Polyominoes with maximally many holes
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Type: 
Papers
Publisher: 

Geombinatorics

Accepted: 
Aug 2018
Publication date: 
2019
Co-author: 
Matthew Kahle
Érika Roldán
Abstract: 
What is the maximum number of holes that a polyomino with $n$ tiles can enclose? Call this number $f(n)$. We show that if $n_k = \left( 2^{2k+1} + 3 \cdot 2^{k+1}+4 \right) / 3$ and $h_k = \left( 2^{2k}-1 \right) /3$, then $f(n_k) = h_k$ for $k \ge 1$. We also give nearly matching upper and lower bounds for all large $n$, showing as a corollary that $f(n) = \left(1/2- o(1) \right) n/2 $.
Notes: 
(to appear)
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