Question

Show the Big O Complexity of the following functions and loop constructions: (Please show work and explain)

a. f(n) = 2n + (blog(n+1))

b. f(n) = n * (log(n-1))/2

c.

int sum = 0;
for (int i=0; i<n; i++)

sum++;
for (int j=n; j>0; j /= 2)

sum++;

d.

int sum = 0;
for (int i=n; i>0; i--)

for (int j=i; j<n; j *= 2)

sum++;

Answer 1

a.

f(n) = 2n + (blog(n+1))

It is also written as (2)n * (b) log(n+1)

Here 2 and b are constant values so it is negligible.

log(n+1) is equivalent to the log(n)

f(n) = n + logn

Here n is bigger than logn for any large value of n so logn is negligible.

f(n) complexity is O(n)

b.

f(n) = n * (log(n-1))/2

It is also written as n * (1/2) log(n-1)

Here 1/2 is constant value so it is negligible.

log(n-1) is equivalent to the log(n)

f(n) complexity is O(nlog n)

c.

int sum = 0; // It runs for 1 time so its complexity is O(1)
for (int i=0; i<n; i++) // It runs for n times so its complexity is O(n)

sum++; // It runs for every for loop so its complexity is O(n)

for (int j=n; j>0; j /= 2) // Here J is reduced to half every time so its complexity is O(log n)

sum++; // It runs for every for loop so its complexity is O(log n)

The complexity of whole program is O(n)

d.

int sum = 0; // It runs for 1 time so its complexity is O(1)
for (int i=n; i>0; i--)   // It runs for n times so its complexity is O(n)

for (int j=i; j<n; j *= 2) // Here j doubles every time so its complexity is O(logn) and also runs for every value of i so overall complexity is O(nlogn)

sum++; // Its complexity is O(nlogn)

sum++;

Overall complexity is O(nlogn)