Update: August 2025

    Much like last year, I spent most of this month busy with things other than puzzles. However, I did still manage to finish the writing project, which I'm feeling proud of despite getting the impression that it's very difficult to deduce a majority of the answers. I also just found out about the Logic Pad online preview branch today, so I've ported all the puzzles from the post there except for the Smores. Progress is going slowly on the second part of my latest puzzle project. At least I've tried to slice it so I don't bite off more than I can chew.

    I'll be planning for October soon. While last year I checked the list a week beforehand out of sudden inspiration, I plan to check it this year as soon as it drops because I expect to be busier and I only barely put out Navigator on time. Hopefully this will also give me more time to come up with a fitting meta since the prompts might not be themed this year.

    This year, I've been prioritizing making ambitious puzzles, but I've been learning that there's not a strong positive correlation between how ambitious a puzzle is and how well it's received. It's very easy for an ambitious puzzle to turn out way too hard or too long and intimidate and discourage lots of potential solvers. I prefer to solve puzzles for the journey, but it's less satisfying when potential progress doesn't feel like it's getting closer to the destination. Maybe I'm just too impatient. (Erratum: I was just too impatient. Things take time.)

    The following crossword was made on 8/19/25, inspired by the suggestion of a nontrivial ∞x∞ crossword. It is diagramless and should be solved with an arbitrarily large solution. There are definitely many solutions due to the open-ended nature of the puzzle.

All Across: This may have been considered a virtually unexpendable source of a classical element of alchemy

All Down: Is this clue number prime?

If you're looking for the intended answer, the solution that I used when constructing this has a size that can be approximated by (2n-2) x (n/2+1.75), where n is the number of entries.