Solving a Recurrence Relation: T(n)=T(n-1)+T(n/2)+n

Question

Solve: T(n)=T(n-1)+T(n/2)+n.

I tried solving this using recursion trees.There are two branches T(n-1) and T(n/2) respectively. T(n-1) will go to a higher depth. So we get O(2^n). Is this idea correct?

Answer 1

This is a very strange recurrence for a CS class. This is because from one point of view: T(n) = T(n-1) + T(n/2) + n is bigger than T(n) = T(n-1) + n which is O(n^2).

But from another point of view, the functional equation has an exact solution: T(n) = -2(n + 2). You can easily see that this is the exact solution by substituting it back to the equation: -2(n + 2) = -2(n + 1) + -(n + 2) + n. I am not sure whether this is the only solution.

Here is how I got it: T(n) = T(n-1) + T(n/2) + n. Because you calculate things for very big n, than n-1 is almost the same as n. So you can rewrite it as T(n) = T(n) + T(n/2) + n which is T(n/2) + n = 0, which is equal to T(n) = - 2n, so it is linear. This was counter intuitive to me (the minus sign here), but armed with this solution, I tried T(n) = -2n + a and found the value of a.

Answer 2

I believe you are right. The recurrence relation will always split into two parts, namely T(n-1) and T(n/2). Looking at these two, it is clear that n-1 decreases in value slower than n/2, or in other words, you will have more branches from the n-1 portion of the tree. Despite this, when considering big-o, it is useful to just consider the 'worst-case' scenario, which in this case is that both sides of the tree decreases by n-1 (since this decreases more slowly and you would need to have more branches). In all, you would need to split the relation into two a total of n times, hence you are right to say O(2^n).