*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:

- 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.
- A collection of (perhaps slightly more than) N cells into which the N (possibly repeated) digits of S need to fit into.
- 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.

I'd like to add something to your view of MSLS as generalised hidden pairs/triples etc. (which I think is a good way of thinking about them). It's not only that the digits of D squeeze out other digits from some cells (as in hidden subsets), but also that the digits of D are excluded from some areas of the grid (as in naked subsets) - a sort of hidden/naked combination working in tandem.

ReplyDeleteIn the Botsu Baku puzzle, the naked side of things is seen in the lower-left 3x3 box. Not only are 1 and 2 squeezed out of the two "Schroedinger Cells", as in a hidden pair, but they are also excluded from the non-orange cells in the box, similarly to a naked pair (except we don't yet know exactly which two orange cells will eventually hold the 1 and 2).

This also happens if you use an MSLS to solve the "Reloaded" puzzle, as in Simon's video. Not only are the odd digits squeezed out of some of the cells in the intersection, but there are some columns where the even digits are excluded from cells outside of the intersection.

Could one way of looking at the the “hidden” vs “naked” part of the deduction be a hint that this is really fundamentally to do with partitions?

DeleteHere we were looking at 12, but maybe there’s a way of looking at this from a partitions point of view, where instead we look at 3456789 squeezing everything else out on the other side of the partition?

Again, not something I’ve fully thought through.

Tom, this is a very insightful comment. I had some thinking about it, and you're exactly right, this deduction *can* be phrased in terms of partitions / "Fred theory", albeit generalised to include boxes (of course!).

DeleteFirst, it's helpful to take a slightly different view of Fred's original insight. Fred's exposition divided the grid into four sets of cells, which he labelled Red, Green, Yellow and Blue. Well, really there's no difference between the Red/Green cells and the Yellow/Blue cells - if we take a different Fred-style partition using the same rows but the complementary columns (or vice-versa), then the Yellow/Blue cells take on the role of the Red/Green cells in the new partition.

Now we'll prove Fred's conclusion about the Red and Green sets using a different argument. It's all about adding and subtracting sets of cells that we know contain a complete 1-9, and so whatever we're left with must be a whole number of 1-9 sets. Add all of the rows that define the Fred-style partition, and subtract away the *complementary* columns from the partition. Some cells cancel out, some cells are included once, and some cells are included -1 (yes, minus one) times, and the result must be a whole number of 1-9 sets.

For example, if we *add* rows 1,2,8 and 9 and *subtract* columns 3,4,5,6 and 7, then we get the following picture, where the Green cells are included once each, and the Red cells are included -1 times each. The conclusion is exactly what Fred originally deduced - that the digits in the Red set must be the same as those in the Green set, plus a copy of 1-9: https://ibb.co/4NBFDv8

But with this new proof, it's clear how we can generalise Fred's "partitions" to include boxes too. The word partitions, though, is probably no longer appropriate once we've done away with Yellow/Blue as we are no longer insisting that the whole grid is covered. Nonetheless, here's how to deal with the Botsu Baku puzzle using this methodology.

*Add* Column 1 and Row 9, and *subtract* Box 1, Box 7 and Box 9. Some cells cancel, some are included once (or -1 times), and r9c1 is even included (2 - 1) times. The picture we're left with is as follows - and this time, I've put the givens in the grid: https://ibb.co/MZtXsMs

The deduction is obvious. Red is equal to Green plus a 1-9 set. But Red already contains two each of 1 and 2, whereas Green has none. Therefore Green must contain a 1 and a 2, and Red cannot have any additional 1s or 2s.

Sam, I think this is exactly how I felt like it might be possible to explain things without ever really getting there myself. Absolutely fabulous stuff, and I was kind of hoping might come out of some of the discussion and a few weeks thought!

DeleteI think it gives an alternative view of phistomefel as well, the one I was trying to get my around when I first saw it. That is very satisfying as it scratches another itch I’d forgotten I even had!

Ie: r1289 and c1288

Is the same thing as

B1379, r37, c37

In fact, if you go back to my first post on the subject where I expressed some doubts about the detail of the sk loop - phistomefel correspondence, which were precisely because somehow we had boxes involved on the one side but not the other. That’s another itch which will probably scratched if I ever get around to making myself comfortable with the notations.

DeleteThis comment has been removed by the author.

ReplyDeleteOh, and maybe I can spell out explicitly here (after doing so implicitly in the comments on the puzzle itself) that I solved your tribute puzzle using a standard Fred/Phistomefel partition on rows/columns 2,3,7 and 8. I am sure that it is unrelated to your exposition of the intended deduction in this post, and that the fact that both approaches are available is mere coincidence. If you tell me otherwise, I think my head will explode.

ReplyDeleteYeah that’s a coincidence as initially intended.

DeleteI suspect it arises from the choice of symmetrical pattern of givens, and the attempt to keep the puzzle impenetrable until you spot the intended deduction. On the other hand, the next few digits I placed were all around the edges of the grid anyway, so perhaps it makes sense. Previous puzzles I’ve seen phistomefel was there, but could be shortcut, nevertheless flowed quite nicely.

I wasn't familiar with the "Schroedinger Cells" idea until the previous post, but viewing it this way as a row and a column covered by a box + two other cells makes me think this is akin to a Mutant Fish (so a Mutant Multifish?). The Schroedinger Cells are effectively the second cover house for the two digits.

ReplyDeleteSeems like this could be extended to three digits pretty easily, yes? Maybe even four. (Or, with two digits this could also be extended to a chain of such connections - a couple digits in r4c23 and r23c9, along with removing the 5r1c9, result in a third cell in r1c9 which must have the same value as r9c5.)

I suspected this might also be related to the SK Loop, and a forum discussion concurred - this is basically a baby SK Loop, on two digits instead of four.

DeleteApparently, Steve K's original name for what came to be called an SK Loop was a "Hidden Pair Loop", which seems particularly appropriate here.

Ok, thinking about it with more digits makes the Phistomefel connection obvious.

ReplyDeleteConsider the nikoli puzzle. Phistomefel tells us that from the 12 pairs in boxes 1 and 9 (in the intersections of r2378 and c2378), we must have two 1s and 2s in r19c14569+r456c19. Eleven of those cells are ruled out by given digits and by being in boxes 1 and 9, so we only have five places for four digits. The final piece is noting that placing three 1s or 2s in r1c45+r6c9 would force another into r2c8 or r3c7, and Phistomefel would then require a fifth, so either way r4c1 and r9c5 must be 1 and 2 (and not the same, for the same reason: it would force that digit into r7c3 or r8c2).

(A non-12 digit in r1c45 would actually give us a second pair of Schroedinger Cells as well.)

This comment has been removed by the author.

DeleteThis "connection" doesn't sit right with me. The Phistomefel only works when considering data in remote parts of the grid, whereas Tom's version works entirely within boxes 1 and 9, column 1 and row 9, and box 7.

DeleteThis is exactly what I referred to in my second comment above.

I think I’m with Sam on this one. Sure that, does seem to work just fine, but I’m curious if there’s any connection by mixing the partitions up a bit:

DeleteSomething like

R9 union C1 / rest of grid

vs.

Box 7 / rest of grid

I haven’t got my head around it just yet!

The setup for the "bidirectional Schrodinger Cells" in Tom's post is that we have a pair of digits (12) in two boxes which are not in the same band or chute (boxes 1 and 9 here), and within those boxes the digits are not in the same row or column. So we have four distinct rows and columns (here 2378 in both cases) intersecting in four boxes (1379). Additionally, we need 12 ruled out of the single cell in a box which sees both of our given 12 boxes (boxes 3 and 7 here) which does not sit on any of the rows and columns in the pattern (so the corner cells r1c9 or r9c1 - in this case, both are filled, by the one we care about is 6r1c9). (At this point, I'm just going to stick with the rows, columns, and boxes in the puzzle above, but the same logic works identically if they are rearranged, so long as the above requirements are met.)

Delete(Note, the Schrodinger Cell technique does *not* require the 12 in box 9. Box 9 is required to make it bidirectional, though - 12 must be locked out of r9c789, though that could be done in other ways.)

Now, at this point, what does Tom's observation tell us, without any other givens? In row 9 and column 1, we must have two 1s and two 2s. Box 7 has one pair. Boxes 1 and 9 are ruled out by givens. That means we must have, at a mininum, a 1 and a 2 somewhere in r9c456+r456c1. (In the case of both puzzles under discussion, we have exactly that minimum, because there are only two cells available, so those two cells must be a 12 pair.)

What does Phistomefel tell us? We have two 1s and two 2s (at a minimum) in the intersections of r2378 and c2378, so we must have two 1s and two 2s (at a minimum) in their counterpart: r19c14569+r456c19. Now there are two cases to consider, for each digit:

a. If there are not two 1s in r1c4569+r456c9, there must be at least one in r9c456 or r456c1.

b. If there two 1s in r1c456+r456c9 (r1c9 can't contain a 1 in this case), it forces a 1 into r23c78 in box 3. But now we have three 1s in our r2378c2378 pattern, so we must have a third 1 in the counterpart. Obviously this can't go in r1 or c9 now, so it must be in r9c456 or r456c1.

The logic is the same for 2s. So Phistomefel proves there must be a 1 and a 2 in r9c456+r456c1 at a minimum, exactly the same as Tom's pattern. (And again in the puzzles under discussion there are only two cells available from those six, so those two cells must be a 12 pair.)

Is Tom's observation simpler? Of course. It's a very elegant argument, easier to see, and a simpler deduction to make (using Phistomefel here is like finding a jellyfish in rows when there is a complimentary x-wing in the columns). But it is also a consequence of Phistomefel, and so it's not a surprise that Phistomefel/Fred/etc. will work here.

(Actually, neither Tom's approach nor the Phistomefel logic requires the 12 to be in different rows/columns in boxes 1 and 9. Nor does the original SC logic, for that matter. But if they are aligned in box 1 column 2, say, that just means we have a hidden pair in r9c35, and r9c3 must be mirrored in r4c1.)

DeleteSmall correction.

Delete"Note, the Schrodinger Cell technique does *not* require the 12 in box 9. Box 9 is required to make it bidirectional, though - 12 must be locked out of r9c789, though that could be done in other ways."

Box 9 is required whether it is bidirectional or not (which should mean that all SC setups are bidirectional). What isn't required is that 1 and 2 appear as givens in r78c789.

It's not as clear, but Phistomefel can still get to the result either way:

Say 1 and 2 do not appear in box 9, and instead r9c789 are other given digits (which will probably result in some other basic deduction, but anyway). We have one of two cases:

a. The digit 1 is in r78c78, in which case the logic is identical to above.

b. The digit 1 is in r78c9. In this case, r456c9 are ruled out. Now, Phistomefel only guarantees a single 1 in our remaining target cells (r19c456+r456c1); however, if 1 appears in r1c456, that again forces a 1 into r23c78, so we need a second 1 and it must appear in r9c456+r456c1.

The same logic holds for 2, so again the result is the same as Tom's approach.

And to emphasize: None of this diminishes the beauty of Tom's pattern! It being a special case of Phistomefel makes it no less useful or elegant. :)

Deleteyou are over killing a basic continuous nice loop: 1/2/4/7 2= r4c1 =1= r7c1 -1- r9c3 =1= r9c5 =2= r9c2 -2- r8c1 =2= r4c1 =1 => r7c3<>1, r8c2<>2, r4c1<>4, r4c1<>7

ReplyDeletewhich reduces the puzzle to basics and singles

alternative is using AHS 2RCC rules

Deletewhich is the inverse function of the more common

ALS 2RCC rules found here http://forum.enjoysudoku.com/almost-locked-rules-for-now-t2510.html

i say inverse as almost locked sets use the "on" data and almost hidden locked sets are comprised of what is off.

Whilst you might use the word “basic“ to describe your notation to describe the equivalent technique, I’m not sure I agree.

DeleteGiven that it also seems to make reference to the digits 4 and 7, I might also choose to use the word “over killing” when comparing and contrasting to the above.

I’m well aware this might be a case of reinventing the wheel - the point for me is that this can be spotted by eye, without making any kind of notes in the grid, and the logic behind it is intuitive to those who may not have been studying advanced technique for years and years!

To clarify a bit, the reference to 47 in the loop is only that you can make eliminations on them (because 1 or 2 must be in r4c1).

DeleteI would write the loop as follows:

1r9c5 = r9c3 - r7c1 = (1-2)r4c1 = r8c1 - r9c2 = (2-1)r9c5

If you're not familiar with the notation, it's basically making the same argument as the original Schroedinger Cell idea.

1r9c5 = r9c3 - r7c1 = 1r4c1

This means the following:

At least one of 1r9c5 and 1r9c3 is true (strong link);

At most one of 1r9c3 and 1r7c1 is true (weak link);

At least one of 1r7c1 and 1r4c1 is true (strong link);

Therefore we have an AIC, so at least one of the ends is true (either 1r9c5 or 1r4c1 - this is equivalent to saying if the 1 is not in r9c5, there must be a 1 in r4c1 as in the SC argument).

(This short chain is also a two-string kite.)

The same short chain/kite exists for the digit 2. What makes the loop is that the ends of both are in the same cells (at most one of 1 and 2 can go in r9c5, and similarly for r4c1). Since we have a continuous loop, alternating nodes must be true, so either:

a. 1r9c5, 1r7c1, 2r4c1, 2r9c2

b. 2r9c5, 1r9c3, 1r4c1, 2r8c1

And we can eliminate any candidate that sees both possibilities: since r4c1 is either 1 or 2, -47r4c1; since 1 is in either r7c1 or r9c3, -1r7c3; since 2 is in either r8c1 or r9c2, -2r8c2. (And additionally since 1 and 2 are in either r9c5 or r4c1, -12r4c5 - these are the eliminations the kites would give.)

I personally find Tom's version much easier to see, but if you've been studying AICs and nice loops and the like as long as strmckr has then it's pretty straightforward.

Sure, it is spotted by eye with no pencil marks which is how i wrote my chain to begin with. Simply put 2,overlaping 2 string kites make the loop. Which is just bilocal digits nothing really fancy required. See pauls notes to what i was aludimg to for why 4/7 is refrenced in the chain.

DeleteSorry phillp

DeleteI get Patrick occasionally for some reason, Paul is a new one. :)

DeletePhillip, thank you so much for your patient explanations, you are making all this stuff in the terms of these old techniques much easier for a layman to understand!

DeleteI expect my next blog pop-culture reference is going to have to be Butch Cassidy. Which I guess makes Strmckr the sundance kid...

You're welcome. :) I'm actually pretty new to sudoku myself, but I kinda got thrown in the deep end on the player's forum and had to get familiar very quickly with the crazy techniques they effortlessly use. So a relatively tame AIC/CNL is quickly becoming familiar.

Deletefor msls definitions and applications as the description in here is faulty see

ReplyDeletehttp://forum.enjoysudoku.com/using-multi-sector-locked-sets-t31222.html?hilit=multi%20sector%20locked%20set

mutifish- comments seem to also be faulty- basic versions are easily the summation of multiple different fish using the same sectors where the eliminations are performed in 1 step.

in basic essence: N sets of N size fish each using 1 digit. where the number of distinct sectors used cannot exceed N.

Personally I find MSLS is a very unfortunate term, a convoluted explanation of something that should be much simpler (and spare a thought for Simon who has to explain MSLS while solving a difficult puzzle). To be fair, I can’t think of a better name either – and I haven’t found time to study Fred’s intersection theory or Sam’s SK loops in much detail.

ReplyDeleteMy preferred explanation is the following: look for two sets that are congruent (same digits, modulo copies of 123456789) and “highly dissonant” (given digits almost imply a contradiction). For instance I could say 25 red squares and 16 blue squares are congruent and highly dissonant because the reds contain only odd digits and blues contain many even digits, etc.

FWIW, I have added further discussion of Tatooine Sunset on Puzzling Stack Exchange (GIYF)

I’d be fascinated by any further insight you have on this Trevor.

DeleteFor what it’s worth I think you understand Fred already, without maybe knowing it.

In your terms, it’s basically a 4-D wonkyfish, rather than a 9-D one, which is why I claim it’s a further generalisation. The logic to fill out the equivalent dual grid is easy enough to make the need to draw out a dual grid somewhat redundant (although you can if you want!)

Trevor, Fred's intersection theory is basically what you are talking about with two sets being congruent (mod sets of 1-9) and dissonant.

DeleteThis should work as a general formula?

Given the following partitions:

Split the 9 rows into sets of size R1 and R2.

Split the 9 columns into sets of size C1 and C2.

Split the 9 digits into sets of size D1 and D2.

Say R1xC1 contains X1 digits from R1. Likewise, R2xC2 contains X2 digits from R2. Then, if X1+X2 = R1*C1*D1/9 + R2*C2*D2/9, the remaining digits of R1xC1 must be from D2, and the remaining digits of R2xC2 must be from D1. (Or more generally: that formula must hold in the solution to the puzzle; if X1+X2 is already equal to rhs in the current state of the puzzle, we can conclude what the remaining digits in the intersections must be.)

So for the specific case of odd digits in 25 red squares and even digits in 16 blue squares, we need 21 total digits to conclude that the remainder must be of the opposite parity.

(Partitioning three ways in each case - as in the Tatooine Sunset puzzle - makes Fred's much messier, which is why the wonkyfish visual really shines there.)

DeleteThe 247 (digit) muti-fish crushs alot on the puzzle.

DeleteNxn+k fish cover set mathematics used by algorthiums is what is being described here.

n base sectors with n cover sectors(+exfra if not all bases are covered)

where by n base cells are covered by n cover sectors exactly n times. Thus all cells in the cover not in the base are excluded. See the ultimate fish guide or obiwans mathmatics on the players forum for refrence or ask and ill link it.

Ok, here's a challenge puzzle derived from Sam's which makes use of the above forumla as an inequality:

Delete7 . 1 . 3 . . . 9

. 8 . 7 . 6 . . .

. . 3 . 5 . 7 . .

. . . 4 . 2 . 9 .

. . . . 7 . 1 . 5

. . . . . 5 . 8 .

1 . . 5 . . . . .

. 2 . . . . . . 4

9 . 5 . . . . 6 .

The goal is not to solve (though I'm sure strmckr will probably post a solution quickly!) - the goal is to place one digit. (Andrew's solver cannot place a digit, but if given that digit is able to solve the puzzle, though it's still Extreme.)

+------------------------+-----------------------+--------------------+

Delete| 7 4(56) 1 | (28) 3 (48) | -8(6) 245 9 |

| 245 8 249 | 7 (1249) 6 | 2345 12345 123 |

| 246 469 3 | 1289 5 1489 | 7 124 68 |

+------------------------+-----------------------+--------------------+

| 3568 367(15) 678 | 4 68(1) 2 | 36 9 367 |

| 23468 3469 24689 | 3689 7 389 | 1 234 5 |

| 2346 134679 24679 | 1369 (169) 5 | 2346 8 2367 |

+------------------------+-----------------------+--------------------+

| 1 347(6) 47(68) | 5 2489(6) 34789 | 2389 237 238 |

| 36(8) 2 67(8) | 13689 (1689) 13789 | 3589 1357 4 |

| 9 347 5 | 1238 1248 13478 | 238 6 1238 |

+------------------------+-----------------------+--------------------+

challenge accepted: however i wont let that 1 digit easily solve it.

Oof. Definitely not what I had, but I follow the logic so looking into chains with you guys is definitely paying off!

DeleteIf 6 is not in r1c7, 6r1c2 (immediately) and:

2r1c4, 4r1c6, 5r4c2, 1r4c5, 9r2c5, 6r6c5

a. 1r4c5, 9r2c5, 6r6c5 => 8r8c5

b. 6r1c2, 6r6c5 => 6r7c3 => 8r8c3

Contradiction, so 6r1c7.

It is possible to solve the tatooine sunset sudoku with intersection theorem, which gives an equivalent solving path than Trevor's one. I explained it there: https://tcollyer.blogspot.com/2020/10/tatooine-sunsets-and-partitions-of.html?showComment=1603315915175#c2274367086054215763

DeleteThanks for that. It's certainly a lovely diagram with the coloring. :)

DeleteInterested what you think of the above challenge as well. It's a little bit trickier of an application; I'm definitely going to have to get it working in a puzzle that solves cleanly afterward.

My answer to Philip's challenge is to place a 6, but I won't add any further spoilers for now

Deletecorrection 247 should read 279 which is this

Delete+------------------------+---------------------------+------------------------+

| 1456-27 12367 13456 | 1345-2 12379 1345-279 | 13468-9 12489 13468 |

| 1456-2 1236 9 | 8 123 1345-2 | 1346 124 7 |

| -14(27) 8 134 | 134(2) 6 -134(79-2) | 134(9) 5 134 |

+------------------------+---------------------------+------------------------+

| -168(29) 5 168 | 16(-2) 4 -18(27) | -168(79) 3 168 |

| 1468 136 7 | 9 138 1358 | 14568 148 2 |

| 1468-29 12369 13468 | 1356-2 12378 1358-27 | 14568-79 14789 14568 |

+------------------------+---------------------------+------------------------+

| 1568 16 2 | 7 138 1348 | 13458 148 9 |

| (79) 4 18 | 13(2) 5 -138(29) | 138(7) 6 138 |

| 3 79 158 | 14 189 6 | 2 1478 1458 |

+------------------------+---------------------------+------------------------+

{3 digit muti fish}

3 digits: 3 fish 3 base & 4 covers

wont completely clear it all but does a lot of the work of a larger msls

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ReplyDeleteI thought at this stage I'd add some further reflections.

ReplyDelete1. I think we're finding our limits with blogger comments unfortunately!

2. It's amazing to have so many of the protagonists I've been following all here on the same page! I think at this stage we're only missing Phistomefel, 胡蒙汀(Hu Meng Ting), and whoever it was at nikoli that came up with that puzzle (Tea.M maybe suggests a collaboration)

3. There is so much gold here in these comments I don't really know where to start. I think Sam's recasting of intersection theory as some kind of multiplicity theory I think is more or less where I was hoping this would all end up at, without ever quite knowing myself where I was hoping this would end up.

4. It's probably worth acknowledging that in one form or another there isn't really anything here that comes as news to the real experts here, if it is dressed up in the appropriate language and notation.

5. I think the real appeal for me with intersection/multiplicity theory is that it very much focuses on cells in the grid, rather than candidates within cells. Once you know that the contents of one group of cells somewhere in the grid is equal to another group of cells, it feels very natural to scan the contents of these two groups and make deductions that way. Sam's diagram [ https://ibb.co/MZtXsMs ] demonstrates this beautifully!

6. I think the contrast with the established theory from where I'm sitting is that it is much more focused on candidates within cells, which for solvers like me is basically not much fun!

7. I wonder how many other techniques that I'd previously considered to be impenetrable might be seen through this more regional lens?

This is not the first time that interesting and deep discussions occur on your blog ;). I remember the fundamental discussion about WSC when you were in charge of it in 2010.

DeleteTom, you've got me experimenting with more exotic partitions now... :)

DeleteFor example, the one I'm playing with at the moment has the red cells as the intersections of r1379 with b1379, and the blue cells as the intersections of r2468 with b24568. Each set has 24 cells, and must contain the same digits. What's especially interesting with this arrangement as it relates to the challenge puzzle above is that boxes 13579 all have six cells of one color, so with a 5/4 split in the digits each of the boxes *must* haso you can immediately get away with only 22 givens matching the partition (ve at least one (or two) of the corresponding digits, even if the are not given.

How long it will take to find an interesting puzzle with something like this, I have no idea...

Sorry: 2014, not 2010...

Delete