SICP/ex-3_73-82.scm

(load "util.scm")

(define (integral integrand initial-value dt)
  (define int
    (cons-stream initial-value
                 (add-streams (scale-stream integrand dt)
                              int)))
  int)

(display "\nex-3.73 - RC\n")

(define (RC R C dt)
  (define (rc-proc i v0)
    (add-streams
      (scale-stream i R)
      (integral (scale-stream i (/ 1 C)) v0 dt)))
  rc-proc)

(define RC1 (RC 5 1 0.5))

(assert (take 5 (RC1 ones 4.2))
        '(9.2 9.7 10.2 10.7 11.2))

(display "\nex-3.74 - zero-crossings\n")

(define sense-data
  (list->stream
                 '(1  2  1.5  1  0.5  -0.1  -2  -3  -2  -0.5  0.2  3  4)))
(define response '(0  0    0  0    0    -1   0   0   0     0    1  0))

(define (sign-change-detector s2 s1)
  (cond
    ((and (< s1 0) (>= s2 0)) 1)
    ((and (>= s1 0) (< s2 0)) -1)
    (else 0)))

(define (make-zero-crossings input-stream last-value)
  (cons-stream
   (sign-change-detector (stream-car input-stream) last-value)
   (make-zero-crossings (stream-cdr input-stream)
                        (stream-car input-stream))))

(define zero-crossings (make-zero-crossings sense-data 0))

(assert (take 12 zero-crossings) response)

(define zero-crossings
  (stream-map sign-change-detector
              sense-data
              (cons-stream 0 sense-data)))

(assert (take 12 zero-crossings) response)

(display "\nex-3.75 - averaged zero-crossings\n")

; Louis' solution uses the average value as last-value which means that the new
; value and the previous average are used for averaging and not simply two
; consecutive values as proposed by Alyssa.

(define (make-zero-crossings-avg input-stream last-value last-avpt)
  (let ((avpt (/ (+ (stream-car input-stream) last-value) 2)))
    (cons-stream (sign-change-detector avpt last-avpt)
                 (make-zero-crossings-avg (stream-cdr input-stream)
                                          (stream-car input-stream)
                                          avpt))))

(display (take 12 (make-zero-crossings-avg sense-data 0 0))) (newline)

(display "\nex-3.76 - smooth zero-crossings\n")

(define (smooth xs)
  (stream-map average xs (stream-cdr xs)))

(define (make-zero-crossings-smoothed input-stream last-value)
  (make-zero-crossings (smooth input-stream) last-value))

(assert (take 11 (make-zero-crossings-smoothed sense-data 0))
        '(0  0    0  0    0    -1   0   0   0     0    1))

(display "\nex-3.77 - lazy-integral\n")

(define (integral delayed-integrand initial-value dt)
  (define int
    (cons-stream initial-value
                 (let ((integrand (force delayed-integrand)))
                   (add-streams (scale-stream integrand dt)
                                int))))
  int)

(define (solve f y0 dt)
  (define y (integral (delay dy) y0 dt))
  (define dy (stream-map f y))
  y)

(assert (stream-ref (solve (lambda (y) y) 1 0.001) 1000) 2.716923932235896)

(define (integral delayed-integrand initial-value dt)
  (cons-stream
    initial-value
    (let ((integrand (force delayed-integrand)))
      (if (stream-null? integrand)
        the-empty-stream
        (integral (delay (stream-cdr integrand))
                  (+ (* dt (stream-car integrand)) initial-value)
                  dt)))))

(assert (stream-ref (solve (lambda (y) y) 1 0.001) 1000) 2.716923932235896)

(display "\nex-3.78 - solve-2nd\n")

; The output stream, modeling y, is generated by a network that contains a loop.
; This is because the value of d2y/dt2 depends upon the values of y and dy/dt and
; both of these are determined by integrating d2y/dt2. The diagram we would like
; to encode is shown in figure 3.35. Write a procedure solve-2nd that takes as
; arguments the constants a, b, and dt and the initial values y0 and dy0 for y
; and dy/dt and generates the stream of successive values of y.

(define (solve-2nd a b dt y0 dy0)
  (define y (integral (delay dy) y0 dt))
  (define dy (integral (delay ddy) dy0 dt))
  (define ddy (add-streams (scale-stream dy a)
                           (scale-stream y b)))
  y)

(assert (stream-ref (solve-2nd 1 0 0.0001 1 1) 10000)
        2.7181459268252266) ; e
(assert (stream-ref (solve-2nd 0 -1 0.0001 1 0) 10472)
        0.5000240628699462) ; cos pi/3 = 0.5
(assert (stream-ref (solve-2nd 0 -1 0.0001 0 1) 5236)
        0.5000141490501059) ; sin pi/6 = 0.5

(display "\nex-3.79 - solve-2nd-general\n")

(define (solve-2nd-general a b) '())

;(display "\nex-3.80\n")
;(display "\nex-3.81\n")
;(display "\nex-3.82\n")