Fibonacci Sequence iteratively? - racket

Question

I'm new to racket and trying to give a function an integer and it should return it's Fibonacci value (iteratively). I don't know what I'm getting wrong with the do and if I incorporate an if statement to catch when the number = 1 everything breaks
(define (fib-iterat n)
(do ((i 1 (+ i 1))
(nextTerm 0 (+ value0 value1))
(value0 0 (+ value1 0))
(value1 1 (+ nextTerm 0)))
((> i n) nextTerm))
)
I get:
1 = 1
2 = 1
6 = 3
9 = 7
12 = 16

Answer 1

;;; correct
(define (fib n)
(do ((i 0 (+ i 1))
(n1 1 (+ n0 n1))
(n0 0 n1))
((= i n) n0)))
;;; still correct
(define (fib n)
(do ((i 0 (+ i 1))
(n0 0 n1)
(n1 1 (+ n0 n1)))
((= i n) n0)))
;;; wrong
(define (fib-iterat n)
(do ((i 0 (+ i 1))
(value1 1 nextTerm)
(value0 0 value1)
(nextTerm 1 (+ value0 value1))) ; first bracket close
((= i n) (list value0 value1 nextTerm)))) ; second bracket
I never read any "do loop" code (but it should be easy to understand). You can think like this way. We separate old state and new state. In first bracket variable calculus only use old state values. So in do loop first bracket order in many case should be irrelevant. We refresh variable value but new values will be use in next loop or second bracket.
try (fib-iterat 2),
we call old state (o1,o2,o3),
i=0≠n,
old state:none,
new state:(value0,value1,nextTerm)=(0,1,1),
i=i+1
i=1≠n,
old state:(0,1,1),
new state:(o2,o3,o2+o3)=(1,1,1),
i=i+1
i=2=n,
old state:(1,1,1),
new state:(o2,o3,o2+o3)=(1,1,2)
we want (1,2,2) so it wrong.
I think the most important thing is don't think it a set! binding. It's more like this.
(define (f n)
(do ([i 0 (+ i 1)]
[n1 1 (+ n1 n3)]
[n2 1 (+ n1 n2)]
[n3 1 (+ n2 n3)])
((= i n) (list n1 n2 n3))))
(define (my-do-loop n)
(local ((define (λ0 i n1 n2 n3) (+ i 1))
(define (λ1 i n1 n2 n3) (+ n1 n3))
(define (λ2 i n1 n2 n3) (+ n1 n2))
(define (λ3 i n1 n2 n3) (+ n2 n3))
(define (aux i n1 n2 n3)
(if (= i n)
(list n1 n2 n3)
(aux (λ0 i n1 n2 n3)
(λ1 i n1 n2 n3)
(λ2 i n1 n2 n3)
(λ3 i n1 n2 n3)))))
(aux 0 1 1 1)))
So it makes us easy understanding why we only use old value and binding order is not important because we don't do any value binding.