Approximate On-line Palindrome Recognition, and Application - PowerPoint Presentation

Slide1

Approximate On-line Palindrome Recognition, and Applications

Amihood AmirBenny PoratSlide2

Moskva RiverSlide3

Confluence of 4 Streams

Palindrome Recognition

Approximate Matching

Interchange Matching

Online Algorithms

CPM 2014Slide4

Palindrome Recognition

- Voz'mi-ka slovo

ropot

, - govoril Cincinnatu ego shurin,

ostriak, -- I prochti obratno. A? Smeshno poluchaetsia?

Vladimir Nabokov,

Invitation to a Beheading (1)

"Take the word

ropot

[murmur]," Cincinnatus' brother-in-law,

the wit, was saying to him, "and read it backwards. Eh? Comes out funny, doesn't it?" [--›

topor: the axe] A palindrome

is a string that is the same whether read from right to left or from left to right: Examples: доход

A man, a plan, a cat, a ham, a yak, a yam, a hat, a canal-Panama! Slide5

Palindrome Example

Ibn Ezra:

Medieval Jewish philosopher, poet, Biblical commentator, and mathematician.

Was asked:

"

אבי אל חי שמך למה מלך משיח לא יבא

"

[ My Father, the Living God, why does the king messiah not arrive?]

His response:

"

דעו מאביכם כי לא בוש אבוש, שוב אשוב אליכם כי בא מועד"[ Know you from your Father that I will not be delayed. I will return to you when the time will come ]Slide6

Palindromes in Computer Science

Great programming exercise in CS 101.

Example of a problem that can be solved by a RAM in

linear time

, but

not

by a 1-tape Turing machine.

(Can be done in linear time by a 2-tape TM)Slide7

Palindrome Concatenation

We may be interested

in finding out whether a string is a concatenation of palindromes of length > 1.

Example:

ABCCBABBCCBCAACB

Why would we be interested in such a funny problem?

– we’ll soon see

Exercise:

Do this in linear time…

ABCCBA

BB

CC

BCAACBSlide8

Stream 2 - Approximations

As in exact matching, there may be errors. Find the

minimum

number of errors that, if fixed, will give a string that is a concatenation of palindromes of length > 1

Example:

ABCCBCBBCCBCABCB

For Hamming distance:

A-Porat [ISAAC 13]:

Algorithm of time

O(n

2

)

ABCCBA

BB

CCBCAACBSlide9

Stream 3 - Reversals

Why is this funny problem interesting?

Sorting by reversals:

In the evolutionary process a substring may “detach” and “reconnect” in reverse:

ABCA

BCDAABC

BAD

CBAADCB

ABCA

BCDAABC

BADSlide10

Sorting by Reversals

What is the

minimum

number of reversals that, when applied to string A, result in string B?

History:

Introduced:

Bafna & Pevzner [95]

NP-hard:

Carpara [97]

Approximations:

Christie [98]

Berman, Hannenhalli, Karpinski [02]

Hartman [03]Slide11

Sorting by Reversals – Polynomial time Relaxations

Signed reversals:

Hannenhalli & Pevzner [99]

Kaplan, Shamir, Tarjan [00]

Tannier & Sagot [04]

. . .

Disjointness:

Swap Matching

Muthu [96]

Two constraints:

The

length

of the reversed substring is limited to 2

. All swaps are disjoint.Slide12

Reversal Distance (RD):The RD between s

1 and s2 is the minimum number k, such that there exist s2’ , where HAM(s1

,s2’) =k, and s

1

reversal match s

2

.

A

B

D

E

A

B

C

D

A

E

CB

A

B

A

A

D

A

S

1

:

S

2

:

RD(S

1

,S

2

) = 2

Pattern Matching with Disjoint ReversalsSlide13

Interleave Strings:

A

B

D

E

A

B

C

D

A

E

D

B

A

B

A

A

D

C

S

1

:

S

2

:

Connection between Reversal Matching and Palindrome Matching

A C D D C A B A A B E A D B B D A ESlide14

On-line Input

Suppose that we get the input a byte at a time:

For the palindrome problem:

A

C

D

A

C

A

B

B

A

A

E

B

B

A

A

A

E

A

D

D

DSlide15

On-line Input

Suppose that we get the input a byte at a time:

For the reversal problem:

AC

CA

BA

AB

EA

BD

A

A

A

AE

DD

DBSlide16

Main Idea – Palindrome Fingerprint

s0,s1

,s2,…sm-1

Φ

R

(S)=r

-1

s

0

+ r

-2

s

1

+… r

-msm-1

mod (p)

Φ(S)=r1s

0+ r2s

1

+… r

m

s

m-1

mod (p)

The Rabin Karp

Fingerprint

If r

m+1

Φ

R

(S) =

Φ

(S) => S is a palindrome.

w.h.p.

The Reversal

FingerprintSlide17

Palindrome Fingerprint

If rm+1ΦR

(S) = Φ(S) => S is a palindrome.

Example:

S =

A B C B A

r

6

Φ

R

(S)=

r

6 (1/r A + 1/r2 B + 1/r3 C + 1/r4 B + 1/r

5 A) =r5 A + r4 B + r3 C + r

2 B + r A = Φ(S)

Φ

R(S)=r

-1s0+ r

-2

s

1

+… r

-m

s

m-1

mod (p)

Φ

(S)=r

1

s

0

+ r

2

s

1

+… r

m

s

m-1

mod (p)Slide18

Simple Online Algorithm for Finding a Palindrome in a Text

t1,t2,t

3, … t

i

,t

i+1

,t

i+2

,

…

t

i+m

, ti+m+1 , …

tn

ΦR=r-1t

i+ r-2ti+1

+… r-mt

i+m mod (p)

Φ

=r

1

t

i

+ r

2

t

i+1

+… r

m

t

i+m

mod (p)

If

not

, then for the next position:

If

r

m+1

Φ

R

=

Φ

=

>

there is a palindrome starting in

the i-th position

.

Φ

=

Φ

+ r

m+1

t

i+m+1

mod (p)

Φ

R

=

Φ

R

+ r

-(m+1)

t

i+m+1

mod (p)

Note:

This algorithm finds online whether the prefix of a text is a permutation. For finding online whether the text is a concatenation of permutations, assume even-length permutations, otherwise, every text is a concatenation of length-1 permutations.Slide19

Palindrome with mismatches

Start with 1 mismatch case.Slide20

1-Mismatch

s0,s1,s2,

… sm-1

S=

Choose

l

prime numbers

q

1

,…,q

l

<

m

such that Slide21

1-Mismatch

s0,s1,s2,

… sm-1

s

0

,s

2

,s

4

…

s

1

s3,s5

…

s0

,s3,s6

…

s

1

,s

4

,s

7

…

s

2

,s

5

,s

8

…

mod 2

mod 3

S=

S

2,0

=

S

2,1

=

S

3,0

=

S

3,1

=

S

3,2

=

For each

q

i

construct

q

i

subsequences of

S

as follows: subsequence

S

q

i

,j

is all elements of S whose index is

j

mod

q

i

.

Examples:

q

1

=2, q

2

=3Slide22

Example

s0,s1,s2, s

3,s4,s

5

s

0

,s

2

,s

4

s

1

s

3

,s

5

s0

,s3

s

1

,s

4

s

2

,s

5

mod 2

mod 3

S=

S

2,0

=

S

2,1

=

S

3,0

=

S

3,1

=

S

3,2

=Slide23

1-Mismatch

We need to compare:We prove that in the partitions strings:

s

0

, s

1

, s

2

,

…

s

m-2

,s

m-1

sm-1, sm-2

, sm-3 … s

1 , s0

S

q,j

= S

R

q,(m-1-j)mod qSlide24

Example

s0,s1,s2,s

3,s4,s

5

s

0

,s

2

,s

4

s

0

,s

3

s

1

,s4

S=

s

5

,s

4

,s

3

,s

2

,s

1

,s

0

S

R

=

s

5

s

3

,s

1

s

5

,s

2

s

4

,s

1

s

0

,s

2

,s

4

s

1

s

3

,s

5

s

0

,s

3

s

1

,s

4

s

2

,s

5

S

2,0

=

S

2,1

=

S

3,0

=

S

3,1

=

S

3,2

=S2,0=SR2,1=S3,0=SR3,2=S3,1=SR3,1=Slide25

Exact Matching

Lemma: S=SR  Sq,j

= SRq,(m-1-j) mod

q

for all q and all

0 ≤ j

≤

q.Slide26

1-Mismatch

Lemma: S is a palindrome with 1-mismatch  for each q, there is exactly

one

j such that:

Φ

(S

q,j

) ≠ r

|Sq,j|

Φ

R(SR

q,(m-1-j)mod q)Slide27

1-Mismatch

Lemma:There is exactly one mismatchThere is exactly one subpattern in each group that does not match.

C.R.TSlide28

Chinese Remainder Theorem

Let n and m two positive integers.

In our case:

if two different indices,

i

and

j

, have an error, and only one subsequence is erroneous, since the product of all q’s > m, it means that

i=j

.Slide29

Complexity

There exists a constant c such that, for any x<m, there are at least

x/log m

prime numbers between

x

and

cx

.

Therefore, choose prime numbers between

log

m

and

c

log

m

.Slide30

Complexity

For each qi we compute 2qi different fingerprints:Overall space

:

Each character participates in

exactly two

fingerprints (the regular and the reverse).

Overall time:Slide31

Online

All fingerprint calculations can be done onlineWe know the m

at every input character, to compute the comparisons.

Conclude:

Our algorithm is online.Slide32

k-Mismatches

Use Group testing…Slide33

k-Mismatches

Group TestingGiven n items with some

positive ones, identify all positive ones by a small number of tests

.

Each test is on a

subset of items

.

Test outcome is positive iff there is a positive item in the subset. Slide34

k-Mismatch

Group: partition of the text.Test: distinguish between:

(using the 1-mismatch algorithm)match 1-mismatch

more then 1-mismatchSlide35

k-Mismatches

s0,s1,s2,

… sm-1

s

0

,s

2

,s

4

…

s

1

s3,s5

…

s0

,s3,s6

…

s

1

,s

4

,s

7

…

s

2

,s

5

,s

8

…

mod 2

mod 3

S=

S

2,0

=

S

2,1

=

S

3,0

=

S

3,1

=

S

3,2

=

Similar to the 1-mismatch algorithm just with more prime numbers…

Each S

q,j

is a

group

in our group testingSlide36

Our tests

We define The reversal pair of Sq,j to be S

Rq,(m-1-j)mod q

Each partition is “tested against” its reversal pair.Slide37

Correctness

s0,s1,s

2, … sj …. s

m-1

For any group of k character i

1

,i

2

,..i

k

There exists a partition where s

j

appears alone

i

2

i

5

i

7

i

9

i

C.R.TSlide38

Correctness

s0,s1,s

2, … sj …. s

m-1

If s

j

invokes a mismatch we will catch it.

i

2

i

5

i

7

i

9

iSlide39

Complexity

Overall space:Overall time:Slide40

Approximate Reversal Distance

Using the palindrome up to k-mismatches algorithm, can be solved in

time, and

space.Slide41

спасибо