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SICP exercise 1.30
The sum procedure given in the text in section 1.3.1 generates a linear recursion. The procedure can be rewritten so that the sum is performed iteratively. Show how to do this by filling in the missing expressions in the following definition:
(define (sum term a next b) (define (iter a result) (if <??> <??> (iter <??> <??>))) (iter <??> <??>))
result is clearly intended to keep a running total of the sum throughout the iteration. We can use lexical scoping to get the value of procedure sum's formal parameter b in the body of the iter procedure, so we don't need to pass it as an argument to iter. Here's an iterative sum procedure:
(define (sum term a next b) (define (iter a result) (if (> a b) result (iter (next a) (+ (term a) result)))) (iter a 0))
Note that the formal parameter a in the body of procedure iter is a different variable than the formal parameter a in the body of procedure sum!
Here's a test using simpsons-rule-integral from exercise 1.29:
(define (simpsons-rule-integral f a b n) (define (inc k) (+ k 1)) (define (helper h) (define (y k) (f (+ a (* k h)))) (define (term k) (cond ((or (= k 0) (= k n)) (y k)) ((even? k) (* 2 (y k))) (else (* 4 (y k))))) (* (sum term 0 inc n) (/ h 3))) (helper (/ (- b a) n))) (define (cube x) (* x x x)) (simpsons-rule-integral cube 0 1 100)