Exercise 1.01

So I had just started reading SICP. It’s this amazing book that was used to teach Comp Sc. to students who had never programmed before at MIT. When I thought that I should write a blog post with a solution for every exercise I attempt. So I am gonna write solutions for all the exercise in SICP. My plan is to finish this book in one year.

You can obtain a copy of SICP for free because of MIT-Open course ware!

My setup

My plan is to use emacs(sorry for the adultery vim!) with slime, and guile. I haven’t setup it up yet, though I will attach a link to the post when I do. I strongly encourage you to setup emacs though because soon the exercises will get really complicated, and the support emacs has for Lisp (Being written in lisp itself) will be really helpful while editing.

So without further delay, Let’s get to the exercise

The Question

Exercise 1.01: 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 presented.

10
(+ 5 3 4)
(- 9 1)
(/ 6 2)
(+ (* 2 4) (- 4 6))
(define a 3)
(define b (+ a 1))
(+ a b (* a b))
(= a b)
(if (and (> b a) (< b (* a b)))
    b
    a)
(cond ((= a 4) 6)
      ((= b 4) (+ 6 7 a))
      (else 25))
(+ 2 (if (> b a) b a))
(* (cond) ((> a b) a)
          ((< a b) b)
          (else -1))
   (+ a 1)

My Thoughts

So far, the authors have been explaining about the basic syntax of scheme. If you are reading this, you most likely also read SICP till this exercise. So I am not gonna speak much about it, but make a table of the syntax:

Answers

So now that we have got out table, let us get started. To check our answers we can use the Scheme shell

10

That will return 10

(+ 5 3 4)

(+ 5 3 4) is basically “sum of 5,3,4” but in prefix notation. That will be 12.

(- 9 1)

9 - 1 is 8. So it will return 8

( / 6 2)

$6 \div 2$ is 3. So 3.

(+ (* 2 4) (- 4 6))

First, $2 \times 4$ is 8 and 4 - 6 is -2. Then 8 + -2 is 6.

(define a 3)

a is now three.

(define b (+ a 1)))

So b will be a + 1 , which is 3 + 1, which equals to 4.

(+ a b (* a b))

First $a \times b = 3 \times 4 = 12$. Then 3 + 4 + 12 = 19.

(= a b)

Now, this test that will output a Boolean. Since 3 is not = 4, it will be False.

(if (and (> b a) (< b (* a b))) b a)

and adds to the test, and now checks if both((> b a) and (< b (* a b) )) return true. Since 4 is greater than 3 and 4 is lesser than 12, this returns True, and 4 is returned.

(cond ((= a 4) 6) ((= b 4) (+ 6 7 a)) (else 25))

b = 4 is the first thing that came true so, what will be returned will be 6 + 7 + 3 = 10.

(* (cond ((> b a) b a)) ((< a b) b) (else -1)) (+ a 1))

This just multiplies whatever cond returns by 4. cond returns 4, so $4^{2} = 16$

That’s it! Tell us about in the comments below about your experience with SICP so far!