Homework 3
Due by 11:59pm on Tuesday, 9/13
Instructions
Download hw03.zip.
The vitamin problems can be found in the vitamin
directory and the homework problems can be found in the problems
directory. You must run
python3 ok --submit
twice: once inside the vitamin
directory, and once
inside the problems
directory.
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:
Several doctests use the construct_check
module, 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.)
Vitamins
For this set of problems, you must run ok
from within the vitamin
directory.
While homework questions may be completed with a partner, please remember that
vitamin questions must be completed alone.
Question 1: Has Seven
Write a function has_seven
that takes a positive integer n
and
returns whether n
contains the digit 7. Do not use any assignment
statements - use recursion instead:
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
"""
"*** YOUR CODE HERE ***"
Use OK to test your code:
python3 ok -q has_seven
Question 2: Summation
Write a recursive implementation of summation
, which takes a positive integer
n
and a function term
. It applies term
to every number from 1
to n
including n
and returns the sum of the results.
def summation(n, term):
"""Return the sum of the first n terms in the sequence defined by term.
Implement using recursion!
>>> summation(5, lambda x: x * x * x) # 1^3 + 2^3 + 3^3 + 4^3 + 5^3
225
>>> summation(9, lambda x: x + 1) # 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
54
>>> summation(5, lambda x: 2**x) # 2^1 + 2^2 + 2^3 + 2^4 + 2^5
62
>>> # Do not use while/for loops!
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'summation',
... ['While', 'For'])
True
"""
assert n >= 1
"*** YOUR CODE HERE ***"
Use OK to test your code:
python3 ok -q summation
Required questions
For this set of problems, you must run ok
from within the problems
directory. Remember that you may choose to work with a partner on homework
questions.
Several doctests refer to these one-argument functions:
def square(x):
return x * x
def triple(x):
return 3 * x
def identity(x):
return x
def increment(x):
return x + 1
Question 3: Accumulate
Show that both summation
and product
are instances of a more
general function, called accumulate
:
from operator import add, mul
def accumulate(combiner, base, n, term):
"""Return the result of combining the first n terms in a sequence and base.
The terms to be combined are term(1), term(2), ..., term(n). combiner is a
two-argument commutative function.
>>> accumulate(add, 0, 5, identity) # 0 + 1 + 2 + 3 + 4 + 5
15
>>> accumulate(add, 11, 5, identity) # 11 + 1 + 2 + 3 + 4 + 5
26
>>> accumulate(add, 11, 0, identity) # 11
11
>>> accumulate(add, 11, 3, square) # 11 + 1^2 + 2^2 + 3^2
25
>>> accumulate(mul, 2, 3, square) # 2 * 1^2 * 2^2 * 3^2
72
"""
"*** YOUR CODE HERE ***"
accumulate(combiner, base, n, term)
takes the following arguments:
term
andn
: the same arguments as insummation
andproduct
combiner
: a two-argument function that specifies how the current term combined with the previously accumulated terms. You may assume thatcombiner
is commutative, i.e.,combiner(a, b) = combiner(b, a)
.base
: value that specifies what value to use to start the accumulation.
For example, accumulate(add, 11, 3, square)
is
11 + square(1) + square(2) + square(3)
Implement accumulate
and show how summation
and product
can both be
defined as simple calls to accumulate
:
def summation_using_accumulate(n, term):
"""Returns the sum of term(1) + ... + term(n). The implementation
uses accumulate.
>>> summation_using_accumulate(5, square)
55
>>> summation_using_accumulate(5, triple)
45
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'summation_using_accumulate',
... ['Recursion', 'For', 'While'])
True
"""
"*** YOUR CODE HERE ***"
return _______
def product_using_accumulate(n, term):
"""An implementation of product using accumulate.
>>> product_using_accumulate(4, square)
576
>>> product_using_accumulate(6, triple)
524880
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'product_using_accumulate',
... ['Recursion', 'For', 'While'])
True
"""
"*** YOUR CODE HERE ***"
return _______
Use OK to test your code:
python3 ok -q accumulate
python3 ok -q summation_using_accumulate
python3 ok -q product_using_accumulate
Question 4: Filtered Accumulate
Show how to extend the accumulate
function to allow for filtering the
results produced by its term
argument, by implementing the
filtered_accumulate
function in terms of accumulate
:
def filtered_accumulate(combiner, base, pred, n, term):
"""Return the result of combining the terms in a sequence of N terms
that satisfy the predicate PRED. COMBINER is a two-argument function.
If v1, v2, ..., vk are the values in TERM(1), TERM(2), ..., TERM(N)
that satisfy PRED, then the result is
BASE COMBINER v1 COMBINER v2 ... COMBINER vk
(treating COMBINER as if it were a binary operator, like +). The
implementation uses accumulate.
>>> filtered_accumulate(add, 0, lambda x: True, 5, identity) # 0 + 1 + 2 + 3 + 4 + 5
15
>>> filtered_accumulate(add, 11, lambda x: False, 5, identity) # 11
11
>>> filtered_accumulate(add, 0, odd, 5, identity) # 0 + 1 + 3 + 5
9
>>> filtered_accumulate(mul, 1, greater_than_5, 5, square) # 1 * 9 * 16 * 25
3600
>>> # Do not use while/for loops or recursion
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'filtered_accumulate',
... ['While', 'For', 'Recursion'])
True
"""
def combine_if(x, y):
"*** YOUR CODE HERE ***"
return accumulate(combine_if, base, n, term)
def odd(x):
return x % 2 == 1
def greater_than_5(x):
return x > 5
filtered_accumulate(combiner, base, pred, n, term)
takes
the following arguments:
combiner
,base
,term
andn
: the same arguments asaccumulate
.pred
: a one-argument predicate function applied to the values ofterm(k)
,k
from 1 ton
. Only values for whichpred
returns a true value are combined to form the result. If no values satisfypred
, thenbase
is returned.
For example, filtered_accumulate(add, 0, is_prime, 11, identity)
would be
0 + 2 + 3 + 5 + 7 + 11
for a suitable definition of is_prime
.
Implement filtered_accumulate
by defining the combine_if
function. Exactly
what this function does is something for you to discover. Do not write any
loops or recursive calls to filtered_accumulate
.
Use OK to test your code:
python3 ok -q filtered_accumulate
Question 5: Repeated
Previously, you implemented the function repeated(f, n, x)
, where:
f
was a one-argument functionn
was a non-negative integerx
was an argument forf
repeated(f, n, x)
returned the result of composing f
n
times on x
, i.e.,
f(f(...f(x)...))
. Now let's write a higher-order version of this function,
repeated(f, n)
.
The new repeated
, instead of returning the result directly, returns function
that, when given the argument x
, will compute f(f(...f(x)...))
. For example,
repeated(square, 3)(42)
evaluates to square(square(square(42)))
. Yes, it
makes sense to apply the function zero times! See if you can figure out a
reasonable function to return for that case.
def repeated(f, n):
"""Return the function that computes the nth application of f.
>>> add_three = repeated(increment, 3)
>>> add_three(5)
8
>>> repeated(triple, 5)(1) # 3 * 3 * 3 * 3 * 3 * 1
243
>>> repeated(square, 2)(5) # square(square(5))
625
>>> repeated(square, 4)(5) # square(square(square(square(5))))
152587890625
>>> repeated(square, 0)(5)
5
"""
"*** YOUR CODE HERE ***"
For an extra challenge, try defining
repeated
usingcompose1
andaccumulate
in a single one-line return statement.
Use OK to test your code:
python3 ok -q repeated
Extra questions
Extra questions are not worth extra credit and are entirely optional. They are designed to challenge you to think creatively!
Question 6: Quine
Write a one-line program that prints itself, using only the following features of the Python language:
- Number literals
- Assignment statements
- String literals that can be expressed using single or double quotes
- The arithmetic operators
+
,-
,*
, and/
- The built-in
print
function - The built-in
eval
function, which evaluates a string as a Python expression - The built-in
repr
function, which returns an expression that evaluates to its argument
You can concatenate two strings by adding them together with +
and repeat a
string by multipying it by an integer. Semicolons can be used to separate
multiple statements on the same line. E.g.,
>>> c='c';print('a');print('b' + c * 2)
a
bcc
Hint: Explore the relationship between single quotes, double quotes, and the
repr
function applied to strings.
A program that prints itself is called a Quine. Place your solution in the multi-line string named quine
.
Note: No tests will be run on your solution to this problem.
Question 7: Church numerals
The logician Alonzo Church invented a system of representing non-negative integers entirely using functions. The purpose was to show that functions are sufficient to describe all of number theory: if we have functions, we do not need to assume that numbers exist, but instead we can invent them.
Your goal in this problem is to rediscover this representation known as Church
numerals. Here are the definitions of zero
, as well as a function that
returns one more than its argument:
def zero(f):
return lambda x: x
def successor(n):
return lambda f: lambda x: f(n(f)(x))
First, define functions one
and two
such that they have the same behavior
as successor(zero)
and successsor(successor(zero))
respectively, but do
not call successor
in your implementation.
Next, implement a function church_to_int
that converts a church numeral
argument to a regular Python integer.
Finally, implement functions add_church
, mul_church
, and pow_church
that
perform addition, multiplication, and exponentiation on church numerals.
def one(f):
"""Church numeral 1: same as successor(zero)"""
"*** YOUR CODE HERE ***"
def two(f):
"""Church numeral 2: same as successor(successor(zero))"""
"*** YOUR CODE HERE ***"
three = successor(two)
def church_to_int(n):
"""Convert the Church numeral n to a Python integer.
>>> church_to_int(zero)
0
>>> church_to_int(one)
1
>>> church_to_int(two)
2
>>> church_to_int(three)
3
"""
"*** YOUR CODE HERE ***"
def add_church(m, n):
"""Return the Church numeral for m + n, for Church numerals m and n.
>>> church_to_int(add_church(two, three))
5
"""
"*** YOUR CODE HERE ***"
def mul_church(m, n):
"""Return the Church numeral for m * n, for Church numerals m and n.
>>> four = successor(three)
>>> church_to_int(mul_church(two, three))
6
>>> church_to_int(mul_church(three, four))
12
"""
"*** YOUR CODE HERE ***"
def pow_church(m, n):
"""Return the Church numeral m ** n, for Church numerals m and n.
>>> church_to_int(pow_church(two, three))
8
>>> church_to_int(pow_church(three, two))
9
"""
"*** YOUR CODE HERE ***"
Use OK to test your code:
python3 ok -q church_to_int
python3 ok -q add_church
python3 ok -q mul_church
python3 ok -q pow_church