An Approx. Algo. For Alignment MSA using Motif Discovery

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An Approx. Algo. For Alignment MSA using Motif
Discovery

• B.J. Chen
• M.W. Chang
• C.H. Tsai
• H.Y. Chuang
• NTU

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MSA

• Multiple Sequence Alignment
• Given N sequences, align these sequences,
  possibly with gaps, that brings out the best
  commonality of these N sequences.
• Usually measure the alignment by penalizing the
  mis-matches and gaps, and rewarding the matches.

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Previous works

• Alt89, CHM94, CL88, GKS95etc. works best
  for small values of N (about 2-6).
• ZZ96, Vih98 handle the case of
• N gt2 by first applying pairwise alignment
  techniques.

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Motivation of this work

• For larger values of N, we need additional
  constraints to give biological meaningful
  alignment.
• MOTIF A common patterns across sequences.

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Motivation of this work

• Alignment number K (2 K N)
• A user controlled parameter constrains the
  alignment to have at least K sequences agree on a
  character, whenever possible, in the alignment.
• The commonality across most sequences is required
  to be detected.

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Problem Description

• Align multiple strings with gaps
• Hope Brings out the best commonality
• Parameter K , the alignment number
• The alignment should has at least K sequences
  agree on a character, whenever possible, in the
  alignment.
• K 1 has no meaning

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Problem Definition

• Given N sequences , each have length ni
• Let matrix A be an alignment , the size of A is N
  x T ( T is the length after alignment, T gt ni )
• Let Ai be i th row in A,throw all gaps in Ai
  will become original sequence_i

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Problem Definition(2)

• Define EA(a,i,j) 1 if Ai,j a
• EA(a,i,j) 0 o.w
• (a,j) Si EA(a,i,j)
• a in column j is bad if (a,j) lt K
• Define
• S (a,j)
• a (a,j) gt 0
• (a,j) lt K

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Problem Definition(3)
A A B B B B C C
X X O O O O X X

• Example Given K 4
• A B
• 4

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Problem Definition(4)

• Minimize whole badness
• K-MSA
• The paper proposes the proof that
• this problem is MAX SNP hard (Ill report this
  part later)

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Stage 1 Motif Discovery

• We begin by defining a motif in a sequence
• Given a string s on alphabet ? and an integer K,
  2 K s
• Definition 1 K-motif
• A string m on ??. is K-motif with location list
  Lm ( l1,l2,,lp), if all of the following
  hold
• m0, mm-1 belong to ?
• p K
• Every dont care character at position j in m,
  there exist at least two distinct occurrences li1
  and li2, 1 i1, i2 p , such that s li1j ?
  s li2j

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Definition 2

• Maximal Motif
• Let p1, p2, , pk be all the motifs in the
  sequence s. A motif pi is maximal if and only if
• there is no pj (j ? i) such that pi is a
  substring of pj, or
• if pi is a substring of pj, then there exists at
  least one occurrence of pi in s that is not
  covered by pj in s

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Definition 3

• Redundant motif
• A maximal motif m, with location list Lm, is
  redundant if there exist maximal motifs mi, 1 i
  p , such that Lm Lm1 ? Lm2 ? ? Lmp
• A motif that is not redundant is called an
  irredundant motif

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Example 1

• Let the input string s have the following form
• ac1c2c3baXc2c3bYac1Xc3bYYac1c2Xb
• _________________________________________
• ac1c2c3b aXc2c3bY ac1Xc3bYY
  ac1c2Xb
• ab
  
• a..c3b
• a.c2.b
  
• ac1..b
  
• a.c2c3b
• ac1.c3b
• ac1c2.b
  

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Motif Discovery

• The motifs of interest to the sequence alignment
  problem are the irredundant motifs lemma 4
• There exists a polynomial time algorithm to
  extract them from the input
• In the rest of the paper we use motifs that occur
  in multiple sequences, i.e., occur in each
  sequence exactly once.
• If there exists a motif that occurs more than
  once in a sequence, then for the arguments that
  follow, each occurrence is treated as a distinct
  motif

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Lemma 4

• Lemma 4 if p is a redundant motif, then using
  the motif p does not improve the cost of the
  K-MSA optimization problem
• Proof. Let p be rendered redundant by motifs p1,
  p2, , pn, n 1. By definition, motif p has less
  number of solid-characters than each of pi, 1 i
  n. Thus if an alignment can use motif p, it can
  certainly use all the motifs p1, p2, ,pn, giving
  a larger number of solid-characters thus a
  higher cost for the K-MSA optimization problem

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Stage 2 Sequence Alignment

• Definition 4
• two irredundant motifs pi and pj, if there
  exists a sequence s containing both these motifs
• Let ni and nj be the sizes of the motifs pi and
  pj and let li and lj be the locations (offsets)
  in a sequence s respectively
• Motifs Overlap if the intervals li, li ni
  and lj, lj nj have a non-empty intersection

p1
p2
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Definition 5

• Pairwise Compatible Motifs Two motifs, p1 and
  p2, are pairwise feasible if there exists an
  alignment of the sequences that does not
  introduce gaps in the motifs p1 and p2

p1
p2
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Lemma 1

• Two irredundant motifs pi and pj are pairwise
  feasible if and only if none of the following
  hold
• (domain crossing mismatch) if pi and pj do not
  overlap in all the sequence, then pi is to the
  left pj, without loss of generality
• (overlap mismatch) if pi and pj overlap in any
  sequence, than pi is at some fixed distance d to
  left of pj, without loss of generality

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Mismatch
Motif A
Motif B
Overlap mismatches
(i)
(ii)
Domain-crossing mismatches
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Definition 6

• Motif alignment Given a set M of motifs, a motif
  alignment of the sequences, s1, s2, ,sm, is the
  alignment such that in all the sequences, without
  breaking the motifs in M, the motifs are aligned
  (in all the sequences they appear)
• Feasible set If such an alignment exists, the
  set is called a feasible set

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Definition 7

• Linear ordering of motifs given a set of
  feasible motifs, a consistent ordering of the
  motifs such that in every sequence, the set of
  motifs that are present in the sequence appear in
  the left to right order is called the linear
  ordering

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Example 2

• H I A J G L B
• A M C N B Q D
• C O P E D R F
• H I S J G E T U F
• H I V J G
• AB in sequence 1 and 2
• CD in sequence 2 and 3
• E..F in sequence 3 and 4
• HI.JG in sequence 1, 4 and 5

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Example 2

• (iv) (i) (ii) (iii)
• H I A J G L B -- -- -- --
• -- -- A M C N B Q D -- --
• -- -- -- -- C O P E D R F
• H I S J G -- -- E T U F
• H I V J G -- -- -- -- -- --

i
ii
iii
iv
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Example 3

• H I A J G L B -- -- -- --
• -- -- A M C N B Q D -- --
• -- -- -- -- C O P E D R F
• -- -- -- -- -- H I E J G F
• -- -- -- -- -- H I K J G --
• No alignment due to overlap mismatch
• There is no linear ordering of the motifs

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Example 4

• A G H D X Y
• A I C D J F
• X Y C D K F
• A..D in sequences 1 and 2
• CD.F in sequences 2 and 3
• XY in sequences 1 and 3
• linear order of the motifs is (i), (iii), (ii)
• no alignment due to crossing mismatch

i
iii
ii
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Definition 8

• Domain crossing error Given a set of motifs, m1,
  m2,, mn, a domain crossing error is said to
  occur if there exists a linear ordering of the
  motifs mi1, mi2, , min, yet there exists no
  alignment that respects all the n motifs

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Lemma 2

• A set of irredundant motifs p1, p2, , pn is
  feasible if and only if none of the following
  holds
• There exist distinct motifs pi and pj such that
  pi and pj are pairwise infeasible
• There exists a non-empty subset of the motifs
  without a linear ordering
• There exists a non-empty subset of the motifs
  that demonstrate domain crossing error

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The Graph-theoretic Formulation

• Construct a directed graph G ( V, E) where every
  motif pi corresponds to a vertex vi, thus n
  V. The directed edge are introduced as follows
• There is no edge between two vertices where the
  two corresponding motifs do not occur
  simultaneously in any sequence
• If pi is to the left of pj in every sequence that
  the two motifs are present, then a directed edge
  is placed from vi to vj
• The edges are labeled as follows
• Forbidden if the motifs are not pairwise
  feasible
• Overlap if the motifs corresponding to v1 and v2
  overlap
• Non-overlap if the motifs do not overlap

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Handling Domain Crossing Mismatches - intro

• Introduce consistent graph w.r.t. a vertex.
• First, define distance on edges.
• Dv1,v2 minimum distance between v1 and v2
  in every sequence that both them appear in.

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Consistent Graph w.r.t. a Vertex

• G (V,E) for each edge (u,v) label?
  forbidden, overlap, nonoverlap weight Du,v
  
• Definition
• Valid path contains no forbidden edge
• Overlap-path all edges in the path are
  overlap
• Weight Dp ? De, e in p (path)

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Consistent Graph w.r.t. a Vertex

• p ?V The graph is consistent w.r.t. p if
• ?q ?V, for all pair of vertex-disjoint valid
  paths from p to q, P1 and P2,
• 1. Dp1 Dp2, if P1 and P2 are both
  overlap-paths, or,
• 2. Dp1 ? Dp2, if P1 is an overlap-path and P2 is
  not.

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An Example

• Consider a previous example.
• AGHDXY
• AICDJF
• XYCDKF
• A..D
• CD.F
• XY

P1 v1 -gt v2, D 2 P2 v1 -gt v3 -gtv2, D
6 Not consistent!
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Graph Consistent v.s. Domain Crossing
Mismatches

• Use Graph Consistent property to deal with the
  domain crossing mismatches problem.
• If the induced graph is consistent w.r.t every
  vertex, then there is no domain crossing
  mismatches.
• Moreover, it may rule out overlap mismatches.

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Lemma Rewritten

• Given a subset of vertices(motifs) v1vn,
  construct graph as previously defined. This
  induced subgraph on v1vn is feasible, if the
  following holds
• 1. there is no edge labled forbidden in the
• subgraph,
• 2. the induced subgraph is acyclic, and,
• 3. the subgraph is consistent w.r.t. every vi.

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Algorithm

• Idea an infeasible set to a feasible set

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Algorithm

• Detect the basic infeasible subsets.
• Eliminate motifs to obtain a feasible set that
  maximizes the cost.
• Render the alignment.

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Algorithm

• Before we construct the basic infeasible sets, we
  have one more definition
• Given a graph G and two cycles C1 and C2 on G,
  if all vertices defining C1 are also in C2, then
  C2 is redundant with respect to C1.

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Algorithm Step1

• Compute the following sets
• 1. Fi contains the 2 endpoints of i-th
  forbidden edge.
• 2. Cj contains vertices that form a directed
  non-redundant cycle in the graph.
• 3. Pk contains vertices that a non-redundant
  path in the graph.

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Algorithm Step1

• The number of the total basic infeasible sets we
  obtained in Step1 is bounded.
• More precisely, there would be no more than .
• It is easy to prove, since the number would be
  bounded by the number of cycles or closed path.

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Algorithm Step2

• Mapping to SETCOVER problem.
• v1, v2vn F1 ? Fnf ? C1Cnc ? P1Pnp
• U F1Fnf, C1Cnc, P1Pnp
• Si Fl vi ?Fl, 1? l ? nf ?
• Cl vi ?Cl, 1? l ? nc ?
• Pl vi ?Pl, 1? l ? np
• costi solid_char in motifi sequences
• containing motifi
• AB..CD ? 4

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Algorithm Step3

• Find feasible subgraph containing only overlap
  edges.
• For each sequence, find an ordering of motifs in
  that subgraph. Say, p1, p2pj.
• Align sequences.

p1 p2 p4
p1
p2 p4
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Example

• A set of 3 sequences
• GFPCQFSAG
• GFPCQFSGG
• GPCQSAGK
• K2

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Example

• Irredundant motifs (By Teiresias)
• PCQ in Seq. 1,2,3
• GFPCQFS.G in Seq. 1,2
• PCQ..G in Seq. 2,3
• SAG in Seq. 1,3
• GFPCQFSAG
• GFPCQFSGG
• GPCQSAGK

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Example

• Sequences to Graph

2 (0.V.)
1
2
0 (O.V.)
2 (O.V.)
3 (non.)
4
3
3 (O.V.)
6 (O.V.)
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Example

• Basic infeasible sets
• No F
• No C
• Closed path set P,
• P1 214
• P2 123
• P3 134
• P4 234

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Example

• Make feasible set by S.C. problem
• G P1,P2,P3,P4
• S1 P1,P2,P3 w1 3
• S2 P1,P2,P4 w2 9
• S3 P2,P3,P4 w3 6
• S4 P1,P3,P4 w4 3
• Solution S1,S4

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Example

• Get aligned blocks from overlap graphs
• GFPCQFS.G
• PCQ..G

2
2 (O.V.)
3
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Example

• Result ---
• G F P C Q F S a G
• G F P C Q F S G G
• - g P C Q s a G K

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Demonstration

• IBM Bioinformatics Group
• http//cbcsrv.watson.ibm.com/Tmsa.html

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Approximation Analysis

• Q1 Is it polynomial time?
• Recall this algorithm---
• 1. Convert to graph.
• 2. Detect the basic infeasible subsets.
• 3. Eliminate motifs to obtain a feasible set that
  maximizes the cost.
• 4. Render the alignment.

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AnalysisQ1

• Step1---Covert to graph
• one irredundant motif ? one vertex
• one relation between two irredundant motifs ?
  one edge
• of irredundant motifs is linear in the input
  size. (by Par98)

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AnalysisQ1

• Step2--- Compute the following sets
• 2.1 Fi contains the 2 endpoints of i-th
  forbidden edge.
• Simply scan all the edges and collecting the
  end points of the edges labeled Forbidden.
• In G(V,E),Vn
• E n(n-1)/2O(n2)

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AnalysisQ1

• 2.2 Cj contains vertices that form a
  directed non-redundant cycle in the graph.
• Do DFS of the graph to capture all the
  cycles. (If you do a DFS, you have a cycle iff
  you have a back edge. )
• Dont need to traverse along a whole path.
• DFS is polynomial.

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AnalysisQ1

• 2.3 Pk contains vertices that a non-redundant
  path in the graph.
• Do BFS rooted at each vertex. This search is
  also partial as in the last step.
• n(BFS) is still polynomial.

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AnalysisQ1

• Step3--- Mapping to SETCOVER problem.
• Use V. Chvatal.
• A greedy heuristic for the set covering
  problem.
• Math. Oper. Res.,4233-235,1979

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AnalysisQ1

• Step4---
• Find a polynomial algorithm for each block.
• After summing up these 3 steps, its still
  polynomial.

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Approximation Analysis

• Q2Is it feasible?
• Yes. Because of the blocking alignment, the
  solution is an alignment which doesnt break any
  needed motifs.

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Approximation Analysis

• Q3 What is approximation factor?
• Using the result of Chv79,the set cover
  problem is approximable within 1 lnX, where X
  is the number of the elements of the ground set.
  We have shown that the maximum number of
  non-redundant basic infeasible sets is no more
  than (nN)6. Thus X (nN)6 . The approximation
  factor is 1 6ln(nN).

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Conclusion

• Two stages one is identifying the local
  similarities and the other is aligning the
  similarities appropriately.
• We dont think its a good approximation.

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MAX SNP

• Definition
• The class of problems having constant-factor
  approximation algorithms, but no approximation
  schemes unless PNP.
• In other words, no PTAS exits.
• Fact Maximum Cut is MAX SNP hard

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Proof Procedure

• First , the paper prove that the max-version
  K-MSA is MAX SNP hard.
• Second , the paper prove that K-MSA is MAX SNP
  hard

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MAX Version K-MSA

• Define
• K-MSA (max)

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Proof Procedure(2)

• The procedure to prove K-MSA(max) is MAX SNP hard
• Give a reduction from Maximum Cut to Bipartite
  Maximum Cut
• Then give another reduction from BMC to K-MSA(max)

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Maximum Cut

• Given graph (V,E),
• Output Two vertex set V1,V2
• Max The cardinality of the cut, i.e., the
  number of edges with one end point in V1 and one
  endpoint in V2.

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Bipartite Maximum Cut

• Given graph (Vtop,Vdown,E),
• any e in E is between Vtop,Vdown
• value(e) 1 or -1
• Output Two vertex set V1,V2
• Max The cardinality of the cut, i.e., the
  number of edges with one end point in V1 and one
  endpoint in V2.

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Proof Procedure(3)
Some instance of Bipartite Maximum Cut problem
Any instance of Maximum Cut problem
Some instance of K-MSA(max) problem

• Whole procedure from MC to K-MSA(max)
• some trouble in middle stage

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Procedure Overview(1)

• MC to BMC
• Given a cut of BMC , a cut of MC is constructed

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Procedure Overview(2)

• BMC to K-MSA(max)

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Procedure Overview(3)

• If we have those two reduction, then
• Assume that we have a PTAS for K-MSA(max)
• gt
• We will have a PTAS for Maximum Cut !

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MC to BMC

• A basic idea for mc to bmc
• Split a vertex into top and bottom
• Much more complex because K-MSA
• Given a Graph. Construct an instance of BMC. For
  each vi , construct 3di 6 vertex.

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MC to BMC (2)
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MC to BMC(3)
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MC to BMC(4)
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MC to BMC(5)

• So Complicated? Dont Worry
• Main vertex
• only these point connect to outside
• only negative edge
• only postive edge
• Banalce to zero

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Construct the Cut of MC

• Define

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Construct the Cut of MC(2)

• In any cut of BMC , we can modify the cut such
  that
• without decreasing the cost.
• Then we assign vi ( in MC) according to
  ffff .In other words, if
  then
• vi belongs to S1(in MC) ,too

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Construct the Cut of MC(3)

• Cost(BMC) cost_in_square
  cost_out_square
• Total cost_in_squre S (3di 2)
• each vertex in MC
• 6e 2n

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Construct the Cut of MC(4)

• Cost_out_square cost from
• cost from
• Assume we will have cost(MC) K
• total cost from 2K
• total cost from S(1 (di ci))
• ci is the cut from vi

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Construct the Cut of MC(5)

• Cost(BMC) 6e 2n 2K S(1 (di ci)) 3n
  4e 4K 3n 4e 4cost(MC)
• 4Cost(MC) cost(BMC) 3n - 4e
• Claim MC has solution K iff BMC has solution 4K
  4e 3n

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BMC to K-MSA(max)

• This part I cant understand very well.
• This paper lacks definition of some terms.
• May have error in this part?
• In Thesis , BSC problem

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BMC to K-MSA(max)

• Make reduction from BMC to restricted MSA
  problem.
• Every sequence has the same length.
• Gap can only be inserted in left end of right
  end.
• Only one gap can be inserted.

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Construct KSA problem
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Construct KSA problem
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Construct the cut of BMC

• Key observation
• Every edge link to ui is ????
• If ui change from S1 to S2 than
• the cost will shift!

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Construct the cut of BMC

• If one row i is right alignment , then
  corresponding ui is belong to S1, else S2.
• If one column j has more one than column j 1,
  then corresponding vi belong to S1, else S2 (not
  sure)
• (??)Cluster negative edge in one Set.

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Complain

• The structure of the paper is hard to read and
  understand.
• Some proof is bypass .
• Some information lacked.
• But to appear in JCO.?
• But the reduction is interesting.