now consider a function that calculates partial

Question

Now, consider a function that calculates partial sum of an integer n. int psum(int n)

{

int i, partial_sum;

partial_sum = 0;                                           /* Line 1 */

for (i = 1; i <= n; i++) {                                /* Line 2 */

partial_sum = partial_sum + i*i;            /* Line 3 */

}

return partial_sum;                                                 /* Line 4 */

}

This function returns the sum by i = 1 to n of i squared, which means p sum = 1² + 2²+ 3²

+ .............  + n² .

Ø  As we ought to determine the running time for each of statement in this program, we ought to count the number of statements which are executed in this process. The code at line 1 & line 4 are one statement each. Actually the for loop on line 2 are 2n+2 statements:

• i = 1; statement: simple assignment, therefore one statement.
• i <= n; statement is executed once for each value of i from 1 to n+1 (until the condition becomes false). The statement is executed n+1 times.
• i++ is executed once for each of execution of body of the loop. It is executed for n times.

Therefore, the sum is equal to 1+ (n+1) + n+1 = 2n+ 3 times.

In terms of big-O notation described above, this function is O (n), since if we choose c=3, then we notice that c_n > 2n+3. As we have already illustrious earlier, big-O notation only provides a upper bound to the function, it is also O(nlog(n)) & O(n²), since n² > nlog(n) > 2n+3. However, we will select the smallest function which describes the order of the function and it is O (n).

Through looking at the definition of Omega notation & Theta notation, it is also apparent that it is of Θ(n), and thus ?(n) too. Because if we select c=1, then we see that cn < 2n+3, therefore ?(n) . Since 2n+3 = O(n), & 2n+3 = ?(n), this  implies that 2n+3 = Θ(n) , too.

Again it is reiterated here that smaller order terms and constants may be avoided while describing asymptotic notation. For instance, if f(n) = 4n+6 rather than f(n) = 2n +3 in terms of big-O, ? and Θ, It does not modify the order of the function. The function f(n) = 4n+6 = O(n) (through choosing c appropriately as 5); 4n+6 = ?(n) (through choosing c = 1), and thus 4n+6 = Θ(n). The spirit of this analysis is that in these asymptotic notation, we may count a statement as one, and should not worry regarding their relative execution time that may based on several hardware and other implementation factors, as long as this is of the order of 1, that means O(1).