[英]Find solution to recurrence: T(N) = 2 T(N/4 + √N) + (√10) N
While solving a complex recurrence equation like this T(N) = 2 T(N/4 + √N) + (√10) N ;T(1) = 1
在求解像这样的复杂递推方程时T(N) = 2 T(N/4 + √N) + (√10) N ;T(1) = 1
I tried to make some change of variables to make it easy and solve it by master theorem but i failed ,so i take the dominant one so it will be: T(N) = 2 T(N/4) + (√10) N
so it is T(N)=Θ(N)
.我试图对变量进行一些更改以使其更容易并通过主定理解决它但我失败了,所以我选择了占主导地位的变量,因此它将是: T(N) = 2 T(N/4) + (√10) N
所以它是T(N)=Θ(N)
。 Is that true or not ?这是真的还是假的?
Trying to unroll recursion or to make a substitution left me nowhere.试图展开递归或进行替换让我无处可去。 So the only thing I was able to do is to see that所以我唯一能做的就是看到 for any sufficiently large n
(above 64).对于任何足够大的n
(大于 64)。 You can select any number (not just 8), bigger than 4.您可以选择大于 4 的任何数字(不仅仅是 8)。
So you end up with所以你最终得到
Solving this with master's theorem you see that it falls in the first case with用大师定理解决这个问题,你会发现它属于第一种情况 . .
Therefore the solution is Θ(N)
which is the same as you wondered.因此,解决方案是Θ(N)
,这与您想知道的相同。
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