Concrete Abstractions: Chapter 6

Exercise 6.23
A three-dimensional (3D) vector has x, y, and z coordinates, which are numbers. 3D vectors can be constructed and accessed using the following abstract interface:

(make-3D-vector x y z)
(x-coord vector)
(y-coord vector)
(z-coord vector)

a. Using this abstract interface, define procedures for adding two vectors, finding the dot-product of two vectors, and scaling a vector by a numerical scaling factor.
b. Choose a representation for vectors and implement make-3D-vector, x-coord, y-coord, and z-coord.

Solution: a.

b.

This was a quite short chapter. However chapter 7 is rather long. This exercise above covers pretty much the whole chapter, it was just about cons, car and cdr.

Concrete Abstractions: Chapter 5

Exercise 5.6 Write a general purpose procedure, that when given two integers, low and high, and a procedure for computing a function f , will compute $f(text{low}) + f(text{low} + 1) + f (text{low} + 2) + ... + f (text{high}).$ Show how it can be used to sum the squares of the integers from 5 to 10 and to sum the square roots of the integers from 10 to 100.

Solution: I wrote a iterative version of it because you have to directly apply the function to each number, i.e. it isn’t commutative.

Exercise 5.11 Write a procedure make-verifier, which takes f and m as its two arguments and returns a procedure capable of checking a number. The argument f is itself a procedure, of course. Here is a particularly simple example of a verifier being made and used:

The value #t is the “true” value; it indicates that the number is a valid ISBN.
As we just saw, for ISBN numbers the divisor is 11 and the function is simply $f(i, d_i ) = i * di.$ Other kinds of numbers use slightly more complicated functions, but you will still be able to use make-verifier to make a verifier much more easily than if you had to start from scratch.

Solution:

Exercise 5.12 For UPC codes (the barcodes on grocery items), the divisor is 10, and the function $f (i, d_i )$ is equal to $d_i$ itself when i is odd, but to $3d_i$ when i is even. Build a verifier for UPC codes using make-verifier, and test it on some of your groceries. (The UPC number consists of all the digits: the one to the left of the bars, the ones underneath the bars, and the one on the right.) Try making some mistakes, like switching or changing digits. Does your verifier catch them?

Solution: Yeah, the mistakes were caught. By the way you can see here how powerful high-order functions are.

Exercise 5.13 Credit card numbers also use a divisor of 10 and also use a function that yields $d_i$ itself when i is odd. However, when i is even, the function is a bit fancier: It is $2d_i$ if $latex d_i < 5$, and $latex 2d_i + 1$ if $latex d_i geq 5.$ Build a verifier for credit card numbers. In the dialog at the beginning of this section, did the order taker really mistype the number, or did the customer read it incorrectly? Solution: The credit card number was correct.

Exercise 5.13 The serial number on U.S. postal money orders is self-verifying with a divisor of 9 and a very simple function: $f (i, d_i ) = di$, with only one exception, namely, $f (1, d_1 ) = -d_1.$ Build a verifier for these numbers, and find out which of these two money orders is mistyped: 48077469777 or 48077462766.
Actually, both of those money order numbers were mistyped. In one case the error was that a 0 was replaced by a 7, and in the other case two digits were reversed. Can you figure out which kind of error got caught and which didn’t? Does this help explain why the other kinds of numbers use fancier functions?

Solution: The reversal of digits, if the digit isn’t the first one, isn’t caught by this verification function. This is a reason for using fancier verification functions.

Exercise 5.17 Suppose you have a function and you want to find at what integer point in a given range it has the smallest value. For example, looking at the following graph of the function $f (x) = x^2 - 2x$, you can see that in the range from 0 to 4, this function has the smallest value at 1. We could write a procedure for answering questions like this; it could be used as follows for this example:

Here is the procedure that does this; fill in the two blanks to complete it:

Solution: I find it actually pretty hard to do fill in questions. Anyhow, here’s the solution:

Exercise 5.18 Consider the following definitions:

For the following questions, be sure to indicate how you arrived at your answer:
a. What is the value of (mystery 4)?
b. What is the value of the procedural call ((make-scaled 2 (make-scaled 3 add-one)) 4)?

Solution: a. You can see that mystery gets a function from (make-scaled 3 add-one). This produces (lambda(x) (* 3 (add-one x))) which equals (lambda(x) (* 3 (+ 1 x))). Therefore the value is 15.
b. Let’s take this call apart. (make-scaled 2 f) produces (lambda(x) (* 2 (f x)). Here f is (make-scaled 3 add-one) which produces the known (lambda(x) (* 3 (+ 1 x))) = 15 for x = 4. If you substitute this into (make-scaled 2 f) it becomes (lambda (x) (* 2 15)) = 30.

Exercise 5.20 Suppose the following have been defined:

For each of the following expressions, indicate whether an error would be signaled, the value would be a procedure, or the value would be a number. If an error is signaled, explain briefly the nature of the error. If the value is a procedure, specify how many arguments the procedure expects. If the value is a number, specify which number.
a. f
b. g
c. (* (f 3 2) 7)
d. (g 6)
e. (f 6)
f. ((f 4 7) 5)

Solution:
a. f is a procedure with two arguments.
b. g is a procedure returned by f with one argument.
c. (* (f 3 2) 7) evals to (* (lambda (x) (+ (* 3 x) 2) 7) = (* 23) which throws an error because * needs at least two arguments.
d. Here we got another value: (g 6) = ((f 3 2) 6) = (+ (* 3 6) 2) which equals 20.
e. f needs two arguments. So this throws an error.
f. (+ (* 4 5) 7) which equals 27.

Exercise 5.24 Consider the following procedure:

a. Write a mathematical formula involving n that tells how many times this procedure uses the procedure it is given as its second argument. Justify your answer.
b. Give a simple $Theta$ order of growth for the quantity you determined in part a. Justify your answer.
c. Suppose you were to rewrite this procedure to make it more efficient. What (in terms of n) is the minimum number of times it can invoke f and still always determine the correct answer? Justify your answer. (You are not being asked to actually rewrite the procedure.)

Solution: a. If next > n then loop returns directly smallest-so-far, so there’s no call of f. If next <= n then there are two calls in the if header. Therefore there are $2(n-1)$ function calls of f.
b. We can see that $2(n-1) = 2n - 2.$ $2n$ is the biggest term and 2 is just a factor, therefore $\Theta(n).$
c. We can save (f where-smallest-so-far) by adding it in the parameter list of loop. In the ideal case we just need to calculate (f where-smallest-so-far) one time and (f next-to-try) (n – 1) times. However, in the worst case there’s no improvement.

Concrete Abstractions: Chapter 4

Exercise 4.2 In this exercise you will show that this version of mod-expt does $\Theta (e)$ multiplications, as we claimed.
(1) e is a nonnegative integer, (mod-expt b e m) does at least e multiplications.
(2) When e is a positive integer, (mod-expt b e m) does at most 2e – 1 multiplications.

Solution: At first I wrote the function mathematically. $mod^* = m^*(x, y) = xy \text{ mod } m.$
$\text{mod-expt} = m(b, e, m)$
$m(b, 0, m) = 1$
$m(b, e, m) = \begin{cases} m^*(m(b, e/2, m), m(b, e/2, m) & \text{ if e is even}\\ m^*(m(b, e-1, m), b) & \text{ if e is odd }\\ \end{cases}$

(1) Ok, now let’s start with the first assumption: m(b, e, m) does at least e multiplications. For $e=0$ follows that $m(b, 0, m) = 1.$ So that’s true. Therefore we can assume that this holds to $e+1.$
If $e+1$ is odd:
$m(b, e+1, m) = m^*(m(b, e, m), b)$, we assumed that $m(b, e, m)$ does $e$ multiplications. If we look back at the definition of $m^*(x,y)$, we can see that there’s one multiplication. Therefore there are $e + 1$ multiplications. That was the best case, let’s look at the worst one.

(2) The assumption is that $m(b, e, m)$ needs $2e-1$ multiplications at most. Let’s check the base case for $e=1.$ $m(b, 1, m) = m^*(m(b, 0, m), b)$. Here we get one multiplication from $m^*$, so the maximum limit holds. If $e+1$ is even, it follows that:
$m(b, e+1, m) = m^*(m(b, (e+1)/2, m), m(b, (e+1)/2, m)).$ Each nested $m(b, (e+1)/2, m)$ needs at most $2* \frac{(e+1)}{2} -1$ multiplications and we got another from $m^*$. That is, there are $2 * (2*\frac{(e+1)}{2} -1) + 1 = 2(e+1) - 1$ multiplications.

You can check the even case in (1) and the odd case in (2) for yourself and see that these are bigger and respectively smaller than the other ones.

Exercise 4.11 Consider the following procedures:

a. Give a formula for how many multiplications the procedure factorial does as a function of its argument n.
b. Give a formula for how many multiplications the procedure factorial-sum1 does (implicitly through factorial) as a function of its argument n.
c. Give a formula for how many multiplications the procedure factorial-sum2 does as a function of its argument n.

Solution:
a. $\text{factorial}(0) = 1, \text{factorial}(n) = n * \text{factorial}(n-1)$
What happens for $n = 1, 2, 3,..$? Let’s see:
$\text{factorial}(1) = 1 * \text{factorial}(0) = 1 * 1$
$\text{factorial}(2) = 2 * \text{factorial}(1) = 2 * 1 * \text{factorial}(0) = 2 * 1 * 1$
$\text{factorial}(3) = 3 * \text{factorial}(2) = 3 * 2 * \text{factorial}(1)$
$= 3 * 2 * 1 * \text{factorial}(0) = 3 * 2 * 1 * 1$
We can see that there are probably $n+1$ multiplications. We can prove that by induction.
$\text{factorial}(n+1) = (n+1) * \text{factorial}(n)$. We assumed that $\text{factorial}(n)$ multiplies $(n+1)$ times. Therefore we have $1 + (n+1) = (n+1) + 1$ multiplications.
b. And again written mathematically:
$\text{factorial-sum1}(0) = 0$
$\text{factorial-sum1}(n) = \text{factorial}(n) + \text{factorial-sum1}(n-1)$
So was happens here here for $n = 1, 2, ..$?
$\text{factorial-sum1}(1) = \text{factorial}(1) + \text{factorial-sum1}(0)$
$= (1 * 1) + 0$
$\text{factorial-sum1}(2) = \text{factorial}(2) + \text{factorial-sum1}(1)$
$= (2 * 1 * 1) + \text{factorial}(1) + \text{factorial-sum1}(0)$
$= (2 * 1 * 1) + (1 * 1) + 0$
For $n=1$ we need 2 multiplications, for $n=2$ we need $2 + 3$ multiplications. In general we need $\sum_{i=1}^n (i+1)$ multiplications. This can be simplified: $\sum_{i=1}^n i + sum_{i=1}^n 1 = \frac{n(n+1)}{2} + n = \frac{n^2 + 3n}{2}.$
c. This is pretty straight forward. You can see that each iteration k is increased by one. It starts at k = 1 and stops if k > n. So it runs n times.

Exercise 4.12 How many ways are there to factor n into two or more numbers (each of which must be no smaller than 2)? We could generalize this to the problem of finding how many ways there are to factor n into two or more numbers, each of which is no smaller than m. That is, we write

Your job is to write ways-to-factor-using-no-smaller-than.

Solution: An easy way to find factors is just to try every number if it divides n, starting with m. If a number divides n, take the division and find its factors.

Exercise 4.13 Consider the following procedure:

How many multiplications (expressed in $\Theta$ notation) will the computation of (bar n) do? Justify your answer. You may assume that n is a nonnegative integer.

Solution: The definition of bar is:
$\text{bar}(0) = 5$
$\text{bar}(1) = 7$
$\text{bar}(n) = n*\text{n-2}$
For $n = 0, 1, 2, 3, ...$ we get:
$\text{bar}(0) = 5 \text{ | (0 multiplications)}$
$\text{bar}(1) = 7 \text{ | (0 multiplications)}$
$\text{bar}(2) = 2 * \text{bar}(0) = 2 * 5 \text{ | (1 multiplications)}$
$\text{bar}(3) = 3 * \text{bar}(1) = 3 * 7 \text{ | (1 multiplications)}$
$\text{bar}(4) = 4 * \text{bar}(2) = 4 * 2 * 5 \text{ | (2 multiplications)}$
$\text{bar}(5) = 5 * \text{bar}(3) = 5 * 3 * 7 \text{ | (2 multiplications)}$
You can see that there are between $\frac{n-1}{2}$ and $\frac{n}{2}$ multiplications. Therefore there are $\Theta(n)$ multiplications at general.

Exercise 4.14 Consider the following procedures:

How many multiplications (expressed in $\Theta$ notation) will the computation of (foo n) do? Justify your answer.

Solution: It’s basically the same schema as the other exercises. If we look at foo it looks like this:
$\text{foo}(n) = \text{fac}(n) + \text{bar}(n, n)$ We already know that $\text{fac}(n)$ needs $(n+1)$ multiplications. If we look at bar, we can see how it works.
$\text{bar}(i,j) = \text{fac}(i) * \text{bar}(i,j-1)$ we can simplify this because foo which calls bar only uses the argument n.
$\text{bar}(n,n) = \text{fac}(n) * \text{bar}(n,n-1).$ Because we already know how many multiplications fac needs, we can concentrate on bar. Let’s see what it does for $n = 1, 2, 3, ...$:
$\text{bar}(0, 0) = 1 \text{ | 0 multipl. }$
$\text{bar}(1, 1) = \text{fac}(1) * \text{bar}(0, 0) text { | 2 + 1 + 0 multipl.}$
$\text{bar}(2, 2) = \text{fac}(2) * \text{bar}(1, 1) \text { | 3 + 1 + (2 + 1) multipl.}$
We can generalize this as $\sum_{k=3}^{n+2} k = \sum_{k=1}^{n+2} - 3$ which can be simplified as: $\frac{(n+2)(n+3)}{2} - 3 = \frac{(n+2)(n+3) - 6}{2} = \frac{n^2 + 5n}{2}$ therefore we have $\Theta(n^2)$ multiplications.

Exercise 4.17 Consider the following enumeration problem: How many ways can you choose k objects from n distinct objects, assuming of course that $0 \leq k \leq n$? For example, how many different three-topping pizzas can be made if you have six toppings to choose from?
The number that is the answer to the problem is commonly written as C(n, k). Here is an algorithm for computing C(n, k): […]
Using this algorithm, write a tree-recursive procedure that calculates the numbers C(n, k) described above.

Solution:

Exercise 4.18 One way to sum the integers from a up to b is to divide the interval in half, recursively sum the two halves, and then add the two sums together. Of course, it may not be possible to divide the interval exactly in half if there are an odd number of integers in the interval. In this case, the interval can be divided as nearly in half as possible.
a. Write a procedure implementing this idea.
b. Let’s use n as a name for the number of integers in the range from a up to b. What is the order of growth (in $\Theta$ notation) of the number of additions your procedure does, as a function of n? Justify your answer.

Solution: a.

b. It’s basically the same as in mod-expt because both split up the calculation in two branches each time. Therefore it needs $\Theta(\text{log } n)$ additions.

Concrete Abstractions: Chapter 3

Let’s call the number of people in the circle n and number the positions from 1 to n. We’ll assume that the killing of every third person starts with killing the person in position number 3. […]
As we saw above, if the person we care about is in position 3, that person is the one killed and hence definitely not a survivor:

Suppose we aren’t interested in the person in position number 3 but rather in some other person—let’s say J. Doe. The person in position number 3 got killed, so now we only have n – 1 people left. Of that smaller group of n – 1, there will still be two survivors, and we still want to know if J. Doe is one of them.

Exercise 3.9 How about the people who were in positions 1 and 2; what position numbers are they in after the renumbering?

Solution: In the example there were 8 persons at the start. After killing #3, the new #1 is #4, the new #2 is #5*, etc. #2 is #(n-1) and #1 is #(n-2).

*: clearly a reference to the Prisoner ;)

Exercise 3.10 Write a procedure for doing the renumbering. It should take two arguments: the old position number and the old number of people (n). (It can assume that the old position number won’t ever be 3, because that person is killed and hence doesn’t get renumbered.) It should return the new position number.

Solution:

Exercise 3.11 Finish writing the survives? procedure, and carefully test it with a number of cases that are small enough for you to check by hand but that still cover an interesting range of situations.

Solution:

I made a table for testing if the program works correctly. Each line represents a new round, i.e. somebody got killed. The killed person is labeled with X.

Exercise 3.15 Consider the following two procedures:

a. Use the substitution model to evaluate each of (f 1), (f 2), and (f 3).
b. Can you predict (f 4)? (f 5)? In general, which arguments cause f to return 0 and which cause it to return 1? (You need only consider nonnegative integers.)
c. Is the process generated by f iterative or recursive? Explain.

Solution: a. I started to define each function mathematically. $f(0) = 0, f(n) = g(n-1)$ and $g(0) = 1, g(n) = f(n-1)$. Now we can see what’s happening.
$f(1) = g(0) = 1$
$f(2) = g(1) = f(0) = 0$
$f(3) = g(2) = f(1) = g(0) = 1$
b. You see that odd numbers lead to $g(0) = 1$ and even numbers lead to $f(0) = 0$.
c. I would say that f’s process iterative because you could stop at any time and just continue later with the saved parameters.

Exercise 3.16 Consider the following two procedures:

a. Use the substitution model to illustrate the evaluation of (f 2), (f 3), and (f 4).
b. Is the process generated by f iterative or recursive? Explain. c. Predict the values of (f 5) and (f 6).

Solution: a. Same trick at last time.
$f(0) = 0, f(n) = 1 + g(n-1)$ and $g(0) = 1, g(n) = 1 + f(n - 1)$
Therefore: $f(2) = 1 + g(1) = 1 + (1 + f(0)) = 1 + 1 + 0 = 2$
$f(3) = 1 + g(2) = 1 + (1 + f(1)) = 1 + (1 + (1 + g(0)))$ $1 + (1 + (1 + 1) = 3$
$f(4) = 1 + g(3) = 1 + (1 + f(2)) = 1 + (1 + (1 + g(1)))$ $= 1 + (1 + (1 + (1 + f(0)))) = 3$

b. The process is recursive because you can’t just stop and continue later. This happens because of the (+ 1 f) which have to be saved on the stack.
c. The values of (f 5) and (f 6) are both 6.

Exercise 3.18 We’ve already seen how to raise a number to an integer power, provided that the exponent isn’t negative. We could extend this to allow negative exponents as well by using the following definition:
$b^n = \begin{cases} 1 & \text{ if } x=0 \\ b^{n-1}*b & \text{ if } n>0 \\ b^{n+1}/b & \text{ if } n<0 \end{cases}$
a. Using this idea, write a procedure power such that (power b n) raises b to the n power for any integer n.
b. Use the substitution model to show how (power 2 -3) would be evaluated. (You can leave out steps that just determine which branch of a cond or if should be taken.) Does your procedure generate a recursive process or an iterative one?

Solution a.

b.

It’s an recursive one because you can’t determine the final solution without knowing the beginning parameters.

Exercise 3.19 Prove that, for all nonnegative integers n and numbers a, the following procedure computes the value $2^n *a$:

Solution Let’s define the function again mathematically. $f(0, a) = a, f(n, a) = f(n-1, 2a)$. The base case is correct:
$f(0, a) = 2^0*a = a$.
We assume that $f(n+1, a) = 2^{n+1}*a = 2^n * 2a$ which can be written as $2^n * 2a = 2^n * b$.
With our assumption it should follow that $f(n+1, a) = f(n, b).$
If we look at the definition, we see that $f(n, a) = f(n-1, 2a)$ which implies $f(n+1, a) = f(n, 2a) = f(n, b).$