Factorial function? what?

#1



https://www.codecademy.com/courses/python-intermediate-en-rCQKw/0/5?curriculum_id=4f89dab3d788890003000096#

The instructions say:
Define a function factorial that takes an integer x as input.

Calculate and return the factorial of that number.

I have been reading the forums and I understand the solution (even though I am curious about the aim of this exercise since I don't recall having learn the factorial function before in previous exercises), but I am very puzzled about something. The forums say that factorial is a recursive function, which I take to mean something like it will automatically repeat itself or something. But I don't understand...why can't we just take factorial(x) straightaway, where x is a number? if factorial is a function recognised by Python why can't we just use it directly?

Many thanks :slight_smile:

#2

Factorial of x can be expressed as x * factorial(x-1), which means that the function calls itself recursively. (and factorial(0) is defined as 1)

#3

I understand what that means, and I also learnt in Math in school that 0! = 1. But I don't understand why you can't apply it directly :frowning: since ultimately you are still having to use the factorial function directly (if you don't do the iteration method)

#4

You don't have to use math.factorial to write a recursive factorial function, no.

Just a single function that takes one argument and returns one value. That function calls itself recursively.

#5

woops, I don't quite mean that either. I mean, why can't you do this directly:

def factorial(x):
return factorial(x)

Thanks for helping :slight_smile:

#6

That would just keep calling itself until Python refuses to keep track of any more call stacks. Try it.

It will not return.

#7

yup I tried it and got an error message, which is why I am trying to find out more about why it doesn't work by asking here on this forum.

I was essentially wondering since you are going to be using the factorial function eventually anyway when doing x * factorial(x - 1), why can't I use it directly with factorial(x) ?

Sorry if this is an ultra beginner question, because that is precisely what I am :frowning:

#8

If a function just calls itself over and over then it will not terminate, that's why

If it doesn't terminate you won't ever get a result.

If you do x-1 untill it reaches 0 and then return 1, then it terminates, and you've got a result.

def factorial(x):
    return factorial(x)

That's just an infinite loop, there's nothing there that does any computing.

#9

ah I see :slight_smile:

But how would a code like this which works (with the correct indentation):

def factorial(x):
if x == 1:
return 1
else:
return x * factorial(x-1)

know when to terminate the function? wouldn't it loop infinitely as well?

#10

Try executing it manually?

And strictly speaking that doesn't run. Your indentation is all messed up in your post.

#11

Okay so:

def factorial(x):
if x == 1:
return 1
else:
return x * factorial(x-1)

factorial(5) #will return 5 * 4!
factorial(4) #will return 4 * 3!
factorial(3) #will return 3 * 2!
factorial(2) #will return 2 * 1!
factorial(1) #will return 1
factorial(0) #will return 0 * (-1)! -> incorrect

so how does it know when to stop?

With the comment in the brackets I meant given that indentations are correct. I know it is wrong in my reply, but I forgot how to do it (I could do it once a few weeks back) and I can't find how. I thought it was triple quotation marks with [python] but it didn't work so yeah messy indentation sorry

#12

It stops when it returns without calling itself first.

And yes, it should handle 0 so your version is slightly incorrect.

#13

You are almost there. Look at the line before the return x and compare it to your last line.

def factorial(x):
if x == 1:
return 1
else:
x = x * factorial(x-1)
return x

VBR Patrick

#14

I think i know what you meant, and i think i managed to solve it: #v2 means version2

def factorial_v2(x):
#indentation result = factorial(x)
#indentation return result

test examples

print factorial_v2(3) #prints 6
print factorial_v2(0) #prints 1
print factorial_v2(1) #prints 1

so looks like you can just use a built in FACTORIAL function, without even telling it that factorial 1 is 1, the only thing is you first have to define your own function and inside this function first ask to store the result in a variable and then return the result of built-in FACTORIAL function, if that makes any sense..?

#15

Are you sure your code works? I just tested it... :stuck_out_tongue:

#16

hmmm... this is just bizzare...

i just pasted only this code and its a fact, it doesnt work on its own, however when i had it written down below the first version of my factorial it did work:

def factorial(x):
if x == 1 or x == 0:
f = 1
return f
else:
f = x * factorial(x - 1)
return f

test examples

print factorial(3) # prints 6
print factorial(4) # prints 24
print factorial(0) # prints 1

def factorial_v2(x):
result = factorial(x)
return result
print factorial_v2(3)
print factorial_v2(0)
print factorial_v2(1)

so it must be that factorial_v2 is using the first factorial function which is useless..

#17

yup and I was just wondering why. That was my original question :slight_smile:

Anyway 2 days ago I realised I accidentally skipped some exercises and jumped to this so I am leaving this alone for now and doing what I haven't done previously and hope that I will understand later on :slight_smile:

#18

How about simply:

import math
print math.factorial(1)

#19

that's won't be following the instructions, that's all :stuck_out_tongue:

#20

For anyone searching for an answer to this, I recommend reading this (which I stumbled upon after searching for quite a while):