On Subdivision Graphs which are 2-steps Hyperhamiltonian graphs and Hereditary non 2-steps Hamiltonian graphs Sin-Min Lee, San Jose State University Hsin-hao Su*, Stonehill College Yung-Chin Wang, Tzu-Hui Institute of Technology 46th SICCGTC at Florida Atlantic University March 4, 2015

Knight's Tour Problem A knight's tour is a sequence of moves of a knight on a chessboard such that the knight visits every square only once. If the knight ends on a square that is one knight's move from the beginning square (so that it could tour the board again immediately, following the same path), the tour is closed, otherwise it is open.

Knight's Tour Problem

Hamiltonian Graphs A path in a graph G that includes every vertex in G is called a Hamiltonian path of G. A cycle in a graph G that includes every vertex in G is called a Hamiltonian cycle of G. If a graph G contains a Hamiltonian cycle, then G is called a Hamiltonian graph.

Hamiltonian Graphs

AL(k)-traversal Graphs Definition: For k ≥ 2, a (p,q)- graph G = (V,E) is said to have k-step traversal if its vertex set can be arranged as {v 1,v 2,…,v p } such that for each i =1,2,…,p-1, the distance for v i and v i+1 is equal to k. A graph admits a k-step traversal is called the AL(k)-traversal graph.

Example: AL(2)-traversal Note that it is not AL(3)-traversal.

k-steps Hamiltonian Graphs Definition: We call a graph k-steps Hamiltonian graph if it has a AL(k)- traversal in G and d(v p,v 1 ) = k. Definition: We call a graph almost k- Hamiltonian graph if it has a AL(k)- traversal in G and d(v p,v 1 ) = k ± 1.

Ex.: 2-steps Hamiltonian

Alexander Nien-Tsu Lee [1] S-M. Lee, A. Nien-Tsu Lee, On Super Edge-Magic Graphs with Many Odd Cycles, Congressus Numeratium 163 (2003),

Known Results Proposition: The cycle C n is k-Hamiltonian if and only if gcd(n,k)=1. Proposition: The graph G is k-steps Hamiltonian if and only if its distance k- graph is Hamiltonian. Proposition: A bipartite graph is not AL(2)- traversal. Thus, it is not 2-steps Hamiltonian.

Bipartite Graphs

k-steps Hyperhamiltonian Definition: A graph G is said to be k-steps hyperhamiltonian if it is k-steps Hamiltonian and for any v in G, the vertex-deleted subgraph G-{v} is also k- steps Hamiltonian.

The Möbius ladder M 8 is 2- steps hyperhamiltonian

Hereditary non k-steps Hamiltonian Definition: A graph property is called hereditary if it is closed with respect to deleting vertices. Definition: A graph G is said to be hereditary non k-steps Hamiltonian if it is not k-steps Hamiltonian and for any v in G, the vertex-deleted subgraph G-{v} is also not k-steps Hamiltonian.

Subdivision of a Graph Definition: Let G be a graph with an edge set E(G). Let S be a subset of E(G) and f be a function from S to the set of natural numbers. The subdivision graph Sub(G,S,f) is constructed from G by inserting f(e)=m vertices into any edge e in S.

Subdivision of a Graph

Subdivision Prism Graphs Theorem: For any k ≥ 2, the subdivision graph Sub(C 2k × K 2,{(u 1,u 2k )},f((u 1,u 2k )) =h) is 2-steps Hamiltonian if and only if h is odd.

Subdivision Prism Graphs Theorem: For any k > 2, the subdivision graph Sub(C 2k+1 × K 2,{(u 1,u 2k )},f((u 1,u 2k )) =h) is 2-steps Hamiltonian if and only if h is even.

Subdivision Wheel Graphs Theorem: The graph Sub(W(n),X,f) is 2-steps Hamiltonian if and only if one of the following conditions is satisfied: n is even and h 1 +h 2 +…+h n-1 is even; n is odd and h 1 +h 2 +…+h n-1 is odd. There is a h i for some i which equals to 2.

Subdivision W(4)

Subdivision Wheel Graphs Proof: When all h i are odd, the graph Sub(W(n),X,f) is bipartite. If n is even and h 1 +h 2 +…+h n-1 is even, then h 1 +h 2 +…+h n-1 +(n-1) is odd. The labeling c, v 11, …(around the cycle to go through every vertex in the cycle), v n-1,h_{n- 1,h_i}, c is a 2-steps Hamiltonian cycle.

Subdivision Wheel Graphs We assume that there is an h i which is equal to 2, w.l.o.g., we can assume that h 1 =2. We would start labeling from c and then v 11 following by v n-1,h_{n-1,h_i} and travel counterclockwise through the cycle by every other vertices until reaching v 2. It is possible because the number of vertices in the cycle part is even and h 1 =2.

Subdivision Wheel Graphs After that, we continue the labeling by jumping to v 1 following by v n-1,h_{n-1,h_i-1} (or v n if h n-1 =1) and travel counterclockwise through the cycle by every other vertices until reaching v 12. Again, it is possible because the number of vertices in the cycle part is even and h 1 =2. Since we have labeled every vertex and the distance between v 12 and c is 2, the graph Sub(W(n),X,f) is 2- steps Hamiltonian.

Shorten a Path Lemma: A graph G with a subgraph P of a path of length 5 or more is 2-steps Hamiltonian if and only if the induced graph H from G by removing two middle vertices from the path P is 2-steps Hamiltonian.

Subdivision Wheel Graphs Finally, there is only one case left, that is, all even h i are greater or equal to 4. By Lemma, we know that if we insert 6 vertices or more on an edge then we can remove even vertices to keep the 2-steps Hamiltoniancy. Thus, we only need to consider then all even h i are equal to 4. With a pair of adjacent h i and h i+1 to be 4, its D 2 graph would have a cycle with two paths of length 4 that have only 2 order 2 vertices in the middle.

Subdivision graphs with perfect matching The four perfect matchings, up to isomorphic, of the C 4 ×K 2 are illustrated below:

Subdivision graphs with perfect matching Theorem: If f(e) = k for any e in P 1, then Sub(C 4 ×K 2, P 1,f: P 1  N) is not 2-steps Hamiltonian for any k. Corollary: Let P be the perfect matching in C 2n ×K 2 containing edges between v i and w i for all i. If f(e) = k for any e in P, then Sub(C 2n ×K 2, P,f: P  N) is not 2-steps Hamiltonian for any k.

Subdivision graphs with perfect matching Theorem: If f(e) = k for any e in P 3, then Sub(C 4 ×K 2, P 3,f: P 3  N) is not 2-steps Hamiltonian for any k. Corollary: Let P be the perfect matching in C 2n ×K 2 containing edges between v 2i-1 and v 2i and between w 2i-1 and w 2i for all 1 ≤ i ≤ n. If f(e) = k for any e in P, then Sub(C 2n ×K 2, P,f: P  N) is not 2-steps Hamiltonian for any k.

Subdivision graphs with perfect matching Theorem: Let P be a perfect matching in C n ×K 2. For any k ≥ 3, if f(e) = k for any e in P, then Sub(C n ×K 2, P,f: P  N) is not 2-steps Hamiltonian.

Subdivision graphs with perfect matching Proof: For any edge in the perfect matching, to reach the vertices in the middle of the path, we need to travel from and through two end vertices on the cycle part. Therefore, all v i and w i where 1 ≤ i ≤ n would be visited in any 2-steps Hamiltonian labeling. But, at the same time, since there are only three distance 2 vertices to any u i1 are u i3, v i+1 and v i-1, to label u i1, we need to come from or go through v i+1 or v i-1.

Subdivision graphs with perfect matching Similarly, since there are only three distance 2 vertices to any u ik are u i,k-2, w i+1 and w i-1, to label u ik, we need to come from or go through w i+1 or w i-1. Thus, since there are 2n vertices in this kind of position, we would visit all v i and w i 2n times. Totally, we would visit all v i and w i 4n times. It is impossible.

Subdivision graphs with perfect matching Theorem: If f(e) = k for any e in P 2, then Sub(C 4 ×K 2, P 2,f: P 2  N) is 2-steps if and only if k is 1.

Subdivision graphs with perfect matching Theorem: If f(e) = k for any e in P 4, then Sub(C 4 ×K 2, P 4,f: P 4  N) is 2-steps if and only if k is 1.

Thank you!