1. 2/25/2016Tucker, Sec. 2.21 Applied Combinatorics, 4 th Ed. Alan Tucker Section 2.2 Hamilton Circuits Prepared by: Nathan Rounds and David Miller

2. 2/25/2016Tucker, Sec. 2.22 Definitions Hamilton Path – A path that visits each vertex in a graph exactly once. B C D E F Possible Hamilton Path: A-F-E-D-B-C AB C D E F

3. 2/25/2016Tucker, Sec. 2.23 Definitions Hamilton Circuit – A circuit that visits each vertex in a graph exactly once. Possible Hamilton Circuit: A-F-E-D-C-B-A A B C D E F

4. 2/25/2016Tucker, Sec. 2.24 Rule 1 If a vertex x has degree 2, both of the edges incident to x must be part of any Hamilton Circuit. A B C D E F Edges FE and ED must be included in a Hamilton Circuit if one exists.

5. 2/25/2016Tucker, Sec. 2.25 Rule 2 No proper subcircuit, that is, a circuit not containing all vertices, can be formed when building a Hamilton Circuit. A B C D E F Edges FE, FD, and DE cannot all be used in a Hamilton Circuit.

6. 2/25/2016Tucker, Sec. 2.26 Rule 3 Once the Hamilton Circuit is required to use two edges at a vertex x, all other (unused) edges incident at x can be deleted. A B C D E F If edges FA and FE are required in a Hamilton Circuit, then edge FD can be deleted in the circuit building process.

7. 2/25/2016Tucker, Sec. 2.27 Example Using rules to determine if either a Hamilton Path or a Hamilton Circuit exists. A D E G F I H C B JK

8. 2/25/2016Tucker, Sec. 2.28 Using Rules Rule 1 tells us that the red edges must be used in any Hamilton Circuit. H A D E G F I C B J K Vertices A and G are the only vertices of degree 2.

9. 2/25/2016Tucker, Sec. 2.29 Using Rules Rules 3 and 1 advance the building of our Hamilton Circuit. A D E G F I H C B JK Since the graph is symmetrical, it doesn’t matter whether we use edge IJ or edge IK. If we choose IJ, Rule 3 lets us eliminate IK making K a vertex of degree 2. By Rule 1 we must use HK and JK.

10. 2/25/2016Tucker, Sec. 2.210 Using Rules All the rules advance the building of our Hamilton Circuit. A D E G F I C B J H K Rule 2 allows us to eliminate edge EH and Rule 3 allows us to eliminate FJ. Now, according to Rule 1, we must use edges BF, FE, and CH.

11. 2/25/2016Tucker, Sec. 2.211 Using Rules Rule 2 tells us that no Hamilton Circuit exists. A D E G F I C B J Since the circuit A-C-H-K-J-I-G-E-F-B-A that we were forced to form does not include every vertex (missing D ), it is a subcircuit. This violates Rule 2. H K

12. 2/25/2016Tucker, Sec. 2.212 Theorem 1 A graph with n vertices, n > 2, has a Hamilton circuit if the degree of each vertex is at least n/2. A B C D E F n = 6 n/2 = 3 Possible Hamilton Circuit: A- B-E-D-C-F-A

13. 2/25/2016Tucker, Sec. 2.213 However, not “if and only if” E B C D F AB C D F Theorem 1 does not necessarily have to be true in order for a Hamilton Circuit to exist. Here, each vertex is of degree 2 which is less than n/2 and yet a Hamilton Circuit still exists.

14. 2/25/2016Tucker, Sec. 2.214 Theorem 2 Let G be a connected graph with n vertices, and let the vertices be indexed x 1,x 2,…,x n, so that deg(x i ) deg(x i+1 ). If for each k n/2, either deg(x k ) > k or deg(x n-k ) n-k, then G has a Hamilton Circuit. n/2 = 3 k = 3,2,or 1 Possible Hamilton Circuit: X 1 -X 5 -X 3 -X 4 -X 2 -X 6 -X 1 X5X5 X1X1 X6X6 X3X3 X4X4 X2X2

15. 2/25/2016Tucker, Sec. 2.215 Theorem 3 Suppose a planar graph G, has a Hamilton Circuit H. Let G be drawn with any planar depiction. Let r i denote the number of regions inside the Hamilton Circuit bounded by i edges in this depiction. Let be the number of regions outside the circuit bounded by i edges. Then numbers r i and satisfy the following equation.

16. 2/25/2016Tucker, Sec. 2.216 Use of Theorem 3 4 44 66 6 6 6 6 Planar Graph G No matter where a Hamilton Circuit is drawn (if it exists), we know that and. Therefore, and must have the same parity and.

17. 2/25/2016Tucker, Sec. 2.217 Use of Theorem 3 Cont’d Eq. (*) Consider the case. This is impossible since then the equation would require that which is impossible since. We now know that, and therefore. Now we cannot satisfy Eq. (*) because regardless of what possible value is taken on by, it cannot compensate for the other term to make the equation equal zero. Therefore, no Hamilton Circuit can exist.

18. 2/25/2016Tucker, Sec. 2.218 Theorem 4 Every tournament has a directed Hamilton Path. Tournament – A directed graph obtained from a (undirected) complete graph, by giving a direction to each edge. AB C D The tournaments (Hamilton Paths) in this graph are: A-D-B-C, B-C-A-D, C-A-D-B, D-B-C-A, and D-C-A-B. (K 4, with arrows)

19. 2/25/2016Tucker, Sec. 2.219 Definition Grey Code uses binary sequences that are almost the same, differing in just one position for consecutive numbers. A=000 B=100 C=110 D=010 F=011 G=111 H=101 I=001 Advantages for using Grey Code: -Very useful when plotting positions in space. -Helps navigate the Hamilton Circuit code. Example of an Hamilton Circuit: 000-100-110-010-011-111-101-001-000

20. 2/25/2016Tucker, Sec. 2.220 Class Exercise Find a Hamilton Circuit, or prove that one doesn’t exist. ABC DE FGH Rule’s: If a vertex x has degree 2, both of the edges incident to x must be part of any Hamilton Circuit. No proper subcircuit, that is, a circuit not containing all vertices, can be formed when building a Hamilton Circuit. Once the Hamilton Circuit is required to use two edges at a vertex x, all other (unused) edges incident at x can be deleted.

21. 2/25/2016Tucker, Sec. 2.221 Solution By Rule One, the red edges must be used Since the red edges form subcircuits, Rule Two tells us that no Hamilton Circuits can exist. ABC DE FGH