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

I think you are asking how one might discover the algorithm naturally, outside of a 'design pattern'.

It was helpful for me to look at the expansion of the f(n) at each n value:

f(0) = 0  |
f(1) = 1  | all known values
f(2) = 2  |

f(3) = f(2) + 2f(1) + 3f(0)
f(4) = f(3) + 2f(2) + 3f(1)
f(5) = f(4) + 2f(3) + 3f(2)
f(6) = f(5) + 2f(4) + 3f(3)

Looking closer at f(3), we see that we can calculate it immediately from the known values. What do we need to calculate f(4)?

We need to at least calculate f(3) + [the rest]. But as we calculate f(3), we calculate f(2) and f(1) as well, which we happen to need for calculating [the rest] of f(4).

f(3) = f(2) + 2f(1) + 3f(0)
            ↘       ↘
f(4) = f(3) + 2f(2) + 3f(1)

So, for any number n, I can start by calculating f(3), and reuse the values I use to calculate f(3) to calculate f(4)...and the pattern continues...

f(3) = f(2) + 2f(1) + 3f(0)
            ↘       ↘
f(4) = f(3) + 2f(2) + 3f(1)
            ↘       ↘
f(5) = f(4) + 2f(3) + 3f(2)

Since we will reuse them, lets give them a name a, b, c. subscripted with the step we are on, and walk through a calculation of f(5):

  Step 1:    f(3) = f(2) + 2f(1) + 3f(0) or f(3) = a1 + 2b1 +3c1

where

a₁ = f(2) = 2,

b₁ = f(1) = 1,

c₁ = 0

since f(n) = n for n < 3.

Thus:

f(3) = a₁ + 2b₁ + 3c₁ = 4

  Step 2:  f(4) = f(3) + 2a1 + 3b1

So:

a₂ = f(3) = 4 (calculated above in step 1),

b₂ = a₁ = f(2) = 2,

c₂ = b₁ = f(1) = 1

Thus:

f(4) = 4 + 2*2 + 3*1 = 11

  Step 3:  f(5) = f(4) + 2a2 + 3b2

So:

a₃ = f(4) = 11 (calculated above in step 2),

b₃ = a₂ = f(3) = 4,

c₃ = b₂ = f(2) = 2

Thus:

f(5) = 11 + 2*4 + 3*2 = 25

Throughout the above calculation we capture state in the previous calculation and pass it to the next step, particularily:

a_step = result of step - 1

b_step = a_{step - 1}

c_step = b_{step -1}

Once I saw this, then coming up with the iterative version was straightforward.