dynamic programming - algorithm

8/6/2019 Dynamic Programming - Algorithm

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CSC2105: AlgorithmsCSC2105: Algorithms

Dynamic ProgrammingDynamic Programming

Mashiour RahmanMashiour RahmanAssistant ProfessorAssistant [email protected]@aiub.edu

American International University BangladeshAmerican International University Bangladesh


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Mashiour Rahman AIUB::CSC2105::Algorithms Dynamic Programming2

gor m es gngor m es gntechniquestechniques

Iterative (brute-force) algorithmsIterative (brute-force) algorithms

Example: insertion sort

Algorithms that use efficient data structuresAlgorithms that use efficient data structuresExample: heap sort

Divide-and-conquer algorithmsDivide-and-conquer algorithms

Example: Binary search, merge sort, quick sortDynamic programmingDynamic programming

Example: Fibonacci numbers, Matrix multiplication optimization, LongestCommon Subsequence

Greedy algorithmsGreedy algorithms

Example: Huffman coding, Minimum Spanning Tree, Dijkstara

Graph AlgorithmsGraph AlgorithmsExample: BFS (Robot moving, Path Finding), DFS( Topological Sorting,Strongly Connected Component)


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Fibonacci NumbersFibonacci Numbers

Leonardo Fibonacci (1202)Leonardo Fibonacci (1202)::A rabbit starts producing offspring during the second yearafter its birth and produces one child each generation

How many rabbits will there be aftern generations?

F(1)=1 F(2)=1 F(3)=1 F(4)=3 F(5)=5 F(6)=8+1F(3)=2 F(4)=2+1 F(5)=3+2 F(6)=5+3


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F(n)= F(n-1)+ F(n-2)F(n)= F(n-1)+ F(n-2)

F(0)F(0)=0, F(1)=0, F(1)

=1=1

0, 1, 1, 2, 3, 5, 8, 13, 21, 34

Straightforward recursive procedure is slow!Straightforward recursive procedure is slow!

FibonacciR(n)

01 if n 1 then return n02 else return FibonacciR(n-1) + FibonacciR(n-2)

FibonacciR(n)

01 if n 1 then return n02 else return FibonacciR(n-1) + FibonacciR(n-2)


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We keep calculating the same value over and over!We keep calculating the same value over and over!Subproblems are overlapping they share sub-subproblems

F(6) = 8

F(5)

F(4)

F(3)

F(1)

F(2)

F(0)

F(1)F(1) F(1)F(1)

F(2)F(2)

F(0)F(0)

F(3)F(3)

F(1)F(1)

F(2)F(2)

F(0)F(0)

F(1)F(1)

F(4)F(4)

F(3)F(3)

F(1)F(1)

F(2)F(2)

F(0)F(0)

F(1)F(1) F(1)F(1)

F(2)F(2)

F(0)F(0)


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How many summations are there in F(How many summations are there in F(nn)?)?

F(n) = F(n1) + F(n 2) + 1

F(n) 2F(n 2) +1 and F(1) = F(0) = 0

Solving the recurrence we get

F(n) 2n/2 1 1.4n

Running time isRunning time is exponentialexponential!!


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We can calculateWe can calculate FF((nn)) inin linearlinear time bytime by

remembering solutionsremembering solutions to the solvedto the solved

subproblems (=subproblems (= dynamic programmingdynamic programming).).

Compute solution in a bottom-up fashionCompute solution in a bottom-up fashion

Trade space for time!Trade space for time!

Fibonacci(n)

01 F[0]0

02 F[1]1

03 for i 2 to n do

04 F[i]

F[i-1]

+ F[i-2]

05 return F[n]

Fibonacci(n)

01 F[0]0

02 F[1]103 for i 2 to n do

04 F[i]

F[i-1]

+ F[i-2]

05 return F[n]




In fact, only two values need to be rememberedIn fact, only two values need to be remembered

at any time!at any time!

FibonacciImproved(n)

01 if n 1 then return n02 Fim2 0

03 Fim1 1

04 for i 2 to n do

05 FiFim1+ Fim206 Fim2 Fim1

07 Fim1 Fi

05 return Fi

FibonacciImproved(n)

01 if n 1 then return n02 Fim2 0

03 Fim1 1

04 for i 2 to n do

05 Fi

Fim1

+ Fim2

06 Fim2 Fim1

07 Fim1 Fi

05 return Fi



HistoryHistory

Dynamic programmingDynamic programming

Invented in the 1957 by Richard BellmanRichard Bellman as a

general method for optimizing multistage decision

processes

The term programming refers to a tabular methodtabular method.Often used foroptimizationoptimization problems.


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Dynamic ProgrammingDynamic Programming

Solves problems bySolves problems by combiningcombining the solutions tothe solutions tosubproblems that contain common sub-sub-problems.subproblems that contain common sub-sub-problems.

Difference between DP and Divide-and-ConquerDifference between DP and Divide-and-Conquer

Using Divide and ConquerDivide and Conquerto solve these problems isinefficientinefficient as the same common sub-sub-problems

have to be solved many timesmany times.

DP will solve each of them onceonce and theiranswersanswers

are stored in a tableare stored in a table for future reference.



Intuitive ExplanationIntuitive Explanation

Given a problemGiven a problem PP, obtain, obtaina sequence of problemsa sequence of problems

QQ00,, QQ11, .,, ., QQmm, where:, where:

You have a solution to QQ00

The solution to a problem QQjj,

j > 0j > 0, can be obtained from

solutions to problems

QQ00...QQkk, kk< j< j, that appear

earlier in the sequence.

El t f D iE t D



Elements of DynamicE ements o Dynam cProgrammingProgramming

DPDP is used to solve problems with the followingis used to solve problems with the following

characteristics:characteristics:

Optimal sub-structureOptimal sub-structure (Principle of Optimality)

an optimal solution to the problem contains within it

optimal solutions to sub-problems.

Overlapping subproblemsOverlapping subproblems

there exist some places where we solve the same

subproblem more than once



xamp e: onaccxamp e: onaccNumbersNumbers

We keep calculating the same value over and over!We keep calculating the same value over and over!Subproblems are overlapping they share sub-subproblems

F(6) = 8

F(5)

F(4)

F(3)

F(1)

F(2)

F(0)

F(1)F(1) F(1)F(1)

F(2)F(2)

F(0)F(0)

F(3)F(3)

F(1)F(1)

F(2)F(2)

F(0)F(0)

F(1)F(1)

F(4)F(4)

F(3)F(3)

F(1)F(1)

F(2)F(2)

F(0)F(0)

F(1)F(1) F(1)F(1)

F(2)F(2)

F(0)F(0)



eps o es gn ng aeps o es gn ng aDynamic ProgrammingDynamic Programming

AlgorithmAlgorithm

1.1. CharacterizeCharacterize optimal sub-structureoptimal sub-structure

2.2. RecursivelyRecursivelydefine the value of andefine the value of anoptimal solutionoptimal solution

3.3. Compute the valueCompute the value bottom upbottom up

4.4. (if needed)(if needed) ConstructConstructan optimal solutionan optimal solution



1.Optimal Sub-Structure1.Optimal Sub-Structure

An optimal solution to the problem contains optimalAn optimal solution to the problem contains optimal

solutions to sub-problemssolutions to sub-problems

Solution to a problem:

Making achoicechoice out of a number of

possibilitiespossibilities (look what

possible choices there can be)

SolvingSolving one or more sub-problemssub-problems that are the result of aresult of a

choicechoice (characterize the space of sub-problems)

Show that solutions to sub-problems must themselves be

optimal for the whole solution to be optimal.



2. Recursive Solution2. Recursive Solution

Write a recursive solution for the value of anWrite a recursive solution for the value of an

optimal solution.optimal solution.

Show that the number of different instancesShow that the number of different instancesof sub-problems is bounded by a polynomial.of sub-problems is bounded by a polynomial.

+=

k

kMM

opt

k

optchoicethemakingwithassociatedcostThe

choicefromresultingssubproblemallforofnCombinatioOptimal

choicesallover



3. Bottom Up3. Bottom Up

ComputeCompute the value of an optimal solution inthe value of an optimal solution in

a bottom-up fashion, so that you always havea bottom-up fashion, so that you always have

the necessarythe necessary sub-results pre-computedsub-results pre-computed

(or use memoization)(or use memoization)

Check if it is possible toCheck if it is possible to reducereduce thethe spacespace

requirements, by requirements, by forgettingforgetting solutions to solutions to

sub-problems that will not be used any moresub-problems that will not be used any more



. ons ruc p ma. ons ruc p maSolutionSolution

Construct an optimal solution fromConstruct an optimal solution fromcomputed informationcomputed information (which(which

records a sequence of choices maderecords a sequence of choices made

that lead to an optimal solution)that lead to an optimal solution)



ev ew: a r xev ew: a r xMultiplicationMultiplication

Matrix-Multiply(A,B):Matrix-Multiply(A,B):

1 if columns[A] != rows[B] then2 error "incompatible dimensions"

3 else{

5 for i = 1 to rows[A] do

6 for j = 1 to columns[B] do{

7 C[i,j] = 0

8 for k = 1 to columns[A] do

9 C[i,j] = C[i,j]+A[i,k]*B[k,j]

10 }

11 }

12 return C

Time complexityTime complexity = O(pqr),

where |A|=pq and |B|=qr



Two matrices,Two matrices,A nA nmmmatrixmatrixandand B mB mkkmatrix, can be multiplied to getmatrix, can be multiplied to get CCwithwith

dimensionsdimensions nnkk,, usingusing nnmmkkscalarscalar

multiplicationsmultiplications

Problem:: ComputeCompute a product of manya product of many

matricesmatrices efficientlyefficiently

Multiplying MatricesMultiplying Matrices

, , ,

1

m

i j i l l j

l

c a b=

= 11 12

1311 12

21 22 22

2321 22

31 32

... ... ...

... ...

... ... ...

a abb b

a a cbb b

a a

=



Matrix Chain Multiplication [MCM]:Matrix Chain Multiplication [MCM]:

The ProblemThe Problem

Input: Matrices: Matrices TT11,T,T22,, T,, Tnn, each, each TTii of sizeof size ddi-1i-1 x dx dii

Output: Fully: Fullyparenthesizedparenthesizedproductproduct TT11TT22TTnn thatthat

minimizes the number of scalar multiplications.minimizes the number of scalar multiplications.

A product of matrices is fully parenthesized if it is eitherA product of matrices is fully parenthesized if it is either

a) a single matrix, or

b) the product of 2 fully parenthesized matrix products surrounded byparentheses.

Example:Example: TT11 TT22 TT33 TT44can be fully parenthesized as:can be fully parenthesized as:

1.1. (T(T11 (T(T22 (T(T33 TT44 )))))) 4.4. ((T((T11 (T(T22 TT33 ))T))T44 ))

2.2. (T(T11 ((T((T22 TT33 )T)T44 )))) 5.5. (((T(((T11 TT22 )T)T33 )T)T44 ))

3.3. ((T((T11 TT22 )(T)(T33 TT44 ))))



(A1( A

2( A

3A4) )) : p

2p3p4+ p

1p2p4+ p

0p1p4

(((A1

A2) A

3)A4) : p

0p1p2+ p

0p2p3+ p

0p3p4

Suppose size ofAiis p

i-1by p

i.

A1

A2A3

A4

p0

p1 p2 p3 p4


The ProblemThe Problem



A1

A2A3

A4

p0

p1 p2 p3 p4

(A(A33 AA44))

p2p3p4 multiplications

A1

A2

p0

p1 p2

A3 A4

p4

A1

A2(A3A4)

p0

p1 p4

(A(A22(A(A33 AA44))))

p1p2p4

multiplications

A1(A2(A3A4))p0

p4

(A(A11(A(A22(A(A33 AA44))))))

p0p1p4multiplications

Total: p2p3p4+ p

1p2p4+ p

0p1p4


(A(A11(A(A22(A(A33 AA44))))))



A1

A2A3

A4

p0

p1

p2

p3

p4

(A1A2)

A3

A4

p0

p2

p3

p4

(A(A11 AA22))p0p1p2 multiplications

(A1A2)A3

A4

p0

p3 p4

((A((A11 AA22)A)A33))

p0p2p3

multiplications

((A1A2)A3)A4p0

p4

(((A(((A11 AA22)A)A33)A)A44))

p0p3p4multiplications

Total: p0p1p2+ p

0p2p3+ p

0p3p4


(((A(((A11 AA22)A)A33)A)A44))



MCM: ParenthesizationMCM: ParenthesizationSome Preliminaries

Matrix multiplication is associativeassociative

(ABAB)CC=AA(BCBC)

The parenthesizationparenthesization matters

ConsiderAABBCCDD, where

AAis 303011, BBis 114040,

CCis 40401010, DDis 10102525Costs:

(((ABAB)CC)DD)

30x1x40

(AB)=30x40

30x40x10

((AB)C)=30x10

30x10x25

(((AB)C)D)=30x25

A=30 x 1 B=1 x 40

C=40x10

D=10x25

=1200 + 12000 7500 = 20700+


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Mashiour Rahman AIUB::CSC2105::Algorithms Dynamic Programming



(ABAB)CC=AA(BCBC)



AAis 303011, BBis 114040,

CCis 40401010, DDis 10102525Costs:

(((ABAB)CC)DD)

= 1200 + 12000 + 7500 = 20700

((ABAB)(CDCD))

40x10x25

(CD)=40x25

30x40x25

((AB)(CD))=30x25

C=40x10 D=10x25

30x1x40

(AB)=30x40

A=30 x 1 B=1 x 40

=1200 + 10000 30000 = 41200+


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(ABAB)CC=AA(BCBC)



AAis 303011, BBis 114040,

CCis 40401010, DDis 10102525Costs:

(((ABAB)CC)DD)

= 1200 + 12000 + 7500 = 20700

((ABAB)(CDCD))

= 1200 + 10000 + 30000 = 41200

(AA((BCBC)DD))

1x40x10

(BC)=1 x10

1x10x25

((BC)D)=1x25

30x1x25

(A((BC)D))=30x25

A=30 x 1

B=1 x40 C=40x10

D=10x25

=400 + 250 750 = 1400+


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Mashiour Rahman AIUB::CSC2105::Algorithms Dynamic Programming2 2



(ABAB)CC=AA(BCBC)



AAis 303011, BBis 114040,

CCis 40401010, DDis 10102525Costs:

(((ABAB)CC)DD)

= 1200 + 12000 + 7500 = 20700

((ABAB)(CDCD))

= 1200 + 10000 + 30000 = 41200

(AA((BCBC)DD))

= 400 + 250 + 750 = 1400

We need to optimally parenthesize

TT11x TT22xTTnn, where TTii is a ddi-1i-1 x ddii matrix.

According to the given example

dd = {30, 1, 40, 10, 25 }

TT

11x

TT

22x

TT

33x

TT

44where

TT11 = dd1-11-1 x dd11= dd00 x dd11 = 30 x 1

TT22 = dd2-12-1 x dd22 = dd11 x dd22 = 1 x 40

TT33 = dd3-13-1 x dd33 = dd22 x dd33 = 40 x 10

TT44 = dd4-14-1 x dd44 = dd33 x dd44 = 10 x 25

Costs:

((TT11 TT22) TT33) TT44= 20700

(TT11 TT22) (TT33 TT44) = 41200

TT11 ((TT22 TT33) TT44) = 1400


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MCM: ParenthesizationMCM: Parenthesization

Let the number of different parenthesizations, P(n).

Then,

Using Generating FunctionGenerating Function we have,P(n) = C(n-1), the (n-1)thCatalan Numberwhere

Exhaustively checking all possible parenthesizations

take exponential time!

( )n4C3/2n2n

n/1)1/(nC(n) =+=

==

=

1-n

1k

2nifk)-P(k)P(n

1nif1P(n)

MCM :: Step 1: Characterize Optimal Sub-MCM :: Step 1: Characterize Optimal Sub-


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MCM :: Step 1: Characterize Optimal Sub-MCM :: Step 1: Characterize Optimal Sub-structurestructure

LetLetMM((i, ji, j))be thebe the minimumminimum number of multiplications necessarynumber of multiplications necessary

to computeto compute TTi..ji..j = T= Tii x x Tx x Tjj

Key observationsKey observations

The outermost parenthesis partitions the chain of matrices ((i, ji, j))at some

kk, (ii k < jk < j): (TTiiTTkk)(TTk+1k+1 TTjj)

The optimal parenthesization of matrices ((i, ji, j))is also optimal on either

side ofkk; i.e., for matrices (i, ki, k) and (k+1, jk+1, j)

Within the optimal parenthesization ofTTi..ji..j

,

(a) the parenthesization ofTTi..ki..k must be optimal

(b) the parenthesization ofTTk+1..jk+1..j must be optimal

Why?

MCM ::MCM :: Step 2: Recurs ve Recurrenceep : ecurs ve ecurrence


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MCM ::MCM :: Step 2: Recurs ve Recurrenceep : ecurs ve ecurrenceFormulationFormulation

Need to find TT1..n1..n

LetM(i, j)M(i, j)= minimum # of scalar multiplications needed to compute TTi..ji..j

Since TTi..ji..j can be obtained by breaking it into TTi..ki..k & TTk+1..jk+1..j , we have

Note: The sizes ofTTi..ki..k is ddi-1i-1 x dx dkk and TTk+1..jk+1..j is ddkk x dx djj,

and TTi..ki..k TTk+1..jk+1..j is ddi-1i-1 x dx djjafterddi-1i-1 ddkk ddjjscalar multiplications.

Let s(i, j)s(i, j) be the value kk where the optimal split occurs

A direct recursive implementation is exponential a lot of duplicated work.

But there are only few different subproblems ((i, ji, j)): one solution for each

choice ofiiandjj (i < ji < j).

{ }


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MCM::MCM::Step 2: Recursive (Recurrence)Step 2: Recurs ve RecurrenceFormulationFormulation

Recursive-Matrix-Chain(Recursive-Matrix-Chain(dd,, ii,, jj))

11 ififii == jjthenthen

22 returnreturn 0;0;

33 MM[i,[i, jj] =] = ;;

44 forforkk == iitotojj-1-1 dodo55 qq= Recursive-Matrix-Chain(= Recursive-Matrix-Chain(dd,, ii,, kk) +) +

Recursive-Matrix-Chain(Recursive-Matrix-Chain(dd,, k+1k+1,, jj) +) +

ddi-1i-1ddkkddjj

66 ififqq


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MCMMCM :: Step 2: Recursive (Recurrence):: Step 2: Recursive (Recurrence)

FormulationFormulationOverlapping SubproblemsOverlapping Subproblems

Let T(n) be the time complexity ofLet T(n) be the time complexity ofRecursive-Matrix-Chain(d, 1, n)Recursive-Matrix-Chain(d, 1, n)

For n > 1, we haveFor n > 1, we have

T(n)= 1 +

a) 1 is used to cover the cost of lines 1-3, and 8

b) 1 is used to cover the cost of lines 6-7

Using substitution, we can show thatUsing substitution, we can show that T(n) 2T(n) 2n-1n-1

HenceHence T(n) = (2T(n) = (2nn))

=

++1n

1k

1)k)-T(n(T(k)

MCMMCM :: tep : omput ng pt ma:: ep : ompu ng p ma


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MCMMCM :: tep : omput ng pt ma:: ep : ompu ng p maCostsCosts

To compute TTi..ji..jwe need only values for subproblems of

length


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MCM :: : uCostsCosts

Idea:Idea: store the optimal coststore the optimal costM(i, j)M(i, j)for eachfor each

subproblem in a 2d arraysubproblem in a 2d array

MM

[[1..n,1..n1..n,1..n

]]

TriviallyM(i, i)M(i, i) = 0= 0, 11 ii nn

To computeM(i, j)M(i, j),, where i ji j == LL, we need only

values ofMMfor subproblems of length


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MCMMCM :: Step 3: Computing the Optimal:: tep : omput ng t e pt maCostsCosts

Matrix-Chain-Order(Matrix-Chain-Order( dd))

0101 nn== lengthlength[[dd]-1]-1 ////dd is the array of matrix sizesis the array of matrix sizes02 for02 for ii = 1 to= 1 to nndodo

0303 MM[[ii,,ii] = 0] = 0 // no multiplication for 1 matrix// no multiplication for 1 matrix

04 for04 for lenlen= 2 to= 2 to nndodo // len is length of sub-chain// len is length of sub-chain

05 for05 for ii = 1 to= 1 to nn--lenlen+1 do+1 do //// ii: start of sub-chain: start of sub-chain

0606 jj ==ii++lenlen-1-1 //// jj: end of sub-chain: end of sub-chain0707 MM[[ii,,jj] =] =

0808 forfor kk== ii toto jj-1 do-1 do

0909 qq==MM[[ii,,kk]+M[]+M[kk+1,+1,jj]+]+ddii-1-1ddkkddjj

1010 ifif qq


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MCMMCM :: Step 3: Computing the Optimal:: Step 3: Computing the OptimalCostsCosts

After the execution:After the execution:MM[[1,n1,n]]contains the value of ancontains the value of an

optimal solution andoptimal solution and ss contains optimal subdivisions (choicescontains optimal subdivisions (choices

ofofkk) of any subproblem into two sub-subproblems) of any subproblem into two sub-subproblems

Let us run the algorithm on the six matrices:Let us run the algorithm on the six matrices:

d={ 30, 35, 15, 5, 10, 20, 25 }d={ 30, 35, 15, 5, 10, 20, 25 }

See CLRS Figure 15.3See CLRS Figure 15.3

MatrixMatrix DimensionDimension

T1T1 30 X 3530 X 35

T2T2 35 X 1535 X 15

T3T3 15 X 515 X 5

T4T4 5 X 105 X 10

T5T5 10 X 2010 X 20

T6T6 20 X 2520 X 25

Simulation-Simulation- 0 1 2 3 4 5 6


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S 1 2 3 4 5 6

1

2

3

4

5

6


0101 nn== lengthlength[[dd]-1]-102 for02 for ii = 1 to= 1 to nndodo

0303 MM[[ii,,ii] = 0] = 004 for04 for lenlen= 2 to= 2 to nndodo05 for05 for ii = 1 to= 1 to nn--lenlen+1 do+1 do0606 jj ==ii++lenlen-1-1

0707 MM[[ii,,jj] =] =

0808 forfor kk== ii toto jj-1 do-1 do0909 qq==MM[[ii,,kk]+M[]+M[kk+1,+1,jj]+]+ddii-1-1 ddkkddjj1010 ifif qq


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S 1 2 3 4 5 6

1

2

3

4

5

6




0707 MM[[ii,,jj] =] =



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S 1 2 3 4 5 6

1 1

2 2

3 3

4 4

5 5

6




0707 MM[[ii,,jj] =] =



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S 1 2 3 4 5 6

1 1

2 2

3 3

4 4

5 5

6




0707 MM[[ii,,jj] =] =



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:: : uSolutionSolution

To get the optimal solutionTo get the optimal solution TT1..61..6 ,, s[]s[] isis

used as follows:used as follows:TT1..61..6

= (T= (T1..31..3 TT4..64..6 ) ;) ;sincesince s[1,6] = 3s[1,6] = 3= ((T= ((T1..11..1 TT2..32..3 )(T)(T4..54..5 TT6..66..6 ))))

;;sincesince s[1,3] =1s[1,3] =1 andand s[4,6]=5s[4,6]=5

=((T=((T11 (T(T22 TT33 ))((T))((T44 TT55 )T)T66 ))))

MCM can be solved inMCM can be solved in O(n3)O(n3) timetime

Matrix Chain MultiplicationMatr x C a n Mu t p cat on


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Matrix Chain MultiplicationMatr x C a n Mu t p cat onProblemProblem

Running timeRunning time

It is easy to see that it is O(n3)

(three nested loops)

It turns out it is also (n3)

Thus, a reduction from exponential time toThus, a reduction from exponential time topolynomial time.polynomial time.

MemoizationMemoization


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MemoizationMemoizationMemoizationMemoization is one way to deal with overlapping subproblemsis one way to deal with overlapping subproblems

After computing the solution to a subproblem, store it in a table

Subsequent calls just do a table lookup

Can modify recursive algorithm to use memoziationCan modify recursive algorithm to use memoziation

If we prefer recursion we can structure our algorithm as aIf we prefer recursion we can structure our algorithm as a

recursive algorithmrecursive algorithm::

Initialize all elements toInitialize all elements to and calland callMemoMCM(i,j)MemoMCM(i,j)

MemoMCM(i,j)

1. if i = j thenreturn 0

2. else if M[i,j] < then return M[i,j]

3. else for k := i to j-1 do

4. q := MemoMCM(i,k)+MemoMCM(k+1,j) + d

i-1dkdj

5. if q < M[i,j] then

6. M[i,j] := q

7. return M[i,j]

MemoMCM(i,j)

1. if i = j thenreturn 0

2. else if M[i,j] < then return M[i,j]

3. else for k := i to j-1 do

4. q := MemoMCM(i,k)+

MemoMCM(k+1,j) + di-1dkdj

5. if q < M[i,j] then

6. M[i,j] := q

7. return M[i,j]

M i tiM i ti


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MemoizationMemoization

Memoization:Memoization:

Solve the problem in a top-downtop-down fashion, but

record the solutions to subproblems in a table.

Pros and cons:Pros and cons:

Recursion is usually slower than loops and usesstack space

Easier to understand

If not all subproblems need to be solved, you aresure that only the necessary ones are solved

Longest Common SubsequenceLongest Common Subsequence


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Longest Common SubsequenceLongest Common SubsequenceThe ProblemThe Problem

Given two sequencesGiven two sequences

X = < x1, x2, , xm>

Z = < z1, z

2, , z

k>

ZZ is ais a subsequencesubsequence ofofXX ifif

zj = xi( j ) for all j = 1, , k

whereis strictly increasing(but not required to

be consecutive)

Examples:Examples:

Let X = < A, B, C, B, D, A, B >

< A >, < B >, < C >, and< D >are subsequences.

< C, A >, < C, B >, < C, B, A, B >are subsequences.

How many possible subsequences for a n-element sequence?



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Longest Common SubsequenceLongest Common SubsequenceProblemProblem

ZZ is ais a commoncommon subsequence of sequencessubsequence of sequences XX andand YY ififZZ is ais a

subsequence of bothsubsequence of both XX andand YY..

Example:Example:

Let X = < A, B, C, B, D, A, B >X = < A, B, C, B, D, A, B >and Y = < B, D, C, A, B, A >Y = < B, D, C, A, B, A >

< A >, , ,< A >, , , andare common subsequences.

< C, A >< C, A >is, but< A, C >< A, C >is not.

< B, C, A>< B, C, A>is a common subsequence

< B, C, B, A >< B, C, B, A >is the longest common subsequence.

Example:Example:

xx = = ssarariieempmpioiollcceewewe

yy = we= wessttiiggmmuuppsasallrtrtee



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Longest Common SubsequenceLongest Common SubsequenceProblemProblem

LCSLCS::

InputInput: two sequences: two sequences x[1..x[1..mm]] andand y[1..y[1..nn]]

OutputOutput: longest common subsequence of: longest common subsequence ofxx andand yy(denoted(denoted

LCS(x,y)LCS(x,y)))

Brute-force algorithm:Brute-force algorithm:

For every subsequence ofxx, check if it is a subsequence ofyy

22mm subsequences ofxx to check against nnelements ofyy : O(O(n 2n 2mm))

LCS O ti l S b t tLCS: Optimal Sub structure


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LCS: Optimal Sub-structureLCS: Optimal Sub-structure

TheThe iiththprefixprefixofofX = < xX = < x11,x,x22,,x,,xmm >>is denotedis denoted XXii

= < x= < x11,x,x22,,x,,xii >>

XX00is the emptyemptysequence

XXmmis the wholewhole sequence XX

Theorem 15.1 (Optimal Sub-structure of LCS)Theorem 15.1 (Optimal Sub-structure of LCS)

LetLet X=, andand

Z=

bebe

LCS(X,Y)LCS(X,Y)

.

.

(1) Ifxxmm= y= ynn, then zzkk = x= xmm = y= ynn and ZZk-1k-1 is LCS(XLCS(Xm-1m-1 ,YYn-1n-1 ))

(2) ifxxmm y ynn, then zzkk x xmmimplies ZZ is LCS(XLCS(Xm-1m-1 ,Y),Y)

(3) ifxxmm y ynn, then zzkk y ynn implies ZZ is LCS(X,YLCS(X,Yn-1n-1 ))


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LCS Optimal S bstr ct reLCS: Optimal Substructure


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LCS: Optimal SubstructureLCS: Optimal Substructure

LCSLCS has an optimal sub-structure defined byhas an optimal sub-structure defined byprefixesprefixesofof

XX andand YY

The sub-problems of findingThe sub-problems of finding LCS(XLCS(Xm-1m-1 ,Y,Ynn)) andand

LCS(XLCS(Xmm,Y,Yn-1n-1 )) share a common sub-sub-problem ofshare a common sub-sub-problem of

findingfinding LCS(XLCS(Xm-1m-1 ,Y,Yn-1n-1 )). They are overlapping.. They are overlapping.

Simplify:Simplify: just worry aboutjust worry about LCSLCS length for nowlength for now

Define c[i, j]c[i, j]= length ofLCS(XLCS(Xii, Y, Yjj))

So c[m, n]c[m, n]= length ofLCS(X, Y)LCS(X, Y)

LCS: RecurrenceLCS: Recurrence


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LCS: RecurrenceLCS: RecurrenceDefineDefine c[i, j]c[i, j] = length of= length ofLCSLCS ofofx[1..i],x[1..i],

y[1..j]y[1..j]

Note that the conditions in the problem restrict sub-Note that the conditions in the problem restrict sub-

problems (ifproblems (ifxxii = y= yii we considerwe considerxxi-1i-1 andand yyi-1i-1, etc), etc)

UseUseb[i, j]b[i, j] to remember where to extract anto remember where to extract an

element of anelement of an LCSLCS..

0 if 0 or 0

[ , ] [ 1, 1] 1 if , 0 and

max{ [ , 1], [ 1, ]} if , 0 and

i j

i j

i j

c i j c i j i j x

c i j c i j i j x

= =

= + >

>



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LCS: RecurrenceLCS: Recurrenceintint lcsRec(lcsRec(intint i,i, intint j) {j) {

ifif (i==0 || j==0) return 0;(i==0 || j==0) return 0;

else ifelse if (x[i] == y[j])(x[i] == y[j])

returnreturn lcsRec(i-1,j-1) + 1;lcsRec(i-1,j-1) + 1;

elseelse

returnreturn max(lcsRec(i-1,j),lcsRec(i,j-1));max(lcsRec(i-1,j),lcsRec(i,j-1));

}}

intint lcsRec(lcsRec(intint i,i, intint j) {j) {

ifif (i==0 || j==0) return 0;(i==0 || j==0) return 0;

else ifelse if (x[i] == y[j])(x[i] == y[j])

returnreturn lcsRec(i-1,j-1) + 1;lcsRec(i-1,j-1) + 1;

elseelse

returnreturn max(lcsRec(i-1,j),lcsRec(i,j-1));max(lcsRec(i-1,j),lcsRec(i,j-1));

}}

Recurrence for time complexity:

T(2n)=T(2n-2)+1 if x[n]=y[n]

T(2n)=2T(2n-1) if x[n]!=y[n]T(2n)=1 if n=0

T(2n)=2T(2n-1)

=2(2T(2n-2))

=4T(2n-2)

=4(2T(2n-3))

=8T(2n-3)

=2iT(2n-i)

= 2

2n

= 4

n



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LCS: RecurrenceLCS: Recurrenceintint lcsMemo(lcsMemo(intint i,i, intint j) {j) { ifif (c[i][j] != -1)(c[i][j] != -1) returnreturn c[i][j]c[i][j]

else ifelse if (x[i] == y[j]) {(x[i] == y[j]) {c[i][j] = lcsMemo(i-1,j-1) + 1c[i][j] = lcsMemo(i-1,j-1) + 1

returnreturn c[i][j]c[i][j]}}

elseelse {{

c[i][j]=max(lcsMemo(i-1,j),lcsMemo(i,j-1))c[i][j]=max(lcsMemo(i-1,j),lcsMemo(i,j-1)) returnreturn c[i][j]c[i][j]}}

}}

intint lcsMemo(lcsMemo(intint i,i, intint j) {j) { ifif (c[i][j] != -1)(c[i][j] != -1) returnreturn c[i][j]c[i][j]

else ifelse if (x[i] == y[j]) {(x[i] == y[j]) {c[i][j] = lcsMemo(i-1,j-1) + 1c[i][j] = lcsMemo(i-1,j-1) + 1

returnreturn c[i][j]c[i][j]}}

elseelse {{

c[i][j]=max(lcsMemo(i-1,j),lcsMemo(i,j-1))c[i][j]=max(lcsMemo(i-1,j),lcsMemo(i,j-1)) returnreturn c[i][j]c[i][j]}}

}}

T(n) = O(n2)c String x

String

y

LCS: Computing LengthLCS: Computing Length


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LCS: Computing LengthLCS: Computing Length

LCS-Length(X, Y, m, n)

1 for i1 to m do2 c[i,0] 0

3 for j0 to n do

4 c[0,j] 0

5 for i1 to m do

6 for j1 to n do

7 if xi= y

jthen

8 c[i,j] c[i-1,j-1]+1

9 b[i,j] copy

10 else if c[i-1,j] c[i,j-

1] then

11 c[i,j] c[i-1,j]

12 b[i,j] skipX

13 else

14 c[i,j] c[i,j-1]

15 b[i,j] skipY

16 return c, b

LCS-Length(X, Y, m, n)

1 for i1 to m do

2 c[i,0] 0

3 for j0 to n do

4 c[0,j] 0

5 for i1 to m do

6 for j1 to n do

7 if xi = yjthen8 c[i,j] c[i-1,j-1]+1

9 b[i,j] copy

10 else if c[i-1,j] c[i,j-

1] then

11 c[i,j] c[i-1,j]12 b[i,j] skipX

13 else

14 c[i,j] c[i,j-1]

15 b[i,j] skipY

16 return c, b

LCS: ExampleLCS: Example


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00 00 00 00 00 00 00 00

aa bb bb cc aa aa cc

aa

cc

cc

bb

cc

cc

aa

00

00

00

00

00

00

00

XX

YY

LCS: ExampleLCS: Example0, if i=0, j=00, if i=0, j=0

c[i,j] = c[i-1,j-1]+1 if i,j>0 and xc[i,j] = c[i-1,j-1]+1 if i,j>0 and xii = y= yjj

max ( c[i,j-1], c[i-1,j] ) if i,j>0 and xmax ( c[i,j-1], c[i-1,j] ) if i,j>0 and xii yyii



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00 00 00 00 00 00 00 00


aa

cc

cc

bb

cc

cc

aa

00

00

00

00

00

00

00






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00 00 00 00 00 00 00 00


aa

cc

cc

bb

cc

cc

aa

00

00

00

00

00

00

00






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0, if i=0, j=00, if i=0, j=0


max(c[i,j-1],c[i-1,j]) if i,j>0 and xmax(c[i,j-1],c[i-1,j]) if i,j>0 and xii yyii


00 00

11

11

11

00 00

11 11

11 11

11 11

11

11

11

11

22 22

22 22

22 22

22 22

00 00

11 11

22 22

22 22

00 00

11 11

22 22

22 33

22 22

33 33

33 33

33 44

22 33

33 33

33 44

44 44


aa

cc

cc

bb

cc

cc

aa

00

00

00

00

00

00

00


Constructing an LCSConstructing an LCS


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Constructing an LCSConstructing an LCSPrint-LCS(b, X, i, j)

1 if i = 0 or j = 0 then

2 return

3 if b[i,j] = copy" then

4 Print-LCS(b,X,i-1,j-1)

5 print x[i]

6 elseif b[i,j]=skipX" then

7 Print-LCS(b,X,i-1,j)

8 else

Print-LCS(b,X,i,j-1)

Print-LCS(b, X, i, j)

1 if i = 0 or j = 0 then

2 return3 if b[i,j] = copy" then

4 Print-LCS(b,X,i-1,j-1)

5 print x[i]

6 elseif b[i,j]=skipX" then7 Print-LCS(b,X,i-1,j)

8 else

Print-LCS(b,X,i,j-1)

length[X] = m, length[Y] = n,

Call Print-LCS(b, X, n, m) to construct LCS

Time complexity: O(m+n).



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0 0

11

1

0 0

1 11 1

1 1

1

1

1

1

2 2

2 2

2 2

2 2

0 0

1 12 2

2 2

0 0

1 12 2

2 3

2 2

3 3

3 3

3 4

2 3

3 3

3 4

4 4

a b b c a a c

ac

c

b

c

c

a

00

0

0

0

0

0



max(c[i,j-1],c[i-1,j]) if i,j>0 and xmax(c[i,j-1],c[i-1,j]) if i,j>0 and xii yyii

onges ommononges ommon


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onges ommononges ommonSubsequenceSubsequence

There is a need to quantify how similar theyThere is a need to quantify how similar they

are:are:Comparing DNA sequences in studies of evolutionof different species

Spell checkers

Editing

References & ReadingsReferences & Readings


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References & ReadingsReferences & Readings

CLRS:CLRS:

15.2, 15.3, 15.4

Exercises: 15.2-1, 15.2-2, 15.2-3, 15.3-2, 15.4-1,

15.4-3, 15.4-5, 15.4-6.

Problems: 15-3.

HSR:HSR:

5.6

dynamic programming - algorithm

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