MIT公開課：電腦科學及編程導論 Python 筆記5 浮點數，逐次逼近法和二分法

標籤：python

Lecture5： Floating point number , successive refinement, finding roots 浮點數和二分法

3wschool 數字

>>> a = 2 ** 1000>>> a10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376L>>> b = 2 ** 999>>> b5357543035931336604742125245300009052807024058527668037218751941851755255624680612465991894078479290637973364587765734125935726428461570217992288787349287401967283887412115492710537302531185570938977091076523237491790970633699383779582771973038531457285598238843271083830214915826312193418602834034688L>>> a/b2L

IEEE 754 floating point
wiki IEEE floating point
wiki IEEE 754 中文

• scientific notation

• a mantissa 尾數（0=< mantissa < 2） and an exponent 指數

  worry about == on float number!

>>> import math>>> a = math.sqrt(2)>>> a1.4142135623730951>>> a * a == 2False

# 大約相等abs(a*a - 2.0) < epsilon

>>> s = 0.0>>> for i in range(10): s += 0.1... >>> s0.9999999999999999

Why might this be a challenge? What are some of the issues?

• might not be an exact answer 沒有確切的值 例如：sqrt(2)
• can’t enumerate all guesses 無法枚舉所有的可能值，因為實數是不可數的
• guess, check, and improve
• successive approximation 逐次逼近法

# 逐次逼近法(虛擬碼)：guess = inital guessfor iter in range(100):    if f(guess) close enough: return guess    else: guess = better guesserror

逐次逼近法求平方根
- 二分法，求x的平方根，epsilon (ε) 是一個足夠小的數，ctr 迭代次數

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

MIT公開課：電腦科學及編程導論 Python 筆記5 浮點數，逐次逼近法和二分法