05-DynamicProgramming

1

Recursive Back Tracking&

Dynamic Programming

Jeff Edmonds York University

COSC 3101Lecture 7

2

Optimization ProblemsA Sequence of DecisionsThe Little Bird & FriendOptimal SubstructureMemoizationSet of Sub-InstancesTracing Dyn. Prog. AlgReversingCodeSpeeding Up Running TimeMultiple Opt SolutionsReviewQuestion for Little BirdReview & Don'ts

Bellman FordBest PathPrinting NeatlyLongest Common SubsequenceKnapsack ProblemThe Event Scheduling ProblemParsingSatisfiability

Techniques Problems

3

• Consider your instance I.• Ask a little question (to the little bird) about its optimal solution.• Try all answers k.

• Knowing k about the solutionrestricts your instance to a subinstance subI.

• Ask your recursive friend for a optimal solution subsol for it. • Construct a solution optS<I,k> = subsol + k

for your instance that is the best of those consistent with the kth bird' s answer.

• Return the best of these best solutions.

Recursive Back TrackingBellman Ford

4

Specification: All Nodes Shortest-Weighted Paths • <preCond>: The input is a graph G (directed or undirected)

with edge weights (possibly negative)• <postCond>: For each u,v, find a shortest path from u to v

Stored in a matrix Dist[u,v].

b

d

cu

k

g i

v

h

40

1 10

2

15

181

2

6

8

1

230325

12

3

For a recursive algorithm, we must give our friend a smaller subinstance.How can this instance be made smaller?Remove a node? and edge?


with ≤l edges

and integer l.

with at most l edge.

l=3

l=4

5

b

d

cu

k

g i

v

h

40

1 10

2

15

181

2

6

8

1

230325

12

3


l=4

• Consider your instance I = u,v,l.• Ask a little question (to the little bird) about its optimal solution.

• “What node is in the middle of the path?”• She answers node k.• I ask one friend subI = u,k, l/2

and another subI = k,v, l/2• optS<I,k> = subsolu,k,l/2

+ k + subsolk,v,l/2 is the best solution for I consistent with the kth bird‘s answer.

• Try all k and return the best of these best solutions.

6

Dynamic Programming Algorithm• Given an instance I,

• Imagine running the recursive alg on it.• Determine the complete set of subI

ever given to you, your friends, their friends, …• Build a table indexed by these subI• Fill in the table in order so that nobody waits.

Recursive Back Tracking

Given graph G, find Dist[uv,l] for l =1,2,4,8,…

b

d

cu

k

g i

v

h

40

1 10

2

15

181

2

6

8

1

230325

12

3

l=4

,n

7

b

d

cu

k

g i

v

h

40

1 10

2

15

181

2

6

8

1

230325

12

3

l=4

Dynamic Programming AlgorithmLoop Invariant: For each u,v,

Dist[u,v,l] = a shortest path from u to v with ≤l edges

Exit

for l = 2,4,8,16,…,2n % Find Dist[uv,l] from Dist[u,v,l/2] for all u,v Verticies Dist[u,v,l] = Dist[u,v,l/2] for all k Verticies Dist[u,v,l] = min( Dist[u,v,l], Dist[u,k,l/2]+Dist[k,v,l/2] )

8

b

d

cu

k

g i

v

h

40

1 10

2

15

181

2

6

8

1

230325

12

3

l=4



When l = 1, Dist[b,c,1] = 10Dist[u,v,1] = ∞

% Smallest Instancesfor all u,v Verticies if u,v Edges Dist[u,v,1] = weight[u,v] else Dist[u,v,1] = ∞

Dist[u,u,1] = 0 (sometimes useful)

9

b

d

cu

k

g i

v

h

40

1 10

2

15

181

2

6

8

1

230325

12

3

l=4



Exit

When to exit? A simple path never uses a node more than once and so does not have more than l=n-1.

for all u,v Verticies Dist[u,v] = Dist[u,v,n]

10

Dynamic Programming Algorithm• Dealing with negative cycles.

b

d

cu

v

-5

1

2253

Dist[u,v,4] = ∞

Dist[u,v,8] = 25+3+2 = 30

Dist[u,v,9] = 25+(3+1-5)+3+2 = 29

Dist[u,v,2] = ∞

Dist[u,v,12] = 25+(3+1-5)2+3+2 = 28Dist[u,v,303] = 25+(3+1-5)300+3+2 = 25-300 = -275Dist[u,v,∞] = 25+(3+1-5)∞+3+2 = 25-300 = -∞

• There is a negative cycle ifDist[u,v,n] > Dist[u,v,2n]

% Check for negative cycles for all u,v Verticies if( Dist[u,v,2n]<Dist[u,v,n] ) Dist[u,v] = ∞

11

Dynamic Programming AlgorithmAlgorithm BellmanFord(G)% Smallest Instancesfor all u,v Verticies if u,v Edges Dist[u,v,1] = weight[u,v] else Dist[u,v,1] = ∞ for l = 2,4,8,16,…,2n % Find Dist[uv,l] from Dist[u,v,l/2] for all u,v Verticies Dist[u,v,l] = Dist[u,v,l/2] for all k Verticies Dist[u,v,l] = min( Dist[u,v,l], Dist[u,k,l/2]+Dist[k,v,l/2] )% Check for negative cycles for all u,v Verticies if( Dist[u,v,2n]==Dist[u,v,n] ) Dist[u,v] = Dist[u,v,n] else Dist[u,v] = ∞

Time = O(n3 logn)

• Don’t actually need to keep old and new values.

12

Dynamic Programming AlgorithmAlgorithm BellmanFord(G)% Smallest Instancesfor all u,v Verticies if u,v Edges Dist[u,v] = weight[u,v] else Dist[u,v] = ∞ for l = 2,4,8,16,…,2n % Find Dist[uv,l] from Dist[u,v,l/2] for all u,v Verticies for all k Verticies Dist[u,v] = min( Dist[u,v], Dist[u,k]+Dist[k,v] )% Check for negative cycles for all u,v Verticies if( changed last iteration ) Dist[u,v] = ∞

• Don’t actually need to keep old and new values.

13

Dynamic Programming

• A hard topic.• I try to provide a unified way to think of it

and a fixed set of steps to follow.• Even if you don’t get the details of the

algorithm correct, at least get the right structure. • I provide analogies (little bird) to make it

hopefully more fun & easier to follow.

14

Optimization Problems• An important and practical class of computational

problems.

• For most of these, the best known algorithm runs in exponential time.

• Industry would pay dearly to have faster algorithms.

• Heuristics

• Some have quick Greedy or Dynamic Programming algorithms

• For the rest, Recursive Back Tracking is the best option.

15

Ingredients:• Instances: The possible inputs to the problem. • Solutions for Instance: Each instance has an

exponentially large set of solutions. • Cost of Solution: Each solution has an easy to

compute cost or value.

Optimization Problems

16

Specification of an Optimization Problem• <preCond>: The input is one instance.• <postCond>:

The output is one of the valid solutions for this instance with optimal cost. (minimum or maximum)

• The solution might not be unique.• Be clear about these ingredients!

Optimization Problems

17

Search Graph For Best Path

We use it because it nicely demonstrates the concepts in a graphical way.

18

Search Graph For Best PathAn instance (input) consists of <G,s,t>.

G is a weighted directed layered graphs source nodet sink node

5

3

4

2 53

3 2

212

4

19

Search Graph For Best PathAn instance (input) consists of <G,s,t>.

The cost of a solution is the sum of the weights.

2+6+3+7=18

A solution for an instance is a path from s to t.

The goal is to find a path with minimum total cost.

4+2+1+5=12

20

Brute Force Algorithm

But there may be an exponential number of paths!

Try all paths, return the best.

21

An Algorithm As A Sequence of Decisions

“Which edge should we take first?”

Some how I decide <s,v3>.

I ask a question about the solution.

“Which edge do we take second?”

Some how he decides <v3,v5>.

My friend asks the next question.

“Which edge do we take third?”

Some how he decided <v5,v8>.

His friend asks the next question.

22


“Which edge should we take first?”I ask a question about the solution.

How do I decide?

The greedy algorithm?

Does not work!

Taking the best first edge.

23

Local vs Global Considerations

• We are able to make local observations and choices.–Eg. Which edge out of s is cheapest?

• But it is hard to see the global consequences –Which path is the overall cheapest?

• Sometimes a local initial sacrifice can globally lead to a better overall solution.

24


“Which edge should we take first?”I ask a question about the solution.

How do I decide?

• But let's skip this partby pretending that we have

a little bird to answer this little question.

• In reality we will try all possible first edges.

25

"Little Bird" Abstraction

Recall: Non-deterministic Finite Automata Non-deterministic Turing Machine

0

The little bird is a little higher power, answering a little question about an optimal solution.

These have a higher power to tell them which way to go.

(It is up to you whether or not to use it)

26

“Which edge should we take first?”

The bird answers <s,v1>.

I ask a question about the solution.

“Which edge do we take second?”

The bird answers <v1,v4>.

My friend asks the next question.

But we don’t want toworry about how our friend

solves his problem.

Little Bird & Friend Alg

27

Sub-Instance for Friend

Our instance is <G,s,t>: Find best path from s to t.Our friend is recursion• i.e. he is a smaller version of ourselves• we can trust him to give us a correct answer • as long as we give him

• a smaller instance• of the same problem.

• What sub-instance do we give him?

28

“Which is the best path from v1 to t?”

I tack on the bird’s edge making the path <s,v1,v6,t>

Friend answers <v1,v6,t> with weight 10.

If I trust the little bird, I take step along edge <s,v1> and ask my friend,

The bird answers <s,v1>.

To get my solution

with weight 10+3=13.

Little Bird & Friend Alg

29

Faulty BirdBut what if we do not have a bird that we trust?

• i.e. the best path from s to tfrom amongst those starting with <s,v1>.

This work is not wasted, because we have found

Define optS<I,k> to be: the optimum solution for instance I consistent with the kth bird' s answer.

• the best solution to our instance from amongst those consistent with this bird' s answer.

30

Faulty BirdBut what if we do not have a bird that we trust?

This work is not wasted, because we have found

In reality we will try all possible first edges, giving …..

• the best solution to our instance from amongst those consistent with this bird' s answer.

• i.e. the best path from s to tfrom amongst those starting with <s,v1>.

31

…the best path from amongst those starting with <s,v1>.

Faulty Bird

32

… and the best path from amongst those starting with <s,v2>.

Faulty Bird

33


Faulty Bird

34


Faulty Bird

35

At least one of these four paths must be an over all best path.

I give the best of the best as the best path.

Faulty Bird

36

Bird/Friend - Best of the Best

A sequence of question to a little bird about a solutionforms a tree of possible answers.

Consider our instance I.

Consider the set of solutions

37


But we only care aboutthe first bird answer.



The answers classifiesthe possible solutions.

Solutions consistent with the kth bird' s answer.

k

k

38




Solutions consistent with the kth bird' s answer.

k

Define optS<I,k> to be: the optimum solution for instance I consistent with the kth bird' s answer.Do this for each k.

optS<I,k>

39




Let kmax be the bird' s answergiving the best optS<I,k>.

Define optS<I,k> to be: the optimum solution for instance I consistent with the kth bird' s answer.Do this for each k.

k

optS<I,k>

kmax

optS[I] = optS<I,k > = Bestk optS<I,k > max

optS[I]

40

Bird/Friend - Best of the Best Constructing optS<I,k> : the optimum solution for instance I consistent with the kth bird' s answer.

Given my instance I.

I ask my little bird foran answer k. I ask my friend for his solution.

I combine them.

41

Recursive backtracking

code always has thissame basic structure.

42

Be clear what are• the instances• it’s solution• the cost of a sol.

43

Loop through the bird answers.

Be clear which is the current one

being tried.

44

Give the bird & friend algorithm

as a comment.(Unless it is in

an earlier question.)

45

What is the bird asked?

What does she answer?

46

Get help from friendBe clear what sub-instance

you give him.

Store the solution& cost

he gives you.

47

How do you formyour solution from

the friend’s and fromthe bird’s?

48

How do you formyour cost from


49

Take the bestof the best

optSolk is a best solutionfor our instance from amongst

those consistent with the bird's

kth answer.

50

Return the solutionand cost for the original instance.

51

Base Cases:Instances that are too small to have smaller

instances to give to friends.

What are these?What are their

solutionsand costs?

52

i.e. for a path from s to t to be optimal,the sub-path from vi to t must optimal.

In order to be able to design a recursive backtracking algorithm for a computational problem,

the problem needs to have a recursive structure,

If shorter from vi to t. shorter to s to t.

Optimal Substructure

53

i.e. for a path from s to t to be optimal,the sub-path from vi to t must optimal.

In order to be able to design a recursive backtracking algorithm for a computational problem,

the problem needs to have an optimal substructure,

And finding such a sub-path is a sub-instance of the same computational problem.


54

Optimal Substructure• Optimal substructure means that

– Every optimal solution to a problem contains...– ...optimal solutions to subproblems

• Optimal substructure does not mean that– If you have optimal solutions to all subproblems...– ...then you can combine any of them to get an optimal

solution to a larger problem.• Example: In Canadian coinage,

– The optimal solution to 7¢ is 5¢ + 1¢ + 1¢, and– The optimal solution to 6¢ is 5¢ + 1¢, but– The optimal solution to 13¢ is not 5¢ + 1¢ + 1¢ + 5¢ + 1¢

• But there is some way of dividing up 13¢ into subsets with optimal solutions (say, 11¢ + 2¢) that will give an optimal solution for 13¢– Hence, the making change problem exhibits optimal

substructure.

55

Don’t all problems have this optimal substructure property?


56

Longest simple path

• Consider the following graph:

• The longest simple path (path not containing a cycle) from A to D is A B C D

• However, the subpath A B is not the longest simple path from A to B (A C B is longer)

• The principle of optimality is not satisfied for this problem

• Hence, the longest simple path problem cannot be solved by a dynamic programming approach

A C D

B

42

31

1


NP-Complete

57

Time?

I try each edge out of s.

Same as the brute force algorithm that tries each path.

A friend tries each edge out of these. A friend tries each edge out of these.

Same as Brute Force Algorithm

58


But there may be an exponential number of paths!

59

Why do all this work with birds & friends?• How else would you iterate through all paths? • But sometimes we can exploit the structure

to speed up the algorithm.

Speeding Up the Time

60

• Perhaps because these solutions are not valid or not highly valued.

Sometimes entire an branch can be pruned off.

• Or because there is at least one optimal solution elsewhere in the tree.• A Greedy algorithm prunes off all branches

except the one that looks best.


61

• Remembers the solutions for the sub-instances so that if ever it needs to be solved again, the answer can be used.

• This effectively prunes off this later branch of the classification tree.


Memoization

62

Exponential TimeRedoing Work


How many friends solve this sub-instance?

63


“Which is the best path from s to t?”

64



65



66



There’s one.

67


“Which is the best path from s to t?”

68



69



70



There’s another.

71



How many friends solve this sub-instance?

Once for each path to v7

Save time by only doing once.

Waste time redoing work

72

Drop bread crumbs

and don’t revisit.

Depth First Search

But we need shortest path

73

Having many friends solving this same sub-instanceis a waste of time.

We allocate one friend to the job.

Dynamic Programming

74

It is my job to learn and remember

the optSol to my sub-Instancei.e. the best path from v7 to t

Dynamic Programming

75

When I need to find the best path from v4 to t

I will ask you forthe best path from v7 to t

I will find my best pathand tell you.

Dynamic Programming

76



I remember my best pathand will tell you.

Dynamic Programming

77



I remember my best pathand will tell you.

Dynamic Programming

78

But I hate to wait for you.Recursion has a lot of overhead

Why don’t you go first?

I will find my best pathand tell you.

Dynamic ProgrammingWhen I need to find

the best path from v2 to tI will ask you for

the best path from v7 to t

Avoid waiting.

79

Dynamic Programming

Before anyone asks me,I will find my best path

and remember it.

80

Set of Sub-InstancesBut what sub-instance need to be solvedand in which order?

Imagine running the recursive algorithm on it.

Determine the complete set of sub-Instances ever given to you, your friends, their friends, …

Given an instance I,

81

Guess the complete set S of sub-Instances.

“Best path from v7 to t?” Yes“Best path from v21 to t?” No

v21

v21 is not a part of ouroriginal instance.

Set of Sub-Instances

82


“Best path from v7 to t?” Yes

“Best path from v3 to v7?” No

“Best path from v21 to t?” No

All paths considered end in t.


83





All paths considered end in t.


84





“Best path from vi to t?” i


Yes

85

Guess the complete set S of sub-Instances is



Assign one friend to each sub-Instance.

86




The set S of sub-Instances needs to:• include our given I

87




• closed under “friend” operation

Integers closed under addition x,y I x+y I

sub-Instance S

subsub-Instance S


88






• each sub-Instance needs to beasked of some friend, friend, …

89







A fine set of sub-instances!

90

The complete set S of sub-Instances is


Order to complete

• in an order such that no friend must wait.

• from “smallest” to “largest”

In what order should they go?

For this problem, the order relies on

the graph being “leveled.”

91






Order to complete

First

Last

Base Case easy

Instance to be solved.

92

Dynamic Programming"Which is the best path from t to t?"

Base Caseeasy

93

Dynamic Programming"Which is the best path from v8 to t?"

easy

94


easy

95


easy

96


Harder

97


Friend gives bestpath <v7,t>.

Little bird suggests first edge <v5,v7>

98




99


Take best of best

100


101




102


Friend gives bestpath <t,t>.

Little bird suggests first edge <v4,t>

103




104


Take best of best

105


106




107




108


Take best of best

109


110




111




112


Take best of best

113


114




115




116




117


Take best of best

118

Dynamic Programming"Which is the best path from s to t?"

Original Problem

119



Little bird suggests first edge <s,v1>

120




121




122




123


Take best of best

DONE

124

Construct a table • for storing an optimal solution & cost • for each sub-instance.

Dynamic Programming

Sub-InstancesMap

Indexes

Cell of table


i ϵ [n], i.e. for each node vi

t, v8, v7, v6, v5, …., s i

“Which is the best path from vi to t?”

125

Fill out a table containing an optimal solution for each sub-instance.


Dynamic Programming

t, v8, v7, v6, v5, …., s Base case Original

126

Communication

optSubSol = optSol[k]

Friend k gives friend i a best path from vk to t.

Recursive BackTracking<optSubSol,optSubCost> = LeveledGraph(<G,vk,t>)

optSol[k] = optSolmin

return(optSolmin,optCostmin) ,

Dynamic Programmingk

i k

i

k

i

optSolk = <vi,vk> + optSol[k]

127

Dynamic Programming


128

Be clear what are• the instances• it’s solution• the cost of

a solution.

129

Dynamic Programsdo not recurse making the instance smaller

and smaller.

Instead, it up frontdetermines the set Sof all sub-instances

that ever need to be solved.

Be clear what sub-instances are.

130


How are they indexed?

Tables indexed by these sub-instances

store an optimal solution and it’s cost.

131

The set Sof sub-instancesare solved from

smallest to largestso that no body waits.



They get solved firstand their solutions

stored.

132

Then we iterate through the remaining

sub-instances.

From smallest to largest.

Each gets solvedand their solutions

stored.

133

Consider yourselfto be a friend

working on oneof these.

Be clear which sub-instance is yours.

Solve this as you did before.

134



being tried.

135




136



137

Be clear what sub-instance

you give your friend.

kkk k

Get help from friend

138

Instead of recursing,we simply look

in the table for the solution.

Because his instance is smaller, he has

already solved it and stored sol in the table.

Get help from friendk

kk k

139



140



141


optSol<i,k> is a best solution for our instance subI[i]

from amongstthose consistent with the bird's

kth answer.

142

Store the solution to our instance subI[i]

in the table.

143



Is this code correct?

144


and smaller.Hence, lets not worry about our instance I

being a base case.

145

But there is a tableof subinstances

that must be solved.

Some of these will bebase cases

and their solutionsmust be stored

in the table.

146





in the table.

nnn n

t=

147



Then we solvethe rest.

148


000 n

s=

149

Order Feels Backwards

Path from t to t.

Path from s to t.

150

An Esthetically Better

Path from s to s.

Path from s to t.

151

Reversing

152

Reversing

Determine the complete set of sub-instances

“Which is the best path from s to vi?” i

153

Reversing


s, v1, v2, v3, v4, …., t

“Which is the best path from s to vi?”

Base case Original

154

155

Running Time

# of Sub-Instances× # of Bird Answers× (1)q

Time =

?= n × d

156

Communication TimeoptSolk = <optSol[k],vi>

Friend k gives best path from s to vk

to friend i, who adds the edge <vk,vi>.

k

i

(1)qTime = ?Size of path =

(q n)Time =

(q n).

157

# of Sub-Instances× # of Bird Answers× size of solution= q(n × d × n)

Time =

Running Time

Store path costs, not paths

Space =# of Sub-Instances× (1)q= (q n)

158

"What is cost of the best path from s to v7?"

Store Path Costs, not Paths

159

Friend gives cost 8of best path <s,v4>.

Little bird suggests last edge <v4,v7> with weight 2.8

Best cost via <v4,v7> is 8+2=10.



160


Little bird suggests last edge <v2,v7> with weight 7.

2




161


Little bird suggests last edge <v5,v7> with weight 5.


6



162

Take best of best:



We also learn the wise little bird’s advice.We will store this in the table too.

9

2 10, 9, 11

163

Running Time

birdsAdvice[i] = kmin

164

Leave these linesas commentsfor extra clarity for the reader

165

Find Optimal Path

Previous algorithm gives:• Cost of the best path

from s to vi, i.• Bird’s advice of

last edge to vi.

We run the bird-friend algorithm again, but with a reliable bird.

166

Find Optimal Path

The bird gives that the last edgeof the best path from s to t is <v8,t>.

167

Find Optimal Path

The bird gives that the last edgeof the best path from s to v8 is <v5,v8>.

168

Find Optimal Path

The bird gives that the last edgeof the best path from s to v5 is <v3,v5>.

169

Find Optimal Path

The bird gives that the last edgeof the best path from s to v3 is <s,v3>.

170

Find Optimal Path

Done!

171

Find Optimal Path

This could be done iteratively.As an exercise, design it.

172

Multiple Optimal Solutions

6

“Which is the last edge?”

I ask the bird:

She could give either answer.

By giving this edge she says “There exists an optimal solution consistent with this answer.”

Similar to greedy proof.

173

Multiple Optimal Solutions

6

“Which is the last edge?”

I ask the bird:

We try all the bird answers.

When we try this bird answer,

we find this best solution.When we try this bird answer,

we find this best solution.

When we take best of best, we choose between them.

174

Designing Recursive Back Tracking Algorithm• What are instances, solutions, and costs? • Given an instance I,• What question do you ask the little bird?• Given a bird answer k [K],

• What instance sub-Instance do your give your friend?• Assume he gives you optSubSol for subI.• How do you produce an optSol for I from

• the bird’s k and • the friend’s optSubSol?

• How do you determine the cost of optSol from • the bird’s k and • the cost of the friend’s optSubSol?

• Try all bird’s answers and take best of best.

Review

175

Recursive Back Tracking Algorithm

Dynamic Programming Algorithm• Given an instance I,• Imagine running the recursive alg on it.• Determine the complete set of sub-Instances

ever given to you, your friends, their friends, …• Build a table indexed by these sub-Instances• Fill in the table in order so that nobody waits.

• the cost of its optimal solution• advice given by the bird

• Run the recursive alg with bird’s advice to find the solution to your instance.

Review

176

Purpose of Little Bird: • An abstraction from which it is

easier to focus on the difficult issues. • Her answers give us a list of things to try.• Temporarily trusting the bird,

helps us focus on the remaining questionhelping us formulate sub-instance for friend.

• Coming up with which question is one of the main creative steps.• Hint: Ask about a local property • There are only so many question that you

might ask so just try them all.

The Question For the Little Bird

177

An instance: Graph, s, and t


I ask the bird:

A solution: a path

s

t

“What is the first edge in the path?”

The Dynamic Programming reverses the recursive backtracking algorithm. Hence, to end up with a “forward order”,we first reverse the recursive backtracking algorithm.

178



I ask the bird:

A solution: a path

s

t

The Dynamic Programming reverses the recursive backtracking algorithm. Hence, to end up with a “forward order”,we first reverse the recursive backtracking algorithm.

“What is the last edge in the path?”

179

A good question for the bird leaves you with a good recursive sub-instance to ask your friend.



I ask the bird:

A solution: a path

s

t


“What is the rest of the path?”

180



I ask the bird:

A solution: a path

s

t

Giving a good follow up question for your friend to ask the bird.


“What is the second last edge in the path?”

181

• You can only ask the bird a little question.– Together with your question, you provide the little

bird with a list A1, A2, …, AK of possible answers. – The little bird answers, k [1..K]. – For an efficient algorithm, K must be small.


number of edges into node t.

K

– K =

• Eg. “What is best last edge?”

s

t

182

• You can only ask the bird a little question.– Together with your question, you provide the little

bird with a list A1, A2, …, AK of possible answers. – The little bird answers, k [1..K]. – For an efficient algorithm, K must be small.


Trying all is the Brute Force algorithm.

– K =

• Eg. “What is an optimal solution?”# of solutions.

183



I ask the bird:

A solution: a path

s

t

“How many edges are in the path?”Bad Question: • it is not a local property• How does this help us solve the problem?• What is a good follow up question for the friend

to ask?

184

A solution: a sequence of objectsZ = a b c d

An instance: ???


“What is the last object in the sequence?”

I ask the bird:

# of answers K = # of possible last objects.

I ask my friend:

“What is the rest of the solution?”

185


“What is the last object in the sequence?”

I ask the bird:

An instance: a sequence of objects

X = a s b e f c h d a

A solution: a subset of these objectsZ = a b c d


# of answers K = # of possible last objects.

Is there a smaller question that we could ask?

186


“Is the last object of the instance included in the optimal solution?”

I ask the bird:

An instance: a sequence of objects


A solution: a subset of these objectsZ = a b c d

# of answers K = 2, Yes or No

187

An instance: ???


“What object is at the root?”

I ask the bird:

A solution: a binary tree of objects38

25

17

4 21

31

28 35

51

42

40 49

63

55 71

I ask my friend:

“What is the left sub-tree?”

I ask a second friend:

“What is the right sub-tree?”

Previous problems had one friend given a bird ans.

188

A solution: a binary tree of objects

An instance: ???


“What object is at a leaf?”

I ask the bird:

38

25

17

4 21

31

28 35

51

42

40 49

63

55 71

Bad Question: • How does this help us solve the problem? • What is a good follow up question for the friend

to ask?

189

Printing Neatly

190

Printing NeatlyAn instance: text to print neatly & # chars per line

“Love life man while there as we be”, 11

The goal is to to print it as “neatly” as possible.

The cost: a measure of how neat,

Love.life.. man.while.. there......as.we.be...

11

A solution: # of words to put on each line.

few blanks on the end of each line.

2263

3 = 83 = 83 = 2163 = 27 259

small punishment

big punishment

191

Brute Force Algorithm

But there may be an exponential number of ways to!

Try all ways to print, return the best.

love.life.. man........ love.......life.man...love.......life.man...love.life.. man........

192

Bird & Friend Algorithm

“How many words on the last line?”I ask the bird:

She may answer 3.

An instance:“Love life man while there as we be”, 11

“Which is the best way to print the remaining n-3 words?”

I ask the friend: I combine bird’s andfriend’s answers.

193

Even if the bird was wrong, this work is not wasted.

This is best way to print from amongst those ending in 3 words.

Bird & Friend AlgorithmAn instance:“Love life man while there as we be”, 11

We try the bird answers words,12345 and take best of best.

194

Time?

I try each # words on last line.

Same as the brute force algorithm that tries each path.

A friend tries # on next.

A friend tries # on next.


195

Memoization

Assign one friend to each sub-instance.

“Which is the best path from vi to t?” i

196



• Imagine running the recursive algorithm on it.

• Determine the complete set of sub-Instances ever given to you, your friends, their friends, …

Determine the complete set of sub-Instances.


197

Set of Sub-InstancesGuess the complete set of sub-Instances.

“Love life man while there”, 11 Yes

“Hi there”, 81 No

“man while”, 11 No

This may appear on a line,but it will never be a sub-Instance for a friend.


198



The set of sub-Instances is the set of prefixes.


sub-Instance S


“Love life man while there”, 11“Love life man while”, 11“Love life man”, 11“Love life”, 11“Love”, 11“”, 11

“Love life man while there as”, 11“Love life man while there as we”, 11

199




sub-Instance S subsub-Instance i


The bird answers words,12345




200









201


“Love life man while there as we be”, 11“Love life man while there as we”, 11“Love life man while there as”, 11“Love life man while there”, 11




The bird answers 1.

“Love life man while”, 11

A fine set of sub-instances!


202



Base Case easyInstance to be solved.




FirstLast




203

Construct a table • for storing the cost of opt sol and bird’s advice. • for each sub-instance.

Sub-InstancesMap

Indexes

Cell of table

i ϵ [n], i.e. for each word.

i

“Which is the best printing of first i words?”

The set of prefixes of words.

The Table

204

Dynamic Programming


“Which is the best printing of first i words?”

Base case Original

205


“Love life man while there”, 11The 5th sub-instance is

5 wordswith 4, 4, 3, 5, 5 letters.

206


“Love life man while there”, 11The 5th sub-instance is

Love.life.. man.while.. there......

Its solution iswith 2,2,1 words on each line.

The bird’s advice is 1 word on last.Solution’s cost is 23 + 23 +63 = 232

207


Assume the table is filled in so far.We will work to fill in the last line

208


be.........

Love.life.. man.while.. there.as.we

209


we.be......

Love.life.. man.while.. there.as...

210


as.we.be...

Love.life.. man.while.. there......

211


there.as.we.be

212


Choose best of the best.

Tried all bird answers.

213

Choose best of the best.


214

Dynamic Programming


Amusingly,when formatting

this code, I had to fight with line breaks to get the height/width ratio

Printing Neatly.

215

Be clear what are• the instances• it’s solution• the cost of

a solution.

216


and smaller.

Instead, it up frontdetermines the set Sof all sub-instances

that ever need to be solved.


217


How are they indexed?

Tables indexed by these sub-instances

store an optimal solution and it’s cost.

218

The set Sof sub-instancesare solved from

smallest to largestso that no body waits.



They get solved firstand their solutions

stored.

219

Then we iterate through the remaining

sub-instances.

From smallest to largest.

Each gets solvedand their solutions

stored.

Actually, we store thebird’s advice instead

of the solution.

220

Consider yourselfto be a friend

working on oneof these.

Be clear which sub-instance is yours.

Solve this as you did before.

221



being tried.

222




223



224

Be clear what sub-instance

you give your friend.

i-ki-k

i-k

Get help from friend

225

Instead of recursing,we simply look

in the table for the solution.

Because his instance is smaller, he has

already solved it and stored sol in the table.

i-ki-k

i-k

226



227



228


optSol<i,k> is a best solution for our instance subI[i]

from amongstthose consistent with the bird's

kth answer.

229

Store the solution toour instance subI[i]

in the table.

Actually, we store thebird’s advice instead

of the solution.

230



Is this code correct?

231


and smaller.Hence, lets not worry about our instance I

being a base case.

232





in the table.

233





in the table.

234



Then we solvethe rest.

235

But actually,we don’t have the solution.

We must rerun it, this time with advice

from the bird.


236

Time =# of Sub-Instances

× # of Bird Answers= q(n × n)

Space =# of Sub-Instances

= (q n)

237

Find Optimal Path

Previous algorithm gives cost and bird’s advice.

We run the bird-friend algorithm again, but with a reliable bird.

238


<2Love.life.. ,1

there……

,2

man.while..

,3>

as we be…

239

Longest Common Subsequence problem

X = a s b e f c h d aY = r t w a b g j c k t f d

240

The goal is to find a longest common subsequence.

The cost: The length of Z.

A solution: A common subsequence.Z = a b c d

Longest Common Subsequence problemAn instance: Two strings

X = a s b e f c h d aY = r t w a b g j c k t f dX = a s b e t c h d aY = r t w a b g j c k t f d

241


I ask the bird:

An instance:X = a s b e t c h d aY = r t w a b g j c k t f d

“Is the last character of either X or Y included in Z?”

She answers one of :• Last of X is not included• Last of Y is not included • Last of X is included• Last of Y is included• Neither are included• Both are included

242


I ask the bird:

I ask my friend:



She answers:• Last of X is not included

The instance:X = a s b e t c h d Y = r t w a b g j c k t f d

243


I ask the bird:

My friend answers:

I combine bird’s andfriend’s answersand give



She answers:• Last of X is not included


Z = a b c d X = a s b e t c h d Y = r t w a b g j c k t f d

the same Z.

244


I ask the bird:

I ask my friend:



She answers:• Last of Y is not included

The instance:X = a s b e t c h d a Y = r t w a b g j c k t f

245


I ask the bird:

My friend answers:




She answers:• Last of Y is not included

The instance:X = a s b e t c h d a Y = r t w a b g j c k t f

Z = a b c X = a s b e t c h d a Y = r t w a b g j c k t f

the same Z.

Not as good as lastbut we need to try.

246


I ask the bird:

An instance:X = a s b e t c h dY = r t w a b g j c k d f d


She answers:• Last of X and last of Y are both included

I ask my friend:The instance:

X = a s b e t c hY = r t w a b g j c k d f

Last chars equal

247


I ask the bird:

An instance:X = a s b e t c h dY = r t w a b g j c k d f d



The instance:X = a s b e t c hY = r t w a b g j c k d f


Zd = abcd.

Last chars equal

Z = a b c X = a s b e t c h Y = r t w a b g j c k d f

My friend answers:

248


I ask the bird:

I politely tell her that she is wrong.




Last chars not equal

249


I ask the bird:

I ask my friend:



She answers:• Last of X is included


250


I ask the bird:





Za = abcda.


Z = a b c d X = a s b e t c h d Y = r t w a b g j c k t f d

My friend answers:

Wrong

251


I ask the bird:

I ask my friend:




The instance:X = a s b e t c h d Y = r t w

252


I ask the bird:

My friend answers:I combine bird’s andfriend’s answersand give




The instance:X = a s b e t c h d Y = r t w

Z = t X = a s b e t c h d Y = r t w Za = ta.

253


I ask the bird:





Given any optSolshe needs to have

a valid answer.

Can we eliminatesome of her answers?

???

254


I ask the bird:





# of answers K = 3

255

Time?

Same as the brute force algorithm that tries each solution.

Same as Brute Force AlgorithmI try each of 3 bird ans.

My friends try 3

His friends try 3

256

Memorization



257




X = a s b e t c h d aY = r t w a b g j c k t f d• Imagine running the recursive alg on it.• Determine the complete set of sub-

Instances ever given to you, your friends, their friends…

X’ = a s b e t c Y’ = r t w a b g j c k

Is this a sub-Instance? Yes

258





Instances ever given to you, your friends, their

friends, …Is this a sub-Instance? NoX’ = b e t

Y’ = a b g j c k

259





Instances ever given to you, your friends, their

friends, …Is this a sub-Instance? Yes

X’ = x1,…xi

Y’ = y1,…,yj

i [0..|X|] j [0..|Y|]|X| × |Y| of these.

260

Set of Sub-InstancesGuess the complete set S of sub-Instances.

i [0..|X|] j [0..|Y|]

Xi = x1,…xi

Yj = y1,…,yj


Yes: i = |X| & j = |Y|

261


i [0..|X|] j [0..|Y|]

Xi = x1,…xi

Yj = y1,…,yj

The set S of sub-Instances needs to:• include our given I• closed under “friend” operation sub-Instance S subsub-Instance S

Xi = x1,…xi

Yj = y1,…,yj

S Xi-1 = x1,…xi-1 Yj = y1,…,yjXi = x1,…xi Yj-1 = y1,…,yj-1Xi-1 = x1,…xi-1 Yj-1 = y1,…,yj-1

S

262


i [0..|X|] j [0..|Y|]

Xi = x1,…xi

Yj = y1,…,yj

The set S of sub-Instances needs to:• include our given I• closed under “friend” operation

sub-Instance S subsub-Instance S• each sub-Instance needs to be

asked of some friend, friend, …We showed this.

This is a fine set of sub-Instances!

263


Sub-InstancesMap

Indexes

Cell of table

i

“LCS of x1,…xi and y1,…,yj ?”

i [0..|X|] j [0..|Y|]

Xi = x1,…xi

Yj = y1,…,yj

j

The Table

264

The Table

265

Table

X

Y Original instance I = <X,Y>

266

Table

Xi = x1,…xi

Yj = y1,…,yj

sub-Instancei,j =

Xi

Yjj=

i=

Cost = length of LCS.

Optimal Solution = Longest Common Subsequence

267

Table

Xi = x1,…xi

Yj = y1,…,yj

sub-Instancei,j =

Xi

Yj


Bird’s Advice• delete xi

268

Table

Xi

Yj



• take both xi and yj

Xi = x1,…xi

Yj = y1,…,yj

sub-Instancei,j =

269

Table

Xi

Yj



• delete yj

• take both xi and yj

Xi = x1,…xi

Yj = y1,…,yj

sub-Instancei,j =

270

Fill in Box

Xi

Yj

Fill in box• Try all bird’s ans.• delete xi

Friend’s sub-InstanceOur cost = friend’s cost5

Xi = x1,…xi

Yj = y1,…,yj

sub-Instancei,j =

271

Xi

Yj

Fill in box• Try all bird’s ans.• delete yj


Fill in Box

Xi = x1,…xi

Yj = y1,…,yj

sub-Instancei,j =

272

Xi

Yj

Fill in box• Try all bird’s ans.• take both xi and yj

Friend’s sub-InstanceOur cost = friend’s cost +1

6

Fill in Box

Xi = x1,…xi

Yj = y1,…,yj

sub-Instancei,j =

273

Xi

Yj

Fill in box• Try all bird’s ans.• Take best of best

6

Fill in Box

Xi = x1,…xi

Yj = y1,…,yj

sub-Instancei,j =

274

Fill in Box

Xi

Yj

Fill in box• Try all bird’s ans.• delete xi


Xi = x1,…xi

Yj = y1,…,yj

sub-Instancei,j =

275

Xi

Yj

Fill in box• Try all bird’s ans.• delete yj


Fill in Box

Xi = x1,…xi

Yj = y1,…,yj

sub-Instancei,j =

276

Xi

Yj

Fill in box• Try all bird’s ans.• take both xi and yj

Sorry bird is wrong.Our cost = -

-

Fill in Box

Xi = x1,…xi

Yj = y1,…,yj

sub-Instancei,j =

277

Xi

Yj

Fill in box• Try all bird’s ans.• Take best of best

Fill in Box

4

Xi = x1,…xi

Yj = y1,…,yj

sub-Instancei,j =

278

Fill in Box

279

Fill in Box

280

Order to fill table:• so that nobody waits

This guy waits for

Order to Fill in Table

281

(later)

Order to Fill in Table

282

Base Cases:• general algorithmdoes not work

• This guy’s friends are

Base Cases

283

Base Cases:• general algorithmdoes not work

Base Cases

284

Base Cases

285

With Advice

286

With Advice

Done

287

Knapsack Problem

Get as much valueas you can

into the knapsack

288

Knapsack Problem

Ingredients:• Instances: The volume V of the knapsack.

The volume and price of n objects <<v1,p1>,<v2,p2>,… ,<vn,pn>>.

• Solutions: A set of objects that fit in the knapsack.• i.e. i S vi V

• Cost of Solution: The total value of objects in set.• i.e. i S pi

• Goal: Get as much value as you can into the knapsack.

289

v=4,p=4

v=4,p=4

V=8

Greedy Algorithm

Most valuable piGreedy Criteria:

v=7,p=5 v=4,p=4 v=4,p=4

V=8

Greedy give 5 Optimal gives 8

290

v=7,p=5V=7

Greedy AlgorithmGreedy Criteria:

v=7,p=5 v=4,p=4 v=4,p=4

V=7

Most dense in valuepi vi

Greedy give 4 Optimal gives 5

V=8

291

Greedy AlgorithmGreedy Criteria:

v=4,p=4

V=7

Greedy give 4

v=7,p=5V=7

v=4,p=4¾ of

+ ¾ × 4 = 7

If fractional solutions are allowed.

= Optimal

Works

Most dense in valuepi vi

Optimal gives 5

Often an Integersolution is MUCH

harder to find.

292


I ask the bird:

My instance:

“What is the last object to take?”

<V:<v1,p1>,<v2,p2>,..........................,<vn,pn>>.

A solution: <<v5,p5>,<v9,p9>,...........,<v82,p82>>.

# of answers K = n

v=7,p=5 v=4,p=4 v=4,p=4V=12

293


I ask the bird:

My instance:

A solution:

# of answers K =

v=7,p=5 v=4,p=4 v=4,p=4V=12

“Do we keep the last object?” 2 Yes & No

<<v5,p5>,<v9,p9>,...........,<v82,p82>>.

<V:<v1,p1>,<v2,p2>,..........................,<vn,pn>>.

294

Bird & Friend AlgorithmMy instance:

v=7,p=5 v=4,p=4 v=4,p=4V=12

Bird says, Yes keep the last object.

Trust her and put it into your knapsack.

I ask my friend:To fill the rest of the knapsack.But what instance do I give him?

<V:<v1,p1>,<v2,p2>,..........................,<vn,pn>>.

295

Bird & Friend AlgorithmHis instance:

v=7,p=5 v=4,p=4 v=4,p=4V=12-4

<V-vn:<v1,p1>,<v2,p2>,.........<vn-1,pn-1>,<vn,pn>>.

His solution: <<v5,p5>,<v9,p9>,...........,<v82,p82>>.

My solution:

My cost: same + pn

<<v5,p5>,<v9,p9>,...........,<v82,p82>,<vn,pn>>

296


My solution: <<v5,p5>,<v9,p9>,...........,<v82,p82>,<vn,pn>>

My instance:

v=7,p=5 v=4,p=4 v=4,p=4V=12

<V:<v1,p1>,<v2,p2>,..........................,<vn,pn>>.

If we trust the bird and friend, this is valid and optimal.

My cost: same +pn

297

Bird & Friend AlgorithmMy instance:

v=7,p=5 v=4,p=4 v=4,p=4V=12

Bird says, No do not keep the last object.

Trust her and delete it.

I ask my friend:To fill the knapsack with the rest.What instance do I give him?

<V:<v1,p1>,<v2,p2>,..........................,<vn,pn>>.

298

Bird & Friend AlgorithmHis instance:

v=7,p=5 v=4,p=4 v=4,p=4V=12

<V:<v1,p1>,<v2,p2>,..........................,<vn,pn>>.

My solution:

My cost: same

His solution: <<v5,p5>,<v9,p9>,...........,<v82,p82>>.

sameIf we trust the bird and friend,

this is valid and optimal.

299

Time?



My friends tries 2

His friends tries 2

300

Memoization



301

Set of Sub-InstancesDetermine the complete set of sub-Instances.

• Imagine running the recursive algorithm on it.• Determine the complete set of sub-Instances

ever given to you, your friends, their friends, …

Given an instance I,<V:<v1,p1>,<v2,p2>,<v3,p3>,<v4,p4>,<v5,p5>,<v6,p6>>.

Is this a sub-Instance?

Yes, if the bird keeps saying “No”.

<V:<v1,p1>,<v2,p2>,<v3,p3>>.

302






No, the set of objects is always a prefix of the original set.

<V:<v1,p1>,<v2,p2>,<v3,p3>,<v4,p4>,<v5,p5>,<v6,p6>>.

303





Quite possibly, if V’ V.

Is this a sub-Instance? <V’:<v1,p1>,<v2,p2>,<v3,p3>>.

It is easier to solve than to determine if it is a sub-instance.

304



The set S of sub-Instances needs to:• include our given I Yes: V’=V & i = n

My instance:

• closed under “friend” operation sub-Instance S subsub-Instance S

<V:<v1,p1>,<v2,p2>,..........................,<vn,pn>>.

i [0..n]<V’:<v1,p1>,<v2,p2>,......,<vi,pi>>. V’ [0..V]

<V’:<v1,p1>,<v2,p2>,......,<vi,pi>> S

YesNo <V’ :<v1,p1>,<v2,p2>,...,<vi-1,pi-1>>

<V’-vi:<v1,p1>,<v2,p2>,...,<vi-1,pi-1>> S

305


Sub-InstancesMap

Indexes

Cell of table

i

“Which of first i objects to put in a knapsack of size v’?”

v’

<V’:<v1,p1>,<v2,p2>,......,<vi,pi>>.

i [0..n] V’ [0..V]

The Table

306

The TableThe complete set S of sub-Instances.

i [0..n]<V’:<v1,p1>,<v2,p2>,......,<vi,pi>>. V’ [0..V]

01

i-1

n

0 1

2

2 V’-vi VV’

iYesNo

OptSol Cost &

Bird’s Advicefor this

sub-Instance

Our cost?

samesame + pi

Take best of best.

307

The TableThe complete set S of sub-Instances.

i [0..n]<V’:<v1,p1>,<v2,p2>,......,<vi,pi>>. V’ [0..V]

01

i-1

n

0 1

2

2 V’-vi VV’

i

OptSol Cost &

Bird’s Advicefor this

sub-Instance

Order to fillso nobody

waits?

308

The Code

309

310

311

Running Time

Running time = ( # of sub-instances × # bird answers ) = ( Vn × 2 ) = ( 2#bits in V × n )

My instance:

Polynomial?

<V:<v1,p1>,<v2,p2>,..........................,<vn,pn>>.The complete set S of sub-Instances is

i [0..n]<V’:<v1,p1>,<v2,p2>,......,<vi,pi>>. V’ [0..V]

YesNo

Exponential in “size” in instance!

312

The Knapsack Problem• Dynamic Programming

Running time = ( V × n ) = ( 2#bits in V × n )• Poly time if size of knapsack is small• Exponential time if size is an arbitrary integer.

313

The Knapsack Problem

If there is a poly-time algorithmfor the Knapsack Problem

For EVERY optimization problem there is a poly-time algorithm.

• Dynamic Programming Running time = ( V × n ) = ( 2#bits in V × n )

• NP-Complete

314

The Knapsack Problem

Likely there is not a poly-time algorithmfor the Knapsack Problem.

Likely there is not a poly-time algorithmfor EVERY optimization problem.

• Dynamic Programming Running time = ( V × n ) = ( 2#bits in V × n )

• NP-Complete

315

The Knapsack Problem• Dynamic Programming

Running time = ( V × n ) = ( 2#bits in V × n )

• NP-Complete• Approximate Algorithm• In poly-time, solution can be found

that is (1+) as good as optimal.

done

316

The Job/Event Scheduling Problem

Schedule as many eventsin your room

as possible

317

Ingredients:• Instances: Events with starting and finishing times

<<s1,f1>,<s2,f2>,… ,<sn,fn>>.• Solutions: A set of events that do not overlap. • Cost of Solution: The number of events scheduled.

• Goal: Given a set of events, schedule as many as possible.

The Job/Event Scheduling Problem

318

Greedy Algorithm

Earliest Finishing TimeSchedule the event which will free up your room for someone else as soon as possible.

Motivation:

Greedy Criteria:

319

Ingredients:• Instances: Events with starting and finishing times

and weights <<s1,f1,w1>,<s2,f2,w2>,… ,<sn,fn,wn>>.

• Solutions: A set of events that do not overlap. • Cost of Solution: Total weight of events scheduled.

• Goal: Given a set of events, schedule max weight

Weighted Event Scheduling

320

Greedy Algorithm

Earliest Finishing TimeSchedule the event which will free up your room for someone else as soon as possible.

Motivation:

Greedy Criteria:

1001 1

321


I ask the bird:

An instance:

“What is the last event to take?”

<<s1,f1,w1>,<s2,f2,w2>,… ,<sn,fn,wn>>.

A solution: <<s5,f5,w5>,<s9,f9,w9>,… ,<s82,f82,w82>>.

# of answers K = n

322


I ask the bird:

An instance:

“Do we keep the last event?”

<<s1,f1,w1>,<s2,f2,w2>,… ,<sn,fn,wn>>.

A solution: <<s5,f5,w5>,<s9,f9,w9>,… ,<s82,f82,w82>>.

# of answers K = 2 Yes & No

323

Bird & Friend AlgorithmAn instance:

I ask the bird:


<<s1,f1,w1>,<s2,f2,w2>,… ,<sn,fn,wn>>.

His solution: <<s5,f5,w5>,<s9,f9,w9>,… ,<s82,f82,w82>>.

I ask my friend:

She answers: No

My solution: same My cost: same

<<s1,f1,w1>,<s2,f2,w2>,… ,<sn,fn,wn>>.

324

Bird & Friend AlgorithmAn instance:

I ask the bird:


<<s1,f1,w1>,<s2,f2,w2>,… ,<sn,fn,wn>>.

His solution: <<s5,f5,w5>,<s9,f9,w9>,… ,<s82,f82,w82>>.

I ask my friend:

She answers: Yes

My solution: same + <sn,fn,wn>. My cost: same +wn

<<s1,f1,w1>,<s2,f2,w2>,… ,<sn,fn,wn>>.

Carefull

325


Bird answers:

“Yes keep the last event.” Give the rest to my friend.

No this solution is not valid!

Here is my best subsolution.

I add to his subsolution the bird’s answer.

last event

326


Bird answers:

“Yes keep the last event.” Then I should politely tell the bird she is wrong

No we trust the bird!

327


Bird answers:

“Yes keep the last event.” No we trust the

bird!

You only tell her she is wrong if you really know.

Eg k words don’t fit on the last lineThe bear does not fit into the knapsack

328


Bird answers:


bird!

Your friend could have just as easily given you this subsolution that does not

conflict with the bird’s answer.

329


Bird answers:


bird!

Or maybe he needs to make a sacrifice when finding his answer in order that the

overall solution is the best.

330


Bird answers:


bird!

Or goal now is to find the best solution to our instance that is consistent with the

bird’s answer.Then we will take the best of best.

331


Bird answers:


bird!

Dude! It is your job to give me the right subinstance so that I give you a

subsolution that does not conflict with the bird

332


Bird answers:

“Yes keep the last event.”

My instance:last event

Cant keep any events that overlap with it.

333


Bird answers:


My instance:last event

I ask my friend:

334


Bird answers:


His instance:His solution

I ask my friend:

335


Bird answers:


I ask my friend:

My solution: same + <sn,fn,wn>. Valid?

My instance:My solution:

Yes

336


Bird answers:


My instance:My solution:

I ask my friend:

My solution: same + <sn,fn,wn>. My cost: same +wn

337

Time?



My friends tries 2

His friends tries 2

338

Memorization



339






340

Every subset of {1,…,9}is a possible sub-Instance.

I.e. could be an exponentialnumber of them.

Hence, running time is exponential.

Greedy algorithm sortedjobs by finishing time.

Let us do that too.

341

Each sub-Instance is prefix.I.e. only n of them.

Hence, running time is polynomial!

342



The set S of sub-Instances needs to:• include our given I Yes: i = n

My instance:

<<s1,f1,w1>,<s2,f2,w2>,................… ,<sn,fn,wn>>.

i [0..n]<<s1,f1,w1>,<s2,f2,w2>,… ,<si,fi,wi>>

• closed under “friend” operation sub-Instance S subsub-Instance S

• each sub-Instance needs to beasked of some friend, friend…

?

Only n sub-InstancesGood enough.

343


sub-Instance =

<<s1,f1,w1>,<s2,f2,w2>,................................,<si,fi,wi>>

last event

Show closed under “friend” operation sub-Instance S subsub-Instance S

Events sorted by finishing time.

344


sub-Instance =

<<s1,f1,w1>,<s2,f2,w2>,................................,<si,fi,wi>>

Bird answers: “Yes keep the last event.”

last event


Delete overlapping events for friend.

345


subsub-Instance =

<<s1,f1,w1>,<s2,f2,w2>,.....,<sj,fj,wj>>

Bird answers: “Yes keep the last event.”


Delete overlapping events for friend.

346


subsub-Instance =

<<s1,f1,w1>,<s2,f2,w2>,.....,<sj,fj,wj>>


subsub-Instance S set of kept jobs is a prefix of events.

typical deleted jobtypical kept job

Event j is kept fj si

347



My instance:

<<s1,f1,w1>,<s2,f2,w2>,................… ,<sn,fn,wn>>.

i [0..n]<<s1,f1,w1>,<s2,f2,w2>,… ,<si,fi,wi>>

Table:

0, 1, 2, 3, 4, …. n Base case Original

348



Running time = # of sub-instances × # bird answers = n × 2

My instance:

<<s1,f1,w1>,<s2,f2,w2>,................… ,<sn,fn,wn>>.

i [0..n]<<s1,f1,w1>,<s2,f2,w2>,… ,<si,fi,wi>>

Done

But to find your friend’s “yes” sub-instanceyou must know how many events overlapwith your last event. This takes time:O(logn) using binary searchfor a total of O(nlogn) time.

349

Input:

Output:

s=6*8+((2+42)*(5+12)+987*7*123+15*54)

Parsing

350

Parsing

Recursive Alg:• GetExp calls GetTerm• GetTerm calls GetFact• GetFact calls GetExp

351

T AB CA TT

A AA BT a

B TA BC b e

C CB AC c d

Parsing

Context Free Grammar (Not look ahead one) For ease, we will assume every non-terminal either goes to two non-terminals or to one terminal

352

T AB CA TT

A AA BT a

B TA BC b e

C CB AC c d

Parsing

Input: start non-terminal = T string to parse = a1a2a3 ..... an = baeaadbda

Output: A parsing

T a1a2a3 ..... an T

C A

A

A

B

A

C

C B

A CTA B

B T

C A

b d a b a e a a d b

353

T AB CA TT

A AA BT a

B TA BC b e

C CB AC c d

Parsing

Recursive Algorithm: GetT does not know whether to call GetA, GetC, or GetT.

Input: T a1a2a3 ..... an T

C A

A

A

B

A

C

C B

A CTA B

B T

C A

b d a b a e a a d b

354

T AB CA TT

A AA BT a

B TA BC b e

C CB AC c d

Parsing

Ask Little Bird:• For first ruleAsk Friend• Parse leftAsk Another Friend• Parse right.


C A

A

A

B

A

C

C B

A CTA B

B T

C A

b d a b a e a a d b

355

T AB CA TT

A AA BT a

B TA BC b e

C CB AC c d

Parsing

Ask Little Bird:• For first ruleInstance to give Friend•?


C A

b d a b a e a a d b

356

T AB CA TT

A AA BT a

B TA BC b e

C CB AC c d

Parsing

Ask Little Bird:• For first ruleWant from Friend:• Left sub-parse tree.Instance to give him:• C baeaadb

Input: T a1a2a3 ..... an

C

A

A

B

A

C

C B

A CTA B

T

A

b d a b a e a a d b

357

T AB CA TT

A AA BT a

B TA BC b e

C CB AC c d

Parsing

Ask Little Bird:• For first ruleHow can we know split?• Ask the Bird!


C A

A

A

B

A

C

C B

A CTA B

B T

C A

b d a b a e a a d b

358

T AB CA TT

A AA BT a

B TA BC b e

C CB AC c d

Parsing

Ask Little Bird:• For first rule• For the split.

# of ans K =# of ans K =

mT = # of rules for T.n = # chars in string.

Total # of ans K = mT × n.


b d a b a e a a d b

T

C A

359

T AB CA TT

A AA BT a

B TA BC b e

C CB AC c d

Parsing

Ask left friend:• Instance: C baeaadb• Solution: Left parsing


b a e a a d b

C

T

b d a

A

A

A

B

A

C

C B

A CTA B

360

T AB CA TT

A AA BT a

B TA BC b e

C CB AC c d

Parsing

Ask right friend:• Instance: A bda• Solution: Right parsing


A

B T

C A

b d a

T

C

b a e a a d b

361

T AB CA TT

A AA BT a

B TA BC b e

C CB AC c d

Parsing

Combine:• Instance: • Bird’s Answer• Left Friend’s Answer• Right Friend’s Answer


A

A

B

A

C

C B

A CTA B

B T

C A

T

b d a b a e a a d b

C A

362

363

Time?



My friends tries 2

His friends tries 2

364

Memoization



365





366


My instance I:

gives:

My left sub-Instance.

gives:

His right sub-Instance.

T a1a2a3 ..... an T

C A

CA

b d a b a e a a d b

367


T’ aiai+1 ..... aj • non-terminals T’ • i,j [1,n]

My instance I:T a1a2a3 ..... an

sub-Instances:

# of sub-Instances = # of non-terminals × n2

C

a d b

aiai+1...aj

T’=

a1...ai-1 aj+1...an

368


Sub-InstancesMap

Indexes

Cell of table

The Table

T’ aiai+1 ..... aj

T’ i,j [1,n]

j

T’ i

369

370

T’ aiai+1 ..... aj non-terminals T’

& i,j [1,n]

sub-Instances:

Running Time

Running time = ( # of sub-instances × # bird answers ) = ( # of non-terminals × n2

gives: First rule and split

× # of rules · n )

Done

371

Find a Satisfying Assignment

c = (x3 or x5 or x6) and (x2 or x5 or x7) and (x3 or x4)An instance (input) consists of a circuit:

A solution is an assignment of the variables.x1 = 0, x2 = 1, x3 = 0, x4 = 0, x5 = 1, x6 = 0, x7 = 1

true falsetrue true

truetrue truetrue

The cost of a solution is • 1 if the assignment satisfies the circuit.• 0 if not.

The goal is to find satisfying assignment.

372


c = (x3 or x5 or x6) and (x2 or x5 or x7) and (x3 or x4)Instance:

Ask the little bird

Value of x1 in an optimal solution or even better

Value of x3 in an optimal solution

For now, suppose she answered x3 = 0.

We will have to try both x3 = 0 and x3 = 1.

373


c = (x3 or x5 or x6) and (x2 or x5 or x7) and (x3 or x4)Instance:

truetrue false

Commit to x3 = 0 and simplify

c = (x2 or x5 or x7) and x4

Sub-Instance:

Friend gives Sub-Solution: x1 = 0, x2 = 1, x4 = 0, x5 = 1, x6 = 0, x7 = 1

Our Solution: x1 = 0, x2 = 1, x3 = 0, x4 = 0, x5 = 1, x6 = 0, x7 = 1

374

In the end, some friend looks at each of the 2n assignments,

Speeding Up the Timex3

0 1x2

0 1

x1

0 1x1

0 1

x2

0 1

x2

0 1

x1

0 1

375

Memoization



376






377





Yes

c = (x1 or y1) and (x2 or y2) and (x3 or y3) and (x4 or y4)

c = (x1) and (x3)

Commit to y1=0, y2=1, y3=0, y4=1, and simplify

True for any subset of the xi. could be an exponential # of different sub-Instances. running time is exponential.

378

In the end, some friend looks at each of the 2n assignments,


0 1x2

0 1

x1

0 1x1

0 1

x2

0 1

x2

0 1

x1

0 1

379


0 1x2

0 1

x1

0 1x1

0 1

x2

0 1

x2

0 1

x1

0 1

But sometimes we can prune off branches.

380


Instance:c = (x2 or x5 or x7) and x3

x3 is forced to x3 = 0

x3

0 1x2

0 1

x1

0 1x1

0 1

x2

0 1

x2

0 1

x1

0 1

381


Instance:c = (x2 or x5 or x7) and x3 and x3

This is trivially unsatisfiable because x3 can’t be both 0 and 1.

x3

0 1x2

0 1

x1

0 1x1

0 1

x2

0 1

x2

0 1

x1

0 1

382

383

384

Designing Recursive Back Tracking Algorithm• What are instances, solutions, and costs? • Given an instance I,• What question do you ask the little bird?• Given a bird answer k [K],

• What sub-Instance do your give your friend?• Assume he gives you optSubSol for sub-Instance.• How do you produce an optSol for I from

• the bird’s k and • the friend’s optSubSol?

• How do you determine the cost of optSol from • the bird’s k and • the cost of the friend’s optSubSol?

• Try all bird’s answers and take best of best.

Review

385

Recursive Back Tracking Algorithm

Dynamic Programming Algorithm• Given an instance I,• Imagine running the recursive alg on it.• Determine the complete set of sub-Instances

ever given to you, your friends, their friends, …• Build a table indexed by these sub-Instances• Fill in the table in order so that nobody waits.

• the cost of its optimal solution• advice given by the bird

• Run the recursive algorithm with bird’s advice to find the solution to your instance.

Review

386

Optimization Problems• Don’t mix up the following

– What is an instance– What are the objects in an instance– What is a solution– What are the objects in a solution– What is the cost of a solution

• Greedy algorithm– What does the algorithm do & know– What does the Prover do & know– What does the Fairy God Mother do & know

• Recursive Backtracking / Dynamic Programming– What does the algorithm do & know– What does the little bird do & know– What does the friend do & know

387

Dynamic ProgrammingDon’ts

• Yes, the code has a basic structure that you should learn.• But don’t copy other code verbatim • Don’t say if(ai = cj)

(i.e. Longest Common Subsequence) when our problem does not have cj

388


• When looping over the sub-instances • be clear what the set of sub-instances are • which is currently being solved,

i.e. which instance is cost(i,j)?• If you know that the set of sub-instances are the

prefixes of the input, i.e. <a1,a2, …, ai>, then don’t have a two dimensional table. Table[1..n,1..n].

• Don’t loop over i and loop over j if j never gets mentioned again.

389


• When trying all bird answers • be clear what the set of bird answers are,• which is currently being tried,• & what it says about the solution being looked for.

• When getting help from your friend,• be clear what the sub-instance is that you are giving him• How do you use the current instance and the bird' s

answer to form his sub-instance?• Don’t simply say cost(i-1,j-1)

390


• Think about what the base cases should be.• Don’t make an instance a base cases

if they can be solved using the general method.• % is used to start a comment.

Don’t put it in front of code.

391

• If a solution is a binary tree of objects,


–“What object is at the root of the tree?”

–“Which key is at the root of the tree?”

38

25

17

4 21

31

28 35

51

42

40 49

63

55 71

• Eg. The Best Binary Search Tree problem,

392

Matrix Multiplication

393

394

395

396

397

All Pairs Shortest Paths

398

399

400

401

402

403

end

404

Construct a table • for storing an optimal solution & cost • for each sub-instance.

Dynamic Programming

Sub-InstancesMap

Index

Cell of table


i ϵ [n], i.e. for each node vi

t, v8, v7, v6, v5, …., s i


Date post:	27-Jan-2016
Category:	Documents
Upload:	nicholas-logan
View:	213 times
Download:	0 times

05-DynamicProgramming

Documents