Efficient Haplotype Inference with Boolean Satisfiability

Transcription

1 Efficient Haplotype Inference with Boolean Satisfiability Joao Marques-Silva 1 and Ines Lynce 2 1 School of Electronics and Computer Science University of Southampton 2 INESC-ID/IST Technical University of Lisbon 22 March 2006

2 The HIPP Problem Assume a set G of n strings over the alphabet {0, 1, 2}, each with m characters Each character j in a string g i represented by g i j Example: g i = 012 A string g i is explained by two strings over the alphabet {0, 1}, h a and h b, iff: If g i j = 0, then h a j = h b j = 0 If g i j = 1, then h a j = h b j = 1 If g i j = 2, then h a j h b j Example: g i = 012 is explained by h a = 010 and h b = 011 Our goal is to compute a minimum-size set H of strings over the alphabet {0, 1} such that every string in G is explained by two strings in H Problem is NP-Hard (it is also APX-Hard)

3 An Example of the HIPP Problem Strings to be explained:

4 An Example of the HIPP Problem Strings to be explained: = = = = = = = = = = = = = = = = = = HIPP solution has size 6

5 In This Talk Relate the problem of haplotype inference from genotype data with the HIPP problem Propose a simple Boolean satisfiability (SAT) model for the HIPP problem Compact model Very efficient solution; by far the most efficient solution to the HIPP problem Modern SAT algorithms are extremely effective on the HIPP problem

6 Outline Haplotype Inference Haplotype Inference by Pure Parsimony (HIPP) ILP Models and Variants SAT-Based HIPP Model Results Conclusions

7 Outline Haplotype Inference Haplotype Inference by Pure Parsimony (HIPP) ILP Models and Variants SAT-Based HIPP Model Results Conclusions

8 Haplotype Inference Single Nucleotide Polimorphism (SNPs): DNA sequence variation, occurs when a nucleotide (typically A, C, G or T) changes among elements of the same species Haplotypes: Encode bi-allelic Single Nucleotide Polimorphisms (SNPs) Each site of a haplotype (describing a SNP) can take value 0 (the wild type) or 1 (the mutant) Genotypes: Typically available instead of haplotypes Each genotype describes two haplotypes Each site of a genotype can take value 0, 1 or 2 If site is 0 or 1, site is homozygous, and the two haplotypes must coincide at this site If site is 2, site is heterozygous, and the two haplotypes must differ at this site Haplotype Inference: Identify set of haplotypes which explain set of genotypes

9 Haplotype Inference N tide Alleles G/A C/A G/A C/G C/T T/C T/C T/C G/A C/G G/A C/T C/A h 1 A C G C C T T T A C G C C h 2 A C G G C C C C G G G C C h 3 G A A C C T T T A C G C C h 4 G C A C C T T T A C G C C h 5 G C A C C T T T G C G C C h 6 G C G C C T T T G C A C A h 7 G C G C C T T T G C A T A h 8 G C A C C T T T A C A C A h 9 A C G C T T T T A C G C C h 10 G C G C C T T T G C A C C h 11 G C G C C T T T G C G C C h 12 A C G G C T T T A C G C C Genotype g = is explained by haplotypes h 7 = and h 8 =

10 Why Haplotype Inference? Haplotype map of the human genome (see HapMap project) Mapping complex disease genes Inferring population histories Designing new drugs

11 Haplotype Inference by Pure Parsimony (HIPP) The pure parsimony criterion: Explain set of genotypes G with the least number of haplotypes Biological motivation; use the least number of entities that are required to explain natural phenomena Example: explain 2120, 2102, and 1221 A possible solution (using 6 haplotypes): 2120 = = = A pure parsimony solution (using 4 haplotypes): 2120 = = =

12 The HIPP Problem Revisited Assume a set G of n genotypes, each with m sites A genotype g i is explained by two haplotypes h a and h b iff: If g i j = 0, then h a j = h b j = 0 If g i j = 1, then h a j = h b j = 1 If g i j = 2, then h a j h b j Our goal is to compute a minimum-size set H of haplotypes such that every genotype in G is explained by two haplotypes in H Problem is NP-Hard (it is also APX-Hard)

13 Outline Haplotype Inference Haplotype Inference by Pure Parsimony (HIPP) ILP Models and Variants SAT-Based HIPP Model Results Conclusions

14 A Naive ILP Model I For each genotype g i enumerate all pairs of haplotypes that can explain g i With each pair of haplotypes associate a Boolean variable y ir, denoting whether the pair is selected for explaining g i Clearly, for each genotype g i one pair must be selected y ir = 1 Associate a Boolean variable x k with each haplotype h k, denoting whether h k is selected Total number of x variables equals number of candidate haplotypes; Can grow exponentially with the number of genotypes r

15 A Naive ILP Model II If a pair of haplotypes is selected (i.e. y ir = 1), then associated haplotypes must be selected If the pair of haplotypes for variable y ir includes h k, then the condition becomes y ir x k Objective is to minimize number of haplotypes selected min k x k Space complexity is exponential on the number of genotypes E.g., explained by pairs of haplotypes There are ILP models with polynomial space complexity

16 An Example g 1 = 212 g 2 = 021 x 1 x 2 (010, 111) (011, 110) x 3 x 4 (001, 011) y 11 y 12 y 21 Constraints: y 11 + y 12 = 1 y 21 = 1 y 11 x 1 y 11 x 2 y 12 x 3 y 12 x 4 y 21 x 5 y 21 x 3 x 5 x 3 Cost function: min 5 i=1 x i

17 Hapar A Branch-and-Bound Solution Also based on enumeration of candidate pairs of haplotypes (as in the naive ILP model) Uses greedy approach used for computing an upper bound on the number of required haplotypes For each genotype select pair of haplotypes each of which explains the largest number of genotypes Computed upper bound usually close to optimum Bounding procedure uses this upper bound A number of pruning techniques for eliminating irrelevant pairs of haplotypes Our results indicate that (until now) Hapar was the most efficient approach for the HIPP problem

18 Outline Haplotype Inference Haplotype Inference by Pure Parsimony (HIPP) ILP Models and Variants SAT-Based HIPP Model Results Conclusions

19 SAT-based Haplotype Inference by Pure Parsimony Clearly, the number of haplotypes lies between 1 and 2 n, 1 H 2 n It is in general possible to find tighter lower and upper bounds on the number of haplotypes The SAT model iteratively considers increasing numbers of candidate haplotypes Starts from the lower bound ub A trivial value is 1 Terminates for a value of k when all genotypes can be explained by k haplotypes Guaranteed to terminate until k = 2 n

20 Model for r Candidate Haplotypes I h 1 h 11 h 1m s a 1i s a ri g i1 g im g i s b 1i h r h r1 h rm s b ri Must select two haplotypes for explaining each genotype g i g variables only needed for heterozygous sites Candidate haplotypes represented with r m Boolean variables (h) Selector variables represented with 2 r n Boolean variables (s)

21 Model for r Candidate Haplotypes II Conditions on sites: (with 1 k r) If g ij = 0, add clauses ( h kj s a ki ) ( h kj s b ki ) If h k is used for explaining g i, then h kj = g ij = 0 If g ij = 1, add clauses (h kj s a ki ) (h kj s b ki ) If h k is used for explaining g i, then h kj = g ij = 1 If g ij = 2: Add variables gij a and gij b, and require gij a gij b : (g a ij g b ij ) ( g a ij g b ij ) Add clauses relating h and g variables: (h kj g a ij s a ki) ( h kj g a ij s a ki) (h kj g b ij s b ki ) ( h kj g b ij s b ki )

22 Model for r Candidate Haplotypes III Conditions on selector variables: Exactly one haplotype (for a and for b) is selected for each genotype g i : ( r ) ( r ) ski a = 1 ski b = 1 k=1 Space complexity of the model: Worst-case (r = Θ(n)): O(n 2 m) In practice (r = O(n)): O(n r m) In practice SAT model significantly more compact than ILP models Performance of the model: Not competitive with the best of previous solution, Hapar Need to develop optimizations to the basic model k=1

23 Optimizations I Structural Simplifications There can exist duplicate genotypes If two genotypes are identical, they can be explained by the same pair of haplotypes Eliminate duplicate genotypes; Reconstruct eliminated genotypes from computed haplotypes There can exist globally duplicated and globally complemented sites: Duplicated column Complemented column Eliminate duplicate/completed columns sites; Reconstruct eliminated columns from computed haplotypes

24 Optimizations II Lower Bounds I Two genotypes g a and g b are incompatible if at a given site j one has value 0 and the other has value 1 g a = 012 is incompatible with g b = 102 Can compute clique of mutually incompatible genotypes E.g., g a = 012, g b = 102 and g c = 110 are mutually incompatible, and form a clique of size

25 Optimizations II Lower Bounds II Use clique for computing a lower bound on the number of required haplotypes If genotype in clique has no heterozygous sites, then contribution to the lower bound is 1 Otherwise, each genotype in clique contributes 2 to the lower bound (2) (2) (1) E.g., for g a = 012, g b = 102 and g c = 110, the computed lower bound is 5 Can further improve lower bound (see paper(s))

26 Optimizations III Breaking Symmetries Problem formulation has key symmetries: Haplotypes h k1 and h k2 Selector variables s a k 1i, sa k 2i, sb k 1i and s b k 2i Boolean valuations v x and v y to the sites of haplotypes h k1 and h k2, e.g. h vx k 1 or h vy k 2 h vx k 1 and h vy k 2, with sk a 1i sa k 2i sb k 1i sb k 2i = 1001, corresponds to h vy k 1 and h vx k 2, with sk a 1i sa k 2i sb k 1i sb k 2i = 0110 h k1 = 0100 and h k2 = 0101 with sk a 1 isk a 2 isk b 1 isk b 2 i = 1001 explains genotype g i = 0102 h k1 = 0101 and h k2 = 0100 with sk a 1 isk a 2 isk b 1 isk b 2 i = 0110 also explains genotype g i = 0102 Symmetries can be eliminated: Enforce an ordering of the Boolean valuations to the haplotypes Require h v 1 < h v 2 <... < h v r for any valuation v

27 Organization of SHIPs SHIPs implemented as a Perl script that accepts list of genotypes to explain Script iteratively generates CNF models for increasing larger numbers of haplotypes Each CNF formula is passed on to a SAT solver Can interface minisat, siege, satz Algorithm terminates by returning the smallest number of haplotypes which can explain all genotypes

28 Outline Haplotype Inference Haplotype Inference by Pure Parsimony (HIPP) ILP Models and Variants SAT-Based HIPP Model Results Conclusions

29 SHIPs vs. Others on Standard Instances Instances solved within 7200 seconds CPU time: Instance #S/Rec #G RTIP Poly Hyb Hapar SHIPs Unif 10/ /15 15/15 15/15 15/15 15/15 10/ /15 15/15 15/15 15/15 15/15 10/ /15 12/15 15/15 15/15 15/ /15 11/15 15/15 15/15 15/ /50 27/50 35/50 50/50 50/ /10 4/10 6/10 9/10 10/ /10 3/10 3/10 9/10 10/10 NonUnif /15 14/15 15/15 15/15 15/ /15 8/15 15/15 15/15 15/ /15 0/15 6/15 14/15 15/ /15 0/15 5/15 6/15 15/ /15 0/15 3/15 4/15 15/15 Hapmap 30:75 7:68 0/24 15/24 15/24 17/24 23/24 TOTAL 92/ / / / /229

30 SHIPs vs. Hapar I hapar basic ships

31 SHIPs vs. Hapar II run time instances hapar basic ships

32 Outline Haplotype Inference Haplotype Inference by Pure Parsimony (HIPP) ILP Models and Variants SAT-Based HIPP Model Results Conclusions

33 Conclusions & Ongoing Research Work SAT-Based Haplotype Inference By far the most efficient solution to the HIPP problem Up to 5 orders of magnitude speedup wrt any other solution to the HIPP problem How precise is pure parsimony? SHIPs allows extensive analysis on real-world data Accuracy of HIPP is similar to the best statistical methods Optimizing the model Reducing the number of variables Additional sorting constraints, on the s variables Further uses of incompatibilities Promising preliminary results (see next slide!) Can use SAT to evaluate alternative criteria for the haplotype inference problem

34 The Near Future SHIPs Improved Model I 10 4 std instances 10 3 basic ships improved ships

35 The Near Future SHIPs Improved Model II 10 4 bio test data 10 3 basic ships improved ships