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Eighty-three sublattices and planarity

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abstract

Let $L$ be a finite $n$-element lattice. We prove that if $L$ has at least $83\cdot 2^{n-8}$ sublattices, then $L$ is planar. For $n>8$, this result is sharp since there is a non-planar lattice with exactly $83\cdot 2^{n-8}-1$ sublattices.

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math.RA 1

years

2019 1

verdicts

UNVERDICTED 1

representative citing papers

One hundred twenty-seven subsemilattices and planarity

math.RA · 2019-06-28 · unverdicted · novelty 6.0

A finite n-element semilattice is planar if it has at least 127 * 2^(n-8) subsemilattices, and this bound is sharp for n > 8 via an explicit non-planar counterexample with one fewer subsemilattice.

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  • One hundred twenty-seven subsemilattices and planarity math.RA · 2019-06-28 · unverdicted · none · ref 6 · internal anchor

    A finite n-element semilattice is planar if it has at least 127 * 2^(n-8) subsemilattices, and this bound is sharp for n > 8 via an explicit non-planar counterexample with one fewer subsemilattice.