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Chapter 7: Sor ting A lgorithms

Mark Allen Weiss: Data Structures and Algorithm Analysis in Java. Chapter 7: Sor ting A lgorithms. Merge Sort. Lydia Sinapova, Simpson College. Merge Sort. Basic Idea Example Analysis Animation. Idea. Two sorted arrays can be merged in linear time with N comparisons only.

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Chapter 7: Sor ting A lgorithms

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  1. Mark Allen Weiss: Data Structures and Algorithm Analysis in Java Chapter 7: Sorting Algorithms Merge Sort Lydia Sinapova, Simpson College

  2. Merge Sort • Basic Idea • Example • Analysis • Animation

  3. Idea Two sorted arrays can be merged in linear time with N comparisons only. Given an array to be sorted, consider separately its left half and its right half, sort them and then merge them.

  4. Characteristics • Recursive algorithm. • Runs in O(NlogN) worst-case running time. Where is the recursion? • Each half is an array that can be sorted using the same algorithm - divide into two, sort separately the left and the right halves, and then merge them.

  5. Example

  6. Merge Sort Code void merge_sort ( int [ ] a, int left, int right) { if(left < right) { int center = (left + right) / 2; merge_sort (a,left, center); merge_sort(a,center + 1, right); merge(a, left, center + 1, right); } }

  7. Analysis of Merge Sort Assumption:N is a power of two. For N = 1 time is constant (denoted by 1) Otherwise: time to mergesort N elements = time to mergesort N/2 elements + time to merge two arrays each N/2 el.

  8. Recurrence Relation Time to merge two arrays each N/2 elements is linear, i.e. O(N) Thus we have: (a) T(1) = 1 (b) T(N) = 2T(N/2) + N

  9. Solving the Recurrence Relation T(N) = 2T(N/2) + N divide by N: (1) T(N) / N = T(N/2) / (N/2) + 1 Telescoping:N is a power of two, so we can write (2) T(N/2) / (N/2) = T(N/4) / (N/4) +1 (3) T(N/4) / (N/4) = T(N/8) / (N/8) +1 ……. T(2) / 2 = T(1) / 1 + 1

  10. Adding the Equations The sum of the left-hand sides will be equal to the sum of the right-hand sides: T(N) / N + T(N/2) / (N/2) + T(N/4) / (N/4) + … + T(2)/2 = T(N/2) / (N/2) + T(N/4) / (N/4) + …. + T(2) / 2 + T(1) / 1 +LogN (LogNis the sum of 1’s in the right-hand sides)

  11. Crossing Equal Terms, Final Formula After crossing the equal terms, we get   T(N)/N = T(1)/1 + LogN T(1) is 1, hence we obtain   T(N) = N + NlogN = (NlogN)  Hence the complexity of the Merge Sort algorithm is(NlogN).

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