python函数写二分法 python用二分法求方程的解

用python二分法求平方根,这个程序为什么错了?首先二分法肯定需要一个“不断”二分python函数写二分法的过程python函数写二分法 ,  你的代码里面连一个循环都没有,肯定是不对的吧python函数写二分法?
其次按照你的代码的思路,如果当前估算值guess的平法比x大,那就往0那边靠,否则就往1那边靠,这个好像也不对吧?
二分法的实现方法应该是,在区间[left, right]里面找x的开方,令估算值为guess等于区间的中点,如果guess比实际的大,那就把区间缩小一半,令到右端点移动到中点,如果guess比实际的?。彩墙渌跣∫话? ,但是是令左端点移动到中点 。这样每次缩小一半的区间 , 直到区间的长度非常非常?。?那就认为区间的两个端点是相等的了,这个时候就得到了答案 。
import math
def main():
x = input('x=')
n = 0
if x = 1 and x = 0:
left = 0.
right = 1.
while right - left = 0.0000001:
guess = (left + right) / 2.
if guess ** 2 - x = 0.0000001:
right = guess
else:
left = guess
#return guess
print 'sqrt(x) is', left
else:
print 'x should be in [0,1]'
if __name__ == '__main__':
main()
我按照你的思路又写了另外一种方法:
def second():
x = input('x=')
n = 0
if x = 1 and x = 0:
【python函数写二分法 python用二分法求方程的解】movelen = (1 + 0) / 4.
guess = (1 + 0) / 2.
while abs(guess ** 2 - x) = 0.0000001:
if (guess ** 2 - x) = 0.0000001:
guess = guess - movelen
else:
guess = guess + movelen
movelen = movelen / 2.
print 'sqrt(x) is', guess
else:
print 'x should be in [0,1]'
谁会用python写个二分法查找的循环#!/usr/bin/env python
import sys
def search2(a,m):
low = 0
high = len(a) - 1
while(low = high):
mid = (low + high)/2
midval = a[mid]
if midvalm:
low = mid + 1
elif midvalm:
high = mid - 1
else:
print mid
return mid
print -1
return -1
if __name__ == "__main__":
a = [int(i) for i in list(sys.argv[1])]
m = int(sys.argv[2])
search2(a,m)
python 二分法解方程def erfen(low,high):
while lowhigh:
mid=(low+high)/2
if f(low)*f(mid)0:
high=mid
elif f(mid)*f(high )0:
low=mid
return mid
这个函数没有结束 检查一下如何设置退出条件
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