Please download homework materials

`hw01.zip`

from our QQ group if you don't have one.

In this homework, you are required to complete the problems described in section 2. The starter code for these problems is provided in `hw01.py`

, which is distributed as part of the homework materials in the `code`

directory.

**Submission**: When you are done, submit your code to our OJ website as instructed in lab00. You may submit more than once before the deadline; only the final submission will be scored. See lab00 for more instructions on submitting assignments.

**Readings**: You might find the following references to the textbook useful:

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

to the absolute value of `b`

, without calling `abs`

. You may **not** modify any of the provided code other than the two blanks.

```
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
>>> # a check that you didn't change the return statement!
>>> import inspect, re
>>> re.findall(r'^\s*(return .*)', inspect.getsource(a_plus_abs_b), re.M)
['return h(a, b)']
"""
if b >= 0:
h = _____
else:
h = _____
return h(a, b)
```

You can use doctest to test your code:

```
$ python -m doctest hw01.py
```

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

```
def two_of_three(x, y, z):
"""Return a*a + b*b, where a and b are the two smallest members of the
positive numbers x, y, and z.
>>> two_of_three(1, 2, 3)
5
>>> two_of_three(5, 3, 1)
10
>>> two_of_three(10, 2, 8)
68
>>> two_of_three(5, 5, 5)
50
>>> # check that your code consists of nothing but an expression (this docstring)
>>> # a return statement
>>> import inspect, ast
>>> [type(x).__name__ for x in ast.parse(inspect.getsource(two_of_three)).body[0].body]
['Expr', 'Return']
"""
return _____
```

Hint: Consider using the

`max`

or`min`

function:

>>> max(1, 2, 3) 3 >>> min(-1, -2, -3) -3

Write a function that takes an integer `x`

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

and evenly divides `x`

.

```
def largest_factor(x):
"""Return the largest factor of x that is smaller than x.
>>> 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."

Let's try to 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`

prints the number `6`

, but `with_if_function`

prints both `5`

and `6`

.

```
def with_if_statement():
"""
>>> result = with_if_statement()
6
>>> print(result)
None
"""
if c():
return t()
else:
return f()
def with_if_function():
"""
>>> result = with_if_function()
5
6
>>> print(result)
None
"""
return if_function(c(), t(), f())
def c():
"*** YOUR CODE HERE ***"
def t():
"*** YOUR CODE HERE ***"
def f():
"*** YOUR CODE HERE ***"
```

Hint: If you are having a hard time identifying how an

`if`

statement and`if_function`

differ, consider the rules of evaluation for`if`

statements and call expressions.

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

- Pick a positive integer
`x`

as the start. - If
`x`

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

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

is 1.

The number `x`

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.

Breaking News(or at least the closest thing to that in math). There has been a recent development in the hailstone conjecture that shows that almost all numbers will eventually get to 1 if you repeat this process. This isn't a complete proof but a major breakthrough.

This sequence of values of `x`

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

, prints out the hailstone sequence starting at `x`

, and returns the number of steps in the sequence:

```
def hailstone(x):
"""Print the hailstone sequence starting at x 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?

Let's write a function `falling`

, which is a "falling" factorial that takes two arguments, `n`

and `k`

, and returns the product of `k`

consecutive numbers, starting from `n`

and working downwards. (This problem is an advanced version of Problem 2, Lab 01.)

```
def falling(n, k):
"""Compute the falling factorial of n to depth k.
>>> falling(6, 3) # 6 * 5 * 4
120
>>> falling(4, 3) # 4 * 3 * 2
24
>>> falling(4, 1) # 4
4
>>> falling(4, 0)
1
"""
"*** YOUR CODE HERE ***"
```

Write a function that takes in a number and determines if the digits contain two adjacent 8s. (Reviewing Problem 4 and 5 in Lab 01 might be helpful here!)

```
def double_eights(n):
"""Return true if n has two eights in a row.
>>> double_eights(8)
False
>>> double_eights(88)
True
>>> double_eights(2882)
True
>>> double_eights(880088)
True
>>> double_eights(12345)
False
>>> double_eights(80808080)
False
"""
"*** YOUR CODE HERE ***"
```

Make sure to submit your code to our OJ website as instructed in lab00.