Merge Sort

Objectives

1. Investigate Merge Sort
2. Further study recurrance relations.

Notes

• MERGE_SORT (A)
     Input: an array from 0 to A.size()-1 
     Output: the ordered array
  
   if A.size > 1
       B,C ← SPLIT (A)
       MERGE_SORT(B)
       MERGE_SORT(C)
       A ← MERGE(B,C)
  
  
  SPLIT(A)
     Input: A single array
     Output: The array split into two pieces
  
   mid ← A.size/2
   B ← A[0 to mid]
   C ← B[mid+1 to A.size() -1]
   Return (B,C)
  
  
  MERGE(A,B)
     Input: Two ordered lists 0 to size-1
     Ouptput: A single ordered array
  
  
   posA ← 0
   posB ← 0
   posC ← 0
   While posA < A.size and posB < B.size
     if (A[posA] < B[posB] 
        C[posC] ← A[posA]
        posA ← posA + 1
     else
        C[posC] ← B[posB]
        posB ← posB + 1
     posC ← posC + 1
   While posA < A.size
     C[posC] ← A[posA]
     posA ← posA + 1
     posC ← posC + 1
   While posB < B.size
     C[posC] ← B[posB]
     posB ← posB + 1
     posC ← posC + 1
   Return C 
   

• Timing

  T(n) = 2T(n/2) + cn
  
         T(n/2) = 2T(n/2/2) + cn/2
                = 2T(n/22) + cn/2
  T(n) = 2[2T(n/22) + cn/2]  + cn
       = 22T(n/22) + cn + cn
       = 22T(n/22) + 2cn 
  
         T(n/22)  = 2T(n/22/2)  + cn/22  
                             = 2T(n/23)  + cn/22  
  T(n) = 22(2T(n/23) + cn/22) + 2cn
       = 23T(n/23) + 22cn/22 + 2cn
       = 23T(n/23) + 3cn
  
  at step k:
  
  T(n) = 2kT(n/2k) + kcn
  
  let n = 2i or log2n = i
  When i = k
  T(n) = 2iT(2i/2i) + icn
       = 2i c + i cn
       = n + cn log2n
  
       Which is O(n log2n)