Question

java problem!

Show step! and explain!

Prove that no compare-based algorithm can build a BST using fewer than lg(N !) ~ N lg N compares.

Answer 1

Assume that the comparison based algorithm is represented as A.

• The number of comparisons in the algorithm A on the input of the array of size n in the worst case is represented as C_A(n).
• The worst-case complexity of the comparison based algorithm is
  Ω(nIgn)
  .

Proof of the complexity is explained below:

• The proof of the algorithm uses the sorting model of decision tree.
• For each of the input n, C_A(n) represent the height of the decision tree.
• On at-least one leaf of the decision tree, each permutation of the input n occurs.
• The number of leaves, as the size of the input is n, having at-least n! number of leaves.
• The decision tree which is a simplified decision tree, then it said to be a binary tree.
• The number of leaves of the binary tree with the height, have at-most 2^h number of leaves.

By combining all the information together, the following is obtained:

n! <= Count of the leaves of the decision tree

    <= 2^{height of the decision tree}

    <= 2^C_A⁽ⁿ⁾

Therefore,

C_A(n) >= lg(n!) =

Ω(nIgn)

Use the following, to obtain the inequality:

n^n/2 <= n! <= nⁿ

The above implies the following:

lg n! >= lg(n^n/2)

        = (n/2) lg n

        =

Ω(nIgn)

Hence, the complexity of the comparison based algorithm is no longer than
Ω(nIgn)
.