CS61A 2022 fall HW 01: Functions, Control

HW01对应的是Textbook的1.1和1.2

Q1: A Plus Abs B

• 题目：

  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_plus_abs_b(-1, 4)
      3
      >>> a_plus_abs_b(-1, -4)
      3
      """
      if b < 0:
          f = _____
      else:
          f = _____
      return f(a, b)
  

• solution

      if b < 0:
          f = sub
      else:
          f = add
      return f(a, b)
  

  在终端里面输入python ok -q a_plus_abs_b

  

  这题其实一开始我没反应过来，愣了一下，模块也能赋值给变量

  • 再验证一下这种用法

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Q2: Two of Three

• 题目

  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(i, j, k):
      """Return m*m + n*n, where m and n are the two smallest members of the
      positive numbers i, j, and k.
  
      >>> 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
      """
      return _____
  

  Hint: Consider using the max or min function:

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

• solution

  呃，写这个的时候得把copilot关掉…不然copilot秒解

  • 看到hint其实思路就比较清晰了

    把三个的平方和加起来，然后减掉最大的那个数的平方

    return i*i + j*j + k*k - max(i, j, k)**2
    

  ---

  或者还有种方法

  return min(i*i+j*j,i*i+k*k,j*j+k*k)
  

• python ok -q two_of_three

  

Q3: 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 ***"
  

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.”

• solution

  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 ***"
      newlis = []
  
      for i in range(1,n):
          if n % i == 0:
              newlis.append(i)
      return newlis[len(newlis)-1]
  

  ---

  也可以反向做

  	factor = n - 1
      while factor > 0:
          if n % factor == 0:
              return factor
         	factor -= 1
  

Q4: Hailstone

这个真是个经典例子， 在Linux C编程里讲到死循环的时候，还有邓俊辉《数据结构》里面讲到什么是算法的时候（考虑有穷性），都用到了这个例子

• 题目

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

  1. Pick a positive integer n as the start.
  2. If n is even, divide it by 2.

3. If n is odd, multiply it by 3 and add 1.
4. 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（类似地）, 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
  >>> b = hailstone(1)
  1
  >>> b
  1
  """
  "*** YOUR CODE HERE ***"
  

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

Note that if n == 1 initially, then the sequence is one step long.
Hint: Recall the different outputs from using regular division / and floor division //

• solution

  一开始我补充了这个代码，好像导致死循环了…?

      length = 1
      while n!=1:
          print(n)
          length += 1
          if n % 2 == 0:
              n /= 2
       
          if n % 2 != 0:
              n = 3 * n + 1
          
      print(1)
      return length
  
  

  

  

  然后我改成了 n //= 2，还是死循环啊啊啊啊啊

  ---

  恍然大悟！下面这两个if 不构成选择结构，还是顺序结构！！！

  如果n执行完第一个if变成奇数的话，他还会执行第二个if…呜呜呜

          if n % 2 == 0:
              n /= 2
       
          if n % 2 != 0:
              n = 3 * n + 1
  

  另外死循环主要是看控制条件哪里出错了，所以原因就应该去n上面找，按照这个思路来

  
  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
      >>> b = hailstone(1)
      1
      >>> b
      1
      """
      "*** YOUR CODE HERE ***"
      length = 1
      while n != 1:
          print(n)
          length += 1
          if n % 2 == 0:
              n //= 2
  
          elif n % 2 != 0:
              n = 3 * n + 1
  
      print(1)
      return length
  
  

这个ok的评测机制应该是用的python的Doctests吧🤔

orz，怎么说呢，这个作业，反正是我在大学没有的体验

• 最后给的这个hailstone猜想的拓展阅读材料挺有意思的
  • https://www.quantamagazine.org/mathematician-proves-huge-result-on-dangerous-problem-20191211
• 更好玩的是它给的一个网站
  • https://www.dcode.fr/collatz-conjecture