Haplotype Inference Constrained by Plausible Haplotype Data

Haplotype Inference Constrained by PlausibleHaplotype Data

Michael R. Fellows, Tzvika Hartman,Danny Hermelin, Gad M. Landau,

Frances Rosamond, andLiat Rozenberg

Abstract—The haplotype inference problem (HIP) asks to find a set of haplotypes

which resolve a given set of genotypes. This problem is important in practical fields

such as the investigation of diseases or other types of genetic mutations. In order

to find the haplotypes which are as close as possible to the real set of haplotypes

that comprise the genotypes, two models have been suggested which are by now

well-studied: The perfect phylogeny model and the pure parsimony model. All

known algorithms up till now for haplotype inference may find haplotypes that are

not necessarily plausible, i.e., very rare haplotypes or haplotypes that were never

observed in the population. In order to overcome this disadvantage, we study in

this paper, a new constrained version of HIP under the above-mentioned models.

In this new version, a pool of plausible haplotypes eH is given together with the set

of genotypes G, and the goal is to find a subset H � eH that resolves G. For

constrained perfect phylogeny haplotyping (CPPH), we provide initial insights and

polynomial-time algorithms for some restricted cases of the problem. For

constrained parsimony haplotyping (CPH), we show that the problem is fixed

parameter tractable when parameterized by the size of the solution set of

haplotypes.

Index Terms—Haplotyping, perfect phylogeny, pure parsimony, polynomial-time

algorithms, parameterized complexity.

Ç

1 INTRODUCTION

GENETIC information in living organisms is encoded in DNAsequences that are organized into chromosomes. Diploid organisms,such as humans, have two copies of every chromosome, which arenot necessarily identical, with each copy called a haplotype.Identifying the common genetic variations that occur in humansis valuable in understanding diseases [1]. The genetic sequences ofthe population are almost totally identical, except from some basesthat differ from one person to another with a frequency of morethan some threshold (one percent for example). Those differencesare the common genetic variations and they are known as singlenucleotide polymorphisms (SNPs).

The data described in each haplotype may be the full DNA, butit is more common to consider only the data of the SNPs, since theother sites are assumed to be identical. A genotype is the descriptionof the two copies (haplotypes) together. When the two haplotypesagree, the site in the genotype has the agreed base. Such a site iscalled a homozygous site. When the two haplotypes disagree, thegenotype has both bases, yet it does not tell which base occurs inwhich haplotype. This type of site is called heterozygous.

Current biological technologies give us an easier and cheaperway to obtain genotype data in comparison to haplotype data.

However, the haplotype information is the one of greater use [19].For this reason, it is necessary to computationally infer thehaplotype information from the genotype data. An importantbiological fact is that almost always there are only two bases at anSNP, which can be marked as 0 and 1. A genotype will have 0 or 1if the two haplotypes both have 0 or 1 in the same site, respectively,or 2 otherwise (see Fig. 1).

In view of that, a set of genotypes and a set of haplotypes can berepresented as matrices. A genotype matrix is a matrix over f0; 1; 2gwhere each row is a genotype and each column represents an SNP,and a haplotype matrix is a matrix over f0; 1g, where each row is ahaplotype and each column represents an SNP.

For the rest of the paper, let gðiÞ represent the data at site i ofgenotype g, and hðiÞ the data at site i of haplotype h. Furthermore,we will write gðijÞ (resp. hðijÞ) instead of gðiÞgðjÞ (resp. hðiÞhðjÞ).Definition 1 (Resolution). A pair of haplotypes fh; h0g is said to

resolve g if for each i: gðiÞ ¼ hðiÞ when hðiÞ ¼ h0ðiÞ, and gðiÞ ¼ 2otherwise. We extend this and say that a set of haplotypes H resolvesa set of genotypes G, if for each g 2 G, there is a pair fh; h0g 2 Hwhich resolves g. The pair fh; h0g is called a resolution of g, and H isa resolution of G.

Definition 2 (Haplotype Inference Problem (HIP)). Given a set of ‘

genotypes G, each of length m, find a resolution of G.

Note that a genotype having d � m heterozygous sites (sitesmarked with 2) has 2d�1 possible resolving pairs of haplotypes. Thegoal is to find the set of pairs which are as close as possible to thereal set of haplotypes that created the genotype. Currently, thereare two models used in practice that give two different biologicallymotivated heuristics on how to determine this:

1. Perfect Phylogeny. The perfect phylogeny model is acoalescent model which assumes no recombination. Thismeans that the history of the haplotypes is represented as atree where two haplotypes from two individuals have atmost one recent common ancestor [19] (see [17], [19], [24],[32] for further information). Formally, a set of haplotypes(binary sequences) of length m defines a perfect phylogeny ifthe haplotypes appear as labels of a rooted tree whichobeys the following properties [12]:

a. Each vertex of the tree is labeled by a binary sequenceof length m representing a possible haplotype;

b. Every edge ðu; vÞ is marked with i, where the base atsite i in sequence u is different from the one insequence v. Every coordinate i labels at most oneedge.

A common way of checking whether a set of haplotypes

defines a perfect phylogeny is to check whether it obeys

the four gamete test, i.e., the corresponding haplotype

matrix does not contain, in any two columns, the forbidden

gamete submatrix

0 11 00 01 1

0BB@

1CCA:

See Fig. 2 for an example of a perfect phylogenetic tree, and

the corresponding haplotype matrix. The Perfect Phylogeny

Haplotyping (PPH) problem is the problem of finding for a

given set of genotypes a resolution which defines a perfect

phylogeny, whenever such a resolution exists.2. Pure Parsimony. The pure parsimony model seeks the

minimum set of haplotypes that resolves a given set ofgenotypes. The biological motivation behind this is the

1692 IEEE/ACM TRANSACTIONS ON COMPUTATIONAL BIOLOGY AND BIOINFORMATICS, VOL. 8, NO. 6, NOVEMBER/DECEMBER 2011

. M.R. Fellows and F. Rosamond are with the Charles Darwin University,Darwin, NT Australia.E-mail: {michael.fellows, fran.rosamond}@cdu.edu.au.

. T. Hartman is with Google, Tel-Aviv, Israel. E-mail: [email protected].

. D. Hermelin is with Max Planck Institute for Informatics, Saarbrucken,Germany. E-mail : [email protected].

. G.M. Landau is with the University of Haifa, Haifa, Israel, and thePolytechnic University, New York. E-mail: [email protected].

. L. Rozenberg is with IBM, Haifa, Israel. E-mail: [email protected].

Manuscript received 16 Aug. 2009; revised 5 Jan. 2010; accepted 3 May 2010;published online 19 Aug. 2010.For information on obtaining reprints of this article, please send e-mail to:[email protected], and reference IEEECS Log Number TCBB-2009-08-0137.Digital Object Identifier no. 10.1109/TCBB.2010.72.

1545-5963/11/$26.00 � 2011 IEEE Published by the IEEE CS, CI, and EMB Societies & the ACM

statistical observation that the number of distinct haplo-types in the population is vastly small [18], [19]. TheParsimony Haplotyping (PH) problem is the problem offinding a resolution of smallest size possible for a given setof genotypes.

In [17], Gusfield showed that the PPH problem is solvable in

Oðnm�ðnmÞÞ time, where � is the inverse Ackerman function.

Gusfield also showed a linear-time algorithm to build, once the

first solution is found, a linear-space data structure that represents

all PPH solutions. However, his work is based on complex graph-

theoretic algorithms which are difficult to implement [19]. In [4],

[12], algorithms fine-tuned to the actual combinatorial structure of

the PPH problem were shown. These algorithms run in Oðnm2Þtime and are easy to understand and implement. They also give a

representation of all PPH solutions. More recent work developed

OðnmÞ time algorithms: In [9], the algorithm is graph-theoretic and

uses a directed rooted graph called a “shadow tree,” and in [30],

the algorithm is based on interdependencies among the pairs of

SNPs, and builds a data structure called “FlexTree” to represent all

PPH solutions. Other works have researched different variations of

PPH [5], [8], [14], [23].The PH problem was first suggested and proved to be NP-hard

by Earl Hubell (unpublished). Gusfield formally introduced the

problem in [18], and proposed an integer linear programming

solution. More integer linear programming solutions following this

were proposed in [6], [18], [21], [27]. Approximation algorithms for

the problem were presented in [27], [28], and in [33], a branch-and-

bound algorithm was proposed. Other theoretical results were

shown in [25], [27], [31]. Most notably is the work of Sharan et al.

[31], who characterized restricted instances of PH under the term

ð�; �Þ-bounded, where � and � stand for the maximum number of

heterozygous sites per row and column of the genotype matrix.

Sharan et al. also showed that the PH problem is fixed parameter

tractable (see [10] for a formal definition) when parameterized by

the number of k haplotypes in the resolution of G. Many other

works have researched different variations of the haplotyping

problem [15], [16], [20], [22], [26], [29].Van Iersel et al. continued in [25] to explore the bounded

instances of the PH problem and of another related problem

Minimum Perfect Phylogeny Haplotyping (MPPH), which looks for

the minimum number of haplotypes that resolve a given set of

genotypes and also define a perfect phylogeny. The MPPH

problem was proved to be NP-hard [3]. The known results up

till now are: For the PH problem, the ð3; �Þ instance is APX-hard

[27], the (4,3) instance is APX-hard [31], the (3,3) instance is APX-

hard [25], the ð2; �Þ instance is polynomial-time solvable [7], [28],

and the ð�; 1Þ instance is also polynomial-time solvable [25]. For the

MPPH problem, the (3,3) instance is APX-hard, and the ð2; �Þ and

ð�; 1Þ instances are polynomial-time solvable [25]. In both pro-

blems, the ð�; 2Þ instance was left as open problem, but it was

shown that it polynomial-time solvable in a special structure of the

genotype matrix [25], [31].All known algorithms up till now for haplotype inference find

resolutions for a given set of genotypes from the superset of all

possible haplotypes (i.e., all m-length binary vectors). However,

these algorithms may find resolutions that include binary vectors

representing haplotypes that do not actually occur in the

population, or are otherwise very rare. It is, therefore, biologically

interesting to force the resolving haplotypes to be chosen only from

a specific pool that contains only plausible haplotypes, i.e.,

haplotypes which have already been observed in relatively high

frequencies in previous experiments. This pool can be determined

by empirically setting up some statistical threshold, or by any other

reasonable method.In view of all this, we study, in this paper, a new constrained

variant of the haplotype inference problem, in which a pool of

plausible haplotypes eH is given alongside the set of genotypes G,

and the goal is to find a resolution of G which is a subset of eH.

Definition 3 (Constrained Haplotype Inference Problem (CHIP)).

Given a set of ‘ distinct genotypes G, each of length m, and a pool of n

distinct plausible haplotypes eH for G, each of length m, find a

resolution H � eH of G.

The constrained perfect phylogeny haplotyping (CPPH) pro-

blem and the constrained parsimony haplotyping (CPH) problem

are defined accordingly. Note that if ‘ > nðn� 1Þ=2 in the above

definition, there is no solution possible, since by taking the entire

pool of n plausible haplotypes, we can resolve at most nðn� 1Þ=2genotypes. On the other hand, there is no inequality necessarily in

the other direction. We therefore assume ‘ � nðn� 1Þ=2 through-

out the paper.Note that CPH can be easily shown to be NP-hard by the

straightforward reduction from ð3; �Þ-bounded PH (shown to be

NP-hard in [27]) which designates eH to be all possible haplotypes

resolving genotypes in G. At the time that this paper was in

review, we did not know the complexity of CPPH. Recently,

Elberfeld and Tantau showed this problem is polynomial [11].

Here, we present polynomial-time algorithms with slightly slower

running times for the ð�; �Þ-bounded cases of ð�; 1Þ, ð2; �Þ, (5,2), and

(3,3). In the second part of the paper, we show that like PH [31],

CPH is fixed-parameter tractable when parameterized by the

number of haplotypes in a minimum size resolution H � eH of G.

The running time of this algorithm was improved recently by

Fleischer et al. [13].

IEEE/ACM TRANSACTIONS ON COMPUTATIONAL BIOLOGY AND BIOINFORMATICS, VOL. 8, NO. 6, NOVEMBER/DECEMBER 2011 1693

Fig. 1. Example of two haplotypes and the corresponding genotype. (b) The samehaplotypes and genotype using the 0, 1, 2 representation. Note that the figuresimplifies matters a bit, since in general each column cannot be relabeled in thesame way.

Fig. 2. Example of a perfect phylogenetic tree for the haplotypes h1 ¼ ð01000Þ,h2 ¼ ð01010Þ, h3 ¼ ð10100Þ, and h4 ¼ ð10001Þ. The matrix H is the haplotypematrix of these four haplotypes, and it obeys the four gamete test.

2 POLYNOMIAL-TIME SPECIAL CASES OF CPPH

In this section, we describe polynomial-time algorithms for CPPH

with genotype matrices of specific structures. In [31], bounded

cases of genotype matrices were introduced in order to explore the

complexity of PH. The bounded cases were defined as follows:

Definition 4 ((�; �)-bounded [31]). A genotype matrix G is (�; �)-

bounded if it has at most � 2’s per row and at most � 2’s per column.

Either � or � might be � which means there is no bound on the number

of 2’s per row or column, respectively.

Here, we use the same term of (�; �)-bounded to present

polynomial-time algorithms for special cases of CPPH. We will

present algorithms for the following cases: ð�; 1Þ, ð2; �Þ, ð5; 2Þ,and ð3; 3Þ.

We begin with the following lemma which lists six matrices that

we can assume G does not contain as submatrices, since including

any one of these implies that all resolutions of G necessarily

include the forbidden gamete submatrix. Note that we allow

permutating rows and columns in each of the submatrices. The

proof of the lemma is left to the reader.

Lemma 1. If G includes one of the following 2� 3 submatrices:

2 00 21 1

0@

1A; 2 1

1 20 0

0@

1A; 2 0

1 20 1

0@

1A; 2 1

0 01 0

0@

1A; or

2 01 10 1

0@

1A;

or the following 2� 2 submatrix 2 02 1

� �, then G does not have a perfect

phylogenic resolution.

As in Eskin et al. [12], we will be working with pairs of columns

in G. Pairs of sites of a genotype can be split into two types,

according to the data in these sites. Type I includes the pairs of

sites that have only one possible resolution. These pairs of sites are

(00), (01), (11), (20), and (21). The resolutions of these sites are

described in the following list:

1:ð00Þ !0 0

0 0

� �2:ð01Þ !

0 1

0 1

� �3:ð11Þ !

1 1

1 1

� �

4:ð20Þ !0 0

1 0

� �5:ð21Þ !

0 1

1 1

� �:

Type II includes pairs of sites with (22) (22-columns). A pair of 22-

columns has two potential resolutions: ð22Þ ! 0 01 1

� �, which will be

called an equal resolution, or ð22Þ ! 0 11 0

� �, which will be called an

unequal resolution.Determining whether there is a perfect phylogenic resolution of

G boils down to deciding the resolution type, equal or unequal, for

every pair of 22-columns. For some 22-columns, the type of the

resolution is determined by the given set of genotypes, for others,

it is determined by the given set of haplotypes, and for the rest we

need algorithms that will find the proper resolution.

2.1 Preprocessing

We next present a preprocessing stage which is performed before

all algorithms regardless of the specific structure of the input sets

of genotypes or haplotypes.A 22-columns ij must be resolved equally if the given set of

genotypes G includes at least one of the following submatrices in

columns ij: 0 01 1

� �, 2 0

1 2

� �, 2 0

1 1

� �, or 2 1

0 0

� �, since any resolution of G must

include the combinations “00” and “11” in this case. For a similar

reason, columns ij must be resolved unequally when G includes at

least one of the submatrices 0 11 0

� �, 2 0

0 2

� �, 2 1

1 2

� �, 2 1

1 0

� �, or 2 0

0 1

� �. We will

call these type of constraints on the resolution type genotype

constraints. In addition, a 22-columns ij must be resolved equally

(unequally) if the haplotype set includes for some genotype only

equal (unequal) resolutions. These type of constraints will be calledhaplotype constraints.

The preprocessing ensures that haplotype pairs which violate theabove constraints will not be chosen. For each genotype gi 2 G, weuse eHðgiÞ to denote all possible resolutions of gi in eH, i.e.,eHðgiÞ ¼ ffh; h0g j h; h0 2 eH; h and h0 resolve gig. The preprocessingstep includes the following four steps:

1. Check whether the genotype matrix G can be resolved in aperfect phylogenic way (use any algorithm from [4], [9],[12], [30]). If not, report there is no solution.

2. For each genotype gi 2 G, 1 � i � ‘, go over all pairs ofhaplotypes from eH and compute eHðgiÞ.

3. For each genotype constraint, delete from the setseHðg1Þ; . . . ; eHðg‘Þ all haplotype pairs that violate the

constraint, i.e., all haplotype pairs that resolve the relevant

sites in a different way than the constraint indicates.4. For each haplotype constraint, delete from the setseHðg1Þ; . . . ; eHðg‘Þ all haplotype pairs that violate it. Note

that the deletion of haplotypes may create new haplotypeconstraints. Repeat Step 4 until there is no change in thehaplotype constraints.

After each step of Steps 2 to 4 in the preprocessing stage, ifone of the haplotypes sets eHðg1Þ; . . . ; eHðg‘Þ becomes empty, itmeans there is no solution, and we are done. Once thepreprocessing is complete, our goal is to find a resolution H �eH of G, by selecting one pair of haplotypes from each eHðgÞ,g 2 G. From here on out, we will only be concerned withresolutions of this type. In other words, we will use eH as the setSg2G

eHðgÞ for the remainder of the section. Note that even afterthe preprocessing stage, not all resolutions in eH will define aperfect phylogeny. This is because we are still left with 22-columns that have yet been resolved, as there might be twodifferent genotypes g and g0 which share a common pair of 22-columns ij, and eHðgÞ and eHðg0Þ includes both resolutions for ij(i.e., equally and unequally). Such a pair of resolutions is said tobe conflicting, and more generally, any pair of resolutions fh1; h

01g

and fh2; h02g are conflicting if fh1; h

01; h2; h

02g does not define a

perfect phylogeny. We have the following two important lemmas:

Lemma 2. Let fh1; h01g; . . . ; fhr; h0rg be resolutions of r (not necessarily

distinct) genotypes in G after the preprocessing stage. Then,fh1; h

01g; . . . ; fhr; h0rg are pairwise nonconflicting iff H ¼S

1�i�rfhi; h0ig defines a perfect phylogeny.

Proof. Let G0 � G denote the subset of r genotypes which H

resolves. If H defines a perfect phylogeny, then clearlyfh1; h

01g; . . . ; fhr; h0rg are pairwise nonconflicting. To prove the

other direction of the lemma, it suffices to show that H does notcontain the forbidden gamete matrix. Suppose by way ofcontraction that this is not the case. Then, the forbidden gametesubmatrix occurs in H at some pair of sites i; j 2 f1; . . . ; mg. Weconsider three possible cases:

. There is no 2 occurring in column i nor column j in G0.But this means that G0 cannot be resolved in a perfectphylogenic way, and so the preprocessing algorithmwould have halted at Step 1 and reported that nosolution can be found.

. G0 has a 2 occurring at one of the two columns i or j,say column i, but there is no pair of 22-columnsoccurring at ij. Let g 2 G0 be a genotype with gðiÞ ¼ 2.Then, either gðjÞ ¼ 1 or gðjÞ ¼ 0, so suppose first thatgðjÞ ¼ 1. We can assume w.l.o.g. that the forbiddengamete submatrix occurs in rows fha; hb; hc; hdg, wherefha; hbg is a resolution in fh1; h

01g; . . . ; fhr; h0rg that

resolves g. Furthermore, we know that hcðijÞ ¼ 10 andhdðijÞ ¼ 00, and moreover, hc and hd belong to two


different resolutions in fh1; h01g; . . . ; fhr; h0rg, as other-

wise fha; hbg and fhc; hdg would be conflicting. Thus,let g0; g00 2 G0 be the two distinct genotypes resolved bythe resolutions in fh1; h

01g; . . . ; fhr; h0rg involving hc and

hd. We know that g0ðijÞ 2 f10; 20; 02g and g00ðijÞ 2f00; 20; 02g. From this, it is not difficult to verify thatG0 either contains the submatrix 2 1

2 0

� �at rows gg0, gg00, or

g0g00, and columns ij, or it contains one of the followingthree submatrices at rows gg0g00 and columns ij:

2 11 20 0

0@

1A; 2 1

0 21 0

0@

1A; or

2 10 01 0

0@

1A:

In either case, any one of these submatrix implies that

G0 has no solution (Lemma 1), and the preprocessing

stage would have halted at its first step. The case where

gðjÞ ¼ 0 is similar.. G0 has a pair of 22-columns at ij occurring in some row

g 2 G0. Again, we can assume w.l.o.g. that the forbiddengamete submatrix occurs in rows fha; hb; hc; hdg, wherefha; hbg is a resolution in fh1; h

01g; . . . ; fhr; h0rg that

resolves g. Suppose first that ha and hb resolve ij

equally, i.e., haðijÞ ¼ 11 and hbðijÞ ¼ 00. Then, hcðijÞ ¼01 and hdðijÞ ¼ 10, and we know that hc and hd bothbelong to two different resolutions in fh1; h

01g; . . . ;

fhr; h0rg. Let g0; g00 2 G0 be the two genotypes that hcand hd resolve. Then, it must be that g0ðijÞ 2 f01; 02; 21gand g00ðijÞ 2 f10; 20; 12g. From this, it is not hard toverify that G0 must include one of the following sixsubmatrices at rows g0g00 and columns ij:

0 1

1 0

� �;

0 2

2 0

� �;

2 1

1 2

� �;

0 1

1 2

� �;

0 1

2 0

� �; or

0 2

1 2

� �:

If G0 includes the last submatrix, then G0 does not have

a perfect phylogenic resolution in the first place

(Lemma 1), and so the preprocessing stage would have

reported “no solution.” If G0 includes one of the first

five submatrices, then there is a genotype constraint on

ij stating that it must be resolved unequally, and so

fha; hbg would have been removed from eHðgÞ at Step 3

of the preprocessing stage. In both cases, we reach a

contradiction. The case where fha; hbg are resolved

unequally is similar.

In all three cases, we have reached a contradiction to the fact thepreprocessing stage did not halt, and so the lemma is proven. tu

Lemma 3. The preprocessing stage takes Oðm4n2 þm2n4Þ time.

Proof. The first step of the preprocessing takes Oð‘mÞ-time byusing one of the algorithms in [9], [30], which is Oðmn2Þ whenrecalling that ‘ < n2. The second step takes Oðmn4Þ time, sincechecking whether a specific haplotype pair resolves a specificgenotype takes OðmÞ time, and there are ‘ ¼ Oðn2Þ genotypesand Oðn2Þ haplotype pairs. The third step takes Oðm2n2Þ-time,since finding one genotype constraint and deleting all violatinghaplotype pairs takes Oð‘þ n2Þ ¼ Oðn2Þ time, and this is donefor each column pair of which we have at most Oðm2Þ of. Thelast step takes Oðm2n2 �minfm2; n2gÞ, since finding all haplo-type constraints and deleting all violating haplotype pairstakes Oðm2n2Þ time, and we repeat this at most minfm2; n2gtimes. This is because each one of the Oðm2Þ column pairs maycreate a new constraint only once, and since we cannot delete

more than n2 haplotype pairs. Thus, the total running time ofthe preprocessing stage is Oðm2n2 �minfm2; n2g þmn4Þ ¼Oðm4n2 þm2n4Þ. tu

2.2 The Dependency Graph

The definition of conflicting resolutions brings us to the notionindependency and dependency between genotypes. Looselyspeaking, a dependency between two genotypes g; g0 2 G ariseswhen the decision on how to resolve g affects the decision on howto resolve g0. This obviously happens when there is a resolutionfh1; h

01g 2 eHðgÞ conflicting with a resolution fh2; h

02g 2 eHðg0Þ. In

this case, we say that g and g0 are directly dependent. If there is noresolution in eHðgÞ conflicting with a solution in eHðg0Þ, we say that gand g0 are not directly dependent.

We next introduce the dependency graph DGðGÞ of our given setof genotypes G, after they have been preprocessed by thealgorithm in the previous section. Later on, we will use theproperties of the dependency graph in our polynomial algorithms.

Definition 5 (dependency graph). The dependency graph DGðGÞof a set of genotypes G is a graph which has a vertex for each genotypeg 2 G, and edge between vertices representing directly dependentgenotypes.

Lemma 4. After the preprocessing stage, two genotypes that do not haveany pair of 22-columns in common are not directly dependent.

Proof. Consider two genotypes g; g0 2 G that don’t have any

common pair of 22-columns. Suppose by way of contradiction

that there is a resolution fhx; hyg 2 eHðgÞ conflicting with a

fhx0 ; hy0 g 2 eHðg0Þ. This means that there is a pair of columns ij

of fhx; hy; hx0 ; hy0 g that has the forbidden gamete matrix. But

this can only happen when G includes (in the rows gg0, and in

columns ij) the forbidden submatrix 2 02 1

� �of Lemma 1. tu

Lemma 5. Let G1 and G2 be two connected components in DGðGÞ afterthe preprocessing stage has been performed. If H1 � eH and H2 � eHare perfect phylogenic resolutions of G1 and G2, respectively, thenH1 [H2 is a perfect phylogenic resolution of G1 [G2.

Proof. Consider any pair of resolutions in H1 [H2 of twogenotypes g; g0 2 G1 [G2. If g and g0 are not both in G1, nor inG2, then there is no edge between them in DGðGÞ, meaning thatthey are not directly dependent. Hence, the pair of resolutionsis nonconflicting by definition. If g; g0 2 G1 or g; g0 2 G2, then byLemma 2, the pair of resolutions is nonconflicting as both H1

and H2 define a perfect phylogeny. It follows that allresolutions in H1 [H2 are pairwise nonconflicting, and soagain by Lemma 2, H1 [H2 defines a perfect phylogeny. tu

In view of Lemma 5, every connected component can beresolved individually, and the union of the chosen haplotypes willgive the desirable resolution of G. As mentioned in Definition 2, agenotype matrix G is (�; �)-bounded when it has at most � 2s perrow and at most � 2s per column, where � or � might be � toindicate there is no such limitation. The following algorithms referto different values of � and �, as the headlines indicate. Note thatthroughout the remaining sections, we assume that the preproces-sing stage has been completed.

2.3 ð�; 1Þ- and ð2; �Þ-Bounded Cases

Let us begin with the simple cases of ð1; �Þ-bounded and ð�; 2Þ-bounded genotype matrices.

Lemma 6. Any pair of distinct genotypes g1 and g2 in a ð�; 1Þ-boundedgenotype matrix is not directly dependent.

Proof. Since there is at most one 2 per column, g1 and g2 cannothave a 2 in the same column. Due to this, g1 and g2 can nevershare a pair of 22-columns, which according to Lemma 4, meansthey cannot be directly dependent. tu


According to Lemma 6, if G is a ð�; 1Þ-bounded genotypematrix, then the dependency graph DGðGÞ has no edges, and itsconnected components are of size one. The algorithm will choosefor H one pair of haplotypes from every set eHðgÞ, g 2 G, and H willdefine a perfect phylogeny according to Lemma 5.

Lemma 7. Any pair of distinct genotypes g1 and g2 in a ð2; �Þ-boundedgenotype matrix is not directly dependent.

Proof. According to Lemma 4, two genotypes are directlydependent only if they have a pair of 22-columns in common.If g1 and g2 are distinct genotypes in a ð2; �Þ-bounded genotypematrix, and they have a pair of 22-columns in common, itmeans that G contains at the forbidden submatrix 2 0

2 1

� �of

Lemma 1, and so at the first step of the preprocessing stage, wewould have determined that there is no solution. tu

According to Lemma 7, if G is a ð2; �Þ-bounded genotypematrix, then the dependency graph DGðGÞ has no edges, and itsconnected components are of size one. The algorithm for solvingthe ð2; �Þ-bounded case is thus identical to the above algorithm forthe ð�; 1Þ-bounded case.

Theorem 1. CPPH with ð�; 1Þ- or ð2; �Þ-bounded genotype matrices issolvable in Oðm4n2 þm2n4Þ time.

Proof. According to the above, in the case of ð�; 1Þ- or ð2; �Þ-boundedgenotype matrices G, one can determine a perfect phylogeneticresolution for G by picking a resolution pair from each eHðgÞ,g 2 G. Since this can be done in Oð‘Þ time, the preprocessingstage dominates the running time of the entire algorithm, whichaccording to Lemma 3 is Oðm4n2 þm2n4Þ. tu

2.4 (5,2)-Bounded Case

It is convenient to mark every edge gg0 in the dependency graphDGðGÞ with the indices of the 2-columns g and g0 share. Observethat in the ð�; 2Þ-bounded case, a specific index will appear only onone edge of the dependency graph, since there are no more thantwo genotypes that have 2 in the same column. Thus, in the ð�; 2Þ-bounded case, each genotype g 2 G can be adjacent to at mostb�=2c other genotypes in DGðGÞ, since according to Lemma 4, eachadjacency means that g shares a pair of 22-columns with hisneighbor genotype. We thus have the following importantproperty in the ð5; 2Þ-bounded case:

Lemma 8. If G is a (5,2)-bounded genotype matrix then DGðGÞ hasmaximum degree 2.

The lemma above implies that if G is a ð5; 2Þ-bounded genotypematrix, then every connected component in DGðGÞ is either a pathor a cycle. In the following lemma, we show that resolving a pathor a cycle in DGðGÞ reduces to solving a 2-SAT instance for whichwe have a well-known polynomial (even linear) time algorithm.

Lemma 9. If G0 is a connected component of DGðGÞ which is either apath or cycle, then one can determine whether G0 has perfectphylogenetic resolution H 0 � eH in Oðsn2Þ time, where s is the sumover all se number of column labels on edges e of DGðG0Þ.

Proof. Let G0 be connected component in DGðGÞ which is either apath or a cycle. As a first step, we get rid of all edges that havemore than two labels by subdividing these edges. That is, if gg0

is an edge in G0 labeled by l1; . . . ; lt, t > 2, then we replace thisedge by the path g1; g2; . . . ; gdt=2e; g

0, where gi; giþ1 is labeled withl2i�1; l2i for i 2 f1; . . . ; dt=2e � 1g, and gdt=2e; g

0 is labeled withlt�1; lt. (Observe that in the ð5; 2Þ-bounded case t � 4, and in factt � 3 if we assume that G0 has more than two genotypes.) Wetherefore henceforth assume that any edge in G0 has two labelscorresponding to a pair of 22-columns.

Let g1; . . . ; gr denote the genotypes in G0 according to theirorder of appearance in G0 (i.e., gi is adjacent to giþ1 for each

i 2 f1; . . . ; r� 1g, and gr is adjacent to g1 if G0 is a cycle). Weconstruct a 2-CNF boolean formula �ðG0Þ corresponding to G0,such that �ðG0Þ is satisfiable iff there is a perfect phylogenicresolution for G0 in eH. For each pair of 22-columns ij appearingon an edge in G0, we assign a boolean variable xij, where weinterpret xij ¼ 1 (resp. xij ¼ 0) to mean that ij should beresolved equally (resp. unequally). The clauses of �ðG0Þ areconstructed as follows: First, if G0 is a path, and ab and cd arethe labels on the edges g1g2 and gr�1gr, respectively, then weadd the clause ðxabÞ (resp. ð:xabÞ) to �ðG0Þ if ab can only beresolved equally (resp. unequally) in eHðg1Þ, and similarly, weadd ðxcdÞ (resp. ð:xcdÞ) if cd can only be resolved equally (resp.unequally) in eHðgrÞ. Next, for every genotype gi of degree twoin DGðG0Þ, where ab and cd be the 22-columns labeling theedges gi�1gi and gigiþ1 (or gig1 in case i ¼ r and G0 is a cycle), weproceed according to the following:

1. If there is no equal resolution for ab in eHðgiÞ, then weadd the clause ð:xabÞ to �ðG0Þ.

2. If there is no unequal resolution for ab in eHðgiÞ, then weadd the clause ðxabÞ to �ðG0Þ.

3. If every equal resolution for ab in eHðgiÞ is an equalresolution for cd, then we add the clause ðxab _ :xcdÞ to�ðG0Þ.

4. If every equal resolution for ab in eHðgiÞ is an unequalresolution for cd, then we add the clause ðxab _ xcdÞ to�ðG0Þ.

5. If every unequal resolution for ab in eHðgiÞ is an equalresolution for cd, then we add the clause ð:xab _ :xcdÞto �ðG0Þ.

6. If every unequal resolution for ab in eHðgiÞ is an equalresolution for cd, then we add the clause ð:xab _ xcdÞ to�ðG0Þ.

It is not difficult to see that �ðG0Þ is satisfiable iff G0 has aperfect phylogenetic resolution H 0 � eH. Furthermore, theconstruction of �ðG0Þ requires Oð‘sÞ ¼ Oðsn2Þ time, since weneed at most Oð‘Þ time for each edge in G0, and there are atmost s such edges. Also observe that �ðG0Þ has at most OðsÞvariables and clauses. Thus, using the linear-time algorithm ofAspvall et al. for determining whether a 2-CNF is satisfiable [2],we obtain the required time bound in the lemma. tu

Theorem 2. CPPH restricted to ð5; 2Þ-bounded genotype matrices is

solvable in Oðm4n2 þm2n4Þ time.

Proof. According to Lemmas 5 and 8, solving CPPH in the ð5; 2Þ-bounded case boils down to resolving connected components inDGðGÞ which are either paths or cycles. According to Lemma 9,

this can be done in Oðsn2Þ time for any such component G0,where s is the total number of column labels that appear on

edges of G0. Since the total number of edge labels in DGðGÞ is atmost m in the ð5; 2Þ-bounded case, the total time required to

resolve all connected components of DGðGÞ is Oðmn2Þ.Accounting also for the time required by our preprocessing

stage, we get the desired time bound of the theorem. tu

2.5 (3,3)-Bounded Case

We next turn to deal with the ð3; 3Þ-bounded case. Similar to theð5; 2Þ-bounded case of the previous section, here every connected

component of DGðG0Þ will have a very simple structure. We havethe following lemma:

Lemma 10. If G is a ð3; 3Þ-bounded genotype matrix, then every

connected component in DGðGÞ is either a path, a cycle, or is of size

at most four.

Proof. Let g be some genotype which is directly dependent withthree other distinct genotypes g1, g2, and g3, and let abc denote

the three columns with gðaÞ ¼ gðbÞ ¼ gðcÞ ¼ 2. Then, the


submatrix of G with rows fg1; g2; g3g and the columns abc mustbe of the following form:

f0; 1g 2 22 f0; 1g 22 2 f0; 1g

0@

1A:

Here, f0; 1g specifies that any one of the two values f0; 1g canappear in the submatrix. It is not difficult to see that since G hasat most three 2’s in each column and each row, and sinceadjacency in DGðG0Þ requires sharing at least a pair of 22-columns (Lemma 4), none of the four genotypes g, g1, g2, and g3

can be adjacent to any other genotype in DGðGÞ. It follows thatany connected component in DGðGÞ with a vertex of degree atleast three has size at most four. The lemma thus follows fromthe fact that every connected component of maximum degreetwo is either a path or a cycle. tu

Theorem 3. CPPH restricted to ð3; 3Þ-bounded genotype matrices issolvable in Oðm4n2 þm2n4Þ time.

Proof. First observe that all connected components ofDGðGÞwhichare paths or cycles can be resolved in the time bound given inthe theorem (including the time required for preprocessing)using a similar analysis as the one used in Theorem 2. Next, notethat since G has at most three 2’s in each row, the number ofpossible resolutions for each genotypes is Oð1Þ, and given agenotype g 2 G, all these resolutions can be found in Oðm‘Þ ¼Oðmn2Þ time by simply going through all haplotypes in eH. Itfollows that one can determine whether a connected componentof size at most four in DGðGÞ can be resolved in a perfectphylogenetic way in the same time complexity, i.e., Oðmn2Þtime. Since there are at most Oð‘Þ ¼ Oðn2Þ components, the timebound in the theorem above follows. tu

3 FPT ALGORITHM FOR CPH

We next consider the CPH problem. In [31], Sharan et al. showedthat PH is fixed parameter tractable when parameterized by thesize k of the minimum size resolution of G. Essentially, this meansthat there is algorithm for PH running in fðkÞ �Nc, where fðÞ is anarbitrary (typically, superpolynomial) computable function, N isthe total input size, and c is a constant independent of k (see [10]for a formal definition of a parameterized algorithm). Observe thatthis is, in general, much faster than the brute-force NkþOð1Þ

algorithm. Here, we show an analogous result for CPH. Ourapproach involves solving a dynamic program to determinewhether there is any H � eH of size � � k which resolves G.Throughout the section, we use Gi, 1 � i � ‘, to denote the subsetof genotypes fg1; . . . ; gig � G.

Probably the first dynamic-programming solution to come tomind for CPH, is to compute all possible resolutions H 0 � eH of Gi

from the resolutions of Gi�1. However, the number of k-subsetsresolving Gi might be �ðnkÞ, which is too much in terms of aparameterized algorithm. We therefore take an alternate route.Instead of computing the actual subsets which resolve Gi, we willcompute abstract “blueprints” of these subsets. But before wedefine these blueprints, let us briefly change our terminology a bit.For presentation purposes, it will be convenient to considerresolutions as ordered sequences rather than sets, where we alsoallow some haplotypes to appear more than once in the sequences.Thus, from here on out, we will think of a resolution of size � as asequence of haplotypes ðh1; . . . ; h�Þ, with possibly hi ¼ hj for somei 6¼ j 2 f1; . . . ; �g. The blueprints of such resolutions, which we call�-plans, are formally defined as follows:

Definition 6 (�-plan). Let � be an integer in f1; . . . ; kg. A �-plan is astring of length i � ‘ over the alphabet ffx; yg j 1 � x � y � �g.

Let H 0 ¼ ðh1; . . . ; h�Þ be a resolution of Gi ¼ fg1; . . . ; gig of size

�. A �-plan p is associated with H 0 if when fx; yg is the jth letter in p,

1 � x � y � � and 1 � j � i, then hx and hy resolve gj. We will say

that p is valid for Gi if there is a resolution of Gi associated with p.

In this way, a valid �-plan does not describe the actual resolution

of Gi, but it does provide all relevant information concerning

which genotypes are resolved using the same haplotypes.

Definition 7 (DP½�; i�;DP½�; i�). Let � be an integer in f1; . . . ; kg, and i

be an integer in f1; . . . ; ‘g. We denote by DP½�; i� the set of all �-

plans of length i, and by DP½�; i� � DP½�; i� the set of all valid �-

plans for Gi ¼ fg1; . . . ; gig.Lemma 11. jDP½�; i�j � jDP½�; i�j � kOðk2Þ for any � � k and i � ‘.Proof. To prove the lemma, recall that ‘ < k2. The number of

distinct strings of length at most k2 over an alphabet of size at

most �2

� �< k2 is bounded by kOðk

2Þ. tu

Our algorithm proceeds by computing DP½�; i� in increasing

values of � and i. The base cases of this computation are

1. DP½�; 1� ¼ DP½�; 1� for all 1 � � � k, and2. DP½1; i� ¼ DP½2; i� ¼ ; for all 2 � i � ‘.

Clearly, G can be resolved using � � k haplotypes if and only if

G‘ ¼ G has at least one valid �-plan. Hence, assuming we can

correctly compute DP½�; i� for all 1 � � � k and 1 � i � ‘, the

correctness of our algorithm is immediate. What remains to be

described is the dynamic-programming step for computing

DP½�; i�.For this, we will first need to introduce some additional

terminology. Let p be some �-plan which is valid for Gi, and let

h; h0 2 eH be some pair of (not necessarily distinct) haplotypes. For

a given x; y 2 f1; . . . ; �g, we say that the assignment of hx ¼ h and

hy ¼ h0 is compatible with p if there is a resolution H 0 ¼ ðh1; . . . ; h�Þof Gi associated with p such that hx ¼ h and hy ¼ h0. We will need

the following lemma.

Lemma 12. Let p be a valid �-plan for Gi, for some i 2 f1; . . . ; ‘g, and let

h; h0 2 eH. Then, given any pair of integers x; y 2 f1; . . . ; �g, one can

determine whether the assignment of hx ¼ h and hy ¼ h0 is

compatible with p in Oðk3mÞ time.

Proof. The algorithm is an iterative process that proceeds as

follows: At each step, the algorithm has a set D of indices

representing all haplotypes which have already been deter-

mined by the algorithm. Thus, at each iteration, we assume that

the algorithm has a specific haplotype hd for each d 2 D. At the

beginning of the algorithm D ¼ fx; yg, where hx ¼ h and

hy ¼ h0.Now, at the start of each iteration, the algorithm goes

through all indices d 2 D, and for each such index, thealgorithm marks all positions j, 1 � j � i, with a letter fd; d0goccurring in p. For each such position j, the algorithm computeshd0 from gj and hd. If d0 62 D, then the algorithm has discovered anew haplotype (possibly identical to a previously discoveredhaplotype), and so it adds d0 to D and starts another iteration. Ifd0 2 D, there are two possible outcomes: Either hd0 is equal tothe haplotype assigned to index d0 by the algorithm in the past,or not. In the first case, the algorithm continues its computationat this iteration, and in the latter the algorithm determinesincompatibility. The algorithm terminates when it has eitherdeclared incompatibility, or when there are no new indicesadded to D in a given iteration. In this case, the algorithmdeclares compatibility.

The correctness of this algorithm is immediate when itdeclares incompatibility. Observe that in case the algorithmdetermines compatibility, it has not necessarily determined allhaplotypes in a resolution corresponding to p. Nevertheless,


since p is known to be a valid a �-plan for Gi, the correctness inthis case follows as well. As for its time complexity, notice thatthe algorithm performs at most � iterations, with each iterationrequiring Oð‘mÞ time. Thus, the total time complexity of thisalgorithm is Oð�‘mÞ, which is Oðk3mÞ since ‘ < k2. tu

We are now in position to present the dynamic-programming

computation of the table DP. The dynamic-programming step for

computing DP½�; i� proceeds as follows:

1. DP½�; i� DP½�� 1; i�.2. For each p 2 DP½�� 2; i� 1�:

a. Concatenate f�; �� 1g to the end of p, and add thisnew �-plan to DP½�; i�.

3. For each h; h0 2 eH resolving gi, for each p 2 DP½�� 1;i� 1�, and for each x 2 f1; . . . ; �� 1g:

a. Set y ¼ x, and check whether the assignment of hy ¼hx ¼ h is compatible with p. If so, concatenate fx; �g tothe end of p, and add this new �-plan to DP½�; i�.

4. For each h; h0 2 eH resolving gi, for each p 2 DP½�; i� 1�,and for each x 6¼ y 2 f1; . . . ; �g:

a. Check whether the assignment of hx ¼ h and hy ¼ h0is compatible with p. If so, concatenate fx; yg to theend of p, and add this new �-plan to DP½�; i�.

Correctness of the dynamic-programming step is straightfor-

ward. Indeed, any valid �-plan p0 of Gi is either a �0-plan of Gi for

some �0 < � (Line 1), or it can be decomposed into either:

. A valid ð�� 2Þ-plan p of Gi�1 concatenated to a new letterf�; �� 1g (Line 2). In this case, in any resolutionðh1; . . . ; h�Þ associated with p0, we know that h� and h��1

resolve gi, and that ðh1; . . . ; h��2Þ resolves Gi�1.. A valid ð�� 1Þ-plan p of Gi�1 concatenated to a new letter

f�; xg for some 1 � x � �� 1 (Line 3). In this case, in anyresolution ðh1; . . . ; h�Þ associated with p0, h� and hx resolvegi, and ðh1; . . . ; h��1Þ resolves Gi�1.

. A valid �-plan p of Gi�1 concatenated to a letter fx; yg forsome 1 � x < y � � (Line 4). In this case, in any resolutionðh1; . . . ; h�Þ associated with p0, hx and hy resolve gi, andðh1; . . . ; h�Þ also resolves Gi�1.

Theorem 4. CPH is solvable in kOðk2Þn2m time, where k is the size of the

required resolution of G.

Proof. The description of the algorithm, and its correctness, is

discussed above. Let us next analyze its time complexity. First,

note that the number of different entries in the dynamic-

programming entries DP isOðk‘Þ, which isOðk3Þ since ‘ < k2. To

compute the entry DP½�; i�, for � 2 f2; . . . ; kg and i 2 f2; . . . ; ‘g,we go through all four steps of the dynamic program described

above. It is not difficult to see that the time required in fourth

step dominates the time required for all the rest of the steps, so

we analyze only this step. In the fourth step, we go through all

pairs of haplotypes h; h0 2 eH that resolve gi, through all �-plans

i n DP½�; i� 1�, a n d t h r o u g h a l l p a i r s o f i n d i c e s

x 6¼ y 2 f1; . . . ; �g, and for each such set of objects we invoke

the algorithm given in Lemma 12 which runs in Oðk3mÞ time.

Since the total number of �-plans in DP½�; i� 1� is bounded by

kOðk2Þ according to Lemma 3, and since checking whether a pair

of haplotypes resolves gi requires OðmÞ time, the total time

complexity of this step is bounded by Oðn2mÞ þ n2kOðk2Þk2 �

k3m ¼ kOðk2Þn2m. Thus, the total time complexity of computing

the entire DP table is Oðk3Þ � kOðk2Þn2m ¼ kOðk2Þn2m. tu

4 CONCLUSIONS AND OPEN PROBLEMS

In this paper, we studied a new variant of the haplotype inference

problem which we termed Constrained Haplotype Inference, where

the haplotypes to be inferred are constrained to be a subset of a

given pool of plausible haplotypes. We focused on two established

models for haplotype inference, the perfect phylogeny model, and

the pure parsimony model, and gave both positive and negative

results for constrained haplotype inference under these models. We

believe that our results could prove to be helpful as subroutines to

quickly solve optimally subinstances where the number of

heterozygous sites are known to be limited, and/or the number

of inferred haplotypes is expected to be relatively small. Also, we

hope our results will provide insights into the general haplotype

inference problem, which has yet to be understood completely.

Finally, as a possible future direction of research, it would be useful

to extend the results here and in succeeding papers [11], [13] to

hold for weighted variants of the problems discussed.

ACKNOWLEDGMENTS

The authors would like to thank an anonymous referee for pointing

out a bug in the proof of Lemma 2 which appeared in an early

version of the paper. Work done while at CRI, Haifa University,

and Department of Computer Science, Bar-Ilan University.

G. Landau is partially supported by the US National Science

Foundation Award 0904246, the Israel Science Foundation grant

347/09, Yahoo, Grant No. 2008217 from the United States-Israel

Binational Science Foundation (BSF), and DFG.

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