Review of Chapter 5

1
Review of Chapter 5

• ???

2
Selection Trees

• Selection trees can merge k ordered sequences
  (assume in non-decreasing order) into a single
  sequence easily.
• Two kinds of selection trees
• Winner trees
• A winner tree is a complete binary tree
• The root represents the smallest node in the
  tree.
• Each leaf node represents the first record in the
  corresponding run.
• Each non-leaf node in the tree represents the
  winner of its right and left subtrees.
• Loser tress
• A loser tree is a complete binary tree.
• Each nonleaf node retains a pointer to the loser
  is called a loser tree.
• Each leaf node represents the first record in the
  corresponding run.
• An additional node, node 0, has been added to
  represent the overall winner of the tournament.

3
Winner Tree (k8)
1
6
2
3
6
8
4
5
7
6
9
6
8
17
8
9
10
11
12
13
14
15
10
9
20
6
8
9
90
17
20
20
15
15
11
95
18
15
38
30
25
50
16
99
20
16
28
run1
run2
run4
run3
run5
run6
run8
run7
4
Winner Tree (k8)
O(nlogk)
1
8
2
3
9
8
4
5
7
6
9
15
8
17
8
9
10
11
12
13
14
15
10
9
20
15
8
9
90
17
20
20
25
15
11
95
18
15
38
30
28
50
16
99
20
16
run1
run2
run4
run3
run5
run6
run8
run7
5
Loser Tree
Overall winner
0
6
1
8
2
3
9
17
4
5
7
6
10
20
9
90
8
9
10
11
12
13
14
15
10
9
20
6
8
9
90
17
20
20
15
15
11
95
18
15
38
30
25
50
16
99
20
16
28
run1
run2
run4
run8
run3
run5
run6
run7
6
Forests

• Definition
• A forest is a set of n 0 disjoint trees.
• When we remove the root of a tree, well get a
  forest.

A
E
G
B
C
D
F
H
I
7
Transforming A Forest Into A Binary Tree

• Definition
• If T1, , Tn is a forest of trees, then the
  binary tree corresponding to this forest, denoted
  by B(T1, , Tn)
• is empty if n 0
• has root equal to root (T1) has left subtree
  equal to B(T11, T12,, T1m), where T11, T12,,
  T1m are the subtrees of root (T1) and has right
  subtree B(T2, , Tn).

8
Transforming A Forest Into A Binary Tree
A
B
E
C
F
G
D
H
I
9
Forest Traversals

• Preorder
• Inorder
• Postorder (not natural)
• Level-order

10
The Satisfiability Problem

• Expression Rules
• A variable is an expression
• If x and y are expressions then
• are expressions
• Parentheses can be used to alter the normal order
  of evaluation, which is not before and before or.
• The satisfiablitity problem for formulas of
  proposition calculus asks if there is an
  assignment of values to the variables that causes
  the values of the expression to be true.
• The satisfiablitity problem is NP-Complete
  problem.

11
The Satisfiability Problem
x3
x1
x3
O(2n)
x2
x1
12
Set Representation

• Trees can be used to represent sets.
• Pairwise Disjoint Sets
• If Si and Sj, i?j, are two sets, then there is
  no element that is in both Si and Sj.
• Set Operations
• Disjoint set union
• If Si and Sj are two disjoint sets, then
  their union Si?Sj all elements x such that x
  is in Si or Sj.
• Find(i)
• Find the set containing element i.

13
Set Representation
n10
0
2
4
6
7
8
3
5
9
1
S1 0,6,7,8
S2 1, 4, 9
S32,3,5
14
Disjoint Set Union
S1 U S2
4
0
OR
4
0
9
1
6
7
8
9
6
7
8
1
15
Data Representation for S1, S2, S3
Set Name
Pointer
S1
0
2
4
S2
6
7
8
3
5
1
9
S3
16
Array Representation of S1, S2, S3
i
0
1
2
3
4
5
6
7
8
9
parent
-1
4
-1
2
-1
2
0
0
0
4
17
Degenerate Tree
n-1
union(0, 1), find(0)
Union operation O(n)
union(1, 2), find(1)
n-2
Find operation O(n2)
union(n-2, n-1), find(n-1)
0
18
Improve the performance of Set Union and Find
Algorithms

• Weighting Rule Weighting rule for union(I, j)
• If the number of nodes in the tree with root i is
  less than the number in the tree with root j,
  then make j the parent of i otherwise make i the
  parent of j.
• Collapsing Rule
• If j is a node on the path from i to its root and
  parenti? root(i), then set parentj to root(i).

19
Set Union with The Weighting Rule
0
1
n-1
0
2
n-1
0
3
n-1
1
1
2
0
0
4
n-1
1
3
2
1
3
2
n-1
20
Set Find with Collapsing Rule
-8
-8
0
0
2
4
2
6
1
7
1
4
3
5
5
6
3
After collapsing
7
Before collapsing
21
Equivalence Class
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
0
1
2
3
4
5
6
7
8
9
10
11
(a) Initial trees
-2
-2
-2
-2
-1
-1
-1
-1
0
3
6
8
2
5
7
11
4
1
10
9
(b) Height-2 trees following 04, 31, 610, and
89
22
Equivalence Class
(C) Trees following 74, 68, 35, and 211
-2
-4
-3
-3
6
2
0
3
11
10
8
7
1
5
4
9
-3
-3
-4
0
3
6
1
5
10
8
4
7
2
11
(d) Thees following 110
9
23
Counting Binary Trees

• Problem
• Determine the number of distinct binary trees
  having n nodes.
• Determine the number of distinct permutations of
  the numbers from 1 through n obtainable by a
  stack.
• Determine the number of distinct ways of
  multiplying n1 matrices.

24
Distinct binary trees

• n0 or n1 ? 1 binary tree
• n2 ? 2 distinct binary trees.
• n3 ? 5 distinct binary trees (?????)

bn
bi
bn-i-1
25
Stack Permutations

• How many permutations can we obtain by passing
  the numbers 1 through n through stack ? See
  chapter 3 about stack.
• For example, the numbers 1, 2, 3
• (1,2,3) (1,3,2) (2,1,3) (2,3,1) (3,2,1) but
  (3,1,2)
• The recursive formula is
• bn b0bn-1 b1bn-2 bn-2b1 bn-1b0

26
Construct The Binary Tree from Preorder and
Inorder Sequence

• Give preorder and inorder sequence as follows.
• preorder sequence A B C D E F G H I
• Inorder sequence B C A E D G H F I
• Problem
• Does such a pair of sequences uniquely define a
  binary tree ? (Can this pair of sequences come
  from more than one binary tree?)
• Conclusion
• Every binary tree has a unique pair of preorder /
  inorder sequences.

27
Construct The Binary Tree from Preorder and
Inorder Sequence
A
A
B, C
D, E, F, G, H, I
B
D
C
F
E
A
I
G
B
D, E, F, G, H, I
H
C
28
Inorder and Preorder Permutations
1
A
2
4
B
D
3
6
5
C
F
E
9
7
I
G
8
H
Preorder 1, 2, 3, 4, 5, 6, 7, 8, 9
Inorder 2, 3, 1, 5, 4, 7, 8, 6, 9
29
Stack Permutations

• Preorder permutation 1, 2, 3

1
1
1
1
1
2
3
2
2
2
2
3
3
3
3
(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 2, 1)
Inorder permutations
30
Stack Permutations
The number of distinct Binary Trees with n nodes

Inorder permutations obtainable from binary trees
having the preorder permutations, 1,2,3,.,n

The number of distinct permutations by passing
1..n through stack
31
Matrix Multiplication

• Computing the product of n matrices
• M1 M2 Mn
• By matrix multiplication associative law, we can
  perform these multiplications in any order.
• For example, n3 (n4 ????)
• (M1 M2) M3
• M1 (M2 M3)
• The number of distinct ways to obtain M1 M2
  Mn

32
Number of Distinct Binary Trees

• Let
• which is the generating function for the
  number of binary trees.
• By the recurrence relation we get