CS481: Bioinformatics Algorithms

Transcription

1 CS481: Bioinformtics Algorithms Cn Alkn EA509

2 EXACT STRING MATCHING

3 Fingerprint ide Assume: We cn compute fingerprint f(p) of P in O(m) time. If f(p) f(t[s.. s+m 1]), then P T[s.. s+m 1] We cn compre fingerprints in O(1) We cn compute f = f(t[s+1.. s+m]) from f(t[s.. s+m 1]), in O(1) f f AALG, lecture 3, Simons Šltenis, 2004

4 Algorithm with Fingerprints Let the lphet ={0,1,2,3,4,5,6,7,8,9} Let fingerprint to e just deciml numer, i.e., f( 1045 ) = 1* * * = 1045 Fingerprint-Serch(T,P) 01 fp compute f(p) 02 f compute f(t[0..m 1]) 03 for s 0 to n m do 04 if fp = f return s 05 f (f T[s]*10 m-1 )*10 + T[s+m] 06 return 1 T[s] new f f T[s+m] Running time 2O(m) + O(n m) = O(n) AALG, lecture 3, Simons Šltenis, 2004

5 Using Hsh Function Prolem: we cn not ssume we cn do rithmetics with m-digits-long numers in O(1) time Solution: Use hsh function h = f mod q For exmple, if q = 7, h( 52 ) = 52 mod 7 = 3 h(s1) h(s2) S1 S2 But h(s1) = h(s2) does not imply S1=S2 For exmple, if q = 7, h( 73 ) = 3, ut Bsic mod q rithmetics: (+) mod q = ( mod q + mod q) mod q (*) mod q = ( mod q)*( mod q) mod q AALG, lecture 3, Simons Šltenis, 2004

6 Preprocessing nd Stepping Preprocessing: fp = P[m-1] + 10*(P[m-2] + 10*(P[m-3]+ + 10*(P[1] + 10*P[0]) )) mod q In the sme wy compute ft from T[0..m-1] Exmple: P = 2531, q = 7, fp =? Stepping: ft = (ft T[s]*10 m-1 mod q)*10 + T[s+m]) mod q 10 m-1 mod q cn e computed once in the preprocessing Exmple: Let T[ ] = 5319, q = 7, wht is the corresponding ft? T[s] new ft AALG, lecture 3, Simons Šltenis, 2004 ft T[s+m]

7 Stepping T = , m = 4, q=7 T 0 = 2531 ft = 2531 mod 7 = 4 T 1 = 5319 ft = ((ft T[s]*(10 m-1 mod q))*10 + T[s+m]) mod q ft = ((ft T[0]*(10 3 mod 7))*10 + T[0+4]) mod 7 = ((4 (2*1000 mod 7)) * 10 + T[4]) mod 7 = ((4-(2*6))*10+6) mod 7 = (-8*10+ 9) mod 7 = -71 mod 7 = mod 7 = 6

8 Rin-Krp Algorithm Rin-Krp-Serch(T,P) 01 q prime lrger thn m 02 c 10 m-1 mod q // run loop multiplying y 10 mod q 03 fp 0; ft 0 04 for i 0 to m-1 // preprocessing 05 fp (10*fp + P[i]) mod q 06 ft (10*ft + T[i]) mod q 07 for s 0 to n m // mtching 08 if fp = ft then // run loop to compre strings 09 if P[0..m-1] = T[s..s+m-1] return s 10 ft ((ft T[s]*c)*10 + T[s+m]) mod q 11 return 1 AALG, lecture 3, Simons Šltenis, 2004

9 Anlysis If q is prime, the hsh function distriutes m-digit strings evenly mong the q vlues Thus, only every q th vlue of shift s will result in mtching fingerprints (which will require compring strings with O(m) comprisons) Expected running time (if q > m): Preprocessing: O(m) Outer loop: O(n-m) All inner loops: Totl time: O(n-m) Worst-cse running time: O(nm) n m m O n m q AALG, lecture 3, Simons Šltenis, 2004

10 Rin-Krp in Prctice If the lphet hs d chrcters, interpret chrcters s rdix-d digits (replce 10 with d in the lgorithm). Choosing prime q > m cn e done with rndomized lgorithms in O(m), or q cn e fixed to e the lrgest prime so tht 10*q fits in computer word. AALG, lecture 3, Simons Šltenis, 2004

11 Serching in n comprisons The gol: ech chrcter of the text is compred only once! Prolem with the nïve lgorithm: Forgets wht ws lerned from prtil mtch! Exmples: T = Tweedledee nd Tweedledum nd P = Tweedledum T = pppppppr nd P = pppr AALG, lecture 3, Simons Šltenis, 2004

12 Finite utomton serch c input stte c P c i T[i] -- c stte (i) Processing time tkes (n). But hve to first construct FA. Min Issue: How to construct FA?

13 Need some Nottion (w) = stte FA ends up in fter processing w. Exmple: () = 4. (x) = mx{k: P k suf x}. Clled the suffix function. Exmples: Let P =. ( ) = 0 (ccc) = 1 (cc) = 2 Note: If P = m, then (x) = m indictes mtch. T: c Sttes: m..m. mtch mtch

14 FA Construction Given: P[1..m] Let Q = sttes = {0, 1,, m}. initil finl Define trnsition function s follows: (q, ) = (P q ) for ech q nd. Exmple: P = c (5, ) = (P 5 ) = () = 4 Intuition: Encountering in stte 5 mens the current sustring doesn t mtch. But, you know this sustring ends with -- nd this is the longest suffix tht mtches the eginning of P. Thus, we go to stte 4 nd continue processing.

15 P=c,c c m=7; Q={0,1,2,3,4,5,6,7) Prefixes c c

16 P=c,c c (1, ) = (P 1 ) = () = () = 1 Prefixes c c

17 P=c,c c c (1, ) = (P 1 ) = () = () = 1 (1, c) = (P 1 c) = (c) = 0 Prefixes c c

18 P=c,c c c c (2, ) = (P 2 ) = () = () = 1 (2, c) = (P 2 c) = (c) = 0 Prefixes c c

19 P=c (fst forwrd & simplified),c c (5, ) = (P 5 ) = () = () = 1 (5, ) = (P 5 ) = () = () = 4 Prefixes c c

20 P=c (finl, simplified),c c

21 Serch,c c T= c Prefixes c c

22 Serch,c c T= c Prefixes c c

23 Serch,c c T= c Prefixes c c

24 Serch,c c T= c Prefixes c c

25 Serch,c c T= c Prefixes c c

26 Serch,c c T= c Prefixes c c

27 Serch,c c T= c Prefixes c c

28 Serch,c c T= c Prefixes c c

29 Serch,c c T= c Accept stte, we re done Prefixes c c

30 Anlysis of FA Serching: O(n) good Preprocessing: O(m ) d Memory: O(m ) d

31 COMBINATORIAL PATTERN MATCHING

32 Genomic Repets Exmple of repets: ATGGTCTAGGTCCTAGTGGTC Motivtion to find them: Genomic rerrngements re often ssocited with repets Trce evolutionry secrets Mny tumors re chrcterized y n explosion of repets

33 Genomic Repets The prolem is often more difficult: ATGGTCTAGGACCTAGTGTTC Motivtion to find them: Genomic rerrngements re often ssocited with repets Trce evolutionry secrets Mny tumors re chrcterized y n explosion of repets

34 l-mer Repets Long repets re difficult to find Short repets re esy to find (e.g., hshing) Simple pproch to finding long repets: Find exct repets of short l-mers (l is usully 10 to 13) Use l-mer repets to potentilly extend into longer, mximl repets

35 l-mer Repets (cont d) There re typiclly mny loctions where n l-mer is repeted: GCTTACAGATTCAGTCTTACAGATGGT The 4-mer TTAC strts t loctions 3 nd 17

36 Extending l-mer Repets GCTTACAGATTCAGTCTTACAGATGGT Extend these 4-mer mtches: GCTTACAGATTCAGTCTTACAGATGGT Mximl repet: TTACAGAT

37 Mximl Repets To find mximl repets in this wy, we need ALL strt loctions of ll l-mers in the genome Hshing lets us find repets quickly in this mnner

38 Hshing DNA sequences Ech l-mer cn e trnslted into inry string (A, T, C, G cn e represented s 00, 01, 10, 11) After ssigning unique integer per l-mer it is esy to get ll strt loctions of ech l- mer in genome

39 Hshing: Mximl Repets To find repets in genome: For ll l-mers in the genome, note the strt position nd the sequence Generte hsh tle index for ech unique l-mer sequence In ech index of the hsh tle, store ll genome strt loctions of the l-mer which generted tht index Extend l-mer repets to mximl repets

40 Hshing: Collisions Deling with collisions: Chin ll strt loctions of l-mers (linked list)

41 Hshing: Summry When finding genomic repets from l-mers: Generte hsh tle index for ech l-mer sequence In ech index, store ll genome strt loctions of the l-mer which generted tht index Extend l-mer repets to mximl repets

42 Pttern Mtching Wht if, insted of finding repets in genome, we wnt to find ll sequences in dtse tht contin given pttern? This leds us to different prolem, the Pttern Mtching Prolem

43 Pttern Mtching Prolem Gol: Find ll occurrences of pttern in text Input: Pttern p = p 1 p n nd text t = t 1 t m Output: All positions 1< i < (m n + 1) such tht the n-letter sustring of t strting t i mtches p Motivtion: Serching dtse for known pttern

44 Exct Pttern Mtching: A Brute-Force Algorithm PtternMtching(p,t) 1 m length of pttern p 2 n length of text t 3 for i 1 to (n m + 1) 4 if t i t i+m-1 = p 5 output i

45 Exct Pttern Mtching: An Exmple PtternMtching lgorithm for: Pttern GCAT Text CGCATC GCAT CGCATC GCAT CGCATC GCAT CGCATC GCAT CGCATC GCAT CGCATC

46 Exct Pttern Mtching: Running Time PtternMtching runtime: O(nm) KMP or BM: O(n+m) Multiply y k if looking for k different ptterns Better solution: suffix trees Cn solve prolem in O(n) time Conceptully relted to keyword trees

47 Keyword Trees: Exmple Keyword tree: Apple Also known s trie

48 Keyword Trees: Exmple (cont d) Keyword tree: Apple Apropos

49 Keyword Trees: Exmple (cont d) Keyword tree: Apple Apropos Bnn

50 Keyword Trees: Exmple (cont d) Keyword tree: Apple Apropos Bnn Bndn

51 Keyword Trees: Exmple (cont d) Keyword tree: Apple Apropos Bnn Bndn Ornge

52 Keyword Trees: Properties Stores set of keywords in rooted leled tree Ech edge leled with letter from n lphet Any two edges coming out of the sme vertex hve distinct lels Every keyword stored cn e spelled on pth from root to some lef

53 Keyword Trees: Threding (cont d) Thred ppel ppel

54 Keyword Trees: Threding (cont d) Thred ppel ppel

55 Keyword Trees: Threding (cont d) Thred ppel ppel

56 Keyword Trees: Threding (cont d) Thred ppel ppel

57 Keyword Trees: Threding (cont d) Thred pple pple

58 Keyword Trees: Threding (cont d) Thred pple pple

59 Keyword Trees: Threding (cont d) Thred pple pple

60 Keyword Trees: Threding (cont d) Thred pple pple

61 Keyword Trees: Threding (cont d) Thred pple pple

62 Multiple Pttern Mtching Prolem Gol: Given set of ptterns nd text, find ll occurrences of ny of ptterns in text Input: k ptterns p 1,,p k, nd text t = t 1 t m Output: Positions 1 < i < m where sustring of t strting t i mtches p j for 1 < j < k Motivtion: Serching dtse for known multiple ptterns

63 Multiple Pttern Mtching: Strightforwrd Approch Cn solve s k Pttern Mtching Prolems Runtime: O(kmn) using the PtternMtching lgorithm k times m - length of the text n - verge length of the pttern

64 Multiple Pttern Mtching: Keyword Tree Approch Or, we could use keyword trees: Build keyword tree in O(N) time; N is totl length of ll ptterns With nive threding: O(N + nm) Aho-Corsick lgorithm: O(N + m)

65 Keyword Trees: Threding To mtch ptterns in text using keyword tree: Build keyword tree of ptterns Thred the text through the keyword tree

66 Keyword Trees: Threding (cont d) Threding is complete when we rech lef in the keyword tree When threding is complete, we ve found pttern in the text Prolem: High memory requirement when N is lrge

67 Suffix Trees=Collpsed Keyword Trees Similr to keyword trees, except edges tht form pths re collpsed Built from text, not ptterns Ech edge is leled with sustring of text All internl edges hve t lest two outgoing edges Leves leled y the index of the pttern.

68 Suffix Tree of Text Suffix trees of text is constructed for ll its suffixes ATCATG TCATG CATG ATG TG G Keyword Tree Suffix Tree

69 Suffix Tree of Text Suffix trees of text is constructed for ll its suffixes ATCATG TCATG CATG ATG TG G Keyword Tree How much time does it tke? Suffix Tree

70 Suffix Tree of Text Suffix trees of text is constructed for ll its suffixes ATCATG TCATG CATG ATG TG G qudrtic Keyword Tree Suffix Tree Time is liner in the totl size of ll suffixes, i.e., it is qudrtic in the length of the text

71 Suffix tree (Exmple) Let s=, suffix tree of s is compressed trie of ll suffixes of s= { }

72 Trivil lgorithm to uild Suffix tree Put the lrgest suffix in Put the suffix in

73 Put the suffix in

74 Put the suffix in

75 Put the suffix in

76 We will lso lel ech lef with the strting point of the corres. suffix Trivil lgorithm: O(n 2 ) time 1 2

77 Suffix Trees: Advntges Suffix trees of text is constructed for ll its suffixes Suffix trees uild fster thn keyword trees ATCATG TCATG CATG ATG TG G qudrtic Keyword Tree liner (Weiner suffix tree lgorithm) Suffix Tree

78 Use of Suffix Trees Suffix trees hold ll suffixes of text i.e., ATCGC: ATCGC, TCGC, CGC, GC, C Builds in O(m) time for text of length m To find ny pttern of length n in text: Build suffix tree for text Thred the pttern through the suffix tree Cn find pttern in text in O(n) time! O(n + m) time for Pttern Mtching Prolem Build suffix tree nd lookup pttern

79 Pttern Mtching with Suffix Trees SuffixTreePtternMtching(p,t) 1 Build suffix tree for text t 2 Thred pttern p through suffix tree 3 if threding is complete 4 output positions of ll p-mtching leves in the tree 5 else 6 output Pttern does not pper in text

80 Suffix Trees: Exmple

81 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 dd specil chr, sy, t the end of s To ssocite ech suffix with unique string in S dd different specil chr to ech s

82 Generlized suffix tree (Exmple) Let s 1 = nd s 2 = here is generlized suffix tree for s 1 nd s 2 # { # 5 4 } # # # 1 3 # 2 # # 3 Mtching pttern ginst dtse of strings

83 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 string depth 3 # 2 # 1 2 # 4 #

84 Multiple Pttern Mtching: Summry Keyword nd suffix trees re used to find ptterns in text Keyword trees: Build keyword tree of ptterns, nd thred text through it Suffix trees: Build suffix tree of text, nd thred ptterns through it