Computational Molecular Biology

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Computational Molecular Biology

• Multiple Sequence Alignment

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Sequence Alignment

• Problem Definition
• Given 2 DNA or protein sequences
• Find Best match between them
• What is an Alignment
• Given 2 Strings S and S
• Goal The lengths of S and S are the same by
  inserting spaces (-- sometimes denote as ?) into
  these strings

A -- T C -- A
-- C T C A A
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Matches, Mismatches and Indels

• Match two aligned, identical characters in an
  alignment
• Mismatch two aligned, unequal characters
• Indel A character aligned with a space

A A C T A C T -- C C T A A C A C T -- -- -- -- C
T C C T A C C T -- -- T A C T T T
10 matches, 2 mismatches, 7 indels
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Basic Algorithmic Problem

• Find the alignment of the two strings that
• max m where m ( matches mismatches indels)
• Or min m where m is the SP-score of an alignment
• m defines the similarity of the two strings, also
  called Optimal Global Alignment
• Biologically a mismatch represents a mutation,
  whereas an indel represents a historical
  insertion or deletion of a single character

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Multiple Sequence Alignment

• Problem Definition
• Similar to the sequence alignment problem but the
  input has more than 2 strings
• Challenges
• NP-hard, MAX-SNP
• Guarantee factor 2 2/k where k is the number
  of the input sequences.
• More work to reduce the time and space complexity

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Sum of Pairs Score (SP-Score)

• Given a finite alphabet and where ? denotes
  a space
• Consider k sequences over that we want to
  align. After an alignment, each sequence has
  length l
• A score d is assigned to each pair of letters

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SP-Score

• The SP-Score of an alignment A is defined as
• Consider a matrix of l columns and k rows where
  the rows represents the sequences and columns
  represent the letters
• SP-Score is the sum of the scores of all columns
• Score of each column is the sum of the scores of
  all distinct unordered pairs of letters in the
  column
• Or we can view as sum of pairwise sequence
  alignment values.
• Find an (optimal) alignment to minimize the
  SP-Score value

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• Proving MSA with SP-Score that is a Metric
  is NP-hard

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Some Notations
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Some Basic Properties

• Lemma 1 Let s1, s2 be two sequences over S such
  that l1s1, l2s2, l2l1 and there are m
  symbols of s1 that are not in s2. Then every
  alignment of the set s1,s2 has at least ml2-l1
  mismatches

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The construction

• Reduce the vertex cover (or node cover) to MSA.
• Vertex cover
• Instance A graph G(V,E) and an integer kV
• Question Is there a vertex cover V1 of G of size
  k or less?
• MSA
• Instance A set Ss1, , sn of finite sequences
  over a fixed alphabet S, an SP-score and an
  integer C
• Question Is there a multiple alignment of the
  sequences in S that is of value C or less?

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SP-Score (alphabet of size 6)
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The Reduction
So, we have , T is a set of C2 sequences t
and X contains C1 sequences x(k), where C1 and C2
will be determined later
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An Example
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Intuition

• By the above construction, an optimal alignment A
  of S is obtained when A satisfies certain
  properties (called standard alignment)
• The value of standard alignment is bounded by a
  given threshold C only where G has a vertex cover
  of size k
• How to obtain
• Force ds of the test sequences to be aligned
  with bs of the edge sequences
• Only one b of each edge sequence can be aligned
  to a d
• The number of such alignment determines the value
  of the alignment

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Standard Alignemnt
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• Let US and US,X denote the upper bounds of D(AS)
  and D(AS,X) respectively
• By Corollary 8 and Lemma 9, we have the standard
  alignment has value not greater than DSD US
  US,X
• where DSD D(AX) D(AT) D(AX,T) D(AS,T)
  over a standard alignment A
• Now, let C1 gt US and C2 gt US US,X, we can prove
  that an optimal alignment must be a standard one

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• Show the NP-hardness of any scoring matrix in
  a broad class M
• Show that there is a scoring matrix M0 such that
  MSA for M0 is MAX-SNP hard

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Interesting Observation

• Via the brute force, optimal MSA contains very
  few gaps
• Suggesting the study of gap limitations
• Have an upper bound of the number of gaps one can
  insert during the alignment
• Special case
• Gap-0 No gap allows, but we can shift the
  strings for an alignment (insert gaps at the
  beginning or at the end of a string)
• Gap-0-1 a gap-0 alignment such that the gaps at
  the beginning or at the end of each string is
  exactly one space

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Problem Definition

• Given a finite alphabet
• Scoring matrix
• For i, j gt 0, si,j represents the penalty for
  aligning ai with aj
• For i gt 0, s0,i and si,0 are called indel
  penalites
• Gap opening penalties (in addition to the indel
  penalties) for aligning ai with the first or last
  ? in the string of ?s

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Generic Scoring Matrix
Where SA,T, x, y, x are fixed nonnegative
numbers and u gt max0, vA, vT holds

• Let M2 be the class of all scoring matrices that
  contain a generic submatrix M
• Let M1 be the class of all scoring matrices that
  contain a sub-matrix isomorphic
• to a generic matrix M with z gt vT.
• Let M be the class of all scoring matrices that
  contain a submatrix isomorphic
• to a generic matrix M with y gt u and z gt vT.
• Theorem 1
• The gap-0-1 multiple alignment problem is NP-hard
  for every scoring matrix M
• in M2.
• (b) The gap-0 multiple alignment problem is
  NP-hard for every M in M1
• (c) The multiple alignment problem is NP-hard for
  every M in M
• Note that M is quite broad and covers most
  scoring schemes used in
• biological applications.

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Reduction

• Reduce the MAX-CUT-B
• Given G(V,E) where kV and each vertex has a
  degree at most B
• Find a partition of V into two disjoint sets such
  that to maximize the number of edges crossing
  these two sets
• Given a graph G(V,E) with k vertices v0, , vk-1
  and l edges e0, , el-1. We will construct a set
  of k2 sequences t0, , tk2-1 as follows

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Reduction

• For each vertex vi, construct a sequence ti such
  that
• for each edge emvh, vi incident at vi, h lt i,
  n lt k5, set
• where ti,j represents the character at the jth
  position in ti.
• For other j, let ti,j T
• For i k, set ti T T T T with length k12l

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An Example
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Proof of Theorem 1(a)

• We will show that a gap-0-1 alignment will
  partition V into two disjoint subsets V0 and V1
• V0 all vertices vi such that ti remains in place
  (a space appends at the end)
• V1 all vertices vi such that ti shifts to the
  right
• Thus, based on the alignment, we can find the
  cut. And vice versa, based on the cut, we can
  find the alignment
• The left part is prove that if k is sufficiently
  large, the optimal gap-0-1 alignment yields a
  partion of V with maximum edge cut.

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Proof of Theorem 1(a)

• Let c denote the cut based on the alignment A
• Consider all the sequences ti after that
  alignment A
• The total indel penalties is of order O(k4)
  (appears at the first and last column in the SP
  score matrix)
• The total number of mismatches before the
  alignment is 3k5l(k2-1)
• To maximally reduce this number
• 1 A-A match reduces 2 A-T mismatches
• For each edge (vh, vi), if there are in different
  subsets (of the partition), then a total of k5
  A-A matches between sequences th and ti are
  created
• No other A-T mismatches can be elimiated
• Thus the SP-score
• k12lvTk2(k2-1)23k5l(u-vT)(k2-1)-ck5(2u-vA-vT)O(k
  4)

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Theorem 2

• Consider the following scoring matrix M0 for the
• alphabet ?0 A,T,C.
• The gap-0-1 MSA problem is MAX-SNP-hard
• The gap-0 MSA problem in MAX-SNP-hard
• The MSA problem in MAX-SNP-hard

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MAX-SNP-hard Proof

• To prove problem A is MAX-SNP-hard, we need to
  L-reduce problem A, which is MAX-SNP-hard to A
• L-reduce
• There are two polynomial-time algorithms f, g and
  constants a, b gt 0 such that for each instance I
  of A
• f produces an instance I f(I) of A such that
  OPT(I) aOPT(I)
• Given any solution of I with cost c, g produces
  a solution of I with cost c such that c-OPT(I)
  bc-OPT(I)

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Proof of Theorem 2

• To prove MSA (with M0 and the scoring matrix
  mentioned before) MAX-SNP-hard
• L-reduce the MAX-CUT-B to another optimization
  problem, called A, which is L-reduce to a scaled
  version of MSA
• Problem A
• Given a graph G(V,E) with bounded degree B. For
  every partition PV0, V1, let cp be the size of
  cut determined by P.
• Find the partition P of V that minimizes dp
  3E-2cp

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Show A is MAX-SNP-hard

• Let f and g be an identity function
• Set a 3B and b 2, we can easily prove the two
  properties of the L-reduction since
• cp E/B and dp 3E - 2 cp 3 E
• Any increase of cp by 1 decrease dp by 2

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Show A L-reduce to scaled MSA
Similar to the above construction, we have
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• Similar to the proof of Theorem 1, we have the
  optimal SP-score where
• If the SP-score is scaled by a factor of k-5/2
  for a MSA of k sequences, then A L-reduce to MSA.

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GENETIC ALGORITHMS
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How do GAs work?

• Create a population of random solutions
• Use natural selection
• crossover and mutation to improve the solutions
• Stop the operation if satisfying some certain
  criteria such as
• No improvement on fitness function
• The improvement is less than some certain
  threshold
• The number of iteration is more than some certain
  threhold

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Terms and Definitions

• Chromosomes
• Potential solutions
• Population
• Collection of chromosomes
• Generations
• Successive populations

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Terms and Definitions

• Crossover
• Exchange of genes between two chromosomes
• Mutation
• Random change of one or more genes in a
  chromosome
• Elitism
• Copy the best solutions without doing crossover
  or mutation.

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Terms and Definitions

• Offspring
• New chromosome created by crossover between two
  parent chromosomes
• Fitness function
• Measures how good a chromosome is.
• Encoding scheme
• How do we represent every chromosome/gene?
• Binary, combination, syntax trees.

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Why are GAs attractive?

• No need for a particular algorithm to solve the
  given problem. Only the fitness function is
  required to evaluate the quality of the
  solutions.
• Implicitly a parallel technique and can be
  implement efficiently on powerful parallel
  computers for demanding large scale problems.

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Basic Outline of a GA

• Initial population composed of random
  chromosomes, called first generation
• Evaluate the fitness of each chromosome in the
  population
• Create a new population
• Select two parent chromosomes from a population
  according to their fitness
• Crossover (with some probability) to form a new
  offspring
• Mutation (with some probability) to mutate new
  offspring
• Place new offspring in a new population
• Process is repeated until a satisfactory solution
  evolves

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Operations

• Mutation Operation
• Modify a single parent
• Try to avoid local minima

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Let's see some running examples

• Minimum of a function
• http//cs.felk.cvut.cz/xobitko/ga/example_f.html
• Elitism
• http//cs.felk.cvut.cz/xobitko/ga/params.html
• The travelling salesman problem
• http//cs.felk.cvut.cz/xobitko/ga/tspexample.html

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Multiple Sequence Alignment

• Fitness function is used to compare the different
  alignments
• Based on the number of matching symbols and the
  number and size of gaps
• Also called the cost function
• Different weights for different types of matches
• Gap costs
• can be simple and count the total matching
  symbols
• can be complicated and consider the type of
  matching symbols, location in the sequence,
  neighboring symbols etc.

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• Approximation Algorithms

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Scoring method

• Score zero for a match or for two opposing spaces
• Score one for a mismatch or for a character
  opposite a space

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Assumptions

• Assume that two opposing spaces have a zero value
• Assume other values satisfies triangle inequality
• s(x,z) s(x,y) s(y,z)
• s(x,z) cost of transforming character x into
  character z

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Objective Functions

• Two objective functions
• SP
• The sum of the values of pairwise alignments
  induced by an alignment A
• TA
• Using the topology of the tree, map the strings
  to the nodes of the tree
• The sum of the selected pairwise alignments is
  called tree alignment

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Center Star Method

• For a set of k strings X
• Choose a center string Xc of X which minimizes
  Sj?cD(Xc,Xj)
• Let M min Sj?cD(Xc,Xj)
• Center star is a star tree of k nodes with the
  center node labeled Xc and each of the k-1
  remaining nodes labeled by a distinct string in X
  \ Xc
• If Xi and Xj are strings labeling adjacent nodes
  of tree T, then alignment of Xi and Xj induced by
  A(T) has value D(Xi,Xj)

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Center Star Method Alg Ac

• Do an optimal alignment for each pair (Xc, Xj)
  for all j ? c
• s0 max number of spaces placed before the first
  char of Xc
• sf max number of spaces placed after the last
  char of Xc
• si max number of spaces placed between Xc(i)
  and Xc(i1)

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Center Star Method Alg Ac

• For Xc, insert s0, si, and sf spaces at the
  beginning, between, and the end of Xc
  respectively. Call Xc
• Then for each Xj, do the optimal alignment
  without modifying Xc

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Analysis

• d(Xi,Xj) D(Xi,Xj)
• V(Ac) Siltjd(Xi,Xj)
• V(Ac) is at most twice the value of the optimal
  multiple alignment of X

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Analysis

• Lemma 3.1 For any 2 strings Xi,Xj, we have
• d(Xi,Xj) d(Xi,Xc) d(Xc,Xj)
• D(Xi,Xc) D(Xc,Xj)
• triangle inequality

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Analysis

• A be the optimal multiple alignment of k strings
  X
• Define V(A) Siltjd(Xi,Xj)

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Analysis

• Theorem 3.1
• V(Ac) / V(A) 2(k-1)/ k lt 2
• Proof

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Disadvantages

• Requires all pairwise alignments
• Computationally expensive
• Faster, Randomized alignments
• Randomly select string Xi
• Build multiple alignment with star centered at Xi
• Select best multiple alignment A from p such
  stars
• At most (k-1)p pairwise alignments need to be
  computed

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Randomized Alignments

• Theorem 3.2For any r gt1, let e(r) be the
  expected number of stars needed to be chosen at
  random before the value of best resulting
  alignment is within a factor of 21/(r-1) of the
  optimal alignment. Then e(r) r.
• e(r) is independent of k and the length of the
  strings.

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Proof of Theorem 3.2

• For r 2, for each string Xi
• define M(i) SjD(Xi,Xj) then M(c) M
• From Theorem 3.1,
• S(i,j)D(Xi,Xj) SjM(i) 2(k-1)M so the Avg
  value of M(i) lt 2 M
• Since min M(i) M, then Median M(i) lt 3M
• Number of centers selected before a selected
  M(i) is less than the median 2

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Proof

• Suppose median is ?M for 1 ? 3
• Then S(i,j)D(Xi,Xj) kM/2 k ? M/2
• Value of the alignment obtained from any below
  median star 2(k-1) ? M
• Therefore, error ratio for this star 2 ? /
  (1/2 ? /2)
• When ? 3, error ratio 3.
• So we have e(2) 2

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Proof

• Now generalize this proof for r gt 2
• At least k/r stars have M(i) less than or equal
  to (2r-1)M/(r-1)
• Minimum M(i) is M
• Mean lt 2M
• expected number of stars to pick with M(i) lt ? M
  is r for 1 ? (2r-1)/(r-1)
• error ratio 2 ? /1/r (r-1) ? /r
• (2r-1)/(r-1)2 1/(r-1)

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Theorem 3.3

• Picking p stars at random, the best resulting
  alignment will have value within a factor of 2
  1/(r-1) of the optimal with probability at least
• 1 (r-1)/rp

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Center Star Method

• Proof
• From theorem 3.2, if Median value was actually 3M
• For half the stars M(i) M and M(i) 3M for the
  other half
• S(i,j)D(Xi,Xj)2kM
• optimal SP alignment can be obtained from any
  center string Xiwith M(i) M
• Probability of selecting such a string is one-half

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Tree Alignment Method

• Typical approach
• first find multiple alignment and then build a
  tree showing the evolutionary derivations
• Another approach (called tree alignment)
• first choose the typology of the tree and then
  map the strings to the nodes of the tree
• Alignment is the pairwise alignments of the
  strings at the ends of the edges of the tree

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Formal Definitions

• Let K be an input set of k strings
• K K be a set of strings containing K
• Evolutionary tree TK for K is a tree
• with at least k nodes
• each string in K labels exactly one node each
  node gets exactly one label in K
• The value of TK V(TK) SD(X,Y)
• the problem is to find a set of strings K and
  T(K) for K which minimizes V (TK)

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• The alignment value D(X,Y ) is interpreted as
  the minimum cost" to transform string X to
  string Y
• The sum of the alignment values of the edges
  gives the evolutionary cost implied by the tree.

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Method

• Let G be a graph with k nodes labeled with a
  distinct string in K
• Each edge (X,Y) has a weight D(X,Y)
• Find the MST of G. This MST is an evolutionary
  tree for K

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Analysis

• T denote the optimal evolutionary tree for K.
• Prove V(MST)/V(T) lt 2OPT
• Let C be a traversal of edges of T which
  traverses everyy edge exactly once in each
  direction
• Let C1, , Ck be the order that C encounters
• Let V(C) D(Ck,C1) SiltkD(Ci,Ci1)

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Analysis
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Analysis

• Corollary 4.1 V(C) 2V(T),
• Let D(Ci,Ci1) be the largest distance of any
  adjacent strings in C traversal
• Lemma(4.2)
• V(MST) V(C) D(Ci,Ci1) V(C) V(C)/K

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Analysis

• Theorem 4.1
• For any set K of k strings, we have
• V(MST)/ V(Tk) 2(k-1)/k lt 2
• Theorem 4.2
• V(MST) / V(Tk) (k-1)/k V(C)/V(Tk) 2 (k-1)/k
• Corollary 4.2
• V(Tk) gt kV(MST)/2(k-1)

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Constrained MSA
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Motivation

• General SP MSA problem
• NP-completeness has already been established
• Appromixation algorithms have been developed
• Heuristics are also avaliable
• Constrained MSA
• Biologists often have additional knowledge of
  data (e.g. active site residues)
• Additional knowledge can specify matches at
  certain locations
• Models allow users to provide additional
  constraints

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Definition of CMSA Problem

• Suppose that P p1p2 . . . pa is a common
  subsequence of S1, S2, . . . , SK
• The constrained multiple sequence alignment of S
  with respect to P is
• an MSA A with the constraints that there are a
  columns in A, c1, c2, . . . , ca with c1 lt c2 lt
  lt ca, such that the characters of column ci, 1
  i a, are all equal to pi.

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Optimal CPSA
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Dynamic Algorithm
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Time and Space Complexities
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CMSA

• The improvement of CPSA in turn improves the time
  space complexity of
• Progressive CMSA from O(akn4) and O(an4) to
  O(ak2n2) and O(an2).
• Optimal CMSA
• This Optimal CMSA algorithm involves the creation
  of a matrix with k1 dimensions.
• (Assume d(x,y) is the distance function and
  satisfies the triangle inequality.)
• Let D(i1, . . . , ik ?) be the optimal CMSA
  score matrix for
• S11..i1, . . . , Sk1..ik where P1..? is
  aligned in ? columns.
• Then optimal alignment score is D(n1, . . . , nk
  a), where ni Si.
• Computing D
• D(0k 0) 0
• Let ej 0 or 1 with ejSjij where j 0
  represents a space, and
• d(x1, . . . , xk) S1iltjkd(xi, xj).
• D(i1, i2, . . . , ik ?) is the minimum of
• if S1i1 . . . Skik P?,
• D(i1 - 1, . . . , ik - 1 ? - 1) d(S1i1, . .
  . , Skik)
• mine?0,1k (D(i1 - e1, . . . , ik - ek ?)
  d(e1S1i1, . . . , ekSkik)).

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CMSA (Center Star)

• The Center-Star method proposed for the general
• MSA problem can be modified to apply to the CMSA
• problem.
• Consider each sequence as the center, Sc.
  Consider each list position that Sc is aligned
  with P.
• Find the minimum star-sum score Sc.
• Create a constrained alignment matrix by merging
  the
• constrained pairwise sequence alignments between
  Sc Sj.

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CMSA (Center Star)

• The recurrence of Thm. 3.1 is only slightly
  modified

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Example