使用主方法求解T(n)= 2T(n / 2)+ n / log n和T(n)= 4T(n / 2)+ n / log n之间的差异 [英] Difference between solving T(n) = 2T(n/2) + n/log n and T(n) = 4T(n/2) + n/log n using Master Method
问题描述
我最近偶然发现一种资源,MM宣布无法解决2T(n / 2)+ n / log n 类型的复发。
I recently stumbled upon a resource where the 2T(n/2) + n/log n type of recurrences were declared unsolvable by MM.
直到今天,当证明另一种资源(在某种意义上是矛盾的)时,我才接受它作为引理。
I accepted it as a lemma, until today, when another resource proved to be a contradiction (in some sense).
根据资源(以下链接) :其中的Q7和Q18是记录。问题7的答案分别是1和2,因此问题7的答案说不能通过给出原因多项式差b / w f(n)和n ^(log a base b)来解决。
相反,答案18使用情况1解决了第二次重现(在这里的问题中)。
As per the resource (link below): Q7 and Q18 in it are the rec. 1 and 2 respectively in the question whereby, the answer to Q7 says it can't be solved by giving the reason 'Polynomial difference b/w f(n) and n^(log a base b)'. On the contrary, answer 18 solves the second recurrence (in the question here) using case 1.
http://www.csd.uwo.ca/~moreno/CS433-CS9624/Resources/master.pdf
有人可以消除混乱吗?
推荐答案
如果您尝试将主定理应用于
If you try to apply the master theorem to
T(n) = 2T(n/2) + n/log n
您考虑 a = 2,b = 2
这意味着 logb(a)= 1
- 您可以应用案例1吗?
0< c < logb(a)= 1
。是n / logn = O(n ^ c)
。不,因为n / logn
的增长无限快于n ^ c
- 您可以申请案例2吗?否。
c = 1
您需要找到一些 k> 0 ,以使n / log n = Theta(n log ^ kn)
- 可以应用案例3吗?
c> 1
,是n / logn = Big Omega(n ^ c)
吗?不,因为它甚至不是Big Omega(n)
- Can you apply case 1?
0 < c < logb(a) = 1
. Isn/logn = O(n^c)
. No, becausen/logn
grow infinitely faster thann^c
- Can you apply case 2? No.
c = 1
You need to find some k > 0 such thatn/log n = Theta(n log^k n )
- Can you apply case 3 ?
c > 1
, isn/logn = Big Omega(n^c)
? No because it is not evenBig Omega(n)
如果您尝试将主定理应用于
If you try to apply the master theorem to
T(n) = 4T(n/2) + n/log n
您考虑 a = 4,b = 2
,这意味着 logb(a)= 2
-
可以应用案例1吗?
c< logb(a)= 2
。是n / logn = O(n ^ 0)
或n / logn = O(n ^ 1)
。是的,确实n / logn = O(n)
。因此,我们有
Can you apply case 1?
c < logb(a) = 2
. isn/logn = O(n^0)
orn/logn = O(n^1)
. Yes indeedn/logn = O(n)
. Thus we have
T(n) = Theta(n^2)
注意:有关0的解释c< 1,案例1
note: Explanation about 0 < c <1, case 1
案例1与分析有关。
f(x) = x/log(x) , g(x) = x^c , 0< c < 1
f(x) is O(g(x)) if f(x) < M g(x) after some x0, for some M finite, so
f(x) is O(g(x)) if f(x)/g(x) < M cause we know they are positive
在这里这不是真的,我们构成 y = log x
This isnt true here We pose y = log x
f2(y) = e^y/y , g2(y) = e^cy , 0< c < 1
f2(y)/g2(y) = (e^y/y) / (e^cy) = e^(1-c)y / y , 0< c < 1
lim inf f2(y)/g2(y) = inf
lim inf f(x)/g(x) = inf
这篇关于使用主方法求解T(n)= 2T(n / 2)+ n / log n和T(n)= 4T(n / 2)+ n / log n之间的差异的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!