Polyomino convolutions and tiling problems

We define a convolution operation on the set of polyominoes and use it to obtain a criterion for a given polyomino not to tile the plane (rotations and translations allowed). We apply the criterion to several families of polyominoes, and show that the criterion detects some cases that are not detectable by generalized coloring arguments.Comment: 8 pages, 8 figures. To appear in \emph{J. of Combin. Theory Ser. A

Enumeration of tilings of diamonds and hexagons with defects

We show how to count tilings of Aztec diamonds and hexagons with defects using determinants. In several cases these determinants can be evaluated in closed form. In particular, we obtain solutions to problems 1, 2, and 10 in James Propp's list of problems on enumeration of matchings

Who witnesses The Witness? Finding witnesses in The Witness is hard and sometimes impossible

We analyze the computational complexity of the many types of pencil-and-paper-style puzzles featured in the 2016 puzzle video game The Witness. In all puzzles, the goal is to draw a simple path in a rectangular grid graph from a start vertex to a destination vertex. The different puzzle types place different constraints on the path: preventing some edges from being visited (broken edges); forcing some edges or vertices to be visited (hexagons); forcing some cells to have certain numbers of incident path edges (triangles); or forcing the regions formed by the path to be partially monochromatic (squares), have exactly two special cells (stars), or be singly covered by given shapes (polyominoes) and/or negatively counting shapes (antipolyominoes). We show that any one of these clue types (except the first) is enough to make path finding NP-complete ("witnesses exist but are hard to find"), even for rectangular boards. Furthermore, we show that a final clue type (antibody), which necessarily "cancels" the effect of another clue in the same region, makes path finding $\Sigma_2$-complete ("witnesses do not exist"), even with a single antibody (combined with many anti/polyominoes), and the problem gets no harder with many antibodies. On the positive side, we give a polynomial-time algorithm for monomino clues, by reducing to hexagon clues on the boundary of the puzzle, even in the presence of broken edges, and solving "subset Hamiltonian path" for terminals on the boundary of an embedded planar graph in polynomial time.Comment: 72 pages, 59 figures. Revised proof of Lemma 3.5. A short version of this paper appeared at the 9th International Conference on Fun with Algorithms (FUN 2018