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Firstly, something that has been playing on my mind a bit recently was the announcement of a Cracking the Cryptic sudoku book via a kickstarter campaign. This does look very exciting, and I've paid up my $19 USD to secure a copy for whenever it's ready. The money they've raised to date is phenomenal and it's exciting to see exactly where Mark and Simon can take Sudoku, particularly when you compare that with what the WPF has been able to achieve. All the best to them!

And yet when I look at the list of the contributors, it's a bit like seeing all the cool kids running off to play their cool kid game, without being invited. Don't get me wrong, it's their book and their channel and they can do whatever they like, I have no problem with that. And I do get that the puzzles I create aren't really the sorts of puzzles that get their viewers' pulses racing, so there's no real reason why I would be included. And the authors on that list certainly include a few of my favourite authors so absolutely no complaints there. I suppose the crux of the matter is simply that I'm kidding myself a bit to look at that list as a "Who's Who" of the best sudoku authors. From that point of view I can console myself with the company of many more of my absolute favourite sudoku authors who didn't make the cut.

I'll leave things there - I do feel a bit better for simply expressing my feelings on the matter to you, dearest reader - I suppose all I'm really saying is that it would have been nice to have been asked. As presumptuous as I'm sure that sounds to even my most dearest of readers.

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Moving on, I've been thinking a little more on some of the recent posts and discussions on this blog, particularly to do with intersection theory. The news thing from that front is that I've created this resource to help any of my interested dearest readers to easily visualise the addition and subtraction of rows, columns and 3x3 boxes.

https://docs.google.com/spreadsheets/d/1dTAJNB-U-5d8bxwFsWwyVHEU9xAGiBf3mjMmUboHzFI/edit?usp=sharing

I've also started fleshing out a list of references which might become the basis of a more permanent article I might put together as and when I get my head around this stuff even more.

Fred's recent comment gives me some hope that there might be some kind of multiplication we can work into this algebra of constraints, but I haven't really given much more thought to it than that. If anyone else gets there before I do I'd be most interested to hear about it!

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The final item of business is simply to reflect that all this discussion holds my interest because it's fundamentally about linking different areas of a sudoku grid to each other in interesting ways. We had started off by thinking about these as partitions of cells, but I did want to discuss something more along the lines of a partition of digits.

At the danger of stating the complete obvious, hidden pairs/triples/etc using a subset of the digits 1-9 are exactly the same thing as naked septuplets/sextuplets/etc, and vice versa. The way I thought about it was like this:

- Suppose you have a naked triple of 3 cells in a row whose possible candidate placements are some of 1,2,3. That means there a 6 cells left in the row, and 6 digits which we haven't been able to place - i.e. we have the very definition of a hidden sextuplet.
- Now suppose you have a hidden triple of digits 1,2,3 - each of which can only go in some or all of 3 different cells in a particular row. This means that for each of the 6 remaining cells, we only have 6 possible digits from which we can choose from - i.e. we have the very definition of a naked sextuplet.

The conclusion is that which ever way you look at things, hidden or naked, you end up making identical deductions.

My next thought is that the naked point of view (i.e. looking at which digits are possibilities for given cells) I've always - perhaps mistakenly - viewed as the basis for some of the more advanced techniques when compared to the hidden point of view (i.e. looking at which cells are possibilities for given digits).

I say perhaps mistakenly because I think I realise now this is probably a misconception I've picked up trying to follow discussions of advanced sudoku solving techniques, where computer assistance is de rigueur and grids are very conveniently displayed with all the possible candidates automatically displayed - an approach to sudoku solving which generally leaves me very cold. When I tend to solve, the more cells in the grid I label with pencilmarks, the less I generally enjoy solving a puzzle.

Anyhow, the link between naked and hidden and partitions and so on reared its head to me again when I came across Bob Hanson's website, discussing:

- Bent naked subsets (in particular Y-wings)
- Intersection ideas briefly
- Intersection ideas in more detail

As I'm sure you can pick up from the somewhat dated web-design, in some sense all of this way of thinking is not particularly new - indeed it's been around nearly as long as sudoku mania (dating back to 2004) itself has. My interest remains in thinking about these things from different perspectives.

There's a bit more there for me to get my head around, particularly with a link to almost locked sets (ALS) and what he describes as almost locked ranges, but I want to finish with the two rules that Bob outlines with relations to bent subsets:

- If a bent naked subset contains one and only one candidate k that is present in both of its nonintersection subdomains, k can be eliminated as a candidate in any cell that sees all the possibilities for k in the subset.
- When there is no common value k in the two nonintersecting regions of a bent naked subset, the subset behaves as a standard naked subset. That is, candidate k can be eliminated from any cell that can "see" all cells of the subset containing k.

What I'd like to try and work out and see if there's a "hidden" point of view for "naked" subsets that span intersections of a row/column with a box, in the same kind of pleasing way that can alternatively view so-called Schroedinger's cells via intersections as a kind of semi-Phistomefel arrangement. Again if anyone else gets there before I do, I'd be most interested to hear about it!

I'm curious what authors you might be referring to. The book is only going to include puzzles which have been covered on the channel, so it's effectively a "Who's who" of the best setters who have appeared on the channel!

ReplyDeleteHi Freddie - I won’t dwell on that part of the post too long, but my personal acid test is whether an author is able to do write decent classics. That seems to have a reasonably good overlap with my favourite (sudoku) authors in general. For me, that includes (but is certainly not limited to!) the likes of Fred Stalder, Thomas Snyder, Ashish Kumar, David McNeill and Wei-Hwa Huang.

ReplyDeleteAnd I suppose if they’d done it purely by puzzles (so a particular author ended up with multiple puzzles) it’d be less disappointing personally. But they seem to have picked the list of authors first, then 1 each from all of them. I guess as a long-term supporter, contributor, and even guest presenter of the channel I might have made the top 25. But I’m sure I’ll get over it. I haven’t really been as involved recently and they certainly don’t seem to be missing anything I can add - they go from strength to strength!

DeleteOn the product of two partitions...

ReplyDeleteIf a one-dimensional "partition" (or whatever we want to call it) is an assignment of an integer coefficient (normally 1, 0 or -1) to each cell so that the (weighted) sum of all cells is a complete 1 to 9 set, then it is natural that an n-dimensional "partition" has the same property but with coefficients in Z^n

The product of an m-dimensional "partition" and an n-dimensional "partition" corresponds to taking the obvious direct product of Z^m and Z^n.

Talking for a moment in two dimensions, like the grouping Fred used for Tattooine Sunset, we of course recover the 1-dimensional "partitions" by taking the projection onto one of the two components in Z^2. But you can get more interesting information by instead changing basis in Z^2 before projecting - this corresponds to performing Gaussian elimination of the defining equations of the two base partitions. And all this applies in more dimensions too.

Small correction - the weighted sum of all cells is a whole number of 1-9 sets, not necessarily just one.

DeleteApologies too for making exactly no effort to tone down the maths jargon. We're getting abstract enough now that it's just much easier if we assume we all understand linear algebra.

Excuse my senior moment. Obviously the change basis then project manoeuvre gets you nothing new, just a linear combination of the component partitions.

DeleteSo sadly we won't get anything totally exotic by considering the products of partitions - only things we could obtain by considering multiple partitions in parallel. Nonetheless it's a good conceptual tool that makes it easier to work with multiple partitions simultaneously.

I’m not sure I fully understood your train of thought there Sam, but admittedly I’m running on fumes a bit recently.

ReplyDeleteI do understand linear algebra, and I like to think I have a certain amount of mathematical proficiency, but I’ve also found I need a good picture in mind to illustrate what the algebra is actually doing - hence why I created that google sheet above.

I think the main thing for me with Fred’s pictures is that he’s taken two easier pictures, which can be expressed as the addition and subtraction of rows and columns, and come up with a third.

The process is something like further dividing up one side of each of the easier pictures in a mutually consistent way. Perhaps that’s not really multiplication in a tensor kind of way, but more like a scalar kind of way. But that doesn’t quite sound right either...

"I think the main thing for me with Fred’s pictures is that he’s taken two easier pictures, which can be expressed as the addition and subtraction of rows and columns, and come up with a third."

DeleteI'm trying to describe the same thing (but in generality), but as you say it's a lot clearer with pictures. I'll rephrase all of the above and ping you an email, pictures included. I'm a bit fed up with trying to explain things clearly in the Blogger comments section :)

Appreciate the email Sam, I suspect there's plenty there which might make for an interesting follow-up post if I ever get round to it. For now, that's not going to be hugely satisfying for anyone else trying to follow the discussion!

DeleteI think the gist of things is that you probably don't want to be somehow multiplying different partition arrangements willy nilly (or else you'll end up with a confused mess that noone can make neither head nor tail out of) - and indeed even if you are trying to combine two together, there are probably some pre-conditions you want to impose so that you still end up with something nice, as Fred does in the comment I linked to above.