The "compact good array" in a "good $N^+$-cycle"

Question

We use $N^+$ to denote the set of positive integers. For any finite array $A:(a_1,b_1),...,(a_k,b_k)$, where every $(a_i,b_i)\in N^+\times N^+$, we call $A$ is good if for every $i\in \{a_1,b_1,...,a_k,b_k\}$, $i$ appears exactly two times in $a_1,b_1,...,a_k,b_k$. For example, $A_1:(1,1)$,$A_2:(1,2),(1,2)$ and $A_3:(1,2),(2,1)$ are all good, but $A_4:(1,2),(1,3)$ is not good. Moreover we call a good array $A$ is compact if there does not exist a proper sub-array of $A$ which is good. For example,$A:(1,2),(3,4),(2,1),(4,3)$ is good but not compact, because $(1,2),(2,1)$ is a proper sub-array of $A$ which is also good.

We define a good $N^+$-cycle $C=a_1$-$a_2$-...-$a_n$-$a_1$ is a "diamond necklace" such that every "diamond" $a_i$ is a positive integer and appears exactly two times in $a_1,a_2,...,a_n$. For example,$1$-$2$-$3$-$1$-$2$-$3$-$1$ is a good $N^+$-cycle. I conjecture that for every $N^+$-cycle $C=a_1$-$a_2$-...-$a_n$-$a_1$, there must exist $1\leq i_1<i_2<...i_k\leq n$ such that

$(1）i_2-i_1\geq 2,...,i_k-i_{k-1}\geq 2,i_1+n-i_k\geq 2$;

$(2)(a_{i_1},a_{i_2-1}),(a_{i_2},a_{i_3-1}）...,(a_{i_{k-1}},a_{i_k-1}),(a_{i_k},a_{i_1-1})$$($when $i_1=1$,let $a_{i_1-1}=a_0=a_n$$)$ is a compact good array.

$($Remark: just like cut the "diamond necklace" into $k$ paragraphs: $(a_{i_1}$-$a_{i_1+1}$-...-$a_{i_2-1})$-$(a_{i_2}$-$a_{i_2+1}$-...-$a_{i_3-1})$-...-$(a_{i_{k-1}}$-$a_{i_{k-1}+1}$-...-$a_{i_k-1})$-$(a_{i_k}$-$a_{i_k+1}$-...-$a_{i_1-1})$-$)$

Is it true?

Answer 1

The conjecture is right. It is equivalent to the claim below:

Let G be a simple graph which is a $2n$-cycle equipped with $n$ chords such that $G$ is $3$-regular, in other words, the set of the $n$ chords is a perfect matching of $G$ (that is, every vertex of $G$ is matched). Then there must exist at least two different $2n$-cycles in $G$.

The proof is given by Tony Huynh, see my other question: Does this graph contain at least two Hamiltonian cycles?.