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Analysis of a Recursive Function

Performing an analysis of a recursive function is not all that different from performing an analysis of a non-recursive function. We are still going to use the same methodology to find a formula that will represent the number of operations required for a given data size. We will then use that formula to find a curve that best describes the resource usage growth rate.

Analysis of Factorial function​

How would we perform an analysis on the following piece of code?

unsigned int factorial(unsigned int n){
    unsigned int rc = 1 ;              //base case result
    if(n > 1) {                 //if n > 1 we  have the recursive case
        rc= n * factorial(n-1);  //rc is n * (n-1)!
    }
    return rc;
}

Start by declaring our mathematical variables and functions

Let nn represent the number we are finding the factorial of

Let T(n)T(n) represent the number of operations it takes to find n! using our recursive function

Similar to iterative version of our code, we will simply count our operations:

unsigned int factorial(unsigned int n){
    unsigned int rc = 1 ;        // 1
    if(n > 1) {                  // 1

        rc= n * factorial(n-1);  // 3 + number of ops done by factorial(n-1)

                                 // There are 3 operators, but it also
                                 // calls factorial(n-1) so we must count
                                 // not just 3, but also all ops done by 
                                 //factorial(n-1)
    }
    return rc;                    //1
}

Now... by definition T(n)T(n) represents the number of ops our function (factorial) requires/does when given nn as its argument. But this also means that T(n−1)T(n-1) would represent the number of ops needed by our function to find (n-1)! (ie number of operations performed by our function when n-1 is passed in as the argument.

Thus, we can actually rewrite our count summary as follows:

unsigned int factorial(unsigned int n){
    unsigned int rc = 1 ;        // 1
    if(n > 1) {                  // 1

        rc= n * factorial(n-1);  // 3 + T(n-1)
    }
    return rc;                   //1
}

Now... how do we solve this though? So, lets start by looking at our base cases. If n <= 1, we do exactly 3 operations:

unsigned int rc = 1 ;
n > 1
return rc;

In otherwords, we can say the following:

T(0)=3T(1)=3T(0) = 3 \\ T(1)=3

Remember that TTis a function that represents how many operations our function requires to find factorial of the argument n. So if argument was 0, it takes 3 ops, if argument was 1, it takes 3 ops also. Notice how the analysis of this coincides with the base case of the recursive function

For n >= 2, we do the following 6 operations + we need count number of ops in the recursive call:

unsigned int rc = 1 ;   <--- 1 op
n > 1                   <--- 1 op
return rc;              <--- 1 op
rc= n * factorial(n-1)  <--- 3 operations, =, * and - ... also need to count
                             number of ops done by factorial(n-1)

Thus, we can write the expression as following for the general T(n)T(n)for n >=2

T(n)=6+T(n−1)T(n) = 6 + T(n-1)

Now... by definition T(n) represents the number of ops needed to find n! using our function. But this also means that T(n-1) would represent the number of ops needed to find (n-1) factorial using our function. Thus,

T(n)=6+T(n−1)T(n) = 6 + T(n-1)

by the same token, we can say that

T(n−1)=6+T(n−2)T(n−2)=6+T(n−3)T(n-1) = 6 + T(n-2) \\ T(n-2) = 6 + T(n-3)

etc...

In other words T(n-1) = 6 + number of operations done by factorial(n-2). Similarly T(n-2) = 6 + number of operations done by factorial(n-3)

The above statement is true for all values of n >= 2. However, we also know that:

T(1)=3T(1) = 3 and T(0)=3T(0) = 3

Thus, what we have is:

T(n)=6+T(n−1)=6+6+T(n−2)=6+6+6+T(n−3)...T(n)=6+6+6+....6+3\begin{aligned} T(n) &= 6 + T(n-1) \\ &= 6 + 6 + T(n-2) \\ &= 6 + 6 + 6+T(n-3) \\ ... \\ T(n) &= 6 + 6 + 6 + .... 6+ 3 \end{aligned}

There are a total of (n-1) 6's.We know this because we need to reach T(1) to get the 3.

Thus:

T(n)=6(n−1)+3=6n−3T(n) = 6(n-1) + 3 = 6n-3

Thus, T(n)T(n) is O(n)O(n)