Flipping letters to minimize the support of a string Giuseppe Lancia, Franca Rinaldi, Romeo Rizzi...

Flipping letters to Flipping letters to minimize minimize

the support of a stringthe support of a stringGiuseppe Lancia, Franca Rinaldi, Romeo Rizzi

University of Udine

Outline of talk:

1. Problem definition2. Parametrized complexity3. Polynomial cases4. NP-hardness5. ILP formulations

1. Problem definition1. Problem definition

We are given a string s and a parameter k (e.g., k = 3)

The string has a set of k-mers, its support, K(s)

010010011

K(s) = { 010 }



010010011

K(s) = { 010, 100 }



010010011

K(s) = { 010, 100, 001}



010010011

K(s) = { 010, 100, 001, 011}



010010011

K(s) = { 010, 100, 001, 011} | K(s) | = 4



010010011

K(s) = { 010, 100, 001, 011} | K(s) | = 4

By flipping some bits, we could reduce the number of k-mers

010010011


K(s) = { 010, 100, 001, 011} | K(s) | = 4


010010011010010010

S’=


K(s) = { 010, 100, 001, 011} | K(s) | = 4


010010011010010010

S’=

K(s’) = { 010, 100, 001} | K(s’) | = 3


The Problem :The Problem :

- A string s over an alphabet

IngredienIngredients:ts:



- A parameter k (k-mer size)





- A budget B





- A budget B


Change at most B letters in s so as resulting s’ has as few distinct k-mers as possible

ObjectiveObjective::




- A budget B


Find a string s’ with d(s,s’) <= B with the smallest number of kmers

ObjectiveObjective::

s

s’

Motivation :Motivation :

Curiosity-driven (it’s a cute combinatorial problem)Real:Real:

Motivation :Motivation :

Curiosity-driven (it’s a cute combinatorial problem)Real:Real:

Analysis of DNA sequencesFictious:Fictious:

atcgattgatccttta

atc, tcg, cga, gat, ….

3-mers are aminoacid codons. Protein complexity relates to # of codons. Mutations may reduce complexity….

Our results:Our results:

The problem has many parameters (|s|, ||, k, B), we study all versions (when possibly some of the parameters are bounded)

- Polynomial special cases (e.g. for B fixed or both k,|| fixed)

- NP-hard special cases (even k=2 or ||=2)

2. Parametrized 2. Parametrized complexitycomplexity

s NO B NO

s YES B NO

s NO B YES

s YES B YES

NO k NO

YES k NO

NO k YES

YES k YES

s NO B NO

s YES B NO

s NO B YES

s YES B YES

We can assume : k <= |s|

NO k NO

YES k NO

NO k YES

YES k YES

s NO B NO

s YES B NO

s NO B YES

s YES B YES

We can assume : B <= |s|

NO k NO

YES k NO

NO k YES

YES k YES

NO k NO

YES k NO

NO k YES

YES k YES

s NO B NO

s YES B NO

s NO B YES

s YES B YES

We can assume : || <= |s| (we don’t need any symbol not already in s)

NO k NO

YES k NO

NO k YES

YES k YES

s NO B NO

s YES B NO

s NO B YES

s YES B YES

NO k NO

YES k NO

NO k YES

YES k YES

s NO B NO

s YES B NO

s NO B YES

s YES B YES

Polynomial cases

)1(O

)|(| )1(2 BsO

|)(| sO

)||( )1(2 BskO

NO k NO

YES k NO

NO k YES

YES k YES

s NO B NO

s YES B NO

s NO B YES

s YES B YES

NP-hard cases

)1(O

)|(| )1(2 BsO

|)(| sO

)||( )1(2 BskO

NP-hard NP-hardfor ||=2

NP-hardfor k=2

3. Polynomial cases3. Polynomial cases

The case |The case || and k fixed:| and k fixed:


NO k NO

YES k NO

NO k YES

YES k YES

s NO B NO

s YES B NO

s NO B YES

s YES B YES )1(O

)|(| )1(2 BsO

|)(| sO

)||( )1(2 BskO


NP-hardfor k=2


We start with this subproblem: SUB(A): Given a set of kmers A, can we correct s within budget so as it has all of its kmers in A?


s = 01000100101110A = 0100, 1001, 0010 , 0001 B = 3



s = 01000100101110A = 0100, 1001, 0010 , 0001 B = 3

0100

1001

0010

0001

0100

1001

0010

0001

0100

1001

0010

0001

……

……

……

0100

1001

0010

0001

0100

1001

0010

0001

0100

1001

0010

0001



s = 01000100101110A = 0100, 1001, 0010 , 0001 B = 3

0100

1001

0010

0001

0100

1001

0010

0001

0100

1001

0010

0001

……

……

……

0100

1001

0010

0001

0100

1001

0010

0001

0100

1001

0010

0001

0100 0 1 ….. 1 1 0



s = 01000100101110A = 0100, 1001, 0010 , 0001 B = 3

0100

1001

0010

0001

0100

1001

0010

0001

0100

1001

0010

0001

……

……

……

0100

1001

0010

0001

0100

1001

0010

0001

0100

1001

0010

0001

0100 0 1 ….. 1 1 0

0

0

0

0

0

3

2

2

0

1

1

0

0

1

1

0

0

1

1

0

0

0

1

1



s = 01000100101110A = 0100, 1001, 0010 , 0001 B = 3

0100

1001

0010

0001

0100

1001

0010

0001

0100

1001

0010

0001

……

……

……

0100

1001

0010

0001

0100

1001

0010

0001

0100

1001

0010

0001

0100 0 1 ….. 1 1 0

0

0

0

0

0

3

2

2

0

1

1

0

0

1

1

0

0

1

1

0

0

0

1

1

Each path corresponds to a string s’ with all its kmers in A



s = 01000100101110A = 0100, 1001, 0010 , 0001 B = 3

0100

1001

0010

0001

0100

1001

0010

0001

0100

1001

0010

0001

……

……

……

0100

1001

0010

0001

0100

1001

0010

0001

0100

1001

0010

0001

0100 0 1 ….. 1 1 0

0

0

0

0

0

3

2

2

0

1

1

0

0

1

1

0

0

1

1

0

0

0

1

1

The length of the path is the Hamming distance d(s’, s)




s = 01000100101110A = 0100, 1001, 0010 , 0001 B = 3

0100

1001

0010

0001

0100

1001

0010

0001

0100

1001

0010

0001

……

……

……

0100

1001

0010

0001

0100

1001

0010

0001

0100

1001

0010

0001

0100 0 1 ….. 1 1 0

0

0

0

0

0

3

2

2

0

1

1

0

0

1

1

0

0

1

1

0

0

0

1

1

SUB(A) has a solution iff the shortest path is <= B



- we can solve SUB(A) in polytime (O|A||||s|) = O(|s|) sincekA 2||



- we can solve SUB(A) in polytime (O|A||||s|) = O(|s|) since

- There are “only” possible subsets A to try…

)1(2 || Ok

problem is solved in polytime O(|s|)

kA 2||

The case of B fixed:The case of B fixed:


NO k NO

YES k NO

NO k YES

YES k YES

s NO B NO

s YES B NO

s NO B YES

s YES B YES )1(O

)|(| )1(2 BsO

|)(| sO

)||( )1(2 BskO


NP-hardfor k=2


For B fixed, we can try all possible solutions. There are

)|(||| BsO

B

s

possible choices of bits to flip. We can try them all, and count the # of k-mers.

Since ||<=|s|, the way to flip them is bounded by )|(| ksO

4. NP-hardness4. NP-hardness

- Theorem: the problem is NP-hard even for k=2.


NO k NO

YES k NO

NO k YES

YES k YES

s NO B NO

s YES B NO

s NO B YES

s YES B YES )1(O

)|(| )1(2 BsO

|)(| sO

)||( )1(2 BskO


NP-hardfor k=2


- Proof: reduction from COMPACT BIPARTITE SUBGRAPH (CBS)

INSTANCE: a bipartite graph G=(U,V;E). Integers n, m

PROBLEM: does there exist a set VUX such that

(i)

(ii)

nX ||

mXGE |])[(|




PROBLEM: does there exist a set such that

(i)

(ii)

nX ||

mXGE |])[(|

n = 6, m = 5

VUX




PROBLEM: does there exist a set such that

(i)

(ii)

nX ||

mXGE |])[(|

n = 6, m = 5

( note that CBS is NP-hard because itincludes MAX BALANCED BIP GRAPH,for n = 2t, m = t^2 )

VUX

CBS: does there exist VUX such that nX || mXGE |])[(|

The reduction:The reduction:and

?

a

b

c

d

e

f

g

h

i

l



?

a

b

c

d

e

f

g

h

i

l

Let = {a,b,c,d,e,f,g,h,i,l} U { , } B = |E| - m



?

a

b

c

d

e

f

g

h

i

l


IDEA: Encode an edge (i,j) as …ij…

and make all k-mers x, x and unavoidable(i.e., insert a LOT of each of them in s)



?

a

b

c

d

e

f

g

h

i

l


The only kmers that can be destroyed are of the form x or x, and this is achieved by “flippling” the into a . This corresponds to removing theedge.

ij.. ...ij





?

a

b

c

d

e

f

g

h

i

l


The set of kmers of type x or xwhich remain define the set X whichcovers at least m edges



The only kmers that can be destroyed are of the form x or x, and this is achieved by “flippling” the into a . This corresponds to removing theedge.

ij.. ...ij



?

a

b

c

d

e

f

g

h

i

l




S =aa. . .aabb. . .bb. . .lll. . .lagbgei

-With similar reductions, we can also prove the

Theorem: the problem is NP-hard even for || = 2

5. Integer Linear 5. Integer Linear ProgrammingProgrammingformulationsformulations

Let K be the set of all possible kmers

hx

iz

Define a 0/1 variable in K

and a 0/1 variable for each position

x

Exponential-sizeExponential-size formulation (formulation (={0,1})={0,1})

K={ 000, 001, 010, 011, 100, 101, 110, 111 }

||,...,1 si

x min


x min

||

1

s

ii Bz


)1()1(),(),(

kzzxrEQLi

irDIFFi

i

x min

||

1

s

ii Bz

1||,...,1 ksr


Exponential n. of variables/constraints

pricing & separation problems

Our P&S strategy is exponential in the general case (polynomial for k fixed)

}0:{ * xA

Exponential n. of variables/constraints

pricing & separation problems

Our P&S strategy is exponential in the general case (polynomial for k fixed)

Fractional solutions may lead to effective heuristics (e.g., solving SUB(A) with

Polynomial-sizePolynomial-size formulation (formulation (={0,1})={0,1})

ijw


In addition to variables ,

there are variables for positions i and j saying

“is the kmer starting at i identical to the kmer starting at j ? “

iz

The variables w depend on s and z via linear constraints

hx

ijw

1iu

To count only kmers that have no identical kmer following them:

if all kmers starting after i are different from kmer at i


In addition to variables ,

there are variables for positions i and j saying

“is the kmer starting at i identical to the kmer starting at j ? “

iz

The variables w depend on s and z via linear constraints


-The formulation (not show) is basically a big boolean formula

-It yields a poor bound compared to the exponential formulation

-It can be improved in many ways (expecially via valid cuts)

-At this stage it’s not clear which method is best for not too small instances

-We’ll run experiments with variants of both formulations

Flipping letters to minimize the support of a string Giuseppe Lancia, Franca Rinaldi, Romeo Rizzi...

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