Recursive Functions
Definition
Recursion has something to do with infinity. I know recursion has something to do with infinity.
I think I know recursion has something to do with infinity. He is sure I think I know recursion
has something to do with infinity. We doubt he is sure I think I know ...
We think that you think that we convinced you now that we can go on forever with this example of a
recursion from natural language. Recursion is not only a fundamental feature of natural language,
but of the human cognitive capacity. Our way of thinking
is based on a recursive thinking processes. Even with a very simple grammar, like "An English sentence
contains a subject and a predicate, and a predicate contains a verb, an object and a complement", we can
demonstrate the infinite possibilities of the natural language. The cognitive scientist and linguist
Stephen Pinker phrases it like this: "With a few thousand nouns that can fill the subject slot and a few
thousand verbs that can fill the
predicate slot, one already has several million ways to open a sentence. The possible combinations
quickly multiply out to unimaginably large numbers. Indeed, the repertoire of sentences is theoretically
infinite, because the rules of language use a trick called recursion. A recursive rule allows a phrase
to contain an example of itself, as in She thinks that he thinks that they think that he knows
and so on, ad infinitum. And if the number of sentences is infinite, the number of possible thoughts
and intentions is infinite too, because virtually every sentence expresses a different thought or
intention."^{1}
We have to stop our short excursion to the use of recursion in natural language to come back to
recursion in computer science and programs and finally to recursion in the programming language Python.
The adjective "recursive" originates from the Latin verb "recurrere", which means "to run back".
And this is what a recursive definition or a recursive function does: It is "running back" or returning
to itself. Most people who have done some mathematics, computer science or read a book about programming
will have encountered the factorial, which is defined in mathematical terms as
n! = n * (n1)!, if n > 1 and f(1) = 1
It's used so often as an example for recursion because of its simplicity and clarity.
We will come back to it in the following.
Definition of Recursion
Recursion is a method of programming or coding a problem, in which a function calls itself one or more times in its body. Usually, it is returning the return value of this function call. If a function definition satisfies the condition of recursion, we call this function a recursive function.Termination condition:
A recursive function has to fulfil an important condition to be used in a program: it has to terminate. A recursive function terminates, if with every recursive call the solution of the problem is downsized and moves towards a base case. A base case is a case, where the problem can be solved without further recursion. A recursion can end up in an infinite loop, if the base case is not met in the calls.
Example:
4! = 4 * 3!
3! = 3 * 2!
2! = 2 * 1
Replacing the calculated values gives us the following expression
4! = 4 * 3 * 2 * 1
Generally we can say: Recursion in computer science is a method where the solution to a problem is based on solving smaller instances of the same problem.
Recursive Functions in Python
Now we come to implement the factorial in Python. It's as easy and elegant as the mathematical definition.def factorial(n): if n == 1: return 1 else: return n * factorial(n1)We can track how the function works by adding two print() function to the previous function definition:
def factorial(n): print("factorial has been called with n = " + str(n)) if n == 1: return 1 else: res = n * factorial(n1) print("intermediate result for ", n, " * factorial(" ,n1, "): ",res) return res print(factorial(5))This Python script outputs the following results:
factorial has been called with n = 5 factorial has been called with n = 4 factorial has been called with n = 3 factorial has been called with n = 2 factorial has been called with n = 1 intermediate result for 2 * factorial( 1 ): 2 intermediate result for 3 * factorial( 2 ): 6 intermediate result for 4 * factorial( 3 ): 24 intermediate result for 5 * factorial( 4 ): 120 120Let's have a look at an iterative version of the factorial function.
def iterative_factorial(n): result = 1 for i in range(2,n+1): result *= i return result
The Pitfalls of Recursion
This subchapter of our tutorial on recursion deals with the Fibonacci numbers. What do have sunflowers, the Golden ratio, fir tree cones, The Da Vinci Code, the song "Lateralus" by Tool, and the graphic on the right side in common. Right, the Fibonacci numbers.The Fibonacci numbers are the numbers of the following sequence of integer values:
0,1,1,2,3,5,8,13,21,34,55,89, ...
The Fibonacci numbers are defined by:
F_{n} = F_{n1} + F_{n2}
with
F_{0} = 0
and F_{1} = 1
The Fibonacci sequence is named after the mathematician Leonardo of Pisa, who is better known as Fibonacci. In his book "Liber Abaci" (publishes 1202) he introduced the sequence as an exercise dealing with bunnies. His sequence of the Fibonacci numbers begins with F_{1} = 1, while in modern mathematics the sequence starts with F_{0} = 0. But this has no effect on the other members of the sequence.
The Fibonacci numbers are the result of an artificial rabbit population, satisfying the following conditions:
 a newly born pair of rabbits, one male, one female, build the initial population
 these rabbits are able to mate at the age of one month so that at the end of its second month a female can bring forth another pair of rabbits
 these rabbits are immortal
 a mating pair always produces one new pair (one male, one female) every month from the second month onwards
The Fibonacci numbers are easy to write as a Python function. It's more or less a one to one mapping from the mathematical definition:
def fib(n): if n == 0: return 0 elif n == 1: return 1 else: return fib(n1) + fib(n2)An iterative solution is also easy to write, though the recursive solution looks more like the definition:
def fibi(n): old, new = 0, 1 if n == 0: return 0 for i in range(n1): old, new = new, old + new return newIf you check the functions fib() and fibi(), you will find out that the iterative version fibi() is a lot faster than the recursive version fib(). To get an idea of how much this "a lot faster" can be, we have written a script, which uses the timeit module, to measure the calls. To do this, we save the function definitions for fib and fibi in a file fibonacci.py, which we can import in the program (fibonacci_runit.py) below:
from timeit import Timer t1 = Timer("fib(10)","from fibonacci import fib") for i in range(1,41): s = "fib(" + str(i) + ")" t1 = Timer(s,"from fibonacci import fib") time1 = t1.timeit(3) s = "fibi(" + str(i) + ")" t2 = Timer(s,"from fibonacci import fibi") time2 = t2.timeit(3) print("n=%2d, fib: %8.6f, fibi: %7.6f, percent: %10.2f" % (i, time1, time2, time1/time2))time1 is the time in seconds it takes for 3 calls to fib(n) and time2 respectively the time for fibi(). If we look at the results, we can see that calling fib(20) three times needs about 14 milliseconds. fibi(20) needs just 0.011 milliseconds for 3 calls. So fibi(20) is about 1300 times faster then fib(20).
fib(40) needs already 215 seconds for three calls, while fibi(40) can do it in 0.016 milliseconds. fibi(40) is more than 13 millions times faster than fib(40).
n= 1, fib: 0.000004, fibi: 0.000005, percent: 0.81 n= 2, fib: 0.000005, fibi: 0.000005, percent: 1.00 n= 3, fib: 0.000006, fibi: 0.000006, percent: 1.00 n= 4, fib: 0.000008, fibi: 0.000005, percent: 1.62 n= 5, fib: 0.000013, fibi: 0.000006, percent: 2.20 n= 6, fib: 0.000020, fibi: 0.000006, percent: 3.36 n= 7, fib: 0.000030, fibi: 0.000006, percent: 5.04 n= 8, fib: 0.000047, fibi: 0.000008, percent: 5.79 n= 9, fib: 0.000075, fibi: 0.000007, percent: 10.50 n=10, fib: 0.000118, fibi: 0.000007, percent: 16.50 n=11, fib: 0.000198, fibi: 0.000007, percent: 27.70 n=12, fib: 0.000287, fibi: 0.000007, percent: 41.52 n=13, fib: 0.000480, fibi: 0.000007, percent: 69.45 n=14, fib: 0.000780, fibi: 0.000007, percent: 112.83 n=15, fib: 0.001279, fibi: 0.000008, percent: 162.55 n=16, fib: 0.002059, fibi: 0.000009, percent: 233.41 n=17, fib: 0.003439, fibi: 0.000011, percent: 313.59 n=18, fib: 0.005794, fibi: 0.000012, percent: 486.04 n=19, fib: 0.009219, fibi: 0.000011, percent: 840.59 n=20, fib: 0.014366, fibi: 0.000011, percent: 1309.89 n=21, fib: 0.023137, fibi: 0.000013, percent: 1764.42 n=22, fib: 0.036963, fibi: 0.000013, percent: 2818.80 n=23, fib: 0.060626, fibi: 0.000012, percent: 4985.96 n=24, fib: 0.097643, fibi: 0.000013, percent: 7584.17 n=25, fib: 0.157224, fibi: 0.000013, percent: 11989.91 n=26, fib: 0.253764, fibi: 0.000013, percent: 19352.05 n=27, fib: 0.411353, fibi: 0.000012, percent: 34506.80 n=28, fib: 0.673918, fibi: 0.000014, percent: 47908.76 n=29, fib: 1.086484, fibi: 0.000015, percent: 72334.03 n=30, fib: 1.742688, fibi: 0.000014, percent: 123887.51 n=31, fib: 2.861763, fibi: 0.000014, percent: 203442.44 n=32, fib: 4.648224, fibi: 0.000015, percent: 309461.33 n=33, fib: 7.339578, fibi: 0.000014, percent: 521769.86 n=34, fib: 11.980462, fibi: 0.000014, percent: 851689.83 n=35, fib: 19.426206, fibi: 0.000016, percent: 1216110.64 n=36, fib: 30.840097, fibi: 0.000015, percent: 2053218.13 n=37, fib: 50.519086, fibi: 0.000016, percent: 3116064.78 n=38, fib: 81.822418, fibi: 0.000015, percent: 5447430.08 n=39, fib: 132.030006, fibi: 0.000018, percent: 7383653.09 n=40, fib: 215.091484, fibi: 0.000016, percent: 13465060.78
What's wrong with our recursive implementation?
Let's have a look at the calculation tree, i.e. the order in which the functions are called. fib() is substituted by f().
We can see that the subtree f(2) appears 3 times and the subtree for the calculation of f(3) two times. If you imagine extending this tree for f(6), you will understand that f(4) will be called two times, f(3) three times and so on. This means, our recursion doesn't remember previously calculated values.
We can implement a "memory" for our recursive version by using a dictionary to save the previously calculated values.
memo = {0:0, 1:1} def fibm(n): if not n in memo: memo[n] = fibm(n1) + fibm(n2) return memo[n]
We time it again to compare it with fibi():
from timeit import Timer from fibo import fib t1 = Timer("fib(10)","from fibo import fib") for i in range(1,41): s = "fibm(" + str(i) + ")" t1 = Timer(s,"from fibo import fibm") time1 = t1.timeit(3) s = "fibi(" + str(i) + ")" t2 = Timer(s,"from fibo import fibi") time2 = t2.timeit(3) print("n=%2d, fib: %8.6f, fibi: %7.6f, percent: %10.2f" % (i, time1, time2, time1/time2))
We can see that it is even faster than the iterative version. Of course, the larger the arguments the greater the benefit of our memoization:
n= 1, fib: 0.000011, fibi: 0.000015, percent: 0.73 n= 2, fib: 0.000011, fibi: 0.000013, percent: 0.85 n= 3, fib: 0.000012, fibi: 0.000014, percent: 0.86 n= 4, fib: 0.000012, fibi: 0.000015, percent: 0.79 n= 5, fib: 0.000012, fibi: 0.000016, percent: 0.75 n= 6, fib: 0.000011, fibi: 0.000017, percent: 0.65 n= 7, fib: 0.000012, fibi: 0.000017, percent: 0.72 n= 8, fib: 0.000011, fibi: 0.000018, percent: 0.61 n= 9, fib: 0.000011, fibi: 0.000018, percent: 0.61 n=10, fib: 0.000010, fibi: 0.000020, percent: 0.50 n=11, fib: 0.000011, fibi: 0.000020, percent: 0.55 n=12, fib: 0.000004, fibi: 0.000007, percent: 0.59 n=13, fib: 0.000004, fibi: 0.000007, percent: 0.57 n=14, fib: 0.000004, fibi: 0.000008, percent: 0.52 n=15, fib: 0.000004, fibi: 0.000008, percent: 0.50 n=16, fib: 0.000003, fibi: 0.000008, percent: 0.39 n=17, fib: 0.000004, fibi: 0.000009, percent: 0.45 n=18, fib: 0.000004, fibi: 0.000009, percent: 0.45 n=19, fib: 0.000004, fibi: 0.000009, percent: 0.45 n=20, fib: 0.000003, fibi: 0.000010, percent: 0.29 n=21, fib: 0.000004, fibi: 0.000009, percent: 0.45 n=22, fib: 0.000004, fibi: 0.000010, percent: 0.40 n=23, fib: 0.000004, fibi: 0.000010, percent: 0.40 n=24, fib: 0.000004, fibi: 0.000011, percent: 0.35 n=25, fib: 0.000004, fibi: 0.000012, percent: 0.33 n=26, fib: 0.000004, fibi: 0.000011, percent: 0.34 n=27, fib: 0.000004, fibi: 0.000011, percent: 0.35 n=28, fib: 0.000004, fibi: 0.000012, percent: 0.32 n=29, fib: 0.000004, fibi: 0.000012, percent: 0.33 n=30, fib: 0.000004, fibi: 0.000013, percent: 0.31 n=31, fib: 0.000004, fibi: 0.000012, percent: 0.34 n=32, fib: 0.000004, fibi: 0.000012, percent: 0.33 n=33, fib: 0.000004, fibi: 0.000013, percent: 0.30 n=34, fib: 0.000004, fibi: 0.000012, percent: 0.34 n=35, fib: 0.000004, fibi: 0.000013, percent: 0.31 n=36, fib: 0.000004, fibi: 0.000013, percent: 0.31 n=37, fib: 0.000004, fibi: 0.000014, percent: 0.29 n=38, fib: 0.000004, fibi: 0.000014, percent: 0.29 n=39, fib: 0.000004, fibi: 0.000013, percent: 0.31 n=40, fib: 0.000004, fibi: 0.000014, percent: 0.29
More about Recursion in Python
If you want to learn more on recursion, we suggest that you try to solve the following exercises. Please do not peer at the solutions, before you haven't given your best. If you have thought about a task for a while and you are still not capable of solving the exercise, you may consult our sample solutions.
In our section "Advanced Topics" of our tutorial we have a comprehensive treatment of the game or puzzle "Towers of Hanoi". Of course, we solve it with a function using a recursive function. The "Hanoi problem" is special, because a recursive solution almost forces itself on the programmer, while the iterative solution of the game is hard to find and to grasp.
Exercises
 Think of a recusive version of the function f(n) = 3 * n, i.e. the multiples of 3
 Write a recursive Python function that returns the sum of the first
n
integers.
(Hint: The function will be similiar to the factorial function!)  Write a function which implements the Pascal's triangle:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

The Fibonacci numbers are hidden inside of Pascal's triangle. If you sum up the coloured
numbers of the following triangle, you will get the 7th Fibonacci number:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
Write a recursive program to calculate the Fibonacci numbers, using Pascal's triangle.  Implement a recursive function in Python for the sieve of Eratosthenes.
The sieve of Eratosthenes is a simple algorithm for finding all prime numbers up to a specified integer. It was created by the ancient Greek mathematician Eratosthenes.
The algorithm to find all the prime numbers less than or equal to a given integer n: Create a list of integers from two to n: 2, 3, 4, ..., n
 Start with a counter i set to 2, i.e. the first prime number
 Starting from i+i, count up by i and remove those numbers from the list, i.e. 2*i, 3*i, 4*i, aso..
 Find the first number of the list following i. This is the next prime number.
 Set i to the number found in the previous step
 Repeat steps 3 and 4 until i is greater than n. (As an improvement: It's enough to go to the square root of n)
 All the numbers, which are still in the list, are prime numbers
You can easily see that we would be inefficient, if we strictly used this algorithm, e.g. we will try to remove the multiples
of 4, although they have been already removed by the multiples of 2.
So it's enough to produce the multiples of all the prime numbers up to the square root of n.
We can recursively create these sets.
 Write a recursive function find_index(), which returns the index of a number in the Fibonacci sequence,
if the number is an element of this sequence and returns 1 if the number is not contained in it, i.e.
fib(find_index(n)) == n
 The sum of the squares of two consecutive Fibonacci numbers is also a Fibonacci number, e.g. 2 and 3
are elements of the Fibonacci sequence and 2*2 + 3*3 = 13 corresponds to Fib(7).
Use the previous function to find the position of the sum of the squares of two consecutive numbers in the Fibonacci sequence.
Mathematical explanation:
Let a and b be two successive Fibonacci numbers with a prior to b. The Fibonacci sequence starting with the number "a" looks like this:0 a 1 b 2 a + b 3 a + 2b 4 2a + 3b 5 3a + 5b 6 5a + 8b
We can see that the Fibonacci numbers appear as factors for a and b. The nth element in this sequence can be calculated with the following formula:
F(n) = Fib(n1)* a + Fib(n) * b
From this we can conclude that for a natural number n, n>1, the following holds true:Fib(2*n + 1) = Fib(n)**2 + Fib(n+1)**2
Solutions to our Exercises

Solution to our first exercise on recursion:
Mathematically, we can write it like this:
f(1) = 3,
f(n+1) = f(n) + 3
A Python function can be written like this:def mult3(n): if n == 1: return 3 else: return mult3(n1) + 3 Towers of Hanoi for i in range(1,10): print(mult3(i))

Solution to our second exercise:
def sum_n(n): if n== 0: return 0 else: return n + sum_n(n1)

Solution for creating the Pacal triangle:
def pascal(n): if n == 1: return [1] else: line = [1] previous_line = pascal(n1) for i in range(len(previous_line)1): line.append(previous_line[i] + previous_line[i+1]) line += [1] return line print(pascal(6))
Alternatively, we can write a function using list comprehension:
def pascal(n): if n == 1: return [1] else: p_line = pascal(n1) line = [ p_line[i]+p_line[i+1] for i in range(len(p_line)1)] line.insert(0,1) line.append(1) return line print(pascal(6))

Producing the Fibonacci numbers out of Pascal's triangle:
def fib_pascal(n,fib_pos): if n == 1: line = [1] fib_sum = 1 if fib_pos == 0 else 0 else: line = [1] (previous_line, fib_sum) = fib_pascal(n1, fib_pos+1) for i in range(len(previous_line)1): line.append(previous_line[i] + previous_line[i+1]) line += [1] if fib_pos < len(line): fib_sum += line[fib_pos] return (line, fib_sum) def fib(n): return fib_pascal(n,0)[1] # and now printing out the first ten Fibonacci numbers: for i in range(1,10): print(fib(i))

The following program implements the sieve of Eratosthenes according to the rules of the task in an iterative way. It will
print out the first 100 prime numbers.
from math import sqrt def sieve(n): # returns all primes between 2 and n primes = list(range(2,n+1)) max = sqrt(n) num = 2 while num < max: i = num while i <= n: i += num if i in primes: primes.remove(i) for j in primes: if j > num: num = j break return primes print(sieve(100))
But this chapter of our tutorial is about recursion and recursive functions, and we have demanded a recursive function to calculate the prime numbers. To understand the following solution, you may confer our chapter about List Comprehension:from math import sqrt def primes(n): if n == 0: return [] elif n == 1: return [] else: p = primes(int(sqrt(n))) no_p = [j for i in p for j in range(i*2, n + 1, i)] p = [x for x in range(2, n + 1) if x not in no_p] return p print(primes(100))

memo = {0:0, 1:1} def fib(n): if not n in memo: memo[n] = fib(n1) + fib(n2) return memo[n] def find_index(*x): """ finds the natural number i with fib(i) = n """ if len(x) == 1: # started by user # find index starting from 0 return find_index(x[0],0) else: n = fib(x[1]) m = x[0] if n > m: return 1 elif n == m: return x[1] else: return find_index(m,x[1]+1)

# code from the previous example with the functions fib() and find_index() print(" index of a  a  b  sum of squares  index ") print("=====================================================") for i in range(15): square = fib(i)**2 + fib(i+1)**2 print( " %10d  %3d  %3d  %14d  %5d " % (i, fib(i), fib(i+1), square, find_index(square)))
The result of the previous program looks like this:index of a  a  b  sum of squares  index ===================================================== 0  0  1  1  1 1  1  1  2  3 2  1  2  5  5 3  2  3  13  7 4  3  5  34  9 5  5  8  89  11 6  8  13  233  13 7  13  21  610  15 8  21  34  1597  17 9  34  55  4181  19 10  55  89  10946  21 11  89  144  28657  23 12  144  233  75025  25 13  233  377  196418  27 14  377  610  514229  29
^{1} Stephen Pinker, The Blank Slate, 2002