Shortness Of Breath With Jelly Earphones

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

Dryer Duct Cleaning Contract

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

Reliable Brand Washer

**.... ** 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.