SICP Exercise 4.15 halts?

Exercise 4.15.  Given a one-argument procedure p and an object a, p is said to ``halt'' on a if evaluating the expression (p a) returns a value (as opposed to terminating with an error message or running forever). Show that it is impossible to write a procedure halts? that correctly determines whether p halts on a for any procedure p and object a. Use the following reasoning: If you had such a procedure halts?, you could implement the following program:

(define (run-forever) (run-forever))

(define (try p)
  (if (halts? p p)
      (run-forever)
      'halted))

Now consider evaluating the expression (try try) and show that any possible outcome (either halting or running forever) violates the intended behavior of halts?.23

R E A S O N I N G

The following is a simple zero-argument procedure that runs forever:

(define (run-forever) (run-forever))

The following is a hypothetical procedure that checks if procedure p halts on input a:

(define (halts? p a)

(if <p halts on a>

true

false

)

)

Assuming that the "halts?" procedure above exists and works, we should be able to write 

another procedure:

(define (try p)

(if (halts? p p)

(run-forever)

'halted

)

)

The "try" procedure takes a procedure as its argument and checks if that procedure halts

on itself. If true, then try runs forever. If false, then try returns.

If p halts on itself, then try runs forever.

If p runs forever, then try halts on p.

So "try" behaves oppositely to the procedure supplied. 

Now if we evaluate (try try), then the logic would work like this:

If try halts on itself, try runs forever.

If try runs forever, try halts on itself.

In both cases, the behaviour of "try" violates the intended behavior of "halts?"

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