I have a Computer Science Midterm tomorrow and I need help determining the complexity of these recursive functions. I know how to solve simple cases, but I am still trying to learn how to solve these harder cases. These were just a few of the example problems that I could not figure out. Any help would be much appreciated and would greatly help in my studies, Thank you!
int recursiveFun1(int n)
{
if (n <= 0)
return 1;
else
return 1 + recursiveFun1(n-1);
}
int recursiveFun2(int n)
{
if (n <= 0)
return 1;
else
return 1 + recursiveFun2(n-5);
}
int recursiveFun3(int n)
{
if (n <= 0)
return 1;
else
return 1 + recursiveFun3(n/5);
}
void recursiveFun4(int n, int m, int o)
{
if (n <= 0)
{
printf("%d, %d\n",m, o);
}
else
{
recursiveFun4(n-1, m+1, o);
recursiveFun4(n-1, m, o+1);
}
}
int recursiveFun5(int n)
{
for (i = 0; i < n; i += 2) {
// do something
}
if (n <= 0)
return 1;
else
return 1 + recursiveFun5(n-5);
}
The time complexity, in Big O notation, for each function:
int recursiveFun1(int n)
{
if (n <= 0)
return 1;
else
return 1 + recursiveFun1(n-1);
}
This function is being called recursively n times before reaching the base case so its O(n)
, often called linear.
int recursiveFun2(int n)
{
if (n <= 0)
return 1;
else
return 1 + recursiveFun2(n-5);
}
This function is called n-5 for each time, so we deduct five from n before calling the function, but n-5 is also O(n)
.
(Actually called order of n/5 times. And, O(n/5) = O(n) ).
int recursiveFun3(int n)
{
if (n <= 0)
return 1;
else
return 1 + recursiveFun3(n/5);
}
This function is log(n) base 5, for every time we divide by 5
before calling the function so its O(log(n))
(base 5), often called logarithmic and most often Big O notation and complexity analysis uses base 2.
void recursiveFun4(int n, int m, int o)
{
if (n <= 0)
{
printf("%d, %d\n",m, o);
}
else
{
recursiveFun4(n-1, m+1, o);
recursiveFun4(n-1, m, o+1);
}
}
Here, it's O(2^n)
, or exponential, since each function call calls itself twice unless it has been recursed n times.
int recursiveFun5(int n)
{
for (i = 0; i < n; i += 2) {
// do something
}
if (n <= 0)
return 1;
else
return 1 + recursiveFun5(n-5);
}
And here the for loop takes n/2 since we're increasing by 2, and the recursion takes n/5 and since the for loop is called recursively, therefore, the time complexity is in
(n/5) * (n/2) = n^2/10,
due to Asymptotic behavior and worst-case scenario considerations or the upper bound that big O is striving for, we are only interested in the largest term so O(n^2)
.
Good luck on your midterms ;)