我如何找到这种递归的渐近紧束缚？ T(n) = T(n-2) + n log (n/2)

Question

The T(n-2) part confuses me a bit because if I find T(n-2) and substitute it into T(n) = T(n-2) + n log(n/2), for this part, does k = 2 or 3?

Also, what happens with T(n-1)?

Do I just not acknowledge it or how am I supposed to apply it to solve this recurrence?

How would I approach this recurrence and solve it? (I approached it using the substitution method, but am not sure how to solve it)

T(n) = T(n-2) + nlog(n/2)

Answer 1

The solution to the recurrence relation consists of two indepenedent parts, the even entries and the odd entries. You need two initial values, one for T_0 and one for T_1.

Consider n even, then T(n) = T(2k) for k=n/2 . Let F(k) = T(2k) . our equation becomes

F(k) = F(k-1) + 2 k log(k)
     = 2 k log(k) + 2 (k-1) log(k-1) + .. 2 1 log(1) + T_0
     = 2 sum i log(i) + T_0

Now we know that within the sum log(i) < log(k) , so we get

F(k) < 2 log(k) sum i + T_0
     = 2 log(k) k * (k+1) / 2 + T0
     = log(k) k * (k+1) + T_0
     = O(k^2 log(k))

so, for even n:

T(n) = O(n^2 log(n))

Proving the same holds for odd n is similar.