# Homework 1

*Due by 11:59pm on Sunday, 6/25*

## Instructions

Download hw01.zip.

**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:

## Homework Questions

### Question 1

Fill in the blanks in the following function definition for adding `a`

to the
absolute value of `b`

, without calling `abs`

.

```
from operator import add, sub
def a_plus_abs_b(a, b):
"""Return a+abs(b), but without calling abs.
>>> a_plus_abs_b(2, 3)
5
>>> a_plus_abs_b(2, -3)
5
"""
if b < 0:
f = _____
else:
f = _____
return f(a, b)
```

Use OK to test your code:

`python3 ok -q a_plus_abs_b`

### Question 2

Write a function that takes three *positive* numbers and returns the
sum of the squares of the two largest numbers. **Use only a single
line for the body of the function.**

```
def two_of_three(a, b, c):
"""Return x*x + y*y, where x and y are the two largest members of the
positive numbers a, b, and c.
>>> two_of_three(1, 2, 3)
13
>>> two_of_three(5, 3, 1)
34
>>> two_of_three(10, 2, 8)
164
>>> two_of_three(5, 5, 5)
50
"""
return _____
```

Use OK to test your code:

`python3 ok -q two_of_three`

### Question 3

Write a function that takes an integer `n`

that is **greater than 1** and
returns the largest integer that is smaller than `n`

and evenly divides `n`

.

```
def largest_factor(n):
"""Return the largest factor of n that is smaller than n.
>>> largest_factor(15) # factors are 1, 3, 5
5
>>> largest_factor(80) # factors are 1, 2, 4, 5, 8, 10, 16, 20, 40
40
>>> largest_factor(13) # factor is 1 since 13 is prime
1
"""
"*** YOUR CODE HERE ***"
```

Hint:To check if`b`

evenly divides`a`

, you can use the expression`a % b == 0`

, which can be read as, "the remainder of dividing`a`

by`b`

is 0."

Use OK to test your code:

`python3 ok -q largest_factor`

### Question 4

Let's write a function that does the same thing as an `if`

statement.

```
def if_function(condition, true_result, false_result):
"""Return true_result if condition is a true value, and
false_result otherwise.
>>> if_function(True, 2, 3)
2
>>> if_function(False, 2, 3)
3
>>> if_function(3==2, 3+2, 3-2)
1
>>> if_function(3>2, 3+2, 3-2)
5
"""
if condition:
return true_result
else:
return false_result
```

Despite the doctests above, this function actually does *not* do the
same thing as an `if`

statement in all cases. To prove this fact,
write functions `c`

, `t`

, and `f`

such that `with_if_statement`

returns the number `1`

, but `with_if_function`

does not (it can do
*anything* else):

```
def with_if_statement():
"""
>>> with_if_statement()
1
"""
if c():
return t()
else:
return f()
def with_if_function():
return if_function(c(), t(), f())
def c():
"*** YOUR CODE HERE ***"
def t():
"*** YOUR CODE HERE ***"
def f():
"*** YOUR CODE HERE ***"
```

To test your solution, open an interactive interpreter

`python3 -i hw01.py`

and try calling `with_if_function`

and `with_if_statement`

to check that one
returns 1 and the other doesn't.

Hint: If you are having a hard time identifying how the if statement and if function differ, first try to get them to

### Question 5

Douglas Hofstadter's Pulitzer-prize-winning book, *GĂ¶del, Escher,
Bach*, poses the following mathematical puzzle.

- Pick a positive integer
`n`

as the start. - If
`n`

is even, divide it by 2. - If
`n`

is odd, multiply it by 3 and add 1. - Continue this process until
`n`

is 1.

The number `n`

will travel up and down but eventually end at 1 (at
least for all numbers that have ever been tried -- nobody has ever
proved that the sequence will terminate). Analogously, a hailstone
travels up and down in the atmosphere before eventually landing on
earth.

This sequence of values of `n`

is often called a Hailstone sequence,
Write a function that takes a single argument with formal parameter
name `n`

, prints out the hailstone sequence starting at `n`

, and
returns the number of steps in the sequence:

```
def hailstone(n):
"""Print the hailstone sequence starting at n and return its
length.
>>> a = hailstone(10)
10
5
16
8
4
2
1
>>> a
7
"""
"*** YOUR CODE HERE ***"
```

Hailstone sequences can get quite long! Try 27. What's the longest you can find?

Use OK to test your code:

`python3 ok -q hailstone`