# Homework 3 Solutions

## Solution Files

You can find solutions for all questions in hw03.py.

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: 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
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'has_seven',
... ['Assign', 'AugAssign'])
True
"""
if k % 10 == 7:
return True
elif k < 10:
return False
else:
return has_seven(k // 10)
```

Use Ok to test your code:

`python3 ok -q has_seven`

The equivalent iterative version of this problem might look something like this:

```
while n > 0:
if n % 10 == 7:
return True
n = n // 10
```

The main idea is that we check each digit for a seven. The recursive solution is similar, except that you depend on the recursive call to check the rest of the number for a seven. All that's left is to check the last digit in the current step.

### Q2: 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
if n == 1:
return term(n)
else:
return term(n) + summation(n - 1, term)
```

Use Ok to test your code:

`python3 ok -q summation`

**Base case:**only one item to sum, so we return that item.**Recursive call:**returns the result of summing the numbers up to`n-1`

using`term`

. All that's missing is the current value`n`

.

```
def square(x):
return x * x
def identity(x):
return x
triple = lambda x: 3 * x
increment = lambda x: x + 1
add = lambda x, y: x + y
mul = lambda x, y: x * y
```

### Q3: Accumulate

Show that both `summation`

and `product`

(from Homework 2) are instances of a
more general function, called `accumulate`

:

```
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
"""
total, k = base, 1
while k <= n:
total, k = combiner(total, term(k)), k + 1
return total
# Recursive solution
# if n == 0:
# return base
# else:
# return combiner(term(n), accumulate(combiner, base, n-1, term))
# Recursive solution using base to keep track of total
# if n == 0:
# return base
# else:
# return accumulate(combiner, combiner(base, term(n)), n-1, term)
```

`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
"""
return accumulate(add, 0, n, term)
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
"""
return accumulate(mul, 1, n, term)
```

Use Ok to test your code:

```
python3 ok -q accumulate
python3 ok -q summation_using_accumulate
python3 ok -q product_using_accumulate
```

Both an iterative and recursive solution were allowed. Note that they are quite similar to the solution for summation! The main differences are:

- Abstracted away the method of combination (either
`+`

or`*`

) - Added in a starting base value, since product behaves poorly if we start with 0

### Q4: 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):
if pred(y):
return combiner(x, y)
else:
return x 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`

The implementation of `filtered_accumulate`

will depend on how you
implemented your `accumulate`

. This is because the `combine_if`

function we
use will always keep the first item (`x`

), and conditionally join with the
second item (`y`

).

The idea behind this combiner goes something like this:

`x`

represents all the things we have successfully combined so far.- If we want to include
`y`

in our combinations, then we return`combiner(x, y)`

. - On the other hand, if we
*don't*want to include`y`

in our combinations, then we don't want to discard our progress so far, so we will just return`x`

as the "result" of combining`x`

and`y`

.

### Q5: Make Repeater

Implement a function `make_repeater`

so that
`make_repeater(f, n)(x)`

returns `f(f(...f(x)...))`

, where `f`

is applied
`n`

times. That is, `make_repeater(f, n)`

returns another function that can then be
applied to another argument. For example,
`make_repeater(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 make_repeater(f, n):
"""Return the function that computes the nth application of f.
>>> add_three = make_repeater(increment, 3)
>>> add_three(5)
8
>>> make_repeater(triple, 5)(1) # 3 * 3 * 3 * 3 * 3 * 1
243
>>> make_repeater(square, 2)(5) # square(square(5))
625
>>> make_repeater(square, 4)(5) # square(square(square(square(5))))
152587890625
>>> make_repeater(square, 0)(5)
5
"""
g = identity
while n > 0:
g = compose1(f, g)
n = n - 1
return g
# Alternatives
def make_repeater2(f, n):
def h(x):
k = 0
while k < n:
x, k = f(x), k + 1
return x
return h
def make_repeater3(f, n):
return accumulate(compose1, lambda x: x, n, lambda k: f)
```

For an extra challenge, try defining

`make_repeater`

using`compose1`

and your`accumulate`

function in a single one-line return statement.

```
def compose1(f, g):
"""Return a function h, such that h(x) = f(g(x))."""
def h(x):
return f(g(x))
return h
```

Use Ok to test your code:

`python3 ok -q make_repeater`

There are many correct ways to implement `make_repeater`

. The first
solution above creates a new function in every iteration of the `while`

statement (via `compose1`

). The second solution shows that it is also
possible to implement `make_repeater`

by creating only a single new
function. That function make_repeaterly applies `f`

.

`make_repeater`

can also be implemented compactly using `accumulate`

, the
third solution.

## Extra questions

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

### Q6: 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.

### Q7: 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)"""
return lambda x: f(x)
def two(f):
"""Church numeral 2: same as successor(successor(zero))"""
return lambda x: f(f(x))
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
"""
return n(lambda x: x + 1)(0)
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
"""
return lambda f: lambda x: m(f)(n(f)(x))
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
"""
return lambda f: m(n(f))
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
"""
return n(m)
```

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
```

Church numerals are a way to represent non-negative integers via
repeated function application. The definitions of `zero`

, `one`

, and
`two`

show that each numeral is a function that takes a function and
repeats it a number of times on some argument `x`

.

The `church_to_int`

function reveals how a Church numeral can be mapped
to our normal notion of non-negative integers using the increment
function.

Addition of Church numerals is function composition of the functions of
`x`

, while multiplication is composition of the functions of `f`

.