MIT公开课：计算机科学及编程导论 Python 笔记5 浮点数，逐次逼近法和二分法

Lecture5： Floating point number , successive refinement, finding roots 浮点数和二分法

>>> a = 2 ** 1000
>>> a
10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376L
>>> b = 2 ** 999
>>> b
5357543035931336604742125245300009052807024058527668037218751941851755255624680612465991894078479290637973364587765734125935726428461570217992288787349287401967283887412115492710537302531185570938977091076523237491790970633699383779582771973038531457285598238843271083830214915826312193418602834034688L
>>> a/b
2L

IEEE 754 floating point

wiki IEEE floating point

wiki IEEE 754 中文

• scientific notation
• a mantissa 尾数（0=< mantissa < 2） and an exponent 指数
  

  worry about == on float number!

>>> import math
>>> a = math.sqrt(2)
>>> a
1.4142135623730951
>>> a * a == 2
False

# 大约相等
abs(a*a - 2.0) < epsilon

>>> s = 0.0
>>> for i in range(10): s += 0.1
...
>>> s
0.9999999999999999

• might not be an exact answer 没有确切的值 例如：sqrt(2)
• can’t enumerate all guesses 无法枚举所有的可能值，因为实数是不可数的
• guess, check, and improve
• successive approximation 逐次逼近法

# 逐次逼近法(伪代码)：
guess = inital guess
for iter in range(100):
    if f(guess) close enough: return guess
    else: guess = better guess
error

逐次逼近法求平方根

- 二分法，求x的平方根，epsilon (ε) 是一个足够小的数，ctr 迭代次数

def squareRootBi(x, epsilon):
    """Assume x >= 0 and epsilon > 0
    Return y s.t. y*y is within epsilon(ε) of x"""

    assert epsilon > 0, ‘epsilon must be postive, not‘ + str(epsilon)
    low = 0
    high = max(x, 1)
    guess = (low + high) / 2.0
    ctr = 1
    while abs(guess ** 2 - x) > epsilon and ctr <= 100:
        # print ‘low:‘, low, ‘high:‘, high, ‘guess:‘, guess if guess**2 < x:
        if guess**2 < x:
            low = guess
        else:
            high = guess
        guess = (low + high) / 2.0
        ctr += 1
    assert ctr <= 100, ‘Iteration count exceeded‘
    print ‘Bi method. Num. iterations:‘, ctr, ‘Estimate:‘, guess
    return guess

def squareRootNR(x, epsilon):
    """Return y s.t. y*y is within epsilon of x"""

    assert epsilon > 0, ‘epsilon must be postive, not‘ + str(epsilon)
    x = float(x)
    guess = x / 2.0
    guess = 0.001
    diff = guess ** 2 - x
    ctr = 1
    while abs(diff) > epsilon and ctr <= 100:
        # print ‘Error:‘, diff, ‘guess:‘, guess guess = guess - diff/(2.0*guess)
        diff = guess ** 2 - x
        ctr += 1
    assert ctr <= 100, ‘Iteration count exceeded‘
    print ‘NR method. Num. iterations:‘, ctr, ‘Estimate:‘, guess
    return guess