Ganon11 3,652
Recognized Expert Specialist
So far, the Articles sections are filled with slow sorting algorithms. Bubble Sort and Selection Sort, while very easy to comprehend, are relatively slow algorithms at Θ(n^2) running time. Here, I will try to explain an algorithm to sort an array of numbers that is fast for small sets of data, though it is still an O(n^2) algorithm; the Insertion Sort.
Suppose we have an array of n-1 elements that is already sorted in increasing order. We then come to the nth element (let us call it x) to be ‘inserted’ into this sorted array. It follows that there is some portion of the array in which the elements are less than x, and another portion in which the elements are greater than x. Further, because we know that the existing array is sorted, we know that the array generally looks like this:
[elements < x][elements>x]
We begin the algorithm by creating a hole at the end of the array:
[elements<x][elements>x][empty hole]
Now, we want to have our array completely sorted for all n elements at the end, so the alert reader may realize that the array must look like this when we are done at this step:
[elements<x][x][elements>x]
It follows that all of the elements greater than x must shift down the array one slot. We can, in fact, do this very simply by moving any element a[j] to a[j+1] as long as a[j] is greater than x. We start this moving at the element just before the hole and continue until the element we check is less than x. Now every element has been shifted one slot and overwritten every previously copied value except for the last to be shifted, of which there remains 2 copies adjacent to each other. We know that this element is greater than x (or else we would not have shifted it) and so we place x in the position that element originally held.
A powerful note to this algorithm is that it can be implemented to use no extra space. If we consider only the first I elements of any array as our sorted array, then creating a hole is as simple as increasing our array’s size, making a copy of the last term (which must be inserted into its sorted position), and following our steps above. The algorithm in pseudocode is presented below: - void insertion_sort(int[] array) {
-
for (Loop i from 1 to array.size-1)
-
int x = array[i]
-
int j = array[i-1]
-
while (j >= 0 && array[j] > x)
-
array[j+1] = array[j]
-
j = j - 1
-
array[j+1] = x
-
}
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Similar topics |
by: N. |
last post by:
Insertion sort is O(N^2), but I figure I can get it in to O(N log2 N)
if the inside loop of the insertion sort is replaced with a binary
search. However, I'm having some implimentation problems...
void insertsort(int a, int n)
{
int i, j, index;
for (i=0; i < n; i++)
{
index = a;
|
by: ashu |
last post by:
can anyone tell me that what is the logic of insertion sort.
thank you
|
by: Franky, Mondestin |
last post by:
Problem description:
I have a text file with 700 lines. Each line has 7 elements separated by
comma. My text file looks something like :
a,b,c,d,e,f,g
d,e,h,k,l,m,n
x,c,v,f,g,t,z
.... and so on
Help needed
I'd like to read this file, place it in an array, sort the array by the
|
by: Am |
last post by:
hi
i came to know that microsoft improved the efficiency of quick sort
by using a cutoff
of 8 elements and continuing with insertion sort then, do anybody have
the details about it
please contant me.
|
by: Am |
last post by:
hi
i came to know that microsoft improved the efficiency of quick sort
by using a cutoff
of 8 elements and continuing with insertion sort then, do anybody have
the details about it
please contant me.
| |
by: Julia |
last post by:
I am trying to sort a linked list using insertion sort. I have seen a
lot of ways to get around this problem but no time-efficient and
space-efficient solution. This is what I have so far:
struct node
{
int x;
struct node *next;
};
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by: tkpmep |
last post by:
I have an ordered list e.g. x = , and given any
positive integer y, I want to determine its appropriate position in
the list (i.e the point at which I would have to insert it in order to
keep the list sorted. I can clearly do this with a series of if
statements:
if y<x:
n = 0
elif y < x:
n = 1
|
by: AhmedGY |
last post by:
Hi, am trying to build an app that uses the insertion sort method to sort numbers entered in a textbox and display them sorted in a label, so i wrote this inside the sort button click event:
label1.Text = Insertion.insert(textBox1.Text);
and here is the insertion class:
class Insertion
{
public static string insert(string x)
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I don't know this list is the right place for newbie questions. I try
to implement insertion sort in pyhton. At first code there is no
problem. But the second one ( i code it in the same pattern i think )
doesn't work. Any ideas ?
------------------------------------------------------------
def insertion_sort(aList):
for i in range(len(aList)):
for j in range(i):
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