Tucker, Applied Combinatorics, Sec. 3.5, Jo E-M

1
Tucker, Applied Combinatorics, Sec. 3.5, Jo E-M

• Big O Notation
• We say that a function is
  if for some
  constant c, when n is large.
• For example, is
  since for n
  gt 3.
• This is described by saying that
  is on the order of .
• Big O notation calls attention to the part of a
  function that grows the fastest, so gives a
  simple estimate of how many steps are required
  for an algorithm to run.

2
Binary Testing Tree
Compare things (objects, subsets, lists,
whatever) in some way with two outcomes.
Keep comparing
Eventually arrive at one of the n! possible orders
Since there are n! leaves on this tree, by
Theorem 3 in section 3.1, the tree has height
. Thus,
whatever binary comparison technique is used, the
worst case will require at least
comparisons.
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Bubble Sort

• For m from 2 to n, do
• For j from n (step 1) to m do
• If Aj lt Aj-1, then interchange Aj and Aj-1.

4
Complexity of the Bubble Sort

• When m 2, j goes from n to 2, so you have to do
  n 1 comparisons.
• When m 3, j goes from n to 3, so you have to do
  n 2 comparisons, and so on.
• Thus, the total number of comparisons is
• Thus, the Bubble sort takes
  comparisons, which is a lot more than the
• theoretical bound of .

5
Merge SortSubdivision Tree
First divide the set repeatedly, roughly in half,
until only single elements are left at the
leaves.
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Merge SortMerging Tree
then merge the sequences in order.
4
7
0
5
4 7
6
1
0 5
2
8
3
9
1 4 7
2 6
0 5 9
3 8
Note This tree is drawn upside down, so that
the root is at the bottom. Thus, the set 0, 5
is at level 3
0 2 5 6 9
1 3 4 7 8
1 2 3 4 5 6 7 8 9
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Complexity of MergingA Simplifying Assumption

• Assume our set has n 2r elements. This means
  that
• We always divide exactly in half in the
  subdivision tree.
• There are r levels
• All the leaves in both the subdivision and the
  merging tree are on level r.
• There are 2k vertices on level k
• The sets on level k all have 2r-k elements in them

n 23


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Complexity of MergingCounting the Comparisons

• At each vertex on level k of the merging tree we
  merge the two children sets each of size 2r-k-1.
• Merging two lists of length L into a single list
  of length 2L requires 2L-1 comparisons (class
  exercise).
• For each vertex on level k, this merging takes 2
  (2r-k-1) 1 2r-k 1 comparisons
• There are 2k vertices on level k, for a total of
  2k (2r-k 1 ) 2r - 2k comparisons on each
  level.
• Since we must do this on each level, the total
  number of comparisons for a Merge Sort is
• Since n 2r, this becomes
  .
• (Still need to account for the subdivision
  preprocessing, but this isnt bad)
• Thus Merge Sort is and
  achieves the theoretical bound of a binary search.

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QUIK Sort
Use the first element in the list to partition
the list, then put it at the end of the left hand
child list. Worst case, , on average
5 0 9 2 6 47 1 3 8
9 6 7 8
0 2 4 1 3 5
6 7 8 9
0
2 4 1 3 5
7 8 9
6
1 2
4 3 5
8 9
3 4
7
5
2
1
8
4
9
3
10
Heap Sort

• A heap is a (nearly) binary tree so that a parent
  is always bigger than its children (root is
  biggest of all).
• Put the root at the beginning of a list, then
  move up the biggest grandchildren.

Preprocessing Need to first construct the heap
(homework)
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Class Exercises

• Prove that it takes at most n 1 comparisons to
  merge two sorted lists into a single sorted list
  of length n.
• Sort 4, 5, 2, 3, 0, 1 using each of a Bubble,
  Merge, QUIK, and Heap Sort.