Analysis of MergeSort

• MergeSort Algorithm:

• MergeSort(A)
  
   If A.size > 1
      Split(A,B,C)
      MergeSort(B)
      MergeSort(C)
      Merge(A,B,C)
   EndIF 
   

• Split is $O(n)$
• As we saw in a previous chapter, Merge is $O(n)$
• The performance of MergeSort is then $$ T(n) = \begin{cases} O(1) & \text{for } n=1 \\ 2T(n/2) + 2n & \text{for } n > 1\\ \end{cases} $$
• Your authors use a $T(n) \lt 2T(\frac{n}{2})+2n$
  • Which is probably true, but we are bounding from above.
  • So I will use =
• They point out $T(n) = T(\lfloor \frac{n}{2}\rfloor) + T(\lceil \frac{n}{2}\rceil) + 2n$
  • But we assume n is even to make this easier to analyze.
• They point out that symbolic algebra packages can solve many of these
  • See Wolfram Alpha's Solution
  • But it is useful to understand how to solve these as it gives insight into the divide and conquer method.
• There are several ways to solve a recurrence relationship.
• As I demonstrated before, I like a method where you continuously expand the equation by substituting in the next solution.
  • $$ \begin{array}{rrl} T(n) = & 2T(\frac{n}{2}) + 2n & \\ & T(\frac{n}{2}) = & 2T(\frac{\frac{n}{2}}{2}) + 2\frac{n}{2} \\ & = & 2T(\frac{n}{2^2}) + 2n\\ T(n) = & 2(2T(\frac{n}{2^2}) + n) + 2n & \\ = & 2^2T(\frac{n}{2^2}) + 2n+ 2n & \\ = & 2^2T(\frac{n}{2^2}) + 2(2n) & \\ & T(\frac{n}{2^2})=& 2T(\frac{\frac{n}{2^2}}{2}) + 2\frac{n}{2^2} \\ & = & 2T(\frac{n}{2^3}) + \frac{n}{2} \\ T(n) = & 2^2(2T(\frac{n}{2^3}) + \frac{n}{2}) + 2(2n)& \\ = & 2^3T(\frac{n}{2^3}) + 2^2\frac{n}{2} + 2(2n)& \\ = & 2^3T(\frac{n}{2^3}) + 2n+ 2(2n)& \\ = & 2^3T(\frac{n}{2^3}) + 3(2n)& \\ \end{array} $$
  • In general we can say, at step k $$ T(n) = 2^kT(\frac{n}{2^k}) + k(2n) $$
  • If we assume $n = 2^k$, we have $\log_2{n} = k$
  • So $$ \begin{array}{rrl} T(n) = & 2^kT(\frac{n}{2^k}) + k(2n) \\ = & 2^{log_2{n}} T(\frac{2^k}{2^k}) + \log_2{n}(2n) \\ = & n T(1) + 2n \log_2{n} \\ = & n + 2n \log_2{n} \\ \end{array}$$
• Give the algorithm and derive the performance of Binary Search.