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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