有没有什么情况下你更喜欢O(log n)时间复杂度而不是O(1)时间复杂度?还是O(n)到O(log n)

你能举个例子吗?


当前回答

我很惊讶没有人提到内存绑定应用程序。

可能存在一种算法具有较少的浮点运算,这要么是因为它的复杂性(即O(1) < O(log n)),要么是因为复杂度前面的常数更小(即2n2 < 6n2)。无论如何,如果较低的FLOP算法的内存限制更大,您可能仍然更喜欢具有更多FLOP的算法。

我所说的“内存受限”是指您经常访问的数据经常超出缓存。为了获取这些数据,在对其执行操作之前,必须将内存从实际内存空间拉到缓存中。这个抓取步骤通常非常慢——比您的操作本身慢得多。

因此,如果你的算法需要更多的操作(但这些操作是在已经在缓存中的数据上执行的[因此不需要读取]),它仍然会在实际的walltime方面以更少的操作(必须在缓存外的数据上执行[因此需要读取])胜过你的算法。

其他回答

我很惊讶没有人提到内存绑定应用程序。

可能存在一种算法具有较少的浮点运算,这要么是因为它的复杂性(即O(1) < O(log n)),要么是因为复杂度前面的常数更小(即2n2 < 6n2)。无论如何,如果较低的FLOP算法的内存限制更大,您可能仍然更喜欢具有更多FLOP的算法。

我所说的“内存受限”是指您经常访问的数据经常超出缓存。为了获取这些数据,在对其执行操作之前,必须将内存从实际内存空间拉到缓存中。这个抓取步骤通常非常慢——比您的操作本身慢得多。

因此,如果你的算法需要更多的操作(但这些操作是在已经在缓存中的数据上执行的[因此不需要读取]),它仍然会在实际的walltime方面以更少的操作(必须在缓存外的数据上执行[因此需要读取])胜过你的算法。

在重新设计程序时,发现一个过程用O(1)而不是O(lgN)进行了优化,但如果不是这个程序的瓶颈,就很难理解O(1) alg。这样就不用用O(1)算法了 当O(1)需要大量的内存而你无法提供时,而O(lgN)的时间可以接受。

并行执行算法的可能性。

我不知道是否有O(log n)和O(1)类的例子,但对于某些问题,当算法更容易并行执行时,您会选择具有更高复杂度类的算法。

有些算法不能并行化,但复杂度很低。考虑另一种算法,它可以达到相同的结果,并且可以很容易地并行化,但具有更高的复杂度类。当在一台机器上执行时,第二种算法速度较慢,但当在多台机器上执行时,实际执行时间越来越短,而第一种算法无法加快速度。

选择大O复杂度高的算法而不是大O复杂度低的算法的原因有很多:

most of the time, lower big-O complexity is harder to achieve and requires skilled implementation, a lot of knowledge and a lot of testing. big-O hides the details about a constant: algorithm that performs in 10^5 is better from big-O point of view than 1/10^5 * log(n) (O(1) vs O(log(n)), but for most reasonable n the first one will perform better. For example the best complexity for matrix multiplication is O(n^2.373) but the constant is so high that no (to my knowledge) computational libraries use it. big-O makes sense when you calculate over something big. If you need to sort array of three numbers, it matters really little whether you use O(n*log(n)) or O(n^2) algorithm. sometimes the advantage of the lowercase time complexity can be really negligible. For example there is a data structure tango tree which gives a O(log log N) time complexity to find an item, but there is also a binary tree which finds the same in O(log n). Even for huge numbers of n = 10^20 the difference is negligible. time complexity is not everything. Imagine an algorithm that runs in O(n^2) and requires O(n^2) memory. It might be preferable over O(n^3) time and O(1) space when the n is not really big. The problem is that you can wait for a long time, but highly doubt you can find a RAM big enough to use it with your algorithm parallelization is a good feature in our distributed world. There are algorithms that are easily parallelizable, and there are some that do not parallelize at all. Sometimes it makes sense to run an algorithm on 1000 commodity machines with a higher complexity than using one machine with a slightly better complexity. in some places (security) a complexity can be a requirement. No one wants to have a hash algorithm that can hash blazingly fast (because then other people can bruteforce you way faster) although this is not related to switch of complexity, but some of the security functions should be written in a manner to prevent timing attack. They mostly stay in the same complexity class, but are modified in a way that it always takes worse case to do something. One example is comparing that strings are equal. In most applications it makes sense to break fast if the first bytes are different, but in security you will still wait for the very end to tell the bad news. somebody patented the lower-complexity algorithm and it is more economical for a company to use higher complexity than to pay money. some algorithms adapt well to particular situations. Insertion sort, for example, has an average time-complexity of O(n^2), worse than quicksort or mergesort, but as an online algorithm it can efficiently sort a list of values as they are received (as user input) where most other algorithms can only efficiently operate on a complete list of values.

考虑一个红黑树。它具有O(log n)的访问、搜索、插入和删除操作。与数组相比,数组的访问权限为O(1),其余操作为O(n)。

因此,对于一个插入、删除或搜索比访问更频繁的应用程序,并且只能在这两种结构之间进行选择,我们更喜欢红黑树。在这种情况下,你可能会说我们更喜欢红黑树更麻烦的O(log n)访问时间。

为什么?因为权限不是我们最关心的。我们正在权衡:应用程序的性能更大程度上受到其他因素的影响。我们允许这种特定的算法受到性能影响,因为我们通过优化其他算法获得了很大的收益。

So the answer to your question is simply this: when the algorithm's growth rate isn't what we want to optimize, when we want to optimize something else. All of the other answers are special cases of this. Sometimes we optimize the run time of other operations. Sometimes we optimize for memory. Sometimes we optimize for security. Sometimes we optimize maintainability. Sometimes we optimize for development time. Even the overriding constant being low enough to matter is optimizing for run time when you know the growth rate of the algorithm isn't the greatest impact on run time. (If your data set was outside this range, you would optimize for the growth rate of the algorithm because it would eventually dominate the constant.) Everything has a cost, and in many cases, we trade the cost of a higher growth rate for the algorithm to optimize something else.