如何求解 T(n) = 2 * T(n/3) + 3/2 * T(n/4) + 5 * T(n/2) + Theta(n^2), p = 2 的递归,574359
[英]How to solve the recurrence of T(n) = 2 * T(n/3) + 3/2 * T(n/4) + 5 * T(n/2) + Theta(n^2), p = 2,574359
问题描述
How to solve the recurrence of T(n) = 2 * T(n/3) + 3/2 * T(n/4) + 5 * T(n/2) + Theta(n^2), p = 2,574359 I don't understand how to solve so complex reccurrence.
如何求解 T(n) = 2 * T(n/3) + 3/2 * T(n/4) + 5 * T(n/2) + Theta(n^2), p = 2 的递归,574359 我不明白如何解决如此复杂的复发。
1 个解决方案
解决方案1
0 已采纳 2020-05-17 14:59:10
You can find an upper-bound like the following:您可以找到如下所示的上限：
T(n) <= (2 + 3/2 + 5) * T(n/2) + Theta(n^2) = 8.5 T(n/2) + Theta(n^2)
And using master theorem, you can say T(n) = O(n^{log(8.5)) ~ O(n^{3.09}) .使用主定理，您可以说T(n) = O(n^{log(8.5)) ~ O(n^{3.09}) 。
Also, you can use akra-bazzi theorem to find a tighter bound.此外，您可以使用akra-bazzi 定理找到更紧密的界限。 First, we need to solve the following equation首先，我们需要求解以下方程
2 * (1/3)^p + 3/2 (1/4)^p + 5 * (1/2)^p = 1
After solving, we know that p ~ 2.57 .求解后，我们知道p ~ 2.57 。 Hence, by the theorem we find that:因此，根据定理，我们发现：
T(n) = Theta(n^{2.57} (1 + int(x^2/x^{3.57} dx, 1, n) ) ) =
Theta(n^{2.57} (1 + int(1/x^{1.57} dx, 1, n) ) )
As int(1/x^{1.57} dx, 1, n) < 3 , we can say T(n) = Theta(n^{2.57}) .作为int(1/x^{1.57} dx, 1, n) < 3 ，我们可以说T(n) = Theta(n^{2.57}) 。
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