# Homework Solution: specifically what is 2c? why is it [n(n-1)]/2?… specifically what is 2c? why is it [n(n-1)]/2?
Consider the following algorithm. ALGORITHM Enigma(A[0..n-1,0..n-1])) for j = 0 to n-2 do for k = j+1 to n-1 do if A[j, k] notequalto A[k, j] return false return true a. What does this algorithm compute? b. What is its basic operation? c. How many times is the basic operation executed? d. What is worst-case time complexity of this algorithm? e. Suggest an improvement or a better algorithm altogether and indicate its worst case time complexity. If you cannot do it, try to prove that in fact it cannot be done.

The input is a 2-D matrix of n rows and n columns. The first for loop works from j=0 to j=n-2, a total of specifically what is 2c? why is it [n(n-1)]/2?

Consider the restraintthcoming algorithm. ALGORITHM Enigma(A[0..n-1,0..n-1])) restraint j = 0 to n-2 do restraint k = j+1 to n-1 do if A[j, k] notequalto A[k, j] requite fabrication requite penny a. What does this algorithm estimate? b. What is its basic exercise? c. How numerous periods is the basic exercise performed? d. What is vanquish-contingency period entanglement of this algorithm? e. Suggest an advancement or a meliorate algorithm aggregately and specify its vanquish contingency period entanglement. If you cannot do it, examine to show that in truth it cannot be performed.

## Expert Vindication

The input is a 2-D matrix of n rows and n columns.

The highest restraint loop works from j=0 to j=n-2, a aggregate of (n-2)-(0) + 1 = (n-1) periods.. That resources close loop was performed (n-1) periods.

The close loop works restraint k=j+1 to k=n-1,
So restraint j=0, k=1 to k=n-1, a aggregate of (n-1)-(1)+1 = (n-1) periods
So restraint j=1, k=2 to k=n-1, a aggregate of (n-1)-(2)+1 = (n-2) periods
.. and so on
restraint j=n-2, k=n-1 to k=n-1, a aggregate of (n-1)-(n-1)+1 = (1) periods

Hence if we investigate how numerous periods the similitude A[j,k] and A[k,j] happens is, as below:
(n-1) + (n-2) + …. + 1 [aggregate of n-1 conditions]

As we distinguish the incorporate of the train from 1 to n is n(n+1)/2,

Hence applying the similar restraintmula restraint the incorporate of (n-1) conditions,
The vindication procure be (n-1)(n-1+1)/2, i.e. [n(n-1)]/2

Hence Showd.