CSC 10400 Discrete Mathematical Structures Lecture 12

Transcription

CSC 10400
Discrete Mathematical Structures
Lecture 12
Instructor: Pavel Rytir
Email: rytirpavel@gmail.com
The City College of New York
Spring 2015
Section 12.1 Tree: Definitions, Properties and Examples
Tree
Properties on Trees
Forest
Tree
Tree
A loop-free undirected graph is a tree if it is connected and
contains no cycles.
Only G1 is a tree. G2 contains a cycle, and G3 is not connected.
FG1 is also a spanning tree of G2 .
Theorem
If u and v are distinct vertices in a tree T , then there is a unique
path that connects u and v .
Theorem
If u and v are distinct vertices in a tree T , then there is a unique
path that connects u and v .
Proof.
Because T is connected, there is at least one path connecting u
and v . If there are more than one path between u and v , then from
two such paths some of the edges would form a cycle. Because T
has no cycles, there is a unique path connecing u and v .
Theorem
If G is an undirected graph, then G is connected if and only if G
has a spanning tree.
Theorem
If G is an undirected graph, then G is connected if and only if G
Proof.
If G has a spanning tree T , then for every pair of vertices u and v ,
there is a path in T connecting them. Thus, G is connected.
If G is connected and G is not a tree, we break each cycle by
deleting an edge in the cycle. The resulting subgraph is connected
and contains all vertices in G . If the resulting subgraph is a tree,
we are done. Otherwise, the resulting subgraph has a cycle, and we
can continue to break the cycle, until we obtain a tree. Hence, G
Theorem
In every tree T = (V , E ), |V | = |E | + 1.
Theorem
In every tree T = (V , E ), |V | = |E | + 1.
Proof.
If |E | = 0, the tree consists of a single isolated vertex, so |V | = 0.
Suppose |E | = |V | + 1 for every |E | ≤ k for some k. When
|E | = k + 1, we delete an edge from T , and obtain two trees
T1 = (V1 , E1 ) and T2 = (V2 , E2 ). Clearly, |V | = |V1 | + |V2 | and
|E | = |E1 | + |E2 | + 1. Thus, we have
|E | = |E1 | + |E2 | + 1 = (|| V1 − 1) + (|V2 | − 1) + 1 = |V | − 1
By mathematical induction, |V | = |E | + 1.
Theorem
For any tree T = (V , E ), if |V | ≥ 2, then T has at least two vertices
with degree 1.
Theorem
For any tree T = (V , E ), if |V | ≥ 2, then T has at least two vertices
with degree 1.
Proof.
Let
P |V | = n ≥ 2. First, we have |E | = |V | − 1 = n − 1. Second,
v ∈V deg (v ) = 2|E | = 2(n − 1). Because T is connected, we
have deg (v ) ≥ 1 for every node v . If there are k vertices with
degree 1, then we have n − k vertices has degree at least 2. Thus
X
k + 2(n − k) ≤
deg (v ) = 2(n − 1)
v ∈V
Solve this inequality, we have k ≥ 2.
Theorem
The following statements are equivalent for a loop-free undirected
graph G = (V , E )
(a) G is a tree.
(b) G is connected, but the removal of any edge from G
disconnects G into two subgraphs that are trees.
(c) G contains no cycles, and |V | = |E | + 1.
(d) G is connected and |V | = |E | + 1.
(e) G contains no cycles and if u, v ∈ V with (u, v ) ∈
/ E , then the
graph obtained by adding the edge (u, v ) to G has exactly one
cycle.
Theorem
graph G = (V , E )
(a) G is a tree.
Proof.
(a) ⇒ (b).
If G is a tree, then G is connected. Let e = {u, v } be any edge of
G . If G − e is connected, there are at least two paths in G from u
to v , contradicting that G is a tree. Thus, G − e is disconnected.
Removing e from G partitions G into two components: one led by
u and the other one led by v .
Theorem
graph G = (V , E )
Proof.
(b) ⇒ (c).
If G has a cycle, then let e = {u, v } be an edge of the cycle. But
G − e will be connected, contradicting the hypothesis in part (b).
So G has no cycle. Because G is a loop-free undirected graph, by
the definition, we know that G is a tree. Consequently, we have
|V | = |E | + 1
Theorem
graph G = (V , E )
Proof.
(c) ⇒ (d).
Suppose G has r components: G1 , G2 , . . . , Gr . For 1 ≤ i ≤ r ,
select a vertex vi ∈ Gi , and add r − 1 edges
{v1 , v2 , }, {v2 , v3 }, . . . , {vr −1 , vr } to G to form a new graph
G 0 = (V , E 0 ). The graph G 0 still has no cycle, i.e., G 0 is a tree.
Since G 0 is a tree, we know that |V | = |E 0 | + 1. But from part (c),
|V | = |E | + 1, so |E | = |E 0 | and r − 1 = 0. Because r = 1, it
follows that G is connected.
Theorem
graph G = (V , E )
/ E , then the
cycle.
Proof.
(d) ⇒ (e).
Let C be a cycle in G with r vertices and r edges. Since G is
connected, the remaning vertices of G can each be connected to a
vertex in C by a path. Each such connection requires at least one
new edge. Thus, |E | ≥ |V |, contradicting |V | = |E | + 1.
Therefore, G has no cycle and connected, i.e., G is a tree.
Theorem
graph G = (V , E )
/ E , then the
cycle.
Proof.
(d) ⇒ (e) (Continue...)
When we add a new edge (u, v ) into G , there are two paths
between u and v : one is the unique path P between u and v ,
because G is a tree; the other one is the edge (u, v ). Therefore,
P ∪ {(u, v )} is a cycle C1 .
Theorem
graph G = (V , E )
/ E , then the
cycle.
Proof.
(d) ⇒ (e) (Continue...)
If the new graph has the second cycle C2 , C2 must contain the
edge (u, v ); otherwise, G would contain a cycle. If C2 6= C1 ,
because C2 = P 0 ∪ {u, v }, where P2 is a path from u to v in G , we
have P 6= P 0 . Thus, there are two distinct paths from u to v in G ,
contradicting that G is a tree. Therefore, there has exactly one
cycle in the new graph.
Theorem
graph G = (V , E )
/ E , then the
cycle.
(a) G is a tree.
Proof.
(e) ⇒ (a)
If G is not connected, let G1 and G2 be the components of G with
u ∈ G1 and v ∈ G2 . Adding an new edge (u, v ) will not result in a
cycle. Thus, G is connected. Because G has no cycle. G is a
tree.
Section 12.2 Rooted Trees
Rooted trees
Tree traversals: preorder, postorder and inorder
Graph traversal: BFS and DFS
Balanced trees
Rooted Tree
A tree is a rooted tree if we specify one node as the root of
the tree.
Figure: A tree with root node r .
Ancestors, Descendents, Parents, Children and Siblings
Let T = (V , E ) be a tree with root node r .
Ancestors
For any node x in the tree, if a node y 6= x locates on the
path from r to x, then y is an ancestor of x.
Ancestors
Descendents
If x is an ancestor of y , then y is a descendent of x.
Ancestors
Descendents
Parents
If x is an ancestor of y , and (x, y ) is an edge in the tree, then
x is the parent of y .
Ancestors
Descendents
Parents
Children
If x is the parent of y , then y is a child of x.
Ancestors
Descendents
Parents
Children
If x is the parent of y , then y is a child of x.
Siblings
If two distinct nodes x and y have the same parent, then x
and y are siblings.
Examples
r ,a and c are
ancestors of g .
f , g , j, k, p and q
are descendants of c.
c is the parent of f
and g .
p and q are children
of j.
h and i are siblings.
Leaves, Internal Nodes, and Levels
Let T = (V , E ) be a tree with root r .
Leaves
A leaf is a node with no child.
Internal nodes
An internal node is a node with children.
Levels
For every vertex v , the level of v is the edges contained in the
path from r to v .
FThe root node has level 0!
Examples
The nodes p, q, k, d,
s, t and u are leaves;
and other nodes are
internal nodes
The node on level 0 is
r.
The nodes on level 1
is a, b.
is c, d, e.
are f , g , h, i.
Subtree
Let T = (V , E ) be a tree. For any node x, the subtree of T
at x is the subgraph induced by x and its descendants.
(a) A tree T
(b) The subtree of T at b
Preorder Traversal
Let T = (V , E ) be a tree with root r . A preorder traversal on T is
defined as follow.
If T only contains the root r , then the preorder traversal is
the root r itself.
Otherwise, let T1 , T2 , . . . , Tk be the subtrees at the children
of r , from left to right. The preorder traversal
First, visits r .
Then, traverses T1 in preorder.
Traverses T2 in preorder and so on until traverses Tk in
preorder.
An Example of Preorder Traversal
Let T be a tree with root 1, shown in the following figure.
An preorder traversal of this tree is:
1, 2, 5, 11, 12, 13, 14, 3, 6, 7, 4, 8, 9, 10, 15, 16, 17
Postorder Traversal
Let T = (V , E ) be a tree with root r . A postorder traversal on T
is defined as follow.
If T only contains the root r , then the postorder traversal is
the root r itself.
Otherwise, let T1 , T2 , . . . , Tk be the subtrees at the children
of r , from left to right. The postorder
First, traverses T1 in postorder.
Then, traverses T2 in postorder and so on until traverses Tk in
postorder.
Finally, visits r .
An Example of Postorder Traversal
Let T be a tree with root 1, shown in the following figure.
An postorder traversal of this tree is:
11, 12, 13, 14, 5, 2, 6, 7, 3, 8, 9, 15, 16, 17, 10, 4, 1
Question
How can computer calculate an expression, e.g., ((7 + 3)/5) ↑ ((8 −
5) ∗ 2), where a ↑ b = ab ?
Question
5) ∗ 2), where a ↑ b = ab ?
Use a tree.
((7 + 3)/5) ↑ ((8 − 5) ∗ 2)
=((7 + 3)/5) ↑ (3 ∗ 2)
=((7 + 3)/5) ↑ 6
=(10/5) ↑ 6
=2 ↑ 6
=64
Polish Notation
Question
5) ∗ 2), where a ↑ b = ab ?
Use the Polish notation, or prefix notation; i.e., presenting a ⊕ b by
⊕ab.
Polish Notation
Question
5) ∗ 2), where a ↑ b = ab ?
Use the Polish notation, or prefix notation; i.e., presenting a ⊕ b by
⊕ab.
Therefore, the mathematical equation
((7 + 3)/5) ↑ ((8 − 5) ∗ 2)
can be represented by Polish notation as
↑ / + 735 ∗ −852
Computation on Polish Notation
↑ / + 735 ∗ −852
= ↑ / + 735∗32
= ↑ /+7356
= ↑ / 10 56
=↑ 26
=64
Preorder Traversal and Polish Notation
The tree corresponding to the equation ((7 + 3)/5) ↑ ((8 − 5) ∗ 2)
is:
Its preorder traversal is: ↑ / + 735 ∗ −852, which is exactly the
polish notation of the equation.
Binary Tree
A tree is a binary tree if every internal node has at most two
children.
A tree is a complete binary tree if every internal node has
exactly two children.
(a) A binary tree T
(b) A complete binary tree T
Inorder Traversal
Let T = (V , E ) be a binary tree with root r . A inorder traversal
on T is defined as follow.
If T only contains the root r , then the inorder traversal is the
root r itself.
Otherwise, let TL and TR be the subtrees at the left child and
the right child R, respectively. The inorder traversal
First, traverses TL in inorder, if TL exists.
Then, visits r .
Finally, traverses TR in inorder, if TR exists.
An Example of Postorder Traversal
Let T be a tree with root r , shown in the following figure.
An postorder traversal of this tree is:
p, j, q, f , c, k, g , a, d, r , b, h, s, m, e, i, t, n, u
Depth-First Search Algorithm
Depth-First search (DFS) algorithm can traverse a loop-free
undirected graph.
undirected graph.
Initial a stack S with only one element v , and initial a
spanning tree T which consists of only one vertex v as its
root.
undirected graph.
root.
While S is not empty
undirected graph.
root.
Pop an element u from the top of the stack S, and mark u as
visited.
undirected graph.
root.
Pop an element u from the top of the stack S, and mark u as
visited.
Find all unvisited vertices adjacent with u, push these vertices
at the top of the stack S, and attach all edges from u to these
vertices into the spanning tree T so that u is the parent of
these vertices.
FThe resulting tree T is exactly a spanning tree of the
graph.
An Example of DFS
Step 1: S = {A}, T = {A}.
Step 2:
S = {B, C , D}, T = {A, B, C , D}.
Step 3: S = {B, C , E , F }, T =
{A, B, C , D, E , F }.
Step 4: S = {B, C , E }, T =
{A, B, C , D, E , F }.
Step 5: S = {B, C }, T =
{A, B, C , D, E , F }.
Step 6:
S = {B}, T = {A, B, C , D, E , F }.
Step 7:
S = ∅, T = {A, B, C , D, E , F }.
An Example of DFS
The spanning tree
generated by DFS is:
Breadth-First Search Algorithm
Breadth-First search (BFS) algorithm can traverse a loop-free
undirected graph.
undirected graph.
Initial a queue Q with only one element v , and initial a
root.
undirected graph.
root.
While Q is not empty
undirected graph.
root.
Pop an element u from the front of the queue Q, and mark u
as visited.
undirected graph.
root.
Pop an element u from the front of the queue Q, and mark u
as visited.
Find all unvisited vertices adjacent with u, push these vertices
at the end of the queue Q, and attach all edges from u to
these vertices into the spanning tree T so that u is the parent
of these vertices.
FThe resulting tree T is exactly a spanning tree of the
graph.
An Example of BFS
Step 1: Q = {A}, T = {A}.
Step 2:
Q = {B, C , D}, T = {A, B, C , D}.
Step 3: Q = {C , D, E }, T =
{A, B, C , D, E }.
Step 4: Q = {D, E , F }, T =
{A, B, C , D, E , F }.
Step 5: Q = {E , F }, T =
{A, B, C , D, E , F }.
Step 6:
Q = {F }, T = {A, B, C , D, E , F }.
Step 7:
Q = ∅, T = {A, B, C , D, E , F }.
An Example of BFS
The spanning tree
generated by BFS is:
Height and Balanced Tree
The height of a tree is the largest level number achieved by a
leaf of the tree.
A rooted tree T of height h is balanced if the level number of
every leaf in T is h − 1 or h.
(a) height = 3, balanced
(b) height = 3, unbalanced
Interview Question
Suppose you are given 8 coins. Exactly one coin is fake. If we knows
that the fake coin is heavier than others, and we have a two pan
scale. How many weighings do we need to find the fake one?
Interview Question
Suppose you are given 8 coins. Exactly one coin is fake. If we knows
that the fake coin is heavier than others, and we have a two pan
scale. How many weighings do we need to find the fake one?
Use the decision tree shown as follows.
Thus, two weighings is sufficient and necessary.
Interview Question
Suppose you are given 12 coins. Exactly one coin is fake. If we do
not know whether the fake coin is heavier or lighter than others, and
we have a two pan scale. How many weighings do we need to find
the fake one?
Interview Question
Suppose you are given 12 coins. Exactly one coin is fake. If we do
not know whether the fake coin is heavier or lighter than others, and
we have a two pan scale. How many weighings do we need to find
the fake one?
This problem is much more difficult than the previous one. The
answer is 3. You can google/wiki it.
Section 12.4 Weighted Trees and Prefix Codes
Weighted trees
Prefix code: Huffman code
Weighted Tree
Let w1 , w2 , . . . , wn be a set of positive numbers called
weights. A tree T with n leaves is a weighted tree if we assign
w1 , . . . , wn to the n leaves.
Prefix Code
A set P of binary sequences is called a prefix code if no
sequence in P is the prefix of any other sequence in P.
Examples
111, 0, 1100, 1101, 10 constitute a prefix code for letters
a, b, c, d, e
111, 0, 110, 1101, 10 is not a prefix code because 110 is a prefix
of 1101.
Use Prefix Code
Suppose that 111, 0, 1100, 1101, 10 is the prefix code
corresponding to letters a, b, c, d, e respectively. The original
message is acdbe. The encoding result is:
11111001101010
Use Prefix Code
Suppose that 111, 0, 1100, 1101, 10 is the prefix code
corresponding to letters a, b, c, d, e respectively. The original
message is acdbe. The encoding result is:
11111001101010
Decoding Method:
First, build a tree corresponding to the encoding rule.
Scan the codes from the beginning to the end; search the tree
until reaching leaves, and translate the substring into a letter.
Question
How to design a prefix code for a given alphabet (e.g., English letters) so that the expected code length for strings is minimum?
Answer
Question
How to design a prefix code for a given alphabet (e.g., English letters) so that the expected code length for strings is minimum?
Answer
Use Huffman code.
Huffman Code
Let w1 , w2 , . . . , wn are weights assigned to n elements. The
Huffman code can be constructed by
Assign the given weights, one each to a set S of n isolated
vertices. Each vertex is a root of complete binary tree of
height 0 with a weight assigned to it.
While |S| > 1
Choose two trees T and T 0 in S with the smallest two root
weights w and w 0 respectively.
Create a new tree T ∗ with root weight w ∗ = w + w 0 and
having T and T 0 as its left and right subtrees, respectively.
Place T ∗ into S and delete T and T 0 from S.
When |S| = 1, this complete binary tree is a tree
corresponding to Huffman code.
An Example of Constructing a Huffman Code
Suppose that letters a, b, c, d, e, f have weight 4, 7, 12, 17, 20, 28
respectively.
The huffman code of a, b, c, d, e, f is 1000, 1001, 101, 00, 01, 11
respectively.

CSC 10400 Discrete Mathematical Structures Lecture 12

Transcription

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