1. The computation of hitting sets: Review and new algorithms
Li Lin , Yunfei Jiang
Department of Mathematics, Jinan University, Guangzhou, PR China
Institute of Computer Software, Sun Yat-sen University, Guangzhou, PR China
From Information Processing Letters 86 (2003)
報告人：張家榮

2. Overview Introduction Backgrounds BHS-tree Boolean algorithm
Empirical results
Some ideas

3. Minimal hitting set
First used by R. Reiter 1987 in “A theory of diagnosis from first principles, Artificial Intelligence ”
Can be used to solve minimum set cover problem, diagnosis problem and teachers and courses problem.

4. Minimal hitting set
Given a collection C={Si │i ∈ N} of sets of elements from some universe U, a hitting set is a set S µ U such that S  Si  ® for all i. “Minimal” means no element in S can be deleted.
For example, C={ {M1, M2, A1}, {M1, A1, A2, M3}}. The minimal hitting sets are {M1}, {A1}, {M2, A2}, {M2, M3}.

5. Minimum set cover problem
Definition: A set of sets whose union has all members of the union of all sets. The set cover problem is to find a minimum size set.
Formal Definition: Given a set S of sets, choose c ⊆ S such that∪ c = ∪ S.

6. Reduced to MHS a b c d e 1 2 1 2 3 3 2 3 5 4 4 5 5 1 2 3 4 5 e c a d a

7. Diagnosis problem
Some components of a system may cease to operate as designed and cause the discrepancy between the expected behavior of the system and the observed behavior

8. Diagnosis problem (cont.)
(1) compute the collection of all minimal conflict sets
(2) transform the conflict sets into diagnoses.
A minimal conflict set is a minimal set of components, such that the assumption that each of these components is behaving correctly is inconsistent with the system description and the observation.

9. How? (1)Conflict sets Not my business (2)Diagnoses
Using minimal hitting set

10. Overview Introduction Backgrounds BHS-tree Boolean algorithm
Empirical results
Some ideas

11. Backgrounds (R. Reiter,1987)
Hitting sets can be computed by HS-trees.
Problem: the size of the tree grows exponentially
Minimal hitting sets can be efficiently obtained by a pruned HS-tree.
Problem: If the construction of the tree starts with an unfavorable set, the hitting sets may be deleted by pruning;

12. Backgrounds (cont.)
Greiner (1989) has revised the HS-tree into an HS-DAG in which the minimal hitting sets are not be deleted.
Separately, Reggia (1983) found the relationship between set-covering problem and hitting sets
Haenni (1997) found that the inversions of the hypergraph are the minimal hitting sets.

13. Backgrounds (cont.)
Vinterbo (2000) presented approximate hitting sets, He also used genetic algorithms.
Wotawa (2001) presented the HST-tree algorithm. The implementation of the algorithm can be done in a straightforward way, so an efficient implementation is possible.

14. Overview Introduction Backgrounds BHS-tree Boolean algorithm
Empirical results
Some ideas

15. Notations Minimal set cluster MCS = {C1,C2,…Cn}
Each node is a tuple (C,H), where C and H are set clusters.
The root node is (MCS, {}).
The left and right children of a node are denoted by (Cl ,Hl ) and (Cr ,Hr ), respectively.
function μ is used for the deletion of non-minimal conflict/hitting sets

16. BHS-tree The tree is defined recursively as follows.
if C = {}, then the BHS-tree is empty
else select any element a ∈ ∪ Ci, (Cl = {Ci -{a} | a ∈ Ci }, Hl = {a}) and (Cr = {Ci | a /∈ Ci }, Hr = {}).

17. Example

18. Algorithm
Step 1. If a node is leaf node, then MHS of this node is H; else run Steps 2 and 3 recursively.
Step 2. Replace every parent node H with { H ,{ ml ∪ mr | ml ∈ Hl , mr ∈ Hr } }.
Step 3. Minimize H at the root node with the functionμ until it comprises all minimal hitting sets

20. Online?
When a new measurement is added to the conflict sets it is not necessary to compute again the old conflict sets, but it is only necessary to add a new branch to the BHS-tree.

21. Add conflict set <5, 6>

22. Overview Introduction Backgrounds BHS-tree Boolean algorithm
Empirical results
Some ideas

23. Boolean algorithm
The conflict sets CS are presented as CNFs where each atom is negative.
Example Suppose CS = {C1,C2, ,Cm} is a conflict set cluster, where Ci = {ei1, ei2, ,ein},
Conflict-set Boolean formula (CSF)=
~e11 ~e12 …~e1n1 + ~e21 ~e22 …~e2n2 +…+
~em1 ~em2 …~emnm

24. Boolean algorithm(cont.)
A hitting set H is presented as a disjunction of its elements.
Example H={h1,h2, ,hn}
Hitting-set Boolean formula (HF)=…
h1h2. …hn
Why?
-Theorem 1
If H is a HS of CS, CSF · HF=0

25. Definition C is a boolean formula H(C) function is defined recursively
H(0)=1,H(1)=0
H( ~e )=e
H( ~e · C ) = e + H( C )
H( ~e + C ) = e · H( C )
else
5) H( C ) = e ·H(C1)+H(C2) where
C1 ⊆ C and ~e /∈ C1 and
C2 ={c | c ∪ {~e} ∈ C} ∪C1

26. Theorem 2
Suppose CSF is a Boolean formula of CS, then H(CSF) is a Boolean formula of HS of CS
The proof can be done by using mathematical introduction over the size of CS, k=|CS|

27. Proof of theorem 2 1)..4) are straightforward
Suppose that, when k · n, situation (5) is proved. Now take k = n + 1. For an arbitrary element e ∈ ∪ S∈CS S, H(CSF) = e ·H(CSF1)+H(CSF2), while CSF1 and CSF2 are as defined.

28. Proof of theorem 2(cont.)
By assumption, H(CSF1) is HS of CS1. CS1 ⊆ CS and ~e /∈ CS1,so e ·H(CSF1) must be HS of CS.
Obviously, H(CSF2) is HS of CS.
So, H(CSF) must be HS of CS
Minimum property can be proved by Boolean absorption properties.

29. Overview Introduction Backgrounds BHS-tree Boolean algorithm
Empirical results
Some ideas

30. Empirical results In ANSI C (UNIX)
(SGI 2200 Origin, CPU 4400 MHz, MIPS R12000 (IP27) processors, main memory 2 GB, OS IRIX 64 Release 6.5.)
HS-tree ( □ )
BHS-tree ( + )
Boolean algebraic ( O )

33. Empirical results (cont.)
Boolean algebraic algorithm needs less memory or running time.
It is not sensitive to the selection sequence of element
Only |CS| and ∪ c∈CS C influence the efficiency.

35. Difference between BHS-tree and Boolean algebraic
(1) BHS-tree: Binary tree structures
Boolean algebraic: list structures
(2) BHS-tree: Two steps
constructing the binary-tree
using it to compute the hitting sets recursively.
Boolean algebraic: One step

36. Overview Introduction Backgrounds BHS-tree Boolean algorithm
Empirical results
Some ideas

37. Approximation Algorithms for the Selection of Robust Tag SNPs
Kui Zhang
Ting Chen
This talk is about how to handle SNP genotyping with missing data.
My name is Yao-Ting Huang and my advisor is Kun-Mao Chao.
And we have two coauthors not here today, They are Prof. Zhang and Prof. Chen.
Yao-Ting Huang
Kun-Mao Chao
Dept. Computer Science & Information Engineering, National Taiwan University
Dept. Biostatistics, University of Alabama at Birmingham, USA
Dept. Biological Sciences, University of Southern California, USA
2019/2/24

38. Transformation Each SNP can distinguish partial pairs of patterns.
S1 can distinguish (P1, P3), (P1, P4), (P2, P3), and (P2, P4).
S2 can distinguish (P1, P4), (P2, P4), (P3, P4). S3 S4 S1 S2
To solve this problem, we first take a closer look at the function of each SNP.
If we pick SNP 1, we can be sure that we can distnguish patterns 1 and 3,
Because they are in different color at this SNP locus.
And we formulate this relation into a bipartite graph.
SNP 1 can also distinguish patterns 1 and pattern 4. and so on.
(1,2)
(1,3)
(1,4)
(2,3)
(2,4)
(3,4)
There are pairs of patterns

39. Observation 1: Tag SNPs The SNPs can form a set of tag SNPs iff
each pair of patterns is covered by at least one edge from the SNPs.
e.g., S1 and S3 can form a set of tag SNPs.
e.g., S1 and S2 can not be tag SNPs. S3 S1 S2
One unanswered question is what kind of SNPs can be tag SNPs.
We can easily answer this question by seeing if the bottom nodes in the graph are all covered by edges from them.
For example, SNPs 1 and 3 are tag SNPs.
And SNPs 1 and 2 are not tag SNPs.
Because patterns 1 and 2 are not covered.
So we can not distinguish patterns 1 and 2.
(1,2)
(1,3)
(1,4)
(2,3)
(2,4)
(3,4)
Each pair of patterns is covered by at least one edge

40. Observation It is the same with the minimal set cover problem .
Minimal set cover problem can be transformed to the minimal hitting set problem.

41. Observation (cont.)
CSF = ( ~S3 ~S4+ ~S1 ~S3 + ~S1 ~S2 ~S4 + ~S1 ~S4 + ~S1 ~S2 ~S3 + ~S2 ~S3 ~S4 )
= ~S3 ~S4 + ~S1 ~S3 + ~S1 ~S4
H(CSF) = S3 ·H( ~S1 ~S4 ) +H(~S4 + ~S1 + ~S1 ~S4 )
= S3 · S1 + S3 · S4 + S4 · S1
So, S1 and S3 can form a set of tag SNPs.
so can S3 and S3 ,S1 and S4

42. Observation 2: Missing DataP1 P2 P3 P4 S3 S1 S2
If a SNP is genotyped as missing data, it is the same as the removal of its node and edges. S4 S3 S4 S1 S2
Another important question is what’s the effect of missing data?
It is easy to tell by this graph because it’s just like removing the node and edges from the graph.
(1,2)
(1,3)
(1,4)
(2,3)
(2,4)
(3,4)
Suppose S4 is genotyped as missing data

43. Problem ReformulationS3 S4 S1 S2
To tolerate m missing tag SNPs, we need to find a set of SNPs such that each pair of patterns is covered by (m+1) edges.
e.g., We wish to find a set of robust tag SNPs that tolerates 1 missing tag SNP. S4 S3 S1
From the above two observations, we claim that if we wanna tolerate m missing data,
We have to guarantee that each bottom node is covered by at least m plus 1 edges.
For example, if we wanna tolerate one missing data, SNPs 1 3 and 4 can be robust tag SNPs.
Because each node is covered by at least two edges.
(1,2)
(1,3)
(1,4)
(2,3)
(2,4)
(3,4)
Each pair of patterns is covered by at least two edges

44. Modification of boolean algorithm
H1(0)=1,H1(1)=0, H0( C ) =1
H1( ~e )=e
Hi( ~e · C ) = e · Hi-1( C ) + Hi ( C )
Hi ( C ) = 0 if there exists a CS ∈ C which has elements less than I
Hi( ~CS+C ) = CS · Hi( C ) if CS has elements equal to I
Hi( C ) = e ·Hi-1(C1)+Hi(C2) where
C1 ⊆ C and ~e /∈ C1 and
C2 ={c | c ∪ {~e} ∈ C} ∪C1

45. Modification of boolean algorithm (cont.)
To tolerate m missing tag SNPs, we can use Hm+1(CSF)
Example
H2(CSF)= S3 S4 ·H2( ~S1 ~S3 + ~S1 ~S4)
= S1 S3 S4 ·H2 (~S1 ~S4)
= S1 S3 S4