Dynamic Programming: String Editing - BGUmichaluz/seminar/Edit1Class.pdf · 2008-05-18 · Finding...
Transcript of Dynamic Programming: String Editing - BGUmichaluz/seminar/Edit1Class.pdf · 2008-05-18 · Finding...
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Dynamic Programming:String Editing
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Outline
• DNA Sequence Comparison: First Success Stories
• Manhattan Tourist Problem
• Longest Paths in Grid Graphs
• Sequence Alignment
• Edit Distance
• Back to Longest Common Subsequence Problem
• Reducing the space requirement
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Genome: The digital backbone of molecular biology
Transcripts: Perform functions
encoded in the genome
The Central Dogma of Molecular Biology
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5
A = c t a c g a g a c
B = a a c g a c g a t
- a c g t
- -1 -1 -1 -1
a -1 1 -1 -1 -1
c -1 -1 1 -1 -1
g -1 -1 -1 1 -1
t -1 -1 -1 -1 1
The Sequence Alignment Problem
Compare two strings A and B and measure their similarity
by finding the optimal alignment between them.
The alignment is classically based on the transformation
of one sequence into the other, via operations of substitutions,
insertions, and deletions (indels).
The Scoring Matrix
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6
A = c t a c g a g a c
B = a a c g a c g a t
A = c t a c g a g a c
B = a a c g a c g a t
Global Alignment.
Local Alignment.
Two Sequence Alignment Problems
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7
A = c t a c g a g a c
B = a a c g a c g a t
A = c t a c g a g a c
B = a a c g a c g a t
Global Alignment.
Local Alignment.
- a c g t
- -1 -1 -1 -1
a -1 1 -1 -1 -1
c -1 -1 1 -1 -1
g -1 -1 -1 1 -1
t -1 -1 -1 -1 1
The Scoring Matrix
Two Sequence Alignment Problems
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8
A = c t a c g a g a c
B = a a c g a c g a t
A = c t a c g a g a c
B = a a c g a c g a t
Global Alignment.
Local Alignment.
- a c g t
- -1 -1 -1 -1
a -1 1 -1 -1 -1
c -1 -1 1 -1 -1
g -1 -1 -1 1 -1
t -1 -1 -1 -1 1
The Scoring Matrix
Two Sequence Alignment Problems
Value: 2
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9
A = c t a c g a g a c
B = a a c g a c g a t
A = c t a c g a g a c
B = a a c g a c g a t
Global Alignment.
Local Alignment.
- a c g t
- -1 -1 -1 -1
a -1 1 -1 -1 -1
c -1 -1 1 -1 -1
g -1 -1 -1 1 -1
t --1 -1 -1 -1 1
The Scoring Matrix
Two Sequence Alignment Problems
Value:
Value: 2
5
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10
A = c t a c g a g a c
B = a a c g a c g a t
- a c g t
- -1 -1 -1 -1
a -1 1 -1 -1 -1
c -1 -1 1 -1 -1
g -1 -1 -1 1 -1
t -1 -1 -1 -1 1
The Sequence Alignment Problem
Compare two strings A and B and measure their similarity
by finding the optimal alignment between them.
The alignment is classically based on the transformation
of one sequence into the other, via operations of substitutions,
insertions, and deletions (indels).
The Scoring Matrix
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11
A = c t a c g a g a c
B = a a c g a c g a t
A = c t a c g a g a c
B = a a c g a c g a t
Global Alignment.
Local Alignment.
- a c g t
- -1 -1 -1 -1
a -1 1 -1 -1 -1
c -1 -1 1 -1 -1
g -1 -1 -1 1 -1
t --1 -1 -1 -1 1
The Scoring Matrix
Two Sequence Alignment Problems
Value:
Value: 2
5
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DNA Sequence Comparison: First Success Story
• Finding sequence similarities with genes of known function is a common approach to infer a newly sequenced gene’s function
• In 1984 Russell Doolittle and colleagues found similarities between cancer-causing gene and normal growth factor (PDGF) gene
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Cystic Fibrosis
• Cystic fibrosis (CF) is a chronic and frequently fatal genetic disease of the body's mucus glands (abnormally high level of mucus in glands). CF primarily affects the respiratory systems in children.
• Mucus is a slimy material that coats many epithelial surfaces and is secreted into fluids such as saliva
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Cystic Fibrosis: Inheritance
• In early 1980s biologists hypothesized that CF is an autosomal recessive disorder caused by mutations in a gene that remained unknown till 1989
• Heterozygous carriers are asymptomatic
• Must be homozygously recessive in this gene in order to be diagnosed with CF
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Cystic Fibrosis: Finding the Gene
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Finding Similarities between the Cystic Fibrosis Gene
and ATP binding proteins
• ATP binding proteins are present on cell membrane and act as transport channel
• In 1989 biologists found similarity between the cystic fibrosis gene and ATP binding proteins
• A plausible function for cystic fibrosis gene, given the fact that CF involves sweet secretion with abnormally high sodium level
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Cystic Fibrosis: Mutation Analysis
If a high % of cystic fibrosis (CF) patients have a certain mutation in the gene and the normal patients don’t, then that could be an indicator of a mutation that is related to CF
A certain mutation was found in 70% of CF patients, convincing evidence that it is a predominant genetic diagnostics marker for CF
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Cystic Fibrosis and the CFTR Protein
•CFTR (Cystic Fibrosis Transmembrane conductance Regulator)protein is acting in the cell membrane of epithelial cells that secrete mucus
•These cells line the airways of the nose, lungs, the stomach wall, etc.
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Mechanism of Cystic Fibrosis
• The CFTR protein (1480 amino acids) regulates a chloride ion channel
• Adjusts the “wateriness” of fluids secreted by the cell
• Those with cystic fibrosis are missing one single amino acid in their CFTR
• Mucus ends up being too thick, affecting many organs
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Manhattan Tourist Problem (MTP)
Imagine seeking a path (from source to sink) to travel (only eastward and southward) with the most number of attractions (*) in the Manhattan grid Sink
*
*
*
*
*
**
* *
*
*
Source
*
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Manhattan Tourist Problem: Formulation
Goal: Find the longest path in a weighted grid.
Input: A weighted grid G with two distinct vertices, one labeled “source” and the other labeled “sink”
Output: A longest path in G from “source” to “sink”
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MTP: An Example
3 2 4
0 7 3
3 3 0
1 3 2
4
4
5
6
4
6
5
5
8
2
2
5
0 1 2 3
0
1
2
3
j coordinatei c
oo
rdin
ate
13
source
sink
4
3 2 4 0
1 0 2 4 3
3
1
1
2
2
2
419
95
15
23
0
20
3
4
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MTP: Greedy Algorithm Is Not Optimal
1 2 5
2 1 5
2 3 4
0 0 0
5
3
0
3
5
0
10
3
5
5
1
2promising start, but leads to bad choices!
source
sink18
22
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MTP: Simple Recursive Program
MT(n,m)
if n=0 or m=0
return MT(n,m)
x MT(n-1,m)+
length of the edge from (n- 1,m) to (n,m)
y MT(n,m-1)+
length of the edge from (n,m-1) to (n,m)
return max{x,y}
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MTP: Simple Recursive Program
MT(n,m)
x MT(n-1,m)+
length of the edge from (n- 1,m) to (n,m)
y MT(n,m-1)+
length of the edge from (n,m-1) to (n,m)
return min{x,y}
What’s wrong with this approach?
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1
5
0 1
0
1
i
source
1
5
S1,0 = 5
S0,1 = 1
• Calculate optimal path score for each vertex in the graph
• Each vertex’s score is the maximum of the prior vertices score plus the weight of the respective edge in between
MTP: Dynamic Programmingj
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MTP: Dynamic Programming(cont’d)
1 2
5
3
0 1 2
0
1
2
source
1 3
5
8
4
S2,0 = 8
i
S1,1 = 4
S0,2 = 33
-5
j
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MTP: Dynamic Programming(cont’d)
1 2
5
3
0 1 2 3
0
1
2
3
i
source
1 3
5
8
8
4
0
5
8
103
5
-5
9
13
1-5
S3,0 = 8
S2,1 = 9
S1,2 = 13
S3,0 = 8
j
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MTP: Dynamic Programming (cont’d)
greedy alg. fails!
1 2 5
-5 1 -5
-5 3
0
5
3
0
3
5
0
10
-3
-5
0 1 2 3
0
1
2
3
i
source
1 3 8
5
8
8
4
9
13 8
9
12
S3,1 = 9
S2,2 = 12
S1,3 = 8
j
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MTP: Dynamic Programming(cont’d)
1 2 5
-5 1 -5
-5 3 3
0 0
5
3
0
3
5
0
10
-3
-5
-5
2
0 1 2 3
0
1
2
3
i
source
1 3 8
5
8
8
4
9
13 8
12
9
15
9
j
S3,2 = 9
S2,3 = 15
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MTP: Dynamic Programming(cont’d)
1 2 5
-5 1 -5
-5 3 3
0 0
5
3
0
3
5
0
10
-3
-5
-5
2
0 1 2 3
0
1
2
3
i
source
1 3 8
5
8
8
4
9
13 8
12
9
15
9
j
0
1
16S3,3 = 16
(showing all back-traces)
Done!
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MTP: Recurrence
Computing the score for a point (i,j) by the recurrence relation:
si, j =max
si-1, j + weight of the edge between (i-1, j) and (i, j)
si, j-1 + weight of the edge between (i, j-1) and (i, j)
The running time is n x m for a n by m grid
(n = # of rows, m = # of columns)
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Manhattan Is Not A Perfect Grid
What about diagonals?
• The score at point B is given by:
sB =max of
sA1 + weight of the edge (A1, B)
sA2 + weight of the edge (A2, B)
sA3 + weight of the edge (A3, B)
B
A3
A1
A2
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Manhattan Is Not A Perfect Grid (cont’d)
Computing the score for point x is given by the recurrence relation:
sx = max
of
sy + weight of vertex (y, x) where
y є Predecessors(x)
• Predecessors (x) – set of vertices that have edges leading to x
•The running time for a grid graph G(V, E) (V
is the set of all vertices and E is the set of all edges) is O(E) since each edge is evaluated once
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Traveling in the Grid•The only hitch is that one must decide on the order in which visit the vertices
•By the time the vertex x is analyzed, the values sy
for all its predecessors y should be computed –otherwise we are in trouble.
•We need to traverse the vertices in some order
•Try to find such order for a directed cycle
???
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DAG: Directed Acyclic Graph• Since Manhattan is not a perfect regular grid, we
represent it as a DAG
• DAG for Dressing in the morning problem
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Topological Ordering
• A numbering of vertices of the graph is called topological ordering of the DAG if every edge of the DAG connects a vertex with a smaller label to a vertex with a larger label
• In other words, if vertices are positioned on a line in an increasing order of labels then all edges go from left to right.
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Topological ordering• 2 different topological orderings of the DAG
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Longest Path in DAG Problem
• Goal: Find a longest path between two vertices in a weighted DAG
• Input: A weighted DAG G with source and sink vertices
• Output: A longest path in G from source to sink
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Manhattan Tourist Problem (MTP)
Imagine seeking a path (from source to sink) to travel (only eastward and southward) with the most number of attractions (*) in the Manhattan grid Sink
*
*
*
*
*
**
* *
*
*
Source
*
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Manhattan Tourist Problem: Formulation
Goal: Find the longest path in a weighted grid.
Input: A weighted grid G with two distinct vertices, one labeled “source” and the other labeled “sink”
Output: A longest path in G from “source” to “sink”
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MTP: Greedy Algorithm Is Not Optimal
1 2 5
2 1 5
2 3 4
0 0 0
5
3
0
3
5
0
10
3
5
5
1
2promising start, but leads to bad choices!
source
sink18
22
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MTP: Dynamic Programming(cont’d)
1 2 5
-5 1 -5
-5 3 3
0 0
5
3
0
3
5
0
10
-3
-5
-5
2
0 1 2 3
0
1
2
3
i
source
1 3 8
5
8
8
4
9
13 8
12
9
15
9
j
0
1
16S3,3 = 16
(showing all back-traces)
Done!
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MTP: Recurrence
Computing the score for a point (i,j) by the recurrence relation:
si, j =max
si-1, j + weight of the edge between (i-1, j) and (i, j)
si, j-1 + weight of the edge between (i, j-1) and (i, j)
The running time is n x m for a n by m grid
(n = # of rows, m = # of columns)
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Manhattan Is Not A Perfect Grid
What about diagonals?
• The score at point B is given by:
sB =max of
sA1 + weight of the edge (A1, B)
sA2 + weight of the edge (A2, B)
sA3 + weight of the edge (A3, B)
B
A3
A1
A2
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Manhattan Is Not A Perfect Grid (cont’d)
Computing the score for point x is given by the recurrence relation:
sx = max
of
sy + weight of vertex (y, x) where
y є Predecessors(x)
• Predecessors (x) – set of vertices that have edges leading to x
•The running time for a grid graph G(V, E) (V
is the set of all vertices and E is the set of all edges) is O(E) since each edge is evaluated once
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Longest Path in DAG Problem
• Goal: Find a longest path between two vertices in a weighted DAG
• Input: A weighted DAG G with source and sink vertices
• Output: A longest path in G from source to sink
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Longest Path in DAG: Dynamic Programming
• Suppose vertex v has indegree 3 and predecessors {u1, u2, u3}
• Longest path to v from source is:
In General:
sv = maxu (su + weight of edge from u to v)
sv = max of
su1 + weight of edge from u1 to v
su2 + weight of edge from u2 to v
su3 + weight of edge from u3 to v
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MTP: An Example
3 2 4
0 7 3
3 3 0
1 3 2
4
4
5
6
4
6
5
5
8
2
2
5
0 1 2 3
0
1
2
3
j coordinatei c
oo
rdin
ate
13
source
sink
4
3 2 4 0
1 0 2 4 3
3
1
1
2
2
2
419
95
15
23
0
20
3
4
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Traversing the Manhattan Grid
• 3 different strategies:
• a) Column by column
• b) Row by row
• c) Along diagonals
a) b)
c)
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Alignment: 2 row representation
Alignment : 2 * k matrix ( k > m, n )
A T -- G T A T --
A T C G -- A -- C
letters of v
letters of w
T
T
A T C T G A TT G C A T A
v :w :
m = 7 n = 6
4 matches 2 insertions 2 deletions
Given 2 DNA sequences v and w:
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Aligning DNA Sequences
V = ATCTGATG
W = TGCATAC
n = 8
m = 7
A T C T G A T G
T G C A T A C
VW
match
deletioninsertion
mismatch
indels
4122
matchesmismatchesinsertions
deletions
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Edit Script: Example
TGCATAT ATCCGAT in 5 steps
TGCATAT (delete last T)
TGCATA (delete last A)
TGCAT (insert A at front)
ATGCAT (substitute C for 3rd G)
ATCCAT (insert G before last A)
ATCCGAT (Done)
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TGCATAT ATCCGAT in 5 steps
TGCATAT (delete last T)
TGCATA (delete last A)
TGCAT (insert A at front)
ATGCAT (substitute C for 3rd G)
ATCCAT (insert G before last A)
ATCCGAT (Done)
What is the minimal edit script?
Edit Script: Example
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Edit Script: Example (cont’d)
TGCATAT ATCCGAT in 4 steps
TGCATAT (insert A at front)
ATGCATAT (delete 6th T)
ATGCATA (substitute G for 5th A)
ATGCGTA (substitute C for 3rd G)
ATCCGAT (Done)
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Edit Script: Example (cont’d)
TGCATAT ATCCGAT in 4 steps
TGCATAT (insert A at front)
ATGCATAT (delete 6th T)
ATGCATA (substitute G for 5th A)
ATGCGTA (substitute C for 3rd G)
ATCCGAT (Done)
Can it be done in 3 steps???
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57
A = c t a c g a g a c
B = a a c g a c g a t
A = c t a c g a g a c
B = a a c g a c g a t
Global Alignment.
Local Alignment.
Two Sequence Alignment Problems
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58
- a c g t
- -1 -1 -1 -1
a -1 1 -1 -1 -1
c -1 -1 1 -1 -1
g -1 -1 -1 1 -1
t -1 -1 -1 -1 1
The Scoring Matrix
The O(n ) time, Classical Dynamic Programming Algorithm2
c
a
c
t
a a c g a c g a0
1
1 2 3 4 5 6 7 8
2
3
4
a
g
a
g5
6
7
c
|B|= n
|A|= n
t9
9
8
0 1 2 3 4 5 6 8 7 9
The Alignment Graph
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59
Computing the Optimal Global Alignment Value
c
a
c
t
a a c g a c g a0
1
1 2 3 4 5 6 7 8
2
3
4
a
g
a
g5
6
7
c
|B|= n
|A|= n
t9
9
8
0 1 2 3 4 5 6 8 7 9
Classical Dynamic Programming: O(n )
Score of = 1
Score of = -1
2
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60
Computing an Optimal Local Alignment Value
c
a
c
t
a a c g a c g a0
1
1 2 3 4 5 6 7 8
2
3
4
a
g
a
g5
6
7
c
|B|= n
|A|= n
t9
9
8
0 1 2 3 4 5 6 8 7 9
Classical Dynamic Programming: O(n )
Score of = 1
Score of = -1
2
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62
Computing the Optimal Global Alignment Value
c
a
c
t
a a c g a c g a0
1
1 2 3 4 5 6 7 8
2
3
4
a
g
a
g5
6
7
c
|B|= n
|A|= n
t9
9
8
0 1 2 3 4 5 6 8 7 9
Classical Dynamic Programming: O(n )
Score of = 1
Score of = -1
2
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Back to LCS
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Edit Graph for LCS Problem
T
G
C
A
T
A
C
1
2
3
4
5
6
7
0i
A T C T G A T C0 1 2 3 4 5 6 7 8
j
Every path is a common subsequence.
Every diagonal edge adds an extra element to common subsequence
LCS Problem: Find a path with maximum number of diagonal edges
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Alignment: Dynamic Programming
si,j = si-1, j-1+1 if vi = wj
max si-1, j+0
si, j-1+0
This recurrence corresponds to the Manhattan Tourist problem (three incoming edges into a vertex) with all horizontal and vertical edges weighted by zero and diagonal edges weighted by one.
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Alignment as a Path in the Edit Graph
Every path in the edit graph corresponds to an alignment:
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Alignment as a Path in the Edit Graph
Old Alignment0122345677
v= AT_GTTAT_w= ATCGT_A_C
0123455667
New Alignment0122345677
v= AT_GTTAT_w= ATCG_TA_C
0123445667
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Alignment as a Path in the Edit Graph
0122345677v= AT_GTTAT_w= ATCGT_A_C
0123455667
(0,0) , (1,1) , (2,2), (2,3),
(3,4), (4,5), (5,5), (6,6),
(7,6), (7,7)
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Alignment: Dynamic Programming
si,j = si-1, j-1+1 if vi = wj
max si-1, j
si, j-1
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Dynamic Programming Example
Initialize 1st row and 1st
column to be all zeroes.
Or, to be more precise, initialize 0th row and 0th
column to be all zeroes.
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Dynamic Programming Example
Si,j = Si-1, j-1
max Si-1, j
Si, j-1
value from NW +1, if vi = wj
value from North (top)
value from West (left)
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Dynamic Programming Example
Si,j = Si-1, j-1
max Si-1, j
Si, j-1
value from NW +1, if vi = wj
value from North (top)
value from West (left)
A T C G T A C
A
T
G
TTAT
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Alignment: Backtracking
Arrows show where the score originated from.
if from the top
if from the left
if vi = wj
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Alignment: Dynamic Programming
si,j = si-1, j-1+1 if vi = wj
max si-1, j
si, j-1
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LCS Algorithm1. LCS(v,w)2. for i 1 to n3. si,0 04. for j 1 to m5. s0,j 06. for i 1 to n7. for j 1 to m8. si-1,j9. si,j max si,j-110. si-1,j-1 + 1, if vi = wj
11. “ “ if si,j = si-1,j• bi,j “ “ if si,j = si,j-1• “ “ if si,j = si-1,j-1 + 1•• return (sn,m, b)
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Now What?
• LCS(v,w) created the alignment grid
• Now we need a way to read the best alignment of v and w
• Follow the arrows backwards from sink
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Printing LCS: Backtracking
1. PrintLCS(b,v,i,j)
2. if i = 0 or j = 0
3. return
4. if bi,j = “ “
5. PrintLCS(b,v,i-1,j-1)
6. print vi
7. else
8. if bi,j = “ “
9. PrintLCS(b,v,i-1,j)
10. else
11. PrintLCS(b,v,i,j-1)
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LCS Runtime
• It takes O(nm) time to fill in the nxm dynamic programming matrix.
• Why O(nm)? The pseudocode consists of a nested “for” loop inside of another “for” loop to set up a nxm matrix.
• Memory? Naively O(nm)