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Analysis of Merge Sort

Note if n = 1, Then T(n) = $\Theta(1)$

1.: DIVIDE computes middle of array in constant time: D(n) = $\Theta(1)$
2.: CONQUER sorts 2 subarrays of size n/2 in time: 2T(n/2)
3.: COMBINE (Merge) procedure: C(n) = $\Theta(n)$

T(n) = $\left\{ \begin{array}{ll} \Theta(1) & {\rm if} \; n = 1 \\ 2T(n/2) + \Theta(1) + \Theta(n) & {\rm if} \; n > 1 \end{array} \right.$

Note that $\Theta(n)$ dominates $\Theta(1)$ .

We will show next class that T(n) = $\Theta(n \lg n)$ , where lg = $\log_2 n$ .

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