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2 Dimensional Parameterized MatchingPowerPoint Presentation

2 Dimensional Parameterized Matching

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2 Dimensional Parameterized Matching

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2 Dimensional Parameterized Matching

Carmit Hazay

Moshe Lewenstein

Dekel Tsur

- 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.

Example:

a

a

b

b

b

s

(a)=x

x

x

y

y

y

t

(b)=y

- Input: Two strings T, P; |T|=n, |P|=m.
- Output: All text locations i,
such that (P)=Ti …Ti+m-1.

- 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-

T

a b c

a a b

b b b

(x)=a

(y)=b

(z)=c

P

x y z

x x y

y y y

2D Parameterized Matching

pattern

‘A horse is a horse,

it ain’t make a difference

what color it is’ John Wayne

- 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].

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

Time Analysis: If done properly – O(n2m2)

- Pair of locations such that the characters disagree parameterized.
- Example,

a a b a a a

x x y x z y

- Encode every text location by its predecessor location.

First a to its left

T

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

1 3 6 13 14 15 16 17 18

Encoded T

0 1 3 6 13 14 15 16 17

- Two p-matching strings have the same encoded texts.

S

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

Encoded S

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

T

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

Encoded T

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

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

S

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

Encoded S

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

T

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

Encoded T

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

- 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

- First idea,
Encode the linearization of text and pattern.

As you will all see this box frames the text 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.

As you will see this box

frames the texts that it

Contains. That is 2D text

All in this little box.

- First idea,
Encode the linearization of text and pattern.

As you will see this box

frames the texts that it

Contains. That is 2D text

All in this little box.

- First idea,
Encode the linearization of text and pattern.

Overflow problem!!

a

b

Different character than b

b

a

b

- 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

- 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?

- 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)

- Observation:
T p-matches P

Every text location and its predecessor are not a mismatch pair

+

# of distinct characters in P and T equal

- 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.

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

+a

+b

- 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).

- 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.

- 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

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

- 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.

- 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).

Checking all predecessors takes linear time.

Total time O(n2).

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

+a

+b

- 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.

- End cases, multiple cases.

Less than

B1

A1

A2

B2

B3

A3

B4

A4

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