Question

所以我遇到了一个我有2个递归调用的情况 - 而不是一个。我知道如何解决一个递归调用，但在这种情况下我不确定我是对还是错。

我有以下问题：

T（n）= T（2n / 5）+ T（3n / 5）+ n

我需要找到最坏情况的复杂性。（仅供参考 - 这是一种增强合并排序）

我的感觉是使用定理中的第一个等式，但我觉得我的想法有问题。任何解释如何解决这样的问题将不胜感激：）

Answer 1

给定递归的递归树将如下所示：

                                Size                        Cost

                                 n                           n
                               /   \
                             2n/5   3n/5                     n
                           /   \     /    \
                       4n/25  6n/25  6n/25  9n/25            n

                         and so on till size of input becomes 1

从根到叶子的简单路径将是n-> 3 / 5n - ＆gt; （3/5）^ 2 n ..直到1

 Therefore  let us assume the height of tree = k

            ((3/5) ^ k )*n = 1 meaning  k = log to the base 5/3 of n

 In worst case we expect that every level gives a cost of  n and hence 

        Total Cost = n * (log to the base 5/3 of n)

 However we must keep one thing in mind that ,our tree is not complete and therefore

 some levels near the bottom would be partially complete.

 But in asymptotic analysis we ignore such intricate details.

 Hence in worst Case Cost = n * (log to the base 5/3 of n)

          which  is O( n * log n )

现在，让我们使用替换方法验证这一点：

 T(n) =  O( n * log n)  iff T(n) < = dnlog(n) for some d>0

 Assuming this to be true:

 T(n) = T(2n/5) + T(3n/5) + n

      <= d(2n/5)log(2n/5) + d(3n/5)log(3n/5) + n

       = d*2n/5(log n  - log 5/2 )  + d*3n/5(log n  - log 5/3) + n

       = dnlog n  - d(2n/5)log 5/2 - d(3n/5)log 5/3  + n

       = dnlog n  - dn( 2/5(log 5/2)  -  3/5(log 5/3)) + n

       <= dnlog n

       as long as d >=  1/( 2/5(log 5/2)  -  3/5(log 5/3) )

运行时复杂性|使用Master's定理进行递归计算

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