Homework 4

Due by 11:59pm on Thursday, 2/15

Instructions

Download hw04.zip. Inside the archive, you will find a file called hw04.py, along with a copy of the ok autograder.

Submission: When you are done, submit with python3 ok --submit. You may submit more than once before the deadline; only the final submission will be scored. Check that you have successfully submitted your code on okpy.org. See Lab 0 for more instructions on submitting assignments.

Using Ok: If you have any questions about using Ok, please refer to this guide.

Readings: You might find the following references useful:

The construct_check module is used in this assignment, which defines a function check. For example, a call such as

check("foo.py", "func1", ["While", "For", "Recursion"])

checks that the function func1 in file foo.py does not contain any while or for constructs, and is not an overtly recursive function (i.e., one in which a function contains a call to itself by name.)

Required questions

Q1: Taxicab Distance

An intersection in midtown Manhattan can be identified by an avenue and a street, which are both indexed by positive integers. The Manhattan distance or taxicab distance between two intersections is the number of blocks that must be traversed to reach one from the other, ignoring one-way street restrictions and construction. For example, Times Square is on 46th Street and 7th Avenue. Ess-a-Bagel is on 51st Street and 3rd Avenue. The taxicab distance between them is 9 blocks (5 blocks from 46th to 51st street and 4 blocks from 7th avenue to 3rd avenue). Taxicabs cannot cut diagonally through buildings to reach their destination!

Implement taxicab, which computes the taxicab distance between two intersections using the following data abstraction. Hint: You don't need to know what a Cantor pairing function is; just use the abstraction.

def intersection(st, ave):
    """Represent an intersection using the Cantor pairing function."""
    return (st+ave)*(st+ave+1)//2 + ave

def street(inter):
    return w(inter) - avenue(inter)

def avenue(inter):
    return inter - (w(inter) ** 2 + w(inter)) // 2

w = lambda z: int(((8*z+1)**0.5-1)/2)

def taxicab(a, b):
    """Return the taxicab distance between two intersections.

    >>> times_square = intersection(46, 7)
    >>> ess_a_bagel = intersection(51, 3)
    >>> taxicab(times_square, ess_a_bagel)
    9
    >>> taxicab(ess_a_bagel, times_square)
    9
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q taxicab

Q2: Squares only

Implement the function squares, which takes in a list of positive integers. It returns a list that contains the square roots of the elements of the original list that are perfect squares. Try using a list comprehension.

You may find the round function useful.

>>> round(10.5)
10
>>> round(10.51)
11
def squares(s):
    """Returns a new list containing square roots of the elements of the
    original list that are perfect squares.

    >>> seq = [8, 49, 8, 9, 2, 1, 100, 102]
    >>> squares(seq)
    [7, 3, 1, 10]
    >>> seq = [500, 30]
    >>> squares(seq)
    []
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q squares

Q3: G function

A mathematical function G on positive integers is defined by two cases:

G(n) = n,                                       if n <= 3
G(n) = G(n - 1) + 2 * G(n - 2) + 3 * G(n - 3),  if n > 3

Write a recursive function g that computes G(n). Then, write an iterative function g_iter that also computes G(n):

def g(n):
    """Return the value of G(n), computed recursively.

    >>> g(1)
    1
    >>> g(2)
    2
    >>> g(3)
    3
    >>> g(4)
    10
    >>> g(5)
    22
    >>> from construct_check import check
    >>> check(HW_SOURCE_FILE, 'g', ['While', 'For'])
    True
    """
    "*** YOUR CODE HERE ***"

def g_iter(n):
    """Return the value of G(n), computed iteratively.

    >>> g_iter(1)
    1
    >>> g_iter(2)
    2
    >>> g_iter(3)
    3
    >>> g_iter(4)
    10
    >>> g_iter(5)
    22
    >>> from construct_check import check
    >>> check(HW_SOURCE_FILE, 'g_iter', ['Recursion'])
    True
    """
    "*** YOUR CODE HERE ***"

Use Ok to test your code:

python3 ok -q g
python3 ok -q g_iter

Q4: Ping pong

The ping-pong sequence counts up starting from 1 and is always either counting up or counting down. At element k, the direction switches if k is a multiple of 7 or contains the digit 7. The first 30 elements of the ping-pong sequence are listed below, with direction swaps marked using brackets at the 7th, 14th, 17th, 21st, 27th, and 28th elements:

1 2 3 4 5 6 [7] 6 5 4 3 2 1 [0] 1 2 [3] 2 1 0 [-1] 0 1 2 3 4 [5] [4] 5 6

Implement a function pingpong that returns the nth element of the ping-pong sequence. Do not use any assignment statements; however, you may use def statements.

Hint: If you're stuck, try implementing pingpong first using assignment and a while statement. Any name that changes value will become an argument to a function in the recursive definition.

def pingpong(n):
    """Return the nth element of the ping-pong sequence.

    >>> pingpong(7)
    7
    >>> pingpong(8)
    6
    >>> pingpong(15)
    1
    >>> pingpong(21)
    -1
    >>> pingpong(22)
    0
    >>> pingpong(30)
    6
    >>> pingpong(68)
    2
    >>> pingpong(69)
    1
    >>> pingpong(70)
    0
    >>> pingpong(71)
    1
    >>> pingpong(72)
    0
    >>> pingpong(100)
    2
    >>> from construct_check import check
    >>> check(HW_SOURCE_FILE, 'pingpong', ['Assign', 'AugAssign'])
    True
    """
    "*** YOUR CODE HERE ***"

You may use the function has_seven, which returns True if a number k contains the digit 7 at least once.

Use Ok to test your code:

python3 ok -q pingpong
def has_seven(k):
    """Returns True if at least one of the digits of k is a 7, False otherwise.

    >>> has_seven(3)
    False
    >>> has_seven(7)
    True
    >>> has_seven(2734)
    True
    >>> has_seven(2634)
    False
    >>> has_seven(734)
    True
    >>> has_seven(7777)
    True
    """
    if k % 10 == 7:
        return True
    elif k < 10:
        return False
    else:
        return has_seven(k // 10)

Q5: Count change

Once the machines take over, the denomination of every coin will be a power of two: 1-cent, 2-cent, 4-cent, 8-cent, 16-cent, etc. There will be no limit to how much a coin can be worth.

A set of coins makes change for n if the sum of the values of the coins is n. For example, the following sets make change for 7:

  • 7 1-cent coins
  • 5 1-cent, 1 2-cent coins
  • 3 1-cent, 2 2-cent coins
  • 3 1-cent, 1 4-cent coins
  • 1 1-cent, 3 2-cent coins
  • 1 1-cent, 1 2-cent, 1 4-cent coins

Thus, there are 6 ways to make change for 7. Write a function count_change that takes a positive integer n and returns the number of ways to make change for n using these coins of the future:

def count_change(amount):
    """Return the number of ways to make change for amount.

    >>> count_change(7)
    6
    >>> count_change(10)
    14
    >>> count_change(20)
    60
    >>> count_change(100)
    9828
    """
    "*** YOUR CODE HERE ***"

Hint: you may find it helpful to refer to the implementation of count_partitions.

Use Ok to test your code:

python3 ok -q count_change

Extra questions

Extra questions are not worth extra credit and are entirely optional. They are designed to challenge you to think creatively!

Q6: Anonymous factorial

The recursive factorial function can be written as a single expression by using a conditional expression.

>>> fact = lambda n: 1 if n == 1 else mul(n, fact(sub(n, 1)))
>>> fact(5)
120

However, this implementation relies on the fact (no pun intended) that fact has a name, to which we refer in the body of fact. To write a recursive function, we have always given it a name using a def or assignment statement so that we can refer to the function within its own body. In this question, your job is to define fact recursively without giving it a name!

Write an expression that computes n factorial using only call expressions, conditional expressions, and lambda expressions (no assignment or def statements). Note in particular that you are not allowed to use make_anonymous_factorial in your return expression. The sub and mul functions from the operator module are the only built-in functions required to solve this problem:

from operator import sub, mul

def make_anonymous_factorial():
    """Return the value of an expression that computes factorial.

    >>> make_anonymous_factorial()(5)
    120
    >>> from construct_check import check
    >>> check(HW_SOURCE_FILE, 'make_anonymous_factorial', ['Assign', 'AugAssign', 'FunctionDef', 'Recursion'])
    True
    """
    return 'YOUR_EXPRESSION_HERE'

Use Ok to test your code:

python3 ok -q make_anonymous_factorial