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So I've encountered a case where I have 2 recursive calls - rather than one. I do know how to solve for one recursive call, but in this case I'm not sure whether I'm right or wrong.

I have the following problem:

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

And I need to find the worst-case complexity for this. (FYI - It's some kind of augmented merge sort)

My feeling was to use the first equation from the Theorem, but I feel something is wrong with my idea. Any explanation on how to solve problems like this will be appreciated :)

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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

从根到叶的最长简单路径是 n-> 3/5n -> (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) )
于 2015-05-04T17:45:42.600 回答