Solutions: You can find the file with solutions for all questions here.

Question 1: Scale

Implement the generator function scale(s, k), which yields elements of the given iterable s, scaled by k.

def scale(s, k):
    """Yield elements of the iterable s scaled by a number k.

    >>> s = scale([1, 5, 2], 5)
    >>> type(s)
    <class 'generator'>
    >>> list(s)
    [5, 25, 10]

    >>> m = scale(naturals(), 2)
    >>> [next(m) for _ in range(5)]
    [2, 4, 6, 8, 10]
    for elem in s:
        yield elem * k

Use OK to test your code:

python3 ok -q scale --local

Question 2: Merge

Implement merge(s0, s1), which takes two iterables s0 and s1 whose elements are ordered. merge yields elements from s0 and s1 in sorted order, eliminating repetition. You may also assume s0 and s1 represent infinite sequences; that is, their iterators never raise StopIteration.

See the doctests for example behavior.

def merge(s0, s1):
    """Yield the elements of strictly increasing iterables s0 and s1 and 
    make sure to remove the repeated values in both. 
    You can also assume that s0 and s1 represent infinite sequences.

    >>> twos = scale(naturals(), 2)
    >>> threes = scale(naturals(), 3)
    >>> m = merge(twos, threes)
    >>> type(m)
    <class 'generator'>
    >>> [next(m) for _ in range(10)]
    [2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
    i0, i1 = iter(s0), iter(s1)
    e0, e1 = next(i0), next(i1)
    while True:
        yield min(e0, e1)
        if e0 < e1:
            e0 = next(i0)
        elif e1 < e0:
            e1 = next(i1)
            e0, e1 = next(i0), next(i1)

Use OK to test your code:

python3 ok -q merge --local

Question 3: Remainder generator

Like functions, generators can also be higher-order. For this problem, we will be writing remainders_generator, which yields a series of generator objects.

remainders_generator takes in an integer m, and yields m different generators. The first generator is a generator of multiples of m, i.e. numbers where the remainder is 0. The second, a generator of natural numbers with remainder 1 when divided by m. The last generator yield natural numbers with remainder m - 1 when divided by m.

def remainders_generator(m):
    Takes in an integer m, and yields m different remainder groups
    of m.

    >>> remainders_mod_four = remainders_generator(4)
    >>> for rem_group in remainders_mod_four:
    ...     for _ in range(3):
    ...         print(next(rem_group))
    def remainder_group(rem):
        start = rem
        while True:
            yield start
            start += m

    for rem in range(m):
        yield remainder_group(rem)

Note that if you have implemented this correctly, each of the generators yielded by remainder_generator will be infinite - you can keep calling next on them forever without running into a StopIteration exception.

Hint: Consider defining an inner generator function. What arguments should it take in? Where should you call it?

Use OK to test your code:

python3 ok -q remainders_generator --local

Question 4: Primes

Write a generator that generates prime numbers. Fill out the is_prime helper function and use that to create your generator.

def is_prime(n):
    Return True if n is prime, false otherwise.

    >>> is_prime(1)
    >>> is_prime(2)
    >>> is_prime(19)
    if n < 2:
        return False
    counter = 2
    while counter <= sqrt(n):
        if n % counter == 0:
            return False
        counter += 1
    return True
def primes():
    An infinite generator that outputs primes. 

    >>> p = primes()
    >>> for i in range(3):
    ...     print(next(p))
    num = 0
    while True:
        if is_prime(num):
            yield num
        num += 1

Use OK to test your code:

python3 ok -q is_prime --local

Use OK to test your code:

python3 ok -q primes --local