# Homework 2

*Due by 11:59pm on Wednesday, 6/29*

## Instructions

Download hw02.zip. Inside the archive, you will find a file called hw02.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. See Lab 0 for 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:

## Required questions

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.)
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 1: Product

The `summation(term, n)`

function from lecture adds up `term(1) + ... + term(n)`

Write a similar `product(n, term)`

function that returns ```
term(1) * ... *
term(n)
```

. Show how to define the
factorial function in terms of
`product`

. *Hint*: try using the `identity`

function for `factorial`

.

```
def product(n, term):
"""Return the product of the first n terms in a sequence.
n -- a positive integer
term -- a function that takes one argument
>>> product(3, identity) # 1 * 2 * 3
6
>>> product(5, identity) # 1 * 2 * 3 * 4 * 5
120
>>> product(3, square) # 1^2 * 2^2 * 3^2
36
>>> product(5, square) # 1^2 * 2^2 * 3^2 * 4^2 * 5^2
14400
"""
"*** YOUR CODE HERE ***"
def factorial(n):
"""Return n factorial for n >= 0 by calling product.
>>> factorial(4)
24
>>> factorial(6)
720
>>> from construct_check import check
>>> check(HW_SOURCE_FILE, 'factorial', ['Recursion', 'For', 'While'])
True
"""
"*** YOUR CODE HERE ***"
return _______
```

Use OK to test your code:

```
python3 ok -q product
python3 ok -q factorial
```

### Question 2: 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 3: 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:

```
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', 'FunctionDef'])
True
"""
"*** YOUR CODE HERE ***"
return _______
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`

with a single **return** statement containing
a call to `accumulate`

. Do not write any loops, **def** statements, or
recursive calls to `filtered_accumulate`

.

Hint: It may be useful to use one line if-else statements, otherwise known as ternary operators. The syntax is described in the Python documentation:The expression

`x if C else y`

first evaluates the condition,`C`

rather than`x`

. If`C`

is true,`x`

is evaluated and its value is returned; otherwise,`y`

is evaluated and its value is returned

Use OK to test your code:

`python3 ok -q filtered_accumulate`

### Question 4: Repeated

In lab 1, we 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)...))`

. Let's write a slightly different version of this function,
`repeated(f, n)`

.

The new `repeated`

, instead of returning the result directly, returns another
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 ***"
```

*Hint*: You may find it convenient to use `compose1`

from the textbook:

```
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 repeated`

## Extra questions

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

### Question 5: 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
```