Lecture 2: Dynamic Programming 主講人 : 虞台文. Content What is Dynamic Programming? Matrix...

Lecture 2: Dynamic Programming

主講人 :虞台文

Content What is Dynamic Programming? Matrix Chain-Products Sequence Alignments Knapsack Problem All-Pairs Shortest Path Problem Traveling Salesman Problem

Conclusion

What is Dynamic Programming?

Dynamic Programming (DP) tends to break the original problem to sub-problems, i.e., in a smaller size

The optimal solution in the bigger sub-problems is found through a retroactive formula which connects the optimal solutions of sub-problems.

Used when the solution to a problem may be viewed as the result of a sequence of decisions.

Properties for Problems Solved by DP

Simple Subproblems– The original problem can be broken into smaller subproblems with the same structure

Optimal Substructure of the problems– The solution to the problem must be a composition of subproblem solutions (the principle of optimality)

Subproblem Overlap– Optimal subproblems to unrelated problems can contain subproblems in common

The Principle of Optimality The basic principle of dynamic programming Developed by Richard Bellman An optimal path has the property that whatever

the initial conditions and control variables (choices) over some initial period, the control (or decision variables) chosen over the remaining period must be optimal for the remaining problem, with the state resulting from the early decisions taken to be the initial condition.

Example: Shortest Path Problem

GoalStart

Start Goal

Recall Greedy Method forShortest Paths on a Multi-stage Graph

Problem– Find a shortest path from v0 to v3

Is the greedy solution optimal?

Recall Greedy Method forShortest Paths on a Multi-stage Graph

Problem– Find a shortest path from v0 to v3

Is the greedy solution optimal?

The optimal path

Example Dynamic Programming

min 1,1 3

min 1,2 3min 0 3

min 1,3 3

min 1,4 3

( , )( , )

( , )( , )( , )

d v vd v v

d v vd v vd v v

Matrix Chain-Products

Matrix MultiplicationC = A × BA is d × e and B is e × f

O(def )

ij ik kjk

Matrix Chain-Products Given a sequence of matrices, A1, A2, …, An, find

the most efficient way to multiply them together. Facts:

– A(BC) = (AB)C– Different parenthesizing may need different numbers of

operation. Example: A:10 × 30, B: 30 × 5, C : 5 × 60

– (AB)C = (10×30×5) + (10×5×60) = 1500 + 3000 = 4500 ops – A(BC) = (30×5×60) + (10×30×60) = 9000 + 18000 = 27000 ops

Matrix Chain-Products Given a sequence of matrices, A1, A2, …, An, find the m

ost efficient way to multiply them together. A Brute-force Approach:

– Try all possible ways to parenthesize A=A1A2…An

– Calculate number of operations for each one– Pick the best one

Time Complexity:– #paranethesizations = #binary trees of n nodes– O(4n)

A Greedy Approach Idea #1:

– repeatedly select the product that uses the most operations.

Counter-example: – A: 10 5, B: 5 10, C: 10 5, and D: 5 10– Greedy idea #1 gives (AB)(CD), which takes

500+1000+500 = 2000 ops– A((BC)D) takes 500+250+250 = 1000 ops

Another Greedy Approach

Idea #2: – repeatedly select the product that uses the least operations.

Counter-example: – A: 101 11, B: 11 9, C: 9 100, and D: 100 999– Greedy idea #2 gives A((BC)D), which takes 109989+9900+108900=228789 ops– (AB)(CD) takes 9999+89991+89100=189090

DP Define Subproblem

OriginalProblem 1 i j nA A A A 1 i j nA AA A

Subproblem (P ij, i j)(P1n)

Suppose #operations for the optimal solution of Pij is Nij #operations for the optimal solution of the original problem P1n is N1n

1:i i iA d d

What is the relation btw Nij (Pij) and N1n (P1

DP Principle of Optimality

1i k k jA A A A

1:i i iA d d

N ik N k+1,n

d i d k+1 d k d j+1

1 1 1,min ik ki jj i k j i k jd dNN N d 11, 1minij ik k i k ji k j j dN N N d d

DP Implementation 11, 1minij ik k i k ji k j j dN N N d d

1 2 j nNij

DP Implementation 11, 1minij ik k i k ji k j j dN N N d d

1 2 j nNij

DP Implementation

1 2 j nNij

11, 1minij ik k i k ji k j j dN N N d d

DP Implementation

1 2 j nNij

DP Implementation

1 2 j nNij

DP Implementation

1 2 j nNij

DP Implementation

1 2 j nNij

DP Implementation

1 2 j nNij

DP Implementation

1 2 j nNij

DP Implementation

1 2 j nNij

DP Implementation

1 2 j nNij

DP Implementation

1 2 j nNij

DP for Matrix Chain-Products

Algorithm matrixChain(S):Input: sequence S of n matrices to be multipliedOutput: number of operations in an optimal parenthesization of Sfor i 1 to n // main diagonal terms are all zero

Ni,i 0

for d 2 to n // each diagonal do followingfor i 1 to nd+1 // do from top to bottom for each diagona

lj i+d1

Ni,j infinity

for k i to j1 // counting minimum Ni,j min(Ni,j, Ni,k +Nk+1,j +di dk+1 dj+1)

Time ComplexityAlgorithm matrixChain(S):Input: sequence S of n matrices to be multipliedOutput: number of operations in an optimal parenthesization of Sfor i 1 to n // main diagonal terms are all zero

Ni,i 0

for d 2 to n // each diagonal do followingfor i 1 to nd+1 // do from top to bottom for each diagona

lj i+d1

Ni,j infinity

for k i to j1 // counting minimum Ni,j min(Ni,j, Ni,k +Nk+1,j +di dk+1 dj+1)

Exercises 11,min i kij ik ji k j k jN N N d d d

1. The matrixChain algorithm only computes #operations of an optimal parenthesization. But, it doesn’t report the optimal parenthesization scheme. Please modify the algorithm so that it can do so.2. Given an example with 5 matrices to illustrate your idea using a table.

Sequence Alignment

QuestionGiven two strings

are they similar?what is their distance?

1 2, , , mX x x x 1 2, , , nY y y y and

Example

Y:applicableplausibly

How similar they are?Can you give them a score?

Example

applica---ble

-p-l--ausibly

X’:Y:applicableplausibly

Three cases:MatchesMismatchesInsertions & deletions (indel)

Example

applica---ble

-p-l--ausibly

(+1)(1)

The values depends on applications. It can be

described using a so-called substitution matrix, to be

discussed shortly.

Example

applica---ble

-p-l--ausibly

(+1)(1)

Score = 5(+1) + 1(1) + 7 (1) = 3

Example

applica---ble

-p-l--ausibly

(+1)(1)

Score = 5(+1) + 1(1) + 7 (1) = 3 Is the alignment optimal?

Sequence Alignment In bioinformatics, a sequence alignment is a way of

arranging the primary sequences of DNA, RNA, or protein to identify regions of similarity that may be a consequence of functional, structural, or evolutionary relationships between the sequences.

Global and Local Alignments

L G P S S K Q T G K G S - S R I W D N

Global alignmentL N - I T K S A G K G A I M R L G D A

- - - - - - - T G K G - - - - - - - -

Local alignment- - - - - - - A G K G - - - - - - - -

L G P S S K Q T G K G S - S R I W D N

Global alignmentL N - I T K S A G K G A I M R L G D A

- - - - - - - T G K G - - - - - - - -

Local alignment- - - - - - - A G K G - - - - - - - -

Global Alignment– attempts to align the entire sequence– most useful when the sequences in the query set are similar and of roughly equal size.– Needleman–Wunsch algorithm (1971).

Local Alignment– Attempts to align partial regions of sequences

with high level of similarity.– Smith-Waterman algorithm (1981)

Needleman–Wunsch Algorithm

Find the best global alignment of any two sequences under a given substitution matrix.

Maximize a similarity score, to give maximum match

Maximum match = largest number of residues of one sequence that can be matched with another allowing for all possible gaps

Based on dynamic programming Involves an iterative matrix method of calculation

Substitution Matrix In bioinformatics, a substitution matrix estimates

the rate at which each possible residue in a sequence changes to each other residue over time.

Substitution matrices are usually seen in the context of amino acid or DNA sequence alignment, where the similarity between sequences depends on the mutation rates as represented in the matrix.

Substitution Matrix (DNA) w/o Gap Cost

A C G TA 2 1 1 1C 1 2 1 1G 1 1 2 1T 1 1 1 2

Substitution Matrix (DNA) w/ Gap Cost

A C G TA 2 1 1 1C 1 2 1 1G 1 1 2 1T 1 1 1 2

A C G T

A 2 1 1 1 2C 1 2 1 1 2G 1 1 2 1 2T 1 1 1 2 2 2 2 2 2 0

Substitution Matrix (3D-BLAST)

Consider two strings, s of length n and t of length m. Let S be the substitution matrix.

Subproblem: Let Pij is defined to be the optimal aligning for the two substrings:

t[1..i] and s[1..j],and let Mij be the matching score.

Original Problem: Pmn (matching score Mmn)

max __

i ji j

ij i j

ap scoregap score

maxi j

ij i j

ExampleStep 1. Create a scoring matrixStep 2. Make an empty table for Mij

Step 3. Initialize base conditionsStep 4. Fill table by

Step 5. Trace back

s :t :ACGGTAGCCTAAG

Mij A C G G T A G

CCTAAG

22 11 1

2 11 1T0 2 2 2

22 11G 2 12 1C 21 12AGCA

22 11 1

2 11 1T0 2 2 2

22 11G 2 12 1C 21 12AGCA

maxi j

ij i j

Step 5. Trace back

s :t :ACGGTAGCCTAAG

Mij A C G G T A G

CCTAAG

22 11 1

2 11 1T0 2 2 2

22 11G 2 12 1C 21 12AGCA

22 11 1

2 11 1T0 2 2 2

22 11G 2 12 1C 21 12AGCA

2 4 6 8 10 12 14

Step 5. Trace back

s :t :ACGGTAGCCTAAG

Mij A C G G T A G

CCTAAG

22 11 1

2 11 1T0 2 2 2

22 11G 2 12 1C 21 12AGCA

22 11 1

2 11 1T0 2 2 2

22 11G 2 12 1C 21 12AGCA

maxi j

ij i j

0 2 4 6 8 10 12 14

1 0 2 4 6 8 10

3 1 1 3 3 5 7

5 1 0 2 1 3 5

4 3 0 1 1 1 1

6 5 2 1 0 1 2

8 7 3 0 0 1 3

Step 5. Trace back

s :t :ACGGTAGCCTAAG

Mij A C G G T A G

CCTAAG

maxi j

ij i j

2 4 6 8 10 12 14

s’ :t’ :

Step 1. Create a scoring matrixStep 2. Make an empty table for Mij

Step 5. Trace back

maxi j

ij i j

for i = 2 to m+1 do for j from 2 to n +1 do

maxi j

ij i j

s’ ””, t’ ”” while i < 1 and j < 1 do if

s’ sj+ s’ t’ ti+ t’ else if

s’ sj+ s’ t’ gap+ t’ else s’ gap+ s’ t’ ti+ t’while i > 1 do t’ gap+ t’while j > 1 do s’ gap+ s’

Step 1. Create a scoring matrixStep 2. Make an empty table for Mij

Step 5. Trace back

maxi j

ij i j

,1, 1 i jij i j t sM SM

, ,1 jij i j sM M S

Local Alignment Problem

Given two strings s = s1……sn,

t = t1…….tmFind substrings s’, t’ whose similarity (optimal global alignment value) is maximum.

Example: Local AlignmentGTAGT CATCAT ATG TGACTGAC G

TC CATDOGCAT CC TGACTGAC A

GTAGT CATCAT ATGCC TGACTGAC G

TC CATDOGCAT CCTACTAC TGACTGAC A

difference block

difference block difference

difference block

Best aligned subsequeces

Global Alignment (Needleman–Wunsch Algorithm)

Local Alignment (Smith-Waterman Algorithm)

Recursive Formulation

max __

i ji j

ij i j

ap scoregap score

maxi j

i jsi j

gap scoregap sco

MMeM r

Exercises3. Find the best local aligned

substrings for the following two DNA strings:GAATTCAGTTA

GGATCGA

You have to give the detail.Hint: start from the left table.

Exercises

4. What is longest common sequence (LCS) problem? How to solve LCS using dynamic programming technique?

Knapsack Problem

Knapsack Problems Given some items, pack the knapsack to get the

maximum total value. Each item has some weight and some benefit. Total weight that we can carry is no more than some fixed capacity.

Fractional knapsack problem– Items are divisible: you can take any fraction of an

item.– Solved with a greedy algorithm.

0-1 knapsack problem– Items are indivisible; you either take an item or not.– Solved with dynamic programming.

Given a knapsack with maximum capacity W, and a set S consisting of n items

Each item i has some weight wi and benefit value bi (all wi and W are integer values)

Problem: How to pack the knapsack to achieve maximum total value of packed items?

0-1 Knapsack Problem

Why it is called a 0-1 Knapsack Problem?

Example: 0-1 Knapsack Problem

Which boxes should be chosen to maximize the amount of money while still keeping the overall weight under 15 kg ?

Example: 0-1 Knapsack Problem

1 {0,1}x

2 {0,1}x

3 {0,1}x

5 {0,1}x

4 {0,1}x

Objective Function

Unknowns or Variables

Constraints's, 1, ,5ix i

, ,0,1 1, 5ix i

1 2 3 4 54 2 10 2 1Maximize

x x x x x

1 2 3 4 512 1Subject t

154 2o

1x x x x x

Formulation: 0-1 Knapsack Problem

Maximize

Subject t

0-1 Knapsack Problem: Brute-Force Approach

Since there are n items, there are 2n possible combinations of items.

We go through all combinations and find the one with maximum value and with total weight less or equal to W

Running time will be O(2n)

DP Define Subproblem Suppose that items are labeled 1,..., n. Define a subproblem, say, Pk as to finding an optimal solution for items in Sk = {1, 2,..., k}.

original problem is Pn. Is such a scheme workable? Is the principle of optimality held?

A Counterexample

1. 2kgs, 3$

2. 3kgs, 4$

3. 4kgs, 5$

4. 5kgs, 8$

5. 9kgs, 10$20 kgs

A Counterexample

Sub-problem

Optimum Value

P1 1 3$P2 1, 2 7$P3 1, 2, 3 12$P4 1, 2, 3, 4 20$P5 1, 3, 4, 5 26$

1. 2kgs, 3$

2. 3kgs, 4$

3. 4kgs, 5$

4. 5kgs, 8$

5. 9kgs, 10$20 kgs

A Counterexample

Sub-problem

Optimum Value

P1 1 3$P2 1, 2 7$P3 1, 2, 3 12$P4 1, 2, 3, 4 20$P5 1, 3, 4, 5 26$

1. 2kgs, 3$

2. 3kgs, 4$

3. 4kgs, 5$

4. 5kgs, 8$

5. 9kgs, 10$20 kgs

Solution for P4 is not part of the solution for P5 !!!

DP Define Subproblem Suppose that items are labeled 1,..., n. Define a subproblem, say, Pk as to finding an optimal solution for items in Sk = {1, 2,..., k}.

original problem is Pn. Is such a scheme workable? Is the principle of optimality held?

DP Define Subproblem Suppose that items are labeled 1,..., n. Define a subproblem, say, Pk,w as to finding an optimal solution for items in Sk = {1, 2,..., k} and with total weight no more than w.

original problem is Pn,W. Is such a scheme workable? Is the principle of optimality held?

New version

Denote the benefit for the optimal solution of Pk,w as Bk,w.

,1, 1,

max , if k

k wk w k w w k k

B w wB

B B b w w

Denote the benefit for the optimal solution of Pk,w as Bk,w.

,1, 1,

max , if k

k wk w k w w k k

B w wB

B B b w w

In this case, it is impossible to include the kth object.

Not includethe kth object

includethe kth object There are two

possible choices.

Example

1. 2kgs, 3$

2. 3kgs, 4$

3. 4kgs, 5$

4. 5kgs, 6$5 kgs

w k 0 1 2 3 4012345

,1, 1,

max , if k

k wk w k w w k k

B w wB

B B b w w

,1, 1,

max , if k

k wk w k w w k k

B w wB

B B b w w

Example

1. 2kgs, 3$

2. 3kgs, 4$

3. 4kgs, 5$

4. 5kgs, 6$5 kgs

w k 0 1 2 3 4012345

0 0 0 0 000000

Step 1. Setup table and initialize base conditions.

,1, 1,

max , if k

k wk w k w w k k

B w wB

B B b w w

Example

1. 2kgs, 3$

2. 3kgs, 4$

3. 4kgs, 5$

4. 5kgs, 6$5 kgs

w k 0 1 2 3 4012345

0 0 0 0 000000

Step 2. Fill all table entries progressively.

,1, 1,

max , if k

k wk w k w w k k

B w wB

B B b w w

Example

1. 2kgs, 3$

2. 3kgs, 4$

3. 4kgs, 5$

4. 5kgs, 6$5 kgs

w k 0 1 2 3 4012345

0 0 0 0 000000

,1, 1,

max , if k

k wk w k w w k k

B w wB

B B b w w

Example

1. 2kgs, 3$

2. 3kgs, 4$

3. 4kgs, 5$

4. 5kgs, 6$5 kgs

w k 0 1 2 3 4012345

0 0 0 0 000000

,1, 1,

max , if k

k wk w k w w k k

B w wB

B B b w w

Example

1. 2kgs, 3$

2. 3kgs, 4$

3. 4kgs, 5$

4. 5kgs, 6$5 kgs

w k 0 1 2 3 4012345

0 0 0 0 000000

,1, 1,

max , if k

k wk w k w w k k

B w wB

B B b w w

Example

1. 2kgs, 3$

2. 3kgs, 4$

3. 4kgs, 5$

4. 5kgs, 6$5 kgs

w k 0 1 2 3 4012345

0 0 0 0 000000

Step 3. Trace back

Pseudo-Polynomial Time Algorithm

The time complexity for 0-1 knapsack using DP is O(Wn).

Not a polynomial-time algorithm if W is large. This is a pseudo-polynomial time algorithm.

All-PairsShortest Path Problem

All-Pairs Shortest Path Problem

Given weighted graph G(V,E), we want to determine the cost dij of the shortest path between each pair of nodes in V.

Floyd's Algorithm Let be the minimum cost of a path from node i to node j, using only nodes in Vk={v1,…,vk}.

i j1kijd

1 1 1min , 0

k k kij ik kjk

d d d kd

The all-pairs shortest path problem is to find all paths with costs

nij ijd d i j

Floyd's AlgorithmInput Parameter: DOutput Parameter: D, nextall_paths(D, next) n = D.NumberOfRows;// initialize next: if no intermediate// vertices are allowed next[i][j] = jfor i = 1 to n

for j = 1 to nnext[i][j] = j;

for k = 1 to n // compute D(k) for i = 1 to n for j = 1 to n if (D[i][k] + D[k][j] < D[i][j]) D[i][j] = D[i][k] + D[k][j];

next[i][j] = next[i][k];

1 1 1min , 0

k k kij ik kjk

d d d kd

Floyd's AlgorithmInput Parameters: next, i, jOutput Parameters: Noneprint_path(next, i, j) { // if no intermediate vertices, just

// print i and j and return if (j == next[i][j]) { print(i + “ ” + j); return; } // output i and then the path from the vertex // after i (next[i][j]) to j print(i + “ ”); print_path(next,next[i][j], j);}

Traveling Salesman Problem

Traveling Salesman Problem (TSP)

How many feasible paths?

n cities

! ( 1)!2 2n nn

Example (TSP)

6 4 7 6

(1234) = 18(1243) = 19(1324) = 23(1342) = 19(1423) = 23(1432) = 18

Subproblem Formulation for TSP

g(i, S) length of the shortest path from i to 1 visiting each city in S exactly once.

g(1, V {1})length of the optimal TSP tour.

Subproblem Formulation for TSP

g(i, S) length of the shortest path from i to 1 visiting each city in S exactly once.

( , ) min ( , { })ijj Sg i S d g j S j

Goal: g(1, V {1})

Example ( , ) min ( , { })ijj S

g i S d g j S j

Goal: g(1, V {1})

2 6 42 7 66 7 54 6 5

(1,{2,3,4})g

(2,{3,4})g

(3,{2, 4})g

(4,{2,3})g

(3,{4})g

(4,{3})g

(2,{4})g

(4,{2})g

(2,{3})g

(3,{2})g

(4, )g

(3, )g

(4, )g

(2, )g

(3, )g

(2, )g

6 4 7 6

Example ( , ) min ( , { })ijj S

g i S d g j S j

Goal: g(1, V {1})

2 6 42 7 66 7 54 6 5

(1,{2,3,4})g

(2,{3,4})g

(3,{2, 4})g

(4,{2,3})g

(3,{4})g

(4,{3})g

(2,{4})g

(4,{2})g

(2,{3})g

(3,{2})g

(4, )g

(3, )g

(4, )g

(2, )g

(3, )g

(2, )g

4 6 4 2 6 2

5 5 6 6 7 7

9 11 10 8 13 97 6 7 5 6 5

16 13 142 6 4

6 4 7 6

DP TSP AlgorithmInput Parameter: DOutput Parameter: P // pathTSP(D) n = Dim(D);for i = 1 to n

g[i, ] = D[i, 1];for k = 1 to n2 // compute g for subproblems for all S V{1} with |S|=k for all i S {1}

g[i, S] = minjS{D[i, j], g[j, S {j}] };P[i, S] = arg minjS{D[i, j], g[j, S {j}] };

// compute the TSP tourg[1, V{1}] = minjV{1}{D[1, j], g[j, V {1, j}]};P[1, V{1}] = arg minjV{1}{D[1, j], g[j, V {1, j}]};

Goal: g(1, V {1})

DP TSP AlgorithmInput Parameter: DOutput Parameter: P // pathTSP(D) n = Dim(D);for i = 1 to n

g[i, ] = D[i, 1];for k = 1 to n2 // compute g for subproblems for all S V{1} with |S|=k for all i S {1}

g[i, S] = minjS{D[i, j], g[j, S {j}] };P[i, S] = arg minjS{D[i, j], g[j, S {j}] };

// compute the TSP tourg[1, V{1}] = minjV{1}{D[1, j], g[j, V {1, j}]};P[1, V{1}] = arg minjV{1}{D[1, j], g[j, V {1, j}]};

Goal: g(1, V {1})

Lecture 2: Dynamic Programming 主講人 : 虞台文. Content What is Dynamic Programming? Matrix...

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