recursive factorial function

Question

How can I combine these two functions into one recursive function to have this result:

factorial(6)

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720

This is the current code for my factorial function:

def factorial(n):
   if n < 1:   # base case
       return 1
   else:
       return n * factorial(n - 1)  # recursive call


def fact(n):
   for i in range(1, n+1 ):
       print "%2d! = %d" % (i, factorial(i))

and the output that this code produces is the following:

fact(6)

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720

As you see, the execution of these two functions gives me correct answers, but I just wanted to simplify the two functions to a single recursive function.

Solution 1

We can combine the two functions to this single recursive function:

def factorial(n):
   if n < 1:   # base case
       return 1
   else:
       returnNumber = n * factorial(n - 1)  # recursive call
       print(str(n) + '! = ' + str(returnNumber))
       return returnNumber

Solution 2

2 lines of code:

def fac(n):
    return 1 if (n < 1) else n * fac(n-1)

Test it:

print fac(4)

Result:

24

Solution 3

def factorial(n):
    result = 1 if n <= 1 else n * factorial(n - 1)
    print '%d! = %d' % (n, result)
    return result

Solution 4

a short one:

def fac(n):
    if n == 0:
        return 1
    else:
        return n * fac(n-1)
print fac(0)

Solution 5

try this:

def factorial( n ):
   if n <1:   # base case
       print "%2d! = %d" % (n, n)
       return 1
   else:
       temp = factorial( n - 1 )
       print "%2d! = %d" % (n, n*temp)
       return n * temp  # recursive call

One thing I noticed is that you are returning '1' for n<1, that means your function will return 1 even for negative numbers. You may want to fix that.

View more solutions

Comments

[FrustratedWithFormsDesigner]

I'm not 100% sure that this is correct, but since OP said it's for an exam, I won't go into any further details...

[Tadeck]

@D.Shawley: This is quite inefficient solution, as you calculate factorial(1) n times, factorial(2) n-1 times, factorial(3) n-2 times and so on...

[mcocdawc]

Mathematically 0! evaluates to 1. So the first part of your conditional should be changed.

[imbatman]

you will have a recursion error as you don't handle 0 and below

[TrebledJ]

Although correct, the while is redundant as return will kick in on the first iteration (and no other iterations will be performed). Changing while to if is much better.

[TrebledJ]

What I meant by redundant was the communicative aspect... other coders seeing the function will see while and think: "Okay, it's factorial by looping"; then one line later they see return and realise it's actually factorial by recursion. (Usually, recursion is a substitute for loops.) And... ah, I see a benchmark. A small difference in performance between while and if, but your new content seems well researched. :-)

[cdlane]

This doesn't work. The second reference to factorial() should instead be f()