To show the difference between the output of the array when it is passed as an array for BUILD-MAX-HEAP and BUILD-MAX-HEAP'.

Given Information: A heap that has n-elements.

Explanation:

Pseudo code for BUILD-MAX-HEAP is given below:

BUILD-MAX-HEAP( A )
A.heapsize = A.length

for i = A.length/2 downto 1
MAX-HEAPIFY( A,i )

Consider an array A= {1, 2, 3, 4, 5, 6}. If the above algorithm is used, A.heapsize =6 and i =6/2=3.

The for-loop will iterate from 3 to 1.

At i=3 , MAX-HEAPIFY will be called with arguments as array A and i=3 . The following pseudo code for MAX-HEAPIFY is as follows:

MAX-HEAPIFY( A,i )
l = LEFT( i )

r = RIGHT( i )
ifl <= A.heapsize and A[l]> A[i]

largest = l
Else

largest = i
ifr <= A.heapsize and A[r]> A[largest]

largest=r
iflargest != i

exchange A [ i ] with A [ largest ]
MAX-HEAPIFY( A,largest )

Currently at i=3 , A [3]=3 and have the following tree:

Introduction to Algorithms, Chapter 6, Problem 1P , additional homework tip 1

At the end of the iteration:

Introduction to Algorithms, Chapter 6, Problem 1P , additional homework tip 2

At i=2 , A [2]=2 and after executing the algorithm, the following tree is evolved:

Introduction to Algorithms, Chapter 6, Problem 1P , additional homework tip 3

At i=1 , A [1]=1 and after executing the algorithm, the following tree is evolved:

Introduction to Algorithms, Chapter 6, Problem 1P , additional homework tip 4

Resultant array after BUILD-MAX-HEAP is A= {6, 5, 3, 4, 2, 1}

However, according to the question, the algorithm for BUILD-MAX-HEAP ', there is an alteration in A.heapsize, which is by default 1. Moreover, the value of i starts from 2 and goes on till A.length that is 6.

At i=2 , A [2]=2. The tree is given below-

Introduction to Algorithms, Chapter 6, Problem 1P , additional homework tip 5

Look at the algorithm of BUILD-MAX-HEAP' . It is using MAX-HEAP-INSERT. So, it becomes-

Introduction to Algorithms, Chapter 6, Problem 1P , additional homework tip 6

At i=3 , A [3]=3 and after using MAX-HEAP-INSERT

Introduction to Algorithms, Chapter 6, Problem 1P , additional homework tip 7

At i=4 , A [4]=4 and after using MAX-HEAP-INSERT

Introduction to Algorithms, Chapter 6, Problem 1P , additional homework tip 8

At i=5 , A [5]=5 and after using MAX-HEAP-INSERT

Introduction to Algorithms, Chapter 6, Problem 1P , additional homework tip 9

Finally at i=6,A [6]=6 and after using MAX-HEAP-INSERT

Introduction to Algorithms, Chapter 6, Problem 1P , additional homework tip 10

Resultant array after BUILD-MAX-HEAP' is A= {6, 4, 5, 1, 3, 2}

Hence, the resultant array after BUILD-MAX-HEAP and BUILD-MAX-HEAP' is not the same.

To determine the worst case scenario for BUILD-MAX-HEAP' while entering n-elements.

The BUILD-MAX-HEAP' has MAX-HEAP-INSERT in it, which has worst case of Θ(log n). The call to this function is done for n-1 times, since it is not considering the parent node here.

The worst case for MAX-HEAP-INSERT happens when an array is sorted in ascending order.

Each insert step takes maximum Θ(log n ) steps, and since it is done for n times, it takes a worst case of Θ(nlog n ). Each iteration in the new block is actually taking more time as the older version as it has more iteration in it. The new block will work in even work in even worse case as it has the usage of other algorithm as well.