Ex-3.50 - 3.52

Some code from book to execute the solutions..

(define (stream-enumerate-interval low high)
 (if (> low high)
     the-empty-stream
     (cons-stream
      low
      (stream-enumerate-interval (+ low 1) high))))

(define (stream-ref s n)
 (if (= n 0)
     (stream-car s)
     (stream-ref (stream-cdr s) (- n 1))))

(define (display-stream s)
 (stream-for-each display-line s))
(define (display-line x)
 (newline)
 (display x))

Ex 3.50

(define (stream-map proc . argstreams)
 (if (stream-null? (car argstreams))
     the-empty-stream
     (cons-stream
      (apply proc (map stream-car argstreams))
      (apply stream-map
             (cons proc (map stream-cdr argstreams))))))

;Test
(define s1 (cons-stream 1 (cons-stream 2 the-empty-stream)))
(define s2 (cons-stream 1 (cons-stream 2 the-empty-stream)))
(define s3 (stream-map + s1 s2))
s3 ;Value 12: (2 . #[promise 13])
(display-stream s3) ;(2,4)

Ex-3.51

(define (show x)
 (display-line x)
 x)
(define x (stream-map show (stream-enumerate-interval 0 10)))
(stream-ref x 5)
(stream-ref x 7)

;Results Printed
1 ]=> (stream-ref x 5)

1
2
3
4
5
;Value: 5

1 ]=> (stream-ref x 7)

6
7
;Value: 7

Explaination: (stream-ref x 5) printed 1,2,3,4,5 because all the show calls were done after calling it and memoized values are stored and hence (stream-ref x 7) just displays 6,7

Ex-3.52

(define sum 0)
;sum = 0

(define (accum x)
 (set! sum (+ x sum))
 sum)
;sum = 0

(define seq (stream-map accum (stream-enumerate-interval 1 20)))
;sum = 1 , only one call to accum is evaluated to get first element of seq

(define y (stream-filter even? seq))
;sum = 6 ; only calls to accum are evaluated to get first element even element in seq  which is 6

(define z (stream-filter (lambda (x) (= (remainder x 5) 0))
                        seq))
;sum = 10 , only calls to accum are evaluated to get first element in seq that is divisible by 5

(stream-ref y 7)
;Value: 136
;sum = 136 ,only calls to accum upto calculating 7th element in y are evaluated

(display-stream z)
;(10 15 45 55 105 120 190 210)
;sum = 210 ,all the calls are evaluated

No, If we do not use memoization then these results will not stay the same as same accum calls will be evaluated multiple times and hence modifying sum multiple times for the same accum calls.