1 String Matching with k Mismatches by Using Kangaroo Method Efficient string with k mismatches,...

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String Matching with k Mismatches by Using Kangaroo Method

Efficient string with k mismatches, Landau, G.M., and Vishkin, U., Theoret. Comput Sci 43, 1986, pp. 239-249

Speaker: C. C. LinAdviser: R. C. T. Lee

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Problem definition:

Input: A text T with length n , a pattern P with

length m and a mismatching threshold k.

Output: All sub-strings of T with length m

matching P with k maximal number of

mismatches.

T = A G C T G C D C A C G I A B...1 4 3 2

P = A G C C

If k = 2k:

P = A G C CP = A G C CP = A G C C

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The concept of the Kangaroo method can be

explained as the following figure.

Assume that it is known before hand there

t1t2…ta=p1p2…pa and ta+1 is not equal to pa+1.

Thus we do not have to examine t1t2…ta+1 with

p1p2…pa+1 and jump directly to match the suffixes

beginning from ta+2 and pa+2.

Text: t1 t2… ta ta+1 ta+2 ta+3…tk…………Pattern: p1p2…pa pa+1 pa+2pa+3...pk…………

mismatch

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T = ABCCABDADBDETADBAADFDAAEERDXTDADCT…P = ETBDBCCDFDC

Kangaroo method will process as follows.

start

k=0

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k=1

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k=2

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k=3

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k=4

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We continue the above process. Whenever we

come to the situation that it is known a

substring of T exactly matching with a substring

of P, we skip this substring. This process is

stopped when k+1 mismatches have been found.

Input: T=ABAABBCCDD, P=ACDCB and k=2.

T=ABAABCCDD

P=ACDCB

k=3, we stop and discard ABAAB, then we start to

compare “BAADB” and “ACDCB”.

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Before we introduce the Kangaroo algorithm,

we shall first introduce the suffix tree and the

lowest common ancestor of two nodes.

The properties of suffix tree and the lowest

common ancestor of two nodes will be used in

Kangaroo algorithm.

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S = ABCDEADDBE

Suffix tree of a string with length n can be constructed in O(n).

Weiner, 1973McCreight, 1976Ukkonen, 1995

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CDEADDBE$

A

B DE

61

924 7 8

105

BCDEADDBE$DDBE$

CDEADDBE$

E$

EADDBE$DBE$

BE$

ADDBE$ $

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The lowest common ancestor of two leaf nodes can be found in O(1) by O(n) preprocessing in constructing time.

Harel and Tarjan, 1984

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CDEADDBE$

A

B DE

61

924 7 8

105

BCDEADDBE$DDBE$

CDEADDBE$

E$

EADDBE$DBE$

BE$

ADDBE$ $

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The Kangaroo method constructs a suffix tree

for text T and pattern P. Let the leaf node

corresponding to the substring starting from the

location be denoted as X. Let the leaf

corresponding to the pattern be denoted as Y.

The Kangaroo Method finds the lowest common

ancestor of X and Y to verify a text location with

k mismatches in O(k).

Let us consider the next page to figure out the

Kangaroo method.

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ANBECF$

ANCEC$

AN

BECF$ CEC$

Two suffix strings:

ANBECF$

ANCEC$

ANBECF$

ANCEC$

Then we can know that they have the same prefix “AN” and a mismatch “B” and “C”.

We now have to find whether there is any mismatches between ECF and EC.

ANBECF$ ANCEC$

mismatches=1

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We get remaining suffix strings:

ECF$

EC$

EC

$F$Then we can know that they have the same prefix “EC” and because we touch $, we finish the verification.

ECF$

EC$

ECF$ EC$

mismatches=1

Thus we could know thatthe mismatches between “ANBECF” and “ANCEC”is 1.

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We will not have to compare all characters by using the finding of the lowest common ancestor of two strings of text and pattern in the suffix tree.

This is useful if there are many equivalent characters between the text and the pattern because we will not have to compare those equivalent characters.

Finding the lowest common ancestor between two suffixes is to find the next mismatch between two strings.

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Input: T=ABCCBDCDBC, P=ABCD and k=2 The suffix tree of T and P is:

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$

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The lowest common ancestor of “ABCD” and

“ABCCBDCDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$

T=ABCCBDCDBC P=ABCD k=1, return “ABCC”.

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“BCCBDCDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$

T=ABCCBDCDBC P=ABCD k=1.

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The lowest common ancestor of “BCD” and

“CCBDCDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$


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The lowest common ancestor of “CD” and

“CBDCDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$

T=ABCCBDCDBC P=ABCD k=3, discard “BCCB”.

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“CCBDCDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$


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“CBDCDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$


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“BDCDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$

T=ABCCBDCDBC P=ABCD k=3, discard “CCBD”.

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“CBDCDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$


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“BDCDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$


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The lowest common ancestor of “D” and

“CDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$

T=ABCCBDCDBC P=ABCD k=3, discard “CBDC”.

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“BDCDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$


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“DCDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$


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“CDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$

T=ABCCBDCDBC P=ABCD k=2, return “BDCD”.

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“DCDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$


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“CDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$


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“DBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$

T=ABCCBDCDBC P=ABCD k=3, discard “DCDB”.

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“CDBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$


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“DBC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$


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“BC”.

ABC

CBDCDBC$

D$

B

D

C

DCDBC$

C

$

CDBC$ BC$

$

CBDCDBC$

BDCDBC$

D

$

CBDCDBC$

ABCCBDCDBC$

ABCD$

BCCBDCDBC$

BCD$

BC$CBDCDBC$

DCDBC$

DBC$

BC$$

CDBC$ CD$

BDCDBC$

D$

C$

CCBDCDBC$

D$

T=ABCCBDCDBC P=ABCD k=3, discard “CDBC”.

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Input: T=ABCCBDCDBC, P=ABCD and k=2.

Output: “ABCC” and “BDCD”.

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In order to use Kangaroo method, we construct

a suffix tree for the text T with the length n and

the pattern p with the length m in O(n+m).

By using Kangaroo method, we take O(1) time

to find one mismatch. We stop when there are

more than k mismatches. Therefore, we take

O(k) time to find at most k mismatches.

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Thus, the time complexity of finding out all

locations of text T with k maximal mismatches

with the pattern P is O(nk).

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References

For Construction of Suffix trees:[M76] McCreight, E.M., A Space-Economical Suffix Tree Construction Algorithm, J. ACM 23 (1976): 262-272.[U95] Ukkonen, E., On-line Construction of Suffix Trees, Algorithmica 41 (1995): 249-260.

For Finding Lowest Common Ancestor:[HT84] Harel, D. and Tarjan, R.E., Fast Algorithms for Finding Nearest Common Ancestor, SIAM Journal on Computing 13 (1984): 338-355.

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References

For String Matching with k Mismatches:

[LV86] Landau, G.M., and Vishkin, U., Efficient string with k mismatches, Theoret. Comput Sci 43 (1986): 239-249.

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Thank you

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