GATE | GATE-CS-2014-(Set-1) | Question 51

Read

Discuss

Consider the following pseudo code. What is the total number of multiplications to be performed?

D = 2
for i = 1 to n do
   for j = i to n do
      for k = j + 1 to n do
           D = D * 3

(A) Half of the product of the 3 consecutive integers.
(B) One-third of the product of the 3 consecutive integers.
(C) One-sixth of the product of the 3 consecutive integers.
(D) None of the above.

Answer: (C)

Explanation: The statement “D = D * 3” is executed n*(n+1)*(n-1)/6 times. Let us see how.
For i = 1, the multiplication statement is executed (n-1) + (n-2) + .. 2 + 1 times.
For i = 2, the statement is executed (n-2) + (n-3) + .. 2 + 1 times
………………………..
……………………….
For i = n-1, the statement is executed once.
For i = n, the statement is not executed at all

So overall the statement is executed following times
[(n-1) + (n-2) + .. 2 + 1] + [(n-2) + (n-3) + .. 2 + 1] + … + 1 + 0

The above series can be written as
S = [n*(n-1)/2 + (n-1)*(n-2)/2 + ….. + 1]

The sum of above series can be obtained by trick of subtraction the series from standard Series S1 = n2 + (n-1)2 + .. 12. The sum of this standard series is n*(n+1)*(2n+1)/6

S1 – 2S = n + (n-1) + … 1 = n*(n+1)/2
2S = n*(n+1)*(2n+1)/6 – n*(n+1)/2
S = n*(n+1)*(n-1)/6

The above solution is relies on a trick to obtain the sum of the series, but there is a much more straightforward way of obtaining the same result :

For i = 1, the multiplication statement is executed (n-1) + (n-2) + .. 2 + 1 times.

For i = 2, the statement is executed (n-2) + (n-3) + .. 2 + 1 times

Therefore, for each i, the number of times the statement is executed is:

$\sum_{k=1}^{n-i}k$

This is the sum of n natural numbers from 1 to n – i. Therefore, it can be simplified to:

$\frac{(n -i) \times (n - i + 1)}{2}$

Multiplying, we will get the following:

$\frac{n^{2}-2ni + n + i^{2} - i}{2}$

Now, this is the number of times the statement will be executed for each i. Therefore, in total, the number of times the statement will be executed is:

$\frac{1}{2}\sum_{i = 1}^{n}\left ( n^{2}-2ni + n + i^{2} - i \right )$