19 KiB
1.2 - Procedures and the Processes They Generate
- Linear Recursion and Iteration
- 1.2.2: Tree Recursion
- 1.2.3: Orders of Growth
- 1.2.4: Exponentiation
- 1.2.5: Greatest Common Divisors
- 1.2.6: Example: Testing for Primality
Linear Recursion and Iteration
;; ===================================================================
;; 1.2.1: Linear Recursion and Iteration
;; ===================================================================
(define (inc n) (+ n 1))
(define (dec n) (- n 1))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?
;; -------------------------------------------------------------------
;; 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)
9Exercise 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)
1024
(A 2 4)
65536
(A 3 3)
65536Consider 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
- 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
;; ===================================================================
;; 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))))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.
;; -------------------------------------------------------------------
;; 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)))Exercise 1.12
The following pattern of numbers is called "Pascal's triangle".
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
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.
;; -------------------------------------------------------------------
;; Exercise 1.12
;; -------------------------------------------------------------------
(define (pascal row column)
(cond ((= column 1) 1)
((= row column) 1)
(else (+ (pascal (- row 1) (- column 1))
(pascal (- row 1) column)))))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
x x
sin x = 3 sin --- - 4 sin^3 ---
3 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)))))- How many times is the procedure `p' applied when `(sine 12.15)' is evaluated?
- 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
;; ===================================================================
;; 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))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.)
;; -------------------------------------------------------------------
;; 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))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 `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 +
- 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)
;; -------------------------------------------------------------------
;; 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)))))1.2.5: Greatest Common Divisors
(define (gcd a b)
(if (= b 0)
a
(gcd b (remainder a b))))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,
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.
(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))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
- 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
(define (expmod base exp m)
(remainder (fast-expt base exp) m))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':
(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))))"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.