Scheme Exercise (Structure and Interpretation book)

Question

I've been working through the Structure and Interpretation of Computer Programs book and have hit a bit of a problem.  Since I chose to use this book as part of my curriculum, my teacher doesn't actually know much of anything about Lisp or its varients, so I was hoping to get a bit of help.
The problem is:
Section 1.2 wrote:Exercise 1.11.  A function f is defined by the rule that f(n) = n if n<3> 3. Write a procedure that computes f by means of a recursive process. Write a procedure that computes f by means of an iterative process.
I have no trouble writing a recursive procedure/recursive process, but I am having trouble writing it iteratively; I'm not familiar enough with Lisp to know how to model this, and I don't know how to do it iteratively with three branching sections.  I'm assuming it'd require some sort of running total and a number of variables to save the state, and I was thinking about possibly trying to work from the bottom up, but I'm not real sure. 
If anyone can help me with an iterative version, it'd be much appreciated; right now I'm just confused.

Answer 1

Let's write down some examples of this function:
f(0) = 0
f(1) = 1
f(2) = 2
f(3) = f(2) + 2*f(1) + 3*f(0)
f(4) = f(3) + 2*f(2) + 3*f(1)
f(5) = f(4) + 2*f(3) + 3*f(2)
If you compute the value of f(5) recursively you do it like this:
f(5) = f(4) + 2*f(3) + 3*f(2)
f(4) = f(3) + 2*f(2) + 3*f(1)
f(3) = f(2) + 2*f(1) + 3*f(0)
f(2) = 2
f(1) = 1
f(0) = 0
f(3) = 2 + 2*1 + 3*0
If you do this iteratively you do it the other way around:
f(0) = 0
f(1) = 1
f(2) = 2
f(3) = f(2) + 2*f(1) + 3*f(0)
f(4) = f(3) + 2*f(2) + 3*f(1)
f(5) = f(4) + 2*f(3) + 3*f(2)
The last three lines have the same structure:
f(n) = previous + 2*before-previous + 3*before-before-previous
If you want to compute f(n) you only need the 3 previous values at any time. To compute f(3) you need f(2), f(1) and f(0). Let's give these short names:
previous = a
before-previous = b
before-before-previous = c
If you want to compute f(3):
a = f(2) = 2
b = f(1) = 1
c = f(0) = 0
Now we compute f(4):
a = f(3) = ?
b = f(2) = 2
c = f(1) = 1
For b and c it's easy to define them in terms of previous values a, b and c:
b = previous-a
c = previous-b
Verify that this is correct.
a is a little tickier because we need the original formula:
a = previous-a + 2*previous-b + 3*previous-c
So this is what we have now:
a = previous-a + 2*previous-b + 3*previous-c
b = previous-a
c = previous-b
Let's rewrite that with arrow notation:
a -> a + 2*b + 3*c
b -> a
c -> b
These arrows mean that you compute them in parallel; the a in b -> a is still the previous a, and not the new a generated by a -> a + 2*b + c.
Computing f(5) with this iterative method:
a -> 2
b -> 1
c -> 0
These are the base-case values because f(n) = n if n <3> 2 + 2*1 + 3*0 = 4
b -> 2
c -> 1
(n = 4)
a -> 4 + 2*2 + 3*1 = 11
b -> 4
c -> 2
(n = 5)
a -> 11 + 2*4 + 3*2 = 25
b -> 11
c -> 4
The answer of f(5) is in a: 25.
The basic structure of the procedure looks like this:
(define (f n)
(define (iter counter a b c)
(if (= counter n)
c
(iter (+ counter 1)
; new a
; new b
; new c
(iter 0 2 1 0))
The rest is an exercise.
Don't shoot me if this isn't correct! (I didn't implement the actual procedure ;-))