Algorithm Theory. 13 Text Search - Knuth, Morris, Pratt, Boyer, Moore. Christian Schindelhauer

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Algorithm Theory 13 Text Search - Knuth, Morris, Pratt, Boyer, Moore Institut für Informatik Wintersemester 2007/08

Text Search Scenarios Static texts Literature databases Library systems Gene databases World Wide Web Dynamic texts Text editors Symbol manipulators 2

Text Search Data type string array of character file of character list of character Operations: (T, P of type string) length: length () i-th character : T [i] concatenation: cat (T, P) T.P 3

... Problem Definition Given: Text t 1 t 2... t n Σ n pattern p 1 p 2... p m Σ m Goal: Find one or all occurences of the pattern in the text, i.e. positions i (0 i n m) such that p 1 = t i+1 p 2 = t i+2 p m = t i+m 4

Problem Definition i i+1 i+m text: t 1 t 2... t i+1... t i+m.. t n pattern: p 1... p m Possible Running Time : 1. Worst Solution: O(n m) # possible alignments: n m + 1 # pattern positions: m 2. Best possible solution: At least one comparison per m consecutive text positions: Ω ( m + n/m ) 5

Naive Method For each possible position 0 i n m check at most m characters pairs. Whenever a mismatch occurs, shift to the next position textsearchbf := proc (T : : string, P : : string) # Input: text T, pattern P # Output: list L of positions i, at which P occurs in T n := length (T); m := length (P); L := []; for i from 0 to n-m do j := 1; while j m and T[i+j] = P[j] do j := j+1 od; if j = m+1 then L := [L [], i] fi; od; RETURN (L) end; 6

Naive Method Running time: 0 0... 0... 0... 0 0... i 0... 0... 0 1 Worst Case: Ω(m n) For real world examples, mismatches occur usually very often. Running time ~ c n 7

Knuth-Morris-Pratt Algorithm (KMP) Let t i and p j+1 be the characters to be compared: t 1 t 2...... t i...... = = = = p 1... p j p j+1... p m If, for a certain alignment, the first mismatch occurs for characters t i and p j+1 then the last j characters compared in T equal the first j characters of P t i p j+1 8

Knuth-Morris-Pratt Algorithm (KMP) Idea: Find j = next[j] < j such that t i can then be compared to p j +1 Find maximum j < j such that P 1...j = P j-j +1...j. Find the longest prefix of P which is a proper suffix of P 1... j. t 1 t 2...... t i...... = = = = p 1... p j p j+1... p m 9

Knuth-Morris-Pratt Algorithm (KMP) Example for determining next[j]: t 1 t 2... 01011 01011 0... 01011 01011 1 01011 01011 1 next[j] = length of the longest prefix of P which is a proper suffix of P 1...j 10

Knuth-Morris-Pratt Algorithm (KMP) for P = 0101101011, next = [0,0,1,2,0,1,2,3,4,5] : 1 2 3 4 5 6 7 8 9 10 0 1 0 1 1 0 1 0 1 1 0 0 1 0 0 1 0 1 0 0 1 0 1 0 1 0 1 1 11

Knuth-Morris-Pratt Algorithm (KMP) KMP := proc (T : : string, P : : string) # Input: text T, pattern P # Output: list L of positions i at which P occurs in T n := length (T); m := length(p); L := []; next := KMPnext(P); j := 0; for i from 1 to n do while j>0 and T[i] <> P[j+1] do j := next [j] od; if T[i] = P[j+1] then j := j+1 fi; if j = m then L := [L[], i-m] ; j := next [j] fi; od; RETURN (L); end; 12

Knuth-Morris-Pratt Algorithm (KMP) Pattern: abrakadabra, next = [0,0,0,1,0,1,0,1,2,3,4] a b r a k a d a b r a b r a b a b r a k... a b r a k a d a b r a next[11] = 4 a b r a k a d a b r a b r a b a b r a k... a b r a k a d a b r a next[4] = 1 13

Knuth-Morris-Pratt Algorithm (KMP) a b r a k a d a b r a b r a b a b r a k... - a b r a k next [4] = 1 a b r a k a d a b r a b r a b a b r a k... - a b r a k next[2] = 0 a b r a k a d a b r a b r a b a b r a k... a b r a k 14

Knuth-Morris-Pratt Algorithm (KMP) Correctness: t 1 t 2...... t i...... = = = = p 1... p j p j+1... p m When starting the for-loop: P 1...j = T i-j...i-1 und j m if j = 0: we are located at the first character of P P if j 0: P can be shifted while j>0 and t i p j+1 15

Knuth-Morris-Pratt Algorithm (KMP) If T[i] = P[j+1], j and i can be increased (at the end of the loop) If P has been compared completely (j = m) an occurrence of P in T has been found and we can shift to the next position. 16

Knuth-Morris-Pratt Algorithm (KMP) Running time: text pointer i will be never decreased text pointer i and pattern pointer j are always incremented together: next[j] < j; j can be decreased only as many times as it has been increased If the next-array is known, KMP runs in time O(n). 17

Computation of Array next next[i] = length of the longest prefix of P which is a proper suffix of P 1... i next[1] = 0 Let next[ i-1] = j: p 1 p 2...... p i...... = = = = p 1... p j p j+1... p m 18

Computation of Array next Consider the following two cases 1) p i = p j+1 next[i] = j + 1 2) p i p j+1 replace j by next[ j ] until p i = p j+1 or j = 0. if p i = p j+1 then next[i] = j + 1 otherwise next[i] = 0. 19

Computation of Array next KMPnext := proc (P : : string) #Input : pattern P #Output : next-array for P m := length (P); next := array (1..m); next [1] := 0; j := 0; for i from 2 to m do while j > 0 and P[i] <> P[j+1] do j := next [j] od; if P[i] = P[j+1] then j := j+1 fi; next [i] := j od; RETURN (next); end; 20

Run Time of KMP The KMP-algorithm runs in time O(n+m). Can text search be any faster? 21

The Boyer-Moore-Algorithm (BM) Idea: For any alignment of the pattern with the text, scan the characters from right to left rather than from left to right Example: he said abrakadabut but but but but but but but but but Large jumps and few comparisons Desired running time: O(m+n/m) 22

BM Last-Occurrence Function For c Σ and pattern P let δ (c) := index of the right-most occurrence of c in P = max {j p j = c} ={ 0 if c P j if c = p and c p, for j < k m How hard is it to compute all values of δ? Let Σ = l: Running time: O(m+l) 23

BM Last-Occurrence Function Let c = the character causing the mismatch j = the index of the current character in the pattern (c p j ) 24

BM Last-Occurrence Function Computation of the pattern shift Case 1 c does not occur in the pattern P (δ(c) = 0) Shift the pattern j characters to the right text pattern i + 1 i + j i + m c p j p m 25

BM Last-Occurrence Function Case 2 c occurs in the pattern. (δ(c) 0) Shift the pattern to the right until the rightmost c in the pattern is aligned with a potential c in the text. Text i + 1 i + j i + m c Pattern j - k c k p j c p m 26

BM Last-Occurrence Function Case 2 a: δ(c) > j text c c pattern p j c no c Shift the rightmost c in the pattern to a potential c in the text shift by 27

BM Last-Occurrence Function Case 2 b: δ(c) < j text c pattern c p j Shift the rightmost c in the pattern to c in the text shift by 28

Boyer-Moore Algorithm Version 1 Algorithm BM-search1 Input: text T and pattern P Output: all positions of P in T 1 n := length(t); m := length(p) 2 compute δ 3 i := 0 4 while i n m do 5 j := m 6 while j > 0 and P[j] = T[i + j] do 7 j := j 1 end while; 8 if j = 0 9 then output position i 10 i := i + 1 11 12 13 else if δ(t[i + j]) > j 14 end while; then i := i + m + 1 - δ[t[i + j]] else i := i + j - δ[t[i + j]] 29

Boyer-Moore Algorithm Version 1 Analysis: Desired running time : O(m + n/m) Worst-case running time: Ω(n m) 0 0... 0 0... 0... 0... i 1 0... 0... 0 30

Match Heuristic Use the information collected before a mismatches p j t i + j occurs. t 1 t 2... t i+1... t i+j... t i+m = = = i p 1... p j... p m gsf[j] = position of the end of the next occurrence of the suffix P j+1... m from the right that is not preceded by character P j (good suffix function) Possible shift: γ[j] = m gsf[j] 31

Example for Computing gsf gsf[j] = position of the end of the closest occurrence of the suffix P j + 1... m from the right that is not preceded by character P j Pattern: banana gsf[j] inspected suffix forbidden character Other matches Position gsf[5] a n banana 2 gsf[4] na a *** bana na 0 gsf[3] ana n banana 4 gsf[2] nana a banana 0 gsf[1] anana b banana 0 gsf[0] banana ε banana 0 32

Example for Computing gsf gsf (banana) = [0,0,0,4,0,2] a b a a b a b a n a n a n a n a = = = b a n a n a b a n a n a 33

Boyer-Moore Algorithm Version 2 Algorithm BM-search2 Input: text T and pattern P Output: shift for all occurrences of P in T 1 n := length(t); m := length(p) 2 compute δ and γ 3 i := 0 4 while i n m do 5 j := m 6 while j > 0 and P[j] = T[i + j] do 7 j := j 1 end while; 8 if j = 0 9 then output position i 10 i := i + γ[0] 11 else i := i + max(γ[j], j - δ[t[i + j]]) 12 end while; 34

Algorithm Theory 13 Text Search - Knuth, Morris, Pratt, Boyer, Moore Institut für Informatik Wintersemester 2007/08