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1 Suffix Arrays: A new method for on-line string searches Udi Manber Gene Myers May 1989 Presented by: Oren Weimann

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2 Introduction - Problem definition “Is W a substring of A?” |A|=N and |W|=P A = a 0 a 1 …a N-1 A i = suffix beginning at index i = a i a i+1 …a N-1 A= abccbbadgfbbcahgjf W= badgfbb A= abccbbadgfbbcahgjf

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3 Introduction – what is a suffix array? Example: Pos Pos[2] = 6 (A 6 = in) A = assassin 01234567

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4 Introduction – what is a suffix array? A lexicographically sorted array- Pos[N], of all the suffixes of A: Pos[k] = i A i is the kth smallest suffix in the set {A 0, A 1, A 2…… A N-1 }

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5 Introduction – what is a suffix tree? Example: A trie that contains all suffixes of A: s a 4 3 s s s s a i n 0 i n 6 i n A = assassin 01234567 s i n a s s i n 2 i n 5 1 a s s i n

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6 The Article Overview 1. A search algorithm In O(P+logN) (assuming we already computed Pos[ ] and the longest common prefix (lcp) information). 2. How to construct Pos[ ] in O(NlogN) time and O(N) space. (assuming lcp info is known) 3. An Algorithm for computing the lcp information in O(NlogN). 4. Algorithms for Expected-time improvement.

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7 The Search algorithm - Definitions For any string u, u p = u 1 u 2 u 3……. u p (or u if |u| p) Let “ “ denote a Lexicographical order, We say u v u p v p Note that for any choice of p: Note that W is a substring of A there is an i such that W

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8 The Search algorithm – how does the array help us know if W is a substring of A? We define a search interval: L W = min {k | W A Pos[k] or k = N} R W = max {k | W A Pos[k] or k = -1} W matches a i a i+1...a i+P-1 i=Pos[k] for some k [L W, R W ]

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9 Example: Pos A = assassin 01234567 Option 1 Option 2 Option 3

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10 Why finding L W, R W == Finding the matches: If L W > R W => W is not a substring of A. Else: there are (R W -L W +1) matches - A Pos[L W ],…, A Pos[R W ] W>A Pos[k] W
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11 The Search algorithm – The easy way - O(PlogN) L MR abcde...abcdf...abd... Pos Log(N) iterations, each iteration sets new L,R bonds (initially L=0, R=N-1) according to a comparison of W with A Pos[M], where M=(L+R)/2. In the end L W R W=“abcx”

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12 The Search algorithm using lcp values in O(P+logN) – Definitions: Speedup using precomputed lcp Values, for now We assume lcp is known. Each iteration We define: – l = lcp(A Pos[L], W) – r=lcp(W, A Pos[R] ) – Llcp[M] = lcp(A Pos[L] A Pos[M] ) – Rlcp[M] = lcp(A Pos[M], A Pos[R] )

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13 The Search algorithm using lcp values in O(P+logN) Example: A=“abcx” l = 3 Llcp[M]=4 Rlcp[M]=2 L MR abcde...abcdf...abd... Pos r = 2 Note that Llcp[M] is well defined because every midpoint M has one L M and one R M

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14 So how do we use l,r,Llcp[M] ? Example: W=abcx abcde...abc... abcdf…abd… l=3r=2 Case 1: Llcp[M] > l (Llcp[M]=4 and l=3 ) W>A Pos[L] W>A Pos[M] Go right l is unchanged = 3 LM R Llcp[M]=4

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15 Example: W=abcx (cont.) Case 2: Llcp[M] < l (Llcp[M]=2 and l=3 ) A Pos[L]
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"name": "15 Example: W=abcx (cont.) Case 2: Llcp[M] < l (Llcp[M]=2 and l=3 ) A Pos[L]

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16 Example: W=abcx (cont.) abcde...abc... abcp…abd… l=3 r=2 Case 3: Llcp[M] = l (Llcp[M]=3 and l=3 ) Compare W l and A Pos[M] l until W l+j A Pos[M] l+j Go right or left according to W l+j, A Pos[M] l+j new l or r = (l+j) Number of comparisons = j+1 LMR Llcp[M]=3

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17 The Search algorithm using lcp values- complexity In each iteration there are maximum j+1 comparisons, when in total Total comparisons (P + #Iterations) O(P+logN) running time Requires only 3N-sized arrays

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18 The Article Overview 1. A search algorithm In O(P+logN) (assuming we already computed Pos[ ] and the longest common prefix (lcp) information). 2. How to construct Pos[ ] in O(NlogN) time and O(N) space. (assuming lcp info is known) 3. An Algorithm for computing the lcp information in O(NlogN). 4. Algorithms for Expected-time improvement.

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19 Construction of suffix array in O(NlogN) Sorting the suffixes in a unique Radix sort – We Will have O(logN) stages (numbered 1,2,4,8,16…) In stage H the suffixes are sorted in buckets called H Buckets, according to the first H characters. (next stage is 2H– thus, in stage H the suffixes are sorted by )

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20 Construction of suffix array – The general idea If A i, A j H-bucket we Sort them by the Next H symbols, but: Their next H symbols = first H symbols of A i+H and A j+H which are already sorted in phase H. abef…abcd…ab…bb...bb…cd… ef… H=2 : AiAi AjAj A j+H A i+H first bucketfourth bucketthird bucketsecond bucket

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21 Construction of suffix array – The general idea (cont.) Let A i be in first H-bucket after stage H A i starts with smallest H-symbol string A i-H should be first in its H-bucket abef…abcd…ab…bb...bb…cdef…cdab…ef… AiAi A i-H H=2 :

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22 Construction of suffix array – The algorithm Go over the suffix array: For each A i : Move A i-H to next available place in its H-bucket The suffixes are now sorted according to -order Go over the array again, and decide which suffix opens a new 2H-bucket, use lcs knowledge (described later)

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23 Construction of suffix array – The algorithm Example: A = assassin 01234567 assinassassininnsinssinsassinssassin H=1 A3A3 A2A2 A i sets A i-1

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24 Construction of suffix array – The algorithm Example: assinassassininnsassinssinsinssassin H=1 A0A0 A = assassin 01234567 A i sets A i-1

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25 Construction of suffix array – The algorithm Example: assinassassininnsassinssinsinssassin H=1 A6A6 A = assassin 01234567 A5A5 A i sets A i-1

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26 Construction of suffix array – The algorithm Example: assinassassininnsassinsinssinssassin H=1 A7A7 A = assassin 01234567 A6A6 A i sets A i-1

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27 Construction of suffix array – The algorithm Example: assinassassininnsassinsinssinssassin H=1 A2A2 A1A1 A = assassin 01234567 A i sets A i-1

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28 Construction of suffix array – The algorithm Example: assinassassininnsassinsinssassinssin H=1 A4A4 A = assassin 01234567 A5A5 A i sets A i-1

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29 Construction of suffix array – The algorithm Example: assinassassininnsassinsinssassinssin H=1 A = assassin 01234567 A1A1 A0A0 A i sets A i-1

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30 Construction of suffix array – The algorithm Example: assassinassininnsassinsinssassinssin H=1 A = assassin 01234567 A4A4 A3A3 A i sets A i-1

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31 Construction of suffix array – The algorithm Example: assassinassininnsassinsinssassinssin H=1 A = assassin 01234567 Go over array to get new 2-buckets lcs(sassin,sin)= 1+ lcs(assin,in)= 1+0=1 so “sin” opens a new 2-bucket back A i sets A i-1

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32 Construction of suffix array – The algorithm Example: assassinassininnsassinsinssassinssin H=2 A = assassin 01234567 A0A0 A i sets A i-2

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33 Construction of suffix array – The algorithm Example: assassinassininnsassinsinssassinssin H=2 A = assassin 01234567 A3A3 A1A1 A i sets A i-2

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34 Construction of suffix array – The algorithm Example: assassinassininnsassinsinssassinssin H=2 A = assassin 01234567 A6A6 A4A4 A i sets A i-2

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35 Construction of suffix array – The algorithm Example: assassinassininnsassinsinssassinssin H=2 A = assassin 01234567 A7A7 A5A5 A i sets A i-2

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36 Construction of suffix array – The algorithm Example: assassinassininnsassinsinssassinssin H=2 A = assassin 01234567 A2A2 A0A0 A i sets A i-2

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37 Construction of suffix array – The algorithm Example: assassinassininnsassinsinssassinssin H=2 A = assassin 01234567 A5A5 A3A3 A i sets A i-2

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38 Construction of suffix array – The algorithm Example: assassinassininnsassinsinssassinssin H=2 A = assassin 01234567 A1A1 A i sets A i-2

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39 Construction of suffix array – The algorithm Example: assassinassininnsassinsinssassinssin H=2 A = assassin 01234567 A4A4 A2A2 A i sets A i-2

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40 Construction of suffix array – The algorithm Example: assassinassininnsassinsinssassinssin H=2 A = assassin 01234567 Go over array to get new 4-buckets A i sets A i-2

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41 Construction of suffix array – The algorithm Example: assassinassininnsassinsinssassinssin H=4 A = assassin 01234567 That’s it, we are sorted!

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42 Construction of suffix array – Complexity Summary Sorting by first char – O(N) O(logN) stages of O(N) operations = O(NlogN) Total - time: O(NlogN) - space: 2 integer arrays of size N back

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43 The Article Overview 1. A search algorithm In O(P+logN) (assuming we already computed Pos[ ] and the longest common prefix (lcp) information). 2. How to construct Pos[ ] in O(NlogN) time and O(N) space. 3. An Algorithm for computing the lcp information in O(NlogN). 4. Algorithms for Expected-time improvement.

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44 How to find Longest Common Prefixes – the general idea We don’t care what is the lcp between suffixes in the same H-bucket. For A p, A q in the same H-bucket but different 2H-buckets: – H lcp(A p, A q ) < 2H – lcp(A p, A q ) = H + lcp(A p+H, A q+H ) – lcp(A p+H, A q+H ) < H that is why A p+H, A q+H Are in different H-buckets, but which ones?

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45 How to find Longest Common Prefixes – the general idea If A p+H and A q+H were in adjacent H-buckets then lcp is known. how?how? If not, Then: lcp(A Pos[i], A Pos[j] ) = {lcp(A Pos[k],A Pos[k+1] )}

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46 How to find Longest Common Prefixes – the general idea lcp(A p+H, A q+H ) = min{1,1,2} = 1 assassinassininnsassinsinssassinssin A q+h A p+h 1 1 2 Notice that if 2 neighbors are in the same H-bucket we can consider there lcp to be H, since lcp(A p+H, A q+H ) < H H=2

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47 How to find lcp – algorithm and data structures – Hgt[] During the construction stage, we build an array Called Hgt[N]: Hgt(i)=lcp(A Pos[i-1], A Pos[i] ), initialized so that Hgt[i]=N+1 for every i. In stage H=1: Hgt(i)=0 for A Pos[i] that are first in their buckets. In stage 2H: we update every Hgt(i) that A Pos[i] is the first in a newly created 2H bucket

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48 How to find lcp – Hgt[] example: H=1 assinassassininnsinssinsassinssassin 000999911 assinassassininnsinssinsassinssassin 00099 H=2 lcp(ssin,sin)=1+lcp(sin,in)=1+min{lcp(in,n),lcp(sin, n)}=1

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49 How to find lcp – Hgt[] example (cont.) 23 assinassassininnsin ssin sassinssassin H=4 00011 lcp( assassin,assin)=2+lcp(sin, sassin)=2+1=3 lcp(ssin, ssassin)=2+lcp(in, assin)=2+0=2

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50 How to find lcp – data structures We need a data structure that will contain lcp(A Pos[j], A Pos[i] ) between any i and j (not just i and i+1 which Hgt[] supplies) Hgt[] will become the leaves of a binary balanced tree called the Interval tree.

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51 How to find lcp – example of Interval tree (2,3)(3,4)(4,5)(5,6)(6,7)(1,2)(0,1) 0 9000 0 0 9 0 99 9 9 11 1 1 3 2

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52 How to find lcp – Complexity Each time a leaf opens a new bucket we change Hgt[i] for that leaf. That change requires O(logN) changes in the interval tree There are O(N) leaves opening new bucket In total we get O(NlogN) to get all lcp values

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53 The Article Overview 1. A search algorithm In O(P+logN) (assuming we already computed Pos[ ] and the longest common prefix (lcp) information). 2. How to construct Pos[ ] in O(NlogN) time and O(N) space. 3. An Algorithm for computing the lcp information in O(NlogN). 4. Algorithms for Expected-time improvement.

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54 Time Expected-case Improvement of the construction of pos[] Assumptions: - All N-symbol strings are equally likely. – Under this assumption: Expected length of longest repeated substring = O(log | | N) This immediately implies that construction of pos[] is reduced to O(NLogLogN). why?why? Next is a way to reduce it to O(N).

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55 Time Expected-case Improvement of the construction of pos[] Let T = We encode each possible T length string to an integer with the isomorphism Int T (u) Map each A P to Int T (A P ) [0,| | T -1] : – Int T (A P ) = a p | | T-1 +

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56 Example of the mapping Int T (A P ) = a p | | T-1 + 2*4^0 + 02 | |= 4, a=0, i=1, n=2, s=3 N=8 T= =1 1*4^0 + 01 3*4^0 + 03 3 0*4^0 + 00 3*4^0 + 03 3 0*4^0 + 00

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57 Time Expected-case Improvement of the construction of pos[] By the definition of Int T (A P ) it takes O(N) to compute all Int T (A P ) values of all suffixes. So now instead of starting with H=1 we start with H= But since the longest repeated substring length is O(log | | N) we will have O(1) stages of the radix sort. Thus, the total time for constructing pos[] = O(N)

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58 So is a suffix array better then a suffix tree? Suffix arraySuffix tree Construction time O(NlogN) - for small | | O(N) – needs additional space O(N) Time Complexity O(P+logN) – good for large alphabets O(Plog| |) Space Complexity requires 2N integers – this is the main advantage. O(N) dependent on | | ? NoYes

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Suffix trees and suffix arrays presentation by Haim Kaplan.

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