Procedureof Need Analysis

The cultural concept of man

Subject: Social Anthropology

Karina Vega MA

Questions:

1 .- that characterized the Middle Ages? R =

the religious sense which direction the way of life and man's ideas

2.-Main substance of medieval man

R = The academic and intellectual
In the religious culture

It

l

3 .- What is the importance in education and human development? R =

be influenced by education and the role of wealth stimulating environment.

4 .- What is socialization?

R = Process through which the human being begins to learn the way of life of their society.

5.-What are the goals of socialization? R = Get

skills, communicate (speak, write, read)

6.-In which is the personality? R =

characteristics and behavioral traits of the person

7 .- Name the first 3 stages of life

significant

R = Children-Children Childhood-Adolescence

Adolescence-13 to 19 years

8 .- What is diversity?

R = A variety of differences that exist between individuals or organizations

9 .- What is culture?

R = the background, costubres and beliefs that contribute to the understanding of the world by one person.

10 .- What is social stratification?

R = is the formation in layers (vertical groups) that help to study the composition of a complex social environment.

if they can not comment, send me an email, to know how many have approximated the questions.

gaby.agata @ live.com.mx.

And Remember to tell others of REVIEW NEXT WEEK, SO DO NOT COME TO MISS THAT DAY.

Greetings.

How Long It Takes Male Pubic Hairs To Grow Back

Existing models for measuring the quality

This class is for the Software Quality field of the MSF.

SOFTWARE QUALITY .- According to the definition of the Institute of Electrical and Electronics Engineers (IEEE Std 610-1990)
"Software quality is the degree to which a system, component or process meets the requirements and needs spécifications or customer or user expectations."

Quality can be defined as "a characteristic or attribute of a thing." In this way one could say that the quality of the products can be measured as a comparison of their characteristics and attributes. Thus, this concept can be applied to any product.


factors determining Software Quality are:
• Correction. Does what I want?
• Reliability. Does it reliably all the time?
• Efficiency. Is it run on my hardware as best you can?
• Security (Integrity). Is it safe?
• Ease of use. Is it designed to be used?.

The implementation of quality systems provide numerous benefits to companies that opt \u200b\u200bfor this strategy. Not only reduce their costs in a reasonable, but also increase their income through greater customer satisfaction and improved employee motivation.

QUALITY MEASUREMENT .- The measurement of software quality was given to improve objectives and improve software quality.
One way to make a quality measure is to observe the differences occurring in the production two identical products.
production of articles of any kind does not ensure that two of them are exactly alike. It may be necessary to make detailed comments to make variations to distinguish from each other, since they may not be obvious.
.

Some models have different metrics to evaluate product quality attributes almost always at the level of design or code.

quality * The latest models are designed to improve processes *

QUALITY MEASURES BASED ON MODELS:
Successful software measurement is related to the acquisition, definition and joint manipulation of two models:


• Empirical models
, are the ones that use direct observation or the results of experiments studied phenomenon. ◦
empirical real-world context
• Numerical models
Formalizing ◦ empiric measures

For software development we must support in empirical models.

Currently there are two most prominent and popular in Europe: ISO 9000 and EFQM model.

STANDARD ISO 9000:

Designates a set of quality standards and continuous quality management, established by the International Organization for Standardization (ISO). Can be applied in any organization or activity aimed at the production of goods or services.
The rules include both the minimum and specific guidelines and tools for implementation, and audit methods.
The ISO 9000 specifies how an organization operates, its standards of quality, delivery times and service levels. There are more than 20 elements in the ISO standards that relate to the way operating systems.


Its implementation, although it means hard work, offers numerous advantages for companies, which have:
-standardize the activities of personnel working within the organization through documentation
-Increase customer satisfaction
-Measure and monitor process performance-Reduce

-re-processes to increase efficiency and / or efficiency of the organization in achieving its objectives
-Continually improve processes, products, efficiency, etc.
-Reduce the cost of production or services




EFQM MODEL:

In 1988 the European Business Excellence Model, known as the EFQM Model by its acronym in English (European Foundation for Quality Managment ) organization that focuses on the models of total quality management (TQM or TQM), strategies designed to optimize resources, reduce costs and improve performance, in order to constantly improve the production process.


The EFQM Model is a non-regulatory model, whose basic concept is the self-evaluation based on a detailed analysis of the performance management system to guide the organization using the criteria of the model.


This is not a contrast to other approaches (application of certain management techniques, ISO, specific industrial standards, etc..), But rather the integration of them into a scheme broad and comprehensive management.


The systematic and regular use of the EFQM model for the management team allows it to establish plans for improvement based on objective facts and the achievement of a common vision about the goals to be achieved and the tools to use . That is, its implementation is based on:


-depth understanding of the model by all levels of company management.
"The assessment of the situation the same in each of the areas.


The EFQM model consists of two parts:
* A set of criteria for business excellence covering all areas of organizational performance.
* A set of rules to evaluate the performance of the organization in each criterion. There are two sets of criteria: _The
Results (Criteria 6 to 9) represent what the organization gets for each of its stakeholders (customers, employees, Company and Investor). _The
Agents (Criteria 1 to 5) are aspects of the management system of the organization. Causes of the results. For each group of criteria is a set of rules-based assessment called RADAR logic. "

How Much Doesjusterini And Brooks

SWIMMERS SOLUTION EXAMPLE ... Hungarian method.

redrafted the problem: competition

A 400-meter relay includes four different swimmers who swim 100 meters on back, breast, butterfly and freestyle. A coach has 6 very fast swimmers whose expected time in seconds in the individual events are given in Table

How should the coach assigned to the relay swimmers to minimize the amount of his time?

timesheets.





The graph of this graph, each swimmer is related to each of the types tioned, which indicates that the graph is overpopulated, and the view has a lot of traffic, which does not help us solve nueestro
problem ...

would be something like, being the vertices of the inzquierda each swimmer, and the right of each type of swim y  en las aristas iria indicado eltiempo que se tarda en llegar cada uno de ellos.















Segun el metodo hungaro, la mejor asignacion encontrada para resolver este problema de relevos, a manera de grafo seria la siguiente.

Como es evidente los nadadores con el numero 4 y 6,
se quedaron sin participar en el evento, pues sus tiempos no eran los ideales para soucionar el problema.
Clasificaciones de tipo de nado.
D= dorso
P= pecho
M= maripoza
L= libre

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CONTINUED ... proposed problem. PROPOSED PROBLEM

Continuacion sobre the previous entry. INITIAL GRAPH
...

This is the original graph, according to the data table
previous entry.

According to the algorithm and the data given.
THE final graph, with its optimal solution would be as follows.

final graph:

This solution is mid to run the algoritmohungaro,
and considering you is the most optimal (not unique, but the most optimal) because
relates the beginnings and endings,
in order to find the tour, produced the lowest cost.

I had doubts about the pairing of vertices, but unfortunately found

information about, related to the method hugaro, fortunately
Dr. Schaeffer me
mentioned in the previous post, and that is how I concluded that for
this solution as you will notice, the final 6 and 9 have no beginning,
and that is the conflict that can be created.

But serious as the informaciion example, that of
swimmers, where there is the same case, where swimmers
simply do not provide good times and overlap, they do not participate.

Snl Christopher Lowell Skin

, Hungarian method

minimum cost solution for this graph, simple and weighted undirected.

STEP BY STEP ...

Notes:

To solve a problem of allocation in which the goal is to maximize the objective function , multiply the payoff matrix by negative one (-1) and solve the problem as one of minimization, which is not the case.

In a big problem, it may be difficult to obtain the minimum number of rows needed to cover all the zeros in the matrix current cost. It can be shown that if j lines are needed to cover all zeroes, then j can be assigned only work at zero cost in the current matrix, which accounts for over when they need more lines.

If the number of rows and columns in the matrix of costs are different, the problem of allocation is unbalanced. The Hungarian method can provide an incorrect solution if the problem is not balanced, because of the above, it should balance any issues first assignment (adding rows or columns ficiticias) before resolving by the Hungarian method. ADD
I8 and I9 to equlibrar space as recommended.
* I = start, F = final. STEP 1 .-

first

Find smallest element in each row of cost matrix m * m;

a new array is constructed on by subtracting from each cost the minimum cost of each row, find for this new matrix, the minimum cost in each column.

Below is built a new matrix (called the reduced cost matrix) by subtracting the cost of each cost minimum in its column.

STEP 2 .- Draw the minimum number of lines (horizontal or vertical or both), which are required to cover all the zeros in the reduced cost matrix; if m lines are needed to cover all zeros, you have an optimal solution between the zeros of the matrix covered . If you require less than m lines to cover all zeroes, continue with step 3 .
Each horizontal line must go through the whole line (row) and each vertical line across the column.

The number of lines that needed was 6, and 6 is less

am, proceed to the next step.

STEP 3 .-

Find the smallest number that is not covered by a line on the cost matrix. Subtract the value of this number to each item covered by a line and add it to each item covered by two lines.

is deve

Finally back to step 2.

AND OUR PARENT AS FOLLOWS:

again when changing vertical and horizontal lines, we realize that the amount is the same as above, 6

and as 6 is less than m, continue with step 3.

AND IS WELL ...

As the number of lines increased by assigning numbers as indicated, now has to m = 7, therefore has an optimal solution between the zeros of the matrix covered.

GRAPH ..

So the matrix is \u200b\u200brepresented above.
Taking into account the costs that occurred at the beginning of example, in the data table.

Barasoain Church Contact Number

PROJECT 5 .- Assignment (matching): Hungarian algorithm

Assignment (matching): Hungarian algorithm

INTRODUCTION

start by describing as assignment (matching) _
Given a graph a matching V-U is a set of edges not adjacent to each other.
We say that a vertex is matched (assigned) if it is contiguous with an edge in the matching.
Otherwise, the vertex is unmatched (free).

A perfect matching is a matching covering all vertices of the graph. That is, each vertex is saturated under the matching.


2. Matchings in bipartite graphs
The bipartite graphs are usually represented graphically with two columns (or rows) of vertices and edges connecting vertices of columns (or rows) different.

The two sets U and V can be thought of as a graph coloring with two colors: if the vertices in U painted blue and the green-dev Verica get a two-colored graph where each edge has one blue and one vertex green. On the other hand, if a graph does not have the property that can be colored with two colors is not bipartisan.

matching problems are often related to bipartite graphs. Finding a maximum bipartite matching (often called maximum cardinality of a bipartite graph) in a bipartite graph is perhaps the simplest problem.
In a weighted bipartite graph, each edge has an associated value.
A bipartite maximum weighted matching is defined as a perfect matching where the sum of the values \u200b\u200bof its arcs in the matching has maximal value. If the graph is not complete bipartite, missing arcs are introduced zero.
Find this matching problem is known as assignment.

The more specialized is the Hungarian algorithm that solves the problem of cost allocation of time.


DEFINITION:
The first known version of the Hungarian method, was invented and published by Harold Kuhn in 1955.

This was reviewed by James Munkres in 1957 and has since been known as the Hungarian algorithm.

porJames Munkres This was reviewed in 1957 and has been known as the Hungarian algorithm, the algorithm deasignación Munkres, or Kuhn-Munkres algorithm.

The algorithm models a problem as a matrix deasignación cost mn × , where each element represents the cost of assigning the n worker m work.

The algorithm performs the minimization on elementos de la matriz como en el caso de un problema de minimización de precios.

Seutiliza el método de eliminación Gaussiana para hacer aparecer ceros (al menos un ceropor línea y por columna). Sin embargo, en el caso de un problema de maximización de  beneficio, el costo de la matriz necesita ser modificada de modo que la minimización de sus elementos resulte maximizar los valores de costo originales.
En un problema de costo infinito, la matriz de costo inicial puede ser remodelada restando cada elemento de cada línea del valor máximo del elemento de esa línea (o la columna respectivamente). En un problema de costo finito, todos los elementos son restados del maximum value of the entire matrix.


PHASES METHOD FOR THE APPLICATION OF HUNGARIAN:
phases for implementation of the Hungarian method are:

first 1 .- Find the smallest element in each row of cost matrix m * m must build a new matrix by subtracting from each cost the minimum cost of each row, find for this new matrix, the minimum cost in each column. Next you must build a new matrix (called matrix reucidos costs) by subtracting from each cost the minimum cost of your spine.

2 .- (In some texts this step is attributed to Flood), consists of drawing the minimum number of lines (horizontal or vertical or both of these ways only), required to cover all the zeros in the matrix of reduced costs, if they need more lines to cover all zeros, you have an optimal solution between covered zeros matrix. If it takes less than m lines to cover all zeroes, continue with step 3. The number of lines to cover the zeros is equal to the number of assignments to date can be performed.

3 .- Find the smallest nonzero element (called k) in the reduced cost matrix, which is not covered by the lines drawn in step 2, then you must subtract k from each uncovered element of matrix reduced costs and add k to each element of the matrix of reduced costs covered by two lines (intersection). Finally it deve return to step 2.

NOTES:
A) To solve an allocation problem in which the goal is to maximize the objective function, multiply the payoff matrix by negative one (-1) and solve the problem as a minimization .
B) If the number of rows and columns in the matrix of costs are different, the problem of allocation is unbalanced. The Hungarian method can provide an incorrect solution if the problem is not balanced, because of the above, it should balance any issues first assignment (adding rows ficiticias or columns) before using the Hungarian method to solve it.

C) In a big problem, it may be difficult to obtain the minimum number of rows needed to cover all the zeros in the current cost matrix. It can be shown that if j lines are needed to cover all zeroes, then j can be assigned only work at zero cost in the current matrix, which accounts for over when they need more lines.

IF SET OPTIMIZATION OF DECISION OR:
The Hungarian algorithm is a combinatorial optimization algorithm that solves problems of allocation in time n (O) 3.


Optimization combinatorial optimization combinatorial
is a branch of optimization in applied mathematics and computer science, related to operations research, algorithm theory and computational complexity theory. It interacts with other fields such as artificial intelligence and software engineering. Combinatorial optimization algorithms solve instances of problems that are believed to be difficult in general, exploring the solution space (usually large) for these instances. Combinatorial optimization algorithms accomplish this by reducing the effective size of the space, and exploring the search space efficiently.


combinatorial optimization algorithms are often implemented in imperative languages \u200b\u200blike C and C + + in software programming languages \u200b\u200bsuch as Prolog, or even multi-paradigm languages \u200b\u200bsuch as Oz.

By studying the theory of computational complexity is possible to understand the importance of combinatorial optimization. Combinatorial optimization algorithms are commonly associated with NP-hard. These general problems are not resolved effectively, however, several approximations of complexity theory suggests that some instances (eg "small" instances) of these problems can be solved efficiently. Such instances often have important practical ramifications.

COMPUTATIONAL COMPLEXITY:

The complexity is O (n3) cubic complexity.
often occur in loops with triple nesting. If n is doubled, the execution time is multiplied by eight. For a large value of n starts to grow dramatically.

computadoracapaz ◦ We have a process data in 10-4 sec. This computer runs an algorithm that reads records from a database, this algorithm has exponential complexity 2n How long does it take to process an input n data?

◦ Now you have the same computer capable of processing data in 10-4 sec. But running an algorithm that does the same work cited above, but this algorithm has cubic complexity n3, how long will it take to process a data entry no?

can say that only an efficient algorithm with a low order of complexity, can handle large volumes of data, it is reasoned that an algorithm is:

"Very efficient if its complexity is order log n
-Efficient if its complexity is of order na
-inefficient if its complexity is of order 2n
"He considers that a problem is tractable if there is an algorithm that solves it with complexity less than 2n, and that is untreatable or devoid of solution otherwise.

Pseudocode of the algorithm:

This is the link where I found detailed information about each step, and what is to be taken into account.
http://www.public.iastate.edu/ ~ ddoty / HungarianAlgorithm.html


asymptotic analysis of the algorithm (logarithmic, polynomial, exponential):
This algorithm is of class P, since there is an algorithm that can be solved in polynomial time, which for reasonable values \u200b\u200bcan be solved on a computer using an appropriate program.


data structure used (arrays, lists, queues, stacks, trees):
The structure used is: queues, it provides a theoretical basis of the type of service we can espeerar of a particular resource, such as the way in which that action can be designed to provide a certain level of service to its customers.

queuing systems are computer models that provide service. As a model can determine any system where customers come looking for work or service of any kind, and leave the service after it has been treated.
We can model such systems, as well as simple lines or as an interconnected system of casting, forming a network of queues. EXAMPLE

(manual, scheduled, busy):
http://www.tu.tv/videos/metodo-hungaro

1. Locate the smallest element of each row and subtract the other parts of the same line. Repeat this procedure for each column where the column minimum is determined after the remaining items.
2. Determine if there is an assignment likely to involve zero costs in the revised cost matrix. If there is such an allocation is optimal. If there continue to step 3.
3. Cover all the zeros in the revised cost matrix with the smallest number of horizontal and vertical lines as possible. Each horizontal line must go through the line and each vertical line across the column. Find the smallest number that is not covered by a line on the cost matrix. Subtract the value of this number to each item covered by a line and add it to each item covered by two lines.
4. Repeat step 2.
EXAMPLE:
A race of 400-meter relay includes four different swimmers who swim 100 meters on back, breast, butterfly and freestyle. A coach has 6 very fast swimmers whose times expected in second in individual events are given in Table How should the coach assigned to the relay swimmers to minimize the amount of his time?

APPLICATIONS (where is the problem):
When there is a dilemma, on the assignment of a person or service to a response or work, so the settlement of the first approach, and be effective and useful for those in need.



IN TO USE THIS ALGORITHM:
This algorithm is used to solve minimization problems, since it is more effective than used to solve the transport problem of the high degree of degeneration that may occur in allocation problems.




* SOME DEFINITIONS number of examples.
* These are some of the licks I did my research *
http://www.itlalaguna.edu.mx/Academico/Carreras/industrial/invoperaciones1/u5.HTML
http: / / www.scribd.com/doc/62765/memmetpp

  http://www.sangakoo.com/es/Divulgacion/divulgacion/157/algoritmos.aspx
http://www.slideshare.net/rezzaca/u1-analisis-algoritmos-complejidad
http://www.pdfqueen.com/html/aHR0cDovL2l0LmNpaWRpdC51YW5sLm14L35lbGlzYS90ZWFjaGluZy9hYS9wZGYvYWEucGRm
http://www.eici.ucm.cl/Academicos/R_Villarroel/descargas/investigacion_operaciones/Asignacion_y_Vendedor_Viajero.pdf

http://www.monografias.com/trabajos27/complejidad-algoritmica/complejidad-algoritmica.shtml
  http://www.scribd.com/doc/7046323/EL-METODO-HUNGARO ...

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PROJECT 4 (Ordering by the method of the bubble)

Toothbrush Receding Gums

representation and manipulation of trees.

This extra task on binary trees, is information from wikipedia, clearing my doubts .

A binary search tree is a particular type of binary tree that has a data structure in a tree used in computing.


all empty tree is a binary search tree.
A nonempty binary tree, rooted in R is a binary search tree if:
• In If you have left subtree, the root R must be greater than the maximum value stored in the hive
left and the left subtree is a binary tree search
.

• If you have the right subtree, the root R must be less than the minimum value stored in the hive
right, and that the right subtree is a binary tree search
.

may have different binary search trees for the same set of elements.

The interest of binary search trees (ABB) is that their tour inorder provides the elements sorted in increasing and that the search for an item is usually very efficient.
Depending on user needs dealing with a structure of this type may allow strict equality in, any or both of the subtrees hanging from the root. Allow the use of equality leads to the appearance of double values \u200b\u200band makes the search more complex. SEARCH
The search is access to the root of the tree, if the element to locate it matches the search has completed successfully, if the item is less you search the left subtree and if it is greater in the right . If a leaf node is reached and the element was not found are not supposed to exists in the tree. Note that the search on this type of tree is very efficient, is a logarithmic function. The maximum number of comparisons would need to determine whether an item is in a binary search tree would be between [log2 (N +1)] and N, N being the number of nodes. The search for an item in an ABB (binary search tree) can be done in two ways, iterative or recursive.



Example iterative version in the C programming language, assuming that we are looking for a key hosted on a node where the relevant "fact" that we need find:


data Buscar_ABB (abb t, key k)

abb {p;
data e;
e = NULL;
p = t;
if (! isEmpty (p))

{while (! isEmpty (p) & & (p-> k! = k))

{if (k \u0026lt; p-> k)

{p = p-> l;

} if (p-> k \u0026lt;k)

{p = p-> r;}


} if (! IsEmpty (p) & & (p-> d! = NULL)) {

copiaDato e = (p-> d);}


} return e;}



Insertion Insertion is similar to the search and can be given an iterative solution both as a resource. If we initially empty tree as a parameter creates a new node as a single content item to insert. If not is, it checks if the given element is less than the initial tree root that is inserted into the left subtree and if more is inserted into the right subtree. In this way the inserts are made in the leaves.


As in the case of the search can be several alternatives when implementing the inclusion in the ADT (abstract data type), and is the decision to take when the item (or key item) to add is already in the tree, this may be modified or ignored it insertion. It is obvious that this operation modifies the losing ABB previous version. PROC

InsertarABB (tree: TABB; data: TElement)


ele VARIABLES: TElement

HOME IF (ABBVacío (tree)) THEN
tree \u0026lt;- NEW (TNodoABB)
tree ^. left \u0026lt;- NULL
tree ^. der \u0026lt;- NULL
tree ^. elem \u0026lt;- data


otherwise


InfoABB ele = (tree)
SI (dato.clave \u0026lt;ele.clave) THEN
InsertarABB (tree ^. Left, data)

otherwise


InsertarABB (tree ^. dch, data)
FINSI
FINSI
FIN

Disposal:
The operation deletion is more complicated, the search and insertion.
There are several cases to consider:
* Remove a node or leaf node with no children, only clears and sets to zero the pointing of his father.
* Remove a child node to a subtree: elnodo clears and is assigned its subtree subtree son as his father. * Remove a child node with two hives: the solution is to replace the value of the node by its predecessor or its successor in the inorder and then delete this node.


inorder Its predecessor is the node to the right of its left subtree (left-most node of the subtree), and its successor node to the left of its right subtree (subtree node lower right).


reviewing and understand a little better.
this information can be found at: wikipedia / / binary tree

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Project 3: Verification of palindromes

RECURSION:
is when a function calls itself (recursion).

FOR SERVING:
is to calculate an easier problem, because when a module calls itself on each call to the module decreases the difficulty until it is not necessary and the problem is resolved.

DO NOT USE WHEN ...
When your program is needed resources to functions that are already in use, I mean when you do not have to go back (when there is like a circle, with one output).

EXAMPLE ...
The calculation of factorials. The factorial of 0 is defined specifically as 1. The factorial of n, an integer greater than 0, is the product of all integers in the range between 1 and n.


TEAMWORK ...
We split the job in half, since we are 4 people, two of iterativity we play it, and the rest of recursion.
For lack of time we get together physically, but we were in touch through the internet,
each team made a program, even though we had differences, everything was afloat.

STRENGTHS:
I understood that to mean the group or team, most of the time things are shared, and the truth is not a team, if not a shared tranajo.
But that is in some respects good, because there people who work better alone than with people around, speaking in the area of \u200b\u200bconcentration.

AREAS OF OPPORTUNITY:
not understood this point well, I see opportunity
teamwork to minimize the delivery time of work, distributed among the peers, and in the end if there inner doubts, among all solve.

CONTRIBUTION TO WORK:
As my contribution to the work I think it was the division of labor.
regard to the project, because in conjunction with Daniela, we propose the sketch of our program, you for then modify it under the major requirements of Dr.

COMPARISON OF MY WORK WITH THE OTHER:
Actually I do not like the comparisons, I think when seeing the work and ask people they did, with the answers is the contribution and the interest that everyone puts, and so is the best way to qualify or to realize who really do things, who understands you and who you copy.

IMPROVE THE FUTURE:
Referring to a team in the future, it would improve choosing instead to my colleagues, that the qualification that will give us, I imagine it will be a group, and always I work there will be more than others, so it is necessary to take into account the schedules of others, to seek a better union as a team.

BLOGS OF MY PARTNERS:
Daniela Aguilar
Hector Tinajero
Salomon Karr

Palindromes

View more presentations from danielaaguilar .

http://www.slideshare.net/danielaaguilar/palindromos

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**.... ** Project 2 Bin packing "Container Packing"

DESCRIPTION:
The container packing problem , the objects must be packed in a finite number of bins of capacity, while minimizing the use of containers.
There many variations of this problem, as in 2D packaging, packaging line, weight, packaging, packing by cost, and so on, it
try to comprehend and make according to the area of \u200b\u200bcontainer declared by me, by calculating the maximum number of items that can be stored safely.

mathematical definition:

V Given the size bin, and a list ... @ n @ 1, sizes of items to pack, find an integer partition AA of

so that

all

OPTIMA SOLUTION IF YOU HAVE B-minimum, B-value, to denote an optimal solution OPT.



examples of instances, optimal solutions:
The processing algorithm is in random order, for each object, attempts to place the object in the first bin that can hold the object. If not found bin, open a new folder and puts the object at the beginning again. Get

approximation factor of 2.


DECISION PROBLEM: is impossible for 2 containers of being in the majority of the half. The reason is that if at any time was a hub at most half which means having at least one field of V / 2, the algorithm does not open a new folder for any item whose size is at most V / 2. Only after the tray is filled with more than V / 2 or if an item with a size greater than V / 2 arrives, the algorithm can open a new folder.


You have to check if the object I need to save, rate less than the amount of container space.
OB \u0026lt;= MDE
OB = object, while MDE is defined as half the space of the container.


DECISION ALGORITHM:
estimated if the total dimension of the items is less than or equal to the total dimension of a container.
asking the user the actual amounts, and making transactions with the appropriate functions


Explain the asymptotic complexity :
The asymptotic upper bound is of great importance in computational complexity theory when defining classes complexity.

f (x) = O (g (x)) Although containers (g (x)) is defined as a group, it is customary to write f (x) = O (g (x)) instead of f (x) ∈ O (g (x)). Also often speak of a function by naming only its expression, such as x ² instead of h (x) = x ², provided it is clear which is the parameter of the function within the expression. This graph gives a schematic example of how it behaves cg (x) with respect to f (x) when x tends to infinity.

The tight asymptotic bound (Θ notation) is related to the asymptotic upper and lower bounds (notation Ω):
f (x) = Θ (g (x)) if and only if f (x) = O (g (x)) f (x) = Ω (x)
This means that you can save the maximum amount of items not exceeding the space to use the container.



desicion The problem belongs to P and NP, since
recuersos algorithms, it can do iterative, with a more optimal solution, however, the iterative, recursive can not be made, and involved more aspectors , and instead of making them easier to understand, become more complex, decreasing the simplicity is being sought.

If, NP-complete is the subset of decision problems in NP such that any problem in NP can be reduced in each of the NP-complete problems. You could say that the NP-complete problems are NP hard problems and very likely not part of the complexity class P.
Reason is that to be a polynomial solution for NP-complete problem, all NP problems would also have a solution in polynomial time (and therefore, it is shown that for an NP-complete problem there is no solution in polynomial time, none of the NP problems have a solution).



desicion There are several answers, but what is most effective, is to link the capacity of containers, the space equivalent of items in order not to saturate the container, and use the lowest possible, ie :
That if there is a total sum of items for a minimum number of containers, items can be placed evenly distributed, at the very minimum number of containers?

argue if NP-hard.
is NP-Hard, because there are several ways to find a good solution, but in some cases this is not the best, the optimal solution, using a fitting algorithm, first given the quick fix, but not optimal, putting each item in the container, and if they do not fit elsewhere.


recommend using a heuristic algorithm for this type of problem, because in this way, items are sorted according to volume, and relate to the container space, defining that all containers have the same capacity.


ALGORITHM FOR THE OPTIMIZATION PROBLEM:
Define the dimension of the containers, taking into account that everyone has the same capacity and dimension.
dimension is requested for each item, if it is more than one type of item it
And the number of items to keep,
According to the primary amount of items to keep (by multiplying by the overall dimensions, with a same type), this result is subtracted from the total dimension of a container, if you subtract the space to the container, and there are more items on hold, choose, to fill the entire container, not triple the weight of all items , the weight of the container.


Explain asymptotic complexity of this algorithm as well. Usually Landu notation used to refer to the superiorly bounded functions, which depend on other variables to be true which is defined as:

contenedoresf
A (x) belongs to items (g (x)) when there is a positive constant c such that from a number of articles x0, f (x) does not exceed a container (x). Means that the function f is less aga from a given value except for a constant factor.

Best Ways To Masterbate For Males

PROJECT 1, PROJECT 1

we chose to do 3 diagrams, but only finished two, the directory, which is what this acontinuación, assuming that the person who will find the number, knows the alphabet.
Our diagram starts asking the user if it is directory, because as look for a number in a directory, if not a directory. We complicate it a bit at the beginning, we did not know if the algorithm had to throw you a result to the user, but Dr. Elisa, told us they were just as instructions, or good but rather that he understood.
already with some clear ideas, we begin to do the job and ended up in the following.

DIAGRAM

the pseudocode:

# include

# include
int dir;
main () {


do {do {
printf ("You have a directory? \\ n1." If \\ n2. "No \\ n");
scanf ("% d", & dir);
if (dir == 2)
printf ("Get a directory \\ n");
} while (dir == 2);


printf ("Find the section of the first letter the name \\ n ");
getch ();
printf (" Get the name you are looking identical to \\ n ");
getch ();
printf (" I find it? \\ n1 .- If \\ n2. "No \\ n");
scanf ("% d", & dir);

if (dir == 1)

{printf ("\\ nExiste more than a name like \\ n1. "If \\ n2." No \\ n ");
scanf ("% d ", & dir);
if (dir == 1
dir = = 2)

{printf ("\\ NCOMP full name \\ n");
getch ();
printf ("Dig \\ n1." If \\ n2. "No \\ n");
scanf ("% d", & dir);
if (dir == 1)
printf ("\\ nFelicidades SBEs and use a \\ n");

else printf ("\\ n \\ nThe person is not registered in this directory \\ n" );}


} else {

printf ("\\ nThe person is not registered in this directory \\ n");}

getch ();
clrscr ();
printf ("\\ want to make another appointment? \\ N1." If \\ n2. "No \\ n");
scanf ("% d", & dir);
} while (dir == 1);}


EXAMPLE 1

Start asking if the user directory, as in this case, if available, it usually starts giving instructions to the user to search the phone number you need, by the holder of that line.
When not find the name of the person, informs the user that this itself is not registered in the directory that you are looking for.
are asked if you want to another query, Notec, which is inactive when
given the instruction not want to see another number.

EXAMPLE 2

In this case the user has no board, so the machine is cycled, telling the user to get a directory, and asking if you already have, giving the necessary instructions until the user holds a directory in order to follow the instructions.
Here the user wants to make another query so that you repeat the instructions to find the next number of the person you are looking for.
The program does not end until the person says no need to make another query.

USER-MACHINE INTERACTION IS GIVEN, PRESS ENTER OR ANY ONE POINT AFTER EACH TRAINING AND GIVING NUMBERS ACCORDING TO THE CHOICES ARE GIVEN A CHOICE.

GERMAN GARCIA GABRIELA 1410319
VILLARREAL JUAN MANUEL CASANOVA
1453829 GROUP OF TUESDAY.