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MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

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MATH 310, FALL 2003(Combinatorial Problem Solving)Lecture 6, Friday, September 12

- Homework (MATH 310#2F):
- Read 2.3. Write down a list of all newly introduced terms (printed in boldface or italic)
- Do Exercises 2.2: 2,4a-g,16,20
- Volunteers:
- ____________
- ____________
- Problem: 16.

- + if you think you do not need the definition on your cheat sheet,
- check (if you need just the term as a reminder),
- - if you need more than just the definition to understand the term.

- A ciruit that visits every vertex of a graph is called a Hamilton circuit.
- A path that visits every vertex of a graph is called a Hamilton path.

- Rule1. If a vertex x has degree 2, both edges incident to x must be part of any Hamilton circuit.
- Rule 2. No proper subcircuit can be formed when building a Hamilton circuit.
- Rule 3. Once the Hamilton circuit is required to use two edges at a vertex x, all other edges incident to x must be removed from consideration.

- Show that the graph on the left has no Hamilton circuit.
- Hint: Apply Rule 1 four times.

- Show that the graph on the left has no Hamilton circuit.
- Hint: Apply Rule 1 twice and use symmetry.

- Show that the graph on the left has no Hamilton circuit.
- Hint: Apply Rule 1 twice and use symmetry.

click

- If a connected graph G contains k vertices whose removal disconnects G into more than k pieces, then G has no Hamilton circuit.

- A graph with n vertices, n > 2, has a Hamilton circuit if the degree of each vertex is at least n/2.

- Let G be a connected graph with n vertices: x1, x2, ..., xn, so that deg(x1) · deg(x2) · ... · deg(xn). If for each k · n/2, either deg(xk) > k or deg(xn-k) ¸ n – k, then G has a Hamilton circuit.

- Suppose a plane graph G has a Hamilton circuit H. Let ri denote the number of regions inside the Hamilton circuit bounded by i edges. Let r’i be the number of regions outside the circuit bounded by i edges. Then the numbers ri and r’i satisfy the equation:
- (3 - 2)(r3 – r’3) + (4 - 2)(r4 – r’4) + (5 - 2)(r5 – r’5) + ... = 0

- Here we have:
- (a) r4 + r’4 = 3
- (b) r6 + r’6 = 6

- By Grinberg we should also have:
- (c) 2(r4 – r’4) + 4(r6 – r’6) = 0.

- r4¹ r’4) r6¹ r’6.
- Hence |r6 – r’6| ¸ 2.
- |r6 – r’6| ¸ 2 \implies |r4 – r’4| ¸ 4.
- Contradiction! The graph in Figure 2.8 has no Hamilton circuit.

- A tournament is a directed graph obtained from a complete graph by giving a direction to each edge.

- Every tournament has a (directed) Hamilton path.
- Proof. By induction on the number of vertices.

Example: n = 3. There are 8 binary sequences:

000

001

010

011

100

101

110

111

There are 2n binary sequences of length n.

An ordering of 2n binary sequences with the property that any two consecutive elements differ in exactly one position is called a Gray code.

- Thegraph with one vertex for each n-digit binary sequence and an edge joining vertices that correspond to sequences that differ in just one position is called an n-dimensional cube or hypercube.
- v = 2n
- e = n 2n-1

0110

0010

0111

1110

0011

1010

1011

1111

0001

1101

1001

0000

0100

1100

1000

- Salvador Dali (1904 – 1998) produced in 1954 the Crucifixion (Metropolitan Museum of Art, New York) in which the cross is a 3-dimensional net of a 4-dimensional hypercube.

010

011

- A Hamilton path in the hypercube produces a Gray code.

111

110

100

101

001

000