No idea how to solve SICP exercise 1.11

Question

Exercise 1.11:

A function f is defined by the rule that f(n) = n if n < 3 and f(n) = f(n - 1) + 2f(n - 2)

Answer 1

You need to capture the state in some accumulators and update the state at each iteration.

If you have experience in an imperative language, imagine writing a while loop and tracking information in variables during each iteration of the loop. What variables would you need? How would you update them? That's exactly what you have to do to make an iterative (tail-recursive) set of calls in Scheme.

In other words, it might help to start thinking of this as a while loop instead of a recursive definition. Eventually you'll be fluent enough with recursive -> iterative transformations that you won't need to extra help to get started.

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For this particular example, you have to look closely at the three function calls, because it's not immediately clear how to represent them. However, here's the likely thought process: (in Python pseudo-code to emphasise the imperativeness)

Each recursive step keeps track of three things:

f(n) = f(n - 1) + 2f(n - 2) + 3f(n - 3) 

So I need three pieces of state to track the current, the last and the penultimate values of f. (that is, f(n-1), f(n-2) and f(n-3).) Call them a, b, c. I have to update these pieces inside each loop:

for _ in 2..n:
    a = NEWVALUE
    b = a
    c = b
return a

So what's NEWVALUE? Well, now that we have representations of f(n-1), f(n-2) and f(n-3), it's just the recursive equation:

for _ in 2..n:
    a = a + 2 * b + 3 * c
    b = a
    c = b
return a

Now all that's left is to figure out the initial values of a, b and c. But that's easy, since we know that f(n) = n if n < 3.

if n < 3: return n
a = 2 # f(n-1) where n = 3
b = 1 # f(n-2)
c = 0 # f(n-3)
# now start off counting at 3
for _ in 3..n:
    a = a + 2 * b + 3 * c
    b = a
    c = b
return a

That's still a little different from the Scheme iterative version, but I hope you can see the thought process now.