Transcript 投影片 1 - ecp168.net

A Hybrid Indexing Method for
Approximate String Matching
Journal of Discrete Algorithms, No. 1, Vol. 1, 2000, pp. 205-239,
Gonzalo Navarro and Ricardo Baeza-Yates
Advisor: Prof. R. C. T. Lee
Speaker: Y. K. Shieh
1
The approximate string matching problem is:
Given a text T of length n, a pattern P of length m
(n > m), and a threshold k to the number of
"errors" in the matches, find all occurrences of a
pattern in a text with k errors.
2
This paper uses an exhaustive searching mechanism.
We open a window T’ in T with size m+k (Rule 2) and
try to determine whether we are sure that every prefix
T’’ of this window T’ has ed(T’’,P) > k.
If the answer is yes, we ignore this window; otherwise,
we use dynamic programming to examine whether
any prefix T’’ of the window T’ has ed(T’’,P) ≦k.
3
We use dynamic programming to compute the edit
distance between two strings.
A matrix C0…|m|,0…|n| is filled, where Cj,i represents the
minimum number of operations need to match T1…i to
P1…j. This is computed as follows
C j , 0  j , C 0 ,i  i
C j ,i  if (Ti  Pj ) then C j 1,i 1
else 1  min(C j 1,i , C j ,i 1 , C j 1,i 1 )
Cj,0 and C0,i represent the edit distance between a
string of length j or i and the empty string.
4
example:
T = surgery
P = survey
k=2
s
0
1
s
1
0
u
2
1
r
3
2
g
4
3
e
5
4
r
6
5
y
7
6
u
r
v
e
2
3
4
5
1
2
3
4
0
1
2
3
1
0
1
2
2
1
1
2
3
2
2
1
4
3
3
2
5
4
4
3
y
6
5
4
3
3
2
2
2
There are only three prefixes of T, namely surge, surger
and surgery, whose edit distances with P=survey are
smaller than or equal to k=2.
5
Let us now see how we can be sure that for a
window T’ with size m+k , for every prefix T’’ of T’,
ed(T’’,P) > k.
We present Lemma 1 of this paper as follows.
6
Lemma 1
Let T’ in T and P be two strings such that ed(T’, P)
≦ k. Let P = P1x1P2x2… xj-1Pj, for strings Pi and xi
and for any j ≧ 1. Then, at least one string Pi appears
in T’ with at most k / j  errors.
Thus, we always divide the pattern into j pieces. We
shall point out how to divide later.
7
To be more precise, we may say that if ed(T’,P) ≦ k,
there exists a Pi in P and a T’’ in T’ such that
ed(Pi,T’’) ≦ k / j  .
8
Lemma 1 tells us that if for all Pi in P and every
substring b in T’, ed(Pi,b) > k / j  , then ed(P,T’) > k.
Suppose that there is a window T’ with size m+k and
for all Pi in P and for every substring b in T’, ed(Pi,b)
> k / j  .
Then, we can be sure that for every prefix T’’of T’ , for
all Pi in P and every substring b in T’’, ed(Pi,b)
> k / j  .
T’
T’’
T
b
P
Pi
9
Let us define the following condition.
Condition A: For all Pi in P and every substring b in
T’, ed(Pi, b) > k / j  .
Thus, if Condition A is satisfied, then for every prefix T’’
of T’, ed(T’’,P)>k.
In such a case, we ignore T’ and shift P one step to the
right.
10
Question, how can we be sure that the above condition
is satisfied.
The approach:
For each Pi, we generate all possible modified strings Pi
whose distances with Pi are smaller than or equal to k.
After generating all possible modified Pi ' s , we
may use the suffix tree of T to find all occurrences
of Pi , for all i, in T with error less than k / j  .
11
We still have the following questions:
• Question 1. How to divide P into j pieces?
• Question 2. How to generate all modified Pi’s?
• Question 3. How to find the occurrences of Pi’s in T
with edit distance less than or equal to k / j  .
12
Question 1: How to divide P into j pieces?
It can be proved that an optimal method is to
partition P into j pieces with
j  ( m  k ) / log  n  , where σ is the
alphabet size. We can get j pieces of P, and the
size of every piece is around logσn.
13
Question 2. How to generate all modified Pi’s?
The generation of all modified strings whose distances
with P can be done trivially. One method can be found in
[HHLS2006] which was reported by C. W. Lu.
Another method can be found in [HM2007] reported
By L. C. Chen.
In this paper, the authors used the second method
mentioned in [HM2007].
14
We can use non-deterministic finite automatons
(NFA).
A NFA is a five-tuple M=(Q, Σ, δ, q0 , F), where
Q is a finite set of states, Σ is a finite input
alphabet, δ is a mapping from Q×(Σ∪ {ε}) into
 state,
the set of subsets of Q, q0 Q is an initial
and F Q is a set 
of final states.
15
Lk ( P)  {Y | Y  , DL ( P, Y )  k}
One matched, no error.
a
0,0
ε
a
1,0
b
1,1
b
b
ε
b
b
One matched, one error.
ε
2,0
a
a
a
ε
a
2,1
a
2,2
a
ε
a
3,0
c
3,1
c
3,2
c
Four matched, no error.
4,0
c
ε
c
one error
4,1
c
ε
c
two errors
4,2
Two matched, two errors.
P = abac, k = 2.
The finite automaton M accepts Lk(P).
Lk(P)={aa, ab, ac, ba, bc, aaa, aab, aac, aba, abb, abc, acc,
baa, bab, bac, bbc, bcc, aaaa, aaab, aaac, aaba, aabc, aaca,
aacb, aacc, abaa, abab, abac, abba, abbb, abbc, abca, abcb,
abcc, baac, babc, bbac, bbbc, bcac}.
16
a
0,0
ε
a
1,0
b
1,1
b
2,0
b
a
a
ε
b
a ε
a
2,1
b
a
ε
a
2,2
ε
a
c
3,0
c
3,1
c
3,2
4,0
c
ε
c
4,1
c
ε
c
4,2
Recognize aa
P = abac, k = 2.
The finite automaton M accepts Lk(P).
Lk(P)={aa, ab, ac, ba, bc, aaa, aab, aac, aba, abb, abc, acc,
baa, bab, bac, bbc, bcc, aaaa, aaab, aaac, aaba, aabc, aaca,
aacb, aacc, abaa, abab, abac, abba, abbb, abbc, abca, abcb,
abcc, baac, babc, bbac, bbbc, bcac}.
17
Full example:
T = GACACAGACCAAAGCAG
P = CAAG
k=1
n = 17
m=4
18
P = CAAG
j = (m + k) / logσn
= (4 + 1) / log317
= 1.9388
Therefore, we partition P into two pieces.
P1 = CA
P2 = AG
According to Lemma 1, at least one piece appears in
substrings of T with at most 1/ 2 = 0 error.
This means that we want to find exact matching of P1 and
P2.
19
NFA with k = 1 of P1 = CA:
0,0
C
C
ε
A
1,0
A
ε
zero error
2,0
A
one error
1,1
2,1
A
NFA with k = 1 of P2 = AG:
0,0
A
εA
1,0
G
zero error
2,0
G εG
one error
1,1
2,1
G
20
T = GACACGGACCAAAGCAG
We construct the suffix tree of T.
A
C
A
G
$
C
G
A
17
11
12
8
$
G
G
A
C
C
A
A
A
G
C
A
G
$
16
13
A
A
G
C
A
G
$
14
6
15
9
7
10
5
2
1
4
3
21
We only need to consider the tree level from root to log 17  = 3 .
3
A
G
C
A
$
C
G
A
17
$
6
11
G
12
1,7
A
14
16
2
13
8
4
15
10
9
5
3
T = GACACGGACCAAAGCAG
22
T = GACACGGACCAAAGCAG
k=1
NFA of P1:
0,0
C
C
ε
A
1,0
A
A
ε
A
2,0
A
1,1
G
C
A
$
C
2,1
G
A
A
17
$
6
11
A
0,0
A
1,0
G
G
12
G εG
1,1
13
8
2,1
14
16
2,0
2
εA
1,7
A
4
15 9
10
5
3
G
NFA of P2
23
T = GACACGGACCAAAGCAG
k=1
0,0
C
C
ε
1,0
A
A
ε
A
2,0
A
1,1
A
$
C
2,1
G
A
A
A
17
(not exact match)
$
6
11
0,0
A
εA
1,0
G
14
16
2,0
A
2
13
8
2,1
1,7
A
G
12
G εG
1,1
G
C
4
15 9
10
5
3
G
(not exact match)
24
T = GACACGGACCAAAGCAG
k=1
0,0
C
C
ε
1,0
A
A
ε
A
2,0
A
1,1
G
C
A
$
C
2,1
G
A
A
17
Out of active states.
$
6
11
0,0
A
εA
1,0
G
A
2
13
8
2,1
14
16
2,0
G εG
1,1
G
12
1,7
A
4
15 9
10
5
3
G
(not exact match)
25
T = GACACGGACCAAAGCAG
13 16
k=1
0,0
C
C
ε
1,0
A
A
ε
A
2,0
A
1,1
G
C
A
$
C
2,1
G
A
A
17
Out of active states.
$
G
0,0
A
1,0
G
11
G
12
1,1
G
(exact match)
13
8
2,1
14
16
2,0
G εG
1,7
A
2
εA
6
4
15 9
10
5
3
We record positions 13 and
16 where AG occurs.
26
T = GACACGGACCAAAGCAG
k=1
C
0,0
C
C
ε
1,0
A
A
ε
A
2,0
A
1,1
G
C
A
$
C
2,1
G
A
A
17
$
6
11
0,0
A
1,0
G
G
12
C
G εG
1,1
13
8
2,1
14
16
2,0
2
εA
1,7
A
4
15 9
10
5
3
G
27
T = GACACGGACCAAAGCAG
k=1
A
0,0
C
C
ε
1,0
A
A
ε
A
2,0
G
C
A
A
1,1
$
C
2,1
A
G
A
(exact match)
17
$
6
11
0,0
A
1,0
G
G
12
G εG
1,1
13
8
2,1
14
16
2,0
2
εA
1,7
A
4
15 9
10
5
3
G
Out of active states.
We record positions 3, 10 and
28
15 where CA occurs.
T = GACACGGACCAAAGCAG
k=1
C
0,0
C
ε
1,0
A
A
ε
G
C
C
A
1,1
A
2,0
A
$
C
2,1
G
A
A
(not exact match)
17
$
6
11
0,0
A
1,0
G
G
12
G εG
1,1
13
8
2,1
14
16
2,0
2
εA
1,7
A
4
15 9
10
5
3
G
Out of active states.
29
T = GACACGGACCAAAGCAG
k=1
C
0,0
1,0
C
ε
A
A
ε
G
A
1,1
A
2,0
G
C
A
$
C
2,1
G
A
A
(not exact match)
17
$
6
11
0,0
A
εA
1,0
G
G
2
13
8
2,1
14
16
2,0
G εG
1,1
G
12
1,7
A
4
15 9
10
5
3
G
(not exact match)
30
T = GACACGGACCAAAGCAG
k=1
0,0
C
C
ε
G
1,0
A
A
ε
A
2,0
A
1,1
G
C
A
$
C
2,1
G
A
A
17
$
6
11
0,0
A
1,0
G
G
12
G
G εG
1,1
13
8
2,1
14
16
2,0
2
εA
1,7
A
4
15 9
10
5
3
G
31
T = GACACGGACCAAAGCAG
k=1
0,0
C
C
ε
1,0
A
A
ε
A
2,0
A
1,1
G
C
A
$
C
2,1
G
A
A
17
Out of active states.
$
6
11
0,0
A
1,0
G
G
12
G εG
1,1
13
8
2,1
14
16
2,0
2
εA
1,7
A
4
15 9
10
5
3
G
Out of active states.
32
T = GACACGGACCAAAGCAG
k=1
C
0,0
C
ε
1,0
A
A
ε
A
2,0
A
1,1
A
$
C
2,1
G
A
A
G
C
A
(not exact match)
17
$
6
11
0,0
A
1,0
G
G
12
G εG
1,1
13
8
2,1
14
16
2,0
2
εA
1,7
A
4
15 9
10
5
3
G
Out of active states.
33
T = GACACGGACCAAAGCAG
k=1
0,0
C
C
ε
1,0
A
A
ε
A
2,0
A
1,1
G
C
A
$
C
2,1
G
A
A
17
Out of active states.
$
6
11
0,0
A
1,0
G
G
12
G εG
1,1
13
8
2,1
14
16
2,0
2
εA
1,7
A
4
15 9
10
5
3
G
Out of active states.
34
T = GACACGGACCAAAGCAG
k=1
C
0,0
1,0
C
ε
A
A
ε
A
2,0
A
1,1
G
C
A
$
C
2,1
G
A
A
17
Out of active states.
$
6
11
0,0
A
εA
1,0
G
G
2
13
8
2,1
14
16
2,0
G εG
1,1
G
12
1,7
A
4
15 9
10
5
3
G
(not exact match)
35
After we find all probable positions in T, we
verify every substring of those positions.
The probable positions of T are: 3, 10, 13, 15, 16
We use the dynamic program to verify whether
any approximate string matching occurs between
T and P at the above locations .
36
k=1
The probable positions of T are 3, 10, 13, 15, 16
m+k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
C
A
A
G
0
1
2
3
4
G
1
1
2
3
3
A
2
2
1
2
3
C
3
2
2
2
3
A
4
3
2
2
3
C
5
4
3
3
3
No approximate
matching with k=1
found.
37
k=1
The probable positions of T are: 3, 10, 13, 15, 16
m+k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
C
A
A
G
0
1
2
3
4
A
1
1
1
2
3
C
2
1
2
2
3
A
3
2
1
2
3
C
4
3
2
2
3
G
5
4
3
3
2
No approximate
matching with k=1
found.
38
k=1
The probable positions of T are: 3, 10, 13, 15, 16
m+k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
C
A
A
G
0
1
2
3
4
C
1
0
1
2
3
A
2
1
0
1
2
C
3
2
1
1
2
G
4
3
2
2
1
G
5
4
3
3
2
CACG is found.
39
The probable positions of T are: 3, 10, 13, 15, 16
m+k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
This window does not include any probable position.
Therefore we can ignore this window.
40
The probable positions of T are: 3, 10, 13, 15, 16
m+k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
The window does not include any probable position.
Therefore we can shift the window directly.
41
k=1
The probable positions of T are: 3, 10, 13, 15, 16
m+k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
C
A
A
G
0
1
2
3
4
G
1
1
2
3
3
G
2
2
2
3
3
A
3
3
2
2
3
C
4
3
3
3
3
C
5
4
4
4
4
No approximate
matching with k=1
found.
42
k=1
The probable positions of T are: 3, 10, 13, 15, 16
m+k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
C
A
A
G
0
1
2
3
4
G
1
1
2
3
3
A
2
2
1
2
3
C
3
2
2
2
3
C
4
3
3
3
3
A
5
4
3
3
4
No approximate
matching with k=1
found.
43
k=1
The probable positions of T are: 3, 10, 13, 15, 16
m+k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
C
A
A
G
0
1
2
3
4
A
1
1
1
2
3
C
2
1
2
2
3
C
3
2
2
3
4
A
4
3
2
2
3
A
5
4
3
2
3
No approximate
matching with k=1
found.
44
k=1
The probable positions of T are: 3, 10, 13, 15, 16
m+k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
C
A
A
G
0
1
2
3
4
C
1
0
1
2
3
C
2
1
1
2
3
A
3
2
1
1
2
A
4
3
2
1
2
A
5
4
3
2
2
No approximate
matching with k=1
found.
45
k=1
The probable positions of T are: 3, 10, 13, 15, 16
m+k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
C
A
A
G
0
1
2
3
4
C
1
0
1
2
3
A
2
1
0
1
2
A
3
2
1
0
1
A
4
3
2
1
1
G
5
4
3
2
1
CAA, CAAA and
CAAAG are found.
46
k=1
The probable positions of T are: 3, 10, 13, 15, 16
m+k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
C
A
A
G
0
1
2
3
4
A
1
1
2
2
3
A
2
2
1
2
3
A
3
3
2
1
2
G
4
4
3
2
1
C
5
4
4
3
2
AAAG is found.
47
k=1
The probable positions of T are: 3, 10, 13, 15, 16
m+k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
C
A
A
G
0
1
2
3
4
A
1
1
1
2
3
A
2
2
1
1
2
G
3
3
2
2
1
C
4
3
3
3
2
A
5
4
3
3
3
AAG is found.
48
k=1
The probable positions of T are: 3, 10, 13, 15, 16
m+k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
C
A
A
G
0
1
2
3
4
A
1
1
1
2
3
G
2
2
2
2
2
C
3
2
3
4
5
A
4
3
2
3
4
G
5
4
3
3
3
No approximate
matching with k=1
found.
49
k=1
The probable positions of T are: 3, 10, 13, 15, 16
m
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
C
A
A
G
0
1
2
3
4
G
1
1
2
3
3
C
2
1
2
3
3
A
3
2
1
2
3
G
4
3
2
2
2
No approximate
matching with k=1
found.
50
k=1
The probable positions of T are: 3, 10, 13, 15, 16
m-k
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
G A C A C G G A C C A A A G C A G
C A A G
C
A
A
G
0
1
2
3
4
C
1
0
1
2
3
A
2
1
0
1
2
G
3
2
1
1
1
CAG is found.
51
Time complexity
• The preprocessing time complexity of constructing
automatons and a suffix tree of T is O(|N|*|m|) and
O(n) respectively, |N| is the number of states in a
NFA and |m| is the length of m.
• The search time obtained using the partitioning
scheme is O(nλlogn), where λ < 1 when error
tolerated α < 1-e/ , where e = 2.718… .
52
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Thank you
57