Merge Algorithm Analysis-Analysis of Algorithms-Lecture Slides, Slides of Design and Analysis of Algorithms

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Merge Algorithm Analysis

• Suppose total number of elements to be merged is n
• Running time is O (n)

Merge Sort Analysis

• Suppose n is a power of two, so that we always split into even halves.
• For n = 1 the time to merge sort is1 otherwise
• The time to merge sort n numbers is equal to the time to do two recursive merge sorts of size n/2, plus the time to merge, which is linear. T(1) = 1 T(n) = 2T(n/2) + n
• Solve this recurrence to find out running time.

Solution by Telescoping Sum

T(n) / n + T(n/2) / n/2 + T(n/4) / n/4 + ….+ T(2) / 2 = T(n/2) / 2 + 1 + T(n/4) / n/4 + 1 + T(n/8) / n/8 + 1 +…..+ T(1) / 1 + 1

• All the terms other than T(n) / n on right hand side cancel the terms on left hand side.
• Since n is a power of 2, these equations add up to lgn (Tree has lgn levels). T(n) / n = T(1) / 1 + lgn T(n) = n + nlgn T(n) = O(nlgn)

Solution by Substitution Method

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

• Substitute n/2 into the main equation 2T(n/2) = 2(2(T(n/4)) + n/2) = 4T(n/4) + n And T(n) = 4T (n/4) + 2n
• Again by substituting n/4, we can see 4T(n/4) = 4(2T(n/8)) + (n/4) = 8T(n/8) + n And T(n) = 8T(n/8) + 3n
• Continuing in this manner, we obtain T(n) = 2k^ T(n/2k) + k.n Using k = lgn T(n) = nT(1) + nlgn = nlgn + n T(n) = O(nlgn)

7

Complexity of Quick Sort

Average-case O(NlogN)

Worst Case: O(N^2 )

This happens when the pivot is the

smallest (or the largest) element.

Then one of the partitions is empty,

and we repeat recursively the

procedure for N-1 elements.

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Complexity of Quick Sort

Best-case O(NlogN)

The pivot is the median of the array,

the left and the right parts have same

size.

There are logN partitions, and to

obtain each partitions we do N

comparisons (and not more than N/

swaps). Hence the complexity is

O(NlogN)

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Worst-Case Analysis

The pivot is the smallest (or the largest) element

T(N) = T(N-1) + cN, N > 1

Telescoping:

T(N-1) = T(N-2) + c(N-1)

T(N-2) = T(N-3) + c(N-2)

T(N-3) = T(N-4) + c(N-3) …………... T(2) = T(1) + c.

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Worst-Case Analysis

T(N) + T(N-1) + T(N-2) + … + T(2) =

= T(N-1) + T(N-2) + … + T(2) + T(1) +

c(N) + c(N-1) + c(N-2) + … + c.

T(N) = T(1) +

c times (the sum of 2 thru N)

= T(1) + c (N (N+1) / 2 -1) = O(N^2 )

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Best-Case Analysis

Telescoping:

T(N) / N = T(N/2) / (N/2) + c

T(N/2) / (N/2) = T(N/4) / (N/4) + c

T(N/4) / (N/4) = T(N/8) / (N/8) + c

……

T(2) / 2 = T(1) / (1) + c

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Best-Case Analysis

Add all equations:

T(N) / N + T(N/2) / (N/2) + T(N/4) / (N/4)

• …. + T(2) / 2 =

= (N/2) / (N/2) + T(N/4) / (N/4) + … + T(1) / (1) + c.logN

After crossing the equal terms:

T(N)/N = T(1) + cLogN T(N) = N + Nc*LogN = O(NlogN)

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Summary

• Divide and Conquer
• Merge-Sort
  • Most of the work done in Merging
  •  (n log(n)) time
  •  (n) space
• Quick-Sort
  • Most of the work done in partitioning
  • Average case takes  (n log(n)) time
  • Worst case takes  (n^2 ) time
  •  (1) space

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Choosing the Pivot

Some fixed element: e.g. the first, the last, the one in the middle.

Bad choice - may turn to be the smallest or the largest element, then one of the partitions will be empty

Randomly chosen (by random generator)

• still a bad choice

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Choosing the Pivot

The median-of-three choice:

take the first, the last and the middle element.

Choose the median of these three elements.