在 Python 中,我编写了一个自定义代码 sqrt(x, delta) 以使用 delta-close 近似值计算给定数字的平方根。它使用 while 循环和类似二分搜索的算法。
编码:
from __future__ import division
def sqrt(x, delta):
start = 0
end = x
while (end-start) > delta:
middle = (start + end) / 2
middle_2 = middle * middle
if middle_2 < x:
start = middle
print "too low"
elif middle_2 > x:
end = middle
print "too high"
else:
return middle
result = (start + end) / 2
return result
它基本上可以工作并且速度非常快,但是在某些情况下它会进入无限的while循环。
例子:
sqrt(1e27, 1/1024) => works fine (returns 'too low's and 'too high's, then returns correct result)
sqrt(1e28, 1/1024) => works fine
sqrt(1e29, 1/1024) => never-ending loop, it keeps printing 'too low' forever
sqrt(1e30, 1/1024) => works fine
sqrt(1e31, 1/1024) => 'too low' forever
sqrt(1e32, 1/1024) => works fine
sqrt(1e33, 1/1024) => works fine (also surprising after the problem with 1e29 and 1e31)
sqrt(1e34, 1/1024) => works fine
sqrt(1e35, 1/1024) => 'too low' forever
sqrt(1e36, 1/1024) => works fine
sqrt(1e37, 1/1024) => 'too high' forever (too high this time!)
sqrt(1e38, 1/1024) => works fine
sqrt(1e39, 1/1024) => works fine (surprising again..)
... 1e40-1e45 ... they all work fine
sqrt(1e46, 1/1024) => 'too low' forever (surprisingly it occurs now with 1e'even number')
...
sqrt(1e200, 1/1024) => works fine
sqrt(1e201, 1/1024) => works fine
...
sqrt(1e299, 1/1024) => 'too low' forever
sqrt(1e300, 1/1024) => 'too high' forever
...
sqrt(1e304, 1/1024) => 'too high' forever
sqrt(1e305, 1/1024) => works fine
... 305-308 ... they allwork fine
sqrt(1e309, 1/1024) => inf (reached some 'infinite' limit?)
我首先认为它的数字超过了限制,比如 1e20.. 但后来它也适用于它们。此外,我认为大约是 1e'odd' 或 1e'even' 数字,但正如我们在示例中看到的那样,事实并非如此。我也尝试使用不同的增量而不是 1/1024,但它们表现出类似的行为。
我将不胜感激任何解释导致这种行为的幕后发生的事情。