（CS61A）Homework 1: Variables & Functions, Control

刚开始的写CS61A作业：

OK程序都不知道在哪，自己开个源文件写（后来才发现要在网站作业下载）

Q2: A Plus Abs B

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)

分析：要求返回一个函数，以实现abs的功能

实现：既然要返回h.那么h就需要是函数

（又因为我们只能在横线修改，故观察题头的add sub函数）

答案就呼之欲出了

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 = add
    else:
        h = sub
    return h(a, b)

Q3: Two of Three

Write a function that takes three positive numbers as arguments 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

 

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 min(x*x+y*y,x*x+z*z,y*y+z*z)

Q4: Largest Factor

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."

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 ***"
    n=x-1
    while n>0:
        if x%n==0:
            return n
        n -= 1

 

Q5: If Function vs Statement

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 cond, true_func, and false_func such that with_if_statement prints the number 47, but with_if_function prints both 42 and 47.

def with_if_statement():
    """
    >>> result = with_if_statement()
    47
    >>> print(result)
    None
    """
    if cond():
        return true_func()
    else:
        return false_func()

def with_if_function():
    """
    >>> result = with_if_function()
    42
    47
    >>> print(result)
    None
    """
    return if_function(cond(), true_func(), false_func())

def cond():
    "*** YOUR CODE HERE ***"

def true_func():
    "*** YOUR CODE HERE ***"

def false_func():
    "*** 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

def with_if_function():
    """
    >>> result = with_if_function()
    42
    47
    >>> print(result)
    None
    """
    return if_function(cond(), true_func(), false_func())

def cond():
    "*** YOUR CODE HERE ***"
    return False


def true_func():
    "*** YOUR CODE HERE ***"
    print(42)

def false_func():
    "*** YOUR CODE HERE ***"
    print(47)

 and call expressions.

 

Q6: Hailstone

Douglas Hofstadter's Pulitzer-prize-winning book, Gödel, Escher, Bach, poses the following mathematical puzzle.

1. Pick a positive integer x as the start.
2. If x is even, divide it by 2.
3. If x is odd, multiply it by 3 and add 1.
4. 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 was a recent development in the hailstone conjecture last year 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?

Watch the hints video below for somewhere to start: 

Use Ok to test your code:

python3 ok -q hailstone

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 ***"
    n=0
    while x>0:
        print("%d"%x)
        n+=1
        if x==1:
            return n
        if x%2==0:
            x/=2
        else:
            x=x*3+1

    return n