I’m pretty sure this is difficult. It’s been a long time since my CS classes, so I think i’ve gotta solve it the old fashioned way. Thanks for this im high as hell and laughing atm

I’m pretty sure covering all green tiles is impossible, but I’d love to see someone prove it either way.

Some interesting things to note about the source code:
-It has references to a “freshice” entity which is never used
-There is no win condition (although I’m not sure how you’d construct one for this in PuzzleScript)

Consider a vertical corridor, which is 11 squares long. Number the squares from top to bottom 1,2,…,11.

Every time you fill an odd-numbered square, you must also fill an even-numbered square in order to get out of that corridor. (Only even-numbered squares are “connected to the outside”.)

Since there are 6 odd-numbered squares and only 5 even-numbered ones, and it is impossible to “reuse” squares, it follows that it is impossible to finish even ONE vertical corridor without being stuck in it forever.

## 3 Comments

I’m pretty sure this is difficult. It’s been a long time since my CS classes, so I think i’ve gotta solve it the old fashioned way. Thanks for this im high as hell and laughing atm

I’m pretty sure covering all green tiles is impossible, but I’d love to see someone prove it either way.

Some interesting things to note about the source code:

-It has references to a “freshice” entity which is never used

-There is no win condition (although I’m not sure how you’d construct one for this in PuzzleScript)

Consider a vertical corridor, which is 11 squares long. Number the squares from top to bottom 1,2,…,11.

Every time you fill an odd-numbered square, you must also fill an even-numbered square in order to get out of that corridor. (Only even-numbered squares are “connected to the outside”.)

Since there are 6 odd-numbered squares and only 5 even-numbered ones, and it is impossible to “reuse” squares, it follows that it is impossible to finish even ONE vertical corridor without being stuck in it forever.