SICP Exercise 3.63 sqrt-stream Louis Reasoner

Exercise 3.63.  Louis Reasoner asks why the sqrt-stream procedure was not written in the following more straightforward way, without the local variable guesses:

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

Alyssa P. Hacker replies that this version of the procedure is considerably less efficient because it performs redundant computation. Explain Alyssa's answer. Would the two versions still differ in efficiency if our implementation of delay used only (lambda () <exp>) without using the optimization provided by memo-proc (section 3.5.1)?

 EXPLANATION

 If we do not use the local variable 'guesses', then every call to procedure sqrt-stream creates a brand-new stream. In Louis Reasoner's implementation of 'sqrt-stream', a call to:

(display-stream (sqrt-stream 100))

will result in generating the stream elements from the first element till the 100th element. Since this will be the first time the stream elements are accessed, all delayed objects will be forced when we display the stream.

If after this we call:

(display-stream (sqrt-stream 101))

a new stream is created so the evaluations to generate all elements from the first
till the 101st element will be performed again. All delayed objects will be forced again since this is a new stream that shares no state with the previous stream.

Coming to the second part of the problem, if our implementation of delay used only (lambda () <exp>) without taking advantage of memo-proc, then the two versions will differ in efficiency only to the small extent that whereas sqrt-stream maintains a single stream, Louis' implementation constructs a new stream every time it is run. Post-construction of the stream, the calls to access the stream elements will not differ in efficiency between the two implementations since in both cases, all delayed objects will be forced every time they are accessed.

 The code is here.

Comments

Post a Comment

Popular posts from this blog

SICP Exercise 2.56 differentiation rule

Image
Exercise 2.56.    Show how to extend the basic differentiator to handle more kinds of expressions. For instance, implement the differentiation rule by adding a new clause to the  deriv  program and defining appropriate procedures  exponentiation? ,  base ,  exponent , and  make-exponentiation . (You may use the symbol  **  to denote exponentiation.) Build in the rules that anything raised to the power 0 is 1 and anything raised to the power 1 is the thing itself. SOLUTION The code and tests are here .
Read more

SICP Exercise 1.28 (Miller-Rabin Test)

Exercise 1.28.    One variant of the Fermat test that cannot be fooled is called the  Miller-Rabin test  (Miller 1976; Rabin 1980). This starts from  an alternate form of Fermat's Little Theorem, which states that if  n  is a prime number and  a  is any positive integer less than  n , then  a  raised to the ( n  - 1)st power is congruent to 1 modulo  n . To test the primality of a number  n  by the Miller-Rabin test, we pick a random number  a < n  and raise  a  to the ( n  - 1)st power modulo  n  using the  expmod  procedure. However, whenever we perform the squaring step in  expmod , we check to see if we have discovered a ``nontrivial square root of 1 modulo  n ,'' that is, a number not equal to 1 or  n  - 1 whose square is equal to 1 modulo  n . It is possible to prove that if such a nontrivial square root of 1 exists, then  n  is not prime. It is also possible to prove that if  n  is an odd number that is not prime, then, for at least half the numbers a < n , co
Read more

SICP Exercise 4.18 a alternative strategy for interpreting internal definitions

Exercise 4.18.   Consider an alternative strategy for scanning out definitions that translates the example in the text to (lambda < vars >   (let ((u '*unassigned*)         (v '*unassigned*))     (let ((a < e1 >)           (b < e2 >))       (set! u a)       (set! v b))     < e3 >)) Here  a  and  b  are meant to represent new variable names, created by the interpreter, that do not appear in the user's program. Consider the  solve  procedure from section  3.5.4 : (define (solve f y0 dt)   (define y (integral (delay dy) y0 dt))   (define dy (stream-map f y))   y) Will this procedure work if internal definitions are scanned out as shown in this exercise? What if they are scanned out as shown in the text? Explain. SOLUTION & DISCUSSION The streams based procedure 'solve' in this exercise exposed several issues in my metacircular evaluator implementation. I discovered that streams are not at all handled properly by my evaluator. Moreover, the &
Read more