Reading:  Insertion Sort

Abstract

Many applications require that data be ordered and maintained. Some examples of humans manually using insertion sort include sorting a deck of cards, or sorting books on a bookshelf.

In C and many other languages, a simple way to structure ordered data is in an array (or vector), such as the following ordered array A of integers.

The Insertion Sort

One common sorting approach is based on code that assumes that the first part of an array is ordered and then adds successive items to this array segment until the entire array is sorted. To understand this approach, we first consider how to add one item to an ordered array segment. We then apply this work to each array element in turn to yield an ordered array.

Maintaining An Ordered Array Segment

Suppose items A[0], ..., A[k-1] are ordered in array A:

To insert an item into the first part of this array and maintain the ordering, we proceed from right to left with elements in the array.

• We compare item to successive elements at the right of the array.

  • If the array element is bigger than item, we move the array element to the right.

• Once we move elements larger than item to the right, there is room to insert item in its place.

For example, the following figure shows the insertion of the number 15 into the array A above.

The following code translates this idea into C. When this code terminates, item is placed into the array, so that items A[0], ..., A[k] become ordered:

   int i = k-1;
   while ((i >= 0) && a[i] > item){
      a[i+1] = a[i];
      i--;
   }
a[i+1] = item;

Using this basic insertion step, an array A can be sorted iteratively according to the following outline:

• Consider A[0] as a sorted initial segment of array A. (That is, in the example, consider k = 0.)
• Insert A[1] into segment A[0] to get ordered segment A[0], A[1].
• Insert A[2] into segment A[0], A[1] to get ordered segment A[0], A[1], A[2].
• Insert A[3] into segment A[0], ..., A[2] to get ordered segment A[0], ..., A[3].
• Insert A[4] into segment A[0], ..., A[3] to get ordered segment A[0], ..., A[4].
• Et cetera.

This outline gives rise the the following code, called an insertion sort.

void insertionSort (int a [], int length) {
// method to sort using the insertion sort
// parameters:  a, the array to be sorted
//              length, the size of the a array
   for (int k = 1; k < length; k++) {
      int item = a[k];
      int i = k-1;
      while ((i >= 0) && a[i] > item){
         a[i+1] = a[i];
         i--;
      }
      a[i+1] = item;
   }
}

It is important to notice that the item being compared is not switching place with each item it compares to. The items that are compared against are shifted; the element being compared is inserted.

Insertion Sort Characteristics

This sorting algorithm has several helpful properties.
The insertion sort is

• relatively simple,

• efficient for small data sets and data sets that are already partially sorted,

• stable in the sense that it does not switch identical elements, and

• only requires a constant amount of additional memory space (that is, the amount of memory for the set of elements, plus memory for one element of the set).

The amount of time this algorithm takes to sort a set is dependent upon the set itself.

• For any data set, each successive array element must be compared to the array element just before it.

• In the best case, each successive array element will already be larger than its predecessor, and no array elements will need to be moved.

• In the worse case, each successive array element will be smaller than its predecessors, and all array elements on the left will need to be moved to the right with each insertion.

• If the array data start in random order, each successive array element will be larger than some elements to its left and smaller than other. On average, we might have to shift about half elements on the left over to make room for successive array elements.

Overall, for a set with n elements, if the set is sorted (or nearly so), the algorithm takes only a single time unit for each element, so it takes n time units. On the other hand, if the set is in reverse order (that is, largest to smallest), the time approaches n-squared. You will see the previous two statements proven in a later course.