Non-Programmer's Tutorial for Python 2.6/Advanced Functions Example

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Some people find this section useful, and some find it confusing. If you find it confusing you can skip it (or just look at the examples.) Now we will do a walk through for the following program:

def mult(a, b):
    if b == 0:
        return 0
    rest = mult(a, b - 1)
    value = a + rest
    return value
print "3 * 2 = ", mult(3, 2)
Output
3 * 2 =  6

Basically this program creates a positive integer multiplication function (that is far slower than the built in multiplication function) and then demonstrates this function with a use of the function. This program demonstrates the use of recursion, that is a form of iteration (repetition) in which there is a function that repeatedly calls itself until an exit condition is satisfied. It uses repeated additions to give the same result as mutiplication: e.g. 3 + 3 (addition) gives the same result as 3 * 2 (multiplication).


RUN 1

Question: What is the first thing the program does?
Answer: The first thing done is the function mult is defined with all the lines except the last one.
function mult defined
def mult(a, b):
    if b == 0:
        return 0
    rest = mult(a, b - 1)
    value = a + rest
    return value
This creates a function that takes two parameters and returns a value when it is done. Later this function can be run.

What happens next?
The next line after the function, print "3 * 2 = ", mult(3, 2) is run.

And what does this do?
It prints 3 * 2 = and the return value of mult(3, 2)

And what does mult(3, 2) return?
We need to do a walkthrough of the mult function to find out.

RUN 2

What happens next?
The variable a gets the value 3 assigned to it and the variable b gets the value 2 assigned to it.

And then?
The line if b == 0: is run. Since b has the value 2 this is false so the line return 0 is skipped.

And what then?
The line rest = mult(a, b - 1) is run. This line sets the local variable rest to the value of mult(a, b - 1). The value of a is 3 and the value of b is 2 so the function call is mult(3,1)

So what is the value of mult(3, 1) ?
We will need to run the function mult with the parameters 3 and 1.
RUN 2
def mult(3, 2):
    if b == 0:
        return 0
    rest = mult(3, 2 - 1)
    rest = mult(3, 1)
    value = 3 + rest
    return value



RUN 3

So what happens next?
The local variables in the new run of the function are set so that a has the value 3 and b has the value 1. Since these are local values these do not affect the previous values of a and b.

And then?
Since b has the value 1 the if statement is false, so the next line becomes rest = mult(a, b - 1).

What does this line do?
This line will assign the value of mult(3, 0) to rest.

So what is that value?
We will have to run the function one more time to find that out. This time a has the value 3 and b has the value 0.

So what happens next?
The first line in the function to run is if b == 0:. b has the value 0 so the next line to run is return 0

And what does the line return 0 do?
This line returns the value 0 out of the function.

So?
So now we know that mult(3, 0) has the value 0. Now we know what the line rest = mult(a, b - 1) did since we have run the function mult with the parameters 3 and 0. We have finished running mult(3, 0) and are now back to running mult(3, 1). The variable rest gets assigned the value 0.

What line is run next?
The line value = a + rest is run next. In this run of the function, a = 3 and rest = 0 so now value = 3.

What happens next?
The line return value is run. This returns 3 from the function. This also exits from the run of the function mult(3, 1). After return is called, we go back to running mult(3, 2).

Where were we in mult(3, 2)?
We had the variables a = 3 and b = 2 and were examining the line rest = mult(a, b - 1).

So what happens now?
The variable rest get 3 assigned to it. The next line value = a + rest sets value to 3 + 3 or 6.

So now what happens?
The next line runs, this returns 6 from the function. We are now back to running the line print "3 * 2 = ", mult(3, 2) which can now print out the 6.

What is happening overall?
Basically we used two facts to calculate the multiple of the two numbers. The first is that any number times 0 is 0 (x * 0 = 0). The second is that a number times another number is equal to the first number plus the first number times one less than the second number (x * y = x + x * (y - 1)). So what happens is 3 * 2 is first converted into 3 + 3 * 1. Then 3 * 1 is converted into 3 + 3 * 0. Then we know that any number times 0 is 0 so 3 * 0 is 0. Then we can calculate that 3 + 3 * 0 is 3 + 0 which is 3. Now we know what 3 * 1 is so we can calculate that 3 + 3 * 1 is 3 + 3 which is 6.

This is how the whole thing works:

3 * 2
3 + 3 * 1
3 + 3 + 3 * 0
3 + 3 + 0
3 + 3
6

Should you still have problems with this example, look at the process backwards. What is the last step that happens? We can easily make out that the result of mult(3, 0) is 0. Since b is 0, the function mult(3, 0) will return 0 and stop.

So what does the previous step do? mult(3, 1) does not return 0 because b is not 0. So the next lines are executed: rest = mult (a, b - 1), which is rest = mult (3, 0), which is 0 as we just worked out. So now the variable rest is set to 0.

The next line adds the value of rest to a, and since a is 3 and rest is 0, the result is 3.

Now we know that the function mult(3, 1) returns 3. But we want to know the result of mult(3,2). Therefore, we need to jump back to the start of the program and execute it one more round: mult(3, 2) sets rest to the result of mult(3, 1). We know from the last round that this result is 3. Then value calculates as a + rest, i. e. 3 + 3. Then the result of 3 * 2 is printed as 6.

The point of this example is that the function mult(a, b) starts itself inside itself. It does this until b reaches 0 and then calculates the result as explained above.

Recursion[edit]

Programming constructs of this kind are called recursive and probably the most intuitive definition of recursion is:


Recursion
If you still don't get it, see recursion.

These last two sections were recently written. If you have any comments, found any errors or think I need more/clearer explanations please email. I have been known in the past to make simple things incomprehensible. If the rest of the tutorial has made sense, but this section didn't, it is probably my fault and I would like to know. Thanks.

Examples[edit]

factorial.py

#defines a function that calculates the factorial
 
def factorial(n):
    if n <= 1:
        return 1
    return n * factorial(n - 1)
 
print "2! =", factorial(2)
print "3! =", factorial(3)
print "4! =", factorial(4)
print "5! =", factorial(5)

Output:

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

countdown.py

def count_down(n):
    print n
    if n > 0:
        return count_down(n-1)
 
count_down(5)

Output:

5
4
3
2
1
0

Commented function_interesting.py

# The comments below have been numbered as steps, to make explanation
# of the code easier. Please read according to those steps.
# (step number 1, for example, is at the bottom)
 
 
def mult(a, b): # (2.) This function will keep repeating itself, because....
    if b == 0:
        return 0
    rest = mult(a, b - 1) # (3.) ....Once it reaches THIS, the sequence starts over again and goes back to the top!
    value = a + rest
    return value # (4.) therefore, "return value" will not happen until the program gets past step 3 above
 
print "3 * 2 = ", mult(3, 2) # (1.) The "mult" function will first initiate here
 
 
# The "return value" event at the end can therefore only happen
# once b equals zero (b decreases by 1 everytime step 3 happens).
# And only then can the print command at the bottom be displayed.
 
# See it as kind of a "jump-around" effect. Basically, all you
# should really understand is that the function is reinitiated
# WITHIN ITSELF at step 3. Therefore, the sequence "jumps" back
# to the top.

Commented factorial.py

# Another "jump-around" function example:
 
def factorial(n): # (2.) So once again, this function will REPEAT itself....
    if n <= 1:
        return 1
    return n * factorial(n - 1) # (3.) Because it RE-initiates HERE, and goes back to the top.
 
print "2! =", factorial(2) # (1.) The "factorial" function is initiated with this line
print "3! =", factorial(3)
print "4! =", factorial(4)
print "5! =", factorial(5)

Commented countdown.py

# Another "jump-around", nice and easy:
 
 
def count_down(n): # (2.) Once again, this sequence will repeat itself....
    print n
    if n > 0:
        return count_down(n-1) # (3.) Because it restarts here, and goes back to the top
 
count_down(5) # (1.) The "count_down" function initiates here