Some people find this section useful, and some find it confusing. If you find it confusing you can skip it. 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))
```

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).

Question: What is the first thing the program does?
Answer: The first thing done is the function mult is defined with the lines:
```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.
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.
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
```

#### Recursion

Programming constructs solving a problem by solving a smaller version of the same problem are called recursive. In the examples in this chapter, recursion is realized by defining a function calling itself. This facilitates implementing solutions to programming tasks as it may be sufficient to consider the next step of a problem instead of the whole problem at once. It is also useful as it allows to express some mathematical concepts with straightforward, easy to read code.

Any problem that can be solved with recursion could be re-implemented with loops. Using the latter usually results in better performance. However equivalent implementations using loops are usually harder to get done correctly.

Probably the most intuitive definition of recursion is:

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

Try walking through the factorial example if the multiplication example did not make sense.

### Examples

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
```

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