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
如果不管输入数据如何,答案都是一样的,那么你就有一个O(0)算法。
或者换句话说——在提交输入数据之前,答案就已经知道了 -这个功能可以优化-所以O(0)
如果根本不运行函数(NOOP)呢?或者使用固定值。这算吗?
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.
你不能低于O(1)但是O(k) k小于N是可能的。我们称之为次线性时间算法。在某些问题中,次线性时间算法只能给出特定问题的近似解。然而,有时,一个近似解就可以了,可能是因为数据集太大了,或者计算所有数据的计算成本太高了。
这里有另一种方法来证明它:为了拥有一个函数,你必须调用这个函数,并且你必须返回一个答案。这需要固定的时间。即使余下的处理过程对较大的输入花费更少的时间,打印出答案(我们可以假设是单个位)至少需要常数时间。