a superglue for string comparison

A superglue for string comparison

Alexander Tiskin

Department of Computer Science, University of Warwickhttp://go.warwick.ac.uk/alextiskin

Alexander Tiskin (Warwick) Semi-local LCS 1 / 164

http://go.warwick.ac.uk/alextiskin

1 Introduction

2 Matrix distance multiplication

3 Semi-local string comparison

4 The seaweed method

5 Periodic string comparison

6 Sparse string comparison

7 Compressed string comparison

8 Parallel string comparison

9 The transposition networkmethod

10 Beyond semi-locality


1 Introduction











Introduction

String matching: finding an exact pattern in a string

String comparison: finding similar patterns in two strings

global: whole string a vs whole string b

semi-local: whole string a vs substrings in b (approximate matching);prefixes in a vs suffixes in b

local: substrings in a vs substrings in b

We focus on semi-local comparison:

fundamentally important for both global and local comparison

exciting mathematical properties


IntroductionOverview

Standard approach to string comparison: dynamic programming

Our approach: the algebra of seaweed matrices/braids

Can be used either for dynamic programming, or for divide-and-conquer

Divide-and-conquer is more efficient for:

comparing dynamic strings (truncation, concatenation)

comparing compressed strings (e.g. LZ-compression)

comparing strings in parallel

The “conquer” step involves a magic “superglue” (efficient multiplicationof seaweed matrices/braids)


IntroductionTerminology and notation

The “squared paper” notation:

Integers {. . .− 2,−1, 0, 1, 2, . . .} x− = x − 12 x+ = x + 1

2

Half-integers{. . .− 3

2 ,−12 ,

12 ,

32 ,

52 , . . .

}={. . . (−2)+, (−1)+, 0+, 1+, 2+

}Planar dominance:

(i , j)� (i ′, j ′) iff i < i ′ and j < j ′ (the “above-left” relation)

(i , j) ≷ (i ′, j ′) iff i > i ′ and j < j ′ (the “below-left” relation)

where “above/below” follow matrix convention (not the Cartesian one!)



A permutation matrix is a 0/1 matrix with exactly one nonzero per rowand per column0 1 0

1 0 00 0 1



Given matrix D, its distribution matrix is made up of ≷-dominance sums:DΣ(i , j) =

∑ı>i ,<j D(i , j)

Given matrix E , its density matrix is made up of quadrangle differences:E�(ı, ) = E (ı−, +)− E (ı−, −)− E (ı+, +) + E (ı+, −)

where DΣ, E over integers; D, E� over half-integers0 1 01 0 00 0 1

Σ

=

0 1 2 30 1 1 20 0 0 10 0 0 0

0 1 2 30 1 1 20 0 0 10 0 0 0

�

=

0 1 01 0 00 0 1

(DΣ)� = D for all D

Matrix E is simple, if (E�)Σ = E ; equivalently, if it has all zeros in the leftcolumn and bottom row




∑ı>i ,<j D(i , j)


where DΣ, E over integers; D, E� over half-integers

0 1 01 0 00 0 1

Σ

=

0 1 2 30 1 1 20 0 0 10 0 0 0

0 1 2 30 1 1 20 0 0 10 0 0 0

�

=

0 1 01 0 00 0 1






∑ı>i ,<j D(i , j)


where DΣ, E over integers; D, E� over half-integers0 1 01 0 00 0 1

Σ

=

0 1 2 30 1 1 20 0 0 10 0 0 0

0 1 2 30 1 1 20 0 0 10 0 0 0

�

=

0 1 01 0 00 0 1





Matrix E is Monge, if E� is nonnegative

Intuition: boundary-to-boundary distances in a (weighted) planar graph

G. Monge (1746–1818)

G (planar)

i

j

0

1 2 3

4

0

1 2 3

4



Matrix E is Monge, if E� is nonnegative

Intuition: boundary-to-boundary distances in a (weighted) planar graph

G. Monge (1746–1818)

i

j

0

1 2 3

4

0

1 2 3

4

blue + blue ≤ red + red



Matrix E is unit-Monge, if E� is a permutation matrix

Intuition: boundary-to-boundary distances in a grid-like graph (inparticular, an alignment graph for a pair of strings)

Simple unit-Monge matrix (sometimes called “rank function”): PΣ, whereP is a permutation matrix

Seaweed matrix: P used as an implicit representation of PΣ0 1 01 0 00 0 1

Σ

=

0 1 2 30 1 1 20 0 0 10 0 0 0


IntroductionImplicit unit-Monge matrices

Efficient PΣ queries: range tree on nonzeros of P [Bentley: 1980]

binary search tree by i-coordinate

under every node, binary search tree by j-coordinate

••••−→ •

•••−→ •

•••

↓

••••−→ •

•••−→ •

•••

↓

••••−→ •

•••−→ •

•••


IntroductionImplicit unit-Monge matrices

Efficient PΣ queries: (contd.)

Every node of the range tree represents a canonical range (rectangularregion), and stores its nonzero count

Overall, ≤ n log n canonical ranges are non-empty

A PΣ query is equivalent to ≷-dominance counting: how many nonzerosare ≷-dominated by query point?

Answer: sum up nonzero counts in ≤ log2 n disjoint canonical ranges

Total size O(n log n), query time O(log2 n)

There are asymptotically more efficient (but less practical) data structures

Total size O(n), query time O( log n

log log n

)[JaJa+: 2004]

[Chan, Patrascu: 2010]


1 Introduction











Matrix distance multiplicationSeaweed braids

Distance semiring (or (min,+)-semiring, special case of tropical algebra):

addition ⊕ given by min, multiplication � given by +

Matrix �-multiplication

A� B = C C (i , k) =⊕

j

(A(i , j)� B(j , k)

)= minj

(A(i , j) + B(j , k)

)Intuition: gluing distances in a composition of planar graphs

i

G ′

j

G ′′

k



Matrix classes closed under �-multiplication (for given n):

general (integer, real) matrices ∼ general weighted graphs

Monge matrices ∼ planar weighted graphs

simple unit-Monge matrices ∼ grid-like graphs

Implicit �-multiplication: define P � Q = R as PΣ � QΣ = RΣ

The seaweed monoid Tn:

simple unit-Monge matrices under �equivalently, permutation (seaweed) matrices under �

Isomorphic to the 0-Hecke monoid of the symmetric group H0(Sn)



P � Q = R can be seen as combing of seaweed braids

P

�

Q

=

R

P

Q

= = R




P

�

Q

=

R

P

Q

=

= R




P

�

Q

=

R

P

Q

= =

R




P

�

Q

=

R

P

Q

= = R



The seaweed monoid Tn:

n! elements (permutations of size n)

n − 1 generators g1, g2, . . . , gn−1 (elementary crossings)

Idempotence:g2i = gi for all i =

Far commutativity:gigj = gjgi j − i > 1 · · · = · · ·

Braid relations:gigjgi = gjgigj j − i = 1 =



Special elements of the seaweed monoid Tn

Identity: 1� x = x

1 = =

• · · ·· • · ·· · • ·· · · •

Zero: 0� x = 0

0 = =

· · · •· · • ·· • · ·• · · ·



Seaweed monoid: g2i = gi ; far comm; braid

Related structures:

classical braid group: gig−1i = 1; far comm; braid

Sn (Coxeter presentation): g2i = 1; far comm; braid

positive braid monoid: far comm; braid

locally free idempotent monoid: g2i = gi ; far comm

nil-Hecke monoid: g2i = 0; far comm; braid

Generalisations:

0-Hecke monoid of a Coxeter group

Hecke–Kiselman monoid

J -trivial monoid



Computation in the seaweed monoid: a confluent rewriting system can beobtained by software (Semigroupe, GAP, Sage)

T3: 1, a = g1, b = g2; ab, ba; aba = 0

aa→ a bb → b bab → 0 aba→ 0

T4: 1, a = g1, b = g2, c = g3; ab, ac, ba, bc, cb, aba, abc, acb, bac,bcb, cba, abac , abcb, acba, bacb, bcba, abacb, abcba, bacba; abacba = 0

aa→ abb → b

ca→ accc → c

bab → abacbc → bcb

cbac → bcbaabacba→ 0

Easy to use as a “black box”, but not efficient


Matrix distance multiplicationSeaweed matrix multiplication

Seaweed matrix �-multiplication

Given permutation matrices P, Q, compute R, such that PΣ � QΣ = RΣ

(equivalently, P � Q = R)

Seaweed matrix �-multiplication: running time

type time

general O(n3) standard

O(n3(log log n)3

log2 n

)[Chan: 2007]

Monge O(n2) via [Aggarwal+: 1987]

implicit unit-Monge O(n1.5) [T: 2006]O(n log n) [T: 2010]



P

Q

R

?



Plo , Phi

Qlo , Qhi

Rlo + Rhi



Plo , Phi

Qlo , Qhi

Rlo + Rhi

R



P

Q

R



Seaweed matrix �-multiplication: Steady Ant algorithm

RΣ(i , k) = minj(PΣ(i , j) + QΣ(j , k)

)Divide-and-conquer on the range of j : two subproblems of size n/2

PΣlo � QΣ

lo = RΣlo PΣ

hi � QΣhi = RΣ

hi

Conquer: tracing a balanced trail from bottom-left to top-right R

Trail invariant: equal number of �-dominated nonzeros in PC ,hi and�-dominating nonzeros in PC ,lo

All such nonzeros are “errors” in the output, must be removed

Must compensate by placing nonzeros on the path, time O(n)

Overall time O(n log n)


Matrix distance multiplicationBruhat order

Bruhat order

Permutation A is lower (“more sorted”) than permutation B in the Bruhatorder (A � B), if B A by successive pairwise sorting (equivalently,A B by anti-sorting) of arbitrary pairs

Permutation matrices: P � Q, if Q P by successive 2× 2 submatrixsorting: ( 0 1

1 0 )→ ( 1 00 1 )

Plays an important role in group theory and algebraic geometry (inclusionorder of Schubert varieties)

Describes pivoting order in Gaussian elimination (matrix Bruhatdecomposition)



Bruhat comparability: running time

O(n2) [Ehresmann: 1934; Proctor: 1982; Grigoriev: 1982]O(n log n) [T: 2013]

O( n log n

log log n

)[Gawrychowski: NEW]



Ehresmann’s criterion (dot criterion, related to tableau criterion)

P � Q iff PΣ ≤ QΣ elementwise1 0 00 0 10 1 0

Σ

=

0 1 2 30 0 1 20 0 1 10 0 0 0

≤

0 1 2 30 1 2 20 0 1 10 0 0 0

=

0 0 11 0 00 1 0

Σ

1 0 00 0 10 1 0

Σ

=

0 1 2 30 0 1 20 0 1 10 0 0 0

6≤

0 1 2 30 1 1 20 0 0 10 0 0 0

=

0 1 01 0 00 0 1

Σ

Time O(n2)



Seaweed criterion

P � Q iff PR � Q = IdR , where PR = counterclockwise rotation of P

Intuition: permutations represented by seaweed braids

P � Q, iff

no pair of seaweeds is crossed in P, while the “corresponding” pair isuncrossed in Q

equivalently, no pair is uncrossed in PR , while the “corresponding”pair is uncrossed in Q

equivalently, PR � Q = IdR

Time O(n log n) by seaweed matrix multiplication


Matrix distance multiplicationBruhat order1 0 0

0 0 10 1 0

�0 0 1

1 0 00 1 0

1 0 0

0 0 10 1 0

R

�

0 0 11 0 00 1 0

=

0 1 00 0 11 0 0

�0 0 1

1 0 00 1 0

=

0 0 10 1 01 0 0

= IdR

P

Q

PR

Q

= IdR



0 0 10 1 0

�0 0 1

1 0 00 1 0

1 0 0

0 0 10 1 0

R

�

0 0 11 0 00 1 0

=

0 1 00 0 11 0 0

�0 0 1

1 0 00 1 0

=

0 0 10 1 01 0 0

= IdR

P Q

PR

Q

= IdR



0 0 10 1 0

�0 0 1

1 0 00 1 0

1 0 0

0 0 10 1 0

R

�

0 0 11 0 00 1 0

=

0 1 00 0 11 0 0

�0 0 1

1 0 00 1 0

=

0 0 10 1 01 0 0

= IdR

P Q

PR

Q

=

IdR



0 0 10 1 0

�0 0 1

1 0 00 1 0

1 0 0

0 0 10 1 0

R

�

0 0 11 0 00 1 0

=

0 1 00 0 11 0 0

�0 0 1

1 0 00 1 0

=

0 0 10 1 01 0 0

= IdR

P Q

PR

Q

= IdR



0 0 10 1 0

6�0 1 0

1 0 00 0 1

1 0 0

0 0 10 1 0

R

�

0 1 01 0 00 0 1

=

0 1 00 0 11 0 0

�0 1 0

1 0 00 0 1

=

0 1 00 0 11 0 0

6= IdR

P

Q

PR

Q

= 6= IdR



0 0 10 1 0

6�0 1 0

1 0 00 0 1

1 0 0

0 0 10 1 0

R

�

0 1 01 0 00 0 1

=

0 1 00 0 11 0 0

�0 1 0

1 0 00 0 1

=

0 1 00 0 11 0 0

6= IdR

P Q

PR

Q

= 6= IdR



0 0 10 1 0

6�0 1 0

1 0 00 0 1

1 0 0

0 0 10 1 0

R

�

0 1 01 0 00 0 1

=

0 1 00 0 11 0 0

�0 1 0

1 0 00 0 1

=

0 1 00 0 11 0 0

6= IdR

P Q

PR

Q

=

6= IdR



0 0 10 1 0

6�0 1 0

1 0 00 0 1

1 0 0

0 0 10 1 0

R

�

0 1 01 0 00 0 1

=

0 1 00 0 11 0 0

�0 1 0

1 0 00 0 1

=

0 1 00 0 11 0 0

6= IdR

P Q

PR

Q

= 6= IdR



Alternative solution: clever implementation of Ehresmann’s criterion[Gawrychowski: 2013]

The online partial sums problem: maintain array X [1 : n], subject to

update(k,∆): X [k]← X [k] + ∆

prefixsum(k): return∑

1≤i≤k X [i ]

Query time:

Θ(log n) in semigroup or group model

Θ( log n

log log n

)in RAM model on integers [Patrascu, Demaine: 2004]

Gives Bruhat comparability in time O( n log n

log log n

)in RAM model

Open problem: seaweed multiplication in time O( n log n

log log n

)?


1 Introduction











Semi-local string comparisonSemi-local LCS and edit distance

Consider strings (= sequences) over an alphabet of size σ

Contiguous substrings vs not necessarily contiguous subsequences

Special cases of substring: prefix, suffix

Notation: strings a, b of length m, n respectively

Assume where necessary: m ≤ n; m, n reasonably close

The longest common subsequence (LCS) score:

length of longest string that is a subsequence of both a and b

equivalently, alignment score, where score(match) = 1 andscore(mismatch) = 0

In biological terms, “loss-free alignment” (unlike efficient but “lossy”BLAST)



The LCS problem

Give the LCS score for a vs b

LCS: running time

O(mn) [Wagner, Fischer: 1974]O(

mnlog2 n

)σ = O(1) [Masek, Paterson: 1980]

[Crochemore+: 2003]

O(mn(log log n)2

log2 n

)[Paterson, Dancık: 1994]

[Bille, Farach-Colton: 2008]

Running time varies depending on the RAM model version

We assume word-RAM with word size log n (where it matters)



LCS computation by dynamic programming

lcs(a, ∅) = 0

lcs(∅, b) = 0lcs(aα, bβ) =

{max(lcs(aα, b), lcs(a, bβ)) if α 6= β

lcs(a, b) + 1 if α = β

∗ D E F I N E

∗ 0

0 0 0 0 0 0

D 0

1 1 1 1 1 1

E 0

1 2 2 2 2 2

S 0

1 2 2 2 2 2

I 0

1 2 2 3 3 3

G 0

1 2 2 3 3 3

N 0

1 2 2 3 4 4

lcs(“DEFINE”, “DESIGN”) = 4

LCS(a, b) can be traced back throughthe dynamic programming table at noextra asymptotic time cost




lcs(a, ∅) = 0




∗ D E F I N E

∗ 0 0 0 0 0 0 0D 0

1 1 1 1 1 1

E 0

1 2 2 2 2 2

S 0

1 2 2 2 2 2

I 0

1 2 2 3 3 3

G 0

1 2 2 3 3 3

N 0

1 2 2 3 4 4






lcs(a, ∅) = 0




∗ D E F I N E

∗ 0 0 0 0 0 0 0D 0 1

1 1 1 1 1

E 0

1 2 2 2 2 2

S 0

1 2 2 2 2 2

I 0

1 2 2 3 3 3

G 0

1 2 2 3 3 3

N 0

1 2 2 3 4 4






lcs(a, ∅) = 0




∗ D E F I N E

∗ 0 0 0 0 0 0 0D 0 1 1

1 1 1 1

E 0

1 2 2 2 2 2

S 0

1 2 2 2 2 2

I 0

1 2 2 3 3 3

G 0

1 2 2 3 3 3

N 0

1 2 2 3 4 4






lcs(a, ∅) = 0




∗ D E F I N E

∗ 0 0 0 0 0 0 0D 0 1 1 1 1 1 1E 0

1 2 2 2 2 2

S 0

1 2 2 2 2 2

I 0

1 2 2 3 3 3

G 0

1 2 2 3 3 3

N 0

1 2 2 3 4 4






lcs(a, ∅) = 0




∗ D E F I N E

∗ 0 0 0 0 0 0 0D 0 1 1 1 1 1 1E 0 1

2 2 2 2 2

S 0

1 2 2 2 2 2

I 0

1 2 2 3 3 3

G 0

1 2 2 3 3 3

N 0

1 2 2 3 4 4






lcs(a, ∅) = 0




∗ D E F I N E

∗ 0 0 0 0 0 0 0D 0 1 1 1 1 1 1E 0 1 2

2 2 2 2

S 0

1 2 2 2 2 2

I 0

1 2 2 3 3 3

G 0

1 2 2 3 3 3

N 0

1 2 2 3 4 4






lcs(a, ∅) = 0




∗ D E F I N E

∗ 0 0 0 0 0 0 0D 0 1 1 1 1 1 1E 0 1 2 2 2 2 2S 0

1 2 2 2 2 2

I 0

1 2 2 3 3 3

G 0

1 2 2 3 3 3

N 0

1 2 2 3 4 4






lcs(a, ∅) = 0




∗ D E F I N E

∗ 0 0 0 0 0 0 0D 0 1 1 1 1 1 1E 0 1 2 2 2 2 2S 0 1 2 2 2 2 2I 0

1 2 2 3 3 3

G 0

1 2 2 3 3 3

N 0

1 2 2 3 4 4






lcs(a, ∅) = 0




∗ D E F I N E

∗ 0 0 0 0 0 0 0D 0 1 1 1 1 1 1E 0 1 2 2 2 2 2S 0 1 2 2 2 2 2I 0 1 2 2 3 3 3G 0

1 2 2 3 3 3

N 0

1 2 2 3 4 4






lcs(a, ∅) = 0




∗ D E F I N E

∗ 0 0 0 0 0 0 0D 0 1 1 1 1 1 1E 0 1 2 2 2 2 2S 0 1 2 2 2 2 2I 0 1 2 2 3 3 3G 0 1 2 2 3 3 3N 0 1 2 2 3 4 4





LCS on the alignment graph: directed grid of match and mismatch cells

B

A

A

B

C

B

C

A

B A A B C A B C A B A C A blue = 0red = 1

score(“BAABCBCA”, “BAABCABCABACA”) = len(“BAABCBCA”) = 8

LCS = highest-score path from top-left to bottom-right



LCS: dynamic programming [WF: 1974]

Sweep cells in any �-compatible order

Cell update: time O(1)

Overall time O(mn)



LCS: micro-block dynamic programming [MP: 1980; BF: 2008]

Sweep cells in micro-blocks, in any �-compatible order

Micro-block size:

t = O(log n) when σ = O(1)

t = O( log n

log log n

)otherwise

Micro-block interface:

O(t) characters, each O(log σ) bits, can be reduced to O(log t) bits

O(t) small integers, each O(1) bits

Micro-block update: time O(1), by precomputing all possible interfaces

Overall time O(

mnlog2 n

)when σ = O(1), O

(mn(log log n)2

log2 n

)otherwise



‘Begin at the beginning,’ the King said gravely, ‘andgo on till you come to the end: then stop.’

L. Carroll, Alice in Wonderland

Dynamic programming: begins at empty strings,proceeds by appending characters, then stops

What about:

prepending/deleting characters (dynamic LCS)

concatenating strings (LCS on compressedstrings; parallel LCS)

taking substrings (= local alignment)



Dynamic programming from both ends:better by ×2, but still not good enough

Is dynamic programming strictly necessary tosolve sequence alignment problems?

Eppstein+, Efficient algorithms for sequence

analysis, 1991



The semi-local LCS problem

Give the (implicit) matrix of O((m + n)2

)LCS scores:

string-substring LCS: string a vs every substring of b

prefix-suffix LCS: every prefix of a vs every suffix of b

suffix-prefix LCS: every suffix of a vs every prefix of b

substring-string LCS: every substring of a vs string b



Semi-local LCS on the alignment graph

B

A

A

B

C

B

C

A

B A A B C A B C A B A C AC A B C A B A blue = 0red = 1

score(“BAABCBCA”, “CABCABA”) = len(“ABCBA”) = 5

String-substring LCS: all highest-score top-to-bottom paths

Semi-local LCS: all highest-score boundary-to-boundary paths


Semi-local string comparisonScore matrices and seaweed matrices

The score matrix H

0 1 2 3 4 5 6 6 7 8 8 8 8 8

-1 0 1 2 3 4 5 5 6 7 7 7 7 7

-2 -1 0 1 2 3 4 4 5 6 6 6 6 7

-3 -2 -1 0 1 2 3 3 4 5 5 6 6 7

-4 -3 -2 -1 0 1 2 2 3 4 4 5 5 6

-5 -4 -3 -2 -1 0 1 2 3 4 4 5 5 6

-6 -5 -4 -3 -2 -1 0 1 2 3 3 4 4 5

-7 -6 -5 -4 -3 -2 -1 0 1 2 2 3 3 4

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 3 4

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3

-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2

-12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1

-13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0

5

a = “BAABCBCA”

b = “BAABCABCABACA”

H(i , j) = score(a, b〈i : j〉)H(4, 11) = 5

H(i , j) = j − i if i > j



Semi-local LCS: output representation and running time

size query time

O(n2) O(1) trivial

O(m1/2n) O(log n) string-substring [Alves+: 2003]O(n) O(n) string-substring [Alves+: 2005]O(n log n) O(log2 n) [T: 2006]

. . . or any 2D orthogonal range counting data structure

running time

O(mn2) naiveO(mn) string-substring [Schmidt: 1998; Alves+: 2005]O(mn) [T: 2006]O(

mnlog0.5 n

)[T: 2006]

O(mn(log log n)2

log2 n

)[T: 2007]



The score matrix H and the seaweed matrix P

H(i , j): the number of matched characters for a vs substring b〈i : j〉j − i − H(i , j): the number of unmatched characters

Properties of matrix j − i − H(i , j):

simple unit-Monge

therefore, = PΣ, where P = −H� is a permutation matrix

P is the seaweed matrix, giving an implicit representation of H

Range tree for P: memory O(n log n), query time O(log2 n)




0 1 2 3 4 5 6 6 7 8 8 8 8 8

-1 0 1 2 3 4 5 5 6 7 7 7 7 7

-2 -1 0 1 2 3 4 4 5 6 6 6 6 7

-3 -2 -1 0 1 2 3 3 4 5 5 6 6 7

-4 -3 -2 -1 0 1 2 2 3 4 4 5 5 6

-5 -4 -3 -2 -1 0 1 2 3 4 4 5 5 6

-6 -5 -4 -3 -2 -1 0 1 2 3 3 4 4 5

-7 -6 -5 -4 -3 -2 -1 0 1 2 2 3 3 4

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 3 4

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3

-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2

-12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1

-13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0

5

a = “BAABCBCA”


H(i , j) = score(a, b〈i : j〉)H(4, 11) = 5

H(i , j) = j − i if i > j

blue: difference in H is 0

red : difference in H is 1

green: P(i , j) = 1

H(i , j) = j − i − PΣ(i , j)




0 1 2 3 4 5 6 6 7 8 8 8 8 8

-1 0 1 2 3 4 5 5 6 7 7 7 7 7

-2 -1 0 1 2 3 4 4 5 6 6 6 6 7

-3 -2 -1 0 1 2 3 3 4 5 5 6 6 7

-4 -3 -2 -1 0 1 2 2 3 4 4 5 5 6

-5 -4 -3 -2 -1 0 1 2 3 4 4 5 5 6

-6 -5 -4 -3 -2 -1 0 1 2 3 3 4 4 5

-7 -6 -5 -4 -3 -2 -1 0 1 2 2 3 3 4

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 3 4

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3

-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2

-12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1

-13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0

5

a = “BAABCBCA”


H(4, 11) =11− 4− PΣ(i , j) =11− 4− 2 = 5



The (combed) seaweed braid in the alignment graph

B

A

A

B

C

B

C

A

B A A B C A B C A B A C AC A B C A B A a = “BAABCBCA”


H(4, 11) =11− 4− PΣ(i , j) =11− 4− 2 = 5

P(i , j) = 1 corresponds to seaweed top i bottom j

Also define seaweeds top right, left right, left bottom

Represent implicitly semi-local LCS for each prefix of a vs b



Seaweed braid: a highly symmetric object (element of H0(Sn))

Can be built by assembling subbraids: divide-and-conquer

Flexible approach to local alignment, compressed approximate matching,parallel computation. . .


Semi-local string comparisonWeighted alignment

The LCS problem is a special case of the weighted alignment problem

Scoring scheme: match score wM , mismatch score wX , gap score wG

LCS score: wM = 1, wX = wG = 0

Levenshtein score: wM = 2, wX = 1, wG = 0

Scoring scheme is rational, if wM , wX , wG are rational numbers

Edit distance: minimum cost to transform a into b by weighted characteredits (insertion, deletion, substitution)

Corresponds to a scoring scheme wM = 0:

insertion/deletion cost −wG

substitution cost −wX



Weighted alignment graph for a, b

B

A

A

B

C

B

C

A

B A A B C A B C A B A C AC A B C A B A blue = 0red (solid) = 2

red (dotted) = 1

Levenshtein(“BAABCBCA”, “CABCABA”) = 11



Reduction: ordinary alignment graph for blown-up a, b

$B

$A

$A

$B

$C

$B

$C

$A

$B $A $A $B $C $A $B $C $A $B $A $C $A$C$A$B$C$A$B$A blue = 0red = 1 or 2

Levenshtein(“BAABCBCA”, “CABCABA”) =lcs(“$B$A$A$B$C$B$C$A”, “$C$A$B$C$A$B$A”) = 11



Reduction: rational-weighted semi-local alignment to semi-local LCS

$B

$A

$A

$B

$C

$B

$C

$A

$B $A $A $B $C $A $B $C $A $B $A $C $A$C$A$B$C$A$B$A

Let wM = 1, wX = µν , wG = 0

Increase × ν2 in complexity (can be reduced to ν)


1 Introduction











The seaweed methodSeaweed combing

B

A

A

B

C

B

C

A

B A A B C A B C A B A C A



B

A

A

B

C

B

C

A




Semi-local LCS: seaweed combing [T: 2006]

Initialise uncombed seaweed braid: mismatch cell = crossing


match cell: skip (keep uncrossed)

mismatch cell: comb (uncross) iff the same seaweed pair alreadycrossed before


Overall time O(mn)

Correctness: by seaweed monoid relations


The seaweed methodMicro-block seaweed combing

B

A

A

B

C

B

C

A




B

A

A

B

C

B

C

A




Semi-local LCS: micro-block seaweed combing [T: 2007]


Sweep cells in micro-blocks, in any �-compatible order

Micro-block size: t = O( log n

log log n

)Micro-block interface:

O(t) characters, each O(log σ) bits, can be reduced to O(log t) bits

O(t) integers, each O(log n) bits, can be reduced to O(log t) bits

Micro-block update: time O(1), by precomputing all possible interfaces

Overall time O(mn(log log n)2

log2 n

)


The seaweed methodCyclic LCS

The cyclic LCS problem

Give the maximum LCS score for a vs all cyclic rotations of b

Cyclic LCS: running time

O(mn2

log n

)naive

O(mn logm) [Maes: 1990]O(mn) [Bunke, Buhler: 1993; Landau+: 1998; Schmidt: 1998]

O(mn(log log n)2

log2 n

)[T: 2007]


The seaweed methodCyclic LCS

Cyclic LCS: the algorithm

Micro-block seaweed combing on a vs bb, time O(mn(log log n)2

log2 n

)Make n string-substring LCS queries, time negligible


The seaweed methodLongest repeating subsequence

The longest repeating subsequence problem

Find the longest subsequence of a that is a square (a repetition of twoidentical strings)

Longest repeating subsequence: running time

O(m3) naiveO(m2) [Kosowski: 2004]

O(m2(log log m)2

log2 m

)[T: 2007]


The seaweed methodLongest repeating subsequence

Longest repeating subsequence: the algorithm

Micro-block seaweed combing on a vs a, time O(m2(log log m)2

log2 m

)Make m − 1 prefix-suffix LCS queries, time negligible

Open question: generalise to longest subsequence that is a cube, etc.


The seaweed methodApproximate matching

The approximate pattern matching problem

Give the substring closest to a by alignment score, starting at eachposition in b

Assume rational scoring scheme

Approximate pattern matching: running time

O(mn) [Sellers: 1980]O(

mnlog n

)σ = O(1) via [Masek, Paterson: 1980]

O(mn(log log n)2

log2 n

)via [Bille, Farach-Colton: 2008]


The seaweed methodApproximate matching

Approximate pattern matching: the algorithm

Micro-block seaweed combing on a vs b (with blow-up), time

O(mn(log log n)2

log2 n

)The implicit semi-local edit score matrix:

an anti-Monge matrix

approximate pattern matching ∼ row minima

Row minima in O(n) element queries [Aggarwal+: 1987]

Each query in time O(log2 n) using the range tree representation,combined query time negligible

Overall running time O(mn(log log n)2

log2 n

), same as [Bille, Farach-Colton: 2008]


1 Introduction











Periodic string comparisonWraparound seaweed combing

The periodic string-substring LCS problem

Give (implicit) LCS scores for a vs each substring of b = . . . uuu . . . = u±∞

Let u be of length p

May assume that every character of a occurs in u (otherwise delete it)

Only substrings of b of length ≤ mp (otherwise LCS score = m)



B

A

A

B

C

B

C

A




Periodic string-substring LCS: Wraparound seaweed combing


Sweep cells row-by-row: each row starts at match cell, wraps at boundary


match cell: skip (keep uncrossed)

mismatch cell: comb (uncross) iff the same seaweed pair alreadycrossed before, possibly in a different period


Overall time O(mn)

String-substring LCS score: count seaweeds with multiplicities



The tandem LCS problem

Give LCS score for a vs b = uk

We have n = kp; may assume k ≤ m

Tandem LCS: running time

O(mkp) naiveO(m(k + p)

)[Landau, Ziv-Ukelson: 2001]

O(mp) [T: 2009]

Direct application of wraparound seaweed combing



The tandem alignment problem

Give the substring closest to a in b = u±∞ by alignment score among

global: substrings uk of length kp across all k

cyclic: substrings of length kp across all k

local: substrings of any length

Tandem alignment: running time

O(m2p) all naive

O(mp) global [Myers, Miller: 1989]

O(mp log p) cyclic [Benson: 2005]O(mp) cyclic [T: 2009]

O(mp) local [Myers, Miller: 1989]



Cyclic tandem alignment: the algorithm

Periodic seaweed combing for a vs b (with blow-up), time O(mp)

For each k ∈ [1 : m]:

solve tandem LCS (under given alignment score) for a vs uk

obtain scores for a vs p successive substrings of b of length kp by LCSbatch query: time O(1) per substring

Running time O(mp)


1 Introduction











Sparse string comparisonSemi-local LCS between permutations

The LCS problem on permutation strings (LCSP)

Give LCS score for a vs b; in each of a, b all characters distinct

Equivalent to

longest increasing subsequence (LIS) in a string

maximum clique in a permutation graph

maximum planar matching in an embedded bipartite graph

LCSP: running time

O(n log n) implicit in [Erdos, Szekeres: 1935][Robinson: 1938; Knuth: 1970; Dijkstra: 1980]

O(n log log n) unit-RAM [Chang, Wang: 1992][Bespamyatnikh, Segal: 2000]



Semi-local LCSP

Give semi-local LCS scores for a vs b; in each of a, b all characters distinct

Equivalent to

longest increasing subsequence (LIS) in every substring of a string

Semi-local LCSP: running time

O(n2 log n) naiveO(n2) restricted [Albert+: 2003; Chen+: 2005]O(n1.5 log n) randomised, restricted [Albert+: 2007]O(n1.5) [T: 2006]O(n log2 n) [T: 2010]



C

F

A

E

D

H

G

B

D E H C B A F G



Semi-local LCSP: the algorithm

Divide-and-conquer on the alignment graph

Divide graph (say) horizontally; two subproblems of effective size n/2

Conquer: seaweed matrix �-multiplication, time O(n log n)

Overall time O(n log2 n)


Sparse string comparisonLongest piecewise monotone subsequences

A k-increasing sequence: a concatenation of k increasing sequences

A (k − 1)-modal sequence: a concatenation of k alternating increasing anddecreasing sequences

The longest k-increasing (or (k − 1)-modal) subsequence problem

Give the longest k-increasing ((k − 1)-modal) subsequence of string b

Longest k-increasing (or (k − 1)-modal) subsequence: running time

O(nk log n) (k − 1)-modal [Demange+: 2007]O(nk log n) via [Hunt, Szymanski: 1977]O(n log2 n) [T: 2010]

Main idea: lcs(idk , b) (respectively, lcs((id id)k/2, b))Alexander Tiskin (Warwick) Semi-local LCS 89 / 164

Sparse string comparisonLongest piecewise monotone subsequences

Longest k-increasing subsequence: algorithm A

Sparse LCS for idk vs b: time O(nk log n)

Longest k-increasing subsequence: algorithm B

Semi-local LCSP for id vs b: time O(n log2 n)

Extract string-substring LCSP for id vs b

String-substring LCS for idk vs b by at most 2 log k instances of�-square/multiply: time log k · O(n log n) = O(n log2 n)

Query the LCS score for idk vs b: time negligible

Overall time O(n log2 n)

Algorithm B faster than Algorithm A for k ≥ log n


Sparse string comparisonMaximum clique in a circle graph

The maximum clique problem in a circle graph

Given a circle with n chords, find the maximum-size subset of pairwiseintersecting chords

S

Standard reduction to an interval model: cut the circle and lay it out onthe line; chords become intervals (here drawn as square diagonals)

Chords intersect iff intervals overlap, i.e. intersect without containment



The maximum clique problem in a circle graph

Given a circle with n chords, find the maximum-size subset of pairwiseintersecting chords

S

Standard reduction to an interval model: cut the circle and lay it out onthe line; chords become intervals (here drawn as square diagonals)

Chords intersect iff intervals overlap, i.e. intersect without containmentAlexander Tiskin (Warwick) Semi-local LCS 91 / 164


Maximum clique in a circle graph: running time

exp(n) naiveO(n3) [Gavril: 1973]O(n2) [Rotem, Urrutia: 1981; Hsu: 1985]

[Masuda+: 1990; Apostolico+: 1992]O(n1.5) [T: 2006]O(n log2 n) [T: 2010]



Maximum clique in a circle graph: the algorithm

Helly property: if any set of intervals intersect pairwise, then they allintersect at a common point

Compute seaweed matrix, build range tree: time O(n log2 n)

Run through all 2n + 1 possible common intersection points

For each point, find a maximum subset of covering overlapping segmentsby a prefix-suffix LCS query: time (2n + 1) · O(log2 n) = O(n log2 n)

Overall time O(n log2 n) + O(n log2 n) = O(n log2 n)



Parameterised maximum clique in a circle graph

The maximum clique problem in a circle graph, sensitive e.g. to

number e of edges; e ≤ n2

size l of maximum clique; l ≤ n

cutwidth d of interval model (max number of intervals covering apoint); l ≤ d ≤ n

Parameterised maximum clique in a circle graph: running time

O(n log n + e) [Apostolico+: 1992]

O(n log n + nl log(n/l)) [Apostolico+: 1992]

O(n log n + n log2 d) NEW



Parameterised maximum clique in a circle graph: the algorithm

For each diagonal block of size d , compute seaweed matrix, build rangetree: time n/d · O(d log2 d) = O(n log2 d)

Extend each diagonal block to a quadrant: time O(n log2 d)

Run through all 2n + 1 possible common intersection points

For each point, find a maximum subset of covering overlapping segmentsby a prefix-suffix LCS query: time O(n log2 d)

Overall time O(n log2 d)


1 Introduction











Compressed string comparisonGrammar compression

Notation: pattern p of length m; text t of length n

A GC-string (grammar-compressed string) t is a straight-line program(context-free grammar) generating t = tn by n assignments of the form

tk = α, where α is an alphabet character

tk = uv , where each of u, v is an alphabet character, or ti for i < k

In general, n = O(2n)

Example: Fibonacci string “ABAABABAABAAB”

t1 = A t2 = t1B t3 = t2t1 t4 = t3t2 t5 = t4t3 t6 = t5t4


Compressed string comparisonGrammar compression

Grammar-compression covers various compression types, e.g. LZ78, LZW(not LZ77 directly)

Simplifying assumption: arithmetic up to n runs in O(1)

This assumption can be removed by careful index remapping


Compressed string comparisonExtended substring-string LCS on GC-strings

LCS: running time (r = m + n, r = m + n)

p t

plain plain O(mn) [Wagner, Fischer: 1974]O(

mnlog2 m

)[Masek, Paterson: 1980]

[Crochemore+: 2003]

plain GC O(m3n + . . .) gen. CFG [Myers: 1995]O(m1.5n) ext subs-s [T: 2008]O(m logm · n) ext subs-s [T: 2010]

GC GC NP-hard [Lifshits: 2005]O(r1.2r1.4) R weights [Hermelin+: 2009]O(r log r · r) [T: 2010]O(r log(r/r) · r

)[Hermelin+: 2010]

O(r log1/2(r/r) · r

)[Gawrychowski: 2012]


Compressed string comparisonExtended substring-string LCS on GC-strings

Substring-string LCS (plain pattern, GC text): the algorithm

For every k, compute by recursion the appropriate part of semi-local LCSfor p vs tk , using seaweed matrix �-multiplication: time O(m logm · n)

Overall time O(m logm · n)


Compressed string comparisonSubsequence recognition on GC-strings

The global subsequence recognition problem

Does pattern p appear in text t as a subsequence?

Global subsequence recognition: running time

p t

plain plain O(n) greedy

plain GC O(mn) greedy

GC GC NP-hard [Lifshits: 2005]



The local subsequence recognition problem

Find all minimally matching substrings of t with respect to p

Substring of t is matching, if p is a subsequence of t

Matching substring of t is minimally matching, if none of its propersubstrings are matching



Local subsequence recognition: running time ( + output)

p t

plain plain O(mn) [Mannila+: 1995]O(

mnlog m

)[Das+: 1997]

O(cm + n) [Boasson+: 2001]O(m + nσ) [Tronicek: 2001]

plain GC O(m2 logmn) [Cegielski+: 2006]O(m1.5n) [T: 2008]O(m logm · n) [T: 2010]O(m · n) [Yamamoto+: 2011]




ı0+

Oı1+

Oı2+

O

M0+

M1+

M2+

0H

n′

HnH

b〈i : j〉 matching iff box [i : j ] not pierced left-to-right

Determined by �-chain of ≷-maximal seaweeds

b〈i : j〉 minimally matching iff (i , j) is in the interleaved skeleton �-chain



0+ 1+ 2+n′ n m+n

−m

0

n′

ı0+

ı1+

ı2+



Local subsequence recognition (plain pattern, GC text): the algorithm

For every k, compute by recursion the appropriate part of semi-local LCSfor p vs tk , using seaweed matrix �-multiplication: time O(m logm · n)

Given an assignment t = t ′t ′′, find by recursion

minimally matching substrings in t ′

minimally matching substrings in t ′′

Then, find �-chain of ≷-maximal seaweeds in time n · O(m) = O(mn)

Its skeleton �-chain: minimally matching substrings in t overlapping t ′, t ′′

Overall time O(m logm · n) + O(mn) = O(m logm · n)


Compressed string comparisonThreshold approximate matching

The threshold approximate matching problem

Find all matching substrings of t with respect to p, according to athreshold k

Substring of t is matching, if the edit distance for p vs t is at most k



Threshold approximate matching: running time ( + output)

p t

plain plain O(mn) [Sellers: 1980]O(mk) [Landau, Vishkin: 1989]

O(m + n + nk4

m ) [Cole, Hariharan: 2002]

plain GC O(mnk2) [Karkkainen+: 2003]O(mnk + n log n) [LV: 1989] via [Bille+: 2010]O(mn + nk4 + n log n) [CH: 2002] via [Bille+: 2010]O(m logm · n) [T: NEW]


(Also many specialised variants for LZ compression)



ı0+

Oı1+

Oı2+

Oı3+

Oı4+

O

M0+

M1+

M2+

M3+

M4+

0H

n′

HnH

Blow up: weighted alignment on strings p, t of size m, n equivalent toLCS on strings p, t of size m = νm, n = νn



0+ 1+ 2+ 3+ 4+n′ n m+n

ı0+

ı1+

ı2+

ı3+

ı4+

−m

0

n′



Threshold approx matching (plain pattern, GC text): the algorithm

Algorithm structure similar to local subsequence recognition

�-chains replaced by m ×m submatrices

Extra ingredients:

the blow-up technique: reduction of edit distances to LCS scores

implicit matrix searching, replaces �-chain interleaving

Monge row minima: “SMAWK” O(m) [Aggarwal+: 1987]

Implicit unit-Monge row minima:

O(m log logm) [T: 2012]O(m) [Gawrychowski: 2012]

Overall time O(m logm · n) + O(mn) = O(m logm · n)


1 Introduction











Parallel string comparisonParallel LCS

Bulk-Synchronous Parallel (BSP) computer [Valiant: 1990]

Simple, realistic general-purpose parallelmodel

COMM. ENV . (g , l)

0

PM

1

PM

p − 1

PM· · ·

Contains

p processors, each with local memory (1 time unit/operation)

communication environment, including a network and an externalmemory (g time units/data unit communicated)

barrier synchronisation mechanism (l time units/synchronisation)



BSP computation: sequence of parallel supersteps

0

1

p − 1

Asynchronous computation/communication within supersteps (includesdata exchange with external memory)

Synchronisation before/after each superstep

Cf. CSP: parallel collection of sequential processes



BSP computation: sequence of parallel supersteps

0

1

p − 1 · · ·

· · ·

· · ·

· · ·

Asynchronous computation/communication within supersteps (includesdata exchange with external memory)

Synchronisation before/after each superstep

Cf. CSP: parallel collection of sequential processes



Compositional cost model

For individual processor proc in superstep sstep:

comp(sstep, proc): the amount of local computation and localmemory operations by processor proc in superstep sstep

comm(sstep, proc): the amount of data sent and received byprocessor proc in superstep sstep

For the whole BSP computer in one superstep sstep:

comp(sstep) = max0≤proc<p comp(sstep, proc)

comm(sstep) = max0≤proc<p comm(sstep, proc)

cost(sstep) = comp(sstep) + comm(sstep) · g + l



For the whole BSP computation with sync supersteps:

comp =∑

0≤sstep<sync comp(sstep)

comm =∑

0≤sstep<sync comm(sstep)

cost =∑

0≤sstep<sync cost(sstep) = comp + comm · g + sync · l

The input/output data are stored in the external memory; the cost ofinput/output is included in comm

E.g. for a particular linear system solver with an n × n matrix:

comp = O(n3/p) comm = O(n2/p1/2) sync = O(p1/2)



BSP software: industrial projects

Google’s Pregel [2010]

Apache Spark (hama.apache.org) [2010]

Apache Giraph (giraph.apache.org) [2011]

BSP software: research projects

Oxford BSP (www.bsp-worldwide.org/implmnts/oxtool) [1998]

Paderborn PUB (www2.cs.uni-paderborn.de/~pub) [1998]

BSML (traclifo.univ-orleans.fr/BSML) [1998]

BSPonMPI (bsponmpi.sourceforge.net) [2006]

Multicore BSP (www.multicorebsp.com) [2011]

Epiphany BSP (www.codu.in/ebsp) [2015]

Petuum (petuum.org) [2015]


hama.apache.org

giraph.apache.org

www.bsp-worldwide.org/implmnts/oxtool

www2.cs.uni-paderborn.de/~pub

traclifo.univ-orleans.fr/BSML

bsponmpi.sourceforge.net

www.multicorebsp.com

www.codu.in/ebsp

petuum.org


The ordered 2D grid dag

grid2(n)

nodes arranged in an n × n grid

edges directed top-to-bottom, left-to-right

≤ 2n inputs (to left/top borders)

≤ 2n outputs (from right/bottom borders)

size n2 depth 2n − 1

Applications: triangular linear system; discretised PDE via Gauss–Seideliteration (single step); 1D cellular automata; dynamic programming

Sequential work O(n2)



Parallel ordered 2D grid computation

grid2(n)

Consists of a p × p grid of blocks, eachisomorphic to grid2(n/p)

The blocks can be arranged into 2p − 1anti-diagonal layers, with ≤ p independentblocks in each layer



Parallel ordered 2D grid computation (contd.)

The computation proceeds in 2p − 1 stages, each computing a layer ofblocks. In a stage:

every block assigned to a different processor (some processors idle)

the processor reads the 2n/p block inputs, computes the block, andwrites back the 2n/p block outputs

comp: (2p − 1) · O((n/p)2

)= O(p · n2/p2) = O(n2/p)

comm: (2p − 1) · O(n/p) = O(n)

n ≥ p

comp O(n2/p) comm O(n) sync O(p)



Parallel LCS computation

The 2D grid algorithm solves the LCS problem (and many others) bydynamic programming

comp = O(n2/p) comm = O(n) sync = O(p)

comm is not scalable (i.e. does not decrease with increasing p) :-(

Can scalable comm be achieved for the LCS problem?



Parallel LCS computation

Solve the more general semi-local LCS problem:

each string vs all substrings of the other string

all prefixes of each string against all suffixes of the other string

Divide-and-conquer on substrings of a, b: log p recursion levels

Conquer phase: assembles substring semi-local LCS from smaller ones byparallel seaweed multiplication

Base level: p semi-local LCS subproblems on np1/2 -strings

Sequential time still O(n2)



Parallel LCS computation (cont.)

Parallel semi-local LCS: running time, comm, sync

comp comm sync

O(n2/p) O(n) O(p) 2D grid

O(n2/p) O(

np1/2

)O(log2 p) [Krusche, T: 2007]

O(n log p

p1/2

)O(log p) [Krusche, T: 2010]

O(

np1/2

)O(log p) [T: NEW]

Based on parallel Steady Ant algorithm



Seaweed matrix �-multiplication: parallel Steady Ant


)Divide-and-conquer on the range of j : two subproblems on n/2-matrices

PΣlo � QΣ

lo = RΣlo PΣ

hi � QΣhi = RΣ

hi

Conquer: tracing a balanced trail from bottom-left to top-right R

Partition n × n index space into a regular p × p grid of blocks

Sample R in block corners: total p2 samples

Steady Ant’s trail passes through ≤ 2p blocks

Partition blocks across processors: comp, comm = O(n/p)

comp = O(n log n

p

)comm = O

(n log pp

)sync = O(log p)



Seaweed matrix �-multiplication: generalised parallel Steady Ant


)Divide-and-conquer on the range of j : pε subproblems on n

pε -matrices

Conquer: tracing pε balanced trails from bottom-left to top-right R

Partition n × n index space into a regular p × p grid of blocks

Sample R in block corners: total p2 samples

Steady Ant’s trails pass through ≤ 2p1+ε blocks

Partition blocks across processors: comp, comm = O(

np1−ε

)comp = O

(n log np + n

εp1−ε

)comm = O

(n

p1−ε

)sync = O(1/ε)




Divide-and-conquer on substrings of a, b: log p recursion levels

In every level, semi-local LCS subproblems on input substrings

Conquer phase: generalised parallel Steady Ant algorithm, ε = 1/3

base level: p = p1/2 × p1/2 subproblems on np1/2 -substrings; one

proc/subproblem; comm = O(

np1/2

)middle level: p1/2 = p1/4 × p1/4 subproblems on n

p1/4 -substrings; p1/2

procs/subproblem; comm = O( n/p1/4

(p1/2)1−1/3

)= O

(n

p7/12

)top level: one subproblem on n-strings; p procs;comm = O

(n

p1−1/3

)= O

(n

p2/3

)sync = O

(1

1/3

)= O(1) in every level




comm dominated by base level:O(

np1/2

)+ . . .+ O

(n

p7/12

)+ . . .+ O

(n

p2/3

)= O

(n

p1/2

)comp = O

(n2

p

)comm = O

(n

p1/2

)sync = O(log p)



Parallel LCS on permutation strings (LCSP = LIS)

Sequential time O(n log n), no good way known to parallelise directly

Solution via the more general semi-local LCSP is no longer efficient:sequential time O(n log2 n)

Divide-and-conquer on substrings/subsequences of a, b

Parallelising normal O(n log n) seaweed multiplication: [Krusche, T: 2010]

comp O(n log2 n

p

)comm O

(n log pp

)sync O(log2 p)

Open problem: can we achieve

comp O(n log n

p

)for LCSP?

comp O(n log2 n

p

), comm O(np ), sync O(log2 p) for semi-local LCSP?


1 Introduction











The transposition network methodTransposition networks

Comparison network: a circuit of comparators

A comparator sorts two inputs and outputs them in prescribed order

Comparison networks traditionally used for non-branching merging/sorting

Classical comparison networks

# comparators

merging O(n log n) [Batcher: 1968]

sorting O(n log2 n) [Batcher: 1968]O(n log n) [Ajtai+: 1983]

Comparison networks are visualised by wire diagrams

Transposition network: all comparisons are between adjacent wires



Seaweed combing as a transposition network

A

C

B

C

A B C A

+7

+1

+5

+5

+3

+7

+1

−5

−1

−3

−3

+3

−5

−1

−7

−7

Character mismatches correspond to comparators

Inputs anti-sorted (sorted in reverse); each value traces a seaweed



Global LCS: transposition network with binary input

A

C

B

C

A B C A

1

1

1

1

1

1

1

0

0

0

0

1

0

0

0

0

Inputs still anti-sorted, but may not be distinct

Comparison between equal values is indeterminate


The transposition network methodParameterised string comparison

Parameterised string comparison

String comparison sensitive e.g. to

low similarity: small λ = LCS(a, b)

high similarity: small κ = distLCS(a, b) = m + n − 2λ

Can also use weighted alignment score or edit distance

Assume m = n, therefore κ = 2(n − λ)



Low-similarity comparison: small λ

sparse set of matches, may need to look at them all

preprocess matches for fast searching, time O(n log σ)

High-similarity comparison: small κ

set of matches may be dense, but only need to look at small subset

no need to preprocess, linear search is OK

Flexible comparison: sensitive to both high and low similarity, e.g. by bothcomparison types running alongside each other



Parameterised string comparison: running time

Low-similarity, after preprocessing in O(n log σ)

O(nλ) [Hirschberg: 1977][Apostolico, Guerra: 1985]

[Apostolico+: 1992]

High-similarity, no preprocessing

O(n · κ) [Ukkonen: 1985][Myers: 1986]

Flexible

O(λ · κ · log n) no preproc [Myers: 1986; Wu+: 1990]O(λ · κ) after preproc [Rick: 1995]



Parameterised string comparison: the waterfall algorithm

Low-similarity: O(n · λ)

0 0 0 0 0 0 0 0

1

1

1

1

1

1

1

1

0 0 1 1 0 0 1 0

0

1

0

1

1

0

1

1

High-similarity: O(n · κ)

0 0 0 0 0 0 0 0

1

1

1

1

1

1

1

1

1 1 1 0 1 1 0 0

0

0

0

0

1

1

1

0

Trace 0s through network in contiguous blocks and gaps


The transposition network methodDynamic string comparison

The dynamic LCS problem

Maintain current LCS score under updates to one or both input strings

Both input strings are streams, updated on-line:

prepending/appending characters at left or right

deleting characters at left or right

Assume for simplicity m ≈ n, i.e. m = Θ(n)

Goal: linear time per update

O(n) per update of a (n = |b|)O(m) per update of b (m = |a|)


The transposition network methodDynamic string comparison

Dynamic LCS in linear time: update models

left right

– app+del standard DP [Wagner, Fischer: 1974]prep app a fixed [Landau+: 1998], [Kim, Park: 2004]prep app [Ishida+: 2005]prep+del app+del [T: NEW]

Dynamic semi-local LCS in linear time: update models

left right

prep+del app+del amortised [T: NEW]


The transposition network methodBit-parallel string comparison

Bit-parallel string comparison

String comparison using standard instructions on words of size w

Bit-parallel string comparison: running time

O(mn/w) [Allison, Dix: 1986; Myers: 1999; Crochemore+: 2001]



Bit-parallel string comparison: binary transposition network

In every cell: input bits s, c ; output bits s ′, c ′; match/mismatch flag µ

¬µ

s

c

s′

c ′

s 0 1 0 1 0 1 0 1c 0 0 1 1 0 0 1 1µ 0 0 0 0 1 1 1 1

s′ 0 1 1 1 0 0 1 1c ′ 0 0 0 1 0 1 0 1

+

∧µ

s

c

s′

c ′

s 0 1 0 1 0 1 0 1c 0 0 1 1 0 0 1 1µ 0 0 0 0 1 1 1 1

s′ 0 1 1 0 0 0 1 1c ′ 0 0 0 1 0 1 0 1

2c + s ← (s + (s ∧ µ) + c) ∨ (s ∧ ¬µ)

S ← (S + (S ∧M)) ∨ (S ∧ ¬M), where S , M are words of bits s, µ



Bit-parallel string comparison: binary transposition network

In every cell: input bits s, c ; output bits s ′, c ′; match/mismatch flag µ

¬µ

s

c

s′

c ′

s 0 1 0 1 0 1 0 1c 0 0 1 1 0 0 1 1µ 0 0 0 0 1 1 1 1

s′ 0 1 1 1 0 0 1 1c ′ 0 0 0 1 0 1 0 1

+

∧µ

s

c

s′

c ′

s 0 1 0 1 0 1 0 1c 0 0 1 1 0 0 1 1µ 0 0 0 0 1 1 1 1

s′ 0 1 1 0 0 0 1 1c ′ 0 0 0 1 0 1 0 1

2c + s ← (s + (s ∧ µ) + c) ∨ (s ∧ ¬µ)

S ← (S + (S ∧M)) ∨ (S ∧ ¬M), where S , M are words of bits s, µAlexander Tiskin (Warwick) Semi-local LCS 141 / 164


High-similarity bit-parallel string comparison

κ = distLCS(a, b) Assume κ odd, m = n

Waterfall algorithm within diagonal band of width κ+ 1: time O(nκ/w)

Band waterfall supported from below by separator matches



High-similarity bit-parallel multi-string comparison: a vs b0, . . . , br−1

κi = distLCS(a, bi ) ≤ κ 0 ≤ i < r

Waterfalls within r diagonal bands of width κ+ 1: time O(nrκ/w)

Each band’s waterfall supported from below by separator matches


1 Introduction











Beyond semi-localityFixed-length substring LCS

Given window length w , let w -window be any substring of length w

The window-substring LCS problem

Give LCS score for every w -window of a vs every substring of b

Window-substring LCS: running time

O(mn3w) naiveO(mnw) multiple seaweedO(mn) [Krusche, T: 2010]



Window-substring LCS: the algorithm

Compute seaweed matrices for canonical substrings of a vs b recursively

Compute seaweed matrices for w -windows of a vs b, using thedecomposition of each w -window into canonical substrings

Both stages use matrix �-multiplication, overall time O(mn)



The window-window LCS problem

Give LCS score for every w -window of a vs every w -window of b

Provides an LCS-scored alignment plot (by analogy with Hamming-scoreddot plot), a useful tool for studying weak conservation in the genome

Window-window LCS: running time

O(mnw2) naiveO(mnw) repeated seaweed combingO(mn) [Krusche, T: 2010]



An implementation of alignment plots [Krusche: 2010]

Based on O(mnw) repeated seaweed combing approach, using some ideasfrom the optimal O(mn) algorithm: resulting time O(mnw0.5)

Weighted alignment (match 1, mismatch 0, gap −0.5) via alignment dagblow-up: ×4 slowdown

C++, Intel assembly (x86, x86 64, MMX/SSE2 data parallelism)

SMP parallelism (currently two processors)



Krusche’s implementation used at Warwick Systems Biology Centre tostudy conservation in plant non-coding DNA

More sensitive than BLAST and other heuristic (seed-and-extend) methods

Found hundreds of conserved non-coding regions between papaya (Caricapapaya), poplar (Populus trichocarpa), thale cress (Arabidopsis thaliana)and grape (Vitis vinifera)

Single processor:

×10 speedup over heavily optimised, bit-parallel naive algorithm

×7 speedup over biologists’ heuristics

Two processors: extra ×2 speedup, near-perfect parallelism



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by Katy Edgington

06 December 2012

We can study

selected

conserved

regions

experimentally

to understand

some of the

most ancient

regulatory

mechanisms

underlying plant

development.

Dr Sascha Ott

Researchers from the University of Warwick have found that hundreds of regionsof non-coding DNA are conserved among various plant species that have beenevolving separately for around 100 million years. The findings, which have beendetailed in the journal The Plant Cell, have led the team to believe that thesesequences, although they are not genes, could provide vital information aboutplant development and functioning.

Computational analysis of the genomes of papaya (Carica papaya), poplar(Populus trichocarpa), thale cress (Arabidopsis thaliana) and grape (Vitis vinifera)showed that all four species contained conserved non-coding sequences (CNSs)which may be important for determining whether neighbouring genes areexpressed in the organism.

Dr Sascha Ott, Associate Professor at Warwick Systems Biology Centre and oneof the report’s co-authors, answered ScienceOmega.com’s questions about thestudy, explaining the rationale behind the researchers’ choice of species, as wellas the implications and potential applications of this research as a short cut forbiologists and crop scientists wishing to address such issues as food security…

Why were papaya, poplar, Arabidopsis and grape the species chosen foranalysis?We initially analysed DNA sequences for only a handful of genes in order toestablish which species would be most amenable to our comparative genomicsapproach. On the one hand, we needed species that were sufficiently distant fromour model organism, Arabidopsis, for non-functional DNA sequences to have lostsequence similarity by accumulating many mutations.

On the other hand, we needed species to be sufficiently close such thatnon-coding functional sequences (which accumulate mutations at a slower rate)can still be detected. The choice of species was further restricted by the availabilityof whole-genome sequences at the time we started our project. Some relevantspecies such as potato, tomato, soybean and cotton are closely related to some ofthe species that we have already analysed and we know that these species shareat least some of the conserved non-coding regions – so these species make greattargets for expanding our work.

Did you expect to find that the non-coding sequences were so oftenconserved?No, we did not. In fact after our initial analyses and after reviewing previouslyexisting literature we were concerned that we may not find very much at all.Luckily, it turned out that a large set of genes have conserved non-codingsequences neighbouring the genes.

How much do we know about the function(s) of these regions of plant DNA?There are examples of non-coding DNA regions that have previously been studiedintensively. For example, Alabadi et al found a short region of DNA that isnecessary and sufficient to drive the time-of-day dependent expression of a keygene called TOC1. We previously found that the region they identified is closelymatching with one of the conserved regions that we can identify with our method.

The general picture we have from studies on other organisms is that conservednon-coding regions often control the spatial, temporal, and condition-specificexpression of genes that they are close to. The function of each of the regions we

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As we are not nocturnal animalsby nature, then we must benefit fromsunlight, in several ways. It then

follows that we can deal withsunlight - maybe we need to be

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Conserved DNA could prove crop science shortcut - Science Omega http://www.scienceomega.com/article/732/conserved-dna-could-prove-...

1 of 2 31/05/2013 00:19



Publications:

[Picot+: 2010] Evolutionary Analysis of Regulatory Sequences (EARS) inPlants. Plant Journal.

[Baxter+: 2012] Conserved Noncoding Sequences Highlight SharedComponents of Regulatory Networks in Dicotyledonous Plants. Plant Cell.

Web service: http://wsbc.warwick.ac.uk/ears

Warwick press release (with video): http://www2.warwick.ac.uk/

newsandevents/pressreleases/discovery_of_100

Future work: genomic repeats; genome-scale comparison


http://wsbc.warwick.ac.uk/ears

http://www2.warwick.ac.uk/newsandevents/pressreleases/discovery_of_100

http://www2.warwick.ac.uk/newsandevents/pressreleases/discovery_of_100

Beyond semi-localityQuasi-local LCS and spliced alignment

Given m (possibly overlapping) prescribed substrings in a

The quasi-local LCS problem

Give LCS score for every prescribed substring of a vs every subsring of b

Quasi-local LCS: running time

O(m2n) repeated seaweed combingO(mn log2 m) [T: unpublished]



Quasi-local LCS: the algorithm

Compute seaweed matrices for canonical substrings of a vs b recursively

Compute seaweed matrices for prescribed substrings of a vs b, using thedecomposition of each prescribed substring into canonical substrings

Both stages use matrix �-multiplication, time O(mn log2 m)



The spliced alignment problem

Give the chain of non-overlapping prescribed substrings in a, closest to bby alignment score

Describes gene assembly from candidate exons, given a known similar gene

Assume m = n; let s = total size of prescribed substrings

Spliced alignment: running time

O(ns) = O(n3) [Gelfand+: 1996]O(n2.5) [Kent+: 2006]O(n2 log2 n) [T: unpublished]O(n2 log n) [Sakai: 2009]


Beyond semi-localityLocal alignment

Assume a fixed set of character alignment scores

The Smith–Waterman local alignment problem

For every prefix of a and every prefix of b, give a pair of respective suffixesof these prefixes that have the highest alignment score against each other

Smith–Waterman local alignment: running time

O(m2n2) [naive]O(mn) [Smith, Waterman: 1981]



Smith–Waterman local alignment [SW: 1981]



Overall time O(mn)

Smith–Waterman local alignment maximises alignment score irrespectiveof length

May not always be the most relevant alignment, e.g. may be

too short (would prefer a longer alignment even with a lower score)

too long (would prefer a significantly shorter alignment with slightlylower score)



The bounded substring alignment problem

For every prefix of string a and every prefix of string b, give a pair ofrespective suffixes of these prefixes of length at least/at most t that havethe highest alignment score against each other

Solved efficiently using window-substring alignment as a subroutine, e.g.for alignments of length at least t:

solve the window-substring alignment problem, obtaining implicitalignment score for every t-window in a against every substring in b;

obtain explicit optimal alignment score for every t-window in aagainst all substrings in b terminating in a given position;

extend locally optimal window-substring alignment scores to optimalsubstring-substring scores as in Smith–Waterman algorithm


1 Introduction











Conclusions and open problems (1)

Implicit unit-Monge matrices:

equivalent to seaweed braids

distance multiplication in time O(n log n)

Semi-local LCS problem:

representation by implicit unit-Monge matrices

generalisation to rational alignment scores

Seaweed combing and micro-block speedup:

a simple algorithm for semi-local LCS

semi-local LCS in time O(mn), and even slightly o(mn)

Sparse string comparison:

fast LCS on permutation strings

fast max-clique in a circle graph



Compressed string comparison:

LCS and approximate matching on plain pattern against GC-text

Parallel string comparison:

comm- and sync- efficient parallel LCS

Transposition networks:

simple interpretation of parameterised and bit-parallel LCS

dynamic LCS

Beyond semi-locality:

window alignment and sparse spliced alignment



Open problems (theory):

real-weighted alignment?

max-clique in circle graph: O(n log2 n) O(n log n)?

compressed LCS: other compression models?

parallel LCS: comm per processor O(n log p

p1/2

) O

(n

p1/2

)?

lower bounds?

Open problems (biological applications):

real-weighted alignment?

good pre-filtering (e.g. q-grams)?

application for multiple alignment? (NB: NP-hard)


Thank you
