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. Week 10: test contains all material from Week 1– Week 8, not including this week. Week 9: String Matching Allowing Errors.
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. • Week 10: test contains all material from Week 1– Week 8, not including this week
Week 9: String Matching Allowing Errors • Problem: Given a short pattern P of length m, a long text T of length n, and a maximum allowed number of errors k, find all the text positions where the pattern occurs with at most k errors.
Dynamic Programming • C[i,j] be the number of errors allowed. • Three kinds of error: mismatch (a, b), insertion( a, )and deletion ( , a).
The matrix The dynamic programming algorithm search ‘survey’ in the text ‘surgery’ with two errors. Bold entries indicate matching positions. Running time O(nm).
String Matching Allowing Errors • O(nk) algorithm, where n is the length of the pattern and k the number of errors allowed. • C[i,q] the number of errors for c[i,i+q], • q=-k, -(k+1), …,0, 1, 2, …, k. • C[i,i+q]= C[i-1, i+q-1] if pi=ti+q. • else C[I,i+q]=1+min{C[i-1,i+q], C[I, i+q-1], C[i-1,i+q-1]}.
An NFA for approximate string matching of the pattern ‘survey’ with two errors. The shaded states are those active after reading the text ‘surgery’. Unlabelled transitions match any character.(Also O(nk) time.)
Regular expressions Matching • Regular expression: • Any letter x in {},is a regular expression, where is the set of all letters. • if A and B are regular expression, then A|B, A.B and (A)* are regular expressions.
Regular expressions Matching • Given an regular expression E and a string T, find all the substrings in T that match E. • Let d(i) be the set of all states in the automaton that can be reached after T1T2…Ti is accepted. • Given d(i), d(i+1) can be computed easily. • There is a starting and final state in the automaton. • Whenever the final state is reach, we find a substring in T that match the expression.
Example: • E=(A|AA).(B|AB). • T=ABBAAAB. • D(1)={a,b,d,c, e, f, g, i} • D(2)={a,b,d, h,l }, • D(3)={a,b,d}. • D(4)={a,b,d,c,e,f,g,i} • D(5)={a,b,d,c,e,f,g, i,j} • D(6)={a,b,d,c,e,f,g,i,j} • D(7)={a, h, k,l}.
Running time • O(n2), where n is the size of the automaton since d(s, i) could contain O(n) states.
Choices of Search Engines • Many search engines exist to compete for users • The results are not necessarily the same • Different users prefer different search engines • Search results may, in the future, be biased towards paid advertisements.
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MetaSearch Engine • Metasearch Engines are designed to increase the coverage of web by forwarding users’ queries to multiple search engines • Users’ requests are sent to multiple search engines such as AlltheWeb, Google, MSN. • Then the results from the individual search engine are combined into a single result set to present to users.
Longest common subsequence • Definition 1: Given a sequence X=x1x2...xm, another sequence Z=z1z2...zk is a subsequence of X if there exists a strictly increasing sequence i1i2...ik of indices of X such that for all j=1,2,...k, we have xij=zj. • Example 1: If X=abcdefg, Z=abdg is a subsequence of X. X=abcdefg,Z=ab d g
Definition 2: Given two sequences X and Y, a sequence Z is a common subsequence of X and Y if Z is a subsequence of both X and Y. • Example 2: X=abcdefg and Y=aaadgfd. Z=adf is a common subsequence of X and Y. X=abc defg Y=aaaadgfd Z=a d f
Definition 3: A longest common subsequence of X and Y is a common subsequence of X and Y with the longest length. (The length of a sequence is the number of letters in the seuqence.) • Longest common subsequence may not be unique. • Example: abcd acbd Both acd and abd are LCS.
Longest common subsequence problem • Input: Two sequences X=x1x2...xm, and Y=y1y2...yn. • Output:a longest common subsequence of X and Y. • A brute-force approach Suppose that mn. Try all subsequence of X (There are 2m subsequence of X), test if such a subsequence is also a subsequence of Y, and select the one with the longest length.
Charactering a longest common subsequence • Theorem (Optimal substructure of an LCS) • Let X=x1x2...xm, and Y=y1y2...yn be two sequences, and • Z=z1z2...zk be any LCS of X and Y. • 1. If xm=yn, then zk=xm=yn and Z[1..k-1] is an LCS of X[1..m-1] and Y[1..n-1]. • 2. If xmyn, then zkxm implies that Z is an LCS of X[1..m-1] and Y. • 2. If xmyn, then zkyn implies that Z is an LCS of X and Y[1..n-1].
The recursive equation • Let c[i,j] be the length of an LCS of X[1...i] and X[1...j]. • c[i,j] can be computed as follows: 0 if i=0 or j=0, c[i,j]= c[i-1,j-1]+1 if i,j>0 and xi=yj, max{c[i,j-1],c[i-1,j]} if i,j>0 and xiyj. Computing the length of an LCS • There are nm c[i,j]’s. So we can compute them in a specific order.
The algorithm to compute an LCS • 1. for i=1 to m do • 2. c[i,0]=0; • 3. for j=0 to n do • 4. c[0,j]=0; • 5. for i=1 to m do • 6. for j=1 to n do • 7. { • 8. if x[I] ==y[j] then • 9. c[i,j]=c[i-1,j-1]=1; • 10 b[i,j]=1; • 11. elseif c[i-1,j]>=c[i,j-1] then • 12. c[i,j]=c[i-1,j] • 13. b[i,j]=2; • 14. else c[i,j]=c[i,j-1] • 15. b[i,j]=3; • 14 }
Constructing an LCS (back-tracking) • We can find an LCS using b[i,j]’s. • We start with b[n,m] and track back to some cell b[0,i] or b[i,0]. • The algorithm to construct an LCS 1. i=m 2. j=n; 3. if i==0 or j==0 then exit; 4. if b[i,j]==1 then { i=i-1; j=j-1; print “xi”; } 5. if b[i,j]==2 i=i-1 6. if b[i,j]==3 j=j-1 7. Goto Step 3. • The time complexity: O(nm).
Shortest common supersequence • Definition:Let X and Y be two sequences. A sequence Z is a supersequence of X and Y if both X and Y are subsequence of Z. • Shortest common supersequence problem: Input: Two sequences X and Y. Output: a shortest common supersequence of X and Y. • Example: X=abc and Y=abb. Both abbc and abcb are the shortest common supersequences for X and Y.
Recursive Equation: • Let c[i,j] be the length of an LCS of X[1...i] and X[1...j]. • c[i,j] can be computed as follows: j if i=0 i if j=0, c[i,j]= c[i-1,j-1]+1 if i,j>0 and xi=yj, min{c[i,j-1]+1,c[i-1,j]+1} if i,j>0 and xiyj.