L4Ka – a New Methodology for System Construction

L4Ka – L4Ka Project
Don’t have much interest in OSes or OS research, but this is an interesting project.

Activating this Blog again.

Use cases vs functional specs

I’ve come to the firm belief that writing detailed functional specs upfront for software projects is a total waste of time. Software is meant oto be used and in order to find out what it’s use is going to be software architects need use cases. writing specs is just moronic when you have no idea on how it’s going to be used

Ex 1.8 – Structure and Interpretation of Computer Programs

; Ex 1.8
(define improve-cube-root
  (lambda (x y)
    (/ (+ (/ x (* y y)) (* 2 y))
       3)))

(define cube-root-iter
  (lambda (guess x )
    (define next-guess ( / ( + (/ x (* guess guess)) (* 2 guess))
                           3))
    (if (good-enough-2? next-guess guess)
        next-guess
        (cube-root-iter next-guess x))))

(define cube-root
  (lambda (x)
    (cube-root-iter 1.0 x)))

Ex 1.7 – Structure and Interpretation of Computer Programs

; Ex 1.7
(define (good-enough-2? guess prev-guess)
  (< (abs (- guess prev-guess)) 0.0001))

(define sqrt-iter-2
  (lambda (guess x)
    (define next-guess (improve guess x))
    (if (good-enough-2? next-guess guess)
        next-guess
        (sqrt-iter-2 next-guess x))))

Ex 1.6 – Structure and Interpretation of Computer Programs

; Ex 1.6
; The function will run forever, because Scheme is applicative order
; so the argument for the else clause will be evaluated causing the
; to routine to run forever.

Ex 1.13 – Structure and Interpretation of Computer Programs

Exercise 1.13 – SICP

We already know that

(1)

We also know that

where

From the hint where

So

So according to the finonacci definition (1)

Using wolfram alpha we can compute

If you plugin the values in the wolfram alpha , you can see that that both expansions at are also the same

Edit hmmm, looks like I missed the point of the problem. The real problem was to prove that Fib(n) approx= phi^n / 2 where phi = (1+sqrt(5))/2. I think I’ll have to redo this proof.