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co.combinatorics – Is each path with this property shorter than one other path with the equivalent endpoints? retort

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co.combinatorics – Is each path with this property shorter than one other path with the equivalent endpoints?

Disclaimer: this isn’t a proof however quite some concepts which can or might not help.

If we establish the pairs $v_i, v_j$ such that there is $u$ with $v_iuv_j in E$ with a phase $[i, j]$, we’re going to maintain at the least $frac{n+1}{2}$ segments.

Now we will quarrel equally to the Vitali masking lemma. We maintain a set of segments, masking $1, ldots, n$ (let’s occupy we weight ranging from $1$ quite then from $0$). We can discover two households of segments ${ I }$ and ${ J }$ such that

  1. Segments in every household are both disjoint or nested
  2. There are disjoint subfamilies $tilde I$ and $tilde J$ such that every phase intersect at most two neighbouring.

Now at the least one of many households has touchstone at the least $frac{n+1}{2}$. occupy wlog that
$$
| cup I_j | geq | cup J_j |
$$

If we bid the segments in $tilde I$ and $tilde J$ in order that $J_k$ intersects precisely two neighbours say $I_{l-1}$ and $I_{l}$ and $|J_k cap I_{l-1}| + |J_k cap I_{l}| = |J_k|$, then it will suffice to stroll trough the nested segments in $I_{l-1}$, then “jump” utilizing $J_k$ to $I_l$ and stroll by means of the nested household in $I_l$.

The technique above assumes that every one $u$‘s are patent.

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