Chapter 4: Orders of Growth and Tree Recursion

Analyzing Merge Sort 4

Since n cards are handled on each of log₂n + 1 passes, the total number of handlings H(n) is:

H(n) = n(log₂n + 1)
= nlog₂n + n

Since n ≤ nlog₂n, then

H(n) ≤ nlog₂n + nlog₂n
≤ 2nlog₂n

But H(n) ≥ nlog₂n, so

nlog₂n ≤ H(n) ≤ 2nlog₂n

Thus the time it takes to merge sort n cards is:

Θ(nlog₂n)

This can also be expressed without a logarithmic base as:

Θ(nlogn)

since a number's logarithms of different bases differ only by a constant factor (log_ca = log_ba × log_cb).