Solution Files

You can find the solutions in hw08.py and hw08.scm

Homework Questions

Some Review: Sets

One really convenient thing about Python sets is that many operations on sets (adding elements, removing elements, checking membership) run in θ(1) (constant) time (usually).

Some of the problems use a utility method called timeit, which takes a parameterless function as argument, executes it, and returns the time required to do so. It's a variation on the function timeit.timeit function in the Python3 library.

Question 1: Missing Value

Write the following function so it (usually) runs in θ(n) time, where n is the length of lst0.

def missing_val(lst0, lst1):
    """Assuming that lst0 contains all the values in lst1, but lst1 is missing
    one value in lst0, return the missing value.  The values need not be
    numbers.

    >>> from random import shuffle
    >>> missing_val(range(10), [1, 0, 4, 5, 7, 9, 2, 6, 3])
    8
    >>> big0 = [str(k) for k in range(15000)]
    >>> big1 = [str(k) for k in range(15000) if k != 293 ]
    >>> shuffle(big0)
    >>> shuffle(big1)
    >>> missing_val(big0, big1)
    '293'
    >>> timeit(lambda: missing_val(big0, big1)) < 1.0
    True
    """
return list(set(lst0) - set(lst1))[0]

Use OK to test your code:

python3 ok -q missing_val

Question 2: Find duplicates k

Write the following function so it runs in O(n) time.

Hint: Sets have an add and a remove method.

def find_duplicates_k(k, lst):
    """Returns True if there are any duplicates in lst that are within k
    indices apart.

    >>> find_duplicates_k(3, [1, 2, 3, 4, 1])
    False
    >>> find_duplicates_k(4, [1, 2, 3, 4, 1])
    True
    >>> find_duplicates_k(4, [1, 1, 2, 3, 3])
    True
    """
prev_set = set() for i, elem in enumerate(lst): if elem in prev_set: return True prev_set.add(elem) if i - k >= 0: prev_set.remove(lst[i - k]) return False

Use OK to test your code:

python3 ok -q find_duplicates_k

More Review: Scheme

Note: Q3 and Q4 (Substitute, Sub All) were extra lab questions from Lab 9. You may check the solutions if you are stuck, but we highly recommend you work through the problem on your own for practice.

Question 3: Substitute

Write a procedure substitute that takes three arguments: a list s, an old word, and a new word. It returns a list with the elements of s, but with every occurrence of old replaced by new, even within sub-lists.

Hint: The built-in pair? predicate returns True if its argument is a cons pair.

(define (substitute s old new)
(cond ((null? s) nil) ((pair? (car s)) (cons (substitute (car s) old new) (substitute (cdr s) old new))) ((equal? (car s) old) (cons new (substitute (cdr s) old new))) (else (cons (car s) (substitute (cdr s) old new))))
)

Use OK to test your code:

python3 ok -q substitute

Question 4: Sub All

Write sub-all, which takes a list s, a list of old words, and a list of new words; the last two lists must be the same length. It returns a list with the elements of s, but with each word that occurs in the second argument replaced by the corresponding word of the third argument.

(define (sub-all s olds news)
(if (null? olds) s (sub-all (substitute s (car olds) (car news)) (cdr olds) (cdr news)))
)

Use OK to test your code:

python3 ok -q sub-all

Streams

Question 5: Scale Stream

Implement the function scale_stream, which returns a stream over each element of an input stream, scaled by k:

def scale_stream(s, k):
    """Return a stream of the elements of S scaled by a number K.

    >>> ints = make_integer_stream(1)
    >>> s = scale_stream(ints, 5)
    >>> stream_to_list(s, 5)
    [5, 10, 15, 20, 25]
    >>> s = scale_stream(Stream("x", lambda: Stream("y")), 3)
    >>> stream_to_list(s)
    ['xxx', 'yyy']
    >>> stream_to_list(scale_stream(Stream.empty, 10))
    []
    """
if s is Stream.empty: return Stream.empty return Stream(s.first * k, lambda: scale_stream(s.rest, k))

Use OK to test your code:

python3 ok -q scale_stream

Question 6: Regular Numbers

Acknowledgements. This exercise is taken from Structure and Interpretation of Computer Programs, Section 3.5.2.

A famous problem, first raised by Richard Hamming, is to enumerate, in ascending order with no repetitions, all positive integers with no prime factors other than 2, 3, or 5. These are called regular numbers. One obvious way to do this is to simply test each integer in turn to see whether it has any factors other than 2, 3, and 5. But this is very inefficient, since, as the integers get larger, fewer and fewer of them fit the requirement. As an alternative, we can build a stream of such numbers. Let us call the required stream of numbers s and notice the following facts about it.

  • s begins with 1 .
  • The elements of scale_stream(s, 2) are also elements of s.
  • The same is true for scale_stream(s, 3) and scale_stream(s, 5).
  • These are all of the elements of s.

Now all we have to do is combine elements from these sources. For this we define a merge function that combines two ordered streams into one ordered result stream, eliminating repetitions.

Fill in the definition of merge, then fill in the definition of make_s below:

def merge(s0, s1):
    """Return a stream over the elements of strictly increasing s0 and s1,
    removing repeats. Assume that s0 and s1 have no repeats.

    >>> ints = make_integer_stream(1)
    >>> twos = scale_stream(ints, 2)
    >>> threes = scale_stream(ints, 3)
    >>> m = merge(twos, threes)
    >>> stream_to_list(m, 10)
    [2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
    """
    if s0 is Stream.empty:
        return s1
    elif s1 is Stream.empty:
        return s0

    e0, e1 = s0.first, s1.first
if e0 < e1: return Stream(e0, lambda: merge(s0.rest, s1)) elif e1 < e0: return Stream(e1, lambda: merge(s0, s1.rest)) else: return Stream(e0, lambda: merge(s0.rest, s1.rest))
def make_s(): """Return a stream over all positive integers with only factors 2, 3, & 5. >>> s = make_s() >>> stream_to_list(s, 20) [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36] """ def rest():
two_three = merge(scale_stream(s, 2), scale_stream(s, 3)) return merge(two_three, scale_stream(s, 5))
s = Stream(1, rest) return s

Use OK to test your code:

python3 ok -q merge

Use OK to test your code:

python3 ok -q make_s

Question 7: Linear Congruential Generator

A common method of producing pseudo-random numbers is by means of the following recurrence relation:

  • R0 = seed value
  • Ri+1 = (a*Ri + c) % n where Ri denotes the ith pseudo-random number in the stream; a, c, and n are constant integers, and seed value is some initial value provided by the user or chosen automatically by the system.

Define a function that returns a stream of random numbers that uses this linear-congruential formula.

from operator import add, mul, mod

def make_random_stream(seed, a, c, n):
    """The infinite stream of pseudo-random numbers generated by the
    recurrence r[0] = SEED, r[i+1] = (r[i] * A + C) % N.  Your solution
    must not use any lambdas or def's that we have not supplied in the
    skeleton.
    >>> s = make_random_stream(25, 29, 5, 32)
    >>> stream_to_list(s, 10)
    [25, 26, 23, 0, 5, 22, 3, 28, 17, 18]
    >>> s = make_random_stream(17, 299317, 13, 2**20)
    >>> stream_to_list(s, 10)
    [17, 894098, 115783, 383424, 775373, 994174, 941859, 558412, 238793, 718506]
    """

astr = constant_stream(a) cstr = constant_stream(c) nstr = constant_stream(n) result = Stream(seed, lambda: combine_streams(mod, combine_streams(add, combine_streams(mul, result, astr), cstr), nstr)) return result

Your solution must use only the functions defined in the skeleton, without defining any additional ones. Likewise, any lambda expressions should contain only calls to these functions.

Use OK to test your code:

python3 ok -q make_random_stream

Extra questions

Extra questions are not worth extra credit and are entirely optional.

Question 8: Stream of Streams

Write the function make_stream_of_streams, which returns an infinite stream, where the element at position i, counting from 1, is an infinite stream of integers that start from i. Your solution should have the form
result = Stream(..., lambda: ...)
return result

and should not require any additional helper functions (i.e., just use recursively defined streams, and any additional functions supplied in your starter file). You may find the map_stream function useful.

def make_stream_of_streams():
    """
    >>> stream_of_streams = make_stream_of_streams()
    >>> stream_of_streams
    Stream(Stream(1, <...>), <...>)
    >>> stream_to_list(stream_of_streams, 3)
    [Stream(1, <...>), Stream(2, <...>), Stream(3, <...>)]
    >>> stream_to_list(stream_of_streams, 4)
    [Stream(1, <...>), Stream(2, <...>), Stream(3, <...>), Stream(4, <...>)]
    """
result = Stream(make_integer_stream(1), lambda: map_stream(lambda x: x.rest, result)) return result

Use OK to test your code:

python3 ok -q make_stream_of_streams

Differentiation

The following problems develop a system for symbolic differentiation of algebraic expressions. The derive Scheme procedure takes an algebraic expression and a variable and returns the derivative of the expression with respect to the variable. Symbolic differentiation is of special historical significance in Lisp. It was one of the motivating examples behind the development of the language. Differentiating is a recursive process that applies different rules to different kinds of expressions:

; derive returns the derivative of EXPR with respect to VAR
(define (derive expr var)
  (cond ((number? expr) 0)
        ((variable? expr) (if (same-variable? expr var) 1 0))
        ((sum? expr) (derive-sum expr var))
        ((product? expr) (derive-product expr var))
        ((exp? expr) (derive-exp expr var))
        (else 'Error)))

To implement the system, we will use the following data abstraction. Sums and products are lists, and they are simplified on construction:

; Variables are represented as symbols
(define (variable? x) (symbol? x))
(define (same-variable? v1 v2)
  (and (variable? v1) (variable? v2) (eq? v1 v2)))

; Numbers are compared with =
(define (=number? expr num)
  (and (number? expr) (= expr num)))

; Sums are represented as lists that start with +.
(define (make-sum a1 a2)
  (cond ((=number? a1 0) a2)
        ((=number? a2 0) a1)
        ((and (number? a1) (number? a2)) (+ a1 a2))
        (else (list '+ a1 a2))))
(define (sum? x)
  (and (list? x) (eq? (car x) '+)))
(define (addend s) (cadr s))
(define (augend s) (caddr s))

; Products are represented as lists that start with *.
(define (make-product m1 m2)
(cond ((or (=number? m1 0) (=number? m2 0)) 0)
      ((=number? m1 1) m2)
      ((=number? m2 1) m1)
      ((and (number? m1) (number? m2)) (* m1 m2))
      (else (list '* m1 m2))))
(define (product? x)
  (and (list? x) (eq? (car x) '*)))
(define (multiplier p) (cadr p))
(define (multiplicand p) (caddr p))

Question 9: Derive Sum

Implement derive-sum, a procedure that differentiates a sum by summing the derivatives of the addend and augend. Use data abstraction for a sum:

(define (derive-sum expr var)
(make-sum (derive (addend expr) var) (derive (augend expr) var)))

Use OK to test your code:

python3 ok -q derive-sum

Question 10: Derive Product

Implement derive-product, which applies the product rule to differentiate products:

(define (derive-product expr var)
(make-sum (make-product (derive (multiplier expr) var) (multiplicand expr)) (make-product (multiplier expr) (derive (multiplicand expr) var))))

Use OK to test your code:

python3 ok -q derive-product

Question 11: Make Exp

Implement a data abstraction for exponentiation: a base raised to the power of an exponent. The base can be any expression, but assume that the exponent is a non-negative integer. You can simplify the cases when exponent is 0 or 1, or when base is a number, by returning numbers from the constructor make-exp. In other cases, you can represent the exp as a triple (^ base exponent).

Hint: The built-in procedure expt takes two number arguments and raises the first to the power of the second.

; Exponentiations are represented as lists that start with ^.
(define (make-exp base exponent)
(cond ((= exponent 0) 1) ((= exponent 1) base) ((and (number? base) (integer? exponent)) (expt base exponent)) (else (list '^ base exponent))))
(define (base exp)
(cadr exp))
(define (exponent exp)
(caddr exp))
(define (exp? exp)
(and (list? exp) (eq? (car exp) '^)))
(define x^2 (make-exp 'x 2)) (define x^3 (make-exp 'x 3))

Use OK to test your code:

python3 ok -q make-exp

Question 12: Derive Exp

Implement derive-exp, which uses the power rule to derive exps:

(define (derive-exp exp var)
(let ((b (base exp)) (n (exponent exp))) (if (number? n) (let ((n-1 (make-sum n -1))) (let ((first (make-product n (make-exp b n-1)))) (make-product first (derive b var)))) 'unknown)))

Use OK to test your code:

python3 ok -q derive-exp