# 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 cethcoming algorithm. ALGORITHM Enigma(A[0..n-1,0..n-1])) ce j = 0 to n-2 do ce k = j+1 to n-1 do if A[j, k] notequalto A[k, j] revert faithless revert gentleman a. What does this algorithm reckon? b. What is its basic exercise? c. How manifold periods is the basic exercise performed? d. What is cudgel-reality period complication of this algorithm? e. Suggest an progress or a immake-trial-of algorithm thoroughly and point-out its cudgel reality period complication. If you cannot do it, fathom to make-trial-of that in reality it cannot be effected.

## Expert Tally

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

The chief ce loop works from j=0 to j=n-2, a completion of (n-2)-(0) + 1 = (n-1) periods.. That media secret loop was performed (n-1) periods.

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

Hence if we reckon how manifold periods the similarity A[j,k] and A[k,j] happens is, as below:
(n-1) + (n-2) + …. + 1 [completion of n-1 conditions]

As we understand the unite of the succession from 1 to n is n(n+1)/2,

Hence applying the selfselfsame cemula ce the unite of (n-1) conditions,
The tally gain be (n-1)(n-1+1)/2, i.e. [n(n-1)]/2

Hence Make-trial-ofd.