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# Explain the different data structures that are avaliable to computer programmers, giving examples of their use, and reasons why they would be chosen instead of others.

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Introduction

Explain the different data structures that are avaliable to computer programmers, giving examples of their use, and reasons why they would be chosen instead of others. Introduction Dats structures are one of the most common principles computer operation, the ability to locate, add or delete data is common and used as soon as you turn on your computer system. The fundamental reason for using data structures is that it uses efficient ways of carrying out the above operations when large amounts of data are involved in the calculations. Lists, string, stacks, queues, arrays & trees are some of the most common data structures. They have been adapted from many pre-computing methods, as a queue in its principal is exactly the same as a queue in a shop for items, for example. Linear List A linear list could be considered a one-dimensional array. The list of numbers form what is called a linear list, ie. "5.1, 1.2, .5.9, .3.6, .4.7". Those numbers on themselves are meaningless data, however with a context it becomes information, for example "5.1 is the 0-60 time of a car" would be a suitable context. The data in the list has to have a numeric amount of >=0. Data can be stored inside computers as a linear list. If an item has to be added, then the item of data in the middle of the list, then all the data after the item needs to be inserted after the item to make way for the new item of data. ...read more.

Middle

The pseudocode to delete some data from a queue could be shown as the following: PROCEDURE INSERT(Size, Start_Pointer, Stop_pointer, Data) (*is queue full?*) IF start_pointer = 1 AND Stop_Pointer = Size OR Start_Pointer = Stop_Pointer + 1 THEN PRINT Queue is already full EXIT PROCEDURE ENDIF (*Check to see if queue is empty*) IF Start_pointer = 0 THEN (*initialise queue*) SET Start_pointer = 1 and SET Stop_pointer = 1 (*Queue not empty, update pointers*) ELSE IF Stop_pointer = Size THEN SET Stop_pointer = 1 (*Put stop_pointer back to beginning*) ELSE SET Stop_pointer = Stop_pointer + 1 (*Update stop_pointer*) ENDIF ENDIF QUEUE(Stop_pointer) = Data (*Store data in queue*) END PROCEDURE Arrays An array is ordering of data elements so that information can be extracted from them. The size of an array depends on the number of rows and columns. Most high level languages allow many more than two or three dimensional arrays, however much memory is consumed for multi-dimensional arrays. Eg. A 5-D array containing 10 elements in each D would require: 10 X 10 X 10 X 10 X 10 = 100,000 locations. (if each number could be scored in one location). Arrays must be represented in computers as a linear list ( a 1-D array). To represent an array in a computer's memory requires mapping of each element of the array to the corresponding locations that will score the array. ...read more.

Conclusion

Although this is easier to quickly locate data in this way, it is more difficult to add and delete nodes compared with that of linear lists. It is usual to use some form of stack, so that the route though the tree can be tracked to the previously visited nodes. Binary Trees These are a kind of the 'parent' node is only allowed two terminal nodes. Binary tree structures are implemented using pointer systems in similar ways to the node pointers used with linked lists. Once the child node from the parent node is chosen, the further choices exist in a 'sub-tree' because the root node is no longer entirely accessible. Hash tables A hash funtion is a set of rules when applied to a five digit key field creates a suitable address in the table. The has function is simmialr to a pointer that is used to point to a location where the necessary data is located. A generalised has function is used simmialry to the following example Address = Hash function (key field) Hash function = key field is squared, then taken right- hand digits and finally add 1. Using the hashing function for the following number: (12345) we get: Original Number|Number squared|Right-hand three digits|Right-hand three digits ADD 1 12345 152399025 025 26 One disadvantage of the hashing function is their ability to create the same address within the table for different key fields. This is known as a collision. ?? ?? ?? ?? 1 2.05 ...read more.

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