前言

觀前提示，筆者寫的代碼答案放在github倉庫中，此處僅記錄過程與心得

正文

首先來看下hw的第一道題

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 f(a, b)']
    """
    if b < 0:
        f = _____
    else:
        f = _____
    return f(a, b)

題目大意，賦予f一個函數滿足在不使用abs函數的情況下，實現a+abs（b）的效果

解題過程，剛開始沒注意到是賦予函數，于是分別填了a - b和a + b的錯誤答案，看了報錯類型才反應過來，然后想起python可以函數賦予，于是將sub和add分別賦予f即可

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 _____

題目大意，返回三個參數中最小兩個的平方和

解題過程，剛開始想了會如何在一行里選出三個參數中最小兩個，然后馬上想到全部一起平方加起來再減去最大的平方不就行了嗎

Q4: Largest Factor

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 ***"

題目大意，找到比整數n小的最大的能整除n的數字

解題過程，直接循環，從1開始窮舉，可以對循環條件除2或取根號2來減少循環次數

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 ***"

題目大意，通過編輯三個函數cond(),true_func(),false_func()，使得with_if_statement和with_if_fuction()產生預想的結果

解題過程，關鍵之關鍵在于了解到python中，將函數作為參數時，會先將函數運行，再將其結果作為參數返回，因此填上對應的數字，并調好條件cond就能成功解題

Q6: Hailstone

Douglas Hofstadter's Pulitzer-prize-winning book, G?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.

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 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 ***"

題目大意，有一種數學規律（尚未完全證明），對一個正整數，若其為偶數，則除2，若其為奇數，則乘3并加1，重復上述步驟，此數定終為1。現有n，輸出n經上述規律到1的過程，并返回計算次數

解題過程，寫循環，循環條件是數為1，每次循環輸出該數，再設一個變量每次循環++即可

小結

雖然題目不難，但仍發現了自己一些問題，比如審題不清，有一定的原因是題目是英語而非中文，不過都要適應嘛