User blog:Wythagoras/Graphs

This blog post is about planar, connected and simple variants of SCG.

This site is a great site for small simple graphs: graphclasses

SSCG(3)
PSSCG(3) = SSCG(3)

This is quite easy to prove. The first graph must be \(K_4\) for an optimal sequence. But \(K_4\) is minor of \(K_{3,3}\), therefore \(K_{3,3}\) cannot appear in the sequence for SSCG(3). \(K_5\) cannot appear by definition beacause it is not subcubic.

SSCG(4)
'''PSSCG(4) =? SSCG(4)'''

This one is harder. The following graphs are candidates for the first graph for an optimal sequence (others are not subcubic or are minor of these): The first four are all minor of \(K_{3,3}\) but fifth isn't(?). It seems unlikely that \(K_4\cup K_1\) is optimal as first graph, but I can't say PSSCG(4) = SSCG(4) for now, so giving formal proof of it remains unsolved problem.
 * \(\bar{P_2\cupP_3}\)
 * \(\bar{\text{chair}}\)
 * \(\bar{P_5}\)
 * \(K_{2,3}\)
 * \(K_4\cup K_1\)

SSCG(5) (earlier by Peng and Deedlit)
SSCG(5) >= PSSCG(6)+1

SSCG(6) (earlier by Peng and Deedlit)
SSCG(6) >= PSSCG(12)+6

SSCG(7) (improved)
Analysis:

1. K3,3 with one vertex on two different edges.

2. K3,3 with three vertices on an edge.

3. K3,3 with two vertices on an edge and one vertex connected to each of them.

4. K3,3 with two vertices on an edge and claw connected to one vertex

5. K3,3 with two vertices on an edge and 4-path connected to one vertex

6. K3,3 with two vertices on an edge and 3-path connected to one vertex and a stick

7, 8, 9. K3,3 with two vertices on an edge and 3-path connected to one vertex and 3,2,1 dots.

10. K3,3 with two vertices on an edge and 3-path connected to one vertex.

11. K3,3 with two vertices on an edge and 2-path connected to one vertex and PSSCG(8) sequence.

PSSCG(8)+10. K3,3 with two vertices on an edge and 2-path connected to one vertex.

PSSCG(8)+11. K3,3 with two vertices on an edge and 1-path connected to one vertex and PSSCG(PSSCG(8)+9) sequence.

PSSCG(PSSCG(8)+9)+PSSCG(8)+10. K3,3 with two vertices on an edge and 1-path connected to one vertex.

PSSCG(PSSCG(8)+9)+PSSCG(8)+11. K3,3 with two vertices on an edge and and PSSCG(x) sequence.

PSSCG(PSSCG(PSSCG(8)+9)+PSSCG(8)+10)+PSSCG(PSSCG(8)+9)+PSSCG(8)+10. K3,3 with two vertices on an edge.

Let the number above be X, then X > PSSCG3(8).

We know that, in Hyp Cos' system, K3,3 is at \(\vartheta(\Omega_\omega)\) level, beacause \(\vartheta(\Omega_\omega)\) is limit of PSSCG (isn't it?) So we can draw some very large graph which level is about \(\vartheta(\Omega_{\text{PSSCG}^3(8)}))\). Then using Hyp Cos' method we can reach \(f_{\theta(\Omega_\omega,\Omega_{\text{PSSCG}^3(8)})}(n)\) for very large n as an lower bound for SSCG(7).