Monday, April 29, 13. Tandem repeats
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- Christal Goodwin
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1 Tandem repeats
2 Tandem repeats A string w is a tandem repeat (or square) if: A tandem repeat αα is primitive if and only if α is primitive A string w = α k for k = 3,4,... is (by some) called a tandem array Note: tandem repeat (and tandem array ) is a string property, i.e. a string can be a tandem repeat... How difficult is it to decide if S[1..n] is a tandem repeat?
3 Periodic and primitive strings Any string S[1..n] can be written in its normal form: 1 ß n S = ( ) 3 Let p = n - ß, then S = w n/p = w n/p w, where w = S[1.. n- n/p p] The normal form of S[1..n] is w n/p*, where p* is the minimum period If a string S[1..n] is not periodic, i.e. p*=n, then it is primitive
4 Tandem repeats A string w is a tandem repeat (or square) if: A tandem repeat αα is primitive if and only if α is primitive A string w = α k for k = 3,4,... is (by some) called a tandem array Note: tandem repeat (and tandem array ) is a string property, i.e. a string can be a tandem repeat... Find minimum period p* in time O(n), if n/p* is 2,4,6..., then yes How difficult is it to decide if S[1..n] is a tandem repeat?
5 Occurrences of tandem repeats An occurrence of a tandem repeat in S: Tandem repeat αα is in S, if it occurs one or more times in S Note that an occurrence can be encoded in space O(1) as (i, α, 2)
6 Occurrences of tandem repeats An occurrence of a tandem repeat in S: Tandem repeat αα is in S, if it occurs one or more times in S Note that an occurrence can be encoded in space O(1) as (i, α, 2) Computational problems: (1) Find all occurrences of tandem repeats in S[1..n] (2) Find all (primitive) tandem repeats in S[1..n]
7 A simple example The strings aaaaaa contains: a a a a a a 5 occurrences of aa 3 occurrences of aaaa 1 occurrence of aaaaaa In general the a n contains n 2 /4 occurrences of n/2 tandem repeats, but there is only one primitive tandem repeat aa...
8 A simple example The strings aaaaaa contains: a a a a a a 5 occurrences of aa 3 occurrences of aaaa 1 occurrence of aaaaaa In general the a n contains n 2 /4 occurrences of n/2 tandem repeats, but there is only one primitive tandem repeat aa or equivalently one maximal repetition (1, 1, n), which we also call a primitive tandem array... Observation: there can be no more than O(n 2 ) occurrences of tandem repeats in S[1..n], but how many are e.g primitive?
9 Fibonacci strings Lemma 3.4.1: The Fibonacci string f n is defined recursively as: f 0 = b, f 1 = a, and f n = f n-1 f n-2 f 0 = b f 1 = a f 2 = ab f 3 = aba f 4 = abaab f 5 = abaababa The length of f n is the nth Fibonacci number f n, i.e. f 0 =1, f 1 =1, and f n = f n-1 +f n-2... Fibonacci strings are highly repetitive...
10 Fibonacci strings Lemma 3.4.1: The Fibonacci string f n is defined recursively as: f 0 = b, f 1 = a, and f n = f n-1 f n-2 f 0 = b f 1 = a f 2 = ab f 3 = aba f 4 = abaab The length of f n is the nth Fibonacci number f n, i.e. f 0 =1, f 1 =1, and f n = f n-1 +f n-2... Fibonacci strings are highly repetitive... f 5 = abaababa Theorem 3.4.8: A Fibonacci string of length n contains Ω(n log n) occurrences of primitive tandem repeats actually f n contains f n log f n + O(f n ) occurrences of primitive tandem repeats [Fraenkel & Simpson 1999]...
11 Fibonacci strings Lemma 3.4.1: The Fibonacci string f n is defined recursively as: f 0 = b, f 1 = a, and f n = f n-1 f n-2 f 0 = b... f 1 since = a Crochemore s The algorithm length of f (sect. n is the 12.1) nth finds Fibonacci all occurrences number f n, of f primitive 2 = ab tandem i.e. repeats f 0 =1, fin 1 =1, S[1..n] and fin n = time f n-1 +f O(n n-2... log n), no string of f length 3 = aba n contains more than O(n log n) occurrences of primitive tandem repeats... Fibonacci strings are highly repetitive... f 4 = abaab f 5 = abaababa Theorem 3.4.8: A Fibonacci string of length n contains Ω(n log n) occurrences of primitive tandem repeats actually f n contains f n log f n + O(f n ) occurrences of primitive tandem repeats [Fraenkel & Simpson 1999]...
12 Algorithms Crochemore 1981 (Smyth ): Find all occurrences of primitive tandem repeats in time O(n log n) and space O(n).
13 Algorithms Crochemore 1981 (Smyth ): Find all occurrences of primitive tandem repeats in time O(n log n) and space O(n). Apostolico and Preparata 1983: Find all occurrences of primitive tandem repeats in time O(n log n) and space O(n). Suffix tree based.
14 Algorithms Crochemore 1981 (Smyth ): Find all occurrences of primitive tandem repeats in time O(n log n) and space O(n). Apostolico and Preparata 1983: Find all occurrences of primitive tandem repeats in time O(n log n) and space O(n). Suffix tree based. Main and Lorentz 1984 (Smyth ) : Find all occurrences of tandem repeats in time O(n log n + output ) and space O(n).
15 Algorithms Crochemore 1981 (Smyth ): Find all occurrences of primitive tandem repeats in time O(n log n) and space O(n). Apostolico and Preparata 1983: Find all occurrences of primitive tandem repeats in time O(n log n) and space O(n). Suffix tree based. Main and Lorentz 1984 (Smyth ) : Find all occurrences of tandem repeats in time O(n log n + output ) and space O(n). Stoye and Gusfield 1998: Find all occurrences of tandem repeats in time O(n log n + output ) and space O(n), can also be adapted to find all occurrences of primitive tandem repeats in time O(n log n). Suffix tree based.
16 Algorithms Crochemore 1981 (Smyth ): Find all occurrences of primitive tandem repeats in time O(n log n) and space O(n). Apostolico and Preparata 1983: Find all occurrences of primitive tandem repeats in time O(n log n) and space O(n). Suffix tree based. Main and Lorentz 1984 (Smyth ) : Find all occurrences of tandem repeats in time O(n log n + output ) and space O(n). Stoye and Gusfield 1998: Find all occurrences of tandem repeats in time O(n log n + output ) and space O(n), can also be adapted to find all occurrences of primitive tandem repeats in time O(n log n). Suffix tree based.
17 Basic observation A branching occurrence of a tandem repeats in S: Lemma: any non-branching occurrence (i, l, 2) of a tandem repeat is the left-rotation of another tandem repeat (i+1, l, 2), starting one position to its right. Example:
18 Suffix trees and tandem repeats Lemma (folklore): Consider two positions i and j of S, 1 i < j n, let l = j - i. Then the following assertions are equivalent: a. (i, l, 2) is an occurrence of a tandem repeat b. i and j occur in the same leaf-list of some node v in T(S) with depth D(v) l
19 Suffix trees and tandem repeats Lemma (folklore): Consider two positions i and j of S, 1 i < j n, let l = j - i. Then the following assertions are equivalent: a. (i, l, 2) is an occurrence of a tandem repeat b. i and j occur in the same leaf-list of some node v in T(S) with depth D(v) l
20 Suffix trees and tandem repeats Lemma (folklore): Consider two positions i and j of S, 1 i < j n, let l = j - i. Then the following assertions are equivalent: a. (i, l, 2) is an occurrence of a tandem repeat b. i and j occur in the same leaf-list of some node v in T(S) with depth D(v) l
21 Suffix trees and tandem repeats Lemma (folklore): Consider two positions i and j of S, 1 i < j n, let l = j - i. Then the following assertions are equivalent: a. (i, l, 2) is an occurrence of a tandem repeat b. i and j occur in the same leaf-list of some node v in T(S) with depth D(v) l
22 Suffix trees and tandem repeats Lemma (folklore): Consider two positions i and j of S, 1 i < j n, let l = j - i. Then the following assertions are equivalent: a. (i, l, 2) is an occurrence of a tandem repeat b. i and j occur in the same leaf-list of some node v in T(S) with depth D(v) l
23 ... and branching tandem repeats Lemma: Consider two positions i and j of S, 1 i < j n, let l = j - i. Then the following assertions are equivalent: a. (i, l, 2) is an occurrence of a branching tandem repeat b. i and j occur in the same leaf-list of a node v in T(S) with depth D(v) = l and not in the same leaf-list of any node with depth > l
24 ... and branching tandem repeats Lemma: Consider two positions i and j of S, 1 i < j n, let l = j - i. Then the following assertions are equivalent: a. (i, l, 2) is an occurrence of a branching tandem repeat b. i and j occur in the same leaf-list of a node v in T(S) with depth D(v) = l and not in the same leaf-list of any node with depth > l
25 ... and branching tandem repeats Lemma: Consider two positions i and j of S, 1 i < j n, let l = j - i. Then the following assertions are equivalent: a. (i, l, 2) is an occurrence of a branching tandem repeat b. i and j occur in the same leaf-list of a node v in T(S) with depth D(v) = l and not in the same leaf-list of any node with depth > l
26 ... and branching tandem repeats Lemma: Consider two positions i and j of S, 1 i < j n, let l = j - i. Then the following assertions are equivalent: a. (i, l, 2) is an occurrence of a branching tandem repeat b. i and j occur in the same leaf-list of a node v in T(S) with depth D(v) = l and not in the same leaf-list of any node with depth > l
27 Basic algorithm Idea: for each node v of T(S), find the positions i in LL(v) where αα = L(v)L(v) occurs as a branching tandem repeat (i, D(v), 2)... Algorithm:
28 Basic algorithm Idea: for each node v of T(S), find the positions i in LL(v) where αα = L(v)L(v) occurs as a branching tandem repeat (i, D(v), 2)... Algorithm: i and j are in leaf-lists of distinct children of v, i.e they are not in the same leaf-list for any node with depth > D(v)...
29 Basic algorithm Idea: for each node v of T(S), find the positions i in LL(v) where αα = L(v)L(v) occurs as a branching tandem repeat (i, D(v), 2)... Algorithm: Analysis: O(n 2 ) time, and O(n) space, if we can test in time O(1)...
30 Testing in constant time Idea: make a depth-first numbering of the leaves in T(S) and a corresponding lookup-table...
31 Testing in constant time Idea: make a depth-first numbering of the leaves in T(S) and a corresponding lookup-table...
32 Testing in constant time Idea: make a depth-first numbering of the leaves in T(S) and a corresponding lookup-table...
33 Testing in constant time Idea: make a depth-first numbering of the leaves in T(S) and a corresponding lookup-table...
34 Testing in constant time Idea: make a depth-first numbering of the leaves in T(S) and a corresponding lookup-table...
35 Testing in constant time Idea: make a depth-first numbering of the leaves in T(S) and a corresponding lookup-table...
36 Testing in constant time Idea: make a depth-first numbering of the leaves in T(S) and a corresponding lookup-table...
37 Speeding up the basic algorithm We cannot afford to spend time O( LL(v) ) at each node,...
38 Speeding up the basic algorithm We cannot afford to spend time O( LL(v) ) at each node,... Recall: Lemma: Consider two positions i and j of S, 1 i < j n, let l = j - i. Then the following assertions are equivalent: a. (i, l, 2) is an occurrence of a branching tandem repeat b. i and j occur in the same leaf-list of a node v in T(S) with depth D(v) = l and not in the same leaf-list of any node with depth > l
39 Speeding up the basic algorithm We cannot afford to spend time O( LL(v) ) at each node,... Recall: Lemma: Consider two positions i and j of S, 1 i < j n, let l = j - i. Then the following assertions are equivalent: a. (i, l, 2) is an occurrence of a branching tandem repeat b. i and j occur in the same leaf-list of a node v in T(S) with depth D(v) = l and not in the same leaf-list of any node with depth > l... i and j must be in leaf-lists of different children of v...
40 Speeding up the basic algorithm v Large v Small For node v let v be the child of v with the largest leaf-list... Large(v) = LL(v ) Small(v) = LL (v) = LL(v) - LL(v ) Observation: There is no branching occurrence (i, D(v), 2) where i and j = i+d(v) are both in Large(v), i.e. either i or i+d(v) are in Small(v)...
41 Speeding up the basic algorithm v Large v Small For node v let v be the child of v with the largest leaf-list... Large(v) = LL(v ) Small(v) = LL (v) = LL(v) - LL(v ) Observation: There is no branching occurrence (i, D(v), 2) where i and j = i+d(v) are both in Large(v), i.e. either i or i+d(v) are in Small(v)... Idea: For every i in Small(v) report: (i, D(v), 2) as branching iff i+d(v) LL(v) and S[i] S[i+2D(v)] (i-d(v), D(v), 2) as branching iff i-d(v) LL(v ) and S[i-D(v)] S[i+D(v)]
42 Speeding up the basic algorithm v Large v Small For node v let v be the child of v with the largest leaf-list... Large(v) = LL(v ) Small(v) = LL (v) = LL(v) - LL(v ) Observation: There is no branching occurrence (i, D(v), 2) where i and j = i+d(v) are both in Large(v), i.e. either i or i+d(v) are in Small(v)... This finds all branching occurrences Idea: For every i in Small(v) report: of L(v)L(v) at positions in Small(v) (i, D(v), 2) as branching iff i+d(v) LL(v) and S[i] S[i+2D(v)] (i-d(v), D(v), 2) as branching iff i-d(v) LL(v ) and S[i-D(v)] S[i+D(v)]
43 Speeding up the basic algorithm v Large v Small For node v let v be the child of v with the largest leaf-list... Large(v) = LL(v ) Small(v) = LL (v) = LL(v) - LL(v ) Observation: There is no branching occurrence (i, D(v), 2) where i and j = i+d(v) are both in Large(v), i.e. either i or i+d(v) are in Small(v)... Idea: For every i in Small(v) report: This finds all branching occurrences (i, D(v), 2) as branching iff i+d(v) LL(v) of L(v)L(v) and S[i] at positions S[i+2D(v)] in Large(v) (i-d(v), D(v), 2) as branching iff i-d(v) LL(v ) and S[i-D(v)] S[i+D(v)]
44 Optimized basic algorithm Idea: for each node v of T(S), find the positions i in LL(v) where (i, D(v), 2) is a branching occurrence by examining Small(v)...
45 Optimized basic algorithm Idea: for each node v of T(S), find the positions i in LL(v) where (i, D(v), 2) is a branching occurrence by examining Small(v)...
46 Optimized basic algorithm Idea: for each node v of T(S), find the positions i in LL(v) where (i, D(v), 2) is a branching occurrence by examining Small(v)... i.e test if i is in LL(v ), i.e. Large(v), and if S[i] S[i+2D(v)]...
47 Optimized basic algorithm Idea: for each node v of T(S), find the positions i in LL(v) where (i, D(v), 2) is a branching occurrence by examining Small(v)... Running time?
48 Optimized basic algorithm Idea: for each node v of T(S), find the positions i in LL(v) where (i, D(v), 2) is a branching occurrence by examining Small(v)... Analysis: O( Small(v) ) time at each node, and O(n) space
49 Smaller half trick v Large v Small For node v let v be the child of v with the largest leaf-list... Large(v) = LL(v ) Small(v) = LL (v) = LL(v) - LL(v ) Smaller half trick: Small(v) = O(nlog n) v Using time O( Small(v) ) at each node, implies time O(n log n) in total
50 v v Smaller half trick Proof: count how many times each leaf can be in Small(v) for any v... Large Small Smaller half trick: Small(v) = O(nlog n) v Using time O( Small(v) ) at each node, implies time O(n log n) in total
51 Smaller half trick v v u Proof: count how many times each leaf can be in Small(v) for any v... l Small If leaf l is in Small(v), then T(u) T(v) /2, otherwise T(v ) wouldn t be large Large I.e. leaf l can be in a Small(v) at most log(n) times a long the path from l to the root. Smaller half trick: Small(v) = O(nlog n) v Using time O( Small(v) ) at each node, implies time O(n log n) in total
52 Optimized basic algorithm Idea: for each node v of T(S), determine if αα = L(v)L(v) is a branching tandem repeat by examining Small(v)... Analysis: O(n log n) time, and O(n) space
53 Putting it all together Algorithm: Start at each occurrence of a branching tandem repeats, and do a series of consecutive left-rotations to find all occurrences of tandem repeats... Example: Analysis: O(n log n + output ) time, and O(n) space
54 Putting it all together Algorithm: Start at each occurrence of a branching tandem repeats, and do a series of consecutive left-rotations to find all occurrences of tandem repeats... Example: Analysis: O(n log n + output ) time, and O(n) space The algorithm can be extended to find all occurrences of primitive tandem repeats in time O(n log n)...
55 Implementation details Make depth first -numbering of leaves, lookup-table, and annotation of internal with interval in an initial depth-first traversal of T(S)... Report (or count) occurrences of tandem repeats in a depthfirst traversal of T(S), report from v when all children have been reported from... Note that the annotation of nodes with intervals makes it easy to determine LL(v) and v... Keep track of leaf-lists in e.g. linked lists which can be concatenated in time O(1)... or do you need explicitly to keep track of the leaf-lists at all? (Hint: You don t)
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