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

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MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12. 2.2 Hamilton Circuits. 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: ____________ ____________ - PowerPoint PPT Presentation

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Page 1: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

MATH 310, FALL 2003(Combinatorial Problem

Solving)Lecture 6, Friday, September 12

Page 2: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

2.2 Hamilton Circuits 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.

On Monday you will also turn in the list of all new terms On Monday you will also turn in the list of all new terms with the following marks with the following marks

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

Page 3: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

Hamilton Circuits and Paths 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.

Page 4: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

The Three Rules 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.

Page 5: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

Example 1 Show that the

graph on the left has no Hamilton circuit.

Hint: Apply Rule 1 four times.

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

Example 2 Show that the

graph on the left has no Hamilton circuit.

Hint: Apply Rule 1 twice and use symmetry.

Page 7: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

Example 3 Show that the

graph on the left has no Hamilton circuit.

Hint: Apply Rule 1 twice and use symmetry.

click

Page 8: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

A Useful Result (not from the textbook)

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

Page 9: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

Theorem 1 (Dirac, 1952) A graph with n vertices, n > 2, has a

Hamilton circuit if the degree of each vertex is at least n/2.

Page 10: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

Theorem 2 (Chvatal, 1972) 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.

Page 11: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

Theorem 3 (Grinberg, 1968) 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

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

Example 4 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.

r4 r’4 ) r6 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.

Page 13: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

Tournament A tournament is a

directed graph obtained from a complete graph by giving a direction to each edge.

Page 14: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

Theorem 4 Every tournament has a (directed)

Hamilton path. Proof. By induction on the number of

vertices.

Page 15: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

Example 5: Gray Code 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.

Page 16: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

Hypercube The graph 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

Page 17: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

4-dimensional Cube.

0000

1000

0010

0100

0001

1100

1110

0110

01110011

1001 1101

11111011

1010

Page 18: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

4-dimensional Cube and a Famous Painting by Salvador Dali

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.

Page 19: MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 6, Friday, September 12

Gray Code - Revisited A Hamilton path in

the hypercube produces a Gray code.

000 001

011010

110

100 101

111