.
• 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).
C[0, j ]  0
C[i ,0 ]  i
C[i , j ]  if ( Pi  T j ) then C[i  1, j  1]
else 1  min( C[i  1, j ], C[i, j  1], C[i  1, j  1])
The matrix
s
u
r
g
e
r
y
0
0
0
0
0
0
0
0
s
1
0
1
1
1
1
1
1
u
2
1
0
1
2
2
2
2
r
3
2
1
0
1
2
2
3
v
4
3
2
1
1
2
3
3
e
5
4
3
2
2
1
2
3
y
6
5
4
3
3
2
2
2
The dynamic programming algorithm search ‘survey’ in
the text ‘surgery’ with two errors. Bold entries indicate
matching positions. Running time O(nm).
String Matching Allowing Errors
• O(nk) 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]}.

s
u
r
v
e
y
no errors
ε
ε
s
ε
u
ε
r
ε
v
ε
e
y
1 error
ε
ε
s
ε
u
ε
r
ε
v
ε
e
y
2 errors
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:
1. Any letter x in {},is a regular
expression, where  is the set of all
letters.
2. 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.
ε
FA
ε
S
f
ε
FB
FA|B
ε
S
FA
ε
FA B
FB
f
ε
S
FA
F(A)*
f
ε
b
A
c
a
A
e
ε
d
A
ε
ε
g
B
ε
h
f
l
ε
i
(AA | B)  (B | AB)
A
j
B
k
ε
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.
GOOGLE: City University
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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 mn. 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 xm yn, then zkxm implies that Z is an LCS
of X[1..m-1] and Y.
• 2. If xm yn, then zkyn 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 xiyj.
Computing the length of an LCS
• There are nm 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.
else
if 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
}
Example 3: X=BDCABA and Y=ABCBDAB.
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.
2.
3.
4.
i=m
j=n;
if i==0 or j==0 then exit;
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:
c[i,j]=
j
if i=0
i
if j=0,
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 xiyj.