1

2 Dimensional Parameterized Matching

Carmit HazayMoshe Lewenstein

Dekel Tsur

2

CPM 2005

3

CPM 2005

4

CPM 2005

5

CPM 2005

6

CPM 2005

7

CPM 2005

8

CPM 2005

9

CPM 2005

10

CPM 2005

11

CPM 2005

12

CPM 2005

13

CPM 2005

14

Parameterized Matching

Input: two strings s and t, |s|=|t|, over alphabets ∑s and ∑t.

s parameterize matches t: if bijection : ∑s ∑t , such that (s) = t.

∃

(a)=x

(b)=y

Π Π

ΠΠ

a ab b b

x xy y y

Example: s

t

15

Parameterized Matching

Input: Two strings T, P; |T|=n, |P|=m.

Output: All text locations i, such that (P)=Ti …

Ti+m-1.Π

16

2D Parameterized Matching

Input: Text T and pattern P; |T|=n*n, |P|=m*m.

Output: All text locations (i,j), such that (P)=Ti,j …Ti+m-1,j+m-1.

Example-

Π

a b ca a bb b bx y z

x x yy y y

(x)=a

(y)=b

(z)=c

ΠΠΠ

T

P

17

2D Parameterized Matching

pattern

‘A horse is a horse,it ain’t make a differencewhat color it is’ John Wayne

18

Parameterized Matching History

Introduced by Brenda Baker [Baker93].

Others: [AFM94], [Bak95], [Bak97].

Two Dimensions: [AACLP03][This work].

Used in scaled matching [ABL99].

Periodicity of parameterized matching [ApostolicoGiancarlo].

Approximate parameterized matching [AEL], [HLS04].

19

Naïve Algorithm

For every location (i,j) of text Check if P parameterized matches at (i,j):

1. For each a alphabet of P, check if all

a’s of P align with same character 2. For each b alphabet of T, check if all b’s of T align with same character

20

Naïve Algorithm

Time Analysis: If done properly – O(n2m2)

21

Mismatch pairs

Pair of locations such that the characters disagree parameterized.

Example,

a a b a a ax x y x z y

22

1D Encoding

Encode every text location by its predecessor location.

a b a d d a b d b c b d a a b d a a a a b b b T

First a to its left

Encoded T

1 3 6 13 14 15 16 17 18

0 1 3 6 13 14 15 16 17

23

1D Encoding

Two p-matching strings have the same encoded texts.

a b b c b a a c b b c b a

x y y z y x x z y y z y x

0 0 2 0 3 1 6 4 5 9 8 10 7

0 0 2 0 3 1 6 4 5 9 8 10 7

S

Encoded S

T

Encoded T

24

1D Encoding

Hence, in order to check whether two strings p-match, enough to compare their encoded strings.

Reduction to exact matching problem.

a b b c b b a c b b c b a

x y y z y x x z y y z y x

0 0 2 0 3 5 6 4 5 9 8 10 7

0 0 2 0 3 1 6 4 5 9 8 10 7

S

Encoded S

T

Encoded T

25

2D Mismatch Pairs

Same as 1D mismatch pairs, but with 2D strings.

Example:

a b a

b a b

b a b

x y x

y y y

y y y

26

First idea,Encode the linearization of text and pattern.

2D Encoding

As you will see this boxframes the texts that it Contains. That is 2D textAll in this little box.

As you will all see this box frames the text that itcontains. That is 2D textall in this little box .

27

First idea,Encode the linearization of text and pattern.

2D Encoding

As you will see this boxframes the texts that it Contains. That is 2D textAll in this little box.

As you will see this box frames the texts that it Contains. That is 2D text All in this little box.

28

First idea,Encode the linearization of text and pattern.

Overflow problem!!

2D Encoding

bb

b

Different character than b

a

a

29

2D Encoding

Second idea, use strips.

Strip – Substring of T of size n*m.

i-th strip of T, is n*m substring T[1:n,i:i+m-1]. i

30

Second Solution

For Pattern P compute predecessors on its linearization.

For each strip of T, compute predecessors on its linearization.

Do Pattern Matching for each strip.

Time – O(n2m).

Can we do better?

31

A Faster Solution

Set into Duel-and-Sweep setting Needs special care for Duel, Sweep Especially difficult: Pattern

preprocessing

Desired Time: O(n2 + poly(m))

We Achieve: O(n2 + m2.5polylog m)

32

Remember…

Observation:

T p-matches P

Every text location and its predecessor are not a mismatch pair

+ # of distinct characters in P and T equal

↔

33

Algorithm Outline

Duel and sweep paradigm Find candidates - Dueling Divide candidates by strips Update predecessors of every new strip Check new predecessors - Sweep

Assume pattern witness table given.

34

Witness

Witness – Mismatch pair between P and its alignment to location (a,b).

+a

+b

35

Set Candidates

Using duel-Every two text locations that has a witness within their alignment can eliminate each other.

Apply algorithm [ABF94] and return list of candidates.

Time – O(n2).

36

Sweep Technique

Observation, All candidates agree with each other.

Hence, Mismatch pair eliminates all candidates

containing it.

Therefore, For every predecessor, enough to find

one candidate that contains it.

37

Sweep Technique

How to find? Create new 2m*2m array A such that,

A[i,j] = largest row among candidates that starts at column j and overlap with row i.

x

38

Sweep Technique

For every predecessor (i,j), (x,y), use range minima query to find highest candidate contain predecessor.

39

Sweep Technique

In case of a mismatch pair,eliminate all candidates containing it.

How?Use mismatch vector.Every mismatch pair translate into range.For new strips, delete old mistakes and add new.

All candidates within this range are eliminated.

40

Sweep Technique Reminder-

T p-matches P

Every text location and its predecessor are not mismatch pair

+ # of distinct characters in P and T equal

Left to do?

Count distinct characters for every candidates. Use algorithm of Amir and Cole, time O(m2).

41

Overview

Checking all predecessors takes linear time.

Total time O(n2).

42

Pattern Preprocessing

Witness – Mismatch pair between P and its alignment to location (a,b).

+a

+b

43

Pattern Preprocessing

Find witness table for P in time O(m2.5 * polylogm).

For every pattern location (i,j), create list of size O( ) pointers.

Pointer i is predecessor in lines above (i,j).

Reduce to exact matching with don’t cares.

m

)1+mi ,mi(

44

Pattern Preprocessing

End cases, multiple cases.

A1

A3 A4

A2B1

B2 B3

B4

Less than m

45

Open Questions

Can the algorithm time complexity be reduced into O(n2+m2)?