MERGING SORTING AND
SEARCHINGUNIT 5
INTRODUCTION The related activities of sorting, searching
and merging are central to many computer applications.
Sorting and merging provide us with a means of organizing information take advantage of the organization of information and thereby reduce the amount of effort to either locate a particular item or to establish that it is not present in a data set.
Sorting algorithms arrange items in a set according to a predefined ordering relation.
INTRODUCTION The two most common types of data are
string information and numerical information.
The ordering relation for numeric data simply involves arranging items in sequence from smallest to largest (or vice versa) such that each item is less than or equal to its immediate successor.
This ordering is referred to as non-descending order.
INTRODUCTION Sorted string information is generally
arranged in standard lexicographical or dictionary order.
Sorting algorithms usually fall into one of two classes: The simpler and less sophisticated
algorithms are characterized by the fact that they require of the order of n2 comparisons (i.e. 0(n2)) to sort items.
INTRODUCTIONThe advanced sorting algorithm take of the
order of n log2n (i.e., O(nlog2n)) comparisons to sort n items of data. Algorithms within this set come close to the optimum possible performance for sorting random data
THE TWO WAY MERGE Problem:
Merge two arrays of integers, both with their elements in ascending order into a single ordered array.
SORTING BY EXCHANGE Problem
Given a randomly ordered set of n numbers sort them into non-descending order using exchange method.
SORTING BY EXCHANGEAlmost all sorting methods rely on
exchanging data to achieve the desired ordering.
This method we will now consider relies heavily on an exchange mechanism.
Suppose we start out with the following random data set:
SORTING BY EXCHANGEWe notice that the first two elements are
“out of order”. If 30 and 12 are exchanged we will have the
following configuration:
After seeing the above result we see that the order in the data can be increased further by now comparing and swapping the second and third elements.
SORTING BY EXCHANGE With this new change we get the
configuration
The investigation we have made suggests that the order in the array can be increased using the following steps:
For all adjacent pairs in the array do If the current pair of elements is not in non-
descending order then exchange the two elements.
After applying this idea to all adjacent pairs in our current data set we get the configuration below;
SORTING BY EXCHANGESince there are n elements in the data this
implies that (n-1) passes (of decreasing length) must be made through the array to complete the sort.
SORTING BY EXCHANGE
SORTING BY EXCHANGE
SORTING BY EXCHANGE Algorithm Description
SORTING BY EXCHANGE Algorithm
ApplicationsOnly for sorting data in which a small
percentage of elements are out of order.
SORTING BY INSERTION Problem
Given a randomly ordered set on n numbers sort them into non-descending order using an insertion method.
SORTING BY INSERTIONThis is a simple sorting algorithm that builds
the final sorted array (or list) one item at a time.
Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list.
Each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, and inserts it there.
It repeats until no input elements remain.
SORTING BY INSERTIONSorting is typically done in-place, by
iterating up the array, growing the sorted list behind it.
At each array-position, it checks the value there against the largest value in the sorted list (which happens to be next to it, in the previous array-position checked).
If larger, it leaves the element in place and moves to the next.
If smaller, it finds the correct position within the sorted list, shifts all the larger values up to make a space, and inserts into that correct position.
SORTING BY INSERTION The resulting array after k iterations has the
property where the first k + 1 entries are sorted ("+1" because the first entry is skipped).
In each iteration the first remaining entry of the input is removed, and inserted into the result at the correct position, thus extending the result:
becomes
with each element greater than x copied to the right as it is compared against x.
SORTING BY INSERTIONTo understand this sorting algorithm lets
take up an example
SORTING BY INSERTION
SORTING BY INSERTION
SORTING BY INSERTIONOur complete algorithm can now be
described as:To perform an insertion sort, begin at the
left-most element of the array and invoke Insert to insert each element encountered into its correct position.
The ordered sequence into which the element is inserted is stored at the beginning of the array in the set of indices already examined.
Each insertion overwrites a single value: the value being inserted.
SORTING BY INSERTION Algorithm description
SORTING BY INSERTION Algorithm
ApplicationsWhere there are relatively small data sets. It is sometimes used for more advanced
quick sort algorithm.
SORTING BY DIMINISHING INCREMENT Problem
Given a randomly ordered set on n numbers sort them into non-descending order using Shell’s diminishing increment insertion method.
SORTING BY DIMINISHING INCREMENT Algorithm development
A comparison of random and sorted data sets indicates that for an array of size n elements need to travel on average a distance of about n/3 places.
This observation suggests that progress towards the final sorted order will be quicker if elements are compared and moved initially over longer rather than shorter distances.
This strategy has the effect (on average) of placing each element closer to its final position earlier in sort.
SORTING BY DIMINISHING INCREMENT
A strategy that moves elements over long distances is to take an array of size n and start comparing elements over a distance of n/2 and then successively over the distances n/4, n/8, n/16 and …. 1.
Consider what happens when the n/2 idea is applied to the dataset below
SORTING BY DIMINISHING INCREMENT
After comparisons and exchanges over the distance n/2 we have n/2 chains of length two that are sorted.
The next step is to compare elements over a distance n/4 and thus produce two sorted chains of length 4.
SORTING BY DIMINISHING INCREMENT
Notice that after the n/4 sort the “amount of disorder” in the array is relatively small.
The final step is to form a single sorted chain of length 8 by comparing and sorting elements distance 1 apart.
Since the relative disorder in the array is small towards the end of the sort (i.e. when we are n/8- sorting in this case) we should choose our method for sorting the chains ( an algorithm that is efficient for sorting partially order data).
SORTING BY DIMINISHING INCREMENT
The insertion short should be better because it does not rely so heavily on exchanges.
The next and most important consideration is to apply insertion sorts over the following distances : n/2, n/4, n/8 , … , 1.
We can implement this as follows
SORTING BY DIMINISHING INCREMENT
The next steps in the development are to establish how many chains are to sorted and for each increment gap and then to work out how to access the individual chains for insertion sorting.
We can therefore expand our algorithm to
SORTING BY DIMINISHING INCREMENT Now comes the most crucial stage of the
insertion sort. In standard implementation the first
element that we try is to insert is the second element in the array .
Here for each chain to be sorted it will need to be second element of each chain
The position of k can be given by:
Successive members of each chain beginning with j can be found using
SORTING BY DIMINISHING INCREMENT Algorithm description
SORTING BY DIMINISHING INCREMENT Algorithm
Shellsort.txt
ApplicationsWorks well on sorting large datasets by
there are more advanced methods with better performance.
SORTING BY PARTITIONING Problem
Given a randomly ordered set of n numbers, sort them into non-descending order using Hoare’s partitioning method.
SORTING BY PARTITIONING Algorithm development
Take guess and select an element that might allow us to distinguish between the big and the small elements.
After first pass we have all big elements in the right half of the array and all small elements in the left half of the array.
SORTING BY PARTITIONING
To achieve this do the followingExtend the two partitions inwards until a
wrongly partitioned pair is encountered.While the two partitions have not crossed
Exchange the wrongly partitioned pair; Extend the two partitions inwards again until
another wrongly partitioned pair is encountered.Applying this ideas to the sample data set
SORTING BY PARTITIONING Element 18 is selected as pivot element
This step has given us two independent sets of elements which can be sorted independently.
The basic mechanism to do sort partitions is :
SORTING BY PARTITIONING
While all partitions are not reduced to size one do: Choose next partition to be processed; Select a new partitioning value from the current
partition; Partition the current partition into two smaller
partially ordered sets.
SORTING BY PARTITIONING
After creating partitions of size one do the following: Choose the smaller partition to be processed
next; Select the element in the middle of the partition
as the partitioning value; Partition the current partition into two partially
ordered sets; Save the larger of the partitions from step c for
later processing.
SORTING BY PARTITIONING Algorithm description
SORTING BY PARTITIONING Algorithm
Applications Internal sorting of large datasets.
BINARY SEARCH Problem
Given an element x and a set of data that is in strictly ascending numerical order find whether or not x is present in the set.
BINARY SEARCH Algorithm Development:
BINARY SEARCH Let us now consider an example in order
to try to find the details of the algorithm needed to implement.
Suppose we are required to search an array of 15 ordered elements to find x= 44 is present . If present then return the position of the array that contains 44.
BINARY SEARCHWe start by examining the middle value in
the array.To get the middle value of size n we can try
middle <- n / 2;For the above problem middle value is 8This gives a[middle] = a[8] =39Since the value we are seeking is greater
than 39 it must be somewhere in the range a[9] … a[15].
That is 9 becomes the lower limit and 15 upper limit.
lower = middle +1
BINARY SEARCHWe then have
To calculate the middle index 9 +15 / 2 =12a[12]=49 > 44 so search in a[9] .. a[11].
BINARY SEARCHUsing the same above procedures calculate
the middle position.
Our middle position is 10 and a[10] contains44 which is matching with our key to be found.
Hence return the position 10 .
BINARY SEARCH Algorithm Description
HASH SEARCHING Problem
Design and implement a hash searching algorithm.
HASH SEARCHING Algorithm Description
HASH SEARCHING
HASH SEARCHING Algorithm
Hashsearch.txt
ApplicationsFast retrieval from both small and large
tables