Skip to main content
Computer Science

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

Required fields*

19
  • $\begingroup$ This is not enough. Your constraints do not stop the solver from setting $x_l$ to true everywhere (apart from the origin). Same goes for $y_l$ - the solver can set values that the boundary is everywhere. You can't use material implication just like that. You need to use equivalence (or sums). Or you need to add constraint that a field is either internal or external. $\endgroup$ Commented Aug 27, 2022 at 7:12
  • $\begingroup$ @user1543037, That doesn't seem right to me. The last constraint ensures that $x_0$ is false (at the origin). Then, the first constraint (about adjacent locations) implies that $x_\ell$ is false at the locations that are immediately adjacent to the origin, and so on, until you hit the boundary -- so the combination of the last constraint and the first constraint imply that $x_\ell$ is false everywhere outside the enclosed region. Am I missing something? $\endgroup$ Commented Aug 27, 2022 at 7:15
  • $\begingroup$ When $x_0$ is zero, then the left side of the implication is zero, so right side can be anything. $\endgroup$ Commented Aug 27, 2022 at 7:17
  • $\begingroup$ Same issue applies to the boundary. You add a constraint which makes sure that there is boundary formed on polyominoes. But there is no constraint making sure that there is no boundary outside of polyominoes. $\endgroup$ Commented Aug 27, 2022 at 7:18
  • $\begingroup$ @user1543037, I don't think that's right. Could you take another look? If $\ell$ is adjacent to the origin and the origin is not part of the boundary, consider the case $\ell'=0$; then the constraint $x_\ell \land \neg y_{\ell'} \implies x_{\ell'}$ (i.e., $x_\ell \land \neg y_0 \implies x_0$) combined with $x_0=\text{False}$ and $y_0=\text{False}$ implies $x_\ell=\text{False}$. This continues onward, implying that every cell in the exterior must have $x_\ell$ be false. Do you see something I'm missing? $\endgroup$ Commented Aug 27, 2022 at 7:23