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In order to calculate the maximum subsequence of a set this algorithm first calculates three values, max_right, max_left and max_spanning. Once these values are computed, two comparisons will yield the maximum subsequence in the set. The cost of computing these three values is what determines the complexity of this algorithm.

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

Because this is a recursive solution, the time to find the maximum subsequence of one set depends on the amount of time spent finding the maximum subsequence in each half. It is for this reason that in the above recurrence relation T(n) is related to 2T(n/2). However the addition of the middle term, C(n) greatly complicates this analysis. C(n), above, represents the cost of calculating the spanning subsequence of maximum value.

This term C(n) involves processing each of the n items. One pointer works right from the center until it reaches the right set limit. At each step we add a value to a running sum and check whether the new sum has a greater value than the max so far. This process involves a total of n/2 comparisons. The same process takes place as another pointer moves left from the center until it reaches the leftmost limit. So, the C(n) procedure is a linear process. This means the recurrence relation of the entire algorithm can be written as:

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

By expanding this relation we get:

T(n) = 2T(n/2) + n + 2 \\
T(n/2) = 2T(n/4) ...
T(n/4) = 2T(n/8) + n/4 + 2 \\
\vdots \\
T(1) = 0

So we know that T(n) can be written as:

T(n) = n + 2 + 2( n/2 + 2 + 2( n/4 + 2 + 2( ... 1) ...) )

That is:

\begin{displaymath}T(n) = \sum_{i=1}^{q} n + 2^i

The term q in the above expression is a bit non-deterministic. If we restrict n to be 2x in order to simplify this exercise, then q = log2 n = x.

\begin{displaymath}T(n) = \sum_{i=1}^{log_2 n} n + 2^i

Writing that out in long format:

T(n) = n + 2 + n + 4 + n + 8 + ... n + 2x-1 + n + 2x

Now we'll invoke cancellation by subtracting T(n) from 2T(n)

2T(n) = & & 2n + 4 &+ 2n + 8& + 2n + ...
...= & -n - 2& +n & +n & +n &+ ... &n & + 2n + 2^{x+1}

(x above is log2 n). This gives us:

T(n) = (n-1) log2 n + 2log2 n + 1


T(n) = (n-1) log2 n + 2n

The dominant term is (n-1) log2 n making this an O(n log2 n) algorithm.

Scott Gasch