sequence alignment algorithm a sub-quadraticmichaluz/seminar/lecture3.pdf · sequence alignment...

1

A Sub-quadratic

Sequence Alignment Algorithm for Unrestricted Scoring Matrices

Maxime Crochemore

Michal Ziv Ukelson

Gad M. Landau

2

A = c t a c g a g a c

B = a a c g a c g a t

- a c g t

- -1 -1 -1 -1

a -1 1 -1 -1 -1

c -1 -1 1 -1 -1

g -1 -1 -1 1 -1

t -1 -1 -1 -1 1

The Sequence Alignment Problem

Compare two strings A and B and measure their similarity

by finding the optimal alignment between them.

The alignment is classically based on the transformation

of one sequence into the other, via operations of substitutions,

insertions, and deletions (indels).

The Scoring Matrix

3





Global Alignment.

Local Alignment.

Two Sequence Alignment Problems

4





Global Alignment.

Local Alignment.

- a c g t

- -1 -1 -1 -1

a -1 1 -1 -1 -1

c -1 -1 1 -1 -1

g -1 -1 -1 1 -1

t -1 -1 -1 -1 1

The Scoring Matrix


5





Global Alignment.

Local Alignment.

- a c g t

- -1 -1 -1 -1

a -1 1 -1 -1 -1

c -1 -1 1 -1 -1

g -1 -1 -1 1 -1

t -1 -1 -1 -1 1

The Scoring Matrix


Value: 2

6





Global Alignment.

Local Alignment.

- a c g t

- -1 -1 -1 -1

a -1 1 -1 -1 -1

c -1 -1 1 -1 -1

g -1 -1 -1 1 -1

t --1 -1 -1 -1 1

The Scoring Matrix


Value:

Value: 2

5

8

- a c g t

- -1 -1 -1 -1

a -1 1 -1 -1 -1

c -1 -1 1 -1 -1

g -1 -1 -1 1 -1

t -1 -1 -1 -1 1

The Scoring Matrix

The O(n ) time, Classical Dynamic Programming Algorithm2

c

a

c

t

a a c g a c g a0

1

1 2 3 4 5 6 7 8

2

3

4

a

g

a

g5

6

7

c

|B|= n

|A|= n

t9

9

8

0 1 2 3 4 5 6 8 7 9

The Alignment Graph

9

Computing the Optimal Global Alignment Value

c

a

c

t

a a c g a c g a0

1

1 2 3 4 5 6 7 8

2

3

4

a

g

a

g5

6

7

c

|B|= n

|A|= n

t9

9

8

0 1 2 3 4 5 6 8 7 9

Classical Dynamic Programming: O(n )

Score of = 1

Score of = -1

2

10

Computing an Optimal Local Alignment Value

c

a

c

t

a a c g a c g a0

1

1 2 3 4 5 6 7 8

2

3

4

a

g

a

g5

6

7

c

|B|= n

|A|= n

t9

9

8

0 1 2 3 4 5 6 8 7 9

Classical Dynamic Programming: O(n )

Score of = 1

Score of = -1

2

11

The O(n ) time, Classical Dynamic Programming Algorithm2

c

a

c

t

a a c g a c g a0

1

1 2 3 4 5 6 7 8

2

3

4

a

g

a

g5

6

7

c

|B|= n

|A|

= n

t9

9

8

0 1 2 3 4 5 6 8 7 9

The Alignment Graph

I1

I2 I3

O

O = max(I + edge[I ,O])x

x = 1

3

x

Can the quadratic complexity of the optimal alignment value

computation be reduced without relaxing the problem?

13

Previous Results [Masek and Paterson 1980]

- An O(n ) / log n time global alignment algorithm.

- Constant size alphabet.

- Scoring Matrix values restricted to rational numbers.

Can the quadratic complexity of the optimal alignment value

computation be reduced without relaxing the problem?

2

Open Problem [Masek and Paterson 1980]

Can a better algorithm be found for the constant alphabet case,

which does not restrict the scoring matrix values?

14

LZ78 Parsing: Each phrase is the longest matching phrase

seen previously, plus one character.


Acceleration by Text Compression:

Compress the sequences in order to speed up the alignment process.

Trie for B

0

15






(0, a)

1

Trie for B

0a

1

16






(0, a) (1, c)

1 2

Trie for B

0a

1c

2

17






(0, a) (1, c) (0, g)

1 2 3

Trie for B

0a

1c

2

g

3

18






(0, a) (1, c) (0, g) (2, g)

1 2 3 4

Trie for B

0a

1c

2

g

3

g

4

19






(0, a) (1, c) (0, g) (2, g) (1, t)

1 2 3 4 5

Trie for B

0a

1c

2

g

3

g

4

5

t

Ziv and Lempel

21




Theorem 1 .[Lempel and Ziv 1976]

Given a sequence S of size n over a constant alphabet.

The maximal number of phrases obtained by any scheme which parses

S into distinct phrases is O(n / log n).



(0, a) (1, c) (0, g) (2, g) (1, t)

1 2 3 4 5

22




Theorem 1 .[Lempel and Ziv 1976]


The maximal number of phrases obtained by any scheme which parses

S into distinct phrases is O(n / log n).



(0, a) (1, c) (0, g) (2, g) (1, t)

1 2 3 4 5

Theorem 2.[Ziv and Lempel 1978]


The number of phrases obtained by LZ78 parsing of S

is O(h n / log n), where h <=1.

For most texts, h is the entropy of the text, which is a measure of how "compressible" the text is.

24

ctacg

ag

a

a a c g a c g

c

O(n ) vertices2

a t

25

ctacg

ag

a

a a c g a c g

c

O(n ) vertices2

a t

26

ctacg

ag

a

a a c g a c g

c

O(n ) vertices2

a t

27

ctacg

ag

a

a a c g a c g

c

O(n ) vertices2 O(n ) vertices2

a t

28

ctacg

ag

a

a a c g a c g

c

O(n ) vertices2

a t

29

ctacg

ag

a

a a c g a c g

c

O(n ) vertices2

a t

30

ctacg

ag

a

a ta a c g a c g

c

O(h n / log n) rows of n vertices +

O(h n / log n) columns of n vertices

ctacg

ag

a

a a c g a c g

c

O(n ) vertices2

a t

31

ctacg

ag

a

a ta a c g a c g

c



Our Results:

O(hn / log n) algorithm for Computing the

Optimal Global Alignment Value

and

Optimal Local Alignment Value.

2

Reminder: h <=1,

scoring matrix entries

may be arbitrary real

numbers.

32

The work for each block.

a a c g a c gctacg

ag

a

a t

5/4

1

1 2 3 4 5

2

3

4

5

6

g

a

ga c

I4 I5 I6I3

I2

I1

O2 O3 O4O1

O6

O5

The work for each block is done in O(t) time.

c

t = |I| = |O| = 6G

33

g

a

ga c

I4 I5 I6I3

I2

I1

O4

a a c g a c gctacg

ag

a

a t

5/4

1

1 2 3 4 5

2

3

4

5

6

c

O4 = the weight of an optimal path from vertex (0,0) to vertex 4 of O.

O4

Computing the score for Output Border Vertex O4

x

34

g

a

ga c

I4 I5 I6I3

I2

I1

O4

a a c g a c gctacg

ag

a

a t

5/4

1

1 2 3 4 5

2

3

4

5

6

c

Ix = the weight of an optimal path from vertex (0,0) to vertex x of I.

x

O4 = the weight of an optimal path from vertex (0,0) to vertex 4 of O.

DIST[x,y] = the weight of an optimal path from vertex Ix to vertex O4 .

Computing the score for Output Border Vertex O 4

O4

Oy

35

g

a

ga c

I4 I5 I6I3

I2

I1

O4

a a c g a c gctacg

ag

a

a t1

1 2 3 4 5

2

3

4

5

6

c

I1 DIST[1,4] I1+DIST[1,4]







O4

36

g

a

ga c

I4 I5 I6I3

I2

I1

O4

a a c g a c gctacg

ag

a

a t1

1 2 3 4 5

2

3

4

5

6

c








O = max(I + DIST[x,3])x 4 x = 0

6

O4

37

g

a

ga c

I4 I5 I6I3

I2

I1

O4

Standard, single-cell DP

I1

I2 I3

O

O = max(I + edge[I ,O])x

x = 1

3

x

O = max(I + DIST[x,3])x 4 x = 0

6

New, extended-cell DP

38

g

a

ga c

3

I3

2

I4

1

I5

3

I6

2

I2

1

I1

O4

?

O = max(I + DIST[x,3])x 4 x = 0

6

Score of = 1

Computing the score for Output Border Vertex O 4

Score of = -1

I[*] + DIST[*,4] = OUT[*,4]

1 -3 -2

2 -1 1

3 1 4

2 0 2

1 0 1

3 -2 1

39

g

a

ga c

3

I3

2

I4

1

I5

3

I6

2

I2

1

I1

O4

?

Score of = 1


Score of = -1

I[*] + DIST[*,4] = OUT[*,4]

1 -3 -2

2 -1 1

3 1 4

2 0 2

1 0 1

3 -2 1

O = max(I + DIST[x,3])x 4 x = 0

6

40

g

a

ga c

3

I3

2

I4

1

I5

3

I6

2

I2

1

I1

O4

4

Score of = 1


Score of = -1

I[*] + DIST[*,4] = OUT[*,4]

1 -3 -2

2 -1 1

3 1 4

2 0 2

1 0 1

3 -2 1

O = max(I + DIST[x,3])x 4 x = 0

6

41

I1 = 1

I2 = 2

I3 = 3

I4 =

I5 = 1

I6 =

OUT[x,j] = Ix + DIST[x,j]

Input I\ DIST Matrix

Output vector O

g

a

ga c

3

I3

2

I4

1

I5

3

I6

2

I2

1

I1

O2 O4O3

O5

O6

O1

41 3 3

2

3

42

I1 = 1

I2 = 2

I3 = 3

I4 =

I5 = 1

I6 =



Output vector O

g

a

ga c

3

I3

2

I4

1

I5

3

I6

2

I2

1

I1

O2 O4O3

O5

O6

O1

41 3 3

2

3

Output Vector O values

are set to

OUT Matrix Column Maxima

43

I1 = 1

I2 = 2

I3 = 3

I4 =

I5 = 1

I6 =



Output vector O

The Main Challenges

44

I1 = 1

I2 = 2

I3 = 3

I4 =

I5 = 1

I6 =



Output vector O

How to compute the

column maxima

of OUT in O(t) time ?

(Utilize the

Total Monotonicity

Property of OUT).

The Main Challenges

45

I1 = 1

I2 = 2

I3 = 3

I4 =

I5 = 1

I6 =



Output vector O

How to compute the

column maxima


(Utilize the

Total Monotonicity

Property of OUT).

How to obtain the DIST

for G in O(t) time ?

(Take advantage of the

incremental nature of

LZ78 parsing).

The Main Challenges

46

a

(0,0)

I

O

b

c d

The Total Monotonicity Property [Aggarwal et al

1987].

For any a < b and c < d

OUT[b,c] >= OUT[a,c] ===> OUT[b,d] >= OUT[a,d]

47

a

(0,0)

I

O

b

c d

The Total Monotonicity Property [Aggarwal et al 1987].



48

a

(0,0)

I

O

b

c d




49

a

(0,0)

I

O

b

c d




50

a

(0,0)

I

O

b

c d




51

a

Y

Z

X

W

(0,0)

I

O

b

c dProof:

OUT[a,c] is optimal, and therefore OUT[a,c] >= X + Z

OUT[b,d] is optimal, and therefore OUT[b,d] >= Y + W

OUT[b,c] = Y + Z >= OUT[a,c] >= X+Z ===> Y+Z >= X+ Z==> Y >= X

Therefore, OUT[b,d] >= Y + W >= X+W = OUT[a,d]

The Total Monotonicity property of OUTFor any a < b and c < d


52

a

b

d

OUT Matrix

How does Total Monotonicity affect Column Maxima behavior?

For all a <b and c < d, OUT[a,c] <= OUT[b,c] OUT[a,d] <= OUT[b,d]

Column maxima row indices are monotonically non-decreasing.

SMAWK Matrix Searching[Aggarwal et-al 87] .

The t column maxima of a Totally Monotone array

can be computed in O(t) time, by querying only O(t) elements.

53

OUT Matrix

The Rectangle Problem

g

a

ga c

54

Complementing the undefined OUT entries

(without introducing new column maxima)

1. Upper Right Triangle. All values are set to .

2. Lower Left Triangle.

Let k denote th maximal absolute value of a score in

the scoring matrix .

OUT[i,j] in the lower left triangle will be set to -(n+i+1)*k.

For all a <b and c < d, OUT[a,c] <= OUT[b,c] OUT[a,d] <= OUT[b,d]

55

Complementing the undefined OUT entries,

without changing its Total Monotonicity property,

and without introducing new column maxima.

OUT Matrix

56

I1 = 1

I2 = 2

I3 = 3

I4 =

I5 = 1

I6 =



Output vector O

How to compute the

column maxima


(Utilize the

Total Monotonicity

Property of OUT).

The Main Challenges

57

I1 = 1

I2 = 2

I3 = 3

I4 =

I5 = 1

I6 =



Output vector O

How to compute the

column maxima


(Utilize the

Total Monotonicity

Property of OUT).





LZ78 parsing).

The Main Challenges

58 G

g

a

ga c

ctacg

ag

a

a t

5/4

1

1 2 3 4 5

2

3

4

5

6

a a c g a c g

c

Utilizing the incremental nature of LZ78 parsing

for efficient DIST construction.

59 left prefix block of G G

g

a

ga c

g

a

a c

ctacg

ag

a

a t

5/4

1

1 2 3 4 5

2

3

4

5

6

a a c g a c g

c

5/2

4 = (2, g)

5/4

55

2

60 G

g

a

ga c

a

3/4

ctacg

ag

a

a t

5/4

1

1 2 3 4 5

2

3

4

5

6

a a c g a c g

c

top prefix block of G

ga c

61

3/4

5/2

3/2

left

prefix

(5/2)

diagonal

prefix

(3,2)

top

prefix

(3,4)

G (5,4)

g

a

ga cga c

g

a

a c

a

a c

a

ctacg

ag

a

a t

5/4

1

1 2 3 4 5

2

3

4

5

6

a a c g a c g

c

62

3/43/2

g

a

ga c ga c

g

a

a c

a

ctacg

ag

a

a t

5/4

1

1 2 3 4 5

2

3

4

5

6

a a c g a c g

c

-3

-1

1

0

0

-2

Only one new DIST column needs to be

computed for each

block, and this DIST

column is computed in

O(t) time.diagonal

prefix

a

a c

5/2

left prefix

left prefix

top prefix

I3

O4

63

3/4

5/2

3/2

left prefix (5,2)

diagonal

prefix (3,2)

top

prefix(3,4)

block G (5,4)

g

a

ga cga c

g

a

a c

a

a c

a

ctacg

ag

a

a t

5/4

1

1 2 3 4 5

2

3

4

5

6

a a c g a c g

c

Accessing a Prefix Block in Constant time.

a c

Trie for A

0

13

5

2

t

4

gg

a

c

g

g

Trie for B

0

31

2

46

5c

t

Presentation

• R89922024 蘇展弘

• B86202049 葉恆青

• R90725054 呂育恩

• R90922001 張文亮

• R90922091 游騰楷

Trie for A

0

31

2

54

g

c

ta

g

Trie for B

0

1 3

2

4

g

ga

c

DIST(5,4)

-3

-1

1

0

0

-2

-2

-2

0

-2

-2

-1

-1

0

-2

0

-1

-2

-2

-1

-2

-1

-1

-3

-2

-1

0

0 1 2 3 4

a a c g a c g a

1 c

2 t

3 a

4c

g

5a

g

a

Data Strucure

Trie for A

0

31

2

54

g

c

ta

g

Trie for B

0

1 3

2

4

g

ga

c

DIST(5,4)

-3

-1

1

0

0

-2

-2

-2

0

-2

-2

-1

-1

0

-2

0

-1

-2

-2

-1

-2

-1

-1

-3

-2

-1

0

0 1 2 3 4

a a c g a c g a

1 c

2 t

3 a

4c

g

5a

g

a

Construction

Time and Space Complexity

• 作new column

Trie for A

0

31

2

54

g

c

ta

g

Trie for B

0

1 3

2

4

g

ga

c

DIST(5,4)DIST(5,4)

-2

0

0

1

-1

-3

-2

0

0

1

-1

-3

-2

-2

0

-2

-2

-2

-2

0

-2

-2

-2

0

-1

-1

-2

0

-1

-1

-2

-1

0

-2

-1

0

-1

-1

-2

-1

-2

-1

-1

-2

-1

-2

0

-1

-2

-3

0

-1

-2

-3

a

a

g5

c

g4

a3

t2

c1

aa c gga ca

43210

a

a

g5

c

g4

a3

t2

c1

aa c gga ca

43210

Data Strucure

• 作DIST vector ( 即找出該DIST matrix所有的

column)

• 用SMAWK從這個DIST(加上input)算出

output maxima。

O ( t )

68

ctacg

ag

a

a ta a c g a c g

c



ctacg

ag

a

a a c g a c g

c

O(n ) vertices2

a t

69

ctacg

ag

a

a ta a c g a c g

c



Our Results:

O(hn / log n) algorithm for Computing the

Optimal Global Alignment Value

and

Optimal Local Alignment Value.

2

Reminder: h <=1,

scoring matrix entries

may be arbitrary real

numbers.

70

I1 = 1

I2 = 2

I3 = 3

I4 =

I5 = 1

I6 =



Output vector O

How to compute the

column maxima


(Utilize the

Total Monotonicity

Property of OUT).





LZ78 parsing).

The Main Challenges

71

Summary of Results:

Global Alignment Problem.

-An O(hn / log n) time and space complexity algorithm for computing

the optimal global alignment value.

-After the optimal value has been computed, an optimal alignment trace

can be recovered in time linear with its size.

Local Alignment Problem.

-An O(hn / log n) time and space complexity algorithm for computing

the optimal local alignment value.

-After the optimal value has been computed, given a vertex whose

score is maximal, an optimal alignment trace ending in the vertex

can be recovered in time linear with its size.

2

2

72

Open Problems:

We showed an O(hn / log n) time and space complexity

algorithm for computing the optimal global and local

alignment values of two strings.

In the paper we show how to reduce the space complexity

to O(h n / log n) .

Can the space requirement of the algorithm be further reduced,

without impairing its sub-quadratic time complexity?

2

2 2 2

73

Thank You !

sequence alignment algorithm a sub-quadraticmichaluz/seminar/lecture3.pdf · sequence alignment...

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