"quadratic time algorithms for finding common intervals in two and more sequences" by t....

"Quadratic time algorithms for finding common intervals in two

and more sequences"

by T. Schmidt and J. Stoye, Proc. 15th Annual Symposium on Combinatorial Pattern Matching, Lecture Notes in Computer

Science 3109, pp. 347-358 (2004).

Presented by Gangman Yi

Overview

Introduction Formal Model Algorithms Assignment

Gene Order & Function in Bacteria:

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

?

Formalization of Gene Clusters:

Genomes: permutations π1, π2 ,…, πk Genes: numbers 1,…,n



1 2 3 4 5 6 7 8

π1

π2

π3

π4



1 2 3 4 5 6 7 8

π1

π2

π3

π4

8 7 6 4 5 2 1 3

3 1 2 5 8 7 6 4

6 7 4 2 1 3 8 5


Genomes: permutations π1, π2 ,…, πk Genes: numbers 1,…,n Gene cluster: common interval subset of numbers occurring

contiguously in all permutations)

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

π1

π2

π3

π4


Genomes: permutations π1, π2 ,…, πk Genes: numbers 1,…,n Gene cluster: common interval subset of numbers occurring

contiguously in all permutations)

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 ≥ 2 permutations in O(kn+|output|) time.

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

π1

π2

π3

7 ?



1 2 3 4 5 6 7 8

π1

π2

π3

? 7



1 2 3 4 5 6 7 8

π1

π2

π3

3 1 2 ? ?



1 2 3 4 5 6 7 8

π1

π2

π3

3 ? 2 1 ?


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

S1

S2

S3

3 1 2 4 8 7 6 1 2

8 7 6 7 5 4 2 1 3

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

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:

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 !

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

Formal Model:

Given: Collection of k strings S* = (S1,...,Sk) over alphabet Σ

Definition: C Σ is a common CS-factor of S* if and only if C has a CS-location in each Sl , 1 ≤ l ≤ k.

Example:

common CS-factor: {1,3,5} => S1: (3,7) ― S2: (2,6) ― S3: (2,5)

0 1 2 3 4 5 6 7

S1 : 3 2 1 3 1 5 1 6

S2 : 4 3 5 5 5 1 4 2 2

S3 : 7 5 1 5 3 6 5 1 2 3 4 5 6 7 8 9

Problem Formulation:

A common CS-factor of k strings represents a gene cluster that occurs 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.

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)

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 37 1 28 1

POS[1] = 2,5POS[2] = 3,7POS[3] = 1,4POS[4] = emptyPOS[5] = 6POS[6] = 8

NUM(i,j) : i j

POS[c] holds all positions where character c occurs in S1.

NUM(i,j) counts the number of unique characters in S1[i,j].

Compute two tables for S1= (3,1,2,3,1,5,2,6)

Preprocessing

Algorithm: While reading S2, mark in S1 the observed character and track maximal intervals of marked characters

S2 : 4 3 5 5 5 1 4 2 21 2 3 4 5 6 7 8

S1 : 3 1 2 3 1 5 2 6

ji


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 37 1 28 1

NUM(i,j) :i

j

Algorithm CI

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 37 1 28 1


S2 : 4 3 5 5 5 1 4 2 21 2 3 4 5 6 7 8

S1 : 3 1 2 3 1 5 2 6

ji

NUM(i,j) :i

jPOS[1] = 2,5POS[2] = 3,7POS[3] = 1,4POS[4] = emptyPOS[5] = 6POS[6] = 8

Algorithm CI

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 37 1 28 1


S2 : 4 3 5 5 5 1 4 2 21 2 3 4 5 6 7 8

S1 : 3 1 2 3 1 5 2 6

ji

NUM(i,j) :i

j

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


Algorithm CI

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 37 1 28 1


1 2 3 4 5 6 7 8

S1 : 3 1 2 3 1 5 2 6


NUM(i,j) :i

j

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

i

S2 : 4 3 5 5 5 1 4 2 2

j

Algorithm CI

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 37 1 28 1


1 2 3 4 5 6 7 8

S1 : 3 1 2 3 1 5 2 6


NUM(i,j) :i

j

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

i

S2 : 4 3 5 5 5 1 4 2 2

j

Algorithm CI

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 37 1 28 1


1 2 3 4 5 6 7 8

S1 : 3 1 2 3 1 5 2 6


NUM(i,j) :i

j

i

S2 : 4 3 5 5 5 1 4 2 2

j

Algorithm CI

Time Complexity

Algorithm CI finds all common CS-factors of S1 and S2 in O(n²) time.

1. for i = 1,...,|S2| do2. j = i3. while j < |S2| and (i,j) is maximal do4. if (c = S2[j]) is seen the first time 5. for each entry in POS(c) do6. mark and track7. end for8. end if9. j = j + 110. end while11. end for

Multiple GenomesGoal : Find all common CS-factors of a collection S*=(S1,S2,...,Sk)

Algorithm : Apply Algorithm CI to all pairs (S1,Sl), 2 ≤ l ≤ kOutput 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)

Saving space : Due to the storage of the table NUM, Algorithm CI requires quadratic space.

Assignment Make a clustering algorithm.

Each sequence S has n unique genes, but the same gene can be in the other sequences. The number of sequences are k. Maximum output size for the cluster has to be m, so each cluster can have at most m genes. Do not consider about the order of genes in each cluster.

S1

S2

S3

Sk

n

ABDCBCDAADCBBCAD

Max. size for the cluster, m = 4

Output Example

EFFEEFFE

Gangman Yi

Email : [email protected]

THANK YOU

"quadratic time algorithms for finding common intervals in two and more sequences" by t....

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