简体繁体 English

朗道符号/大 O 符号

[英]Landau Notation/Big O notation

原文 2022-07-29 20:24:50 6 1 algorithm/ big-o

In our class the following exercise/example was given:在我们的 class 中，给出了以下练习/示例：

Compute n_0 and c from the formal definition of each Landau symbol to show that:根据每个朗道符号的正式定义计算 n_0 和 c 以表明：

2^100n belongs O(n^2). 2^100n 属于 O(n^2)。

Then in the Solution the following was done:然后在解决方案中完成了以下操作：

n_0=2^100 and c=1. n_0=2^100 和 c=1。

Show for each n>n_0: 2^100*n=<n^2.为每个 n>n_0 显示：2^100*n=<n^2。

It is true that: n_0^2=2^100 n_0 and for all n>2^100: n^2-2^100n>n^2 -n n=n^2 - n^2=0.确实：n_0^2=2^100 n_0 并且对于所有 n>2^100：n^2-2^100n>n^2 -n n=n^2 - n^2=0。

I have some questions:我有一些问题：

We are looking for n_0 and c, but somehow we give values to them?我们正在寻找 n_0 和 c，但是我们以某种方式给它们赋值？ And why those values in particular?为什么特别是这些价值观？ Why can't n_0=2?为什么不能 n_0=2？ and c=34? c=34？ Is there a logic behind all of this?这一切背后有逻辑吗？
In the last part, I don't see how that expression proves anything, it looks redundant在最后一部分，我看不出这个表达式是如何证明什么的，它看起来是多余的

1 个解决方案

If you read the definition of big-O notation, it is precisely so that if you can find one such n ₀ and c that the inequality holds for all numbers greater than n ₀ , then the big-O relation holds.如果您阅读大 O 表示法的定义，正是因此，如果您能找到一个这样的n ₀和c不等式适用于所有大于n ₀的数字，那么大 O 关系成立。

Of course you can choose another n ₀ , for example a bigger one.当然，您可以选择另一个n ₀ ，例如更大的一个。 As long as you find one, you have proven the relation.只要你找到一个，你就证明了这种关系。

Although for better help with such questions, I recommend Math.stachexchange or CS.stakhexchange .尽管为了更好地解决此类问题，我推荐Math.stachexchange或CS.stakhexchange 。