以√n为参数求解递归关系

考虑复发

$\qquad\displaystyle T(n) = \sqrt{n} \cdot T\bigl(\sqrt{n}\bigr) + c\,n$

对于$n \gt 2$且具有一些正常数，并且。T （2 ）= $c$$T(2) = 1$

我知道用于解决递归的Master定理，但是我不确定如何使用它来解决这种关系。您如何计算平方根参数？

我们将使用Raphael的建议并展开复发。在下面，所有对数均以2为底。

其中，β（n）是必须从平方根开始以n开始并达到2的次数。结果证明，β（n）=loglogn。你怎么看？考虑：

$$\begin{align*} T(n) &= n^{1/2} T(n^{1/2}) + cn \\ &= n^{3/4} T(n^{1/4}) + n^{1/2} c n^{1/2} + cn\\ &= n^{7/8} T(n^{1/8}) + n^{3/4} c n^{1/4} + 2cn\\ &= n^{15/16} T(n^{1/16}) + n^{7/8} c n^{1/8} + 3cn \\ & \ldots \\ &= \frac{n}{2} T(2) + c n \beta(n) \end{align*}.$$ $\beta(n)$$\beta(n) = \log \log n$ 因此，需要平方根才能达到2的次数是1的解。 $$\begin{align*} n &= 2^{\log n}\\ n^{1/2} &= 2^{\frac{1}{2} \log n} \\ n^{1/4} &= 2^{\frac{1}{4} \log n} \\ \ldots \end{align*}$$ ，这是记录日志Ñ。因此，递归的解决方案是cnloglogn+$\frac{1}{2^t} \log n \approx 1$$\log \log n$。为了使此操作绝对严格，我们应该使用替换方法，并且要小心处理。有时间时，我将尝试将此计算结果添加到我的答案中。$c n \log \log n + \frac{1}{2}n$

In your comment you mentioned that you tried substitution but got stuck. Here's a derivation that works. The motivation is that we'd like to get rid of the $\sqrt{n}$ multiplier on the right hand side, leaving us with something that looks like $U(n) = U(\sqrt{n}) + something$. In this case, things work out very nicely:

$$\begin{align} T(n) &= \sqrt{n}\ T(\sqrt{n}) + n & \text{so, dividing by $n$ we get}\\ \frac{T(n)}{n} &= \frac{T(\sqrt{n})}{\sqrt{n}} + 1 &\text{and letting $n = 2^m$ we have}\\ \frac{T(2^m)}{2^m} &= \frac{T(2^{m/2})}{2^{m/2}} + 1 \end{align}$$ Now let's simplify things even further, by changing to logs (since $\lg \sqrt{n} = (1/2)\lg{n}$). Let $$\begin{align} S(m) &= \frac{T(2^m)}{2^m} & \text{so our original recurrence becomes}\\ S(m) &= S(m/2)+1 \end{align}$$ Aha! This is a well-known recurrence with solution $$S(m)=\Theta(\lg m)$$ Returning to $T(\,)$, we then have, with $n=2^m$ (and so $m=\lg n$), $$\frac{T(n)}{n} = \Theta(\lg\,\lg n)$$ So $T(n) =\Theta(n\,\lg\,\lg n)$.

If you write $m=\log n \space$ you have $T(m) = {m \over 2}\cdot T({m\over 2}) + c\cdot 2^m\space$.

Now you know the recursion tree has hight of order $O(\log m)$, and again it's not hard to see it's $O(2^m)\space$ in each level, so total running time is in: $O((\log m) \cdot 2^m)\space$, which concludes $O(n \cdot \log \log n)\space$ for $n$.

In all when you see $\sqrt n$ or $n^{a \over b}, a<b \space$, is good to check logarithm.

_{P.S: Sure proof should include more details by I skipped them.}

Let's follow Raphael's suggestion, for $n = 2^{2^k}$:

$$\begin{align*} T(n) = T(2^{2^k}) &= 2^{2^{k-1}} T(2^{2^{k-1}}) + c2^{2^k} \\ &= 2^{2^{k-1}+2^{k-2}} T(2^{2^{k-2}}) + c(2^{2^k} + 2^{2^k}) \\ &= \cdots \\ &= 2^{2^{k-1}+2^{k-2}+\cdots+2^0} T(2^{2^0}) + c(2^{2^k} + 2^{2^k} + \cdots + 2^{2^k}) \\ &= 2^{2^k-1} + ck2^{2^k} \\ &= (c\log\log n + 1/2)n. \end{align*}$$

Edit: Thanks Peter Shor for the correction!

Unravel the recurrence once as follows:

$$\begin{align} T(n) &=& \sqrt{n}~ T(\sqrt{n}) + n \\ &=& n^{1/2} \big ( n^{1/4} ~ T(n^{1/4}) + n^{1/2}\big) + n \\ &=& n^{1-1/4}~ T(n^{1/4}) + 2n. \end{align}$$

Continuing the unraveling for $k$ steps, we have that:

$$\begin{align} \label{eqn:unrav} T(n) &=& n^{1-1/2^k} T(n^{1/2^k}) + kn. \end{align}$$

These steps will continue until the base case of $n^{1/2^k} = 2$. Solving for $k$ we have:

$$\begin{align} &&n^{1/2^k} = 2 \\ &\implies& \log n= 2^k \\ &\implies& k= \log \log n. \label{eqn:ksol} \end{align}$$

Substituting $k=\log \log n$ into the unraveled recurrence, we have

$$\begin{align} T(n) = \frac{n}{2} T(2) + n \log \log n. \end{align}$$