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
正如已经指出的,除了null函数可能的例外,不可能有O(1/n)个函数,因为所花费的时间必须接近0。
当然,有一些算法,比如康拉德定义的算法,它们至少在某种意义上应该小于O(1)
def get_faster(list):
how_long = 1/len(list)
sleep(how_long)
If you want to investigate these algorithms, you should either define your own asymptotic measurement, or your own notion of time. For example, in the above algorithm, I could allow the use of a number of "free" operations a set amount of times. In the above algorithm, if I define t' by excluding the time for everything but the sleep, then t'=1/n, which is O(1/n). There are probably better examples, as the asymptotic behavior is trivial. In fact, I am sure that someone out there can come up with senses that give non-trivial results.
其他回答
我相信量子算法可以通过叠加“一次”进行多次计算……
我怀疑这是一个有用的答案。
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。
这是一个简单的O(1/n)算法。它甚至做了一些有趣的事情!
function foo(list input) {
int m;
double output;
m = (1/ input.size) * max_value;
output = 0;
for (int i = 0; i < m; i++)
output+= random(0,1);
return output;
}
O(1/n) is possible as it describes how the output of a function changes given increasing size of input. If we are using the function 1/n to describe the number of instructions a function executes then there is no requirement that the function take zero instructions for any input size. Rather, it is that for every input size, n above some threshold, the number of instructions required is bounded above by a positive constant multiplied by 1/n. As there is no actual number for which 1/n is 0, and the constant is positive, then there is no reason why the function would constrained to take 0 or fewer instructions.
我经常用O(1/n)来描述随着输入变大而变小的概率——例如,在log2(n)次投掷中,一枚均匀硬币背面朝上的概率是O(1/n)。
inline void O0Algorithm() {}