of S, then the pth for S j would not end t lef. S = xx S 1 = xx nd S 4 = x How to void this prolem? Assume tht the lst chrcter of S ppers nowhere else in S. Add new chrcter not in the lphet to the end of S. CG Ron Shmir

16 Exmple: Suffix Tree for S=xx CG Ron Shmir 16

17 Exmple: Suffix Tree for S=xx Query: P = xc P is sustring of S iff P is prefix of some suffix of S CG Ron Shmir

18 Trivil lgorithm to uild Suffix tree S= Put the lrgest suffix in Put the suffix in CG Ron Shmir 18

19 Put the suffix in CG Ron Shmir 19

20 Put the suffix in CG Ron Shmir 20

21 Put the suffix in CG Ron Shmir 21

22 We will lso lel ech lef with the strting point of the corresponding suffix CG Ron Shmir

23 Anlysis Tkes O(m 2 ) time to uild. Cn e done in O(m) time - we will sketch the proof. See the CG clss notes or Gusfield s ook for the full detils. CG Ron Shmir 23

24 Building STs in liner time: Ukkonen s lgorithm CG Ron Shmir 24

25 History Weiner s lgorithm [FOCS, 1973] Clled y Knuth The lgorithm of 1973 First lgorithm of liner time, ut much spce McCreight s lgorithm [JACM, 1976] Liner time nd qudrtic spce More redle Ukkonen s lgorithm [Algorithmic, 1995] Liner time lgorithm nd less spce This is wht we will focus on. CG Ron Shmir 25

26 Esko Ukkonen CG Ron Shmir 26

27 Implicit Suffix Trees Ukkonen s lg constructs sequence of implicit STs, the lst of which is converted to true ST of the given string. An implicit suffix tree for string S is tree otined from the suffix tree for S y removing from ll edge lels removing ny edge tht now hs no lel removing ny node with only one child CG Ron Shmir 27

28 Exmple: Construction of the Implicit ST The tree for xx 3 x 6 5 x {xx, x, x, x,, } x x CG Ron Shmir 28

29 Construction of the Implicit ST: Remove Remove 3 x 6 5 x {xx, x, x, x,, } x x CG Ron Shmir 29

30 Construction of the Implicit ST: After the Removl of 3 x 6 5 x x 2 4 x {xx, x, x, x, } 1 CG Ron Shmir 30

31 Construction of the Implicit ST: Remove unleled edges Remove unleled edges 3 x 6 5 x x 2 4 {xx, x, x, x, } x 1 CG Ron Shmir 31

32 Construction of the Implicit ST: After the Removl of Unleled Edges x x x {xx, x, x, x, } x CG Ron Shmir 32

33 Construction of the Implicit ST: Remove degree 1 nodes Remove internl nodes with only one child x x x {xx, x, x, x, } x CG Ron Shmir 33

34 Construction of the Implicit ST: Finl implicit tree x x x x {xx, x, x, x, } 3 CG Ron Shmir 2 Ech suffix is in the tree, ut my not end t lef. 1 34

35 Implicit Suffix Trees (2) An implicit suffix tree for prefix S[1,i] of S is similrly defined sed on the suffix tree for S[1,i]. I i = the implicit suffix tree for S[1,i]. CG Ron Shmir 35

36 Ukkonen s Algorithm (UA) I i is the implicit suffix tree of the string S[1, i] Construct I 1 /* Construct I i+1 from I i */ for i = 1 to m-1 do /* genertion i+1 */ for j = 1 to i+1 do /* extension j */ Find the end of the pth p from the root whose lel is S[j, i] in I i nd extend p with S(i+1) y suffix extension rules; Convert I m into suffix tree S CG Ron Shmir 36

37 Exmple S = xx (initiliztion step) x (i = 1), i+1 = 2, S(i+ 1 )= extend x to x (j = 1, S[1,1] = x) (j = 2, S[2,1] = ) (i = 2), i+1 = 3, S(i+ 1 )= extend x to x (j = 1, S[1,2] = x) extend to (j = 2, S[2,2] = ) (j = 3, S[3,2] = ) CG Ron Shmir 37

38 S(1 ) S(i) S(i+ 1 ) All suffixes of S[1,i] re lredy in the tree Wnt to extend them to suffixes of S[1,i+1] CG Ron Shmir 38

39 Extension Rules Gol: extend ech S[j,i] into S[j,i+1] Rule 1: S[j,i] ends t lef Add chrcter S(i+1) to the end of the lel on tht lef edge Rule 2: S[j,i] doesn t end t lef, nd the following chrcter is not S(i+1) Split new lef edge for chrcter S(i+1) My need to crete n internl node if S[j,i] ends in the middle of n edge Rule 3: S[j,i+1] is lredy in the tree No updte CG Ron Shmir 39

40 Exmple: Extension Rules Constructing the implicit tree for xx from tree for xx CG Ron Shmir x x 4 5 x 2 x 3 Rule 1: 2: 3: t dd lredy lef lef in node edge tree (nd n interior node) x x 1 40

41 UA for xxc (1) S[1,3]=x CG Ron Shmir E S(j,i) S(i+1) 1 x 2 x 3 41

42 UA for xxc (2) CG Ron Shmir 42

43 UA for xxc (3) CG Ron Shmir 43

44 UA for xxc (4) c CG Ron Shmir 44

45 Oservtions Once S[j,i] is locted in the tree, pplying the extension rule tkes only constnt time Nive implementtion: find the end of suffix S[j,i] in O(i-j) time y wlking from the root of the current tree. => I m is creted in O(m 3 ) time. Mking Ukkonen s lgorithm run in O(m) time is chieved y set of shortcuts: Suffix links Skip nd count trick Edge-lel compression A stopper Once lef, lwys lef CG Ron Shmir 45

46 Ukkonen s Algorithm (UA) I i is the implicit suffix tree of the string S[1, i] Construct I 1 /* Construct I i+1 from I i */ for i = 1 to m-1 do /* genertion i+1 */ for j = 1 to i+1 do /* extension j */ Find the end of the pth p from the root whose lel is S[j, i] in I i nd extend p with S(i+1) y suffix extension rules; Convert I m into suffix tree S CG Ron Shmir 46

47 Suffix Links Consider the two strings β nd x β (e.g., x in the exmple elow). Suppose some internl node v of the tree is leled with xβ (x=chr, β= string, possily ) nd nother node s(v) in the tree is leled with β The edge (v,s(v)) is clled the suffix link of v Do ll internl nodes hve suffix links? (the root is not considered n internl node) CG Ron Shmir pth lel of node v: conctention of the strings leling edges from root to v 47

48 Exmple: suffix links S = ACACACAC AC AC C AC AC AC AC v S(v) AC CG Ron Shmir 48

49 Suffix Link Lemm If new internl node v with pth-lel xβ is dded to the current tree in extension j of some genertion i+1, then either the pth leled β lredy ends t n internl node of the tree, or the internl node leled β will e creted in extension j+1 in the sme genertion i+1, or string β is empty nd s(v) is the root CG Ron Shmir 49

50 Suffix Link Lemm If new internl node v with pth-lel xβ is dded to the current tree in extension j of some genertion i+1, then either the pth leled β lredy ends t n internl node of the tree, or the internl node leled β will e creted in extension j+1 in the sme genertion Pf: A new internl node is creted only y extension rule 2 In extension j the pth leled xβ.. continued with some y S(i+1) => In extension j+1, pth p leled β.. p continues with y only => ext. rule 2 will crete node s(v) t the end of the pth β. p continues with two different chrs. => s(v) lredy exists. CG Ron Shmir xβ xβ 50

51 Corollries Every internl node of n implicit suffix tree hs suffix link from it y the end of the next extension Proof y the lemm, using induction. In ny implicit suffix tree I i, if internl node v hs pth lel xβ, then there is node s(v) of I i with pth lel β CG Ron Shmir 51

52 Building I i+1 with suffix links - 1 Gol: in extension j of genertion i+1, find S[j,i] in the tree nd extend to S[j,i+1]; dd suffix link if needed CG Ron Shmir 52

53 Building I i+1 with suffix links - 2 Gol: in extension j of genertion i+1, find S[j,i] in the tree nd extend to S[j,i+1]; dd suffix link if needed S[1,i] must end t lef since it is the longest string in the implicit tree I i Keep pointer to lef of full string; extend to S[1,i+1] (rule 1) S[2,i] =β, S[1,i]=xβ; let (v,1) the edge entering lef 1: If v is the root, descend from the root to find β Otherwise, v is internl. Go to s(v) nd descend to find rest of β CG Ron Shmir 53

54 Building I i+1 with suffix links - 3 In generl: find first node v t or ove S[j-1,i] tht hs s.l. or is root; Let γ = string etween v nd end of S[j-1,i] If v is internl, go to s(v) nd descend following the pth of γ If v is the root, descend from the root to find S[j,i] Extend to S[j,i]S(i+1) If new internl node w ws creted in extension j-1, y the lemm S[j,i+1] ends in s(w) => crete the suffix link from w to s(w). CG Ron Shmir 54

55 Skip nd Count Trick (1) Prolem: Moving down from s(v), directly implemented, tkes time proportionl to γ Solution: To mke running time proportionl to the numer of nodes in the pth serched CG Ron Shmir 55

56 Skip nd Count Trick (2) counter=0; On ech step from s(v), find right edge elow, dd no. of chrs on it to counter nd if still < γ skip to child After 4 skips, the end of S[j, i] is found. Cn show: with skip & count trick, ny genertion of Ukkonen s lgorithm tkes O(m) time CG Ron Shmir 56

57 Interim conclusion Ukkonen s Algorithm cn e implemented in O(m 2 ) time A few more smrt tricks nd we rech O(m) [see scrie] CG Ron Shmir 57

58 Implementtion Issues (1) When the size of the lphet grows: For lrge trees suffix links llow n lgorithm to move quickly from one prt of the tree to nother. This is good for worst-cse time ounds, ut d if the tree isn't entirely in memory. Thus, implementing ST to reduce prcticl spce use cn e serious concern. The min design issues re how to represent nd serch the rnches out of the nodes of the tree. A prcticl design must lnce etween constrints of spce nd need for speed CG Ron Shmir 58

59 Implementtion Issues (2) sic choices to represent rnches: An rry of size Θ( Σ ) t ech non-lef node v A linked list t node v of chrcters tht pper t the eginning of the edge-lels out of v. If kept in sorted order it reduces the verge time to serch for given chrcter In the worst cse it, dds time Σ to every node opertion. If the numer of children k of v is lrge, little spce is sved over the rry, more time A lnced tree implements the list t node v Additions nd serches tke O(logk) time nd O(k) spce. This lterntive mkes sense only when k is firly lrge. A hshing scheme. The chllenge is to find scheme lncing spce with speed. For lrge trees nd lphets hshing is very ttrctive t lest for some of the nodes CG Ron Shmir 59

60 Implementtion Issues (3) When m nd Σ re lrge enough, good design is proly mixture of the ove choices. Nodes ner the root of the tree tend to hve the most children, so rrys re sensile choice t those nodes. (when there re k very dense levels use lookup tle of ll k-tuples with pointers to the roots of the corresponding sutrees) For nodes in the middle of suffix tree, hshing or lnced trees my e the est choice. CG Ron Shmir 60

61 Applictions of Suffix Trees CG Ron Shmir 61

62 Wht cn we do with it? Exct string mtching: Given Text T, T = n, preprocess it such tht when pttern P, P =m, rrives we cn quickly decide if it occurs in T. We my lso wnt to find ll occurrences of P in T CG Ron Shmir 62

63 Exct string mtching In preprocessing we just uild suffix tree in O(m) time Given pttern P = we trverse the tree ccording to the pttern. CG Ron Shmir 63

64 If we did not get stuck trversing the pttern then the pttern occurs in the text. Ech lef in the sutree elow the node we rech corresponds to n occurrence. By trversing this sutree we get ll k occurrences in O(n+k) time CG Ron Shmir 64

65 Generlized suffix tree Given set of strings S, generlized suffix tree of S is compressed trie of ll suffixes of s S To mke these suffixes prefix-free we dded specil chr, sy, t the end of s To ssocite ech suffix with unique string in S dd different specil chr s to ech s CG Ron Shmir 65

66 Generlized suffix tree (Exmple) Let s 1 = nd s 2 = generlized suffix tree for s 1 nd s 2 : { } # # # # 3 # 2 # 1 2 # 4 # CG Ron Shmir 1 66

67 So wht cn we do with it? Mtching pttern ginst dtse of strings CG Ron Shmir 67

68 Longest common sustring (of two strings) Every node with lef descendnt from string s 1 nd lef descendnt from string s 2 represents mximl common sustring nd vice vers. Find such node with lrgest lel depth 1 3 # 2 # 1 2 # 4 # CG Ron Shmir 68

69 Lowest common ncestors A lot more cn e gined from the suffix tree if we preprocess it so tht we cn nswer LCA queries on it CG Ron Shmir 69

70 Why? The LCA of two leves represents the longest common prefix (LCP) of these 2 suffixes Hrel-Trjn (84), Schieer-Vishkin (88): LCA query in constnt time, with liner pre-processing of the tree. 3 # 2 # 1 2 # 4 # CG Ron Shmir 1 70

71 Finding mximl plindromes A plindrome: cc, cc Wnt to find ll mximl plindromes in string s Let s = c The mximl plindrome with center etween i-1 nd i is the LCP of the suffix t position i of s nd the suffix t position m-i+1 of s r CG Ron Shmir 71

72 Mximl plindromes lgorithm Prepre generlized suffix tree for s = c nd s r = c# For every i find the LCA of suffix i of s nd suffix m-i+2 of s r CG Ron Shmir 72

73 Let s = c then s r = c# 6 c # CG Ron Shmir 73

74 Anlysis O(m) time to identify ll plindromes CG Ron Shmir 74

75 ST Drwcks Suffix trees consume lot of spce It is O(m) ut the constnt is quite ig CG Ron Shmir 75

76 Suffix rrys (U. Mnder, G. Myers 91) We lose some of the functionlity ut we sve spce. Let s = Sort the suffixes lexicogrphiclly:,,, The suffix rry gives the indices of the suffixes in sorted order CG Ron Shmir 76

77 How do we uild it? Build suffix tree Trverse the tree in DFS, lexicogrphiclly picking edges outgoing from ech node nd fill the suffix rry. O(m) time CG Ron Shmir 77

78 How do we serch for pttern? If P occurs in S then ll its occurrences re consecutive in the suffix rry. Do inry serch on the suffix rry Nïve implementtion: O(nlogm) time Cn show: O(n+logm) time CG Ron Shmir 78

79 Exmple Let S = mississippi Let P = iss CG Ron Shmir L M R i ippi issippi ississippi mississippi pi ppi sippi sissippi ssippi ssissippi 79

80 Udi Mner Gene Myers CG Ron Shmir 80