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dev diary: picross on ur phone!

🕑 Added 2023-08-07 19:34:22 +0000 UTC
dev diary: picross on ur phone!

Comments

Blackberry Mochi

Laaaate but as always, 'tis a joy looking through code I cannot actually comprehend but can vaguely recognize the makeup/functionality/process of and going "yeah!!! That makes sense!!! In a conceptual way!!!" Although, admittedly, the coding here does seem the most confounding of the bits you've shared, perhaps because of it being so explicitly technical, pff (what the frick does a double asterisk even mean in an equation???) Still fun though!!!!! But, regardless of the joys of code viewing, CONGRATULATIONS ON OVERCOMING SUCH AN OBSTACLE AND ALLOWING THE PHONE-ONLY MASSES TO ENJOY THESE CHARM STUDIES!!! Hopefully lag fixes can be made later, but, for now, it's just nice that it was done!!! (I do really really like the basic math in the bottom right of the '"scribbles for visualization" bit, tho. Sometimes we need to steadfastly remember: 25 - 24 = 1, 24 - 24 = 0, and 0 + 6 - 1....)

Harry Tsang

For the edges, you can think of the rows and columns as separate indices tileno = row * gridsize + col conversely row = tileno // gridsize # integer division that throws away the remainder col = tileno % gridsize # modulo that only get the remainder For looping back, you can use modulo (the % operator) too so you don't even need if-else imagebutton auto "leftarrow_%s": row = tileno // gridsize col = tileno % gridsize row = (row + gridsize - 1) % gridsize # need to add gridsize so that it doesn't go negative action SetVariable("tileno", row * gridsize + col) so edge detection is simply finding the row and col and comparing it to the row numbers def is_leftedge(tileno): row = tileno // gridsize return row == 0 def is_rightedge(tileno): row = tileno // gridsize return row == (gridsize - 1)


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