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 reasoningthe 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?
Recurrence for recursive insertion sort
algorithm analysis;runtime analysis;recurrence relation;sorting
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.