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#+TITLE: 1.2 - Procedures and the Processes They Generate
* Linear Recursion and Iteration
#+BEGIN_SRC scheme :tangle yes
;; ===================================================================
;; 1.2.1: Linear Recursion and Iteration
;; ===================================================================
(define (inc n) (+ n 1))
(define (dec n) (- n 1))
#+END_SRC
** 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.
#+BEGIN_SRC scheme
(define (+ a b)
(if (= a 0)
b
(inc (+ (dec a) b))))
(define (+ a b)
(if (= a 0)
b
(+ (dec a) (inc b))))
#+END_SRC
Using the substitution model, illustrate the process generated by
each procedure in evaluating `(+ 4 5)'. Are these processes
iterative or recursive?
--------------------------------------------------------------------
#+BEGIN_SRC scheme :tangle yes
;; -------------------------------------------------------------------
;; Exercise 1.9
;; -------------------------------------------------------------------
;; First procedure: Recursive
(+ 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
;; Second procedure: Iterative
(+ 4 5)
(+ 3 6)
(+ 2 7)
(+ 1 8)
(+ 0 9)
9
#+END_SRC
** Exercise 1.10
The following procedure computes a mathematical
function called Ackermann's function.
#+BEGIN_SRC scheme
(define (A x y)
(cond ((= y 0) 0)
((= x 0) (* 2 y))
((= y 1) 2)
(else (A (- x 1)
(A x (- y 1))))))
#+END_SRC
What are the values of the following expressions?
#+BEGIN_SRC scheme
(A 1 10)
1024
(A 2 4)
65536
(A 3 3)
65536
#+END_SRC
Consider the following procedures, where `A' is the procedure
defined above:
#+BEGIN_SRC scheme
(define (f n) (A 0 n))
(define (g n) (A 1 n))
(define (h n) (A 2 n))
(define (k n) (* 5 n n))
#+END_SRC
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.
--------------------------------------------------------------------
`(f n)' computes 2n
`(g n)' computes 2^n
`(h n)' computes 2^h(n - 1)
* 1.2.2: Tree Recursion
#+BEGIN_SRC scheme :tangle yes
;; ===================================================================
;; 1.2.2: Tree Recursion
;; ===================================================================
(define (fib n)
(fib-iter 1 0 n))
(define (fib-iter a b count)
(if (= count 0)
b
(fib-iter (+ a b) a (- count 1))))
#+END_SRC
** Exercise 1.11
A function f is defined by the rule that f(n) = n
if n<3 and f(n) = f(n - 1) + 2f(n - 2) + 3f(n - 3) if n>= 3.
Write a procedure that computes f by means of a recursive process.
Write a procedure that computes f by means of an iterative
process.
--------------------------------------------------------------------
#+BEGIN_SRC scheme :tangle yes
;; -------------------------------------------------------------------
;; Exercise 1.11
;; -------------------------------------------------------------------
(define (f-recursive n)
(if (< n 3)
n
(+ (f-recursive (- n 1))
(* 2 (f-recursive (- n 2)))
(* 3 (f-recursive (- n 3))))))
(define (f-iterative n)
(define (do-iter a b c n)
(if (< n 3)
a
(do-iter (+ a (* 2 b) (* 3 c)) a b (- n 1))))
(if (< n 3)
n
(do-iter 2 1 0 n)))
#+END_SRC
** Exercise 1.12
The following pattern of numbers is called "Pascal's
triangle".
#+BEGIN_EXAMPLE
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
#+END_EXAMPLE
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.(4)
Write a procedure that computes elements of Pascal's triangle by
means of a recursive process.
--------------------------------------------------------------------
#+BEGIN_SRC scheme :tangle yes
;; -------------------------------------------------------------------
;; Exercise 1.12
;; -------------------------------------------------------------------
(define (pascal row column)
(cond ((= column 1) 1)
((= row column) 1)
(else (+ (pascal (- row 1) (- column 1))
(pascal (- row 1) column)))))
#+END_SRC
** Exercise 1.13
Prove that _Fib_(n) is the closest integer to
[phi]^n/[sqrt](5), where [phi] = (1 + [sqrt](5))/2. Hint: Let
[illegiblesymbol] = (1 - [sqrt](5))/2. Use induction and the
definition of the Fibonacci numbers (see section *Note 1-2-2::) to
prove that _Fib_(n) = ([phi]^n - [illegiblesymbol]^n)/[sqrt](5).
--------------------------------------------------------------------
http://www.billthelizard.com/2009/12/sicp-exercise-113-fibonacci-and-golden.html
* 1.2.3: Orders of Growth
** Exercise 1.14
Draw the tree illustrating the process generated
by the `count-change' procedure of section *Note 1-2-2:: 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?
** Exercise 1.15
The sine of an angle (specified in radians) can
be computed by making use of the approximation `sin' xapprox x if
x is sufficiently small, and the trigonometric identity
#+BEGIN_EXAMPLE
x x
sin x = 3 sin --- - 4 sin^3 ---
3 3
#+END_EXAMPLE
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:
#+BEGIN_SRC scheme
(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)))))
#+END_SRC
a. How many times is the procedure `p' applied when `(sine
12.15)' is evaluated?
b. What is the order of growth in space and number of steps (as
a function of a) used by the process generated by the `sine'
procedure when `(sine a)' is evaluated?
* 1.2.4: Exponentiation
#+BEGIN_SRC scheme :tangle yes
;; ===================================================================
;; 1.2.4: Exponentiation
;; ===================================================================
(define (square x) (* x x))
(define (expt b n)
(expt-iter b n 1))
(define (expt-iter b counter product)
(if (= counter 0)
product
(expt-iter b
(- counter 1)
(* b product))))
(define (fast-expt b n)
(cond ((= n 0) 1)
((even? n) (square (fast-expt b (/ n 2))))
(else (* b (fast-expt b (- n 1))))))
(define (even? n)
(= (remainder n 2) 0))
#+END_SRC
** 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 `fast-expt'. (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 a
b^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.)
----------------------------------------------------------------------
#+BEGIN_SRC scheme :tangle yes
;; -------------------------------------------------------------------
;; Exercise 1.16
;; -------------------------------------------------------------------
(define (1-16 b n)
(define (expt-iter b n a)
(cond ((= n 0) a)
((even? n) (expt-iter (square b) (/ n 2) a))
(else (expt-iter b (- n 1) (* a b)))))
(expt-iter b n 1.0))
#+END_SRC
** 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:
#+BEGIN_SRC scheme
(define (* a b)
(if (= b 0)
0
(+ a (* a (- b 1)))))
#+END_SRC
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 `fast-expt' that uses a logarithmic number of steps.
** Exercise 1.18
Using the results of *Note Exercise 1-16:: and
*Note Exercise 1-17::, 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.(4)
** 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 `fib-iter'
process of section *Note 1-2-2::: a <- a + b and b <- 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 nth 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 <- bq + aq +
ap and b <- 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 `fast-expt' procedure. Put this all together
to complete the following procedure, which runs in a logarithmic
number of steps:(5)
#+BEGIN_SRC scheme
;; -------------------------------------------------------------------
;; Exercise 1.19
;; -------------------------------------------------------------------
(define (fib n)
(fib-iter 1 0 0 1 n))
(define (fib-iter a b p q count)
(cond ((= count 0) b)
((even? count)
(fib-iter a
b
<??> ; compute p'
<??> ; compute q'
(/ count 2)))
(else (fib-iter (+ (* b q) (* a q) (* a p))
(+ (* b p) (* a q))
p
q
(- count 1)))))
#+END_SRC
* 1.2.5: Greatest Common Divisors
#+begin_src scheme :tangle yes
(define (gcd a b)
(if (= b 0)
a
(gcd b (remainder a b))))
#+end_src
** 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 normal-order
evaluation, as discussed in section *Note 1-1-5::. (The
normal-order-evaluation rule for `if' is described in *Note
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 normal-order evaluation of `(gcd 206 40)'? In the
applicative-order evaluation?
* 1.2.6: Example: Testing for Primality
** Exercise 1.21
Use the `smallest-divisor' procedure to find the
smallest divisor of each of the following numbers: 199, 1999,
19999.
** Exercise 1.22
Most Lisp implementations include a primitive
called `runtime' that returns an integer that specifies the amount
of time the system has been running (measured, for example, in
microseconds). The following `timed-prime-test' procedure, when
called with an integer n, prints n and checks to see if n is
prime. If n is prime, the procedure prints three asterisks
followed by the amount of time used in performing the test.
#+BEGIN_SRC scheme
(define (timed-prime-test n)
(newline)
(display n)
(start-prime-test n (runtime)))
(define (start-prime-test n start-time)
(if (prime? n)
(report-prime (- (runtime) start-time))))
(define (report-prime elapsed-time)
(display " *** ")
(display elapsed-time))
#+END_SRC
Using this procedure, write a procedure `search-for-primes' that
checks the primality of consecutive odd integers in a specified
range. Use your procedure to find the three smallest primes
larger than 1000; larger than 10,000; larger than 100,000; larger
than 1,000,000. Note the time needed to test each prime. Since
the testing algorithm has order of growth of [theta](_[sqrt]_(n)),
you should expect that testing for primes around 10,000 should
take about _[sqrt]_(10) times as long as testing for primes around
1000. Do your timing data bear this out? How well do the data
for 100,000 and 1,000,000 support the _[sqrt]_(n) prediction? Is
your result compatible with the notion that programs on your
machine run in time proportional to the number of steps required
for the computation?
** Exercise 1.23
The `smallest-divisor' procedure shown at the
start of this section does lots of needless testing: After it
checks to see if the number is divisible by 2 there is no point in
checking to see if it is divisible by any larger even numbers.
This suggests that the values used for `test-divisor' should not
be 2, 3, 4, 5, 6, ..., but rather 2, 3, 5, 7, 9, .... To
implement this change, define a procedure `next' that returns 3 if
its input is equal to 2 and otherwise returns its input plus 2.
Modify the `smallest-divisor' procedure to use `(next
test-divisor)' instead of `(+ test-divisor 1)'. With
`timed-prime-test' incorporating this modified version of
`smallest-divisor', run the test for each of the 12 primes found in
*Note Exercise 1-22::. Since this modification halves the number
of test steps, you should expect it to run about twice as fast.
Is this expectation confirmed? If not, what is the observed ratio
of the speeds of the two algorithms, and how do you explain the
fact that it is different from 2?
** Exercise 1.24
Modify the `timed-prime-test' procedure of *Note
Exercise 1-22:: to use `fast-prime?' (the Fermat method), and test
each of the 12 primes you found in that exercise. Since the
Fermat test has [theta](`log' n) growth, how would you expect the
time to test primes near 1,000,000 to compare with the time needed
to test primes near 1000? Do your data bear this out? Can you
explain any discrepancy you find?
** Exercise 1.25
Alyssa P. Hacker complains that we went to a lot
of extra work in writing `expmod'. After all, she says, since we
already know how to compute exponentials, we could have simply
written
#+BEGIN_SRC scheme
(define (expmod base exp m)
(remainder (fast-expt base exp) m))
#+END_SRC
Is she correct? Would this procedure serve as well for our fast
prime tester? Explain.
** Exercise 1.26
Louis Reasoner is having great difficulty doing
*Note Exercise 1-24::. His `fast-prime?' test seems to run more
slowly than his `prime?' test. Louis calls his friend Eva Lu Ator
over to help. When they examine Louis's code, they find that he
has rewritten the `expmod' procedure to use an explicit
multiplication, rather than calling `square':
#+BEGIN_SRC scheme
(define (expmod base exp m)
(cond ((= exp 0) 1)
((even? exp)
(remainder (* (expmod base (/ exp 2) m)
(expmod base (/ exp 2) m))
m))
(else
(remainder (* base (expmod base (- exp 1) m))
m))))
#+END_SRC
"I don't see what difference that could make," says Louis. "I
do." says Eva. "By writing the procedure like that, you have
transformed the [theta](`log' n) process into a [theta](n)
process." Explain.
** Exercise 1.27
Demonstrate that the Carmichael numbers listed in
*Note Footnote 1-47:: really do fool the Fermat test. That is,
write a procedure that takes an integer n and tests whether a^n is
congruent to a modulo n for every a<n, and try your procedure on
the given Carmichael numbers.
** Exercise 1.28
One variant of the Fermat test that cannot be
fooled is called the "Miller-Rabin test" (Miller 1976; Rabin
1980). This starts from an alternate form of Fermat's Little
Theorem, which states that if n is a prime number and a is any
positive integer less than n, then a raised to the (n - 1)st power
is congruent to 1 modulo n. To test the primality of a number n
by the Miller-Rabin test, we pick a random number a<n and raise a
to the (n - 1)st power modulo n using the `expmod' procedure.
However, whenever we perform the squaring step in `expmod', we
check to see if we have discovered a "nontrivial square root of 1
modulo n," that is, a number not equal to 1 or n - 1 whose square
is equal to 1 modulo n. It is possible to prove that if such a
nontrivial square root of 1 exists, then n is not prime. It is
also possible to prove that if n is an odd number that is not
prime, then, for at least half the numbers a<n, computing a^(n-1)
in this way will reveal a nontrivial square root of 1 modulo n.
(This is why the Miller-Rabin test cannot be fooled.) Modify the
`expmod' procedure to signal if it discovers a nontrivial square
root of 1, and use this to implement the Miller-Rabin test with a
procedure analogous to `fermat-test'. Check your procedure by
testing various known primes and non-primes. Hint: One convenient
way to make `expmod' signal is to have it return 0.
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