MELJUN CORTES ALGORITHM_Coping With the Limitations of Algorithm Power

8/21/2019 MELJUN CORTES ALGORITHM_Coping With the Limitations of Algorithm Power

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Design and Analysis of Algorithm

Coping with the Limitations of Algorithm Power *Property o f STIPage 1 of 18

TOPIC TITLE: Coping with the Limitations of Algorithm Power

Speci f ic Object ives:

At the end of the topic session, the students are expected to:

Cognitive:

Understand the idea of coping up with the limitations ofalgorithm power.

Understand the concept of Backtracking algorithm designtechnique.

Understand the application of Backtracking using algorithmdesign technique N-queen problem.

Know the definition of branch and bound algorithm designtechnique.

Affective:

1. Listen to others with respect.2. Participate in class discussions actively.3. Share ideas to the class.

MATERIALS/EQUIPMENT:

o topic slideso OHP

TOPIC PREPARATION:

o

Have the students research on the following:The importance of coping up with the limitation ofalgorithm power.

Definition of Backtracking algorithm design technique.

Application of Backtracking algorithm design techniqueusing N-queen problem.

Definition of branch and bound algorithm designtechnique.

o Provide sample problems that show how to cope up with thelimitations of algorithm power.

o It is imperative for the instructor to incorporate various kinds ofteaching strategies while discussing the suggested topics. Theinstructor may use the suggested learning activities below to

facilitat e a thorough and creative discussion of the topic.o Prepare the slides to be presented in the class.o Prepare seatwork for the students to apply the lessons learned

under coping with the limitations of algorithm power.

TOPIC PRESENTATION:

The topic will cover how to cope up with the limitations of algorithmpower.

The following is the suggested flow of discussion for the course topic:

Discuss the idea of coping up with the limitation of algorithm

power.Discuss the concept of Backtracking algorithm design


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technique.

Provide example of Backtracking algorithm design techniqueusing N-queen problem.

Discuss the definition of branch and bound algorithm designtechnique.

Coping with the Limitations of Algorithm PowerPage 1 of 47

Copingwith theLimitationsof algorithm Power

Designand Analysisof Algorithm

* Property of STI

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The following are the topics to be discussed under

coping with the limitations of algorithm power:

Understand the idea of coping up with the

limitation of algorithm power

Understand the concept of Backtracking

algorithm design technique.

Understand the application of Backtracking

using algorithm design technique N-queen

problem.

Know the definition of branch and bound

algorithm design technique.

Coping with the Limitations of Algorithm Power

The following are the topics to be discussed under coping with thelimitations of algorithm power:

Understand the idea of coping up with the limitation of algorithmpower

Understand the concept of Backtracking algorithm designtechnique.

Understand the application of Backtracking using algorithmdesign technique N-queen problem.Know the definition of branch and bound algorithm designtechnique.




* Property of STI

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There are problems that are difficult to solve

where algorithms takes too much time for arbitrary

instances of problem.

To cope up with this, it is important to understand

what makes a problem difficult and makes a

problem easy.

Theoretically, a problem will turn out to be difficult

if there is only one instance of the problem that

demands an exponential time to solve by any

algorithm.

Coping with the Limitations of Algorithm Power

We are aware that there are problems that are difficult to solve wherealgorithms take too much time for arbitrary instances of problem.Sometimes, we need to consider a number of approaches to solve

problems that occur to be intractable like problems that do not haveefficient solutions. To cope up with this, it is important to understandwhat makes a problem difficult and easy. We can say that a problem iseasy if and only if there is at least one algorithm that solves everyinstance of the problem efficiently. Theoretically, a problem will turn outto be difficult if there is only one instance of the problem that demandsan exponential time to solve by any algorithm.

It is evident that there is infinite number of instances for any problem, sothe subsistence of some problem instances needing exponential timewould be difficult to perceive. However, a fixed percentage of allproblem instances usually require exponential time. The implication ofthis statement is that even for difficult problems, there are percentages

of problem instances that can be efficiently solved.


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Problems can be solved using

back-tracking

branch-and-bound

Both strategies are improvements over exhaustive

Backtracking and branch-and-bound algorithm design techniques areapplied in creating a problem-search-tree whose nodes mirror specificchoices prepared for a solution’s component. Both techniques terminatea node as soon as it can be guaranteed that no solution to the problemcan be achieved by considering choices that correspond to the nodes’successors. The techniques vary in the nature of problems these can

be applied to.

[Coping with the Limitations of Algorithm Power, Pages 1-3 of 47]

BacktrackingPage 4 of 47



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Backtracking: is a refinement of the exhaustive

search strategy where solutions to a problem are

developed and evaluated one step

at a time.

Each partial solution is evaluated for feasibility

and is said to be feasible if it can be developed by

further choices without violating any of the

problem’s constraints.




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The following are the steps in evaluating a

problem thru backtracking:

Backtracking

Backtracking is a refinement of the exhaustive search (brute force)strategy where solutions to a problem are developed and evaluated onestep at a time. Each partial solution is evaluated for feasibility and issaid to be feasible if it can be developed by further choices withoutviolating any of the problem’s constraints. In addition, partial solution issaid to be infeasible if there are no legitimate options for any remainingchoices.

The following are the steps in evaluating a problem throughbacktracking:

Figure 17.1 Illustration of backtracking algorithm


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Copingwith theLimitations of algorithm Power


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Example: N-Queen problem

A standard chess board is an eight-by- eight arrayconsisting eight rows of eight columns each.

To further understand the constraints on this

problem, we shall examine the simpler 4-queens

problem that is set on a four-by -four board.

Backtracking

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The possible moves for this queen are the

following:




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Solution to the problem can be attained thru the

following:

Example: N-Queen Problem

A standard chess board is an eight-by-eight array consisting of eightrows and eight columns. The eight-queen problem is set in the contextof this standard chess board and its main objective is to place eightqueens on the board subject to the constraint that no two queens can

attack each other.

To further understand the constraints of this problem, we shall examinethe simpler 4-queens problem that is set on a four-by-four board.

.

Figure 17.2 Illustration of four-by-four board

In the given figure, the board is represented as either a number ofhorizontal rows or vertical columns. One should be able to show it interms of diagonals. In the real chess games, we consider the queen asthe most powerful piece, which is allowed to move horizontally,vertically, or along either of the two diagonals on which it is placed. Inthis example, the queen is located in row 2 and column 1, or position (2,1).

The possible moves for this queen are the following:

1. Any other position on row 2: (2, 2), (2, 3), or (2, 4)

2. Any other position on column 1: (1, 1), (3, 1), or (4, 1)3. Any other position on the NE diagonal: (1, 2)4. Any other position on the SE diagonal: (3, 2) or (4, 3)

If any other queen is in one of the signified locations, the two queenscan attack each other and the solution is not allowed. On the other hand,the solution to the problem can be attained using the following:

Figure 17.3 Illustration of problem solution

Usually, one has a structured way to place the queens. It is easier torepresent the one that is needed only at the start with two initialpositions, either in column 1 or row 1. For simplicity, let us go by thesolution that begins with placing in a column. It is easier to envision thatthere are two starting positions for this problem. We can simply say that

other solutions can be generated by symmetry arguments. Once wehave one applicable solution, we can use symmetry to get the others, asillustrated below:


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It can be argued that other solutions can be

generated by symmetry arguments, once we have

one applicable solution we shall use symmetry to

get the others as illustrated below:




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The empty board is considered as the root node of

the tree which is the starting position of the

problem. The figure below are the first two levels

of the tree:

Backtracking

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Each of the two leaf nodes in the partial tree is the

root node of a tree with the following leaf count:

Figure 17.4 Illustration of the solution through symmetry arguments

Afterwards, we will set these two possible initial positions within thestate-space-tree. The empty board is considered as the root node of thetree which is the starting position of the problem. The figure below arethe first two levels of the tree:

Figure 17.5 Illustration of the first two level of the tree

This time, we will position the queens by columns, thus the first potentialsolution would observe that there are four different positions in eachcolumn. The fact that no two queens can be in the same row yields 3positions for column 2, 2 positions for column 3, and only 1 for column 4.Each of the two leaf nodes in the partial tree above is the root node of atree with the following leaf count:

Figure 17.6 Illustration of leaf count of the tree

In the given figure, the simplest solution for 43 is equal to 64 leaf nodesand no two n-queens in a row. That is why we have the factor of 3,which is 6 leaf nodes. In this case, we can see immediately that a simpleobservation will trim the number of solutions that needs to be

considered.

We now consider the subtree on the left and see backtracking in action.


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We now consider the subtree on the left and see

backtracking in action. Think an incorrect solution

that places another queen in position (2, 2) as

illustrated below:

You will observe that we have a diagonal conflict

since the first queen at position (1, 1) can attack

the second queen at position (2, 2) by using a

diagonal move.




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This method of stating every node in the subtree

to violate a constraint and therefore dropping the

set of solutions is called pruning the state tree.

It is simple to see that there are 6! = 720 trial

solutions with one queen at (1, 1) and the other at

(2, 2).




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Again, we will look at the other subtree of the

solution space, beginning with the node illustrated

below:

Think of an incorrect solution that places another queen in position (2,2), as illustrated below:

Figure 17.7 Illustration of placing the queen in position (2, 2)

In the given figure, you will observe that we have a diagonal conflictsince the first queen at position (1, 1) can attack the second queen at

position (2, 2) by a diagonal move. As a result, we deny this partialsolution, without generating any more of this subtree. This method ofstating every node in the subtree to violate a constraint and dropping theset of solutions is called pruning the state tree. At this point, the pruning

of the tree keeps consideration of only two needless solutions. It issimple to see that there are 6! = 720 trial solutions with one queen at (1,1) and the other at (2, 2). Therefore, none of these solutions require tobe considered as each defies a diagonal constraint. The completion ofthe first subtree, drawn horizontally in order to save space is illustratedin the following figure.

Figure 17.5 Illustration of completion of the first subtree

Again, we will look at the other subtree of the solution space, beginningwith the node, as illustrated in the figure below:

Figure 17.6 Illustration of the solution of the other subtree


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Only possible placement is at (2, 4), Let’s follow

the search tree for the (2, 4) placement, then

again, we draw it horizontally as illustrated below:




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Remember that there is only one alternative for

positioning the queen in the third column because

of the following reasons:




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Alternative for placing the queen in the fourth

column since:

In reality, the student will note immediately that the options for placing incolumn 2 are quite constrained placing at (2, 1) gives a diagonal conflict,placing at (2, 2) gives a row conflict, and placing at (2, 3) gives adiagonal conflict. Therefore, the only possible placement is at (2, 4).

Let’s follow the search tree for the (2, 4) placement, then again, we drawit horizontally, as illustrated in the figure below:

Figure 17.7 Illustration of the solution of the other subtree

In the given illustration, we should remember that there is only onealternative for positioning the queen in the third column since:

Figure 17.8 Illustration of the reason why there is only one alternative for positioningthe queen in the third column

Similarly, there is only one alternative for placing the queen in the fourthcolumn, since :

Figure 17.9 Illustration of the reason why there is only one alternative for positioning

the queen in the third column

[Backtracking, Pages 4-17 of 47]

•positioning a queen at(4, 1) will result a row

conflict with Q31

•positioning a queen at(4, 2) will result a row

conflict with Q12

•positioning a queen at

(4, 4) will result a row

conflict with Q23


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Branch and BoundPage 18 of 47



* Property of STI

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Branch and Bound is most commonly applied to

optimization problems.

Bounding solution does not satisfy the constraints

but can be shown to be a bound for any optimal

solution.

Branch and Bound

Branch and bound is most commonly applied to optimization problems,although it can be applied to any problem generated by stages in whichthere are easy methods to rank the solutions.

Since the branch-and-bound technique is generally applied tooptimization problems, it would be helpful to remember some of theterminologies. Basically, we are presented with a problem in which wewant to optimize some target function subject to constraints. On theother hand, a bounding solution does not satisfy the constraints but canbe shown to be a bound for any optimal solution. For instance, in theassignment problem, it calls for creating assignments in a way tominimize cost. We can easily come up at a bounding solution for thisproblem that gives a value that will be considered as a strong lowerbound on the objective function for any optimal solution. Though, it ispossible that the bounding solution will suit the constraints, consequentlyit is not a feasible solution and therefore unacceptable as an optimalsolution.




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Assessing the Problem State Space Tree

Branch-and-bound algorithm design technique

is useful to a problem in which a solution can

be viewed as generating nodes in a state-

space tree.

The leaf nodes can represent a solution to the

problem, which may not be a reasonable

solution.

Usually, the root node denotes the problem

prior to any decisions made, like the chess

board before any queens has been placed or

the job status before anybody has been

assigned a job.

Assessing the Problem State-Space Tree

The branch-and-bound algorithm design technique is useful to aproblem in which a solution can be viewed as generating nodes in astate-space tree. The leaf nodes can represent a solution to theproblem, which may not be a reasonable solution. On the other hand,the internal nodes of the tree correspond to partial solutions to theproblem. This will then, produce the state-space tree one node at a time,starting with the root node, which is used to specify the starting point forall solutions. Usually, the root node denotes the problem prior to anydecisions made, like the chess board before any queens has beenplaced or the job status before anybody has been assigned a job.




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Usually, in the state-space tree the previously

generated node must be able to assess the node

and specifically we ask the following questions.

1. How does the partial solution compare to the

best existing solution?

2. Is the solution represented by this node a

feasible solution?

In the state-space tree, the previously generated node must be able toassess the node and specifically we ask the following questions.

1. How does the partial solution compare to the best existingsolution?

2. Is the solution represented by this node a feasible solution?


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To make the branch-and-bound work, it is

necessary to assess each node just generated as

the root node of a tree that represents all

solutions that can be generated based on the

partial solution we have just generated.

Therefore, we “prune” the search tree if any of the

following holds:

1. It can prove that no solution in the subtree will

gratify the problem constraints.

2. It can prove that no solution in the subtree will

be better than the best solution produced

thus far.

3. The node created symbolizes an absolute

feasible solution that can be compared to the

best solution so far generated and possibly

replace it.

To make the branch-and-bound work, it is necessary to assess eachnode generated as the root node of a tree that represents all solutionsand can be produced based on the partial solution we have justgenerated. Therefore, we “prune” the search tree if any of the followingoccurs:

1. It can prove that no solution in the subtree will gratify theproblem constraints.

2. It can prove that no solution in the subtree will be better than thebest solution produced thus far.

3. The node created symbolizes an absolute feasible solution thatcan be compared to the best solution so far generated andpossibly replace it.




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Live node: is a node that can possibly represent a

partial solution leading to an optimal solution of

the whole problem.

If we can assess the live nodes as to the best

solution that can be associated with the node’s

subtree, we can apply best first branch and bound

The key approach to branch-and-bound is to view

each newly generated node as the root node of a

subtree leading to a set of solutions.




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We first choose the assignment problem, which

amounts to assigning each of N workers to N jobs

so that the total costs of the jobs are minimized.

We are given an N-by-N matrix describing the cost

for each worker to do each job

It should be noted that this method can be used to

assign N workers to M jobs, with M < N.

This is done by creating (N – M) dummy jobs, each

with a cost representing the expense of having a

worker do nothing.

We shall assume one job per worker.

Live node is a node that can possibly represent a partial solution leadingto an optimal solution of the whole problem.

If we assess the live nodes as to the best solution that can beassociated with the node’s subtree, we can apply best-first branch-and-bound , which is a deviation that repeatedly examines the best nodegenerated so far. There are different ways of evaluating a node, forinstance, suppose we are looking for a least-cost solution. Since costsare typically non-negative and are additive, we can assess the cost of aninterior node and its subtree as being the cost of the solution up to thatpoint. On the other hand, it may be possible to estimate a remainingcost coupled with producing the rest of the subtree and add that to thevertex cost.

Here, we make an obvious and almost-useless observation. One way toassess any solution in a subtree rooted at an interior node is to expandthat subtree and assess all of its leaf nodes. We are looking for a muchfaster way to assess subtrees.

Thus, the key approach to branch-and-bound is to view each newlygenerated node as the root node of a subtree leading to a set ofsolutions. We want a reasonably fast way to make a useable evaluationof each of these subtrees, so that we may chase the best options.

To further understand the branch-and-bound algorithm technique, wewill examine an example. We first choose the assignment problem,which amounts to assigning each N worker to N jobs so that the totalcosts of the jobs are minimized. We are given an N-by-N matrixdescribing the cost for each worker to do each job.

It should be noted that this method can be used to assign N workers toM jobs, with M < N. This is done by creating (N – M) dummy jobs, eachwith a cost representing the expense of having a worker do nothing. Weshall assume one job per worker.


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Consider the matrix of costs. E ach row in the

matrix represents a worker and each column

represents a job. The constraints of a feasible

solution are:

1. We pick one entry from each row,

corresponding to the assignment of exactly

one worker to a job.

2. We pick one entry from each column, so that

each job has someone working on it.

Consider the matrix of costs. Each row in the matrix represents aworker and each column represents a job. The constraints of a feasiblesolution are:

1. We pick one entry from each row, corresponding to theassignment of exactly one worker to a job.

2. We pick one entry from each column, so that each job has

someone working on it.




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Set another method, a feasible solution

corresponds to the selection of one entry in each

row of the matrix so that no two selected elements

are in the same column.

An optimal solution is that feasible solution of

lowest cost.

Set another method, a feasible solution corresponds to the selection ofone entry in each row of the matrix so that no two selected elements arein the same column.

The table below shows the matrix representing the Worker problem:

Job 1 Job 2 Job 3 Job 4

Worker A 9 2 7 8

Worker B 6 4 3 7

Worker C 5 8 1 8

Worker D 7 6 9 4

Table 17.1 Sample problem table

We first describe the bounding solutions. It is obvious that the total

cost of the jobs must include a cost for each worker and this cost cannotbe less than what each worker is working as cheaply as possible. Thus,the sum of the lowest cost in each row is a lower bound on the cost ofthe job. One may likewise make a column argument, saying that each

job has to be done at a cost by a worker.





For the first bounding solution, find the minimum

cost per row and add it up.

For the first bounding solution, find the minimum cost per row and add itup.

Job 1 Job 2 Job 3 Job 4 Minimum

Worker A 9 2 7 8 2 Worker B 6 4 3 7 3

Worker C 5 8 1 8 1

Worker D 7 6 9 4 4

10

Table 17.2 First bounding solution


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For the second bounding solution consider the

column minimum.

Remember that the first bounding solution is not a feasible solution, asJob 3 is assigned to two workers. Nevertheless, this is a lower boundon the cost. For the second bounding solution, consider the columnminimum.


Worker A 9 2 7 8 2 Worker B 6 4 3 7 0

Worker C 5 8 1 8 6

Worker D 7 6 9 4 4

Minimum 5 2 1 4 12

Table 17.3 Second bounding solution

Again, the solution is not feasible since Worker C has two jobs to do.Nevertheless, we have a better lower bound on the cost. Given twoequally valid lower bounds on a cost function, we may choose the higherone as our best guess – that is 12.




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We need something generated quickly and

guaranteed to be feasible, so we pick as follows,

just moving down the diagonal of the matrix.

We now generate a feasible solution to be used as a guide to ourbranch-and-bound solution. We need something generated quickly andguaranteed to be feasible, so we will select the following jobs by movingdown the diagonal of the matrix.

Job 1 Job 2 Job 3 Job 4

Worker A 9 2 7 8

Worker B 6 4 3 7

Worker C 5 8 1 8

Worker D 7 6 9 4

Table 17.4 Diagonal matrix

The total cost associated with this feasible solution is 9 + 4 + 1 + 4 = 18.Thus, we know that the limits on any optimal solution are 12 and 18.

We can either assign by rows (jobs to workers) or by columns (workersto jobs). The root of the state-space tree is the start node, indicatingthat no assignment has been made. This has 4 child nodes (in general,it has N child nodes), one for each of the 4 jobs that can be assigned tothe first worker.





For the first bounding solution, find the minimum

cost per row and add it up.


Copingwith theLimitationso f algorithm Power



For the second bounding solution consider the

column minimum.

The total cost associated with this feasible

solution is 9 + 4 + 1 + 4 = 18. Thus we know that

the limits on any optimal solution are 12 and 18


Copingwith theLimitationso f algorithm Power



We need something generated quickly and

guaranteed to be feasible, so we pick as follows,

just moving down the diagonal of the matrix.


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We now generate a version of the search tree and

consider how to improve it.

We label each node with the total cost of the

allocation so far, here just the cost of the job.

We now generate a version of the search tree and consider how toimprove it. We label each node with the total cost of the allocation sofar, the figure below includes the cost of each job.

Figure 17.10 Illustration that shows the cost of each job




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Let’s return to the examination of minimum costs

per row. The table is repeated here for easy

reference.

The lower bound for the rest of the workers, after

Worker A has been assigned a job, is 3 + 1 + 4 =

8.

Although this is a valid approach, we can improve it considerably.Consider again the table on the examination of minimum costs per row.


Worker A 9 2 7 8 2

Worker B 6 4 3 7 3

Worker C 5 8 1 8 1

Worker D 7 6 9 4 4

Table 17.5 Illustrates the minimum cost of each job

Note that the lower bound for the rest of the workers, after Worker A hasbeen assigned a job, is 3 + 1 + 4 = 8. Thus, we may immediately

generate a lower bound on any partial solution: the total cost of thepartial solution plus the minimum cost of the remaining assignments.With that in mind, we arrive at a new start to the search tree (refer to thefigure below).




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Thus, we may immediately generate a lower bound

on any partial solution:

The total cost of the partial solution plus the

minimum cost of the remaining assignments.

With that in mind, we arrive at a new start to the

search tree.

Figure 17.11 Illustrates the cost of the partial solution plus the minimum cost

These lower bounds on cost are an improvement, and considered as thenode labeled “A gets job 3”. Under that assignment, no other workercan get job 3, so the row minimal must select from columns 1, 2, and 4.


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Here are the adjusted matrices to be used in

assessing the subtrees:

Here are the adjusted matrices to be used in

assessing the subtrees.

A gets job1 with cost = 9

Here are the adjusted matrices to be used in assessing the subtrees.

A gets job 1 with cost = 9

Job 2 Job 3 Job 4 Minimum

Worker B 4 3 7 3

Worker C 8 1 8 1 Worker D 6 9 4 4

Table 17.6 Resulting table if A gets job 1 with cost = 9

The subtree assesses at 8, so the total lower bound is 17, as shownabove.




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The subtree assesses again at 8, so the total

lower limit is 10



Worker B 6 3 7 3

Worker C 5 1 8 1

Worker D 7 9 4 4


The subtree assesses again at 8, so the total lower limit is 10, as shownabove.




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Here we have a new assessment for the subtree –

a new lower limit of 13, so that the total lower limit

evaluation of the node is 20



Worker B 6 4 7 4

Worker C 5 8 8 5

Worker D 7 6 4 4


Here, we have a new assessment for the subtree – a new lower limit of13, so the total lower limit evaluation of the node is 20.


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Here we also have a new assessment for the

subtree – a new lower limit of 10, so that the total

lower limit evaluation is 18.



Worker B 6 4 3 3

Worker C 5 8 1 1

Worker D 7 6 9 6


Here, we also have a new assessment for the subtree – a new lowerlimit of 10, so the total lower limit evaluation is 18.




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Given these more computationally intensive

calculations, we have the following for the

problem state-space tree.

Given these intensive calculations, we have the following for theproblem state-space tree.

Figure 17.12 Problem-state space tree




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One of the nodes – “A gets job 3” – has a lower

bound exceeding the cost of a known feasible

solution.

For this reason it is pruned and the remaining tree

is as follows.

Note what this complex and accurate lower bound calculation has givenus. One of the nodes – “A gets job 3” – has a lower bound exceedingthe cost of a known feasible solution. With this reason, it is pruned andthe remaining tree is as follows:

Figure 17.13 Pruning of tree

Let’s now begin working on the best looking node “A gets job 2” andconsider Worker B. If Worker A gets job 2, then Worker B can have jobs


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1, 3, or 4.




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Here are the three possibilities for this subtree.

A gets job 2 B gets job 1

The remaining jobs assess at 5, so the node

assesses at a value of 13.

Here are the three possibilities for this subtree.

A gets job 2, B gets job 1

The total cost is 2 + 6 = 8. The matrix for the rest of the jobs andworkers is:

Job 3 Job 4 Minimum

Worker C 1 8 1

Worker D 9 4 4

Table 17.10 Resulting table after getting the total cost

The remaining jobs assess at 5, so the node assesses at value 13.






The total cost is 2 + 3 = 5.

The matrix for the rest of the jobs and workers is


assesses at a value of 14.



Job 1 Job 4 Minimum

Worker C 5 8 5

Worker D 7 4 4

Table 17.11 Resulting table if A gets job 2, B gets job 3





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The total cost is 2 + 7 = 9.

The matrix for the rest of the jobs and workers is


assesses at a value of 17



Job 1 Job 3 Minimum

Worker C 5 1 1

Worker D 7 9 7




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Here is the state-tree after the node “A gets job 2”

has been expanded.

At this point, we have five live nodes to consider.

These are labeled as

1) “A gets job 1”,

2) “B gets job 1”,

3) “B gets job 3”,

4) “B gets job 4”, and

5) “A gets job 4”.

The figure below shows the state-tree after the node “A gets job 2” hasbeen expanded.

Figure 17.14 Illustrates the state-space tree

At this point, we have five live nodes to consider. These are labeled as:1) “A gets job 1”,2) “B gets job 1”, 3) “B gets job 3”, 4) “B gets job 4”, and 5) “A gets job 4”.





The best looking node is the one labeled “B gets

job 1” with a lower bound of 13.

We consider two possibilities for the subtree

rooted at this partial solution.

Recall our matrix of choices for workers and D.


The total cost is 2 + 6 = 8. The matrix for workers

C and D is

The best node is the one labeled “B gets job 1” with a lower bound of 13.We consider two possibilities for the subtree rooted at this partialsolution. Recall our matrix of choices for workers C and D.


The total cost is 2 + 6 = 8. The matrix for workers C and D is:

Job 3 Job 4 Minimum

Worker C 1 8 1

Worker D 9 4 4



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* Property of STI

Page46of 48

Here is the status: we have two feasible solutions,

one with total cost of 13.

We have four non-terminal live nodes, none of

which have a lower bound less than 14.

Figure 17.15 Illustrates the state-space tree





So we have a solution: A gets job 2, B gets job 1,

C gets job 3, and D gets job 4.

The solution generated 9 of the possible (4 + 12 +

24) 40 nodes – an 75% saving.

We have four non-terminal live nodes, none of which have a lowerbound less than 14. So we have a solution: A gets job 2, B gets job 1,C gets job 3, and D gets job 4.

The solution generated 9 of the possible (4 + 12 + 24) 40 nodes – a 75%saving.

[Branch and Bound, Pages 18-47 of 47]

GENERALIZATION:

o It is important to understand what makes a problem difficult andeasy. We can say that a problem is easy if and only if there is atleast one algorithm that solves every instance of the problemefficiently.

o Backtracking is a refinement of the exhaustive search strategywhere solutions to a problem are developed and evaluated onestep at a time.

o Branch and bound is commonly applied to optimizationproblems, although it can be applied to any problem generatedby stages in which there are easy method to rank the solutions.


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REFERENCES:

§ http://www.csc.lsu.edu/~karki/DA-08/DA21.pdf§ http://csc.colstate.edu/Bosworth/cpsc5115/LectureNotes/CPSC

5115_Ch11.htm

§ Anany Levitin,(2007), The design and analysis of algorithm (2nd

ed.), Pearson Education Inc.

MELJUN CORTES ALGORITHM_Coping With the Limitations of Algorithm Power

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