Notes from Structure and interpretation of computer programs.

Note1:Normal Order vs Applicative order

Applicative order: To apply a compound procedure to arguments, evaluate the body of the procedure with each formal parameter replaced by the corresponding argument.

example:(sum-of-squares (+ a 1) (* a 2))

Then we replace the formal parameter a by the argument 5:

(sum-of-squares (+ 5 1) (* 5 2))

Thus the problem reduces to the evaluation of a combination with two operands and an operator sum-of-squares. Evaluating this combination involves three subproblems. We must evaluate the operator to get the procedure to be applied, and we must evaluate the operands to get the arguments. Now (+ 5 1) produces 6 and (* 5 2) produces 10, so we must apply the sum-of-squares procedure to 6 and 10. These values are substituted for the formal parameters x and y in the body of sum-of-squares, reducing the expression to

(+ (square 6) (square 10))

If we use the definition of square, this reduces to

(+ (* 6 6) (* 10 10))

which reduces by multiplication to

(+ 36 100)

and finally to

136


This alternative ``fully expand and then reduce'' evaluation method is known as normal-order evaluation. Instead it would first substitute operand expressions for parameters until it obtained an expression involving only primitive operators, and would then perform the evaluation. If we used this method, the evaluation of

(f 5)

would proceed according to the sequence of expansions

(sum-of-squares (+ 5 1) (* 5 2))

(+    (square (+ 5 1))      (square (* 5 2))  )

(+    (* (+ 5 1) (+ 5 1))   (* (* 5 2) (* 5 2)))

followed by the reductions

(+         (* 6 6)             (* 10 10))

(+           36                   100)

                    136

This gives the same answer as our previous evaluation model, but the process is different. In particular, the evaluations of (+ 5 1) and (* 5 2) are each performed twice here, corresponding to the reduction of the expression

(* x x)

with x replaced respectively by (+ 5 1) and (* 5 2).

 in contrast to the ``evaluate the arguments and then apply'' method that the interpreter actually uses, which is called applicative-order evaluation. It can be shown that, for procedure applications that can be modeled using substitution (including all the procedures in the first two chapters of this book) and that yield legitimate values, normal-order and applicative-order evaluation produce the same value. 


Exercise 1.5.  Ben Bitdiddle has invented a test to determine whether the interpreter he is faced with is using applicative-order evaluation or normal-order evaluation. He defines the following two procedures:

(define (p) (p))

(define (test x y)
  (if (= x 0)
      0
      y))

Then he evaluates the expression

(test 0 (p))

What behavior will Ben observe with an interpreter that uses applicative-order evaluation? What behavior will he observe with an interpreter that uses normal-order evaluation? Explain your answer. (Assume that the evaluation rule for the special form if is the same whether the interpreter is using normal or applicative order: The predicate expression is evaluated first, and the result determines whether to evaluate the consequent or the alternative expression.)

Using applicative-order evaluation, the evaluation of (test 0 (p)) never terminates, because (p) is infinitely expanded to itself:

 (test 0 (p)) 
  
 (test 0 (p)) 
  
 (test 0 (p)) 
... and so on.

Using normal-order evaluation, the expression evaluates, step by step, to 0:

 (test 0 (p)) 
  
 (if (= 0 0) 0 (p)) 
  
 (if #t 0 (p)) 
  
 0