wiki/2020-06-25-23-24-24.org

Structure and interpretation of computer programs

• Elements of programming
• Higher-order functions
• Exercise solutions
  • Exercise 1.2
  • Exercise 1.3
  • Exercise 1.5
  • Exercise 1.6
  • Exercise 1.7
  • Exercise 1.8
  • Exercise 1.9
  • Exercise 1.10
  • Exercise 1.30
  • Exercise 1.31a
  • Exercise 1.31b
  • Exercise 1.32a

This is just my personal notes on Structure and interpretation of computer programs. I also studied with the Brian Harvey's SICP lectures because I am a scrub. ;p

Before you can use this document, you need to do some prerequisite installation of Racket and SICP package.

Elements of programming

Programming often requires the following:

• Simple expressions with atomic value.
• A way to combine procedures into complex expressions.
• A way to define procedures for abstractions of higher-level functions.

In order to do programming, we must have a programming language. A programming language often requires the following to have an effective way of expressing code:

• Expressions which varies from primitive expressions (e.g., 42, 1.683, 53819492184) to compound expressions (e.g., (+ 53 20), (- 464 254)).
• An environment of objects which you can refer by name either with values (e.g., (define x 10), (define pi 3.14), (define e 2.71828)) or procedures (e.g., (define (square x) (* x x)), (define (my-formula height weight length) (* 23 (/ height weight) (+ 3500 length)))).
• An evaluation model for expressions since certain procedures can have different output from the order of operations.

A programming language lets us abstract procedures as a black box. Here's an example of implementing the square root given a number.

(define (square x) (* x x))
(define (improve guess x)
  (/ (+ guess (/ x guess)) 2))

(define (good-enough? guess x)
  (< (abs (- (square guess) x)) 0.001))

(define (sqrt-iter guess x)
  (if (good-enough? guess x)
      guess
      (sqrt-iter (improve guess x) x)))

(define (sqrt x)
  (sqrt-iter 1.0 x))

(sqrt 4)
(sqrt 100)
(sqrt 45.65)

2.0000000929222947
10.000000000139897
6.756478442187127

In order to do the square root extraction in our implementation, we define multiple procedures with each solving a part of the procedure: one procedure for indicating whether the guess is good enough, one for creating an improved guess for the next iteration, and one for actually doing the square root extraction.

In general cases, we don't implement things as one thing as it will result in a messy state of code. Instead, we modularize those functions. We classify these procedures as a procedural abstraction.

Higher-order functions

Functions and data are often separated similarly to verbs and subjects. We tend to think of them as different things relying on each other to do things: functions need data to manipulate while data are raw information to be arranged by a function. However, the reality is that there is a blurry line to how distinct both of them are. Functions can be treated similarly to data and vice versa.

The lesson of higher-order functions proves this. It is one of the foundations of functional programming. In order to learn about it, you need to know the key: generalizing patterns.

For example, say we have different functions for knowing the area of a shape.

(define pi 3.14)

(define (square-area r) (* r r))
(define (circle-area r) (* pi r r))
(define (hexagon-area r) (* (sqrt 3) 1.5 r r))

This could pass as a decent code if each area function is distinct from each other. However, all of the given area functions involves squaring the given parameter (r). We can separate that step in a function like the following.

(define pi 3.14)
(define (area shape r) (* shape r r))

(define square 1)
(define circle pi)
(define hexagon (* (sqrt 3) 1.5))

Exercise solutions

Exercise 1.2

(/ (+ 5 4
    (- 2
        (- 3
          (+ 6
              (/ 1 3)))))
  (* 3
    (- 6 2)
    (- 2 7)))

Exercise 1.3

(define (square x) (* x x))
(define (sum-of-squares x y z)
  (define sum (+ (square x) (square y) (square z)))
  (- sum (square (min x y z))))

Exercise 1.5

If the interpreter evaluates with applicative-order, it will never evaluate the if condition since (p) is now endlessly being evaluated. (Applicative-order evaulates each argument before passing on the function.) Meanwhile, if it's evaluated at normal order, it would simply expand then start to evaluate them in order. It would go evaluate the if condition and proceed to return 0 (since it returns true).

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.

The reason why if needs a special form is because of applicative-order evaluation. Scheme (or rather Racket with the SICP package) interprets with applicative-order evaluation which it means it has to evaluate all of the arguments first before proceeding to evaluate the procedure. As new-if is a procedure that we defined, it would cause an infinite loop of Racket trying to evaluate sqrt-iter inside of our new-if procedure.

Exercise 1.7

The good-enough? test used in computing square roots will not be very effective for finding the square roots of very small numbers. Also, in real computers, arithmetic operations are almost always performed with limited precision. This makes our test inadequate for very large numbers. Explain these statements, with examples showing how the test fails for small and large numbers. An alternative strategy for implementing good-enough? is to watch how guess changes from one iteration to the next and to stop when the change is a very small fraction of the guess. Design a square-root procedure that uses this kind of end test. Does this work better for small and large numbers?

For Exercise 1.7, I'm afraid I cannot easily answer it since the results from the example implementation is already accurate due to the interpreter.

For this exercise, let's pretend the interpreter is not great. For example, (sqrt 0.0001) results in .03230844833048122 (should be 0.01). ¹

The reason varies from a combination of interpreter, hardware configurations, and implementation of arithmetics. This is especially true with floating points arithmetics.

In implementing our improved square root implementation from the question, we start with editing the improve function.

(define (square x) (* x x))
(define (improve guess x)
  (/ (+ guess (/ x guess)) 2))

(define (good-enough? guess old-guess tolerance)
  (<= (abs (- guess old-guess)) tolerance))

(define (sqrt-iter guess old-guess x)
  (if (good-enough? guess old-guess 0.0000001)
      guess
      (sqrt-iter (improve guess x) guess x)))

(define (sqrt x)
  (sqrt-iter 1.0 0.0 x))

(sqrt 4)
(sqrt 1)
(sqrt 0.0001)
(sqrt 0.00001)
(sqrt 123456789000000)

2.000000000000002
1.0
0.01
0.0031622776602038957
11111111.060555555

I've modified the good-enough? function by making the tolerance as an argument. Tested on the MIT Scheme v10.1.10, the results are more accurate closer to modern systems like Julia. Bigger numbers are also calculated quicker than the previous implementation (for some reason that I don't know).

Exercise 1.8

Newton's method for cube roots is based on the fact that if y is an approximation to the cube root of x, then a better approximation is given by the value

\begin{equation*} \frac{x / y^2 + 2y}{3} \end{equation*}

Use this formula to implement a cube-root procedure analogous to the square-root procedure. (In section 1.3.4 we will see how to implement Newton's method in general as an abstraction of these square-root and cube-root procedures.)

(define (square x) (* x x))
(define (improve guess x)
  (/ (+ (- x (square guess)) (* guess 2)) 3))

(define (good-enough? guess x)
  (< (abs (- (square guess) x)) 0.001))

(define (cbrt-iter guess x)
  (if (good-enough? guess x)
      guess
      (cbrt-iter (improve guess x) x)))

(define (cbrt x)
  (cbrt-iter 1.0 x))

(cbrt 9)

3.000163135454436

Exercise 1.9

Each of the following two procedures defines a method for adding two positive integers in terms of the procedures inc, which increments its argument by 1, and dec, which decrements its argument by 1.

(define (+ a b)
  (if (= a 0)
      b
      (inc (+ (dec a) b))))

(define (+ a b)
  (if (= a 0)
      b
      (+ (dec a) (inc b))))

Using the substitution model, illustrate the process generated by each procedure in evaluating (+ 4 5). Are these processes iterative or recursive?

For the first definition, the resulting evaluation would have to look something like the following:

(+ 4 5)
(inc (+ 3 5))
(inc (inc (+ 2 5)))
(inc (inc (inc (+ 1 5))))
(inc (inc (inc (inc (+ 0 5)))))
(inc (inc (inc (inc 5))))
(inc (inc (inc 6)))
(inc (inc 7))
(inc 8)
9

Based from the visualization, it seems it is a recursive process.

As for the second definition, the resulting evaluation would look like the following:

(+ 4 5)
(+ 3 6)
(+ 2 7)
(+ 1 8)
(+ 0 9)
9

As each iteration does not result in embedding procedures in one big procedure, I think it is considered as an iterative process.

Exercise 1.10

The following procedure computes a mathematical function called Ackermann's function.

(define (A x y)
  (cond ((= y 0) 0)
        ((= x 0) (* 2 y))
        ((= y 1) 2)
        (else (A (- x 1)
                 (A x (- y 1))))))

What are the values of the following expressions?

(A 1 10)

(A 2 4)

(A 3 3)

Consider the following procedures, where A is the procedure defined above:

(define (f n) (A 0 n))

(define (g n) (A 1 n))

(define (h n) (A 2 n))

(define (k n) (* 5 n n))

Give concise mathematical definitions for the functions computed by the procedures f, g, and h for positive integer values of $n$. For example, (k n) computes $5n^2$.

For the sake of completeness, here is the function in question along with the given example usage (and its results in the following block):

(define (A x y)
  (cond ((= y 0) 0)
        ((= x 0) (* 2 y))
        ((= y 1) 2)
        (else (A (- x 1)
                 (A x (- y 1))))))

(A 1 10)
(A 2 4)
(A 3 3)

1024
65536
65536

As for notating f, g, and h into mathematical definitions:

• f is $2n$.
• g is $2^n$.
• h is $2^n^2$.

Exercise 1.30

The sum procedure above generates a linear recursion. The procedure can be rewritten so that the sum is performed iteratively. Show how to do this by filling in the missing expressions in the following definition:

(define (sum term a next b)
  (define (iter a result)
    (if (> a b)
        result
        (iter (next a) (+ result a))))
  (iter a 0))

Exercise 1.31a

The sum procedure is only the simplest of a vast number of similar abstractions that can be captured as higher-order procedures. Write an analogous procedure called product that returns the product of the values of a function at points over a given range. Show how to define factorial in terms of product. Also use product to compute approximations to π using the formula.

\begin{equation*} \frac{\pi}{4} = \frac{2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdot 8 \cdots}{3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot 7 \cdots} \end{equation*}

(define (product term a next b)
  (if (> a b)
      term
      (product (* (next a) term) (+ a 1) next b)))

(define (factorial term)
  (product 1  1 (lambda (a) a) term))

(define (wallis_prod term)
  (* 4 (product 1 1
                (lambda (a) (*
                             (/ (* 2 a) (+ (* 2 a) 1))
                             (/ (+ (* 2 a) 2) (+ (* 2 a) 1))))
                term)))

(factorial 1) ; should return 1
(factorial 5) ; should return 120
(factorial 10) ; should return 3628800
(factorial 20) ; should return 20!

; With larger values should return closer to the value of pi.
(wallis_prod 1)
(wallis_prod 5)
(wallis_prod 10)
(wallis_prod 20)

1
120
3628800
2432902008176640000
32/9
524288/160083
274877906944/85530896451
302231454903657293676544/95064880114531295493525

Notwithstanding related to solving the entire problem, I'll include note on how I was able to create a procedure for the Pi value computation since it gave me the hardest time. In order to start creating a procedure, I've simply observed the given formula with the induction that it can be separated into pairs like the following. (I also simply didn't observe that each pair is also an iteration of a function.)

\begin{equation*} \frac{\pi}{4} = \left(\frac{2}{3} \cdot \frac{4}{3} \right) \cdot \left(\frac{4}{5} \cdot \frac{6}{5} \right) \cdot \left(\frac{6}{7} \cdot \frac{8}{7} \right) \end{equation*}

We can then observed that it has a generalized pattern. Each iteration, in isolation, can be summarized as such.

\begin{equation*} \left(\frac{2n}{2n+1} \cdot \frac{2n+2}{2n+1}\right) \end{equation*}

With simple algebra, you can get the approximation of Pi by simply multiplying the equation with $4$. Here is the finalized equation to my solution.

\begin{equation*} f(j) \approx \pi \approx 4 \cdot \prod_{n=1}^j \left(\frac{2n}{2n+1} \cdot \frac{2n+2}{2n+1}\right) \end{equation*}

With larger values, the result would be closer to the value of π.

Exercise 1.31b

If your product procedure generates a recursive process, write one that generates an iterative process. If it generates an iterative process, write one that generates a recursive process.

Based from my answer in Exercise 1.31a, we can simply see whether we have created an iterative or recursive process simply with the trace function.

(require racket/trace)
(define (product total fn a b)
  (if (> a b)
      total
      (product (* total (fn a)) fn (+ a 1) b)))

(define (factorial term)
  (product 1 (lambda (a) a) 1 term))

(trace product)
(factorial 5)

>(product 1 #<procedure:...v3/ob-PoMhn9.rkt:10:13> 1 5)
>(product 1 #<procedure:...v3/ob-PoMhn9.rkt:10:13> 2 5)
>(product 2 #<procedure:...v3/ob-PoMhn9.rkt:10:13> 3 5)
>(product 6 #<procedure:...v3/ob-PoMhn9.rkt:10:13> 4 5)
>(product 24 #<procedure:...v3/ob-PoMhn9.rkt:10:13> 5 5)
>(product 120 #<procedure:...v3/ob-PoMhn9.rkt:10:13> 6 5)
<120
120

With our implementation, we can see it is an iterative process. The following code block is its recursive equivalent along with the stack trace for comprehension.

(require racket/trace)
(define (product total fn a b)
  (if (> a b)
      1
      (* (fn a) (product total fn (+ a 1) b))))

(define (factorial term)
  (product 1 (lambda (a) a) 1 term))

(trace product)
(factorial 5)

>(product 1 #<procedure:...v3/ob-lY382a.rkt:10:13> 1 5)
> (product 1 #<procedure:...v3/ob-lY382a.rkt:10:13> 2 5)
> >(product 1 #<procedure:...v3/ob-lY382a.rkt:10:13> 3 5)
> > (product 1 #<procedure:...v3/ob-lY382a.rkt:10:13> 4 5)
> > >(product 1 #<procedure:...v3/ob-lY382a.rkt:10:13> 5 5)
> > > (product 1 #<procedure:...v3/ob-lY382a.rkt:10:13> 6 5)
< < < 1
< < <5
< < 20
< <60
< 120
<120
120

Exercise 1.32a

Show that sum and product (exercise 1.31) are both special cases of a still more general notion called accumulate that combines a collection of terms, using some general accumulation function:

(accumulate combiner null-value term a next b)

accumulate takes as arguments the same term and range specifications as sum and product, together with a combiner procedure (of two arguments) that specifies how the current term is to be combined with the accumulation of the preceding terms and a null-value that specifies what base value to use when the terms run out. Write accumulate and show how sum and product can both be defined as simple calls to accumulate.

(define (accumulate combiner null-value term a next b)
  (if (> a b)
      term
      (accumulate combiner null-value (combiner (next a) term) (next a) next b)))

---

¹

You can test how it really goes with the MIT Scheme interpreter.