Example-Fibonacci Numbers

In this section we will compare the asymptotic running times of two different programs that both compute Fibonacci numbers.

The Fibonacci numbers are the series of numbers

, ..., given by

Fibonacci numbers are interesting because they seem to crop up in the most unexpected situations. However, in this section, we are merely concerned with writing an algorithm to compute

given n.

Fibonacci numbers are easy enough to compute. Consider the sequence of Fibonacci numbers

The next number in the sequence is computed simply by adding together the last two numbers--in this case it is 55=21+34. Program

is a direct implementation of this idea. The running time of this algorithm is clearly O(n) as shown by the analysis in Table

statement time

5 O(1)

6 O(1)

7a O(1)

7b

7c

9

10

11

13 O(1)

TOTAL O(n)

Table: Computing the running time of Program .

**Table:** Computing the running time of Program .
statement	time
5	O(1)
6	O(1)
7a	O(1)
7b
7c
9
10
11
13	O(1)
TOTAL	O(n)

Recall that the Fibonacci numbers are defined recursively:

. However, the algorithm used in Program

is non-recursive --it is iterative . What happens if instead of using the iterative algorithm, we use the definition of Fibonacci numbers to implement directly a recursive algorithm ? Such an algorithm is given in Program

and its running time is summarized in Table

time

statement
n<2

5 O(1) O(1)

6 O(1) --

8 -- T(n-1)+T(n-2)+O(1)

TOTAL O(1) T(n-1)+T(n-2)+O(1)

Table: Computing the running time of Program .

**Table:** Computing the running time of Program .
	time
statement	n<2
5	O(1)	O(1)
6	O(1)	--
8	--	T(n-1)+T(n-2)+O(1)
TOTAL	O(1)	T(n-1)+T(n-2)+O(1)

From Table

we find that the running time of the recursive Fibonacci algorithm is given by the recurrence

It turns out that there is a simple trick we can use to solve a recurrence containing big oh expressions as long as we are only interested in an asymptotic bound on the result. Simply drop the

s from the recurrence, solve the recurrence, and put the

back! In this case, we need to solve the recurrence

In the previous chapter, we used successfully repeated substitution to solve recurrences. However, in the previous chapter, all of the recurrences only had one instance of

on the right-hand-side--in this case there are two. As a result, repeated substitution won't work.

There is something interesting about this recurrence: It looks very much like the definition of the Fibonacci numbers. In fact, we can show by induction on n that

for all

So, we can now say with certainty that the running time of the recursive Fibonacci algorithm, Program

, is

. But is this good or bad? The following theorem shows us how bad this really is!

Inductive Hypothesis Suppose that Equation

holds for

for some

. First, we make the following observation:

Theorem

gives us that

where

and

. Consider

. A couple of seconds with a calculator should suffice to convince you that

. Consequently, as n gets large,

is vanishingly small. Therefore,

. In asymptotic terms, we write

. Now, since

, we can write that

Returning to Program

, recall that we have already shown that its running time is

. And since

, we can write that

. That is, the running time of the recursive Fibonacci program grows exponentially with increasing n. And that is really bad in comparison with the linear running time of Program

Figure

shows the actual running times of both the non-recursive and recursive algorithms for computing Fibonacci numbers.

Because the largest C# int is 2147483647, it is only possible to compute Fibonacci numbers up to

before overflowing.

The graph shows that up to about n=35, the running times of the two algorithms are comparable. However, as n increases past 40, the exponential growth rate of Program

is clearly evident. In fact, the actual time taken by Program

to compute

was in excess of two and a half minutes!