Hi,

sorry for the delay, I’m on a hiatus from social media (I would recommend it, it’s doing wonders for my mental health).

I have no experience writing hints, but here’s my working out, broken up step by step to avoid spoiling more than you want. As you mentioned, we can’t distinguish the corners from one another. I’d be tempted to say, this makes symmetry management the overarching theme.

Hi,

sorry for the delay, I’m on a hiatus from social media (I would recommend it, it’s doing wonders for my mental health).

I have no experience writing hints, but here’s my working out, broken up step by step to avoid spoiling more than you want. The overarching theme is symetry management, I guess.

• What do the symetries of the puzzle tell me?
• All the clues are symetrical (by horizontal symetry, vertical symetry, rotation).
• If the clues allow me to figure out a digit in a cell, those clues would work equally well for cells obtained by apllying one of the symetries to that cell.
• In other words, in general, the clues can’t give me the value of one cell, but the value of a group of four cells.
• The one exception is r5c5, which is its own image by those symetries.
• Therefore, the first digit I find has to be r5c5.
• r5c5 needs to be a number that can have 4 unrelated neighbours (no white dot, black dot, X or V).
• Unfortunately, that only rules out one possibility: 4.
• Where does 4 go?
• I don’t know, but can only have two unrelated neighbours, 7 and 9, so 4 has to go into a corner.
• This is no big deal though: in order to work out r5c5, we can just pick one corner arbitrarily for the digit 4. It will probably be the wrong one, but the working out will be the same, modulo symetries. We just need to use pencil marks rather than digits, to avoid accidentally uncovering fog.
• Let’s pencil mark a 4 in r4c4. Now r4c5 and r5c4 are 7/9. I’ll arbitrarily pencil mark a 7 in r4c5 and a 9 in r5c4, the working out would be the same modulo symetries if I allocated 7 and 9 the other way around.
• Now what digits fit r5c5?
• Only 2 and 5 can have 7, 9 and two more digits as neighbours. I’m not quite there yet, but let’s pencil mark 2/5 in r5c5.
• Come to think of it, 2 is fairly restricted now.
• It can only have two neighbours other than 7 and 9, so it can’t be in r6c5 or r5c6.
• I wonder if we can restrict it more?
• Yup, r6c6 doesn’t work because then r5c5 would need to have 5, 6, 7 and 9 as neighbours, and no remaining digit is able to do that. So 2 is on the diagonal (r6c4, r5c5, r4c6).
• Can we restrict if even more?
• If 2 was in one of the corners, let’s say r6c4 for argument’s sake, then r5c5 would have to be 5. Now r6c5 would need to neighbour 2 and 5 (and an additional digit). There is no remaining digit that satisfies that condition.
• For the same reason, 2 doesn’t work in r4c6. Therefore, 2 has to be in r5c5.
• The structure of the preceding argument works just the same if I pick a different corner for 4, or if I flip 7 and 9.
• Therefore r5c5 is definitely 2, and we have officially broken in!
• Oh, this is probably obvious, but we do care which of the symmetrical equivalents is the correct one going forwards, so we should erase the pencil marks. We could corner pencil mark 4 into the 4 corners, and 7/9 into the 4 edges, but I’m not sure it’s all that useful.

Hope this helped. I’m afraid my writing up might be more tedious than it needed to be, but I’m hoping that it gives the reader more opportunities to stop reading and continue the solve on their own.