Notes on: Structure and Interpretation of Computer Programs (Abelson & Sussman, 2002)  WIP
The book can be found here.
Building Abstractions with Procedures
The Elements of Programming
A programming language is our mental framework for organising ideas about process. It provides three mechanisms for combining simple ideas such that they together form more complex ideas:
 primitive expressions, which represent the simplest entities the language is concerned with,
 means of combination, by which compound elements are built from simpler ones, and
 means of abstraction, by which compound elements can be named and manipulated as units.
Expressions
Expressions such as these, formed by delimiting a list of expressions within parentheses in order to denote procedure application, are called combinations. the leftmost element in the list is called the operator , and the other elements are called operands . The value of a combination is obtained by applying the procedure speciﬁed by the operator to the arguments that are the values of the operands.
Placing the operator to the left of the operands is called prefixnotation.
Let's take a look at the nesting of expressions:
(+ (* 3
(+ (* 2 4)
(+ 3 5)))
(+ ( 10 7)
6))
If we align the operands vertically as above we prettyprint our code.
Naming and the Environment
Every programming language uses names which identify a variable whose value
is the object. In the Scheme dialect of list we use define
. In Lisp every
expression has a value.
Lisp programmers know the value of everything but the cost of nothing (Alan Perlis)
Here is an example of how to use define
:
(define pi 3.14159)
(define radius 10)
(* pi (* radius radius))
(define circumference (* 2 pi radius))
circumference
In order to keep track of the nameobject pairs, the interpreter maintains a memory called the (global) environment.
Evaluating Combinations
Let us consider the following recursive evaluation rule:
To evaluate a combination, do the following:
 Evaluate the subexpressions of the combination.
 Apply the procedure that is the value of the leftmost subexpression (the operator) to the arguments that are the values of the other subexpressions (the operands).
Hence, the following code
(* (+ 2 (* 4 6))
(+ 3 5 7))
can be represented in the following tree strucure:
This “percolating upwards” is called tree accumulation. This evaluation rule
does not apply to socalled special forms, such as define
, which each have
their own evaluation rule.
Compound Procedures
Any programming language must have:
 Numbers and arithmetic operations are primitive data and procedures. Nesting
 of combinations provides a means of combining opera tions. Deﬁnitions that
 associate names with values provide a limited means of abstraction.
Next, we need procedure definitions which open a whole new realm of possibility.
Let's define a compound procedure called square
:
(define (square x) (* x x))
Now we can easily define another procedure that makes use of square
:
(define (square x) (* x x))
(define (sumofsquares x y)
(+ (square x) (square y)))
(sumofsquares 3 4)
which evaluates to 25
. We can take this even further:
(define (square x) (* x x))
(define (sumofsquares x y)
(+ (square x) (square y)))
(define (f a)
(sumofsquares (+ a 1) (* a 2)))
(f 5)
which gives us 136
.
The Substitution Model for Procedure Application
Let us consider the combination from above to illustrate the subsitution model:
(f 5)
;; retrieve the body of f and replace parameters with the arguments
(sumofsquares (+ 5 1) (* 5 2))
;; apply sumofsquare to 6 and 10
(+ (square 6) (square 10))
;; reduce the expression by using the definition of square
(+ (* 6 6) (* 10 10))
(+ 36 100)
NB: This is not how the interpreter really works, as we'll see later. The subsitution model serves the purpose of providing an entry point to thinking about procedure application.
Applicative vs. Normal Order
The “first evaluate arguments and then apply procedures” way of doing things that we used above (applicativeorder evaluation) is not the only way.
The other evaluation model is the “fully expand and then reduce” model, which is called normalorder evaluation and illustrated below:
(f 5)
(sumofsquares (+ 5 1) (* 5 2))
(+ (square (+ 5 1)) (square (* 5 2)) )
(+ (* (+ 5 1) (+ 5 1)) (* (* 5 2) (* 5 2)))
(+ (* 6 6) (* 10 10))
(+ 36 100)
136
Conditional Expression and Predicates
Often, we want to do different things depending on the result of a test (case
analysis). In Lisp we use cond
to do that. The first expression in each pair
is called the predicate (either true or false) and the second one is the
consequent expression (value returned if predicate is true).
(define (abs x)
;; (<p> <e>)
(cond ((> x 0) x)
((= x 0) 0)
((< x 0) ( x))))
;; use else when to specify what to return if all clauses have been bypassed
(define (abs x)
(cond ((< x 0) ( x))
(else x)))
;; use if if you have exactly two cases in the case analysis
(define (abs x)
(if (< x 0)
( x)
x))
Of course, we should be able to construct compound predicates also with logical composition operations and not purely numerical ones:
;; specify a number range: 5 < y < 10
(and (> x 5) (< x 10))
;; greater than or equal
(define (>= x y)
(or (> x y) (= x y)))
;; alternatively:
(define (>= x y)
(not (< x y)))
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)
;; a
(define b (+ a 1))
;; b
(+ 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:
\[ \frac{5+4+\left(2\left(3\left(6+\frac{4}{5}\right)\right)\right)}{3(62)(27)} \]
(/ (+ 5 4
( 2
( 3
(+ 6
(/ 4 5)))))
(* 3
( 6 2)
( 2 7)))
;; 37/150
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 (sq x)
(* x x))
(define (ssq x y)
(+ (sq x) (sq y)))
(define (max3 x y z)
(cond ((> x y) (cond ((> x z) x)
(z)))
((> y z) y)
(z)))
(define (max2 x y)
(if (> x y) x y))
(define (largerssq x y z)
(cond ((= (max3 x y z) x) (ssq x (max2 y z)))
((= (max3 x y z) y) (ssq y (max2 x z)))
((= (max3 x y z) z) (ssq z (max2 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 (aplusabsb a b)
;; the ifexpression evaluates to a "+" or "" depending on the clause (> b 0)
((if (> b 0) + ) a b))
Exercise 1.5
Ben Bitdiddle has invented a test to determine whether the interpreter he is faced with is using applicativeorder evaluation or normalorder evaluation. He deﬁnes the following two procedures:
(define (p) (p))
(define (test x y)
(if (= x 0) 0 y))
(test 0 (p))
Using normalorder evaluation, the last expression evaluates to 0
as the
infiniteloopproducing procedure p
is not evaluated. This is not true for
applicativeorder evaluation, where the arguments are evaluated first. Here, the
process ends in an infinite loop.
Example: Square Roots by Newton's Method
There is a difference between a mathematical function of a square root (which can be used to recognise a square root or derive some interesting insights about it) and a procedure to generate a squre root.
For generating sqaure roots, we can use Newton's method of approximation:
(define (sqrtiter guess x)
(if (goodenough? guess x)
guess
(sqrtiter (improve guess x)
x)))
(define (improve guess x)
(average guess (/ x guess)))
(define (average x y)
(/ (+ x y) 2))
(define (goodenough? guess x)
(< (abs ( (square guess) x)) 0.001))
(define (sqrt x)
(sqrtiter 1.0 x))
The sqrtiter
procedure also underlines that iteration can be achieved using
no special construct but the ability to call a procedure
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 (newif predicate thenclause elseclause)
(cond (predicate thenclause)
(else elseclause)))
;; demo
(newif (= 2 3) 0 5)
;; 5
(newif (= 1 1) 0 5)
;; 0
Now, Alyssa wants to use newif for the squareroot program:
(define (sqrtiter guess x)
(newif (goodenough? guess x)
guess
(sqrtiter (improve guess x)
x)))
What happens when Alyssa aempts to use this to compute square roots? Explain.
The interpreter returns the following error message:
;Aborting!: maximum recursion depth exceeded
This is due to the fact that the newif
procedure does not share the property
of the if
special form to only evaluate the consequence when the predicate
evaluates to #t
. Hence, infinite recursion whenever we call newif
and there
one of the consequents is a function call.
Exercise 1.7
The goodenough?
test used in computing square roots will not be very
effective for ﬁnding the square roots of very small numbers. Also, in real
computers, arith metic operations are almost always performed with lim ited
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 goodenough?
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 squareroot procedure that uses this
kind of end test. Does this work beer for small and large numbers?
(define (sqrtiter guess x)
(if (goodenough? guess (improve guess x) x)
guess
(sqrtiter (improve guess x)
x)))
(define (abs x)
(if (< x 0)
( x)
x))
(define (improve guess x)
(average guess (/ x guess)))
(define (average x y)
(/ (+ x y) 2))
(define (goodenough? cur next x)
(< (/ (abs ( cur next)) x) 0.0000001))
(define (sqrt x)
(sqrtiter 1.0 x))
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 beer approximation is given by the value
\[ \frac{x / y^{2}+2 y}{3} \]
Use this formula to implement a cuberoot procedure analogous to the squareroot procedure
(define (cbrtiter guess x)
(if (goodenough? guess (improve guess x) x)
guess
(cbrtiter (improve guess x)
x)))
(define (abs x)
(if (< x 0)
( x)
x))
(define (square x) (* x x))
(define (improve guess x)
(/ (+ (/ x (square guess)) (* 2 guess)) 3))
(define (goodenough? cur next x)
(< (/ (abs ( cur next)) x) 0.0000001))
(define (cbrt x)
(cbrtiter 1.0 x))
Procedures as BlackBox Abstractions
The procedure definition binds its formal parameters such that they become bound variables. If variables are not bound, they are free. The set of expressions for which there is a binding defines its name is called scope of that name.
Often it can be useful to “hide” or localise the subprocedures of a given
procedure by utilising what is called a block structure. In the case of our
sqrt
function, we could write:
(define (sqrt x)
(define (goodenough? guess x)
(< (abs ( (square guess) x)) 0.001))
(define (improve guess x)
(average guess (/ x guess)))
(define (sqrtiter guess x)
(if (goodenough? guess x)
guess
(sqrtiter (improve guess x) x)))
(sqrtiter 1.0 x))
As can be inspected above, x
is a free variable in the internal procedure
definitions. This discipline is called lexical scoping, which the authors
define as follows:
Lexical scoping dictates that free variables in a procedure are taken to refer to bindings made by enclosing procedure deﬁnitions; that is, they are looked up in the environment in which the procedure was deﬁned.
Procedures and the Processes They Generate
Our situation is now analogous to someone who knows the rules of how pieces move in chess but knows nothing of openings, tactics or strategy. We don't know any patterns yet.
A procedure is a pattern for the local evolution of a computational process. It specifies how each stage of the process is built upon the previous stage
Linear Recursion and Iteration
Consider the factorial function:
\[ n !=n \cdot(n1) \cdot(n2) \cdots 3 \cdot 2 \cdot 1 \]
Another way to write this is:
\[ n !=n \cdot[(n1) \cdot(n2) \cdots 3 \cdot 2 \cdot 1]=n \cdot(n1) ! \]
From the latter, we can define the following procedure to generate the factorial of \(n\):
(define (factorial n)
(if (= n 1)
1
(* n (factorial ( n 1)))))
The authors visulise the resulting recursion of \(6!\) as follows:
We can also iterate by defining a counter that increases by one each step and is multiplied with the product of the last iteration. So, \(n!\) is the value of the product when the counter exceeds \(n\).
(define (factorial n)
(define (iter product counter)
(if (> counter n)
product
(iter (* counter product)
(+ counter 1))))
(iter 1 1))
This can be visualised as follows:
Here some important distinctions are to be made. The authors clarify that a recursive process is different from a recursive procedure:
When we describe a procedure as recursive, we are referring to the syntactic fact that the procedure deﬁnition refers (either directly or indirectly) to the procedure itself. But when we describe a process as following a pattern that is, say, linearly recursive, we are speaking about how the process evolves, not about the syntax of how a procedure is written.
Scheme is tailrecursive, i.e. it executes an iterative process in constant
space, even if it is described by a recursive procedure. This means, that in
Scheme we don't need any special iteration constructs such as for
, while
,
until
etc. They are only useful as sytactic sugar.
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))))
;; (inc (+ 3 5))
;; (inc (inc (+ 5 2)))
;; (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
;; > recursive
(define (+ a b)
(if (= a 0)
b
(+ (dec a) (inc b))))
;; (+ 3 6)
;; (+ 2 7)
;; (+ 1 8)
;; (+ 0 9)
;; 9
;; > iterative process:
Exercise 1.10
The following procedure computes a mathematical function called Ackermann's
function. What are the values of the expression below the procedure definition.
Also, 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 \(5n^2\).
(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)
;; 1024
(A 2 3)
;; 65536
(A 3 3)
;; 65536
(define (f n) (A 0 n))
;; 2n
(define (g n) (A 1 n))
;; 2^n
(define (h n) (A 2 n))
;; 2^2^2^2 : n times
;; Knuth's uparrow notation
(define (k n) (* 5 n n))
;; 5n^2
Tree Recursion
To understand tree recursion, consider the Fibonacci sequence:
\[0,1,1,2,3,5,8,13,21, …\]
In general, the Fibonacci numbers can be defined by the rule:
\[ \mathrm{Fib}(n)= \left\{\begin{array}{ll}0 & \text { if } n=0 \ 1 & \text { if } n=1 \ \mathrm{Fib}(n1)+\mathrm{Fib}(n2) & \text { otherwise } \end{array}\right. \]
Let's translate that into Lisp:
(define (fib n)
(cond ((= n 0) 0)
((= n 1) 1)
(else (+ (fib ( n 1))
(fib ( n 2))))))
This process looks like a tree:
This is pretty bad as the number of times that the procedure will compute is precisely Fib\((n + 1)\), e.g. in the case above exactly eight times. Thus, the process uses a number of steps that grows exponentially with the input. The space, however, only grows linearly with the input as we only need to keep track of the nodes above the current one at any point during computation.
Let's define an iterative procedure to do the same thing:
(define (fib n)
(fibiter 1 0 n))
(define (fibiter a b count)
(if (= count 0)
b
(fibiter (+ a b) a ( count 1))))
The authors summarise:
The difference in number of steps required by the two methods — one linear in n, one growing as fast as Fib(n) itself — is enormous, even for small inputs.
However, treerecursive processes aren't useless. Often, they are easier to design and understand. Apparently, the Scheme interpreter itself evaluates expression using a treerecursive process.
Example: Counting Change
(define (countchange amount)
(cc amount 5))
(define (cc amount kindsofcoins)
(cond ((= amount 0) 1)
((or (< amount 0) (= kindsofcoins 0)) 0)
(else (+ (cc amount
( kindsofcoins 1))
(cc ( amount
(firstdenomination kindsofcoins))
kindsofcoins)))))
;; takes kinds of coins available
;; returns denomination of first kind
(define (firstdenomination kindsofcoins)
(cond ((= kindsofcoins 1) 1)
((= kindsofcoins 2) 5)
((= kindsofcoins 3) 10)
((= kindsofcoins 4) 25)
((= kindsofcoins 5) 50)))
(countchange 10)
Exercise 1.11
A function f is defined by the rule that
\[ f(n)=\left\{\begin{array}{ll}n \quad \text { if } n<3 \ f(n1)+2 f(n2)+3 f(n3) & \text { if } \quad n \geq 3\end{array}\right. \]
Write a procedure that computes \(f\) by means of a recursive process. Write a procedure that computes \(f\) by means of an iterative process.
;; recursive
(define (frecur n)
(cond ((< n 3) n)
(else (+ (frecur ( n 1))
(* 2 (frecur ( n 2)))
(* 3 (frecur ( n 3)))))))
(frecur 10)
;; 1892
;; iterative version 1
(define (fiter1 n)
(define (fiter a b c count)
(cond ((< count 0) count)
((= count 0) a)
((= count 1) b)
((= count 2) c)
(else (fiter b c (+ c (* 2 b) (* 3 a)) ( count 1)))))
(fiter 0 1 2 n))
(fiter1 4)
;; fiter (1 2 (+ 2 (* 2 1) (* 3 0)) ( 3 1))
;; fiter (1 2 (+ 2 2 0)) 2)
;; fiter (1 2 4 2)
;; > 4
;; iterative version 2
(define (fiter2 n)
(define (fiter a b c count)
(cond ((< n 3) n)
((<= count 0) a)
(else (fiter (+ a (* 2 b) (* 3 c)) a b ( count 1)))))
(fiter 2 1 0 ( n 2)))
(fiter2 10)
;; 1892
Exercise 1.12
The following pattern of numbers is called Pascal’s triangle:
The numbers at the edge of the triangle are all 1, and each number inside the triangle is the sum of the two numbers above it. Write a procedure that computes elements of Pascal's triangle by means of a recursive process.
(define (pascal r c)
(if (or (= c 1) (= c r))
1
(+ (pascal ( r 1) ( c 1))
(pascal ( r 1) c))))
(pascal 5 3)
;; 6
;; c
;; r 1
;; 1 1
;; 1 2 1
;; 1 3 3 1
;; 1 4 6 4 1
;; ...
Exercise 1.13
Proposition
For all \(n \in \mathbb{N}\) let \(P(n)\) be the proposition:
\(Fib(n)=\frac{\varphi^{n}{\psi}^{n}}{\sqrt{5}}\)
Basis for induction
\(P(0)\) is true, as this shows:
\[ \dfrac {\varphi^0  \psi^0} {\sqrt 5} = \dfrac {1  1} {\sqrt 5} = 0 = Fib(0) \]
Induction hypothesis
\(\forall 0 \le j \le k + 1: Fib(j) = \dfrac {\varphi^j  \psi^j} {\sqrt 5}\)
Thus, we need to show:
\(Fib(k + 2) = \dfrac {\varphi^{k + 2}  \psi^{k + 2} } {\sqrt 5}\)
Induction step
We have the following two identities:
\[ \varphi^2 = (\frac {1 + \sqrt 5} 2)^2 = \frac 1 4 ({6 + 2 \sqrt 5}) = \frac {3 + \sqrt 5} 2 = 1 + \varphi \]
\[ \psi^2 = (\frac {1 + \sqrt 5} 2)^2 = \frac 1 4 ({6 + 2 \sqrt 5}) = \frac {3 + \sqrt 5} 2 = 1 + \psi \]
Hence:
\[ \varphi^{k+2}\psi^{k+2} = (1+\varphi) \varphi^{k}(1+\psi) \psi^{k} \]
\(= (\varphi^{k}\psi^{k})+(\varphi^{k+1}\psi^{k+1})\)
\[ = \sqrt{5}(Fib(k)+Fib(k+1)) = \sqrt{5} Fib(k+2) \]
The result follows by the principle of mathematical induction.
Therefore:
\(\forall n \in \mathbb{N}: Fib(n) = \frac {\varphi^n  \psi^n} {\sqrt 5}\)
Orders of Growth
Let \(R(n)\) be the amount of resources the process requires for a problem of size \(n\).
The authors make some further important definitions
We say that \(R(n)\) has order of growth \(\theta(f(n))\), written $R(n) = θ(f(n)) (pronounced “theta of \(f(n)\)”), if there are positive constants \(k_1\) and \(k_2\) independent of \(n\) such that \(k_1f(n) \leq R(n) \leq k_2 f(n)\) for any sufficiently large value of \(n\). (In other words, for large \(n\), the value \(R(n)\) is sandwiched between \(k_1 f (n)\) and \(k_2 f (n)\).)
Exercise 1.14
Draw the tree illustrating the process generated by the aforementioned
countchange
procedure of in making change for 11 cents. What are the orders
of growth of the space and number of steps used by this process as the amount to
be changed increases?
(countchange 11)

(cc 11 5)__
 \
(cc 11 4) (cc 39 5)
 \___
 \
(cc 11 3) (cc 14 4)
 \_______________________________________________________
 \
(cc 11 2) (cc 1 3)
 \_________________________  \__
 \  \
(cc 11 1) (cc 6 2) (cc 1 2) (cc 9 3)
 \___  \__  \__
 \  \  \
(cc 11 0) (cc 10 1) (cc 6 1) (cc 1 2) (cc 1 1) (cc 4 2)
__/  __/   \__  \__
/  /   \  \
(cc 10 0) (cc 9 1) (cc 6 0) (cc 5 1) (cc 1 1) (cc 4 2) (cc 1 0) (cc 0 1)
__/  __/   \__
/  /   \
(cc 9 0) (cc 8 1) (cc 5 0) (cc 4 1) (cc 1 0) (cc 0 1)
__/  __/ 
/  / 
(cc 8 0) (cc 7 1) (cc 4 0) (cc 3 1)
__/  __/ 
/  / 
(cc 7 0) (cc 6 1) (cc 3 0) (cc 2 1)
__/  __/ 
/  / 
(cc 6 0) (cc 5 1) (cc 2 0) (cc 1 1)
__/  __/ 
/  / 
(cc 5 0) (cc 4 1) (cc 1 0) (cc 0 1)
__/ 
/ 
(cc 4 0) (cc 3 1)
__/ 
/ 
(cc 3 0) (cc 2 1)
__/ 
/ 
(cc 2 0) (cc 1 1)
__/ 
/ 
(cc 1 0) (cc 0 1)
The space requirement of cc
is proportional to the maximum height of the
recursion tree, because at any given point in the recursive process, the
interpreter must only keep track of the nodes that lead to the current root.
Since the maximum height of the tree is dominated by the branch that contains
the most successive calls, i.e. the leftmost one in the graph, it is growing
linearly with \(n\) (amount
). In other words, \(\theta(n)\).
The time requirement can be deduced as follows:
(cc amount 1)
= \(\theta(n)\)(cc amount 2)
=(cc amount 1)
+(cc ( amount 5) 2))
 Here, we have \(\theta(n^2)\) when
kindsofcoins
is 2.  Hence, we get \(\theta(n^k)\) (\(k\) being
kindsofcoins
) forcc(amount kindsofcoins)
since every 2nd branch is \(\theta(k)\), and the first branch is called \(\theta(n)\) times.
Exercise 1.15
The sine of an angle (speciﬁed in radians) can be computed by making use of the approximation \(x \approx x\) if \(x\) is sufficiently small, and the trigonometric identity
\(\sin x=3 \sin \dfrac{x}{3}4 \sin ^{3} \dfrac{x}{3}\)
to reduce the size of the argument of \(sin\). (For purposes of this exercise an angle is considered “sufficiently small” if its magnitude is not greater than 0.1 radians.) These ideas are incorporated in the following procedures:
(define (cube x) (* x x x))
(define (p x) ( (* 3 x) (* 4 (cube x))))
(define (sine angle)
(if (not (> (abs angle) 0.1))
angle
(p (sine (/ angle 3.0)))))
(sine 12.15)
;;(p (sine 4.05))
;;(p (p (sine 1.35)))
;;(p (p (p (sine 0.45))))
;;(p (p (p (p (sine 0.15)))))
;;(p (p (p (p (p (sine 0.05))))))
How many times is the procedure
p
applied when(sine 12.15)
is evaluated?As can be seen above the procedure is applied five times.
What is the order of growth in space and number of steps (as a function of \(a\) or
angle
) used by the process generated by thesine
procedure when(sine a)
is evaluated?So, the basic intuition is that
sine
is applied as many times asangle
can be divided by three until the absolute result is smaller than0.1
. To describe this mathematically, we need the notion of a ceiling (as we want to output an integer). So, we can write\[\dfrac{12.15}{3^{n}}< 0.1 \\\
= 3 \times 12.15^{n} < 0.1 \]\[ \log(3) \times n \times log(12.15) < log(0.1) \]
Thus, we can write the number of required computations as
\(\Bigg\lceil\dfrac{\log\dfrac{12.15}{0.1}}{\log{3}}\Bigg\rceil = 5\)
or more generally
\(\Bigg\lceil\dfrac{\log\dfrac{a}{0.1}}{\log{3}}\Bigg\rceil\)
Hence, the order of growth in space is \(\theta(log(x))\).
Exponentiation
This is a recursive definition of the exponent \(n\) for a given integer \(b\):
\begin{array} {l}b^{n}=b \cdot b^{n1} \ b^{0}=1 \end{array}
In Scheme this linearly recursive process looks as such:
(define (expt b n)
(if (= n 0)
1
(* b (expt b ( n 1)))))
This requires \(\theta(n)\) steps and \(\theta(n)\) space. The corresponding iterative definition of the process would be:
(define (expt b n)
(exptiter b n 1))
(define (exptiter b counter product)
(if (= counter 0)
product
(exptiter b
( counter 1)
(* b product))))
This requires \(\theta(n)\) steps and \(\theta(1)\) space. We can be faster, however if we make use of the following:
\begin{array} {l}b^{2}=b \cdot b \ b^{4}=b^{2} \cdot b^{2} \ b^{8}=b^{4} \cdot b^{4} \end{array}
We can thus amend our process such that it runs even faster:
(define (fastexpt b n)
(cond ((= n 0) 1)
((even? n) (square (fastexpt b (/ n 2))))
(else (* b (fastexpt b ( n 1))))))
(define (even? n)
(= (remainder n 2) 0))
How fast exactly? Well, computing \(b^{2n}\) using fastexpt
requires only one
more computation than computing \(b^{n}\).
Exercise 1.16
Design a procedure that evolves an iterative exponentiation process that uses
successive squaring and uses a logarithmic number of steps, as does fastexpt
.
(Hint: Using the observation that \((b^{n/2})^{2} = (b^{2})^{n/2}\) , keep, along
with the exponent n
and the base b
, an additional state variable a
, and
deﬁne the state transformation in such a way that the product \(ab^n\) is
unchanged from state to state. At the beginning of the process a
is taken to
be 1, and the answer is given by the value of a
at the end of the process. In
general, the technique of deﬁning an invariant quantity that remains unchanged
from state to state is a powerful way to think about the design of iterative
algorithms.)
(define (iterfastexpt b n)
(define (iter b n a)
(cond ((= 0 n) a)
((even? n) (iter (square b) (/ n 2) a))
(else (iter b ( n 1) (* b a)))))
(iter b n 1))
(iterfastexpt 5 6)
;; 15625
Exercise 1.17
The exponentiation algorithms in this section are based on performing exponentiation by means of repeated multiplication. In a similar way, one can perform integer multiplication by means of repeated addition. The following multiplication procedure (in which it is assumed that our language can only add, not multiply) is analogous to the expt procedure:
(define (* a b)
(if (= b 0)
0
(+ a (* a ( b 1)))))
This algorithm takes a number of steps that is linear in b
. Now suppose we
include, together with addition, operations double
, which doubles an integer,
and halve
, which divides an (even) integer by 2. Using these, design a
multiplication procedure analogous to fastexpt that uses a logarithmic number
of steps.
(define (double k)
(+ k k))
(define (halve k)
(/ k 2))
(define (multiply a b)
(cond ((or (= a 0) (= b 0)) 0)
((even? a) (multiply (halve a) (double b)))
(else (+ b (multiply ( a 1) b)))))
(multiply 300 5001)
;; 1500300
Exercise 1.18
Using the result of the previous two exercises, devise a procedure that generates an iterative process for multiplying two integers in terms of adding, doubling, and halving and uses a logarithmic number of steps.
(define (double k)
(+ k k))
(define (halve k)
(/ k 2))
(define (fastmultiply a b)
(define (iter a b s)
(cond ((= a 0) s)
((even? a) (iter (halve a) (double b) s))
(else (iter ( a 1) b (+ b s)))))
(iter a b 0))
(fastmultiply 601 3)
;; 1803
Exercise 1.19
There is a clever algorithm for computing the Fibonacci numbers in a logarithmic
number of steps. Recall the transformation of the state variables \(a\) and \(b\) in
the fibiter
process of earlier: \(a \rightarrow a + b\) and \(b \rightarrow a\).
Call this transformation \(T\), and observe that applying \(T\) over and over again
\(n\) times, starting with 1 and 0, produces the pair \(Fib(n + 1)\) and \(Fib(n)\).
In other words, the Fibonacci numbers are produced by applying \(T^{n}\) , the
\(n^{th}\) power of the transformation \(T\), starting with the pair (1, 0). Now
consider \(T\) to be the special case of \(p = 0\) and \(q = 1\) in a family of
transformations \(T_{pq}\), where \(T_{pq}\) transforms the pair (a, b) according to
\(a \rightarrow bq + aq + ap\) and \(b \rightarrow bp + aq\). Show that if we apply
such a transformation \(T_{pq}\) twice, the effect is the same as using a single
transformation \(T_{p'q’}\) of the same form, and compute \(p’\) and \(q’\) in terms
of \(p\) and \(q\). This gives us an explicit way to square these transformations,
and thus we can compute \(T^{n}\) using successive squaring, as in the fastexpt
procedure. Put this all together to complete the following procedure, which runs
in a logarithmic number of steps:
(define (fib n)
(fibiter 1 0 0 1 n))
(define (fibiter a b p q count)
(cond ((= count 0) b)
((even? count)
(fibiter a
b
(+ (* q q) (* p p))
(+ (* 2 (* q p)) (* q q))
(/ count 2)))
(else (fibiter (+ (* b q) (* a q) (* a p))
(+ (* b p) (* a q))
p
q
( count 1)))))
(fib 10)
The intuition here is the following. Observe that we can write the linea \(T_{pq}\) as a matrix:
\[ \begin{pmatrix} p+q & q \ q & p \end{pmatrix} \begin{pmatrix} a \ b \end{pmatrix} = \begin{pmatrix} bp + aq + ap \ bp + aq \end{pmatrix} \]
Now, we are told, we can just apply the matrix on the left twice (square) such that we get a single transformation \(T_{p'q’}\):
\[ \begin{pmatrix}
p+q & q \\\
q & p
\end{pmatrix}
\begin{pmatrix}
p+q & q \\\
q & p
\end{pmatrix} =
\begin{pmatrix}
… & … \\\
p’ & q’
\end{pmatrix} =
\begin{pmatrix}
… & … \\\
q^{2} + 2pq & q^{2} + p^{2}
\end{pmatrix} \]
Greatest Common Divisors
The greatest common divisor (GCD) of two integers \(a\) and \(b\) is deﬁned to be the largest integer that divides both \(a\) and \(b\) with no remainder. For example, the GCD of 16 and 28 is 4.
Euclid's Algorithm is really smart. Let r
be the remainder of the division
of a
by b
. Then GCD(a, b) = GCD(b, r)
. In Scheme this looks as follows:
(define (gcd a b)
(if (= b 0)
a
(gcd b (remainder a b))))
This code represents an iterative process whose number of steps grows as the logarithm of the numbers involved.
Exercise 1.20
The process that a procedure generates is of course dependent on the rules used
by the interpreter. As an example, consider the iterative gcd
procedure given
above. Suppose we were to interpret this procedure using normalorder
evaluation, as discussed before (The normalorderevaluation rule for if is
described in Exercise 1.5.) Using the substitution method (for normal order),
illustrate the process generated in evaluating (gcd 206 40)
and indicate the
remainder operations that are actually performed. How many remainder operations
are actually performed in the normalorder evaluation of (gcd 206 40)
? In the
applicativeorder evaluation?
;; normalorder
;; NB: fully expand, then reduce
(gcd 40 (remainder 206 40))
(if (= (remainder 206 40) 0)
40
(gcd (remainder 206 40) (remainder 40 (remainder 206 40))))
;; 1 in ifcondition
(gcd (remainder 206 40) (remainder 40 (remainder 206 40)))
(if (= (remainder 40 (remainder 206 40)) 0)
(remainder 206 40)
(gcd (remainder 40 (remainder 206 40))
(remainder (remainder 206 40) (remainder 40 (remainder 206 40)))))
;; 2 in ifcondition
(gcd (remainder 40 (remainder 206 40))
(remainder (remainder 206 40) (remainder 40 (remainder 206 40))))
(if (= (remainder (remainder 206 40) (remainder 40 (remainder 206 40))) 0)
(remainder 40 (remainder 206 40))
(gcd (remainder (remainder 206 40) (remainder 40 (remainder 206 40)))
(remainder (remainder 40 (remainder 206 40))
(remainder (remainder 206 40)
(remainder 40 (remainder 206 40))))))
;; 4 in ifcondition
(gcd (remainder (remainder 206 40) (remainder 40 (remainder 206 40)))
(remainder (remainder 40 (remainder 206 40))
(remainder (remainder 206 40)
(remainder 40 (remainder 206 40)))))
(if (= (remainder (remainder 40 (remainder 206 40))
(remainder (remainder 206 40)
(remainder 40 (remainder 206 40)))) 0)
;; the condition is finally met so we now evaluate the line below
(remainder (remainder 206 40) (remainder 40 (remainder 206 40)))
(gcd (remainder (remainder 206 40) (remainder 40 (remainder 206 40)))
(remainder (remainder 40 (remainder 206 40))
(remainder (remainder 206 40)
(remainder 40 (remainder 206 40))))))
;; 7 in ifcondition
(remainder (remainder 206 40) (remainder 40 (remainder 206 40)))
;; 4 in final reduction
;; 18 in total with normalorder evaluation
;; applicativeorder evaluation
;; NB: first evaluate, then apply procedures
(gcd 206 40)
(gcd 40 (remainder (206 40)))
(gcd 40 6)
(gcd 40 (remainder (40 6)))
(gcd 6 4)
(gcd 40 (remainder (6 4)))
(gcd 4 2)
(gcd 40 (remainder (4 2)))
(gcd 2 0)
;; 4 in total with
;; applicativeorder evaluation
Testing for Primality
This first procedure leverages the fact that \(n\) is prime if and only if \(n\) is its smalles divisor:
(define (smallestdivisor n)
(finddivisor n 2))
(define (finddivisor n testdivisor)
(cond ((> (square testdivisor) n) n)
((divides? testdivisor n) testdivisor)
(else (finddivisor n (+ testdivisor 1)))))
(define (divides? a b)
(= (remainder b a) 0))
(define (prime? n)
(= n (smallestdivisor n)))
(prime? 12307926403)
;; #t
The steps required by this procedure has order of growth \(\theta(\sqrt{n})\)
Another procedure leverages Fermat's little theorem which is worth of being stated:
If \(n\) is \(a\) prime number and \(a\) is any positive integer less than \(n\), then \(a\) raised to the \(n^{th}\) power is congruent to \(a\) modulo \(n\).
NB Two numbers are congruent modulo \(n\) if they both have the same remainder when divided by \(n\). The remainder of a number \(a\) when divided by \(n\) is also referred to as the remainder of \(a\) modulo \(n\), or simply as \(a\) modulo \(n\).)
This is the code in Lisp:
;; compute the exponential of number modulo another number
(define (expmod base exp m)
(cond ((= exp 0) 1)
((even? exp)
(remainder (square (expmod base (/ exp 2) m))
m))
(else
(remainder (* base (expmod base ( exp 1) m))
m))))
;; perform the Fermat test by
;; 1. choosing a random number a between 1 and n1 and
;; 2. checking whether the remainder modulo n of a^n is equal to a
(define (fermattest n)
(define (tryit a)
(= (expmod a n n) a))
(tryit (+ 1 (random ( n 1)))))
;; here we can specify how many times we want to run the test, i.e.
;; how sure we want to be
(define (prime? n times)
(cond ((= times 0) true)
((fermattest n) (prime? n ( times 1)))
(else false)))
(prime? 12307926403 10)
Exercise 1.21
Use the smallestdivisor
procedure to ﬁnd the smallest divisor of each of the
following numbers: 199, 1999, 19999.
(define (smallestdivisor n)
(define (finddivisor n testdivisor)
(cond ((> (square testdivisor) n) n)
((divides? testdivisor n) testdivisor)
(else (finddivisor n (+ testdivisor 1)))))
(define (divides? a b)
(= (remainder b a) 0))
(finddivisor n 2))
(smallestdivisor 199)
;;199
(smallestdivisor 1999)
;;1999
(smallestdivisor 19999)
;;7
WIP
Bibliography
</home/lino/org/exocortex/biblio/library.bib>