SICP section 1.1 The Elements of Programming

Exercise 1.1. Below is a sequence of expressions. What is the result printed by the interpreter in response to each expression? Assume that the sequence is to be evaluated in the order in which it is presented.
10 ;; 10 (+ 5 3 4) ;; 12 (- 9 1) ;; 8 (/ 6 2) ;; 3 (+ (* 2 4) (- 4 6)) ;; 6 (define a 3) ;; <nothing is printed> (define b (+ a 1)) ;; <nothing is printed> (+ a b (* a b)) ;; 19 (= a b) ;; #f (if (and (> b a) (< b (* a b)) ) b a) ;; 4 (cond ((= a 4) 6) ((= b 4) (+ 6 7 a)) (else 25) ) ;; 16 (+ 2 (if (> b a) b a)) ;; 6 (* (cond ((> a b) a) ((< a b) b) (else -1)) (+ a 1) ) ;; 16
Exercise 1.2. Translate the following expression into prefix form

(/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5))))) (* 3 (- 6 2) (- 2 7)) )
Exercise 1.3. Define a procedure that takes three numbers as arguments and returns the sum of the squares of the two larger numbers. (define (smallest x y z) (cond ((and (>= x z) (>= y z)) z) ((and (>= y x) (>= z x)) x) ((and (>= z y) (>= x y)) y) ) ) (define (square x) (* x x)) (define (sum_of_squares x y) (+ (square x) (square y) ) ) (define (sum_of_squares_of_larger_two x y z) (cond ((= (smallest x y z) x) (sum_of_squares y z)) ((= (smallest x y z) y) (sum_of_squares x z)) ((= (smallest x y z) z) (sum_of_squares x y)) ) )
Exercise 1.4. Observe that our model of evaluation allows for combinations whose operators are compound expressions. Use this observation to describe the behavior of the following procedure: (define (a-plus-abs-b a b) ((if (> b 0) + -) a b))
Answer: The procedure a_plus_abs_b takes two arguments, a and b. It evaluates the expression (operator a b) where operator is chosen as either + or - depending on whether b is greater than zero or not. The key is that operator itself is the value of evaluating the if expression.

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.)

Answer: If applicative order is used for evaluation the arguments to test are evaluated first. This means (p) is evaluated, leading to an infinite loop. On the other hand with normal order evaluation (p) is not evaluated until it is actually needed. Since the if is designed in such a way that this never happens, there is no infinite loop and the call returns 0.

Exercise 1.6. Alyssa P. Hacker doesn't see why if needs to be provided as a special form. ``Why can't I just define it as an ordinary procedure in terms of cond?'' she asks. Alyssa's friend Eva Lu Ator claims this can indeed be done, and she defines a new version of if: (define (new-if predicate then-clause else-clause) (cond (predicate then-clause) (else else-clause)))
Eva demonstrates the program for Alyssa: (new-if (= 2 3) 0 5) 5 (new-if (= 1 1) 0 5) 0
Delighted, Alyssa uses new-if to rewrite the square-root program: (define (sqrt-iter guess x) (new-if (good-enough? guess x) guess (sqrt-iter (improve guess x) x)))
What happens when Alyssa attempts to use this to compute square roots? Explain.

Answer: Normally if evaluates the alternative only if the predicate is false. This is why it is a special form. Since Alyssa's version is not a special form the alternative is evaluated irrespective of the result of the predicate. In the case of square roots this means calling sqrt-iter again, leading to an infinite loop.