Think of arranging cards for a rummy match. The cards are distributed evenly among players and each players arranges the card sequentially from a pile of card by hand.

You pick the first card and consider it as sorted.

Next, you pick another card and insert it in right position. Similarly u continue inserting the cards. Note that, mostly you take the top most card from the pile in every pick.

Implement the above methodology for sorting numbers.

So, you have,

- List of numbers fully in random order
- pick the first number by default
- take the second number and check where it lies with the first number
- now, the first 2 numbers in the list are sorted
- continue till the end of the list and always create a sorted list in each iteration

**Algorithm:**

You can refer the pseudo code from "Introduction to Algorithms - By Cormen, Rivest". Just jotting down my understanding on the on-place algorithm given on the book.

__Important points:__

* When you sort a list on-place without using extra space, you should create a hole so that numbers can be shifted. So, store the number you pick from unsorted list in a temporary variable and use that as a hole to shift.

* MOST IMPORTANT PROPERTY OF INSERTION SORT, on each iteration you get the numbers sorted at one side of the list. that is, in 1st iteration, first 2 numbers in the list will be sorted,

2nd, first 3 numbers will be sorted

3rd, first 4 numbers will be sorted

So, if you just want partially sorted list(that is only 1/2 list is required), you can stop at the required point.

* Note that the above tells that, every iteration creates a sorted list always. This is the invariant condition. An INVARIANT is an assertion that doesn't change in executing a piece of code. A LOOP INVARIANT is something that holds at the beginning of a loop, and continues to hold after each iteration. Usually, an invariant is used to prove correctness of algorithm

* We need 2 loops, one for taking a number from the list(j). another is for inserting the number at the right position by rearranging (i)

* “i” will run in reverse (exits when it becomes zero) and reorders the array to insert the element at right position.

* “i” will be set to the “j” as “j” will be always the end of the sorted list after every iteration.

**Code**

1: void insertion_sort(int *arr, int n)

` 2: {`

3: for(int j=1; j<n;j++)

` 4: {`

5: int key = arr[j];

6: int i=j-1;

7: while(i>=0) {

8: if(arr[i] < key)

` 9: {`

` 10: arr[i+1] = arr[i];`

` 11: i = i - 1;`

` 12: }`

13: else

14: break;

` 15: }`

` 16: arr[i+1] = key;`

` 17: }`

` 18: }`

` 19: `

20: void main()

` 21: {`

22: int test_arr[] = {5,3,2,6,8,9};

` 23: insertion_sort(test_arr, 6);`

` 24: }`

**Exercise problems from the book:**- Linear search problem - find the loop invariant and prove the algorithm

<http://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15211/fall.97/www/lectures/lect0916>

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