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主定理 f(n) = cn^k

[英]Master theorem where f(n) = cn^k

原文 2021-01-27 17:36:25 5 1 master-theorem

I'm wondering about a specific case for the master theorem that arises when f(n) = xn ^k and n^(log _b a) = n ^k where x is an integer greater than 1. For this case f(n) is larger than n^(log _b a) however it is not polynomially larger than it so case 3 can not arise.我想知道当 f(n) = xn ^k和 n^(log _b a) = n ^k时出现的主定理的特定情况，其中 x 是大于 1 的 integer。对于这种情况，f(n) 是大于 n^(log _b a) 但它不是多项式大于它，因此不会出现情况 3。

For a case like this I would assume you use case 2 as the big-O of them both are the same but that doesn't seem to fit with the equation that I can find.对于这样的案例，我假设您使用案例 2，因为它们的大 O 是相同的，但这似乎不符合我能找到的等式。 It seems possible that I'm making a mistake in taking f(n) directly out of the original recursive relation rather than it's big-O as that seems to make sense to me yet I can't find any clarification on this or any examples where the space of f(n) in the equation is not already it's own big-O.似乎我在将 f(n) 直接从原始递归关系中取出而不是大 O 时犯了一个错误，因为这对我来说似乎很有意义，但我找不到任何关于这个或任何例子的澄清其中方程中 f(n) 的空间还不是它自己的大 O。

Edit: When I say "the equation that I can find" what I mean is that assumption doesn't fit with the master theorem as I can work it out.编辑：当我说“我能找到的方程”时，我的意思是假设不符合主定理，因为我可以解决它。 As I have it the master theorem for case 2 which I am talking about looks like f(n) = Θ(n^(log _b a)) .因为我有它，我正在谈论的案例 2 的主定理看起来像f(n) = Θ(n^(log _b a)) 。 I think the important bit really is whether out of an equation ending in + xn ^k I pull out f(n) = xn ^k or f(n) = n ^k .我认为重要的一点是，是否从以 + xn ^k结尾的等式中提取 f(n) = xn ^k或 f(n) = n ^k 。 Apologies for the poor wording.为糟糕的措辞道歉。

1 个解决方案

I think the important bit really is whether out of an equation ending in + xnk I pull out f(n) = xnk or f(n) = nk.我认为真正重要的是，我是否从以 + xnk 结尾的等式中提取 f(n) = xnk 或 f(n) = nk。

Normally you should take f(n) = x * n ^k .通常你应该采取 f(n) = x * n ^k 。 Because master theorem defines T(n) to be in the form aT(n/b) + f(n) .因为主定理将 T(n) 定义为 aT(n/b) + f(n)的形式。 But in your example, it doesn't really matter.但在你的例子中，这并不重要。

Growth of f(n) and x * f(n) are the same, if x is a positive constant.如果 x 是一个正常数，f(n) 和 x * f(n) 的增长是相同的。 In the case of f(n) = xn ^k , They are both Θ(n ^k ).在 f(n) = xn ^k的情况下，它们都是 Θ(n ^k )。 (Or you could say they are both Θ(x * n ^k ). This is the same set as Θ(n ^k ).) （或者你可以说它们都是 Θ(x * n ^k )。这与 Θ(n ^k ) 的集合相同。）

Since f(n) = Θ(n ^{log _b a} ), case 2 of master theorem should be used here.由于 f(n) = Θ(n ^{log _b a} )，这里应该使用主定理的情况 2。 The theorem says T(n) = Θ(n ^{log _b a} * lgn) in this case.在这种情况下，定理说 T(n) = Θ(n ^{log _b a} * lgn)。

Again, it doesn't matter here if you write Θ(n ^{log _b a} * lgn) or Θ(5 * n ^{log _b a} * lgn) or Θ(x * n ^{log _b a} * lgn) .同样，如果你写Θ(n ^{log _b a} * lgn)或Θ(5 * n ^{log _b a} * lgn)或Θ(x * n ^{log _b a} * lgn)在这里也没关系。 Multiplying a function with a positive constant doesn't change its asymptotic bounds.将 function 与正常数相乘不会改变其渐近界。 Master theorem gives you just the asymptotical bounds of the function, not its exact value.主定理只给你 function 的渐近界限，而不是它的确切值。