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.
Lattices with many congruences are planar
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abstract
Let $L$ be an $n$-element finite lattice. We prove that if $L$ has strictly more than $2^{n-5}$ congruences, then $L$ is planar. This result is sharp, since for each natural number $n\geq 8$, there exists a non-planar lattice with exactly $2^{n-5}$ congruences.
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math.RA 1years
2019 1verdicts
UNVERDICTED 1representative citing papers
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One hundred twenty-seven subsemilattices and planarity
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.