给定一个递归算法可以解决递归关系，并给出最坏情况下的时间复杂度，对吗？

Question

Find the time complexity, in worst case, in function of n = 2N , N >=0.

Find the recurrence relation and solve it.

 public static void xpto(v, n){
    if (n <= 1) 
        return;
    n=n/2;
    for(i=0;i<n;i=i+1) 
        v[i] = v[2i] + v[2i +1];
    xpto(v, n);
}

T(1) = 1

recurrence equation by substituiton:

T(n) = 1 + 1 + (n + 1) + n + T(n/2)

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

T(n/2) = 3(2) + 2n(2) + T(n/4)

T(n/4) = 3(3) + 2n(3) + T(n/8)

T(n/8) = 3(4) + 2n(4) + T(n/16)

pattern found

T(n/8) = 3(4) + 2n(4) + T(n/2^4)

general recurrence in terms of k:

T(n) = 3(k) + 2n(k) + T(n/2^k)

if T(1) = 1 and T(n/2^k) we need to change 2^k by n, this means:

2^k = n

T(n) = 3(log n) + 2n(log n) + 1

The recurrence relation is solved.

Time complexity, in worst case is O(log(n))

Questions:

• Am I doing this right?
• What function of n = 2N , N >=0 means?

Answer 1

I am not sure how you got the constants, but let's assume for simplicity, that the operation v[i] = v[2i] + v[2i +1]; is of cost 1, and everything else is free. (It can be adjusted easily without harming the concept of the following calculations).

Based on that,

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

Based on that, we can use master theorem case 1 with c=1, a=1,b=2 , and conclude T(n) is in Theta(n^1)=Theta(n)

Answer 2

Firstly, if you got: T(n) = 3(log n) + 2n(log n) + 1 as your final solution, then the worst case complexity would not be log n but rather n(log n) because of the term 2n(log n) .

From your initial recurrence relation: T(n) = 3 + 2n + T(n/2) i did the following:

Assume n = 2^k and g(k) = T(n) such that:
g(k) = g(k-1) + 2*2^k + 3 (from simply substituting n=2^k and change of function)
g(k) = sum(i=1 to k) of (2*2^i + 3)
g(k) = 2 * (sum(i=1 to k) of (2^i)) + 3k

Using geometric progression, common ratio = 2:
g(k) = 2 * (2(1-2^k) / (1-2)) + 3k
g(k) = -4 + 4*2^k + 3k

Since we initially assumed n = 2^k, this means k = log n:
T(n) = -4 + 4n + 3(log n)
Hence the worst case complexity is O(n)

For the second part of your question:

n = 2N where N >= 0 simply means n is a set of even numbers since any positive integer multiplied by 2 will be even.