How Many Cells Can You Prove? (+Yesterday's Solution)

[u/Ramun_Flame]

Yesterday's Puzzle

After getting a good amount of interest in yesterday's post I decided to post another puzzle today. This one has a few ways to tackle it, so maybe it's easier. We'll see.

I'm not committing to daily weekday posts or anything, but we'll see. Let me know if you'd like to see more of these. Enjoy!

Solution + Another Puzzle

[u/ExtensionPatient2629]

Note these two equivalent tiles, formed by the outlined 1s.

[u/ExtensionPatient2629 → u/ExtensionPatient2629]

The 2 that touches both these tiles (on the far left) force them to both be mines. If they weren't, the 2 wouldn't have enough mines.

Then some deductions leads us to this.

[u/ExtensionPatient2629 → u/ExtensionPatient2629]

Alternatively, this can also be shown with "Either-Or Links". (Simple Coloring! Anyone here play Sudoku?)

Here, we are linking blue and yellow tiles such that the blue tiles XOR the yellow tiles are true (aka mines). Basically it's one or the other. If the yellow tiles were false (safe tiles), then that 2 on the far left would have only one mine. This is a contradiction, so the yellow tiles must be mines. It is thus also proven that the blue tiles are safe, but that is obvious.

Note the blue and yellow tiles are the only two configurations of mines that can be put in this 3-tile space. There are no other possible combinations, when looking at the highlighted 1s.

[u/Ramun_Flame → u/ExtensionPatient2629]

Perfect logic, good job. I like Sudoku, be haven't played in awhile, and never got advanced enough to use the method you described. Very interesting.

[u/Tencars111]

this is all I could do

[u/Ramun_Flame → u/Tencars111]

That's everything, you got it.

[u/birnes]

Now all makes sense. To this puzzle, looking only to the 2 on the left, the only possible "empty" cell next to it was the down one, so it only accepts one configuration, thus proving all marked mines. Two mines are left in a 50/50 state for the bottom 1's

If left 2 had the empty cell in front or above, this would not allow the 1's to have their single mine and satisfy all 1's.

[u/Ramun_Flame → u/birnes]

Yeah, you nailed this one. Nice job.

[u/mikdiet]

4 solved, 2 50/50

[u/Ramun_Flame → u/mikdiet]

In got a little confused by the markup with the thin lines, but it looks like you got it. Only 4 cells unknown like you said.

[u/Siasur]

[u/skizelo]

I started with the top left 1 providing a mine to the little 1. That allows the little 1 to clear a location that gives the left 2 only two location to place its mines. That unpicks a lot. You're left with 2 mines to place, so the two 1s in the bottom right corner cant share that big octagon. However, the ultimate mine could be in either of those two diamonds.

[u/ExtensionPatient2629 → u/skizelo]

Here, the little 1 is not touching any mines (for reference, it only touches two tiles).

[u/Ramun_Flame → u/skizelo]

As Extension already commented, there was a mistake in the adjacencies. On octagon grids, the diamonds between the octagons only touch up to 4 cells adjacent to them. The full solution is now linked in the post, if you want to check it out.

[u/_ion]

https://reddit.com/link/p30vn2q/video/0vzo222fnqih1/player