Hamiltonian cycle part

A cubic bipartite graph G is nice if any edge in G is nice. q p An edge is nice if (1) there exists a Hamiltonian path of G-{a,b} joining u to v for any u in {p,q} and any v in {r,s} and (2) There are 2 possible ordered pairs (u,v,w,x) from {(p,r,q,s), (p,s,q,r), (p,q,r,s)} such that there exist two disjoint spanning paths P and Q of G-{a,b} such that P joins from u to v and Q joins from w to x. A cubic bipartite graph G is nice if any edge in G is nice. Thus, any path of length 3 can be extended into a hamiltonian cycle.

All Two of three

An edge is super if (1) Any path of length 4 with (a,b) as the second edge can be extended into a Hamiltonian cycle (2) Let t be any neighbor of p in G-{a,r,s}. There are two partitions that divides {p,q,r,s} into two pairs {{u,v},{w,x}} such that exists two disjoint spanning paths P and Q including the edges (p,t) of G-{a,b} such that P joins u to v and Q joins w to x. A cubic bipartite graph G is super if any edge in G is super.

All Two of three

Lemma 1: A super graph is a nice graph Lemma 1: A super graph is a nice graph. A nice graph is 1 edge fault tolerant Hamiltonian. Proof: By definition, any super graph is a nice graph. By definition, any path of length 3 in a nice graph can be extended into a Hamiltonian cycle. Thus, any nice graph is 1-edge fault tolerant hamiltonian.

Theorem 1. Let G is a nice graph and H is a super graph Theorem 1. Let G is a nice graph and H is a super graph. Then the edge join of G and H is a nice graph. Proof of Theorem 1 need the following definition.

Definition: Let e and f be two edges of G Definition: Let e and f be two edges of G. We say a sequence of path e=e0, e1, e2,…, ek=f such that ei and ei+1 has a same a same endpoint for 0  I  k-1 as an edge path. The length of e=e0, e1, e2,…, ek=f is k. Then we define the distance between two edges e and f is defined as the shortest edge paths joins e to f.

Example of distance between edges 1 1 2 3 2 2 2 1 2 1 We will prove Theorem by edge distance between the joined edge. Three cases: distance at least 3, distance 1, and cross edges.

However, the proof is easy. Just using the standard technique. We need to prove the two properties of nice edge remain hold in the large graph. First, we assume that the distance between e and the joined edge is at least 3. However, the proof is easy. Just using the standard technique. Proof of Theorem 1. e e e

Again, both Properties 1 and 2 can be obtained by standard technique. Then the distance is 2. Again, both Properties 1 and 2 can be obtained by standard technique. Consider the solid purple edge As the dashed purple edge 2 2

Now cross edges. Nice Super Suppose that we want to prove the yellow path can be extended into a Hamiltonian cycle. First, extend the solid green path into a Hamiltonian cycle of the left graph. (The graph is nice) Second: extend the solid purple path into a Hamiltonian cycle for the right graph. (The graph is super)

Nice Super Check 2 path spanning paths First, Find the two spanning paths for the left graph (The graph is nice) Second: Find the two spanning paths for the right graph. (The graph is super) Pigeon Hole principle.

With Theorem 1, we know how to obtain a nice graph. However, we need a method to obtain a super graph. We need good edge join instead of edge join.

A good edge join of the graph G and H to be a graph K is defined as an edge join such that pi,pj’,ri,rj’ is a 4-cycle in K if a,b,ri,pi is a 4-cycle in G and a’,b’,rj’,pj’ is a 4-cycle in H for any i,j in {1,2}. I mean “both sides (in G and H) are in 4-cycle” remain “4-cycle” (in K). good bad

Theorem 2: The good edge join of any two super graphs is super.

Proof of Theorem 2

However, the proof is easy. Just using the standard technique. We need to prove the two properties of super edge remain hold in the large graph. First, we assume that the distance between e and the joined edge is at least 3. However, the proof is easy. Just using the standard technique. e

Again, both Properties 1 and 2 can be obtained by standard technique. Then the distance is 2. Again, both Properties 1 and 2 can be obtained by standard technique. 2

Cross edges

General idea First, Find a Hamiltonian cycle for the right graph by extend the green path. After the I/O edge of the right graph settled. Find a Hamiltonian cycle for the left graph by extend the yellow path.

Special case 1 Not A, B A B

Left: 2 spanning path (The graph is super.) Right: Extend the solid green path into a Hamiltonian cycle.

Special case 2 Both sides extend the solid brown path (of length 4) into a Hamiltonian cycle.

Now, we prove the are two path spanning paths We first consider general solution.

Again, solve the right part. After the I/O edge settled, solve the left part.

Special case: 1 Both sides use 2 path spanning Not A, B A’ A B B’

Special case 2 Not A, B A A’ B Left: 2 spanning paths Right: Hamiltonian cycle by extend the green dashed path

Special case 3: Not A, B A Left: 2 spanning paths B Right: Hamiltonian cycle by extended the green dashed path

Special case 4: C C’ Not B’ Not B A’ Not C’ A B’ B Left: 2 spanning paths Right: Hamiltonian cycle by extended the green dashed path

Special case 5: Key idea C C’ X E E’ The drawing of case 4 is rather complicate. I need to explain the location of X and Y more carefully. First, consider they are floating. Since D,B,F, E is a 4-cycle, D’,B’F’E’ is a 4-cycle to fit the definition of good edge join. Otherwise, A=B. Thus, Y=B’. Similarly, X=A. Therefore, the actual drawing is the lower graph. We will use this concept in the following cases. Y D’ D A A’ B’ F’ F B

Special case 5: Continue Just construct Both sides extend the purple path into Hamiltonian cycle.

Special case 6: C C’ Not B’ Not B A’ A B’ B C C’ Not B’ Not B Since it is a good edge join, (A,C) is an edge. Hamiltonian cycle extended by the purple path C C’ A’ A C B Not B C’ B’ Not B’ Not B’ Not B A’ A B’ B

Special case7: A’ A C B C’ B’ Must B and B’ for good edge join A’ A C Since it is a good edge join, (A,C) is an edge. Again, we have (B,C), (B’. C’). Actually, both sides are K 3,3.

New property: 1p fault tolerant Hamiltonian. Theorem 3. Let G be a nice graph with 1p fault tolerant hamiltonian and H be a super graph with 1p fault tolerant hamiltonian. Then the edge join of G and H is a nice graph K with 1p fault tolerant hamiltonian. Proof. By Theorem 1, K is a nice graph. We need to prove K is 1p fault tolerant hamiltonian.

Standard technique Both sides 1p fault tolerant hamiltonian

Now, we discuss the properties such that the edge twist of two cubic bipartite graph to be 1-fault tolerant hamiltonian. e a s r q p b Add new property: Call property A Property A: For any edge e in G is faulty, there are two disjoint spanning paths P and Q of G-{a,b}-e such that one of the following three cases holds P joins p to q and Q joins r to s. P joins p to r and Q joins q to s. P joins p to s and Q joins q to r. e a s r q p b e a s r q p b

Theorem 4. Let G and H are super graph with 1p fault tolerant hamiltonian and property A. Then the edge join of G and H is a nice graph K, 1p fault tolerant hamiltonian and property A. Proof. By theorem 1, K is nice. By Theorem 3, K is 1p fault tolerant hamiltonian. Thus, we need the property A

Case 2 Case 1 b a a b

Case 1: faulty edge distribution 1.3 1.1 1.2

a b e Standard technique. 1.1

1.2 a b e a b e or Left: property A Right: property A a b e similar

a b e 1.3 Left: property A. Right: hamiltonian cycle extend from a path of length 3 (purple).

Case 2: faulty edge distribution 2.2 2.3 2.1

2.1 a b e Left: property A. Right: super property: 2 path spanning (purple edge)

2.2 a b e Left: property A. Right: super property (purple edge)

2.3 a b e Both side, 1p fault tolerant hamiltonian. Actually, here I just use any adjacent pair of vertices is faulty.

Theorem 5. Let G and H are nice graph with 1p fault tolerant hamiltonian and property A. Then the edge twist of G and H is 1-fault tolerant Hamiltonian (non-bipartite graph). Edge twist

1-edge fault tolerant hamiltonian Case B Case C Case A

Case A e e Left: property A Right: the joined edge is nice

Case B and C e Both side extend a path of length 3 into a hamiltonian cycle.

1 vertex fault tolerant hamiltonian Left: 1p fault tolerant hamiltonian Right: extend the solid green path into a hamiltonian cycle

In the next slide, I will discuss the Hamiltonian laceable part. Please check this part first.