Less Than Matching

Orgad Keller

Modified by Ariel Rosenfeld

Less Than Matching

Algorithms 2 2

Less Than Matching

Input: A text , a pattern

over alphabet with order relation . Output: All locations where

Can we use the regular methods?

i

0 1, j i jj m p t

0 1... nT t t 0 1... mP p p

i jt

jp

iT

P

Algorithms 2 3

Transitivity

Less Than Matching is in fact transitive, but that is not enough for us:

does not imply anything about the relation between and .

,a c b c a b

Algorithms 2 4

Approach

A good approach for solving Pattern Matching problems is sometimes solving:The problem for a binary alphabet .The problem for a bounded alphabet .The problem for an ubounded alphabet .

In that order.

0,1

Algorithms 2 5

Binary Alphabet

The only case that prevents a match at location is the case where:

This is equivalent to:

So how can we solve this case?

0 1, 1 0j i jj m p t

i

0 1, 1 1j i jj m p t

Algorithms 2 6

Binary Alphabet

So if , there is no match at .

We can calculate Then we’ll calculate (P reverse) using FFT.We’ll return all locations where

1

0

0m

j i jj

p t

i

0 1... nT t t RT P

( )[ 1] 0RT P i m i

Example

P=0101 T=0101001110

PR = 1010 T! = 1010110001

Algorithms 2 7

Algorithms 2 8

RT P

Algorithms 2 9

P=0101T=0101001110

What just happened?

Algorithms 2 10

T= !

PR=

Complexity

Time:

Algorithms 2 11

( log )O n m

Algorithms 2 12

Bounded Alphabet

We need reductions to binary alphabet. For each we’ll define:

We notice are binary.

0 1

1

0

...

ii

i

n

tt

t

T t t

0 1

1

0

...

ii

i

m

pp

p

P p p

,T P

Algorithms 2 13

Bounded Alphabet

Theorem: (less than) matches at location if and only if , (less than) matches at location .

Proof: does not match at iff .

that is true iff , meaning that does not (less than) match at location .

PP T

iT i

P T i, j i jj p t

1 0j i jp t

P

iT

Algorithms 2 14

Bounded Alphabet

So for each , we’ll run the binary alphabet algorithm on .

We’ll return only the locations that matched in all iterations.

Time:

,T P

(min , log )O m n m

Algorithms 2 15

Problem

Can be worse than the naïve algorithm. What about unbounded alphabet? We present an improvement on the next

slides.

(min , log )O m n m

16

The Trick

We’ll split the text into overlapping segments of size like this:

So every match in the text must appear in whole in one of the segments.

n

m

n

m2m

2m 2m 2m 2m 2m 2m

2m2m 2m 2m 2m 2m

2m

Algorithms 2 17

First, use the segment splitting trick. Therefore we can assume .

For each location in text, we’ll produce a triplet: , where .

For each location in pattern, we’ll produce a triplet: , where .

We now have triplets all together.

Abrahamson-Kosaraju Method

2T m

( , ' ', )a T ii

ip bi

( , ' ', )b P i

3m

it a

Algorithms 2 18


We’ll hold all triplets together. Sort all triplets according to symbol. We’ll define a symbol that has more than

triplets as a “frequent symbol”. There are frequent symbols. Put all frequent symbols’ triplets aside.

m

( )O m

Algorithms 2 19


Split non-frequent symbols’ triplets to groups of size in the following manner:

2m S m

2 1

3 2

Group 1

1 3

2 4

( , ' ', 4), ( , ' ',7),..., ( , ' ',300) , ( , ' ',3),..., ( , ' ', 200) ,

( , ' ',5),..., ( , ' ',1000) , ( , ' ',5),..., ( , ' ',150)

m m

m m

a T a T a P b T b T

d P d T g P g T

Group 2

,...

Algorithms 2 20


The rule is that there can’t be two triplets of the same symbol in different groups.

2 1

3 2

Group 1

1 3

2 4

( , ' ', 4), ( , ' ',7),..., ( , ' ',300) , ( , ' ',3),..., ( , ' ', 200) ,

( , ' ',5),..., ( , ' ',1000) , ( , ' ',5),..., ( , ' ',150)

m m

m m

a T a T a P b T b T

d P d T g P g T

Group 2

,...

Algorithms 2 21


For each such group, choose the symbol of the first triplet in group as the group’s representative.

For instance, on previous example, group 1’s representative is and group 2’s representative is .

There are representatives all together.

ad

( )O m

Algorithms 2 22


To sum up: frequent symbols. representatives of non-frequent

symbols. We’ll swap each non-frequent symbol in

pattern and text with its representative. Now our text and pattern are over

sized alphabet.

( )O m

( )O m

( )O m

Algorithms 2 23


We want to run our algorithm over the new text and pattern to count the mismatches between symbols of different groups.

But we have a problem:Let’s say is a frequent symbol, but:

1 3

2 4

Group 2

..., ( , ' ',5),..., ( , ' ',1000) , ( , ' ',5),..., ( , ' ',150) ,...

m m

d P d T g P g T

f

Algorithms 2 24


The representative of group 2 is , which is smaller than , but the group also contains which is greater than .

1 3

2 4

Group 2

..., ( , ' ',5),..., ( , ' ',1000) , ( , ' ',5),..., ( , ' ',150) ,...

m m

d P d T g P g T

ff

d

g

Algorithms 2 25


In that case we’ll split group 2 to two groups with their own representatives.

Since we performed at most such splits, we still have representatives.

1 3

2 4

Group 2.1 Group 2.2

..., ( , ' ',5),..., ( , ' ',1000) , ( , ' ',5),..., ( , ' ',150) ,...

m m

d P d T g P g T

( )O m

( )O m

Algorithms 2 26


We can now run our algorithm over the new text and pattern in .

But we still haven’t handled comparisons between two non-frequent symbols that are in the same group.

( log )O mm m

Algorithms 2 27


We’ll do so naively in each group:For each triplet in the group

For each triplet of the form in the group, if , then add an error at location

.

Time: ( )O m m

( , ' ', )P j ( , ' ', )T k

i k j

ktjp

iT

P

j kp t

i j

Algorithms 2 28

Running Time

For one segment:Sorting the triplets and representatives:

.Running the algorithm: .Correcting results (Adding in-group errors):

. Overall for one segment: . Overall for all segments: .

( log )O m m

( log )O mm m

( )O m m

( log )O m m m

( log )O n m m

Algorithms 2 29

Running Time

We can improve to .Left as an exercise.

( log )O n m m

Less Than Matching

Documents

Transcript of Less Than Matching