Recurrence for recursive insertion sort

Question Detail: 

I tried this problem from CLRS (Page 39, 2.3-4)

We can express insertion sort as a recursive procedure as follows. In order to sort A[1... n], we recursively sort A[1... n-1] and then insert A[n] into the sorted array A[1... n-1]. Write a recurrence for the running time of this recursive version of insertion sort.

The recurrence I formed was

$$ T(n) = \begin{cases}\Theta(1) & \textrm{if } n = 1,\\ T(n-1) + \Theta(n) & \textrm{if } n > 1. \end{cases} $$

My reasoning

• the base case of $n = 1$ the list is sorted so there is no work hence constant time.
• For all other cases the time depends on sorting the sequence A[1...n-1] and then insertion into that sequence. Hence it should be their sum, i.e., $T(n-1) + \Theta(n)$.

I wanted to know whether the recurrence relation is correct. If not what are the mistakes and how to correctly formulate a recurrence relation?

Asked By : Aseem Bansal

Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/13168

Answered By : Yahav

According to the description you provided the recurrence is correct.
you might think it should be
T(n)=\begin{Bmatrix} 1 & ,\ n=1\\ T(n-1)\ +\ \Theta(log\ n) & ,\ otherwise \end{Bmatrix}
because you can find the insertion place by using Binary-Search, however in order to actually insert the element you'll have to move away all the elements in the worst case.