28 Schmidt

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Quadratic Time Algorithms

for Finding Common Intervals

in Two and More Sequences

Thomas Schmidt

Jens Stoye

CPM 2004, Istanbul

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Thomas Schmidt: Quadratic Time Algorithms for Finding Common Intervals in Two and More Sequences 3

•

Observations:- Gene order in bacterial genomes is weakly conserved

- Some genes tend to cluster together even in unrelated species

- Functional association of genes inside a cluster

Gene Order and Function in Bacteria:

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•





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?


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Are there more clusters ?


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Task:

• Establish a model and search for gene clusters


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Formalization of Gene Clusters:

Genomes: permutations π

1

, π 2

,…,

π

k Genes: numbers 1 ,…,n

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1

, π 2

,…,

π


π 1

π 2

π 3

π 4

1 2 3 4 5 6 7 8

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1

, π 2

,…,

π


π 1

π 2

π 3

π 4

1 2 3 4 5 6 7 8

8 7 6 4 5 2 1 3

3 1 2 5 8 7 6 4

6 7 4 2 1 3 8 5

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1 2 3 4 5 6 7 8

8 7 6 4 5 2 1 3

3 1 2 5 8 7 6 4

6 7 4 2 1 3 8 5

π 1

π 2

π 3

π 4


1

, π 2

,…,

π


Gene cluster: common interval subset of numbers occurring

contiguously in all permutations)

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1 2 3 4 5 6 7 8

8 7 6 4 5 2 1 3

3 1 2 5 8 7 6 4

6 7 4 2 1 3 8 5

π 1

π 2

π 3

π 4


1

, π 2

,…,

π




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Algorithms:- Uno & Yagiura, Algorithmica 2000 : Find all common intervals of

two permutations in O(n+|output|) time.

- Heber & Stoye, CPM 2001: Find all common intervals of k ≥ 2permutations in O(kn+|output|) time.

Genomes: permutations π 1 , π 2 ,…, π k

Genes: numbers 1 ,…,n



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Modeling multiple copies of a gene (paralogs):

Problem:

- Gene duplication results in multiple copies of a gene inside

a genome

- Difficult to assign the correct gene pair

1 2 3 4 5 6 7 8

π 1

π 2

π 3

7 ?

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Problem:


a genome


1 2 3 4 5 6 7 8

π 1

π 2

π 3

? 7

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Problem:


a genome


1 2 3 4 5 6 7 8

π 1

π 2

π 3

3 1 2 ? ?

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Problem:


a genome


1 2 3 4 5 6 7 8

π 1

π 2

π 3

3 ? 2 1 ?

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Solution:

- Do not distinguish between paralogous gene copies

- Each paralogous copy of a gene gets the same number

Consequence:

- Genomes are modeled as sequences instead of

permutations

1 2 3 4 5 6 7 8

S 1

S 2

S 3

3 1 2 4 8 7 6 1 2

8 7 6 7 5 4 2 1 3

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Overview:

• Introduction- Comparative genomics

- Common Intervals and Gene Clusters

• Formal Model

• Algorithms- Simple Data Structure: Quadratic Space

- Saving Space

• Results

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Formal Model:

Given: String S over a finite alphabet Σ

Notation: S [i] = the i-th character of S

S [i,j] = substring of S starting at index i and ending at j

Definition: The character set CS (S [i,j]) := {S [k ] | i ≤ k ≤ j} is

the set of all characters occurring in the substring

S [i,j].

Example:

CS (S [2,5]) := {1,2,3}

1 2 3 4 5 6 7 8

S : 3 1 2 3 1 5 2 6

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Formal Model:

Given: Subset C Σ

Definition: (i, j) is a CS-location of C in S , iff CS (S [i,j]) = C

left-maximal = S [i-1] C

right-maximal = S [ j+1] C maximal = both left- and right-maximal

Example:S : 3 1 2 3 1 5 2 6

1 2 3 4 5 6 7 8

The pair (3,5) is a CS-location of the set C={1,2,3},

because CS (S [3,5]) = {1,2,3}, but it is not left-

maximal !

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Formal Model:

Given: Collection of k strings S* = (S 1 ,...,S k ) over alphabet Σ

Definition: C Σ is a common CS-factor of S* if and only if

C has a CS-location in each S l , 1 ≤ l ≤ k .

Example:

0 1 2 3 4 5 6 7

S 1 : 3 2 1 3 1 5 1 6

S 2 : 4 3 5 5 5 1 4 2 2

S 3: 7 5 1 5 3 6 5

1 2 3 4 5 6 7 8 9

common CS-factor: {1,3,5} => S 1: (3,7) ― S 2: (2,6) ― S 3: (2,5)

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Problem Formulation:

A common CS-factor of k strings represents a gene cluster thatoccurs in each of the k genomes.

Given a collection of k strings S*:

Problem 1: Find all common CS-factors in S*.

Problem 2: For each common CS-factor find all its maximal

CS-locations in each of the strings.

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Overview:

• Introduction

• Formal Model

• Algorithms

- Simple Data Structure: Quadratic Space

- Saving Space

• Results

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Algorithm "Connecting Intervals" (CI)

• Algorithm CI solves Problem 1 and Problem 2 for two sequences

• Input: Two sequences of length up to n with characters drawn

from Σ = {1,...,m}, m ≤ 2n

• Output: Pairs of CS-locations of all common CS-factors

• Time & Space complexity: O(n²)

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Preprocessing

POS[1] = 2,5

POS[2] = 3,7

POS[3] = 1,4

POS[4] = empty

POS[5] = 6

POS[6] = 8

1 2 3 4 5 6 7 81 1 2 3 3 3 4 4 52 1 2 3 3 4 4 53 1 2 3 4 4 54 1 2 3 4 5

5 1 2 3 46 1 2 37 1 28 1

NUM(i, j) : i j

POS[c] holds all positions where character c occurs in S 1.

NUM(i, j) counts the number of different characters in S 1[i, j].

Compute two tables for S 1= (3,1,2,3,1,5,2,6)

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Algorithm CI

Algorithm: While reading S 2, mark in S 1 the observed character

and track maximal intervals of marked characters

S 2 : 4 3 5 5 5 1 4 2 21 2 3 4 5 6 7 8

S 1 : 3 1 2 3 1 5 2 6

ji

POS[1] = 2,5

POS[2] = 3,7POS[3] = 1,4

POS[4] = empty

POS[5] = 6

POS[6] = 8

1 2 3 4 5 6 7 8

1 1 2 3 3 3 4 4 52 1 2 3 3 4 4 53 1 2 3 4 4 54 1 2 3 4 55 1 2 3 46 1 2 3

7 1 28 1

NUM(i, j) : i j

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1 2 3 4 5 6 7 8

1 1 2 3 3 3 4 4 52 1 2 3 3 4 4 53 1 2 3 4 4 54 1 2 3 4 55 1 2 3 46 1 2 3

7 1 28 1


Algorithm CI



S 2 : 4 3 5 5 5 1 4 2 21 2 3 4 5 6 7 8

S 1 : 3 1 2 3 1 5 2 6

ji

NUM(i, j) : i j

POS[1] = 2,5

POS[2] = 3,7POS[3] = 1,4

POS[4] = empty

POS[5] = 6

POS[6] = 8

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1 2 3 4 5 6 7 8

1 1 2 3 3 3 4 4 52 1 2 3 3 4 4 53 1 2 3 4 4 54 1 2 3 4 55 1 2 3 46 1 2 3

7 1 28 1


Algorithm CI



1 2 3 4 5 6 7 8

S 1 : 3 1 2 3 1 5 2 6

POS[1] = 2,5

POS[2] = 3,7POS[3] = 1,4

POS[4] = empty

POS[5] = 6

POS[6] = 8

NUM(i, j) : i j

Output: ((2,2)-(1,1)) ((2,2)-(4,4))

i

S 2 : 4 3 5 5 5 1 4 2 2

j

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1 2 3 4 5 6 7 81 1 2 3 3 3 4 4 52 1 2 3 3 4 4 53 1 2 3 4 4 54 1 2 3 4 55 1 2 3 46 1 2 3

7 1 28 1


Algorithm CI



1 2 3 4 5 6 7 8

S 1 : 3 1 2 3 1 5 2 6

POS[1] = 2,5

POS[2] = 3,7POS[3] = 1,4

POS[4] = empty

POS[5] = 6

POS[6] = 8

NUM(i, j) : i j

Output: ((2,2)-(1,1)) ((2,2)-(4,4))

i

S 2 : 4 3 5 5 5 1 4 2 2

j

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1 2 3 4 5 6 7 81 1 2 3 3 3 4 4 52 1 2 3 3 4 4 53 1 2 3 4 4 54 1 2 3 4 55 1 2 3 46 1 2 3

7 1 28 1


Algorithm CI



1 2 3 4 5 6 7 8

S 1 : 3 1 2 3 1 5 2 6

POS[1] = 2,5

POS[2] = 3,7POS[3] = 1,4

POS[4] = empty

POS[5] = 6

POS[6] = 8

NUM(i, j) : i j

Output: ((2,2)-(1,1)) ((2,2)-(4,4)) ((1,5)-(4,6))

i

S 2 : 4 3 5 5 5 1 4 2 2

j

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1 2 3 4 5 6 7 81 1 2 3 3 3 4 4 52 1 2 3 3 4 4 53 1 2 3 4 4 54 1 2 3 4 55 1 2 3 46 1 2 3

7 1 28 1


Algorithm CI



1 2 3 4 5 6 7 8S 1 : 3 1 2 3 1 5 2 6

POS[1] = 2,5

POS[2] = 3,7POS[3] = 1,4

POS[4] = empty

POS[5] = 6

POS[6] = 8

NUM(i, j) : i j

i

S 2 : 4 3 5 5 5 1 4 2 2

j

(i,j) not left-maximal !

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1. for i = 1,...,|S 2| do

2. j = i

3. while j < |S 2| and (i,j) is maximal do

4. if (c = S 2[ j]) is seen the first time5. for each entry in POS(c) do

6. mark and track

7. end for

8. end if

9. j = j + 110. end while

11. end for

Time Complexity

Algorithm CI finds all common CS-factors of S 1 and S 2 in O(n²) time.

POS[1] = 1,4

POS[2] = 2,6

POS[3] = 0,3

POS[4] = emptyPOS[5] = 5

POS[6] = 7

S 2 : 4 3 5 5 5 1 4 2 2

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Multiple Genomes

Goal : Find all common CS-factors of a collection S*=(S 1 ,S 2 ,...,S k )

Algorithm :

1. Apply Algorithm CI to all pairs (S 1,S l ), 2 ≤ l ≤ k

2. Output only the common CS-factor detected in all pairs

Time complexity : O(kn²)

Space complexity : O(kn²) with redundant output, O(n²) otherwise

Further extension : Find all common CS-factors appearing in at

least k' of k strings of S*

Time complexity : O(k( 1+k-k')n²)

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Saving Space

• Due to the storage of the table NUM , Algorithm CI requiresquadratic space.

• An algorithm presented by Didier, WABI 2003, detects all common

CS-factors of two sequences in O(n² log n) time and linear space

• In a modified version, replacing a binary search by a constant time

Range Maximum Query, it is possible to reduce the time complexity

to O(n²) staying still linear in space.

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Overview:

• Introduction

- Comparative genomics

- Common Intervals and Gene Clusters

• Formal Model

• Algorithms- Simple Data Structure: Quadratic Space

- Saving Space

• Results

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Results on real data

• Data set:

- 43 bacterial genome sequences from NCBI

- All classified in the "Clusters of Orthologous Groups of Proteins"

database (COG)

- Genes are identified by their COG number

- Computation time: approx. 5 -10 minutes on a standard PC

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Results on real data (k'= 2) all 43 genomes

cluster size ≥ 3

without closely related genomes (k = 32)

cluster size ≥ 2

cluster size ≥ 3

cluster size ≥ 2

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