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Applied Combinatorics, 4 th Ed. Alan Tucker

Applied Combinatorics, 4 th Ed. Alan Tucker. Section 3.4 Tree Analysis of Sorting Algorithms. Prepared by: Nathan Rounds and David Miller. “Big O” Notation. − I f we have a function f ( x ) defined f ( x ) = 5 x ³ + x – 4,

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Applied Combinatorics, 4 th Ed. Alan Tucker

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  1. Applied Combinatorics, 4th Ed.Alan Tucker Section 3.4 Tree Analysis of Sorting Algorithms Prepared by: Nathan Rounds and David Miller Tucker, Sec. 3.4

  2. “Big O” Notation − If we have a function f (x) defined f (x) = 5x³ + x – 4, we can approximate its value as a constant multiple of x³ for large values of x, since 5x³ is the dominating term in the function (it grows the fastest as x gets large). − For x = 5, f (x) = 626 and the constant multiple 5(x³) = 625 which is a very close estimate. − Therefore, we say that f (x) in “Big O” notation is O(x³). Tucker, Sec. 3.4

  3. Theorem • In the worst case, the number of binary comparisons required to sort n items is at least O(nlog2n). • This number is a constant multiple of (nlog2n) which approximates the height of a binary testing tree (log2n!) for large values of n. Tucker, Sec. 3.4

  4. Bubble Sort For m from 2 to n do: For i from n (step-1) to m do: If Ai Ai-1, then interchange items Ai and Ai-1. 4 6 3 1 5 2 1 4 2 6 3 5 4 6 3 1 2 5 4 6 1 3 2 5 1 4 6 2 3 5 4 1 6 3 2 5 1 2 4 6 3 5 1 4 6 3 2 5 i = 1 i = 2 i = 3 i = 4 i = 5 i = 6 This technique moves the smallest number to the top, then the next smallest number below that until the set of numbers is in numerical order. m = 2 i = 3 m = 3 i = 3 m = 3 i = 4 m = 3 i = 5 m = 3 i = 2 m = 2 i = 2 m = 2 i = 4 m = 2 i = 5 m = 2 i = 6 m = 3 i = 6 { Tucker, Sec. 3.4

  5. Bubble Sort Comparisons • When m=2, I goes from n to 2, which means you have to do (n-1) comparisons • When m=3, I goes from n to 3, which means you have to do (n-2) comparisons • Continuing this trend, the total number of comparisons must be: This shows that the Bubble Sort method will require comparisons, which is more then the theoretical bound of . Tucker, Sec. 3.4

  6. 5 4 0 9 2 6 7 1 3 8 67 1 3 8 5 4 0 9 2 0 2 1 8 9 3 5 4 6 7 4 6 7 5 Merge Sort – Setup • Roughly splits the set in half, until only single elements are left. 5 4 0 9 2 6 7 1 3 8 Tucker, Sec. 3.4

  7. 5 4 6 7 3 8 1 0 9 2 4 5 6 7 3 8 2 9 0 4 5 1 6 7 0 2 4 5 9 1 3 6 7 8 0 1 2 3 4 5 6 7 8 9 Merge Sort - Ordering • The set then gets “merged” into numerical order. Note: The root of the tree is on the bottom Tucker, Sec. 3.4

  8. n = 24 Level 0 ++++++++++++++++ Level 1 ++++++++ ++++++++ Level 2 ++++ ++++ ++++ ++++ Level 3 ++ ++ ++ ++ ++ ++ ++ ++ Level 4 + + + + + + + + + + + + + + + + Complexity of Merging • Assume that every set has n = 2r elements • There are r levels • There are 2k vertices on level k • On level k, each set will have 2r-k elements Tucker, Sec. 3.4

  9. Merge Sort Comparisons • In general, at each vertex on level k two ordered sublists of 2r-k-1 items are merged into an ordered sublist of 2r-kitems • This merging requires 2r-k-1 comparisons • The two ordered sublists on level one merge into the root • The number of comparisons needed at level k is 2k (2r-k – 1 ) = 2r - 2k since there are 2k different vertices • Totaling the number of comparisons needed for a Merge Sort is: • With n = 2r this becomes: • This shows that Merge Sort method requires comparisons, which achieves the theoretical bound of a binary search. Tucker, Sec. 3.4

  10. Puts the first element at the end of the left list 7 6 8 9 4 0 1 2 3 5 Once an element is alone there is no need to continue to divide it 0 1 2 3 4 5 6 7 8 9 2 1 3 4 7 6 1 2 9 3 4 8 3 4 1 2 0 QUIK Sort • Takes the first element and compares it with the other elements to divide the list, putting smaller elements into a set on the left and the larger elements into a set on the right. 5 4 0 7 1 6 8 2 3 9 Tucker, Sec. 3.4

  11. Heap Sort • A binary tree so that the parent is always bigger than its children Put the root at the beginning of a list, then move up the next biggest grandchildren. 6 LIST: 5 3 4 2 0 1 Tucker, Sec. 3.4

  12. Class Exercise • Sort the following set of numbers using a bubble sort and then do it using a merge sort. • Which sort technique is more efficient for this set? Tucker, Sec. 3.4

  13. Bubble Sort Solution 5 3 8 23 -6 0 2 66 5 3 8 -6 23 0 2 66 5 3 -6 8 23 0 2 66 5 -6 3 8 23 0 2 66 -6 5 3 8 23 0 2 66 -6 0 5 3 2 8 23 66 -6 0 5 2 3 8 23 66 -6 0 2 5 3 8 23 66 -6 0 2 3 5 8 23 66 . . . Tucker, Sec. 3.4

  14. Merge Sort Solution 5 3 8 23 -6 0 2 66 5 3 8 23 -6 0 2 66 -6 0 2 66 8 23 5 3 3 23 -6 0 66 5 8 2 Tucker, Sec. 3.4

  15. Merge Sort Solution cont’d 3 23 -6 0 66 5 8 2 3 5 8 23 -6 0 2 66 3 5 8 23 -6 0 2 66 -6 0 2 3 5 8 23 66 Tucker, Sec. 3.4

  16. Something to Ponder • How does one create an initial heap to perform a heap sort on? • This will be on the homework.  Tucker, Sec. 3.4

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