MIT public class: Introduction to Computer science and programming Python Note 5 floating-point numbers, successive approximation and dichotomy

Lecture5:floating point number, successive refinement, finding roots float and dichotomy

3wschool Digital

>>>A =2** +>>>A107150860718626732094842504906000181056140481170553360744375038837035105112493612249319837881569585812759467291755314 6825187145285692314043598457757469857480393456777482423098542107460506237114187795418215304647498358194126739876755916554 3946077062914571196477686542167660429831652624386837205668069376L>>>b =2**999>>>B535754303593133660474212524530000905280702405852766803721875194185175525562468061246599189407847929063797336458776573 4125935726428461570217992288787349287401967283887412115492710537302531185570938977091076523237491790970633699383779582771 973038531457285598238843271083830214915826312193418602834034688L>>>A/b2L

IEEE 754 Floating Point
Wiki IEEE floating point
Wiki IEEE 754 Chinese

• Scientific notation

• A mantissa mantissa (0=< mantissa < 2) and an exponent index

  worry about = = on float number!

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

# 大约相等2.0) < epsilon

>>> 0.0>>> forin range(100.1... >>> s0.9999999999999999

Why might this is a challenge? What is some of the issues?

• Might not being an exact answer has no exact value for example: sqrt (2)
• Can ' t enumerate all guesses cannot enumerate all possible values, because real numbers are not counted
• Guess, check, and improve
• Successive approximation method of successive approximation

# 逐次逼近法(伪代码)：guess = inital guessforin range(100):    ifreturn guess    else: guess = better guesserror

Finding square root by successive approximation method
-Two method, the square root of X, epsilon (ε) is a small enough number, Ctr iteration number

 def squarerootbi(x, epsilon):    "" " assume x >= 0 and epsilon > 0 Return y s.t. Y*y is within epsilon (ε) of X" ""    assertEpsilon >0,' Epsilon must is postive, not '+ STR (epsilon) low =0High = max (x,1) guess = (low + high)/2.0CTR =1     whileABS (GUESS * *2-X) > Epsilon andCtr <= -:# print ' Low: ', Low, ' High: ', high, ' guess: ', guess if guess**2 < x:        ifguess**2< X:low = GuessElse: High = guess guess = (low + high)/2.0CTR + =1    assertCtr <= -,' iteration count exceeded '    Print ' Bi method. Num. Iterations: ', CTR,' Estimate: ', GuessreturnGuess

 def squarerootnr(x, epsilon):    "" " Return y s.t. Y*y is within epsilon of X" "    assertEpsilon >0,' Epsilon must is postive, not '+ str (epsilon) x = float (x) guess = x/2.0Guess =0.001diff = Guess * *2-X ctr =1     whileABS (diff) > Epsilon andCtr <= -:# print ' Error: ', diff, ' Guess: ', guess guess = guess-diff/(2.0*guess)diff = Guess * *2-X ctr + =1    assertCtr <= -,' iteration count exceeded '    Print ' NR method. Num. Iterations: ', CTR,' Estimate: ', GuessreturnGuess

MIT public class: Introduction to Computer science and programming Python Note 5 floating-point numbers, successive approximation and dichotomy