Question

(15 points) Consider the algorithm for insertion sort shown below. The input to this algorithm is an earray A. You must assume that indexing begins at 1. 1: for j = 2: A.length do key = A i=j-1 while i > 0 and A[i] > key do Ali + 1] = Ai i=i-1 7: A[i+1] = key (a) Follow this algorithm for A[1..4) =< 7,9,6,8 >. Specifically, please indicate the contents of the array after each iteration of the outer for loop. (b) Prove that the outer loop for this algorithm is correct, using the following steps. You may assume for simplicity that the inner while loop is correct (i.e., it inserts the element A[] in its proper position of the sub-array A[1..j]). Given the following pre-condition, loop invariant, and post-condition: • Pre-condition: A[1..n) is an array of integers. • Loop invariant: at the start of each iteration, the sub-array A[1..j - 1] consists of the elements originally in Aſi..j-1], but in sorted order. • Post-condition: the array A[1..n consists of the elements originally in A[l..n], but in sorted order. Prove the following: • Initialization: If the pre-condition is true, then the loop invariant is initially true (prior to the first iteration of the loop). • Maintenance: If the loop invariant is true before an iteration, it remain true before the next iteration • Termination: The loop will eventually terminate. And when it does, the loop invariant implies the post-condition. (20 points) Using the same procedure of the previous question, prove that the following algorithm finds the minimum integer in an array A of n integers (this time, you are asked to write the pre-condition, loop invariant, and post-condition yourself). min = 1 for j = 2 ton if A[D] < A[min] min -

Answer 1

Pre-condition:A[1...n] is an array of integers

Loop invariant:During each iteration the 'min' index value wil be compared with the adjacent value to find the smallest value.

Post condition:'min'wil return the index of the smallest value in the given array A[1..n].

hiven algorithm is min = 1 for s=2 ton 13 A[j] <A (min min =j Let the Asray A be 1716 819] M a ↑ min 1 2 3 4 min = 1 when g=2 A[2]=6 A [min] =7 6< 7 Il condition is tone. min - 2 dej= 3 a 191°W when j = 3 8<6 min=. min A [3] = 8 A [min] = 6 Il condition fails min=2. [7 168 / 9] A [2] =9 A [min] = 6 // conscition fails Hence inden of smallest vakre 1. min=2 min whon j = 4. 1 = 4 946 Scanned with CamScanner