# 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`

and`n`

: the same arguments as in`summation`

and`product`

`combiner`

: a two-argument function that specifies how the current term combined with the previously accumulated terms. You may assume that`combiner`

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`

and`n`

: the same arguments as`accumulate`

.`pred`

: a one-argument predicate function applied to the values of`term(k)`

,`k`

from 1 to`n`

. Only values for which`pred`

returns a true value are combined to form the result. If no values satisfy`pred`

, then`base`

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 function`n`

was a non-negative integer`x`

was an argument for`f`

`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`

using`compose1`

and`accumulate`

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
```