Exercise 3.68 Louis Reasoner pairs.rkt

#lang racket

; Exercise 3.68.  Louis Reasoner thinks that building a stream of pairs from three parts
; is unnecessarily complicated. Instead of separating the pair (S0,T0) from the rest of
; the pairs in the first row, he proposes to work with the whole first row, as follows:

; (define (pairs s t)
;   (interleave
;    (stream-map (lambda (x) (list (stream-car s) x)) t)
;   (pairs (stream-cdr s) (stream-cdr t))))

; Does this work? Consider what happens if we evaluate (pairs integers integers)
; using Louis's definition of pairs.

; E X P L A N A T I O N

; Louis Reasoner's procedure will not work. It will run into an infinite loop. Not having
; an outer 'stream-cons', there is no delayed operation to prevent the program from
; getting into an infinite loop. 

; When 'louis-reasoner-pairs' executes, 'interleave' needs to be run. But before that, it's
; arguments need to be evaluated. The second argument is a call to louis-reasoner-pairs
; again. This puts the program into an infinite loop.

(define (louis-reasoner-pairs s t)
	(interleave
		(stream-map (lambda (x) (list (stream-first s) x)) t)
		(louis-reasoner-pairs (stream-rest s) (stream-rest t))
	)
)

(define (all-pairs s t)
	(stream-cons
		(list (stream-first s) (stream-first t))
		(interleave
			(stream-map (lambda (x) (list (stream-first s) x)) (stream-rest t))
			(interleave
				(stream-map (lambda (x) (list x (stream-first t))) (stream-rest s))
				(all-pairs (stream-rest s) (stream-rest t))
			)
		)
	)
)

(define (pairs s t)
	(stream-cons
		(list (stream-first s) (stream-first t))
		(interleave
			(stream-map (lambda (x) (list (stream-first s) x)) (stream-rest t))
			(pairs (stream-rest s) (stream-rest t))
		)
	)
)

(define (interleave s1 s2)
	(if (stream-empty? s1)
		s2
		(stream-cons
			(stream-first s1)
			(interleave s2 (stream-rest s1))
		)
	)
)

(define (partial-sums S)
	(stream-cons (stream-first S) (add-streams (partial-sums S) (stream-rest S)))
)

(define (ln2-summands n)
	(stream-cons
		(/ 1.0 n)
		(stream-map - (ln2-summands (+ n 1)))
	)
)

(define ln2-stream
	(partial-sums (ln2-summands 1))
)

(define (sqrt x tolerance)
	(stream-limit (sqrt-stream x) tolerance)
)

(define (stream-limit s t)
	(let ((s0 (stream-ref s 0)) (s1 (stream-ref s 1)))
			(if (< (abs (- s1 s0)) t)
			s1
			(stream-limit (stream-rest s) t)
		)
	)
)

(define (sqrt-stream x)
	(define guesses
		(stream-cons
			1.0
			(stream-map
				(lambda (guess) (sqrt-improve guess x))
				guesses
			)
		)
	)
	guesses
)

(define (sqrt-improve guess x)
	(average guess (/ x guess))
)

(define (average x y) (/ (+ x y) 2))
(define (square x) (* x x))

(define (euler-transform s)
	(let ((s0 (stream-ref s 0))          ; Sn-1
		 (s1 (stream-ref s 1))           ; Sn
		 (s2 (stream-ref s 2)))          ; Sn+1
		(stream-cons
			(-
				s2
				(/ (square (- s2 s1)) (+ s0 (* -2 s1) s2))
			)
			(euler-transform (stream-rest s))
		)
	)
)

(define (make-tableau transform s)
	(stream-cons
		s
		(make-tableau transform (transform s))
	)
)

(define (accelerated-sequence transform s)
	(stream-map stream-first (make-tableau transform s))
)

(define (scale-stream stream factor)
	(stream-map (lambda (x) (* x factor)) stream)
)

(define (stream-map proc . argstreams)
	; (displayln "Entered stream-map")
	(if (stream-empty? (car argstreams))
		empty-stream
		(stream-cons
			(apply proc (map stream-first argstreams))
			(apply stream-map (cons proc (map stream-rest argstreams)))
		)
	)
)

(define (stream-ref s n)
	; (display "Entered stream-ref with n = ")
	; (display n)
	; (newline)
	(if (= n 0)
		(stream-first s)
		(stream-ref (stream-rest s) (- n 1))
	)
)

; This procedure produces a stream with a maximum number of count elements from
; the supplied stream
(define (truncate-stream s count)
	(if (<= count 0)
		empty-stream
		(stream-cons
			(stream-first s)
			(truncate-stream (stream-rest s) (- count 1))
		)
	)
)

; This procedure displays a finite number of elements from the supplied stream
; as specified by 'count'
(define (display-stream-elements count s)
	(if (= 0 count)
		(begin
			(newline)
			'done
		)
		(begin
			(newline)
			(display (stream-first s))
			(display-stream-elements (- count 1) (stream-rest s))
		)
	)
)

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

(define (display-line x)
	(newline)
	(display x)
)

(define (indent-and-display pair)
	(display-spaces (car pair))
	(display pair)
	(newline)
)

(define (display-spaces n)
	(if (<= n 0)
		(void)
		(begin
			(display " ")
			(display-spaces (- n 1))
		)
	)
)

(define (div-series dividend-series divisor-series)
	(cond
		((= 0 (stream-first divisor-series))
			(error "Denominator should not have a zero constant: " (stream-first divisor-series))
		)
		(else
			(mul-series
				dividend-series
				(invert-unit-series (scale-stream divisor-series (/ 1 (stream-first divisor-series))))
			)
		)
	)
)

(define (invert-unit-series s)
	(stream-cons
		1
		(mul-series
			(scale-stream (stream-rest s) -1)
			(invert-unit-series s)
		)
	)
)

(define (mul-series s1 s2)
	(stream-cons
		; (a0 * b0) (The following is the constant term of the series resulting from
		; the multiplication)
		(* (stream-first s1) (stream-first s2))
		; The following is the rest of the series starting with the x^1 term
		(add-streams
			; {a0 * (b1x + b2x^2 + b3x^3 + ...)} +
			; {b0 * (a1x + a2x^2 + a3x^3 + ...)} +
			(add-streams
				(scale-stream (stream-rest s2) (stream-first s1))
				(scale-stream (stream-rest s1) (stream-first s2))
			)
			; {(a1x + a2x^2 + a3x^3 + ...) * (b1x + b2x^2 + b3x^3 + ...)}
			(stream-cons
				; 0 needs to be prepended to this stream so that the first term of the stream
				; is the x^1 term. Only then the outer add-streams will add like terms in the
				; two series supplied to it
				0
				(mul-series (stream-rest s1) (stream-rest s2))
			)
		)
	)
)

(define exp-series
	(stream-cons 1 (integrate-series exp-series))
)

; the integral of negative sine is cosine
(define cosine-series
	(stream-cons 1 (integrate-series (scale-stream sine-series -1)))
)

; the integral of cosine is sine
(define sine-series
	(stream-cons 0 (integrate-series cosine-series))
)

(define tan-series 
	(div-series sine-series cosine-series)
)

(define (integrate-series s)
	(div-streams s integers)
)

(define (add-streams s1 s2)
	(stream-map + s1 s2)
)

(define (div-streams s1 s2)
	(stream-map / s1 s2)
)

(define ones (stream-cons 1 ones))

(define integers (stream-cons 1 (add-streams ones integers)))

; Test Driver

(define (run-test return-type proc . args)

	(define (print-item-list items first-time?)
		(cond
			((not (pair? items)) (void))
			(else
				(if (not first-time?)
					(display ", ")
					(void)
				)
				(print (car items))
				(print-item-list (cdr items) false)
			)
		)
	)

	(display "Applying ")
	(display proc)
	(if (not (null? args))
		(begin
			(display " on: ")
			(print-item-list args true)
		)
		(void)
	)
	(newline)
	(let ((result (apply proc args)))
		(if (not (eq? return-type 'none))
			(display "Result: ")
			(void)
		)
		(cond
			((procedure? result) ((result 'print)))
			; ((eq? return-type 'deque) (print-deque result))
			((eq? return-type 'none) (void))
			(else
				(print result)
				(newline)
			)
		)
	)
	(newline)
)

(define (execution-time proc . args)
	(define start-time (current-milliseconds))
	; (display start-time)
	; (display " ")
	(apply proc args)
	(define end-time (current-milliseconds))
	; (display end-time) 
	(display "Execution time of ")
	(display proc)
	(display ": ")
	(- end-time start-time)
)

; Tests

; Test Results

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Language: racket, with debugging; memory limit: 1024 MB.
> (louis-reasoner-pairs integers integers)

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