Algorithms in Bioinformatics: A Practical Introduction. Suffix tree

Transcription

1 Alorithms in Bioinformtis: A Prtil Introdution Suffix tree

2 Overview Wht is suffix tree? Simple pplition of suffix tree Liner time lorithm for onstrutin suffix tree Suffix rry FM-index 1-mismth serh

3 Suffix Trie E.. onsider the strin S = Suffix Trie: ties of ll possile suffies of S Suffix

4 Suffix Tree Pth-lel of node v is Denoted s α(v) Suffix tree for S=: mere nodes with only one hild 7 v S= is n ede lel This is lef ede

5 Size of Suffix Tree (I) How i is suffix tree? Suffix tree hs extly n leves nd t most 2n-1 edes The totl lenth of ll ede lels is O(n 2 ). Cn we store suffix tree usin o(n 2 ) it spe? S=

6 Size of Suffix Tree (II) Suffix tree hs extly n leves nd t most 2n-1 edes Note tht eh ede lel n e represented usin 2 indies Thus, suffix tree n e represented usin O(n lo n) its 7 4,7 7,7 6,7 2,3 6, ,1 6, ,7 2,3 6,7 6 S= Note: The end index of every lef ede should e 7, the lst index of S. Thus, for lef edes, we only need to store the strt index.

7 Property of suffix tree Ft: For ny internl node v in the suffix tree, if the pth lel of v is α(v)=p, then there exists nother node w in the suffix tree suh tht α(w)=p. Proof: Skip the proof. Definition of Suffix Link: For ny internl node v, define its suffix link sl(v) = w.

8 Suffix Link exmple S=

9 Generlized suffix tree Build suffix tree for two or more strins E.. S 1 = t#, S 2 = t 6 # 4 t t t t t t # # # # # t

10 Applitions of Suffix Tree

11 Ext strin mthin prolem To find ll ourrenes of Q in S (serhin) Serh for the node x in the suffix tree whih represent Q All the leves in the sutree rooted t x re the ourrenes Time: O( Q + o) where o is the totl no. of ourrenes E.. S = Q = Ourrenes: 1, 3

12 Lonest repeted sustrin prolem To find the lonest repeted sustrin in S Find the deepest internl node Time: O(n) E.. S = The lonest repet is.

13 Lonest ommon sustrin prolem To find the lonest ommon sustrin of two or more sequenes Note: 1970, Don Knuth onjetured tht liner time lorithm for this prolem is impossile Now, we know tht it n e solved in liner time. E.. onsider two strin S 1 nd S 2, 1. Build enerlized suffix tree for S 1 # nd S 2 2. Then, mrk eh internl node with leves representin suffixes of oth S 1 nd S Report the deepest mrked node

14 Exmple for the lonest ommon sustrin E.. S 1 = t#, S 2 = t The lonest ommon sustrin is. Its lenth is 2. 6 # t 4 t t t t # t # # t # #

15 Lonest ommon prefix (I) Given strin S. For ny i, j, Denote lp(i, j) e the lenth of the lonest ommon prefix of suffix i nd j of S S= The lonest ommon prefix of suffix 1 nd suffix 3 is! lp(1, 3) = 3

16 Lonest ommon prefix (II) Note tht the lowest ommon nestor(l) of leves i nd j identifies the lonest ommon prefix. lp(i, j) = α(l(i, j)). A well-know result: Consider tree of size n, fter n O(n) time preproessin, the l for ny two nodes n e returned in O(1) time. First otined y Hrel nd Trjn (SIAM J. Comp. 1984) Simplified y Shieer nd Vishkin (SIAM J. Comp. 1988) Bsed on the ove result, After n O(n) time preproessin, For ny suffix i nd suffix j, we n ompute the lonest ommon prefix of them in O(1) time.

17 Findin Plindrome (I) Given strin S, plindrome is sustrin u of S s.t. u = u r E.. ACAGACA Consider plindrome u=s[i..i+ u -1], u is lled mximl plindrome if S[i..j ] is not plindrome for ny [i..j ] [i..i+ u -1]. Note tht every plindrome is ontined in mximl plindrome. Thus, mximl plindromes re ompt wy to represent ll plindromes. Complemented Plindrome is strin u s.t. u = ū r E.. ACAUGU Mximl omplemented plindrome is defined similrly.

18 Findin Plindrome (II) Rell tht restrition enzyme usully is in the form of omplemented plindrome. This motivtes the followin two prolems: The plindrome prolem: Given strin S (representin the enome) of lenth n, the prolem is to lote ll mximl plindromes in S. The omplemented plindrome prolem: Given strin S (representin the enome) of lenth n, the prolem is to lote ll mximl omplemented plindromes in S.

19 Properties of plindrome (I) If S[i..i+k-1]=S r [n-i+1..n-i+k], then u=s[i-k+1..i+k-1] is n odd lenth plindrome 1 i-k+1 i i+k-1 n S 1 n-i+1 n S r

20 Properties of plindrome (II) If S[i..i+k-1]=S r [n-i+2..n-i+k+1], then u=s[i-k..i+k-1] is n even lenth plindrome S 1 i-k i i+k-1 n S r 1 n-i+2 n

21 Solution to the plindrome prolem Preproess S nd S r so tht ny lonest ommon prefix query n e nswered in onstnt time. For i=1 to n, Find the lonest ommon prefix for (S i, S r n-i+1). If the lonest prefix is k, we find n odd lenth mximl plindrome S[i-k+1..i+k-1]. Find the lonest ommon prefix for (S i, S r n-i+2). If the lonest prefix is k, we find n even lenth mximl plindrome S[i-k..i+k-1].

22 Extrtin emedded suffix tree from enerlized suffix tree 6 Input: The enerlized suffix tree T of K strins S 1,, S K. Aim: Compute the suffix tree T i of the strin S i. r # t w x y 4 z t t t t t t # # # # # T r # w 6 t # t t t t # # # # T 1 S 1 = t#, S 2 = t

23 6 Extrtin emedded suffix tree from enerlized suffix tree Oservtion: T i is sutree of T suh tht The leves of T i re the leves of T orrespondin to S i. The internl nodes of T i re the lowest ommon nestors of some leves for S i. The edes of T i n e inferred from the nestor desendent reltionship mon those nodes. r # t w x y 4 z t t t t t t # # # # # T r # w 6 t # t t t t # # # # T 1 S 1 = t#, S 2 = t

24 Extrtin emedded suffix tree from enerlized suffix tree r t t t t # w t # 5 # # 6 # # r t t t # w t # # 6 # # r t t w t # # 6 # # r 4 1 t w t # 6 # # r 1 t 6 # # r 6 #

25 Common sustrins of more thn 2 strins (I) Given set of strins (protein or DNA sequenes), we wnt to know wht sustrins re ommon to lre numer of these strins? Why this question is importnt? DNA nd protein sequenes will evolve. If sustrin our ommonly in wide rne of speies. This my men tht the sustrin is ritil for the orret funtionlity.

26 Common sustrins of more thn 2 strins (II) Given K strins whose totl lenth is n. For every 2 k K, define l(k) e the lenth of the lonest sustrin ommon to t lest k of these strins. The prolem is to ompute l(k) for ll k.

27 Common sustrins of more thn 2 strins (III) Exmple: Consider set of 5 strins { sndollr, sndlot, hndler, rnd, pntry } Then, we hve k l(k) orrespondin sustrin 2 4 snd 3 3 nd 4 3 nd 5 2 n

28 Common sustrins of more thn 2 strins (IV) Illustrtin the solution y exmple: S 1 =, S 2 = #, S 3 = %. (K=3) 1. Build enerlized suffix tree T for the K strins in O(n) time. 4 % 5 3 # 5 % 1 # # % 3 1 # 2 % 4 2 # 3

29 Common sustrins of more thn 2 strins (V) 2. By trversin T, for eh internl node v, ompute its strin depth. In totl, O(n) time. 0 4 % 5 3 # 5 % # # 2 % 3 1 # 2 1 % 4 2 # 3

30 Common sustrins of more thn 2 strins (VI) 3. By trversin T, for eh internl node v, ompute C(v). [C(v) is defined s the numer of distint termintion symols in the sutree rooted t v] This step tkes O(Kn) time. 3 4 % 5 3 # 5 % # # 3 % 3 1 # 2 3 % 4 2 # 3

31 Common sustrins of more thn 2 strins (VII) 4. Trverse T nd visit every internl node v. For eh v, if V(C(v)) < strin-depth of v, set V(C(v)) = strin-depth of v. [After step 4, V(k) = the lenth of the lonest sustrin ommon to extly k of these strins.] 5. l(k)=v(k). For i=k-1 downto 2, l(i)=mx{l(i+1), V(i)}. This two steps tke O(n) time. For our exmple, V(2) = 3, V(3) = 2. Thus, l(3) = 2, l(2) = 3. In totl, this lorithm tkes O(Kn) time. Atully, we n improve this lorithm to O(n) time y men of lp!

32 Liner time lorithm for onstrutin suffix tree

33 Strihtforwrd onstrution of suffix tree Consider S = s 1 s 2 s n where s n = Alorithm: Initilize the tree with only root For i = n to 1 Inludes S[i..n] into the tree Time: O(n 2 )

34 Exmple of onstrution S= Init For-loop I 5 I 4 I 3 I 2 I 1

35 Constrution of enerlized suffix tree S = # Init For-loop # # # I 1 J 2 J 1

36 Cn we onstrut suffix tree in o(n 2 ) time? Yes. We n onstrut it in O(n) time. Weiner s lorithm [1973] Liner time for onstnt size lphet, ut muh spe MGreiht s lorithm [JACM 1976] Liner time for onstnt size lphet, qudrti spe Ukkonen s lorithm [Alorithmi, 1995] Online lorithm, liner time for onstnt size lphet, less spe Frh s lorithm [FOCS 1997] Liner time for enerl lphet Hon,Sdkne, nd Sun s lorithm [FOCS 2003] O(n) it spe O(n lo e n) time for 0<e<1 O(n) it spe O(n) time for suffix rry onstrution We will disuss Frh s lorithm lter.

37 Ide Build Odd Suffix Tree nd Even Suffix Tree Then, mere odd nd even suffix tree Even Suffix Tree Odd Suffix Tree

38 Ide Input: strin S of lenth n 1. Reursively ompute the suffix tree T o of ll suffixes einnin t the odd positions. T o is of size n/2. 2. From T o, ompute T e whih is the suffix tree for ll suffixes einnin t the even positions. 3. Mere T o nd T e to form the suffix tree for S.

39 Ste 1: Construtin odd suffix tree Given strin S[1..n], we enerte new strin S [1..n/2] s follows. we mp pirs of hrters into sinle hrters s follows: S[1..2], S[3..4], S[5..6],, S[n-1..n]. Remove the duplites from the pirs of hrters nd sort them y rdix sort. S [i] = rnk of S[2i-1..2i] in the sorted list, for i=1, 2,, n/2. By reursion, we et the suffix tree T for S Convert T to the odd suffix tree T o.

40 Exmple (I) S = S[1..2]=, S[3..4]=, S[5..6]=, S[7..8]=, S[9..10]=, S[11..12]=. By stle sort, < < <. Rnk()=1, Rnk()=2, Rnk()=3, Rnk()=4. So, S =

41 Exmple (II) By reursion, onstrut the suffix tree T for S :

42 Exmple (III) Convert T to the odd tree: 13 5 i 2i This is not suffix tree

43 Exmple (IV) Refine the odd tree T o :

44 Time omplexity for uildin the odd tree Let Time(n) e the time to uild suffix tree for strin of lenth n. Stle sortin nd refinement of the odd trees tke O(n) time. Build suffix tree for S tkes Time(n/2). So, Ste 1 tkes Time(n/2)+O(n) time.

45 Ste 2: Build the even tree 1. Generte the lex-orderin of the leves in T e. 2. For ny two djent leves 2i nd 2j, we find lp(2i, 2j). 3. Construt the even tree T e from left to riht (ordin to the lex-orderin).

46 Build the even tree (Step 1) We et the lex-orderin of the leves in T o. Generte the lex-orderin of the leves in T e. For eh lef i in T o, et the preedin hrter =S[i-1] nd form pir (,i). Eh pir represents even suffix i-1. Perform stle sortin on those pirs. We et the lex-orderin of the leves in T e.

47 Exmple S = Lex-orderin of the leves in T o : 13 < 1 < 7 < 3 < 11 < 9 < 5 The pirs re: (,13), (,1), (,7), (, 3), (, 11), (, 9), (, 5). After stle sortin, we hve (, 1), (, 13), (, 3), (, 11), (, 7), (, 9), (, 5). Hene, the lex-orderin of the leves of T e : 12 < 2 < 10 < 6 < 8 < 4

48 Build the even tree (Step 2) For ny two djent leves 2i nd 2j, we first find lp(2i, 2j). Oservtion: lp(2i, 2j) = lp(2i+1, 2j+1)+1 if S[2i]=S[2j] 0 otherwise Proof: If S[2i] S[2j], lp(2i,2j)=0. Otherwise, lp(2i,2j)=1+lp(2i+1,2j+1).

49 Exmple Rell tht the lexorderin of leves: 12 < 2 < 10 < 6 < 8 < 4. By the previous oservtion, we hve lp(8,4)=lp(9,5)+1=2 Similrly, we hve lp(12,2)=1, lp(2,10)=1, lp(10,6)=0, lp(6,8)=1, lp(8,4)=

50 Build the even tree (Step 3) Construt the even tree T e from left to riht

51 Build the even tree (Step 3)

52 Time omplexity for uildin the even tree Step 1: O(n) time Step 2: O(n) time Step 3: O(n) time

53 Ste 3: Mere odd nd even trees We n mere T o nd T e y DFS. However, it tkes O(n 2 ) time Odd tree Even tree ,11 1,1,1,1,1,1,1,2,1,10 10,4,9,1,1,1,1,2 11 9, ,2,2 8,5 5,8 4

54 Ste 3: Mere odd nd even trees We mere T o nd T e y DFS. We mere two edes s lon s they strt with the sme hrter. The mere is ended when one ede is loner thn the other

55 Mere odd nd even trees We mere T o nd T e y DFS. We mere two edes s lon s they strt with the sme hrter. The mere is ended when one ede is loner thn the other.,1,1,1 13,1,2,1,1 12 2,1 1,11,1 10,4 7,9,1,3 11 9,3 6 3,4 8,3 5,8 4

56 Mere odd nd even trees The merin my over-mered some nodes. To orret the tree, we need to unmere some nodes.,1,1,1 13,1,2,1,1 12,11,1 10,1,3,4 8,8 2,1,4 7,9 11 9,3,

57 Definition of L() nd d() For every node u whih my e over-mered, there exist two leves 2i nd 2j-1 suh tht u=l(2i, 2j-1). Denote L(u) e the orret depth of u, tht is, lp(2i,2j-1). Note tht lp(2i,2j-1) = 1+lp(2i+1,2j) if S[2i]=S[2j-1]; 0 otherwise. Let v e l(2i+1,2j). Denote d(u) = v. Note tht d() is equivlent to suffix link!

58 Exmple of d(),1,1,1 13,1,2,1,1 d() 12,11,1 10,1,3,4 8,8 2,1 1,4 7,9 11 9,3 6 3,3 5 4

59 Reltionship etween L() nd d() Suppose u = l(2i, 2j-1). Note1: if u is not the root, then S[2i]=S[2j-1]. Note2: lp(2i,2j-1) = 1+lp(2i+1,2j) if S[2i]=S[2j-1]; 0 otherwise Note3: d(u) = lp(2i+1,2j) Hene, L(u) = 1 + L(d(u)) if u is not the root. Otherwise, L(u)=0. Lemm: L(u) = the lenth of the purple pth from u to the root.

60 Exmple of L() 13 2,1 12,1 L( )=1,1,1,11,1 10,2,4 7,1 L( )=2,9 L( )=1,1,1,1,3 11 9, L( )=2 L( )=2 L( )=3,4 8,3 5 L( )=2,8 L( )=4 4

61 Unmere the order nodes sed on L() (I) ,1,1 L( )=1,1,1,11,1 10,2,4 7,1 L( )=2,9 L( )=1,1,1,1,3 11 9, L( )=2 L( )=2 L( )=3,4 8,3 5 L( )=2,8 L( )=4 4

62 Unmere the order nodes sed on L() (II),1,1, ,11,1,1,1,2,1,10 10,4,9,1,1,1,1,2 11 9, ,2,2 8,5 5,8 4

63 Time omplexity for merin Mere the tree usin DFS tkes O(n) time. Compute the links d() tkes O(n) time. Compute L() tkes O(n) time. Unmere tkes O(n) time.

64 Totl time omplexity of Frh s lorithm Ste 1: Time(n/2)+O(n) Ste 2: O(n) Ste 3: O(n) Thus, Time(n) = Time(n/2)+O(n). By solvin the eqution, Time(n)= O(n).

65 Disdvnte of suffix tree Suffix tree is spe ineffiient. It requires O(n Σ lo n) its. Mner nd Myers (SIAM J. Comp 1993) proposes new dt struture, lled suffix rry, whih hs similr funtionlity s suffix tree. Moreover, it only requires O(n lo n) its.

66 Suffix Arry (I) It is just sorted suffixes. E.. onsider S = Suffix Position SA[i] Suffix => 3 4 Sort Suffix rry is n rry of n indies. Thus, it tkes O(n lo n) its.

67 Oservtion The leves of suffix tree is in suffix rry order SA[i] Suffix

68 Liner time onstrution of suffix rry from suffix tree Rell tht the suffix tree T of S[1..n] n e onstruted in O(n) time. Then, y lexil depth-first trversl of T, the suffix rry of S is otined. This tkes O(n) time. However, the spe used durin onstrution is the sme s tht for suffix tree! This defets the purpose of suffix rry. Tody, we n uild suffix rry usin O(n) it spe nd O(n) time.

69 rne(t,q) For pttern Q, its ourrenes in T form onseutive SA rne. Exmple: For T=, ours in SA[5] nd SA[6]. Definition: We lled rne(t,q)=[st..ed] if Q is prefix of every T j for j=sa[st], SA[st+1],, SA[ed] where T j = j suffix of T = T[j..n]. Exmple: rne(t,)=[5..6] SA[i] Suffix

70 Find ourrene of query Q in strin S usin suffix rry Input: (1) the suffix rry of strin T of lenth n nd (2) query Q of lenth m Aim: hek if Q ours in T Ide: inry serh!

71 Alorithm

72 Exmple Consider T = Pttern Q = L=1 R=7 M=(L+R)/2=4 i SA[i] Suffix

73 Exmple Consider T = Pttern Q = L=1 R=7 M=(L+R)/2=4 suffix-sa[m] > Q. Set R=M=4. i SA[i] Suffix

74 Exmple Consider T = Pttern Q = L=1 R=4 M=(L+R)/2=2 suffix-sa[m] < Q. Set L=M=2. i SA[i] Suffix

75 Exmple Consider T = i SA[i] Suffix Pttern Q = L=2 R=4 M=(L+R)/2= The pttern Q is found t SA[M]=3.

76 Cn we do etter? Durin eh step of inry serh, we need to ompre Q with suffix usin O(m) time, whih is time onsumin. Cn we do etter? We hve the followin oservtion. Suppose LCP(Q, suffix-sa[l]) is l nd LCP(Q, suffix-sa[r]) is r. Then, LCP(Q, suffix-sa[m]) > min{l,r}. Below, we desrie how to utilize this oservtion to speedup the omputtion.

77 Alorithm

78 Exmple Consider T = Pttern Q = L=1, l=0 R=7, r=0 mlr = min(l,r)=0 M=(L+R)/2=4 i SA[i] Suffix

79 Exmple Consider T = Pttern Q = L=1, l=0 R=7, r=0 mlr = min(l,r)=0 M=(L+R)/2=4, m=1 The (m+1) hr of suffix-sa[m] is. The (m+1) hr of Q is. So, suffix-sa[m] > Q. Set R=M=4 nd r=m=1. i SA[i] Suffix

80 Exmple Consider T = Pttern Q = L=1, l=0 R=4, r=1 mlr = min(l,r)=0 M=(L+R)/2=2, m=3 The (m+1) hr of suffix-sa[m] is. The (m+1) hr of Q is. So, suffix-sa[m] < Q. Set L=M=2 nd l=m=3. i SA[i] Suffix

81 Exmple Consider T = Pttern Q = L=2, l=3 R=4, r=1 mlr = min(l,r)=1 M=(L+R)/2=3, m=4 i SA[i] Suffix The pttern Q is found t SA[M]=3.

82 Time nlysis Binry serh will perform lo n omprisons Eh omprison tkes t most O(m) time In the worst se, O(m lo n) time. Myers nd Mner report tht, in prtie, the time is O(m + lo n).

83 Suffix rry nd suffix tree We show one exmple of replin suffix tree y suffix rry Note tht most pplitions relted to suffix tree n e solved usin suffix rry with some time low up! When spe is limited, replin suffix tree y suffix rry is ood hoie.

84 The size is still too i! Why? DNA sequenes n e very lon! E.. Fly: ~100M ses, Humn: ~3G ses, Tree: ~9G ses Store to store indexin dt struture for humn enome Suffix Tree: ~40G ytes Suffix Arry: ~13G ytes Cn we further redue the spe?

85 Solution Grossi, Vitter (STOC2000) Compressed suffix rry (CSA) Ferrine, Mnzini (FOCS2000) FM-index Both of them n e stored in O(n) it spe For Humn Genome Both CSA nd FM-index n e stored within 2G ytes.

86 FM-index Consider text T= FM-index stores: A. The strin BW= B. C[x] = totl no. of ourrenes of eh symol less thn x. E.. C[]=1, C[]=4, C[]=6, C[t]=7 C. A dt-struture o(x, i) whih tells us the numer of ourrenes of x in BW[1..i] usin O(1) time. SA[i] Suffix T[SA[i]-1]

87 Dt-struture for nswerin the o(x, i) query? BW 1 lo 2 n n/lo 2 n lo n lo n 00x0xxx0x0 Given the text BW[1..n], we divide BW[1..n] into ukets of size lo 2 n. For eh uket i = 1,, n/lo 2 n, we store P[i] = numer of x s in BW[1.. ilo 2 n]. Eh uket is further sudivided into lo n su-ukets of size lo n. For eh su-uket j of the uket i, we store Q[i][j] = numer of x s in the first j su-ukets.

88 Dt-struture for nswerin the o(x, i) query? We lso need lookup tle rnk(,k) is ny strin of lenth (lo n)/2 1 k (lo n)/2 rnk(,k) = numer of x in the first k hrters of (lo n)/ x0x xx x Numer of = 2 lon 2 Eh entry tkes O(lo lo n) its Totl spe = O( lo 2 n entries = n lo 2 n lo n lo lo n) = o( n) its n

89 Spe omplexity of the o() dt-struture P[1..n/lo 2 n] uses O(n/lo n) its Q[1..n/lo 2 n][1..lo n] uses O(n lo lo n / lo n) its rnk(,k) uses 2 lon/2 (lo n/2) = o(n) its In totl, we use O(n lo lo n / lo n) its.

90 How to ompute o(x,i)? BW 1 lo 2 n n/lo 2 n lo n lo n 00x0xxx0x0 Suppose lo n = 10. To ompute o(x, 327), The result is P[3]+Q[4][2]+rnk(00x0x,5)+rnk(xx0x0,2) Hene, O(1) time to ompute o(x,i).

91 Size of FM-index Struture A n e store in 2n its Struture B n e store in O(lo n) its Struture C n e store in O(n lo lo n/lo n) its In totl, the size of FM-index is O(n) its.

92 Oservtion C[x]+o(x,i) is the numer of suffixes smller thn of xt SA[i]. Exmple: C[]=4 o(, 6)=2 Numer of suffixes smller thn T[SA[6]..n]= is 6. SA[i] Suffix T[SA[i]-1]

93 Lemm Suppose rne(t,q) is [st..ed]. Then, rne(t,xq) = [p..q] where p = C[x] + o(x,st-1) + 1 q = C[x] + o(x,ed) Proof: p = 1+numer of suffixes stritly smller thn xq. The ltter term = numer of suffixes smller thn or equl to xt SA[st-1]. q = numer of suffixes smller thn or equl to xt SA[ed].

94 Bkwrd serh Given the text T nd the FM-index, we wnt to determine if Q exists in T. Alorithm BW_exist(Q[1..m]) 1. x=q[m], i=m-1; 2. /* find rne(t,q[m]) */ st = C[x]+1, ed = C[x+1]; 3. while (st ed nd i>1) { /* find rne(t, Q[i-1..m]) */ x = Q[i-1]; st = C[x] + o(x, st-1) + 1; ed = C[x] + o(x, ed); i = i 1; } 4. if st > ed, then pttern not found else pttern found.

95 Exmple T = Q = Q[3..3]= sp=c[] +1 =1+1=2 ep=c[] =4 SA[i] Suffix T[SA[i]-1]

96 Exmple T = Q = Q[2..3]= st=c[]+o(,st old -1)+1 =4+0+1=5 ed=c[]+o(,ed old ) =4+2=6 SA[i] Suffix T[SA[i]-1]

97 Exmple T = Q = Q[1..3]= st=c[]+o(,st old -1)+1 =1+0+1=2 ed=c[]+o(,ed old ) =1+2=3 SA[i] Suffix T[SA[i]-1]

98 Exmple T = Q = Q[1..3]= st=c[]+o(,st old -1)+1 =1+0+1=2 ed=c[]+o(,ed old ) =1+2=3 Q ourrenes in T! SA[i] Suffix T[SA[i]-1]

99 Time omplexity of Bkwrd Serh To find pttern Q[1..m] Step 1, 2, nd 4 n e omputed in O(1) time. For step 3, We need to iterte the loop for m-1 times. Eh itertion of the loop n e omputed in O(1) time. The loop tkes O(m) time. In totl, O(m) time for kwrd serh.

100 Conlusion Suffix tree is powerful dt-struture whih hs lot of pplitions in Computtionl Bioloy. Prolems: Suffix tree is too i! Cn e solved usin CSA nd FM-index Suffix tree n only solve ext mth prolem! (Most of the ioloy prolems re pproximte mth!) Mny works hve een done on this re! But still not prtil. One of the importnt re to explore!

101 1-mismth prolem

102 1-mismth prolem Index: the suffix tree of text T[1..n] For ny pttern P[1..m], the 1-mismth prolem finds ll ourrenes of P in T tht hs hmmin distne t most 1. Exmple: P = ACGT T = AACGTGGCCAACTTGGA

103 Nïve solution Index: rete the suffix tree for T Alorithm for query: Generte ll possile 1-mismth ptterns of P. Find ourrenes of every 1-mismth pttern Runnin time: There re Σ m possile 1-mismth ptterns. Usin suffix tree, it tkes O(m) time to find ourrenes of eh 1-mismth pttern In totl, O(m 2 +o) time where o is totl numer of ourrenes.

104 Any other solutions? Cos [CPM 1995] O(n lo n) it index, O(m 2 +o) query time. Amir et l [Journl of Alorithm 2000] O(n lo 3 n) it index, O(m lo n lolo n + o) query time. Buhsum et l [ESA 2000] O(n lo 2 n) it index, O(m lo lo n + o) query time. Cole et l [SODA 2004] O(n lo 2 n) it index, O(m + lo lo n + o) query time. Trinh et l [CPM 2004] O(n lo n) it index, O(m lo n + o) query time. Lm et l [ISAAC 2005] O(n sqrt(lo n)) it index, O(m lo lo n + o) query time. Chn et l [ESA 2006] O(n lo n) it index, O(m + lo lo n + o) query time. Tody, we hve look of the solution of Trinh et l.

105 Index Suffix rry of T SA[1..n] Inversed suffix rry of T SA -1 [1..n] where SA[SA -1 [i]] = i. Definition: We lled rne(t,p)=[st..ed] if SA[i] 7 Suffix P is prefix of every T j for j=sa[st], SA[st+1],, SA[ed] where T j = j suffix of T = T[j..n]. Exmple: For T=, rne(t,)=[5..6] i

106 Lemm 1 (Forwrd serh) Assume [st..ed]=rne(t,p). We n ompute [st..ed ]=rne(t,p) in O(lo n) time. Proof: inry serh on SA.

107 Lemm 2 Assume [st 1..ed 1 ]=rne(t,p 1 ) nd [st 2..ed 2 ]=rne(t,p 2 ). We n ompute [st..ed]=rne(t,p 1 P 2 ) in O(lo n) time. Proof: Let the lenth of P 1 e k. Note tht T SA[st1],T SA[st1+1],,T SA[ed1] re lexiorphilly inresin. Hene, T SA[st1]+k,T SA[st1+1]+k,,T SA[ed1]+k re lexiorphilly inresin. Thus, SA -1 [SA[st 1 ]+k] < SA -1 [SA[st 1 +1]+k] < < SA -1 [SA[ed 1 ]+k]. To find st nd ed, we need to find the smllest st suh tht st 2 < SA -1 [SA[st]+k]<ed 2 nd the lrest ed suh tht st 2 < SA -1 [SA[ed]+k] < ed 2. This n e done y inry serh.

108 Exmple T= P 1 =, P 2 =. SA[i] 7 rne(t,p 1 )=[2..4], rne(t,p 2 )=[5..6] 7 To find rne(t,p 1 P 2 ), we do the followin: Note tht T SA[2] <T SA[3] <T SA[4] re einnin with. So, T SA[2]+1 <T SA[3]+1 <T SA[4]+1. Note tht 6 SA -1 [SA[2]+1]=5, SA -1 [SA[3]+1]=6, SA -1 [SA[4]+1]=7. Hene, T SA[5] <T SA[6] <T SA[7]. Amon the three suffixes, we need to identify suffix einnin with P 2. Sine rne(t,p 2 )=[5..6], oth T SA[5] nd T SA[6] ontin P 2. As SA -1 [SA[2]+1]=5 nd SA -1 [SA[3]+1]=6, we hve rne(t,p 1 P 2 ) = [2..3]. i Suffix

109 Lemm 3 (Bkwrd serh) Assume [st..ed]=rne(t,p). We n ompute [st..ed ]=rne(t,p) in O(lo n) time. Proof: Let P 1 =, P 2 =P. By Lemm 2, rne(t,p 1 P 2 ) n e omputed in O(lo n) time.

110 Alorithm 1. For j=m,m-1,,1, By kwrd serh, find rne(t,p[j..m]). 2. For j=1,2,,m, By forwrd serh, find rne(t,p[1..j]). 3. Report ll ourrenes of rne(t,p[1..m]) 4. For j=1,2,,m, Let P 1 =P[1..j-1], P 2 =P[j+1..m] For every hrter P[j], By forwrd serh, find rne(t,p 1 ) By Lemm 2, find rne(t,p 1 P 2 ) Report ll ourrenes of rne(t,p 1 P 2 )

111 Time nlysis Inorin the O(o) reportin time! Steps 1 nd 2 tke O(m lo n) time. Step 4 tries ll possile mismthes. There re in totl Σ m mismthes. For eh mismth, it tke O(lo n) time. So, Step 4 tkes O( Σ m lo n) time. In totl, the runnin time is O( Σ m lo n + o). Note tht this solution n e enerlized to hndle k-mismth or k-differene.