lecture-15-dynamic-programming.pdf

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Weighted Interval Scheduling Segmented Least Squares RNA Secondary Structure Sequence Alignment Shortest Paths in Graphs

Dynamic Programming

T. M. Murali

October 14, 19, 21, 26, 28, 2009

T. M. Murali October 14, 19, 21, 26, 28, 2009 CS 4104: Dynamic Programming


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Algorithm Design Techniques

1. Goal: design efficient (polynomial-time) algorithms.



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2. GreedyPro: natural approach to algorithm design.Con: many greedy approaches to a problem. Only some may work.Con: many problems for which no greedy approach is known.



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3. Divide and conquer

Pro: simple to develop algorithm skeleton.Con: conquer step can be very hard to implement efficiently.Con: usually reduces time for a problem known to be solvable in polynomialtime.



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3. Divide and conquer

Pro: simple to develop algorithm skeleton.Con: conquer step can be very hard to implement efficiently.Con: usually reduces time for a problem known to be solvable in polynomialtime.

4. Dynamic programmingMore powerful than greedy and divide-and-conquer strategies.Implicitly explore space of all possible solutions.Solve multiple sub-problems and build up correct solutions to larger and largersub-problems.Careful analysis needed to ensure number of sub-problems solved is polynomialin the size of the input.



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History of Dynamic Programming

Bellman pioneered the systematic study of dynamic programming in the

1950s.



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Review: Interval Scheduling

Interval Scheduling

INSTANCE: Nonempty set {(s i , f i ), 1 i n} of start and nish timesof n jobs.SOLUTION: The largest subset of mutually compatible jobs.

Two jobs are compatible if they do not overlap.

Greedy algorithm: sort jobs in increasing order of nish times. Add next jobto current subset only if it is compatible with previously-selected jobs.




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Weighted Interval Scheduling


INSTANCE: Nonempty set {(s i , f i ), 1 i n} of start and nish timesof n jobs and a weight v i 0 associated with each job.SOLUTION: A set S of mutually compatible jobs such that i S v i ismaximised.




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INSTANCE: Nonempty set {(s i , f i ), 1 i n} of start and nish timesof n jobs and a weight v i 0 associated with each job.SOLUTION: A set S of mutually compatible jobs such that i S v i ismaximised.

Greedy algorithm can produce arbitrarily bad results for this problem.




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g g g q y q g p

Approach

Sort jobs in increasing order of nish time and relabel: f 1 f 2 . . . f n .

Request i comes before request j if i < j . p(j) is the largest index i < j such that job i is compatible with job j .p ( j ) = 0 if there is no such job i .

Jobs at indices {p ( j ) + 1 , p ( j ) + 2 , . . . , j 1} are incompatible with job j .

We will develop optimal algorithm from obvious statements about theproblem.




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g g g q y q g p

Detour: a Binomial Identity




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Pascals triangle:Each element is a binomial co-efficient.Each element is the sum of the twoelements above it.




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nr

=n 1r 1

+n 1

r




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nr

=n 1r 1

+n 1

r

Proof: Consider any subset S of r elements.Case 1 S contains the nth element:

n 1r 1 such subsets.

Case 2 S does not contain the nthelement: n 1r such subsets.




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Sub-problems

Let O be the optimal solution. Let us reason about O.

Case 1 job n is not in O.

Case 2 job n is in O.




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Sub-problems


Case 1 job n is not in O. O must be the optimal solution for jobs{1, 2, . . . , n 1}.

Case 2 job n is in O.



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Sub-problems



Case 2 job n is in O.O cannot use incompatible jobs{p (n) + 1 , p (n) + 2 , . . . , n 1}.Remaining jobs in O must be the optimal solution for jobs{1, 2, . . . , p (n)}.

O must be the best of these two choices!




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Sub-problems



Case 2 job n is in O.O cannot use incompatible jobs{p (n) + 1 , p (n) + 2 , . . . , n 1}.Remaining jobs in O must be the optimal solution for jobs{1, 2, . . . , p (n)}.

O must be the best of these two choices! Suggests nding optimal solution for sub-problems consisting of jobs{1, 2, . . . , j 1, j }, for all values of j .




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Recursion Let O j be the optimal solution for jobs {1, 2, . . . , j } and OPT ( j ) be the valueof this solution (OPT(0) = 0).




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We are seeking On with a value of OPT(n).




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We are seeking On with a value of OPT(n). To compute OPT( j ):

Case 1 j O j :




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Case 1 j O j : OPT( j ) = OPT( j 1).




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Case 1 j O j : OPT( j ) = OPT( j 1).Case 2 j O j :



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Case 1 j O j : OPT( j ) = OPT( j 1).Case 2 j O j : OPT( j ) = v j + OPT( p ( j ))

Final recurrence:

OPT( j ) = max( v j + OPT( p ( j )) , OPT( j 1))




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Final recurrence:


To compute O j : when does request j belong to O j ?




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Final recurrence:


To compute O j : when does request j belong to O j ? If and only if v j + OPT( p ( j )) OPT( j 1).



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Example of Recursive Algorithm

OPT(6) =OPT(5) =OPT(4) =OPT(3) =OPT(2) =OPT(1) =OPT(0) = 0




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OPT(6) = max( v 6 + OPT( p (6)) , OPT(5)) = max(1 + OPT(3) , OPT(5))OPT(5) =OPT(4) =OPT(3) =OPT(2) =OPT(1) =OPT(0) = 0



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l f i l i h


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OPT(6) = max( v 6 + OPT( p (6)) , OPT(5)) = max(1 + OPT(3) , OPT(5))OPT(5) = max( v 5 + OPT( p ( j )) , OPT(4)) = max(2 + OPT(3) , OPT(4))OPT(4) = max( v 4 + OPT( p (4)) , OPT(3)) = max(7 + OPT(0) , OPT(3))OPT(3) =OPT(2) =OPT(1) =OPT(0) = 0



E l f R i Al i h


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OPT(6) = max( v 6 + OPT( p (6)) , OPT(5)) = max(1 + OPT(3) , OPT(5))OPT(5) = max( v 5 + OPT( p ( j )) , OPT(4)) = max(2 + OPT(3) , OPT(4))OPT(4) = max( v 4 + OPT( p (4)) , OPT(3)) = max(7 + OPT(0) , OPT(3))OPT(3) = max( v 3 + OPT( p (3)) , OPT(2)) = max(4 + OPT(1) , OPT(2))OPT(2) =OPT(1) =OPT(0) = 0



E l f R i Al ith


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OPT(6) = max( v 6 + OPT( p (6)) , OPT(5)) = max(1 + OPT(3) , OPT(5))OPT(5) = max( v 5 + OPT( p ( j )) , OPT(4)) = max(2 + OPT(3) , OPT(4))OPT(4) = max( v 4 + OPT( p (4)) , OPT(3)) = max(7 + OPT(0) , OPT(3))OPT(3) = max( v 3 + OPT( p (3)) , OPT(2)) = max(4 + OPT(1) , OPT(2))OPT(2) = max( v 2 + OPT( p (2)) , OPT(1)) = max(4 + OPT(0) , OPT(1))OPT(1) =OPT(0) = 0



E l f R i Alg ith


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OPT(6) = max( v 6 + OPT( p (6)) , OPT(5)) = max(1 + OPT(3) , OPT(5))OPT(5) = max( v 5 + OPT( p ( j )) , OPT(4)) = max(2 + OPT(3) , OPT(4))OPT(4) = max( v 4 + OPT( p (4)) , OPT(3)) = max(7 + OPT(0) , OPT(3))OPT(3) = max( v 3 + OPT( p (3)) , OPT(2)) = max(4 + OPT(1) , OPT(2))OPT(2) = max( v 2 + OPT( p (2)) , OPT(1)) = max(4 + OPT(0) , OPT(1))OPT(1) = v 1 = 2OPT(0) = 0





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OPT(6) = max( v 6 + OPT( p (6)) , OPT(5)) = max(1 + OPT(3) , OPT(5))OPT(5) = max( v 5 + OPT( p ( j )) , OPT(4)) = max(2 + OPT(3) , OPT(4))OPT(4) = max( v 4 + OPT( p (4)) , OPT(3)) = max(7 + OPT(0) , OPT(3))

OPT(3) = max( v 3 + OPT( p (3)) , OPT(2)) = max(4 + OPT(1) , OPT(2))OPT(2) = max( v 2 + OPT( p (2)) , OPT(1)) = max( 4 + OPT(0) , OPT(1)) = 4OPT(1) = v 1 = 2OPT(0) = 0





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OPT(6) = max( v 6 + OPT( p (6)) , OPT(5)) = max(1 + OPT(3) , OPT(5))OPT(5) = max( v 5 + OPT( p ( j )) , OPT(4)) = max(2 + OPT(3) , OPT(4))OPT(4) = max( v 4 + OPT( p (4)) , OPT(3)) = max(7 + OPT(0) , OPT(3))

OPT(3) = max( v 3 + OPT( p (3)) , OPT(2)) = max( 4 + OPT(1) , OPT(2)) = 6OPT(2) = max( v 2 + OPT( p (2)) , OPT(1)) = max( 4 + OPT(0) , OPT(1)) = 4OPT(1) = v 1 = 2OPT(0) = 0





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OPT(6) = max( v 6 + OPT( p (6)) , OPT(5)) = max(1 + OPT(3) , OPT(5))OPT(5) = max( v 5 + OPT( p ( j )) , OPT(4)) = max(2 + OPT(3) , OPT(4))OPT(4) = max( v 4 + OPT( p (4)) , OPT(3)) = max( 7 + OPT(0) , OPT(3)) = 7




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OPT(6) = max( v 6 + OPT( p (6)) , OPT(5)) = max(1 + OPT(3) , OPT(5) ) = 8OPT(5) = max( v 5 + OPT( p ( j )) , OPT(4)) = max( 2 + OPT(3) , OPT(4)) = 8OPT(4) = max( v 4 + OPT( p (4)) , OPT(3)) = max( 7 + OPT(0) , OPT(3)) = 7






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Optimal solution is





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Optimal solution is job 5, job 3, and job 1.



Running Time of Recursive Algorithm


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What is the running time of the algorithm?



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g g

What is the running time of the algorithm? Can be exponential in n. When p ( j ) = j 2, for all j 2: recursive calls are for j 1 and j 2.



Memoisation


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Store OPT( j ) values in a cache and reuse them rather than recompute them.



Memoisation


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Store OPT( j ) values in a cache and reuse them rather than recompute them.



Running Time of Memoisation


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g

Claim: running time of this algorithm is O (n) (after sorting).





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Claim: running time of this algorithm is O (n) (after sorting). Time spent in a single call to M-Compute-Opt is O (1) apart from time spent inrecursive calls.

Total time spent is the order of the number of recursive calls to M-Compute-Opt . How many such recursive calls are there in total?





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Claim: running time of this algorithm is O (n) (after sorting). Time spent in a single call to M-Compute-Opt is O (1) apart from time spent inrecursive calls.

Total time spent is the order of the number of recursive calls to M-Compute-Opt . How many such recursive calls are there in total? Use number of lled entries in M as a measure of progress. Each time M-Compute-Opt issues two recursive calls, it lls in a new entry in M . Therefore, total number of recursive calls is O (n).



Computing O in Addition to OPT (n )


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Explicitly store O j in addition to OPT( j ).





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Explicitly store O j in addition to OPT( j ). Running time becomes O (n2 ).



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From Recursion to Iteration


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Unwind the recursion and convert it into iteration. Can compute values in M iteratively in O (n) time. Find-Solution works as before.



Basic Outline of Dynamic Programming


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To solve a problem, we need a collection of sub-problems that satisfy a fewproperties:

1. There are a polynomial number of sub-problems.2. The solution to the problem can be computed easily from the solutions to the

sub-problems.3. There is a natural ordering of the sub-problems from smallest to largest.4. There is an easy-to-compute recurrence that allows us to compute the solution

to a sub-problem from the solutions to some smaller sub-problems.



Basic Outline of Dynamic Programming


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To solve a problem, we need a collection of sub-problems that satisfy a fewproperties:

1. There are a polynomial number of sub-problems.2. The solution to the problem can be computed easily from the solutions to the

sub-problems.3. There is a natural ordering of the sub-problems from smallest to largest.4. There is an easy-to-compute recurrence that allows us to compute the solution

to a sub-problem from the solutions to some smaller sub-problems. Difficulties in designing dynamic programming algorithms:

1. Which sub-problems to dene?2. How can we tie together sub-problems using a recurrence?3. How do we order the sub-problems (to allow iterative computation of optimal

solutions to sub-problems)?



Least Squares Problem


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Given scientic or statistical dataplotted on two axes.

Find the best line that passesthrough these points.





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How do we formalise the problem?





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How do we formalise the problem?Least SquaresINSTANCE: Set P = {(x 1 , y 1), (x 2 , y 2 ), . . . , (x n , y n )} of n points.SOLUTION: Line L : y = ax + b that minimises

Error(L, P ) =n

i =1

(y i ax i b )2 .





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How do we formalise the problem?Least SquaresINSTANCE: Set P = {(x 1 , y 1), (x 2 , y 2 ), . . . , (x n , y n )} of n points.SOLUTION: Line L : y = ax + b that minimises

Error(L, P ) =n

i =1

(y i ax i b )2 .

Solution is achieved by

a = n i x i y i ( i x i ) ( i y i )

n i x 2i ( i x i )

2 and b =i y i a i x i

n



Segmented Least Squares


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Want to t multiple lines through P . Each line must t contiguous set of x -coordinates. Lines must minimise total error.





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INSTANCE: Set P = {p i = ( x i , y i ), 1 i n} of n points,x 1 < x 2 < < x n and a parameter C > 0.SOLUTION: A integer k , a partition of P into k segments{P 1 , P 2 , . . . , P k }, k lines L j : y = a j x + b j , 1 j k that minimise

k

j =1Error(L j , P j ) + Ck .

A subset P of P is a segment if 1 i < j n exist such thatP = {(x i , y i ), (x i +1 , y i +1 ), . . . , (x j 1 , y j 1), (x j , y j )}.



Formulating the Recursion I

Ob i i f i h i l l i Thi


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Observation: p n is part of some segment in the optimal solution. Thissegment starts at some point p i .

Let OPT (i ) be the optimal value for the points {p 1 , p 2 , . . . , p i }. Let e i , j denote the minimum error of any line that ts {p i , p 2 , . . . , p j }. We want to compute OPT( n).

If the last segment in the optimal partition is {p i , p i +1 , . . . , p n }, then

OPT( n) = e i , n + C + OPT( i 1)



Formulating the Recursion II

C id th b bl th i t { }


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Consider the sub-problem on the points {p 1 , p 2 , . . . p j } To obtain OPT( j ), if the last segment in the optimal partition is{p i , p i +1 , . . . , p j }, then

OPT( j ) = e i , j + C + OPT( i 1)



Formulating the Recursion II

Consider the sub problem on the points {p p p }


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Consider the sub-problem on the points {p 1 , p 2 , . . . p j } To obtain OPT( j ), if the last segment in the optimal partition is{p i , p i +1 , . . . , p j }, then

OPT( j ) = e i , j + C + OPT( i 1)

Since i can take only j distinct values,OPT( j ) = min

1 i j e i , j + C + OPT( i 1)

Segment {p i , p i +1 , . . . p j } is part of the optimal solution for this sub-problem

if and only if the minimum value of OPT( j ) is obtained using index i .



Dynamic Programming Algorithm


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OPT( j ) = min1 i j

e i , j + C + OPT( i 1)





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OPT( j ) = min1 i j

e i , j + C + OPT( i 1)

Running time is O (n3), can be improved to O (n2 ). We can nd the segments in the optimal solution by backtracking.



RNA Molecules RNA is a basic biological molecule. It is single stranded.


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g g RNA molecules fold into complex secondary structures.

Secondary structure often governs the behaviour of an RNA molecule. Various rules govern secondary structure formation:





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1. Pairs of bases match up; each basematches with 1 other base.

2. Adenine always matches with Uracil.

3. Cytosine always matches with Guanine.

4. There are no kinks in the folded

molecule.5. Structures are knot-free.





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Problem: given an RNA molecule, predict its secondary structure.





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RNA molecules fold into complex secondary structures.







Problem: given an RNA molecule, predict its secondary structure. Hypothesis: In the cell, RNA molecules form the secondary structure with thelowest total free energy.



Formulating the Problem

An RNA molecule is a string B = b 1b 2 . . . b n ; each b i {A, C , G , U }.


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g 1 2 n ; i { , , , } A secondary structure on B is a set of pairs S = {(i , j )}, where 1 i , j nand



Formulating the Problem

An RNA molecule is a string B = b 1b 2 . . . b n ; each b i {A, C , G , U }.


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g ; { } A secondary structure on B is a set of pairs S = {(i , j )}, where 1 i , j nand

1. (No kinks.) If (i , j ) S , then i < j 4.2. (Watson-Crick) The elements in each pair in S consist of either {A, U } or

{C , G } (in either order).3. S is a matching : no index appears in more than one pair.4. (No knots) If (i , j ) and ( k , l ) are two pairs in S , then we cannot have

i < k < j < l .

The energy of a secondary structure the number of base pairs in it.



Dynamic Programming Approach

OPT ( j ) is the maximum number of base pairs in a secondary structure for


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( j) p yb 1b 2 . . . b j .

T. M. Murali October 14, 19, 21, 26, 28, 2009 CS 4104: Dynamic Programming Weighted Interval Scheduling Segmented Least Squares RNA Secondary Structure Sequence Alignment Shortest Paths in Graphs




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b 1b 2 . . . b j . OPT( j ) = 0, if j 5.

T. M. Murali October 14 19 21 26 28 2009 CS 4104: Dynamic Programming Weighted Interval Scheduling Segmented Least Squares RNA Secondary Structure Sequence Alignment Shortest Paths in Graphs




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b 1b 2 . . . b j . OPT( j ) = 0, if j 5.

In the optimal secondary structure on b 1b 2 . . . b j

T M Murali October 14 19 21 26 28 2009 CS 4104: Dynamic Programming


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b 1b 2 . . . b j . OPT( j ) = 0, if j 5.

In the optimal secondary structure on b 1b 2 . . . b j 1. if j is not a member of any pair, use OPT( j 1).2. if j pairs with some t < j 4,

T M Murali October 14 19 21 26 28 2009 CS 4104: Dynamic Programming Weighted Interval Scheduling Segmented Least Squares RNA Secondary Structure Sequence Alignment Shortest Paths in Graphs


OPT ( j ) is the maximum number of base pairs in a secondary structure forb b b O ( j) 0 if j


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b 1b 2 . . . b j . OPT( j ) = 0, if j 5.

In the optimal secondary structure on b 1b 2 . . . b j 1. if j is not a member of any pair, use OPT( j 1).2. if j pairs with some t < j 4, knot condition yields two independent

sub-problems!



OPT ( j ) is the maximum number of base pairs in a secondary structure forb b b OPT( j) 0 if j 5


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b 1b 2 . . . b j . OPT( j ) = 0, if j 5.


sub-problems! OPT( t 1) and ???



OPT ( j ) is the maximum number of base pairs in a secondary structure forb b b OPT( j) 0 if j 5


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b 1b 2 . . . b j . OPT( j ) = 0, if j 5.


sub-problems! OPT( t 1) and ???

Insight: need sub-problems indexed both by start and by end.


Correct Dynamic Programming Approach

OPT (i , j ) is the maximum number of base pairs in a secondary structure forb b b


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b i b 2 . . . b j .

T M Murali October 14 19 21 26 28 2009 CS 4104: Dynamic Programming


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OPT (i , j ) is the maximum number of base pairs in a secondary structure forbi b2 bj OPT( i j) = 0 if i j 4


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b i b 2 . . . b j . OPT( i , j ) = 0, if i j 4.

In the optimal secondary structure on b i b 2 . . . b j





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b i b 2 . . . b j . OPT( i , j ) = 0, if i j 4.

In the optimal secondary structure on b i b 2 . . . b j 1. if j is not a member of any pair, compute OPT( i , j 1).

OPT( i , j ) = max OPT( i , j 1),





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b i b 2 . . . b j . OPT( i , j ) 0, if i j 4.

In the optimal secondary structure on b i b 2 . . . b j 1. if j is not a member of any pair, compute OPT( i , j 1).2. if j pairs with some t < j 4, compute OPT( i , t 1) and OPT( t + 1 , j 1).

OPT( i , j ) = max OPT( i , j 1),

T M M li O t b 14 19 21 26 28 2009 CS 4104 D i P g i g Weighted Interval Scheduling Segmented Least Squares RNA Secondary Structure Sequence Alignment Shortest Paths in Graphs


OPT (i , j ) is the maximum number of base pairs in a secondary structure forbi b2 . . . bj OPT( i, j) = 0 if i j 4


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b i b 2 . . . b j . OPT( i , j ) 0, if i j 4.


Since t can range from i to j 5,

OPT( i , j ) = max OPT( i , j 1),

T M M li O t b 14 19 21 26 28 2009 CS 4104 D i P i Weighted Interval Scheduling Segmented Least Squares RNA Secondary Structure Sequence Alignment Shortest Paths in Graphs


OPT (i , j ) is the maximum number of base pairs in a secondary structure forbi b2 . . . bj . OPT( i, j) = 0, if i j 4.


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b i b 2 . . . b j . OPT( i , j ) 0, if i j 4.



OPT( i , j ) = max OPT( i , j 1), maxt 1 + OPT( i , t 1) + OPT( t + 1 , j 1)




OPT (i , j ) is the maximum number of base pairs in a secondary structure forb i b 2 . . . b j . OPT( i , j ) = 0, if i j 4.


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i 2 j ( , j) , j



OPT( i , j ) = max OPT( i , j 1), maxt 1 + OPT( i , t 1) + OPT( t + 1 , j 1)

In the inner maximisation, t runs over all indices between i and j 5 thatare allowed to pair with j .



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OPT( i , j ) = max OPT( i , j 1), maxt

1 + OPT( i , t 1) + OPT( t + 1 , j 1)


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There are O (n2 ) sub-problems. How do we order them from smallest to largest? Note that computing OPT( i , j ) involves sub-problems OPT(l , m) wherem l < j i .





1 + OPT( i , t 1) + OPT( t + 1 , j 1)


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1 + OPT( i , t 1) + OPT( t + 1 , j 1)


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Running time of the algorithm is O (n3).



Example of Algorithm


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Sequence Similarity

Given two strings, measure how similar they are. Given a database of strings and a query string, compute the string most


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similar to query in the database. Applications:

Online searches (Web, dictionary).Spell-checkers.Computational biologySpeech recognition.Basis for Unix diff .



Dening Sequence Similarity


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Dening Sequence Similarity


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Edit distance model: how many changes must you to make to one string totransform it into another?

Changes allowed are deleting a letter, adding a letter, changing a letter.



Edit Distanceo-currance o-curr-ance

occurrence occurre-nce


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Proposed by Needleman and Wunsch in the early 1970s. Input: two strings x = x 1x 2x 3 . . . x m and y = y 1y 2 . . . y n . Sets {1, 2, . . . , m} and {1, 2, . . . , n} represent positions in x and y .






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Proposed by Needleman and Wunsch in the early 1970s. Input: two strings x = x 1x 2x 3 . . . x m and y = y 1y 2 . . . y n . Sets {1, 2, . . . , m} and {1, 2, . . . , n} represent positions in x and y . A matching of these sets is a set M of ordered pairs such that

1. in each pair (i , j ), 1 i m and 1 j n and2. no index from x (respectively, from y ) appears as the rst (respectively,

second) element in more than one ordered pair. An index is not matched if it does not appear in the matching.






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second) element in more than one ordered pair. An index is not matched if it does not appear in the matching. A matching M is an alignment if there are no crossing pairs in M : if (i , j ) M and (i , j ) M and i < i then j < j .






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Cost of an alignment is the sum of gap and mismatch penalties:Gap penalty Penalty > 0 for every unmatched index.

Mismatch penalty Penalty x i y j > 0 if (i , j ) M and x i = y j .






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Cost of an alignment is the sum of gap and mismatch penalties:Gap penalty Penalty > 0 for every unmatched index.

Mismatch penalty Penalty x i y j > 0 if (i , j ) M and x i = y j . Output: compute an alignment of minimal cost.




Consider index mx and index n y . Is (m, n) M ?


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Consider index mx and index n y . Is (m, n) M ? Claim: (m, n) M mx not matched or ny not matched.


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OPT (i , j ): cost of optimal alignment between x = x 1x 2x 3 . . . x i andy = y 1y 2 . . . y j .(i , j ) M :






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OPT (i , j ): cost of optimal alignment between x = x 1x 2x 3 . . . x i andy = y 1y 2 . . . y j .(i , j ) M : OPT( i , j ) = x i y j + OPT( i 1, j 1).






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OPT (i , j ): cost of optimal alignment between x = x 1x 2x 3 . . . x i andy = y 1y 2 . . . y j .(i , j ) M : OPT( i , j ) = x i y j + OPT( i 1, j 1).i not matched:






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OPT (i , j ): cost of optimal alignment between x = x 1x 2x 3 . . . x i andy = y 1y 2 . . . y j .(i , j ) M : OPT( i , j ) = x i y j + OPT( i 1, j 1).i not matched: OPT( i , j ) = + OPT( i 1, j ).



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OPT (i , j ): cost of optimal alignment between x = x 1x 2x 3 . . . x i andy = y 1y 2 . . . y j .(i , j ) M : OPT( i , j ) = x i y j + OPT( i 1, j 1).i not matched: OPT( i , j ) = + OPT( i 1, j ). j not matched: OPT( i , j ) = + OPT( i , j 1).

OPT( i , j ) = min ` x i y j + OPT( i 1, j 1) , +OPT( i 1, j ), +OPT( i , j 1)(i , j ) M if and only if minimum is achieved by the rst term.

What are the base cases?






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OPT (i , j ): cost of optimal alignment between x = x 1x 2x 3 . . . x i andy = y 1y 2 . . . y j .(i , j ) M : OPT( i , j ) = x i y j + OPT( i 1, j 1).i not matched: OPT( i , j ) = + OPT( i 1, j ). j not matched: OPT( i , j ) = + OPT( i , j 1).

OPT( i , j ) = min ` x i y j + OPT( i 1, j 1) , +OPT( i 1, j ), +OPT( i , j 1)(i , j ) M if and only if minimum is achieved by the rst term.

What are the base cases? OPT( i , 0) = OPT(0 , i ) = i .




OPT( i , j ) = min x i y j + OPT( i 1, j 1), + OPT( i 1, j ), + OPT( i , j 1)


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Running time is O (mn). Space used in O (mn).






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Can compute OPT( m, n) in O (mn) time and O (m + n) space (Hirschberg1975, Chapter 6.7).






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Can compute OPT( m, n) in O (mn) time and O (m + n) space (Hirschberg1975, Chapter 6.7).

Can compute alignment in the same bounds by combining dynamicprogramming with divide and conquer.



Graph-theoretic View of Sequence Alignment


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Grid graph G xy :Rows labelled by symbols in x and columns labelled by symbols in y .Edges from node ( i , j ) to ( i , j + 1)), to ( i + 1 , j ), and to ( i + 1 , j + 1).Edges directed upward and to the right have cost .

Edge directed from ( i , j ) to ( i + 1 , j + 1) has cost x i +1 y j +1 .



Graph-theoretic View of Sequence Alignment


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Grid graph G xy :Rows labelled by symbols in x and columns labelled by symbols in y .Edges from node ( i , j ) to ( i , j + 1)), to ( i + 1 , j ), and to ( i + 1 , j + 1).Edges directed upward and to the right have cost .

Edge directed from ( i , j ) to ( i + 1 , j + 1) has cost x i +1 y j +1 . f(i, j): minimum cost of a path in G XY from (0, 0) to ( i , j ). Claim: f (i , j ) = OPT( i , j ) and diagonal edges in the shortest path are thematched pairs in the alignment.



Motivation Computational nance:

Each node is a nancial agent.The cost c uv of an edge (u , v ) is the cost of a transaction in which we buy

from agent u and sell to agent v


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from agent u and sell to agent v .Negative cost corresponds to a prot.

Internet routing protocolsDijkstras algorithm needs knowledge of the entire network.Routers only know which other routers they are connected to.Algorithm for shortest paths with negative edges is decentralised.We will not study this algorithm in the class. See Chapter 6.9.



Problem Statement Input: a directed graph G = ( V , E ) with a cost function c : E R , i.e., c uv is the cost of the edge (u , v ) E .

A negative cycle is a directed cycle whose edges have a total cost that isnegative


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negative. Two related problems:

1. If G has no negative cycles, nd the shortest s-t path : a path of from source s to destination t with minimum total cost.

2. Does G have a negative cycle ?



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Approaches for Shortest Path Algorithm

1. Dijsktras algorithm.


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2. Add some large constant to eachedge.



Approaches for Shortest Path Algorithm

1. Dijsktras algorithm. Computesincorrect answers because it isgreedy


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greedy.2. Add some large constant to each

edge. Computes incorrect answersbecause the minimum cost pathchanges.




Assume G has no negative cycles. Claim: There is a shortest path from s to t that is simple (does not repeat anode)


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Assume G has no negative cycles. Claim: There is a shortest path from s to t that is simple (does not repeat anode) and hence has at most n 1 edges.

How do we dene sub-problems?


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How do we dene sub problems?







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pShortest s -t path has n 1edges: how we can reach t using i edges, for different values of i ?We do not know which nodes will

be in shortest s -t path: how wecan reach t from each node in V ?







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pShortest s -t path has n 1edges: how we can reach t using i edges, for different values of i ?We do not know which nodes will

be in shortest s -t path: how wecan reach t from each node in V ?

Sub-problems dened by varying thenumber of edges in the shortestpath and by varying the starting

node in the shortest path.



Dynamic Programming Recursion

OPT (i , v ): minimum cost of a v -t path that uses at most i edges. t is not explicitly mentioned in the sub-problems.

Goal is to compute OPT(n 1, s ).


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p ( )







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p ( )

Let P be the optimal path whose cost is OPT( i , v ).







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Let P be the optimal path whose cost is OPT( i , v ).1. If P actually uses i 1 edges, then OPT( i , v ) = OPT( i 1, v ).2. If rst node on P is w , then OPT( i , v ) = c vw + OPT( i 1, w ).







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Let P be the optimal path whose cost is OPT( i , v ).1. If P actually uses i 1 edges, then OPT( i , v ) = OPT( i 1, v ).2. If rst node on P is w , then OPT( i , v ) = c vw + OPT( i 1, w ).

OPT( i , v ) = min OPT( i 1, v ), minw V

c vw + OPT( i 1, w )



Example of Dynamic Programming RecursionOPT( i , v ) = min OPT( i 1, v ), min

w V c vw + OPT( i 1, w )

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lecture-15-dynamic-programming.pdf

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