Determining complexity for recursive functions (Big O notation)

Solution 1:

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 ;)

Solution 2:

For the case where n <= 0, T(n) = O(1). Therefore, the time complexity will depend on when n >= 0.

We will consider the case n >= 0 in the part below.

1.

T(n) = a + T(n - 1)

where a is some constant.

By induction:

T(n) = n * a + T(0) = n * a + b = O(n)

where a, b are some constant.

2.

T(n) = a + T(n - 5)

where a is some constant

By induction:

T(n) = ceil(n / 5) * a + T(k) = ceil(n / 5) * a + b = O(n)

where a, b are some constant and k <= 0

3.

T(n) = a + T(n / 5)

where a is some constant

By induction:

T(n) = a * log5(n) + T(0) = a * log5(n) + b = O(log n)

where a, b are some constant

4.

T(n) = a + 2 * T(n - 1)

where a is some constant

By induction:

T(n) = a + 2a + 4a + ... + 2^(n-1) * a + T(0) * 2^n 
     = a * 2^n - a + b * 2^n
     = (a + b) * 2^n - a
     = O(2 ^ n)

where a, b are some constant.

5.

T(n) = n / 2 + T(n - 5)

where n is some constant

Rewrite n = 5q + r where q and r are integer and r = 0, 1, 2, 3, 4

T(5q + r) = (5q + r) / 2 + T(5 * (q - 1) + r)

We have q = (n - r) / 5, and since r < 5, we can consider it a constant, so q = O(n)

By induction:

T(n) = T(5q + r)
     = (5q + r) / 2 + (5 * (q - 1) + r) / 2 + ... + r / 2 +  T(r)
     = 5 / 2 * (q + (q - 1) + ... + 1) +  1 / 2 * (q + 1) * r + T(r)
     = 5 / 4 * (q + 1) * q + 1 / 2 * (q + 1) * r + T(r)
     = 5 / 4 * q^2 + 5 / 4 * q + 1 / 2 * q * r + 1 / 2 * r + T(r)

Since r < 4, we can find some constant b so that b >= T(r)

T(n) = T(5q + r)
     = 5 / 2 * q^2 + (5 / 4 + 1 / 2 * r) * q + 1 / 2 * r + b
     = 5 / 2 * O(n ^ 2) + (5 / 4 + 1 / 2 * r) * O(n) + 1 / 2 * r + b
     = O(n ^ 2)

Solution 3:

One of the best ways I find for approximating the complexity of the recursive algorithm is drawing the recursion tree. Once you have the recursive tree:

Complexity = length of tree from root node to leaf node * number of leaf nodes

1. The first function will have length of n and number of leaf node 1 so complexity will be n*1 = n

2. The second function will have the length of n/5 and number of leaf nodes again 1 so complexity will be n/5 * 1 = n/5. It should be approximated to n

3. For the third function, since n is being divided by 5 on every recursive call, length of recursive tree will be log(n)(base 5), and number of leaf nodes again 1 so complexity will be log(n)(base 5) * 1 = log(n)(base 5)

4. For the fourth function since every node will have two child nodes, the number of leaf nodes will be equal to (2^n) and length of the recursive tree will be n so complexity will be (2^n) * n. But since n is insignificant in front of (2^n), it can be ignored and complexity can be only said to be (2^n).

5. For the fifth function, there are two elements introducing the complexity. Complexity introduced by recursive nature of function and complexity introduced by for loop in each function. Doing the above calculation, the complexity introduced by recursive nature of function will be ~ n and complexity due to for loop n. Total complexity will be n*n.

Note: This is a quick and dirty way of calculating complexity(nothing official!). Would love to hear feedback on this. Thanks.