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Solution using max precision and fix for cube-root of -2 problem. There are still the cube-root of 0 (and cube-root of 100) convergence problem, where the answer approaches 0 (or 100) but never reaches it. (Based on Ex1.7 solution)

(define (square guess) (* guess guess)) ;not used ;(define (average x guess) ; (/ (+ x guess) 2)) ;improve square root ;(define (improve guess x) ; (average guess (/ x guess))) ;cube root improve formula used as is (define (improve guess x) (/ (+ (/ x (square guess)) (* 2 guess)) 3)) ;original test ;(define (good-enough? guess x) ; (< (abs (- (square guess) x)) 0.001)) ;iterates until guess and next guess are equal, ;automatically produces answer to limit of system precision (define (good-enough? guess x) (= (improve guess x) guess)) (define (3rt-iter guess x) (if (good-enough? guess x) guess (3rt-iter (improve guess x) x))) ;<<<expression entry point>>> ;change initial guess to 1.1 to prevent an anomalous result for ;cube root of -2 (define (3root x) (3rt-iter 1.1 x))

Welcome to DrRacket, version 7.5 [3m]. Language: R5RS; memory limit: 128 MB. > (3root 5) 1.709975946676697 > (3root -2) -1.2599210498948732 > (3root 27) 3.0 > (3root 0) 4.9406564584125e-324 > (3root 100000000000000.0001) 46415.88833612779 >

The solution presented here is based on the solution for sicp-ex-1.7 and, similarly, uses the alternative strategy for the `good-enough?` predicate.

;; ex 1.8. Based on the solution of ex 1.7. (define (square x) (* x x)) (define (cube-root-iter guess prev-guess x) (if (good-enough? guess prev-guess) guess (cube-root-iter (improve guess x) guess x))) (define (improve guess x) (average3 (/ x (square guess)) guess guess)) (define (average3 x y z) (/ (+ x y z) 3)) ;; Stop when the difference is less than 1/1000th of the guess (define (good-enough? guess prev-guess) (< (abs (- guess prev-guess)) (abs (* guess 0.001)))) (define (cube-root x) (cube-root-iter 1.0 0.0 x)) ;; Testing (cube-root 1) (cube-root -8) (cube-root 27) (cube-root -1000) (cube-root 1e-30) (cube-root 1e60) ;; this fails for -2 due to zero division :( ;; Fix: take absolute cuberoot and return with sign ;;(define (cube-root x) ;; ((if (< x 0) - +)(cube-root-iter (improve 1.0 (abs x)) 1 (abs x))))

(define (cube x) (* x x x)) (define (improve guess x) (/ (+ (/ x (square guess)) (* 2 guess)) 3)) (define (good-enough? guess x) (< (abs (- (cube guess) x)) 0.001)) (define (cube-root-iter guess x) (if (good-enough? guess x) guess (cube-root-iter (improve guess x) x))) (define (cube-root x) (cube-root-iter 1.0 x))

(define (cube-root x) (cube-root-iter 1.0 x)) (define (cube-root-iter guess x) (if (good-enough? guess x) guess (cube-root-iter (improve guess x) x))) (define (good-enough? guess x) (< (relative-error guess (improve guess x)) error-threshold)) (define (relative-error estimate reference) (/ (abs (- estimate reference)) reference)) (define (improve guess x) (average3 (/ x (square guess)) guess guess)) (define (average3 x y z) (/ (+ x y z) 3)) (define error-threshold 0.01)

This solution makes use of the fact that (in LISP) procedures are also data.

(define (square x) (* x x)) (define (cube x) (* x x x)) (define (good-enough? guess x improve) (< (abs (- (improve guess x) guess)) (abs (* guess 0.001)))) (define (root-iter guess x improve) (if (good-enough? guess x improve) guess (root-iter (improve guess x) x improve))) (define (sqrt-improve guess x) (/ (+ guess (/ x guess)) 2)) (define (cbrt-improve guess x) (/ (+ (/ x (square guess)) (* 2 guess)) 3)) (define (sqrt x) (root-iter 1.0 x sqrt-improve)) (define (cbrt x) (root-iter 1.0 x cbrt-improve))

Use the improved `good-enough?`:

(define (cube-roots-iter guess prev-guess input) (if (good-enough? guess prev-guess) guess (cube-roots-iter (improve guess input) guess input))) (define (good-enough? guess prev-guess input) (> 0.001 (/ (abs (- guess prev-guess)) input))) ;; this should be (abs input) to handle negative inputs. Example: (cube-roots -1) should be -1. Before change, output was 0.33. After fix, output is corrected to -1.000000001794607. (define (improve guess input) (/ (+ (/ input (square guess)) (* 2 guess)) 3)) (define (square x) (* x x)) ;;to make sure the first input of guess and prev-guess does not pass the predicate accidentally, use improve here once: ;;to make sure float number is implemented, use 1.0 instead of 1: (define (cube-roots x) (cube-roots-iter (improve 1.0 x) 1 x))

Chan : I just added one procedure. (But I just made this procedure with low precision. I think you can fix this.) Give me a feedback please.

(define (cube-root-iter guess x) (if (good-enough? guess x) guess (cube-root-iter (improve guess x) x))) (define (improve guess x) (average (/ x (square guess)) (* 2 guess))) (define (average x y) (/ (+ x y) 3)) (define (square x) (* x x)) (define (good-enough? guess x) (< (abs (- (cube guess) x)) (* guess 0.001))) (define (cube x) (* x x x)) (define (cube-root x) (if (< x 0) (* -1 (cube-root-iter 1.0 (abs x))) (cube-root-iter 1.0 x))) (cube-root 27) 3.0000005410641766 (cube-root -27) -3.0000005410641766

I think the following hack may solve the infinite loop problem. I noticed that the reason why calculating the cube root of 100 results in an infinite loop is because once the initial guess has been improved a sufficient number of times it toggles between 4.641588833612779 and 4.641588833612778, which I think are represented identically in binary. I don't think it's possible for more than two numbers to have the same binary representations, so "skipping over" every other number by comparing the guess to a twice improved guess should take care of the issue. Maybe my solution has other issues I'm not aware of or is otherwise flawed but I appear to get the same results as with the original solution.

```
(define (cube x)
(define (cube-iter guess x)
(define (improve guess x)
(/ (+ (/ x (square guess))
(* 2 guess))
3))
(define (good-enough? guess x)
(= (improve (improve guess x) x) guess))
(define (square x) (* x x))
(if (good-enough? guess x)
guess
(cube-iter (improve guess x) x)))
(cube-iter 1.1 x))
```