Homework 3

Due Monday, Oct 22, 2018 at 8pm

A PDF version of this document is located here.

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  1. Decorators and memoization. Memoization is an optimization in recursive algorithms that have repeated computations, where the arguments and result of a function call are stored when a function is first called with that set of arguments. Then if the function is called again with the same arguments, the stored value is looked up and returned rather than being recomputed.

    For this problem, implement a memoize decorator in Python that takes in a function and returns a version that performs memoization. The decorator should work on functions that take in any number of non-keyword arguments. You may not use the functools module in your solution.

    Hint: We recommend using a dictionary to store previously computed values, and variable argument lists to handle functions with any number of parameters.

    def memoize(func):
        """Return a version of func that memoizes computation.
        Returns a function that computes the same result as func, but
        if a given set of arguments has already been seen before,
        returns the previously computed result instead of repeating
        the computation. Assumes func is a pure function (i.e. has no
        side affects), and that all arguments to func are hashable.
        >>> @memoize
        ... def sum_to_n(n):
        ...     return 1 if n == 1 else n + sum_to_n(n - 1)
        >>> try:
        ...     sum_to_n(300)
        ...     sum_to_n(600)
        ...     sum_to_n(900)
        ...     sum_to_n(1200)
        ... except RecursionError:
        ...     print('recursion limit exceeded')
        >>> @memoize
        ... def sum_k_to_n(k, n):
        ...     return k if n == k else n + sum_k_to_n(k, n - 1)
        >>> try:
        ...     sum_k_to_n(2, 300)
        ...     sum_k_to_n(2, 600)
        ...     sum_k_to_n(2, 900)
        ...     sum_k_to_n(2, 1200)
        ... except RecursionError:
        ...     print('recursion limit exceeded')
        # add your code below
  2. Chain. Recall that the compose() higher-order function takes in two single-parameter functions as arguments and returns the composition of the two functions. Thus, compose(f, g)(x) computes f(g(x)).

    The following is a definition of compose() in Python:

    def compose(f, g):
        return lambda x: f(g(x))

    Composition can be generalized to function chains, so that chain(f, g, h)(x) computes f(g(h(x))), chain(f, g, h, k)(x) computes f(g(h(k(x)))), and so on. Implement the variadic chain() function in Python.

    def chain(*funcs):
        """Return a function that is the compositional chain of funcs.
        If funcs is empty, returns the identity function.
        >>> chain()(3)
        >>> chain(lambda x: 3 * x)(3)
        >>> chain(lambda x: x + 1, lambda x: 3 * x)(3)
        >>> chain(lambda x: x // 2, lambda x: x + 1, lambda x: 3 * x)(3)
        # add your code below

    Hint: Your solution should use recursion.

  3. Generators.

    1. Implement a scale() generator that, given an iterable of numbers, scales them by a constant to produce a new iterable. For example, given a generator naturals() for the natural numbers, scale(naturals(), 2) produces an iterable of the even natural numbers.

      def scale(items, factor):
          """Produce an iterable of the elements in items scaled by
          Consumes the elements from items.
          >>> def naturals():
          ...     num = 0
          ...     while True:
          ...         yield num
          ...         num += 1
          >>> values = scale(naturals(), 3)
          >>> [next(values) for i in range(5)]
          [0, 3, 6, 9, 12]
          # add your code below
    2. Implement a merge() generator that, given two infinite iterables whose elements are in strictly increasing order, produces a new iterable that contains the items from both input iterables, in increasing order and without duplicates.

      def merge(items1, items2):
          """Produce an ordered iterable that is the merge of the inputs.
          The resulting iterable contains the elements in increasing
          order from items1 and items2, without duplicates. Requires
          each of items1 and items2 to be infinite iterables in
          strictly increasing order. Consumes the elements from
          items1 and items2.
          >>> def naturals():
          ...     num = 0
          ...     while True:
          ...         yield num
          ...         num += 1
          >>> values = merge(naturals(), naturals())
          >>> [next(values) for i in range(5)]
          [0, 1, 2, 3, 4]
          >>> values2 = merge(scale(naturals(), 2), scale(naturals(), 3))
          >>> [next(values2) for i in range(10)]
          [0, 2, 3, 4, 6, 8, 9, 10, 12, 14]
          # add your code below
    3. A famous problem, first raised by R. Hamming, is to enumerate, in ascending order with no repetitions, all positive integers with no prime factors other than 2, 3, or 5. 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 an iterable of such numbers using a generator. Let us call the required iterable of numbers s and notice the following facts about it.

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

      All that is left is to combine the elements from these sources, which can be done with the merge() generator above. Fill in the make_s() generator that produces the required iterable s.

      def make_s():
          """Produce iterable of ints that only have factors in {2, 3, 5}.
          >>> values = make_s()
          >>> [next(values) for i in range(18)]
          [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30]
          # add your code below
  4. Lambda calculus. For this question, you may use the symbol \(\lambda\) or the capital letter L to signify a \(\lambda\).

    1. Evaluate the \(\lambda\)-calculus term below until it is in normal form. Show each \(\alpha\)-reduction and \(\beta\)-reduction step, as in the following:

      \begin{align*} &(\lambda x.~ x)~ (\lambda x.~ x)\\ \rightarrow~ &(\lambda x.~ x)~ (\lambda y.~ y)~ &(\alpha\textrm{-reduction})\\ \rightarrow~ &\lambda y.~ y~ &(\beta\textrm{-reduction}) \end{align*}

      In plain text:

      (L x. x) (L x. x)
      -> (L x. x) (L y. y)              (alpha-reduction)
      -> L y. y                         (beta-reduction)

      Term to evaluate:

      \begin{equation*} (\lambda x.~ \lambda y.~ x~ \lambda x.~ y~ x)~ (\lambda w.~ w)~ (\lambda x.~ x~ x)~ a \end{equation*}

      Hint: Evaluating this term requires a total of five \(\beta\)-reductions and one \(\alpha\)-reduction.

    2. The following function maps a pair containing numbers \((m, n)\) to a pair containing \((m+1, m)\):

      \begin{equation*} pairincr~ =~ \lambda p.~ pair~ (incr~ (first~ p))~ (first~ p) \end{equation*}

      Applying it to \(pair~ m~ n\) produces:

      \begin{align*} pairincr~ (pair~ m~ n)~ &=~ (\lambda p.~ pair~ (incr~ (first~ p))~ (first~ p))~ (pair~ m~ n)\\ &\rightarrow~ pair~ (incr~ (first~ (pair~ m~ n))~ (first~ (pair~ m~ n)))\\ &\rightarrow~ pair~ (incr~ m)~ (first~ (pair~ m~ n))\\ &\rightarrow~ pair~ (incr~ m)~ m \end{align*}

      Using \(pairincr\), define a \(decr\) function that decrements a Church numeral:

      \begin{equation*} decr~ =~ \lambda n.~ \textrm{[fill in your solution]} \end{equation*}

      Hint: What is the result when \(pairincr\) is applied to the pair \((0, 0)\)? What about when it is applied twice?


Place your solutions to questions 1-3 in the provided hw3.py file. Write your answers to question 4 in a PDF file named hw3.pdf. Make sure to list any other students with whom you discussed the homework, as per course policy in the syllabus. Submit hw3.py to the autograder before the deadline. Submit hw3.pdf to Gradescope before the deadline.