算法分析：选择最佳算法
[英]Algorithm analysis: Choosing the best algorithm
问题描述
I'm trying to solve a question about comparing 2 algorithms in terms of their worst case running times and finding the input size for which one algorithm would have a faster running time than the other. 
我正在尝试解决一个问题，即根据最坏情况的运行时间比较两种算法，并找到一种算法比另一种算法运行时间更快的输入大小。
The two algorithms are: 这两种算法是：
A 1 = 2n log 10 n A 1 = 2n对数10 n
A 2 = 0.1n 2 A 2 = 0.1n 2
Basically, I am trying to solve the following inequality for n: 基本上，我试图解决n的以下不等式：
2n log 10 n < 0.1n 2 2n log 10 n <0.1n 2
Can anyone please point me in the right direction? 谁能指出我正确的方向？
I have managed to get up to: 我设法起床了：
log 10 n < 0.05n ==> n < 10 0.05n log 10 n <0.05n ==> n <10 0.05n
But I have no idea what to do from here (or perhaps I have gone about the wrong way trying to solve it). 但是我不知道从这里开始该怎么做（或者我已经尝试解决错误的方法了）。
Thank you for your help in advance! 提前谢谢你的帮助！
2 个解决方案
解决方案1
4 已采纳 2013-08-06 01:05:50
Actually, you are trying to solve the inequality 实际上，您正在尝试解决不平等问题
because the 因为 algorithm is only going to be faster for a very short time, and then for any larger values of n, the
算法只会在很短的时间内变得更快，然后对于更大的n值， algorithm will be faster.
算法会更快。
Ignore the case n <= 0, multiply by 10, and divide by n to produce: 忽略n <= 0，乘以10并除以n的情况，得出：
Then divide by 20 and exponentiate both sides with a base of 10: 然后除以20，并以10为底对两边取幂：
Use a numeric solver to find the zeros of 使用数值解算器找到的零 on the interval [1, 40] since clearly 40 is an upper bound (because
在区间[1，40]上，因为显然40是一个上限（因为 ).
）。
For instance, in Matlab: 例如，在Matlab中：
>> fzero(@(x) 10^(x/20)- x, 20)
ans =
29.3531
So for any n an integer up through 29, the 因此，对于任何n直至29的整数， algorithm is faster, and for n > 29, the
算法更快，并且对于n> 29， algorithm wins.
算法胜出。
解决方案2
1 2013-08-06 01:17:01
只需使用鼠尾草绘制功能的图像：plot（0.1 * n * n-2 * n * log（n，10），n，0，50）
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