Sometimes it turns out that you aren't always everyone's flavour of the month. As always, I suppose I'll have to make my own fun, and as always I am very grateful for the company of my dearest readers.
Anyhow all I'll say with this puzzle is that if you're looking for dead-or-alive ambiguity you'll know it when you see it, dearest reader. After that the puzzle will flow quite nicely, even if I do say so myself. Enjoy!
#353 Sudoku – rated 9/10 [Very Hard]All puzzles © Tom Collyer 2009-20.
Clever idea - I enjoyed figuring this one out! I ended up proving that e2p8 naq e8p2 jrer rssrpgviryl n 1/2 cnve, which reduced the puzzle to what I'd call "Moderate". I was a bit roundabout in coming to that conclusion though, so I'd love to see what you actually had in mind.
ReplyDeleteQuickest thing to say now is take a look at the pattern showing in the google sheet at the moment
ReplyDeletehttps://docs.google.com/spreadsheets/d/1dTAJNB-U-5d8bxwFsWwyVHEU9xAGiBf3mjMmUboHzFI/edit
It’s essentially an extension of the schroedingers cell pattern. Doing the individual case analysis is quite fun, it’s a bit chain-y (indeed a scanraid is capable of finding this as an AIC in much the same way it also finds the schroedinger example as an AIC) - but actually the intersection theory view proves it obviously and immediately.
That’s probably too technical to count as a spoiler :) I doubt any of that is understandable unless you’ve already found the logic!
And yes, the intention was to do this just-one-cell style so that once you got all that it was easy to the end. I did stumble upon a couple of harder variations which would take you a bit further before needing some further xyz wings and the like that I thought weren’t really worth the bother, given the disjointed solving flow.
Very nice. I think that this is a genuinely novel pattern. I couldn't quite find the right combination to make it work smoothly.
ReplyDeleteRe difficulty ratings - 9 means I’d just about feel happy putting this in a competition round, which roughly guessing might equate to 3/5 for LMD?
ReplyDeleteI suspect more than 2/3s of what appears on cracking the cryptic I’d rate as 10, and probably a good 1/3 of it I’d probably turn up to 11!
Thanks for sharing the document. I could solve it now. But in reality, I need more practice and its hard to find these patterns without having some knowledge about these things :p. Hoping to improve with practice.
ReplyDeleteVery nice pattern, Tom!
ReplyDeleteThere is a "standard" row/column/digit partition for this, but it's very tough to spot:
d12 in r1345679c1469 (Red)
d3456789 in r28c23578 (Blue)
Red = Blue + two sets of 1-9
There are three 1s and three 2s in Red (after singles in r5), so we need at least one of each in Blue, and of the three empty Blue cells, r8c7 is ruled out by the 12 in b9. So r2c8 and r8c2 are a 12 pair (among some 12 eliminations in Red), and stte.