穷举法, 二分法 与牛顿-拉夫逊方法求解平方根的优劣,从左到右依次递优。
经过测试,穷举法基本超过 1 分钟,还没有出数据;
二分法只要区区1秒不到就出结果了。
牛顿-拉夫逊是秒出,没有任何的停顿。
numberTarget =int(input("Please enter a number:"))
numberSqureRoot = 0
while(numberSqureRoot<abs(numberTarget)):
if numberSqureRoot**2 >= abs(numberTarget):
break
numberSqureRoot = numberSqureRoot + 1
if numberSqureRoot**2 != numberTarget:
print("Your number %s is not a perfect squre, the square root is %s " % ( numberTarget,numberSqureRoot) )
else:
if numberTarget < 0 :
numberSqureRoot = -numberSqureRoot
print("Your number %s is a perfect squre, the square root is %s " % ( numberTarget, numberSqureRoot))
print("now we begin to calculate the binary search...")
numberTarget=int(input("Please enter the number for binary search..."))
numberSqureRoot = 0
lowValue = 0.0
highValue=numberTarget*1.0
epsilon = 0.01
numberSqureRoot = (highValue + lowValue)/2
while abs(numberSqureRoot**2 - numberTarget) >=epsilon:
print("lowValue:%s, highValue:%s, currentValue:%s"%(lowValue,highValue,numberSqureRoot))
if numberSqureRoot**2<numberTarget:
lowValue=numberSqureRoot
else:
highValue=numberSqureRoot
numberSqureRoot = (lowValue+highValue) /2
print("The number %s has the squre root as %s " %(numberTarget,numberSqureRoot))
print("now we begin to calculate the newTon search...")
numberTarget=int(input("Please enter the number for newTon search..."))
numberSqureRoot = 0
epsilon = 0.01
k=numberTarget
numberSqureRoot = k/2.0
while( abs(numberSqureRoot*numberSqureRoot - k)>=epsilon):
numberSqureRoot=numberSqureRoot-(((numberSqureRoot**2) - k)/(2*numberSqureRoot))
print("squre root of %s is %s " %(numberTarget,numberSqureRoot))