有没有O(1/n)种算法?
或者其他小于O(1)的数?
有没有O(1/n)种算法?
或者其他小于O(1)的数?
当前回答
It may be possible to construct an algorithm that is O(1/n). One example would be a loop that iterates some multiple of f(n)-n times where f(n) is some function whose value is guaranteed to be greater than n and the limit of f(n)-n as n approaches infinity is zero. The calculation of f(n) would also need to be constant for all n. I do not know off hand what f(n) would look like or what application such an algorithm would have, in my opinion however such a function could exist but the resulting algorithm would have no purpose other than to prove the possibility of an algorithm with O(1/n).
其他回答
我猜小于O(1)是不可能的。算法所花费的任何时间都称为O(1)。但是对于O(1/n)下面的函数呢。(我知道这个解决方案中已经出现了许多变体,但我猜它们都有一些缺陷(不是主要的,它们很好地解释了这个概念)。这里有一个,只是为了方便讨论:
def 1_by_n(n, C = 10): #n could be float. C could be any positive number
if n <= 0.0: #If input is actually 0, infinite loop.
while True:
sleep(1) #or pass
return #This line is not needed and is unreachable
delta = 0.0001
itr = delta
while delta < C/n:
itr += delta
因此,随着n的增加,函数将花费越来越少的时间。此外,如果输入实际为0,则函数将永远返回。
有人可能会说,这将受到机器精度的限制。因此,由于c eit有一个上界,它是O(1)。但我们也可以绕过它,通过在字符串中输入n和C。加法和比较是对字符串进行的。用这个方法,我们可以把n减小到任意小。因此,即使忽略n = 0,函数的上限也是无界的。
我也相信我们不能说运行时间是O(1/n)。我们应该写成O(1 + 1/n)
It may be possible to construct an algorithm that is O(1/n). One example would be a loop that iterates some multiple of f(n)-n times where f(n) is some function whose value is guaranteed to be greater than n and the limit of f(n)-n as n approaches infinity is zero. The calculation of f(n) would also need to be constant for all n. I do not know off hand what f(n) would look like or what application such an algorithm would have, in my opinion however such a function could exist but the resulting algorithm would have no purpose other than to prove the possibility of an algorithm with O(1/n).
O(1)仅仅表示“常数时间”。
当你给循环[1]添加一个早期退出时,你(在大O符号中)把一个O(1)算法变成了O(n)算法,但使它更快。
诀窍是一般情况下,常数时间算法是最好的,线性算法比指数算法好,但对于n很小的时候,指数算法可能更快。
1:假设这个例子的列表长度是静态的
在数值分析中,近似算法在近似公差范围内应具有次常数的渐近复杂度。
class Function
{
public double[] ApproximateSolution(double tolerance)
{
// if this isn't sub-constant on the parameter, it's rather useless
}
}
那么这个呢:
void FindRandomInList(list l)
{
while(1)
{
int rand = Random.next();
if (l.contains(rand))
return;
}
}
随着列表大小的增加,程序的预期运行时间会减少。