1. Applied Combinatorics, 4th Ed. Alan Tucker
Section 3.4
Tree Analysis of Sorting Algorithms
Prepared by: Nathan Rounds and David Miller
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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).
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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.
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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. 1 4 2 6 3 5 1 4 6 2 3 5 4 1 6 3 2 5 4 6 3 1 2 5 1 4 6 3 2 5 4 6 1 3 2 5 4 6 3 1 5 2 1 2 4 6 3 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 = 3
i = 2
m = 3
i = 4
m = 3
i = 3
m = 3
i = 6
m = 3
i = 5
m = 2
i = 3
m = 2
i = 6
m = 2
i = 2
m = 2
i = 5
m = 2
i = 4 {
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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
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6. Merge Sort – Setup
Roughly splits the set in half, until only single elements are left.
9 2
67 1
3 8
5 4 9 2
6 7 1 3 8 5 4 6 7
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7. Note: The root of the tree
Merge Sort - Ordering
The set then gets “merged” into numerical order. 5 4 9 2 6 7 1 3 8
4 5
6 7
2 9
3 8
0 4 5
1 6 7
Note: The root of the tree
is on the bottom
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8. 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
n = 24
++++ ++ +
Level 0
Level 1
Level 2
Level 3
Level 4
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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-k items
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.
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Tucker, Sec. 3.4

10. 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.
Puts the first element
at the end of the left list 5
6 7
8 9
Once an element is alone
there is no need to continue
to divide it 6 7 8 9
1 2
3 4 1 2 3 4
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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 1
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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?
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13. Bubble Sort Solution . . . 5 3 8 23 -6 2 66 5 3 8 -6 23 2 66 5 3 -6 82 66 5 3 8 -6 23 2 66 5 3 -6 8 23 2 66 5 -6 3 8 23 2 66 -6 5 3 8 23 2 66 -6 5 3 2 8 23 66 -6 5 2 3 8 23 66 -6 2 5 3 8 23 66 -6 2 3 5 8 23 66
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14. Merge Sort Solution 5 3 8 23 -6 0 2 66 5 3 8 23 -6 0 2 66 -6 0 2 66
-6 0
2 66
8 23
5 3 3 23 -6 66 5 8 2
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15. Merge Sort Solution cont’d3 23 -6 66 5 8 2
3 5
8 23
-6 0
2 66
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16. Something to Ponder
How does one create an initial heap to perform a heap sort on?
This will be on the homework. 
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Tucker, Sec. 3.4