MIT公開課：計算機科學及編程導論 Python 筆記4 函數分解抽象與遞歸

Lecture4：Decomposition and abstraction through functions；introduction to recursion 函數分解抽象與遞歸

Functions 函數

• block up into modules 分解爲模塊
• suppress detail 忽略細節
• create “new primitives” 創建原語的思考方式
  w3school Python函數

#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 關鍵字
• function_name (formal parameters) 函數名（形參）
• return 關鍵字
• 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:
本地綁定（局部變量）不會影響 全局綁定（變量）：

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

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

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Farmyard problem 農場問題：

• 20 heads， 56 legs
• numPig + numChicken = 20 豬的數量+雞的數量 = 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

農場主又養了蜘蛛：

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

改進：輸出所有的解決方案：

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.'

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recursion 遞歸

• base case – break problem into simplest possible solution把問題分解成最簡單的解決方案
• inductive step, or the recursive step 歸納、遞歸步驟： break problem into a simpler version of the same problem and some other steps

# 字符串是否是迴文 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])

# 付波納切fibonacci數列
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)

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