MIT公开课：计算机科学及编程导论 Python 笔记4 函数分解抽象与递归

Lecture4：Decomposition and abstraction through functions；introduction to recursion 函数分解抽象与递归

Functions 函数

• block up into modules 分解为模块
• suppress detail 忽略细节
• create “new primitives” 创建原语的思考方式
  w3school Python函数

#example code for finding square roots before
x = 16
ans = 0
if x >= 0:
    while ans*ans < x:
        ans = ans + 1
        print ‘ans =‘, ans
    if ans*ans != x:
        print x, ‘is not a perfect square‘
    else: print ans
else:  print x, ‘is a negative number‘

• def 关键字
• function_name (formal parameters) 函数名（形参）
• return 关键字
• None -special value

#example code for finding square roots

def sqrt(x):
  """Returns the square root of x, if x is a perfect square.
       Prints an error message and returns None otherwise"""
  ans = 0
  if x >= 0:
      while ans*ans < x: ans = ans + 1
      if ans*ans != x:
          print x, ‘is not a perfect square‘
          return None
      else: return ans
  else:
        print x, ‘is a negative number‘
        return None

local binding do not affect global binding:
本地绑定（局部变量）不会影响 全局绑定（变量）：

def f(x): x=x+1
return x

>>>x=3
>>>z = f(x) 
>>>print x
3 
>>>print z
4

Farmyard problem 农场问题：

• 20 heads， 56 legs
• numPig + numChicken = 20 猪的数量+鸡的数量 = 20
• 4 *numPig + 2*numChicken = 56

def solve(numLegs, numHeads):
    for numChicks in range(0, numHeads + 1):
        numPigs = numHeads - numChicks
        totLegs = 4*numPigs + 2*numChicks
        if totLegs == numLegs:
            return (numPigs, numChicks)
    return (None, None)

def barnYard():
    heads = int(raw_input(‘Enter number of heads: ‘))
    legs = int(raw_input(‘Enter number of legs: ‘))
    pigs, chickens = solve(legs, heads)
    if pigs == None:
        print ‘There is no solution‘
    else:
        print ‘Number of pigs:‘, pigs
        print ‘Number of chickens:‘, chickens

农场主又养了蜘蛛：

def solve1(numLegs, numHeads):
    for numSpiders in range(0, numHeads + 1):
        for numChicks in range(0, numHeads - numSpiders + 1):
            numPigs = numHeads - numChicks - numSpiders
            totLegs = 4*numPigs + 2*numChicks + 8*numSpiders
            if totLegs == numLegs:
                return (numPigs, numChicks, numSpiders)
    return (None, None, None)

def barnYard1():
    heads = int(raw_input(‘Enter number of heads: ‘))
    legs = int(raw_input(‘Enter number of legs: ‘))
    pigs, chickens, spiders = solve1(legs, heads)
    if pigs == None:
        print ‘There is no solution‘
    else:
        print ‘Number of pigs:‘, pigs
        print ‘Number of chickens:‘, chickens
        print ‘Number of spiders:‘, spiders

改进：输出所有的解决方案：

def solve2(numLegs, numHeads):
    solutionFound = False
    for numSpiders in range(0, numHeads + 1):
        for numChicks in range(0, numHeads - numSpiders + 1):
            numPigs = numHeads - numChicks - numSpiders
            totLegs = 4*numPigs + 2*numChicks + 8*numSpiders
            if totLegs == numLegs:
                print ‘Number of pigs: ‘ + str(numPigs) + ‘,‘,
                print ‘Number of chickens: ‘+str(numChicks)+‘,‘,
                print ‘Number of spiders:‘, numSpiders
                solutionFound = True
    if not solutionFound: print ‘There is no solution.‘

recursion 递归

• base case – break problem into simplest possible solution把问题分解成最简单的解决方案
• inductive step, or the recursive step 归纳、递归步骤： break problem into a simpler version of the same problem and some other steps

# 字符串是否是回文 eg. "abcba"
def isPalindrome(s):
    """Returns True if s is a palindrome and False otherwise"""
    if len(s) <= 1: return True
    else: return s[0] == s[-1] and isPalindrome(s[1:-1])

# 付波纳切fibonacci数列
def fib(x):
    """Return fibonacci of x, where x is a non-negative int"""
    if x == 0 or x == 1: return 1
    else: return fib(x-1) + fib(x-2)