Puzzle 353 Sudoku

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.

So here the intended idea is a little bit obscured by having to add in a couple more givens than I'd have liked, in order to guarantee uniqueness.  This unfortunately gives a few easy-ish placements to start off, maybe lulling you into a false sense of security.  Which is a pity for the flow of the puzzle, but such is life.

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.

Puzzle 352 Sudoku

So here's a puzzle with an idea on a recent theme.  I'd say the execution is reasonably good, but as I'm working mostly by hand it doesn't quite feel polished to its true potential.  It's still doable if you don't get the idea, but fair warning, it might be a bit of a slog.  Failing that it's probably very guessable.  But I have more faith in you than that, dearest reader.  Enjoy!

   #352 Sudoku – rated 8/10 [Very Hard]

Update: I might downgrade this from a 9 to an 8 as I've just realised the intended pattern used is very similar to the original Phistomefel pattern, which perhaps is a bit easier to spot and make use of...

All puzzles © Tom Collyer 2009-20.

Reflections on Intersections, Naked and Hidden.

There's no puzzle in this post I'm afraid, but a few random thoughts.  Some more to do with me, and some more generally on Sudoku Theory.  I'm afraid if you're more interested in one things rather than the other, dearest reader, then you'll just have to bear with me this time.

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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:

1. 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.
2. 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!

The MSLS: Revolutions

Wait. 

I've seen this. 

I stand here, right here, and I'm supposed to say something.

I say, "Every grid with an intersection of partitions deduction has a multi sector locked set, Tom"

A little while back I published a difficult puzzle as a kind of tribute to what I consider one of the greatest Sudoku puzzles of all time.  This puzzle was first published in 2008 in the Botsu Baku section of what was nikoli.com.  Sadly, nikoli.com and its many, many gems are no longer with us, but I do have a screenshot or two in my private records so that there is at least some archive of these puzzles.

To make initial progress with this puzzle, you can place a 3 in R9C6 and 9's in R3C5 and R5C8. And then you grind to a halt.

I suppose it should come as no surprise that this is a puzzle that Cracking the Cryptic has picked up on previously, as wonderfully meticulous as they are.  The position after these three placed digits seems to be where they've picked up on this.  [As an aside, I can't say why they picked up on this non-symmetrical version of the puzzle, to which there isn't the usual attribution - but I'd be interested to hear more about that].  Simon goes on to describe the technique required as a so-called Schroedinger's cell.

R4C1 is identified as the Schroedinger's cell, which seems the natural choice as the ensuing what-if analysis makes use of the fact that the only candidates in R9C5 are 1 and 2.  This mirrors the original explanation given by nikoli in the corresponding solution replay back in 2008, as it happens.  My interpretation of the logic is something like:

• If R9C5 is not a 1, then this means the 1 in Row 9 must go in R9C3.
• This implies the 1 in Column 1 must go in R4C1.
• Equally, if R9C5 is not a 2, this means that R4C1 must be a 2.
• Therefore the only candidates for the Schroedinger's cell R4C1 have to be 1 and 2

The puzzle then continues by noting that the only place for a 4 in Column 1 is at R2C1.

One observation with my interpretation is that we didn't use the fact that the only candidates for R9C5 were 1 and 2.  If R9C5 was neither 1 nor 2, then we'd have to conclude that R4C1 was simultaneously a 1 and 2.  Which, despite all this Schroedinger adjectival being thrown about, is still a contradiction, and so we also deduce that the only candidates for R9C5 are 1 and 2 anyway.

So why not observe that R9C5 is also a Schroedinger's cell?  The logic is something like:

• If R4C1 is not a 1, then this means the 1 in Column 1 must go in R1C7.
• This implies the 1 in Row 9 must go in R9C5.
• Equally, if R4C1 is not a 2, this means that R9C5 must be a 2.
• Therefore the only candidates for the Schroedinger's cell R9C5 have to be 1 and 2

The symmetry is beautiful!  But either way there is still something of the if-this-then-that, or even trial-and-error about this deduction.  Maybe we could make this slightly more satisfying:

 

Let's consider Row 9 and Column 1 simultaneously, which I've circled purple.  Within these 17 cells, we need to place two 1's and two 2's (noting that we already have a 6 placed at the intersection).

Now let's think about where we can possibly place this set of four digits within the confines of the purple cells?

• We can have at most one each in the two cells I've circled red
• We can have at most two cells in the cells I've circled orange (noting that these four cells are contained in one box)
• We can't have a 1 or 2 anywhere else!

Therefore, we can conclude that the "at mosts" are also "at leasts", and subsequently that the only two candidates in each red circled cell have to be 1 or 2.  If either cell was anything other than a 1 or a 2, then we could not fill up Row 9 and Column 1 with their full complements of 1s and 2s.  Put another way, this little trick that unlocks a masterpiece from 2008, is nothing other than an example of MSLS, as first described in 2012.  Which I think is rather extraordinary!

So dearest reader, what are we to make of all of this?  All the recent discussion regarding Phistomefel's theorem, its further generalisation as Fred's intersection theory, its further generalisation as Trevor Tao's "9D Wonkyfish" - as well as Sam Cappelman Lynes' exposition of this in the language of SK Loops and MSLS has certainly got me thinking.

Sam himself said that there is no formal definition of MSLS that is agreed upon, so I thought (at the risk of probably reinventing the wheel) it might be worth sticking my own neck out, and having a fumble in the dark with my own interpretation.

I see MSLS as being a generalisation of hidden pairs/triples etc.  The idea being that we are trying to find areas of the grid where we have:

1. A subset D of the digits 1-9, considered with a multiplicity M, which then makes up a set of (at most) M copies of D.  Let's call that repeated set S, and we'll say that it has a total of N digits to be placed.
2. A collection of (perhaps slightly more than) N cells into which the N (possibly repeated) digits of S need to fit into.
3. The idea then is that the digits in D squeeze out all the other candidates from those cells because there is no room for anything else.

To highlight the generalisation, suppose you have a row where you are looking to place the digits 1,2,3 - and you know that the 1 can only go in C4,9, the 2 can only go in C2,4,9 and the 3 can only go in C2,4.  Then you have a set of three digits (1,2,3), and only three cells in the row in which you can collectively place them.  The hidden triple deduction then lets you say that the possible candidates in C2,4,9 can only be 1,2,3 - and that any other digit can be eliminated as a possibility.

This is easy enough to see in the confines of a single Row (or Column, or 3x3 Box come to think of it) where we don't have the complication of repeated digits.  After all a hidden triple, whilst sometimes very difficult to see when solving by hand, is nevertheless viewed as an elementary sudoku solving technique.  What I find fascinating is that Phistomefel, Fred and Trevor have helped teach us that there are deeper structures within Sudoku, inherited from Latin squares, that link together multiple instances of different rows and columns, and to which we can apply the same kind of thinking.  These structures go beyond the naive understanding that we are simply looking to place each digit from 1-9 both at least once in each Row, Column and 3x3 Box in isolation, and indeed at most once in each Row, Column and 3x3 Box in isolation.

The examples from the Tatooine puzzles (we discuss sunset here), as well as Sam's puzzle (as discussed in Reloaded) involve rather large sets of "hidden" digits with multiplicity that somehow need to be squeezed rather tightly into similarly sized sets of cells within the grid, and both the digits and the cells can be hard to see on the face of it.  The main point of all the recent progress we have made, dearest reader, is to help better visualise when we might be in a situation to uncover one of these hidden multi-pairs, multi-triples or multi-quads.  

I don't think we will ever need anything bigger than quads, in much the same that we never really any thing beyond the 4-way x-wing, a.k.a. "Jellyfish".  In fact, I think the correct way of thinking about all of this is that x-wings, swordfish and jellyfish are basically hidden multi-singles.

I'll begin to conclude.  Hopefully with Revolutions, I have been able to demonstrate that hidden multi-sets don't always have to be big and intimidating beasts.  Once I revisited the nikoli puzzle with the MSLS framework in mind, I was surprised by the simplicity of the deduction - potentially even easier than x-wings and swordfish, and certainly no more difficult.  

Equally interesting is that within this simplicity lies something else that I haven't been able to grasp in terms of intersection theory alone - and I think it's because now we're mixing in 3x3 boxes alongside rows and columns.  As I've hinted previously, I do have other ideas about how this might equivalently end up, but on that note, dearest reader, I'm going to leave things there and simply say I much prefer Sudoku to Column Dance.

The MSLS: Reloaded

"And now here I stand, because of you Mr. Stalder, because of you.  I'm no longer an agent of this system, because of you.  I've changed.  I'm unplugged.  A new man, so to speak, like you, apparently free."

"Congratulations."

"Thank you."

So I was originally going to add one more post on this topic, but this morning it's become clear there's the opportunity to add two related posts, and add in more neat nods to pop culture I enjoy.

This one was prompted by last night's video on cracking the cryptic, which is very much presenting the MSLS interpretation of the discussion.  I'm not entirely sure whether Simon actually reads this blog, or gets the information via separate conversations he may (or may not!) be having with Sam, but in the videos he is vaguely referencing conversations between myself and Sam.  My discussions with Sam have only happened here on this blog, so one way or the other they have to be the source.  I haven't heard back from Simon since pointing him in the direction of Trevor Tao's wonderful video, which I'm sad to say has not (to date) even had a vague reference on cracking the cryptic, so I can't say for sure whether it's a conscious decision to present this theory from one angle, or whether he is simply unaware of the extent to which all this discussion is influenced by many different contributions, Trevor’s certainly not withstanding.

Anyhow, I hope at least that my dearest readers will agree that all this stuff is far too interesting to not explore in plenty of detail and from multiple different angles. 

On to business.  Sam has come up with a wonderful puzzle which I'm pretty sure has been designed to completely bamboozle scanraid.  And it does.  For reference:

• The video is here: https://www.youtube.com/watch?v=uvTV_Wivxmo
• You can play the puzzle here: https://app.crackingthecryptic.com/webapp/dqTM27jm3P

However, I’d like to present it to you again with a bit of colouring in:

As I'm sure all my dearest readers are aware, this looks very much like one of Fred Stalder's examples as discussed in a previous post.  To recap:

• Here's my post: https://tcollyer.blogspot.com/2020/09/puzzle-350-sudoku.html
• Here's Fred's original post: http://forum.logic-masters.de/showthread.php?tid=1811&pid=32019#pid32019

The punchline to all of this is that instead of doing all the elegant MSLS stuff that Simon does in the video, you can instead make the following (simpler!) observations:

1. Thanks to Fred, we know that the digits placed in the blue cells plus a copy of (1-9) are precisely the same digits that need to be placed in the red cells.
2. There are 15 red cells with odd digit givens, and 10 red cells with no givens
3. There are 6 blue cells with even digit givens, and 4 further even digits in the copy of (1-9)
4. Therefore the 10 empty red cells must contain 10 even digits.

As Sam has pointed out on a youtube comment discussion, Fred gives you even more than that, but the observations above are all you need to serenely solve a diabolical-rated sudoku.

That's all for now folks - I still have one further follow up on this subject, again linking together some of the subject matter I've been discussing recently.