Algorithm Design & AnalysisChapter

8/13/2019 Algorithm Design & AnalysisChapter

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Algorithm Design & Analysis

Chapter -04(Dynamic Programming & Greedy

Algorithms )T.E(Computer)

ByI.S.Borse

SSVPS BSD COE ,DHULE3/8/2009 1ADA Unit -4 I.S Borse


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Outline Chapter 4

1. Dynamic Programming ( Introduction)i) Multistage Graphii) Optimal Binary Search Tree(OBST)

iii) 0/1 Knapsack Problemiv) Travelling Salesman Problem2. Greedy Algorithms ( Introduction)i) Job Sequencingii) Optimal Merge Patterns

3/8/2009 2

ADA Unit -4 I.S Borse


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Dynamic Programming

Dynamic Programming is an algorithmdesign method that can be used when

the solution to a problem may beviewed as the result of a sequence ofdecisions

3/8/2009 3ADA Unit -4 I.S Borse


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The General MethodDynamic programming

An algorithm design method that can be used when the solutioncan be viewed as the result of a sequence of decisions

Some solvable by Greedy method under the conditionCondition : an optimal sequence of decisions can be found bymaking the decisions one at a time and never making anerroneous decision

For many other problems

Not possible to make stepwise decisions (based only on localinformation) in a manner like Greedy method

3/8/2009 ADA Unit -4 I.S Borse 4


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The General Method Enumeration vs. dynamic programming

Enumeration Enumerating all possible decision sequences and picking

out the best prohibitive time and storage requirements Dynamic programming

Drastically reducing the time and storage by avoiding somedecision sequences that cannot possibly be optimal

Making explicit appeal to the principle of optimality Definition [Principle of optimality] The principle of optimality

states that an optimal sequence of decisions has the property thatwhatever the initial state and decision are, the remainingdecisions must constitute an optimal decision sequence withregard to the state resulting from the first decision.



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Principle of optimality Principle of optimality: Suppose that in solving a

problem, we have to make a sequence ofdecisions D 1, D2, , Dn. If this sequence isoptimal, then the last k decisions, 1 k n mustbe optimal.

e.g. the shortest path problemIf i, i1, i2, , j is a shortest path from i to j, then i 1,i2, , j must be a shortest path from i 1 to j

In summary, if a problem can be described by amultistage graph, then it can be solved bydynamic programming.



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The General Method

Greedy method vs. dynamic programming Greedy method

Only one decision sequence is ever generated

Dynamic programming Many decision sequences may be generated But sequences containing suboptimal subsequences

discarded because they cannot be optimal due to theprinciple of optimality



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The General Method Notation and formulation for the principle

Notation and formulation S0: initial problem state n decisions d i, 1in have to be made to solve the

problem and D 1={r1,r2,,r j} is the set of possibledecision values for d1

Si is the problem state when r i chosen, and i is anoptimal sequence wrt S

i Then, when the principle of optimality holds, an

optimal sequence wrt S 0 is the best of the decisionsequences r i,i, 1in



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Forward approach and backward approach:

Note that if the recurrence relations are formulatedusing the forward approach then the relations aresolved backwards . i.e., beginning with the lastdecision

On the other hand if the relations are formulatedusing the backward approach , they are solvedforwards.

To solve a problem by using dynamic programming: Find out the recurrence relations. Represent the problem by a multistage graph.

Dynamic Programming



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Steps for Dynamic Programming1) The problem can be divided into stages with decision

required at each stage2) Each state has a number of states associated with it

3) The decision at one stage transform one stage to next

4) Given current state the optimal decision for each of theremaining state does not depend on the previous stateor decision

5) There exists a recursive relationship that identifies theoptimal solution for stage j , given the stage j+1 hasalready been solved

6) One final stage must be solvable by itself.



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The shortest path

To find a shortest path in a multi-stage graph

Apply the greedy method :

the shortest path from S to T :

1 + 2 + 5 = 8

S A B T

3

4

5

2 7

1

5 6



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The shortest path in multistage graphs

e.g.

The greedy method can not be applied to this case:

(S, A, D, T) 1+4+18 = 23. The real shortest path is:(S, C, F, T) 5+2+2 = 9.

S T132

B E

9

A D4

C F2

1

5

11

5

16

18

2



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Multistage Graphs Definition: multistage graph G(V,E)

A directed graph in which the vertices arepartitioned into k2 disjoint sets V i, 1ik

If E, then u Vi and v Vi+1 for some I,1i


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Multistage Graphs



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Multistage Graphs


DP formulation

Every s to t path is the result of a sequence of k-2 decisionsThe principle of optimality holds (Why?)p(i, j) = a minimum-cost path from vertex j in V i to vertex tcost(i, j) = cost of path p(i , j)

cost(k-1,j) = c(j,t) if E, otherwise

Then computing cost(k-2,j) for all j Vk-2Then computing cost(k-3,j) for all j Vk-3Finally computing cost(1,s)

)},1(cos),({min),(cos,

1

l it l jc jit E l j

V l i


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Multistage Graphs


(k=5)Stage 5

cost(5,12) = 0.0Stage 4

cost(4,9) = min {4+cost(5,12)} = 4

cost(4,10) = min {2+cost(5,12)} = 2cost(4,11) = min {5+cost(5,12)} = 5Stage 3

cost(3,6) = min {6+cost(4,9), 5+cost(4,10)} = 7cost(3,7) = min {4+cost(4,9), 3+cost(4,10)} = 5cost(3,8) = min {5+cost(4,10), 6+cost(4,11)} = 7


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Multistage Graphs

Stage 2 cost(2,2) = min {4+cost(3,6), 2+cost(3,7), 1+cost(3,8)} = 7 cost(2,3) = min {2+cost(3,6), 7+cost(3,7)} = 9 cost(2,4) = min {11+cost(3,8)} = 18

cost(2,5) = min {11+cost(3,7), 8+cost(3,8)} = 15 Stage 1

cost(1,1) = min {9+cost(2,2), 7+cost(2,3), 3+cost(2,4), 2+cost(2,5)} = 16 Important notes: avoiding the recomputation of cost(3,6), cost(3,7), and

cost(3,8) in computing cost(2,2)



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Multistage Graphs void Fgraph (graph G, int k, int n, int p[] ) // The input is a k-stage graph G = (V,E) with n vertices indexed in order

// of stages. E is a set of edges and c[i][j] is the cost of . // p[1 : k] is a minimum-cost path. { float cost[MAXSIZE]; int d[MAXSIZE], r; cost[n] = 0.0; for (int j=n-1; j >= 1; j--) { // Compute cost[j]. let r be a vertex such that is an edge of G and c[j][r] + cost[r] is minimum; cost[j] = c[j][r] + cost[r]; d[j] = r; } // Find a minimum-cost path. p[1] = 1; p[k] =n ; for ( j=2; j


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Multistage Graphs


)} j,l(c)l,1i(tcos b{min) j,i(tcos b

El, j

Vl 1i

Backward approach


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The shortest path in multistage graphs

e.g.

The greedy method can not be applied to this case:

(S, A, D, T) 1+4+18 = 23. The real shortest path is:(S, C, F, T) 5+2+2 = 9.

S T132

B E

9

A D4

C F2

1

5

11

5

16

18

2



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Dynamic programming approach

Dynamic programming approach

d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C, T)}

S T2

B

A

C

1

5 d(C, T)

d(B, T)

d(A, T)

A

T

4

E

D

11d(E, T)

d(D, T)d(A,T) = min{4+d(D,T), 11+d(E,T)}= min{4+18, 11+13} = 22.



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Dynamic programming

d(B, T) = min{9+d(D, T), 5+d(E, T), 16+d(F, T)}= min{9+18, 5+13, 16+2} = 18.

d(C, T) = min{ 2+d(F, T) } = 2+2 = 4

d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C, T)}= min{1+22, 2+18, 5+4} = 9.

The above way of reasoning is called

backward reasoning.

B T5

E

D

F

9

16 d(F, T)

d(E, T)

d ( D , T )



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Backward approach (forward reasoning)

d(S, A) = 1d(S, B) = 2

d(S, C) = 5

d(S,D)=min{d(S, A)+d(A, D),d(S, B)+d(B, D)}= min{ 1+4, 2+9 } = 5

d(S,E)=min{d(S, A)+d(A, E),d(S, B)+d(B, E)}

= min{ 1+11, 2+5 } = 7d(S,F)=min{d(S, A)+d(A, F),d(S, B)+d(B, F)}

= min{ 2+16, 5+2 } = 7



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Optimal binary search trees

n identifiers : a 1


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Identifiers : 4, 5, 8, 10, 11,12, 14

Internal node : successfulsearch, P i

External node : unsuccessful

search, Q i

10

14

E 7

5

11

12E 4

4

E 0 E 1

8

E 2 E 3

E 5 E 6

The expected cost of a binary tree:

The level of the root : 1

n

0nii

n

1nii 1))(level(EQ)level(aP



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The dynamic programming approach

Let C(i, j) denote the cost of an optimal binary searchtree containing a i, ,a j . The cost of the optimal binary search tree with a k as its

root :

ak

a1

. . .ak -1

ak + 1

. . .an

P 1 ...P k- 1Q 0 ...Q k- 1

P k+1 ...P nQ k ...Q n

C(1,k-1) C (k+1,n)

n1,k CQPQ1k 1,CQPQPminn)C(1,n

1k iiik

1k

1iii0k

nk 1



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j

immm1-i

jk i

j

1k mmmk

1k

immm1-ik jk i

QPQ j1,k C1k i,Cmin

j1,k CQPQ

1k i,CQPQPmin j)C(i,

General formula

ak

a1

. . .ak -1

ak + 1

. . .an

P1...P

k- 1

Q 0 ...Q k- 1

Pk+1

...Pn

Q k ...Q n

C(1,k-1) C(k+1,n)3/8/2009 30ADA Unit -4 I.S Borse


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0/1 knapsack problem

n objects , weight W 1, W2, ,Wnprofit P 1, P2, ,Pn capacity M

maximize subject to Mxi = 0 or 1, 1 i n e. g.

ni

ii x P

1 niii xW

1

i W i P i

1 10 402 3 203 5 30

M=10



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The multistage graph solution The 0/1 knapsack problem can be described

by a multistage graph.

S T

0

10

10

00

01

100

010

011

000

001

0

0

0

0

00

40

020

0

30

0

0

30

x1=1

x1=0

x2=0

x2=1

x2=0

x3=0

x3=1

x3=0

x3=1

x3=0



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The Dynamic Programming Approach

The longest path represents the optimalsolution:x1=0, x2=1, x3=1

= 20+30 = 50 Let f i(Q) be the value of an optimal solution to

objects 1,2,3, ,i with capacity Q.

f i(Q) = max{ f i-1(Q), f i-1(Q-W i)+Pi } The optimal solution is f n(M).

ii x P



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All Pairs Shortest Paths Problem definition

Determine a matrix A such that A( i, j) is the length of a shortestpath from i to j

One method using the algorithm ShortestPaths Each vertex requiring O(n 2) time, so total time is O(n 3)

Restriction: no negative edge allowed DP algorithm

The principle of optimality (Does it hold? Why?) (weaker) restriction: no cycle with negative length



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All Pairs Shortest Paths


1k )}, j,k (A)k ,i(A), j,i(Amin{) j,i(A 1k 1k 1k k

DP algorithm (Continued)

Ak(i , j) = the length of a shortest path from i to j goingthrough no vertex of index greater than k


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All Pairs Shortest Paths Algorithm

1. void AllPaths(float cost[][SIZE], float A[][SIZE], int n)2. // cost[1:n][1:n] is the cost adjacency matrix of 3. // a graph with n vertices; A[i][j] is the cost4. // of a shortest path from vertex i to vertex j.5. // cost[i][i] = 0.0, for 1


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The Traveling Salesperson Problem Problem

TSP is a permutation problem (not subset problem) Usually the permutation problem is harder than the subset one Because n! > 2n

Given a directed graph G(V,E)

cij= edge cost A tour is a directed simple cycle that includes every vertex in V The TSP is to find a tour of minimum cost

Many applications

1. Routing a postal van to pick up mail from mail boxes located at n differensites 2. Planning robot arm movements to tighten the nuts on n different positions 3. Planning production in which n different commodities are manufactured

on the same sets of machines3/8/2009 ADA Unit -4 I.S Borse 38


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The Traveling Salesperson Problem


}k}){1,Vg(k,c{min 1k nk 2

}{j})Sg(j,c{min ijS j

DP formulation Assumption: A tour starts and ends at vertex 1The principle of optimality holds (Why?)g(i, S) = length of a shortest path starting at vertex i, going through allvertices in S, and terminating at vertex 1

g(1,V-{1}) = -

g(i, S) =

Solving the recurrence relationg(i, ) = c i1, 1inThen obtain g(i, S) for all S of size 1

Then obtain g(i, S) for all S of size 2Then obtain g(i, S) for all S of size 3Finally obtain g(1,V-{1})


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For |S|=0

g(2, )=c21=5 g(3, )=c31=6 g(4, )=c41=8

For |S|=1 g(2,{3})=c 23+g(3, )=15

g(2,{4})=c 24+g(4, )=18 g(3,{2})=c 32+g(2, )=18 g(3,{4})=c 34+g(4, )=20 g(4,{2})=c 42+g(2, )=13 g(4,{3})=c 43+g(3, )=15


0988

12013610905

2012100110

3

2

412

5

208

9

12

6

8

139

10



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22

0

2)1(2

)1( n

n

k

nk

nn

For |S|=2g(2,{3,4}) = min {c 23 +g(3,{4}), c 24 +g(4,{3})} = 25g(3,{2,4}) = min {c 32 +g(2,{4}), c 34 +g(4,{2})} = 25g(4,{2,3}) = min {c 42 +g(2,{3}), c 43 +g(3,{2})} = 23

For |S|=3g(1,{2,3,4} = min {c 12 +g(2,{3,4}), c 13 +g(3,{2,4}),

c14 +g(4,{2,3})} =

= min {35, 40, 43} = 35Time complexity

Let N be the number of g(i, S) that have to be computed before canbe used to compute g(1,V-{1})

N =

Total time = O(n 22n)This is better than enumerating all n! different tours


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The traveling salesperson (TSP) problem

e.g. a directed graph :

Cost matrix:

12

3

2

44

2

56

7

104

8

39

1 2 3 4

1 2 10 5

2 2 9

3 4 3 4

4 6 8 7



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A multistage graph can describe all possible tours of a directedgraph.

Find the shortest path:(1, 4, 3, 2, 1) 5+7+3+2=17

(1) (1,3)

(1,2)

(1,4)

2

5

10

(1,2,3)

(1,2,4)

(1,3,2)

(1,3,4)

(1,4,2)

(1,4,3)

9

3

4

8

7

(1,2,3,4)

(1,2,4,3)

(1,3,2,4)

(1,3,4,2)

(1,4,2,3)

(1,4,3,2)

4

7

8

9

3

1

4

6

6

2

4

2

The multistage graph solution



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The representation of a node

Suppose that we have 6 vertices in the graph. We can combine {1, 2, 3, 4} and {1, 3, 2, 4} into one node.

(3),(4,5,6) means that the last vertex visited is 3 and theremaining vertices to be visited are (4, 5, 6).

combine

(2), (4,5,6)

(b)

(3), (4,5,6)

(4), (5,6)

(1,3,2)

(1,2,3) (1,2,3,4)

(1,3,2,4)

(a)



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Review: Dynamic Programming

Dynamic programming is another strategy fordesigning algorithms

Use when problem breaks down into recurringsmall subproblems



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Review: Dynamic Programming Summary of the basic idea:

Optimal substructure: optimal solution to problemconsists of optimal solutions to subproblems

Overlapping subproblems: few subproblems in total,

many recurring instances of each Solve bottom-up, building a table of solved

subproblems that are used to solve larger ones

Variations: Table could be 3 -dimensional, triangular, a tree, etc.



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Greedy Algorithms



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Overview Like dynamic programming, used to solve

optimization problems. Problems exhibit optimal substructure (like DP). Problems also exhibit the greedy-choice

property. When we have a choice to make, make the one that

looks best right now . Make a locally optimal choice in hope of getting a

globally optimal solution .



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Greedy Strategy The choice that seems best at the moment is the

one we go with. Prove that when there is a choice to make, one

of the optimal choices is the greedy choice.

Therefore, its always safe to make the greedychoice.

Show that all but one of the subproblems

resulting from the greedy choice are emp ty.



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Elements of Greedy Algorithms

Greedy-choice Property. A globally optimal solution can be arrived at by

making a locally optimal (greedy) choice.

Optimal Substructure.


G d Al i h


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Greedy Algorithms

A greedy algorithm always makes the choicethat looks best at the moment My everyday examples:

Walking to the Corner Playing a bridge hand

The hope: a locally optimal choice will lead to aglobally optimal solution

For some problems, it works Dynamic programming can be overkill; greedy

algorithms tend to be easier to code3/8/2009 52ADA Unit -4 I.S Borse


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Review: The Knapsack ProblemAnd Optimal Substructure

Both variations exhibit optimal substructure To show this for the 0-1 problem, consider the

most valuable load weighing at most W pounds If we remove item j from the load, what do we know

about the remaining load? A: remainder must be the most valuable load

weighing at most W - w j that thief could take frommuseum, excluding item j



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Solving The Knapsack Problem The optimal solution to the fractional knapsack

problem can be found with a greedy algorithm How?

The optimal solution to the 0-1 problem cannot

be found with the same greedy strategy Greedy strategy: take in order of dollars/pound Example: 3 items weighing 10, 20, and 30 pounds,

knapsack can hold 50 pounds Suppose item 2 is worth $100. Assign values to the other

items so that the greedy strategy will fail



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Greedy Choice Property Dynamic programming? Memorize? Yes, but Activity selection problem also exhibits the

greedy choice property: Locally optimal choice globally optimal soln Them 17.1: if S is an activity selection problem

sorted by finish time, then optimal solution A S such that {1} A

Sketch of proof: if optimal solution B that does notcontain {1}, can always replace first activity in B with {1}(Why? ). Same number of activities, thus optimal.



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Review: The Knapsack Problem

The famous knapsack problem : A thief breaks into a museum. Fabulous paintings,sculptures, and jewels are everywhere. The thief hasa good eye for the value of these objects, and knows

that each will fetch hundreds or thousands of dollarson the clandestine art collectors market. But, thethief has only brought a single knapsack to the sceneof the robbery, and can take away only what he cancarry. What items should the thief take to maximizethe haul?



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Algorithm Design & AnalysisChapter