1. CS 6293 Advanced Topics: Current Bioinformatics Lecture 5 Exact String Matching Algorithms

2. Overview Sequence alignment: two sub-problems: –How to score an alignment with errors –How to find an alignment with the best score Today: exact string matching –Does not allow any errors –Efficiency becomes the sole consideration Time and space

3. Why exact string matching? The most fundamental string comparison problem Often the core of more complex string comparison algorithms –E.g., BLAST Often repeatedly called by other methods –Usually the most time consuming part –Small improvement could improve overall efficiency considerably

4. Definitions Text: a longer string T (length m) Pattern: a shorter string P (length n) Exact matching: find all occurrences of P in T T P length m length n

5. The naïve algorithm

6. Time complexity Worst case: O(mn) Best case: O(m) e.g. aaaaaaaaaaaaaa vs baaaaaaa Average case? –Alphabet A, C, G, T –Assume both P and T are random –Equal probability –In average how many chars do you need to compare before giving up?

7. Average case time complexity P(mismatch at 1 st position): ¾ P(mismatch at 2 nd position): ¼ * ¾ P(mismatch at 3 nd position): (¼) 2 * ¾ P(mismatch at k th position): (¼) k-1 * ¾ Expected number of comparison per position: p = 1/4  k (1-p) p (k-1) k = (1-p) / p *  k p k k = 1/(1-p) = 4/3 Average complexity: 4m/3 Not as bad as you thought it might be

8. Biological sequences are not random T: aaaaaaaaaaaaaaaaaaaaaaaaa P: aaaab Plus: 4m/3 average case is still bad for long genomic sequences! Especially if this has to be done again and again Smarter algorithms: O(m + n) in worst case sub-linear in practice

9. How to speedup? Pre-processing T or P Why pre-processing can save us time? –Uncovers the structure of T or P –Determines when we can skip ahead without missing anything –Determines when we can infer the result of character comparisons without doing them. ACGTAXACXTAXACGXAX ACGTACA

10. Cost for exact string matching Total cost = cost (preprocessing) + cost(comparison) + cost(output) Constant Minimize Overhead Hope: gain > overhead

11. String matching scenarios One T and one P –Search a word in a document One T and many P all at once –Search a set of words in a document –Spell checking (fixed P) One fixed T, many P –Search a completed genome for short sequences Two (or many) T’s for common patterns Q: Which one to pre-process? A: Always pre-process the shorter seq, or the one that is repeatedly used

12. Pre-processing algs Pattern preprocessing –Knuth-Morris-Pratt algorithm (KMP) –Aho-Corasick algorithm Multiple patterns –Boyer – Moore algorithm (discuss only if have time) The choice of most cases Typically sub-linear time Text preprocessing –Suffix tree Very useful for many purposes

13. Algorithm KMP: Intuitive example 1 Observation: by reasoning on the pattern alone, we can determine that if a mismatch happened when comparing P[8] with T[i], we can shift P by four chars, and compare P[4] with T[i], without missing any possible matches. Number of comparisons saved: 6 abcxabc T abcxabcde P mismatch abcxabc T abcxabcde Naïve approach: abcxabcde ?

14. ? Intuitive example 2 Observation: by reasoning on the pattern alone, we can determine that if a mismatch happened between P[7] and T[j], we can shift P by six chars and compare T[j] with P[1] without missing any possible matches Number of comparisons saved: 7 abcxabc T abcxabcde P mismatch abcxabc T abcxabcde Naïve approach: abcxabcde Should not be a c abcxabcde ?

15. KMP algorithm: pre-processing Key: the reasoning is done without even knowing what string T is. Only the location of mismatch in P must be known. t t’ P t x T y t P y z z Pre-processing: for any position i in P, find P[1..i]’s longest proper suffix, t = P[j..i], such that t matches to a prefix of P, t’, and the next char of t is different from the next char of t’ (i.e., y ≠ z) For each i, let sp(i) = length(t) ij ij

16. KMP algorithm: shift rule t t’ P t x T y t P y z z Shift rule: when a mismatch occurred between P[i+1] and T[k], shift P to the right by i – sp(i) chars and compare x with z. This shift rule can be implicitly represented by creating a failure link between y and z. Meaning: when a mismatch occurred between x on T and P[i+1], resume comparison between x and P[sp(i)+1]. ij ijsp(i)1

17. Failure Link Example P: aataac aataac sp(i) 010020 aaataaat aat aac If a char in T fails to match at pos 6, re-compare it with the char at pos 3 (= 2 + 1)

18. Another example P: abababc abababc Sp(i) 0000040 ababa ababc If a char in T fails to match at pos 7, re-compare it with the char at pos 5 (= 4 + 1) abababab abab

19. KMP Example using Failure Link aataac aataac ^^* T: aacaataaaaataaccttacta aataac.* aataac ^^^^^* aataac..* aataac.^^^^^ Time complexity analysis: Each char in T may be compared up to n times. A lousy analysis gives O(mn) time. More careful analysis: number of comparisons can be broken to two phases: Comparison phase: the first time a char in T is compared to P. Total is exactly m. Shift phase. First comparisons made after a shift. Total is at most m. Time complexity: O(2m) Implicit comparison

20. KMP algorithm using DFA (Deterministic Finite Automata) P: aataac 123450 aa taac 6 a t If the next char in T is t after matching 5 chars, go to state 3 aataac If a char in T fails to match at pos 6, re-compare it with the char at pos 3 a Failure link DFA a All other inputs goes to state 0.

21. DFA Example T: aacaataataataaccttacta Each char in T will be examined exactly once. Therefore, exactly m comparisons are made. But it takes longer to do pre-processing, and needs more space to store the FSA. 1201234534534560001001 123450 aa taac 6 a t a DFA a

22. Difference between Failure Link and DFA Failure link –Preprocessing time and space are O(n), regardless of alphabet size –Comparison time is at most 2m (at least m) DFA –Preprocessing time and space are O(n |  |) May be a problem for very large alphabet size For example, each “char” is a big integer Chinese characters –Comparison time is always m.

23. Boyer – Moore algorithm Often the choice of algorithm for many cases –One T and one P –We will talk about it later if have time –In practice sub-linear

24. The set matching problem Find all occurrences of a set of patterns in T First idea: run KMP or BM for each P –O(km + n) k: number of patterns m: length of text n: total length of patterns Better idea: combine all patterns together and search in one run

25. A simpler problem: spell-checking A dictionary contains five words: –potato –poetry –pottery –science –school Given a document, check if any word is (not) in the dictionary –Words in document are separated by special chars. –Relatively easy.

26. Keyword tree for spell checking O(n) time to construct. n: total length of patterns. Search time: O(m). m: length of text Common prefix only need to be compared once. What if there is no space between words? p o t a t o e t r y t e r y s c i e n c e hoo l 1 2 3 4 5 This version of the potato gun was inspired by the Weird Science team out of Illinois

27. Aho-Corasick algorithm Basis of the fgrep algorithm Generalizing KMP –Using failure links Example: given the following 4 patterns: –potato –tattoo –theater –other

28. Keyword tree p o t a t o t e r 0 t h e r 1 2 3 4 a t t o o h a t e

29. p o t a t o t e r 0 t h e r 1 2 3 4 a t t o o h a t e potherotathxythopotattooattoo

30. Keyword tree p o t a t o t e r 0 t h e r 1 2 3 4 a t t o o h a t e O(mn) m: length of text. n: length of longest pattern potherotathxythopotattooattoo

31. Keyword Tree with a failure link p o t a t o t e r 0 t h e r 1 2 3 4 a t t o o h a t e potherotathxythopotattooattoo

32. Keyword Tree with a failure link p o t a t o t e r 0 t h e r 1 2 3 4 a t t o o h a t e potherotathxythopotattooattoo

33. Keyword Tree with all failure links p o t a t o t e r 0 t h e r 1 2 3 4 a t t o o h a t e

34. Example p o t a t o t e r 0 t h e r 1 2 3 4 a t t o o h a t e potherotathxythopotattooattoo

35. Example p o t a t o t e r 0 t h e r 1 2 3 4 a t t o o h a t e potherotathxythopotattooattoo

36. Example p o t a t o t e r 0 t h e r 1 2 3 4 a t t o o h a t e potherotathxythopotattooattoo

37. Example p o t a t o t e r 0 t h e r 1 2 3 4 a t t o o h a t e potherotathxythopotattooattoo

38. Example p o t a t o t e r 0 t h e r 1 2 3 4 a t t o o h a t e potherotathxythopotattooattoo

39. Aho-Corasick algorithm O(n) preprocessing, and O(m+k) searching. –n: total length of patterns. –m: length of text –k is # of occurrence. Can create a DFA similar as in KMP. –Requires more space, –Preprocessing time depends on alphabet size –Search time is constant A: Where can this algorithm be used in previous topics? Q: BLAST –Given a query sequence, we generate many seed sequences (k- mers) –Search for exact matches to these seed sequences –Extend exact matches into longer inexact matches

40. Suffix Tree All algorithms we talked about so far preprocess pattern(s) –Boyer-Moore: fastest in practice. O(m) worst case. –KMP: O(m) –Aho-Corasick: O(m) In some cases we may prefer to pre-process T –Fixed T, varying P Suffix tree: basically a keyword tree of all suffixes

41. Suffix tree T: xabxac Suffixes: 1.xabxac 2.abxac 3.bxac 4.xac 5.ac 6.c a b x a c b x a c c c x a b x a c c 1 2 3 4 5 6 Naïve construction: O(m 2 ) using Aho-Corasick. Smarter: O(m). Very technical. big constant factor Difference from a keyword tree: create an internal node only when there is a branch

42. Suffix tree implementation Explicitly labeling sequence end T: xabxa$ a b x a b x a x a b x a 1 2 3 a b x a b x a x a b x a 1 2 3 $ $ $ $ $ 4 5 One-to-one correspondence of leaves and suffixes |T| leaves, hence < |T| internal nodes

43. Suffix tree implementation Implicitly labeling edges T: xabxa$ a b x a b x a x a b x a 1 2 3 $ $ $ $ $ 4 5 2:2 3:$ 1 2 3 $ $ 4 5 1:2 3:$ |Tree(T)| = O(|T| + size(edge labels))

44. Suffix links Similar to failure link in a keyword tree Only link internal nodes having branches x a b c d e f g h i j a b c d e f g h i j P: xabcf f

45. ST Application 1: pattern matching Find all occurrence of P=xa in T –Find node v in the ST that matches to P –Traverse the subtree rooted at v to get the locations a b x a c b x a c c c x a b x a c c 1 2 3 4 5 6 T: xabxac O(m) to construct ST (large constant factor) O(n) to find v – linear to length of P instead of T! O(k) to get all leaves, k is the number of occurrence. Asymptotic time is the same as KMP. ST wins if T is fixed. KMP wins otherwise.

46. ST Application 2: set matching Find all occurrences of a set of patterns in T –Build a ST from T –Match each P to ST a b x a c b x a c c c x a b x a c c 1 2 3 4 5 6 T: xabxac P: xab O(m) to construct ST (large constant factor) O(n) to find v – linear to total length of P’s O(k) to get all leaves, k is the number of occurrence. Asymptotic time is the same as Aho-Corasick. ST wins if T fixed. AC wins if P’s are fixed. Otherwise depending on relative size.

47. ST application 3: repeats finding Genome contains many repeated DNA sequences Repeat sequence length: Varies from 1 nucleotide to millions –Genes may have multiple copies (50 to 10,000) –Highly repetitive DNA in some non-coding regions 6 to 10bp x 100,000 to 1,000,000 times Problem: find all repeats that are at least k- residues long and appear at least p times in the genome

48. Repeats finding at least k-residues long and appear at least p times in the seq –Phase 1: top-down, count label lengths (L) from root to each node –Phase 2: bottom-up: count # of leaves descended from each internal node (L, N) For each node with L >= k, and N >= p, print all leaves O(m) to traverse tree

49. Maximal repeats finding 1.Right-maximal repeat –S[i+1..i+k] = S[j+1..j+k], –but S[i+k+1] != S[j+k+1] 2.Left-maximal repeat –S[i+1..i+k] = S[j+1..j+k] –But S[i] != S[j] 3.Maximal repeat –S[i+1..i+k] = S[j+1..j+k] –But S[i] != S[j], and S[i+k+1] != S[j+k+1] acatgacatt 1.cat 2.aca 3.acat

50. Maximal repeats finding Find repeats with at least 3 bases and 2 occurrence –right-maximal: cat –Maximal: acat –left-maximal: aca 5:e 2 4 1234567890 acatgacatt 5:e 5 c a t t 7 c a t t 6 a 3 1 t 8 t t t 9 10 $

51. Maximal repeats finding How to find maximal repeat? –A right-maximal repeats with different left chars 5:e 2 4 1234567890 acatgacatt 5:e 5 c a t t 7 c a t t 6 a 3 1 t 8 t t t 9 10 $ Left char = [] gcc aa

52. ST application 4: word enumeration Find all k-mers that occur at least p times –Compute (L, N) for each node L: total label length from root to node N: # leaves –Find nodes v with L>=k, and L(parent) =p –Traverse sub-tree rooted at v to get the locations L<k L>=k, N>=p L = K L=k This can be used in many applications. For example, to find words that appeared frequently in a genome or a document

53. Joint Suffix Tree (JST) Build a ST for more than two strings Two strings S 1 and S 2 S* = S 1 & S 2 Build a suffix tree for S* in time O(|S 1 | + |S 2 |) The separator will only appear in the edge ending in a leaf (why?)

54. Joint suffix tree example S1 = abcd S2 = abca S* = abcd&abca$ a b c d & a b c a bcd&abcabcd&abca c d & a b c d d & a b c d & a b c d a a a $ 1,1 2,1 1,2 1,3 1,4 2,2 2,3 2,4 (2, 0) useless Seq ID Suffix ID

55. To Simplify We don’t really need to do anything, since all edge labels were implicit. The right hand side is more convenient to look at a b c d & a b c a bcd&abcabcd&abca c d & a b c d d & a b c d & a b c d a a a $ 1,1 2,1 1,2 1,3 1,4 2,2 2,3 2,4 useless a b c d bcdbcd c d d a a a $ 1,1 2,1 1,2 1,3 1,4 2,2 2,3 2,4

56. Application 1 of JST Longest common substring between two sequences Using smith-waterman –Gap = mismatch = -infinity. –Quadratic time Using JST –Linear time –For each internal node v, keep a bit vector B –B[1] = 1 if a child of v is a suffix of S1 –Bottom-up: find all internal nodes with B[1] = B[2] = 1 (green nodes) –Report a green node with the longest label –Can be extended to k sequences. Just use a bit vector of size k. a b c d bcdbcd c d d a a a $ 1,1 2,1 1,2 1,3 1,4 2,2 2,3 2,4

57. Application 2 of JST Given K strings, find all k-mers that appear in at least (or at most) d strings Exact motif finding problem L< k L >= k B = BitOR(1010, 0011) = 1011 cardinal(B) = 3 3,x 4,x B = 0011 1,x B = 1010 cardinal(B) >= 3

58. Application 3 of JST Substring problem for sequence databases –Given: A fixed database of sequences (e.g., individual genomes) –Given: A short pattern (e.g., DNA signature) –Q: Does this DNA signature belong to any individual in the database? i.e. the pattern is a substring of some sequences in the database Aho-Corasick doesn’t work –This can also be used to design signatures for individuals Build a JST for the database seqs Match P to the JST Find seq IDs from descendents a b c d bcdbcd c d d a a a $ 1,1 2,1 1,2 1,3 1,4 2,2 2,3 2,4 Seqs: abcd, abca P1: cd P2: bc

59. Application 4 of JST Detect DNA contamination –For some reason when we try to clone and sequence a genome, some DNAs from other sources may contaminate our sample, which should be detected and removed –Given: A fixed database of sequences (e.g., possible cantamination sources) –Given: A DNA just sequenced (e.g., DNA signature) –Q: Does this DNA contain longer enough substring from the seqs in the database? Build a JST for the database seqs Scan T using the JST a b c d bcdbcd c d d a a a $ 1,1 2,1 1,2 1,3 1,4 2,2 2,3 2,4 Contamination sources: abcd, abca Sequence: dbcgaabctacgtctagt

60. Suffix Tree Memory Footprint The space requirements of suffix trees can become prohibitive –|Tree(T)| is about 20|T| in practice Suffix arrays provide one solution.

61. Suffix Arrays Very space efficient (m integers) Pattern lookup is nearly O(n) in practice –O(n + log 2 m) worst case with 2m additional integers –Independent of alphabet size! Easiest to describe (and construct) using suffix trees –Other (slower) methods exist a b x a b x a x a b x a 1 5 3 $ $ $ $ $ 4 2 52341 abxa$ a$ bxa$xa$xabxa$ 1. xabxa 2. abxa 3. bxa 4. xa 5. a

62. Suffix array construction Build suffix tree for T$ Perform “lexical” depth-first search of suffix tree –output the suffix label of each leaf encountered Therefore suffix array can be constructed in O(m) time.

63. Suffix array pattern search If P is in T, then all the locations of P are consecutive suffixes in Pos. Do binary search in Pos to find P! –Compare P with suffix Pos(m/2) –If lexicographically less, P is in first half of T –If lexicographically more, P is in second half of T –Iterate!

64. Suffix array pattern search T: xabxa$ P: abx a b x a b x a x a b x a 1 5 3 $ $ $ $ $ 4 2 52341 abxa$ a$ bxa$xa$xabxa$ L RM R M

65. Suffix array binary search How long to compare P with suffix of T? –O(n) worst case! Binary search on Pos takes O(n log m) time Worst case will be rare –occur if many long prefixes of P appear in T In random or large alphabet strings –expect to do less than log m comparisons O(n + log m) running time when combined with LCP table –suffix tree = suffix array + LCP table

66. Summary One T, one P –Boyer-Moore is the choice –KMP works but not the best One T, many P –Aho-Corasick –Suffix Tree (array) One fixed T, many varying P –Suffix tree (array) Two or more T’s –Suffix tree, joint suffix tree Alphabet independent Alphabet dependent

67. Boyer – Moore algorithm Three ideas: –Right-to-left comparison –Bad character rule –Good suffix rule

68. Boyer – Moore algorithm Right to left comparison x y y Skip some chars without missing any occurrence. Resume comparison here

69. Bad character rule 0 1 12345678901234567 T:xpbctbxabpqqaabpq P: tpabxab *^^^^ What would you do now?

70. Bad character rule 0 1 12345678901234567 T:xpbctbxabpqqaabpq P: tpabxab *^^^^ P: tpabxab

71. Bad character rule 0 1 123456789012345678 T:xpbctbxabpqqaabpqz P: tpabxab *^^^^ P: tpabxab * P: tpabxab

72. Basic bad character rule charRight-most-position in P a6 b7 p2 t1 x5 tpabxab Pre-processing: O(n)

73. Basic bad character rule charRight-most-position in P a6 b7 p2 t1 x5 T: xpbctbxabpqqaabpqz P: tpabxab *^^^^ P: tpabxab When rightmost T(k) in P is left to i, shift pattern P to align T(k) with the rightmost T(k) in P k i = 3 Shift 3 – 1 = 2

74. Basic bad character rule charRight-most-position in P a6 b7 p2 t1 x5 T: xpbctbxabpqqaabpqz P: tpabxab * P: tpabxab When T(k) is not in P, shift left end of P to align with T(k+1) k i = 7Shift 7 – 0 = 7

75. Basic bad character rule charRight-most-position in P a6 b7 p2 t1 x5 T: xpbctbxabpqqaabpqz P: tpabxab *^^ P: tpabxab When rightmost T(k) in P is right to i, shift pattern P by 1 k i = 55 – 6 < 0. so shift 1

76. Extended bad character rule charPosition in P a6, 3 b7, 4 p2 t1 x5 T: xpbctbxabpqqaabpqz P: tpabxab *^^ P: tpabxab Find T(k) in P that is immediately left to i, shift P to align T(k) with that position k i = 55 – 3 = 2. so shift 2 Preprocessing still O(n)

77. Extended bad character rule Best possible: m / n comparisons Works better for large alphabet size In some cases the extended bad character rule is sufficiently good Worst-case: O(mn) –Expected time is sublinear

78. 0 1 123456789012345678 T:prstabstubabvqxrst P: qcabdabdab *^^ P: qcabdabdab According to extended bad character rule

79. (weak) good suffix rule 0 1 123456789012345678 T:prstabstubabvqxrst P: qcabdabdab *^^ P: qcabdabdab

80. (Weak) good suffix rule t x t y t’ t y Preprocessing: For any suffix t of P, find the rightmost copy of t, denoted by t’. How to find t’ efficiently? T P P

81. (Strong) good suffix rule 0 1 123456789012345678 T:prstabstubabvqxrst P: qcabdabdab *^^

82. (Strong) good suffix rule 0 1 123456789012345678 T:prstabstubabvqxrst P: qcabdabdab *^^ P: qcabdabdab

83. (Strong) good suffix rule t x t y t’ t y In preprocessing: For any suffix t of P, find the rightmost copy of t, t’, such that the char left to t ≠ the char left to t’ T P P z z z ≠ y

84. Example preprocessing qcabdabdab charPositions in P a9, 6, 3 b10, 7, 4 c2 d8, 5 q1 q c a b d a b d a b 1 2 3 4 5 6 7 8 9 10 0 0 0 0 2 0 0 2 0 0 dab cab Bad char rule Good suffix rule dabdab cabdab Where to shift depends on T Does not depend on T Largest shift given by either the (extended) bad char rule or the (strong) good suffix rule is used.

85. Time complexity of BM algorithm Pre-processing can be done in linear time With strong good suffix rule, worst-case is O(m) if P is not in T –If P is in T, worst-case could be O(mn) –E.g. T = m 100, P = m 10 –unless a modification was used (Galil’s rule) Proofs are technical. Skip.

86. How to actually do pre-processing? Similar pre-processing for KMP and B-M –Find matches between a suffix and a prefix –Both can be done in linear time –P is usually short, even a more expensive pre-processing may result in a gain overall t t’ P yx KMP t y t’ P x B-M i i j j For each i, find a j. similar to DP. Start from i = 2

87. Fundamental pre-processing Z i : length of longest substring starting at i that matches a prefix of P –i.e. t = t’, x ≠ y, Z i = |t| –With the Z-values computed, we can get the preprocessing for both KMP and B-M in linear time. aabcaabxaaz Z = 01003100210 How to compute Z-values in linear time? t t’ P i x y i+z i -1zizi 1

88. Computing Z in Linear time t t’ P l x y rk We already computed all Z- values up to k-1. need to compute Zk. We also know the starting and ending points of the previous match, l and r. t t’ P l x y rk We know that t = t’, therefore the Z-value at k-l+1 may be helpful to us. 1 k-l+1

89. Computing Z in Linear time No char inside the box is compared twice. At most one mismatch per iteration. Therefore, O(n). P k The previous r is smaller than k. i.e., no previous match extends beyond k. do explicit comparison. P l x y rk Z k-l+1 <= r-k+1. Z k = Z k-l+1 No comparison is needed. 1 k-l+1 Case 1: Case 2: P l rk Z k-l+1 > r-k+1. Z k = Z k-l+1 Comparison start from r 1 k-l+1 Case 3:

90. Z-preprocessing for B-M and KMP Both KMP and B-M preprocessing can be done in O(n) t t’ i x y j = i+z i -1 zizi 1 t t’ yx KMP t y t’ x B-M i j Z j i j For each j sp’(j+z j -1) = z(j) Use Z backwards