The New Sudoku Players' Forum

by **ArkieTech** » Wed May 01, 2019 11:04 am

Code: Select all: *-----------* |5..|.3.|..4| |..2|1.5|7..| |.6.|...|.9.| |---+---+---| |6..|...|..9| |.8.|6.4|.3.| |...|...|...| |---+---+---| |..5|.7.|1..| |.7.|...|.2.| |...|.1.|...| *-----------*

by **Leren** » Wed May 01, 2019 11:26 am

Code: Select all: *---------------------------------------------------------------------* | 5 19 1789 | 2789 3 26789 | 268 168 4 | | 3489 349 2 | 1 4689 5 | 7 68 368 | | 13478 6 13478 | 2478 248 278 |d2358 9 c25-138 | |----------------------+---------------------+------------------------| | 6 12345 1347 | 23578 258 12378 | 48-25 14578 9 | | 1279 8 179 | 6 259 4 |a25 3 b1257 | | 123479 123459 13479 | 235789 2589 123789 | 468-25 145678 b125678 | |----------------------+---------------------+------------------------| | 23489 2349 5 | 23489 7 23689 | 1 468 368 | | 13489 7 134689 | 34589 45689 3689 | 34689-5 2 3568 | | 23489 2349 34689 | 234589 1 23689 | 34689-5 45678 35678 | *---------------------------------------------------------------------*

M Ring Type B : (5=2) r5c7 - r56c9 = (2-5) r3c9 = (5) r3c7 Loop => - 138 r3c9, - 25 r46c7, - 5 r89c7; lclste, or can be seen as a

Code: Select all: *---------------------------------------------------------------------* | 5 19 1789 | 2789 3 26789 |a268 c1-68 4 | | 3489 349 2 | 1 4689 5 | 7 c68 c368 | | 13478 6 13478 | 2478 248 278 |a2358 9 25-138 | |----------------------+---------------------+------------------------| | 6 12345 1347 | 23578 258 12378 | 48-25 14578 9 | | 1279 8 179 | 6 259 4 |b25 3 b1257 | | 123479 123459 13479 | 235789 2589 123789 | 468-25 145678 b125678 | |----------------------+---------------------+------------------------| | 23489 2349 5 | 23489 7 23689 | 1 468 368 | | 13489 7 134689 | 34589 45689 3689 | 34689-5 2 3568 | | 23489 2349 34689 | 234589 1 23689 | 34689-5 45678 35678 | *---------------------------------------------------------------------*

Sue de Coq: Base Cells r1c7 {268} r3c7 {2358} Pincer Cells r5c7 {25} + r1c8 {168} r2c8 {68} r2c9 {368} => - 68 r1c8 - 138 r3c9, - 25 r46c7, - 5 r89c7; lclste

Leren

by **SpAce** » Wed May 01, 2019 11:38 am

Code: Select all: .------------------------.-----------------------.-----------------------------. | 5 19 1789 | 2789 3 26789 | a268 a68(1) 4 | | 3489 349 2 | 1 4689 5 | 7 a68 368 | | 13478 6 13478 | 2478 248 278 | c2358 9 c23(5)8-1 | :------------------------+-----------------------+-----------------------------: | 6 12345 1347 | 23578 258 12378 | 2458 14578 9 | | 1279 8 179 | 6 259 4 | b25 3 1257 | | 123479 123459 13479 | 235789 2589 123789 | 24568 145678 125678 | :------------------------+-----------------------+-----------------------------: | 23489 2349 5 | 23489 7 23689 | 1 468 368 | | 13489 7 134689 | 34589 45689 3689 | 345689 2 3568 | | 23489 2349 34689 | 234589 1 23689 | 345689 45678 35678 | '------------------------'-----------------------'-----------------------------'

ALS-H3-Wing:

(1=682)b3p251 - (2=5)r5c7 - (5)r3c[7=9] => -1 r3c9; lclste

...or a double-kraken:

Code: Select all: .-------------------------.---------------------------.-------------------------------. | 5 Be19 1789 | e2789 3 e26789 | 268 Cf168 4 | | 3489 349 2 | 1 d4689 5 | 7 68 368 | | 13478 6 13478 | 2478 248 278 | Dgz23(5)8 9 Dgz12358 | :-------------------------+---------------------------+-------------------------------: | 6 A12345 1347 | a23578 a258 a12378 | x2458 14578 9 | | 1279 8 179 | 6 bc2(5)9 4 | 2-5 3 y1257 | | 123479 B123459 13479 | 235789 2589 123789 | 24568 145678 y125678 | :-------------------------+---------------------------+-------------------------------: | 23489 2349 5 | 23489 7 23689 | 1 468 368 | | 13489 7 134689 | 34589 45689 3689 | 345689 2 3568 | | 23489 2349 34689 | 234589 1 23689 | 345689 45678 35678 | '-------------------------'---------------------------'-------------------------------' Kraken Row (2)r4c24567 & Kraken Cell (259)r5c5 (2-1|5)r4c2 = (51)r61c2 - r1c8 = (15)r3c97 || (2)r4c456 - (2)r5c5 || || || (5)r5c5 || || || (9)r5c5 - r2c5 = (91)r1c462 - r1c8 = (15)r3c97 || (2)r4c7 - r56c9 = (25)r3c97 => -5 r5c7; stte

...or as an AIC for lols:

Code: Select all: .-------------------------.--------------------------.--------------------------------. | 5 ch19 1789 | h2789 3 h26789 | 268 bi168 4 | | 3489 349 2 | 1 g4689 5 | 7 68 368 | | 13478 6 13478 | 2478 248 278 | 2358 9 acj(12)-358 | :-------------------------+--------------------------+--------------------------------: | 6 de12345 1347 | e23578 e258 e12378 | e2458 14578 9 | | 1279 8 179 | 6 f259 4 | gh25 3 bdi1257 | | 123479 d123459 13479 | 235789 2589 123789 | 24568 145678 bdi125678 | :-------------------------+--------------------------+--------------------------------: | 23489 2349 5 | 23489 7 23689 | 1 468 368 | | 13489 7 134689 | 34589 45689 3689 | 345689 2 3568 | | 23489 2349 34689 | 234589 1 23689 | 345689 45678 35678 | '-------------------------'--------------------------'--------------------------------'

by **Cenoman** » Wed May 01, 2019 3:15 pm

Code: Select all: +-----------------------------+----------------------------+-----------------------------+ | 5 19 1789 | 2789 3 26789 | b268 168 4 | | 3489 349 2 | 1 4689 5 | 7 68 368 | | 13478 6 13478 | 2478 248 278 |da2358 9 ea25-138 | +-----------------------------+----------------------------+-----------------------------+ | 6 12345 1347 | 23578 258 12378 | 2458 14578 9 | | 1279 8 179 | 6 259 4 | c25 3 1257 | | 123479 123459 13479 | 235789 2589 123789 | 24568 145678 125678 | +-----------------------------+----------------------------+-----------------------------+ | 23489 2349 5 | 23489 7 23689 | 1 468 368 | | 13489 7 134689 | 34589 45689 3689 | 345689 2 3568 | | 23489 2349 34689 | 234589 1 23689 | 345689 45678 35678 | +-----------------------------+----------------------------+-----------------------------+

Almost hidden pair:
(25)r3c79 = (2)r1c7 - (2=5)r5c7 - r3c7 = (5)r3c9 => -138 r3c9; lclste

by **SpAce** » Wed May 01, 2019 4:02 pm

Cenoman wrote:Almost hidden pair:
(25)r3c79 = (2)r1c7 - (2=5)r5c7 - r3c7 = (5)r3c9 => -138 r3c9; ste

Nice! (Should be "lclste", though.)

Btw, how would you express my kraken and the horror AIC as matrices? For the kraken I came up with this:

Code: Select all: 5r3c7 5r3c9 1r3c9 1r1c8 2r3c9 2r56c9 1r1c2 9r1c2 9r1c46 9r2c5 1r1c2 15r46c2 2r4c7 2r4c2 2r4c456 5r5c5 9r5c5 2r5c5 => -5 r5c7; stte

Is it sort of ok? How would you improve it? What about the AIC? I have no idea what to do with it.

Added. This is my attempt for the AIC, but I suspect it has problems:

Code: Select all: 1r3c9 1r1c8 1r1c2 9r1c2 1r1c2 15r46c2 9r1c46 9r2c5 9r5c5 2r5c5 5r5c5 5r5c7 2r5c7 2r4c2 2r4c456 2r4c7 2r3c9 2r56c9 => -358 r3c9; stte

Is the fifth row wrong because it adds two terms?

by **Cenoman** » Wed May 01, 2019 11:09 pm

SpAce wrote:Nice! (Should be "lclste", though.)

Btw, how would you express my kraken and the horror AIC as matrices? For the kraken I came up with this:

Is it sort of ok? How would you improve it? What about the AIC? I have no idea what to do with it.

Added. This is my attempt for the AIC, but I suspect it has problems:

Is the fifth row wrong because it adds two terms?

Thank you for spotting the lclste finish !

As regards your matrices, I need time to analyse. Comments will come asap.

by **SpAce** » Wed May 01, 2019 11:54 pm

Cenoman wrote:As regards your matrices, I need time to analyse. Comments will come asap.

Great! Take your time. If it helps, I think this double-kraken matches with the AIC and is probably easier to decipher:

Code: Select all: (2-1|5)r4c2 = (51)r61c2 - r1c8 = (1)r3c9 || (2)r4c456 - (2)r5c5 || || || (5)r5c5 - (5=2)r5c7 - r56c9 = (2)r3c9 || || || (9)r5c5 - r2c5 = r1c46 - (9=1)r1c2 - r1c8 = (1)r3c9 || (2)r4c7 - r56c9 = (2)r3c9 -> (1|2)r3c9 => -358 r3c9

Btw, I think the AIC might actually have other eliminations in its pocket as well, but I'm not sure which ones. It seems kind of loopish, but since it's really a net in disguise, I wouldn't know which loop eliminations are safe (all applicable seem to work but that's not proof of anything). My preliminary attempt to write it as truths and links yielded this (with at least three additional Rank 0 eliminations, I think):

Alien 9-Fish (Mixed Rank 1/0 (*): {2R4 15C2 2C9 1B3 9B2 1N2 5N57} \ {19r1 5r5 9c5 2b56 3n9 46n2}
=> -358 r3c9, -1 r1c3, -2 r6c7, -5 r5c9 (all Rank 0)

(*) set triple: 1r1c2 => Rank 0 along link 1r1 extending to 3n9 (i.e. r3c9), 2b6, 5r5; Rank 1 elsewhere

All Rank 0 PEs: -19 r1c3, -358 r3c9, -34 r4c2, -2349 r6c6, -2 r6c456, -9 r68c5, -2 r6c7, -5 r5c9

All of them happen to be valid eliminations (because 1r1c2 is false), but I'm quite sure our pattern doesn't prove them all (because it's only partly Rank 0). However, are the ones I listed ok? Any easy way to see that, I wonder?

by **Cenoman** » Thu May 02, 2019 3:44 pm

Space wrote:Btw, how would you express my kraken and the horror AIC as matrices? For the kraken I came up with this:
Code: Select all
Kraken Row (2)r4c24567 & Kraken Cell (259)r5c5 Matrix (2-1|5)r4c2 = (51)r61c2 - r1c8 = (15)r3c97 5r3c7 5r3c9 || 1r3c9 1r1c8 (2)r4c456 - (2)r5c5 2r3c9 2r56c9 || || 1r1c2 9r1c2 || (5)r5c5 9r1c46 9r2c5 || || 1r1c2 15r46c2 || (9)r5c5 - r2c5 = (91)r1c462 - r1c8 = (15)r3c97 2r4c7 2r4c2 2r4c456 || 5r5c5 9r5c5 2r5c5

The matrix is fully OK to me.
Side remark 1: note the presence of 1r1c2 twice in column 3. No trouble in a TM (Triangular Matrix) But this prevents to consider it as a PM (Pigeonhole Matrix), since 1r1c2 isn't in a weak link with itself.
I'd have written the first chain in your kraken consistently with the matrix: (2)r4c2 - (15)r46c2 = (1)r1c2- r1c8 = (15)r3c97 :lol:

Space wrote:What about the AIC? I have no idea what to do with it.
... This is my attempt for the AIC, but I suspect it has problems:
Code: Select all
1r3c9 1r1c8 1r1c2 9r1c2 1r1c2 15r46c2 9r1c46 9r2c5 9r5c5 2r5c5 5r5c5 5r5c7 2r5c7 2r4c2 2r4c456 2r4c7 2r3c9 2r56c9

Is the fifth row wrong because it adds two terms?

...and in a further message,

Space wrote:I think this double-kraken matches with the AIC and is probably easier to decipher:
Code: Select all
(2-1|5)r4c2 = (51)r61c2 - r1c8 = (1)r3c9 || (2)r4c456 - (2)r5c5 || || || (5)r5c5 - (5=2)r5c7 - r56c9 = (2)r3c9 || || || (9)r5c5 - r2c5 = r1c46 - (9=1)r1c2 - r1c8 = (1)r3c9 || (2)r4c7 - r56c9 = (2)r3c9 -> (1|2)r3c9 => -358 r3c9

I guess you understand why I have not even tried to decipher your AIC. Full stop: let's avoid any restart of a past tough discussion.
So, before you posted your equivalent double kraken, I tried to decipher your logic from the strong links in your draft matrix. I got this:

Code: Select all: Double kraken cell (259)r5c5 & Row (2)r4c24567 Triangular matrix (2)r5c5 - (2)r4c456 1r3c9 1r1c8 || || 1r1c2 9r1c2 || (2)r4c2 - (15)r46c2 = r1c2 - r1c8 = (1)r3c9 1r1c2 15r46c2 || || 9r1c46 9r2c5 || (2)r4c7 - r56c9 = (2)r3c9 2r3c9 2r56c9 (5)r5c5 - (5=2)r5c7 - r56c9 = (2)r3c9 2r5c7 5r5c7 || 9r5c5 5r5c5 2r5c5 (9)r5c5-r2c5=r2c12-(9=1)r1c2-r1c8=(1)r3c9 2r4c2 2r4c7 2r4c456

The concern with your draft matrix was, at a first glance, that it was not triangular. The flaw is not fatal. We could be facing a case which requires a BTM (Block Triangular Matrix). I rarely used BTMs. In the present case, as long as there were only two 3-SIS (2=5=9)r5c5 & (2)r4c2=r4c456=r4c7, I suspected a double kraken logic that can always be put into a TM. Fortunately, I found the same double kraken as yours...

Note that my TM has exactly the same lines as yours. I just re-ordered lines and columns.

Edit, from here to the end, demonstrating the eliminations.
Now, there is more to say about it. Noticing (as you already did), that the first column is also a weak inference, some additional eliminations can be anticipated, as for a loop. The weak link in the first column allow it to be any internal column in the matrix. Exchanging the first and second column is trivial. But from the third column, we enter into trouble, because of the double presence of 1r1c2 in the second. I failed to imagine any row-column re-ordering that could keep 1r1c8 above 1r1c2 (and meeting all TM requirements)
As mentioned above, 1r1c2 present twice in column 2 prevents the matrix to be also a Pigeonhole Matrix. Were it a PM, then so good !
In the special case of a PM, when the first column is also a weak inference set, then all columns are proven to be strong inference sets. Steve K. called this a "symmetric pigeonhole matrix", without using the expression "rank 0 logic" though, but it would be appropriate.
I have tried to imagine an arrangement cancelling the presence of 1r1c2 in two rows by concatenating rows 2&3 and columns 3&4, but it doesn't work better.

Hidden Text: Show

This arrangement doesn't improve the issue, since the third column is now the troublemaker.

So, keeping the above matrix, here is the best result I could get:

Code: Select all: 1r3c9 1r1c8 1r1c2 9r1c2 1r1c2 15r46c2 9r1c46 9r2c5 2r3c9 2r56c9 2r5c7 5r5c7 9r5c5 5r5c5 2r5c5 2r4c2 2r4c7 2r4c456 -358r3c9 -1r1c3 -2r6c7 -5r5c9

Bottom line: eliminations that are demonstrated by exchanging c1 with c2, c6, c7 resp. Exchanging c1 vs c3, c4, c5 can't be carried out without bringing 1r1c2 as an element of the upper diagonal. Exchanging c1 vs c8 can't give a valid TM nor PM.
Note: exchanging c1 vs c2, c6, c7 while getting a valid TM has been checked comprehensively. Matrices here.

Hidden Text: Show

The demonstrated eliminations are the same as those from your Alien 9-Fish.

That was the expected conclusion, wasn't it ?

by **SpAce** » Fri May 03, 2019 2:13 pm

Many thanks for your comments, Cenoman! That was very educational! I've been certain for a while now that matrices are a good tool, but unfortunately there are very few resources available to learn them. I sort of understand the principles, but so far I haven't had a clear idea about the differences between the matrix types or how to write more complex cases. A lot of it is still unclear, but this was definitely a step forward! I really appreciate it!

Cenoman wrote:The [kraken] matrix is fully OK to me.

Somewhat surprising, but good to hear!

Side remark 1: note the presence of 1r1c2 twice in column 3.

I noticed and wondered about that but didn't really know its significance or how to circumvent it. You answered both of those questions! First this:

No trouble in a TM (Triangular Matrix) But this prevents to consider it as a PM (Pigeonhole Matrix), since 1r1c2 isn't in a weak link with itself.

I wouldn't have known what matrix type to call it, but I guess it's a TM, then? This is the first time I think I understood the significant differences between TM and PM, so thanks for that! Sure, it's all said here, but it doesn't really explain why the differences matter. I've also needed to rewrite SteveK's examples to be able to read them.

I'd have written the first chain in your kraken consistently with the matrix: (2)r4c2 - (15)r46c2 = (1)r1c2- r1c8 = (15)r3c97

Yes, that's better. I just wrote it the way I saw it, but should have cleaned it up a bit.

I guess you understand why I have not even tried to decipher your AIC.

Yes, I understand that perfectly. It was more fun to write than to read

I kind of enjoy those kinds of exercises because it takes a bit of creativity to come up with a valid AIC, even if the end result is horrible. In this case it also opened up some new views and questions. Without it I wouldn't have noticed the partial Rank 0 potential. (I first attempted to translate the original kraken into an AIC but it was too difficult to synchronize the split chains, so I ended up with a different elimination -- first it was just -5r3c9 with the endpoints 5r3c7 and (1|2)r3c9 but then I noticed that the endpoint 5r3c7 was unnecessary, which resulted in the more interesting variant that spawned these questions.)

The concern with your draft [AIC] matrix was, at a first glance, that it was not triangular.

That was my concern too, though again, I didn't really understand the significance. I just suspected it was a problem. So, I guess it means my (AIC) matrix is neither PM (because of the double column entry of 1r1c2) nor TM (because it's not triangular)?

The flaw is not fatal. We could be facing a case which requires a BTM (Block Triangular Matrix).

So, is my matrix a valid Block Triangular Matrix if left as written? I don't really understand what it means in practice, as SteveK's explanation (and example) is not exactly easy. (The same is even more true about the Mixed BM which I haven't even tried to understand yet.) I guess this is the significant bit about BMs:

SteveK wrote:If there are two top entries in a single row, then they each form a block. Each of these blocks must form triangular matrices.

Does my fifth row have "two top entries" and thus two "blocks"? How do we see and use those if we don't convert it into a TM like you did (or if it's not possible)?

Note that my TM has exactly the same lines as yours. I just re-ordered lines and columns.

Nice! That part is still difficult for me. It felt more natural to have the result on the last line to be able to read the progression from top to bottom, so I didn't even consider your ordering. I think I see now why it works, though: since in a TM the top node of a column has a weak link with everything below (and vice versa, of course) we can also read it from bottom to top when necessary? I guess that can be done in a PM, too, except with more degrees of freedom since every entry in a column is weakly linked with each other?

Now, there is more to say about it. Noticing (as you already did), that the first column is also a weak inference, some additional eliminations can be anticipated, as for a loop.
...
The demonstrated eliminations are the same as those from your Alien 9-Fish.

That was the expected conclusion, wasn't it ?

Well, I'm speechless! I need to analyze this much more and see if I can understand it, but at least I'm happy that my conclusions based on the "Alien Fish" view seem to have been correct. I wasn't at all certain about that either because seeing the boundaries of the Rank 0 region was more of a case by case analysis (the set triplet 1r1c2 both on and off) than anything I could see directly (even with Allan Barker's triplet rules). But, if it can be seen with such matrix transformations, then it gives hope that there may be ways to see it more easily and with more certainty. Thanks for that too! In general, it seems that matrices would be a good tool to visualize truths and links since they have a pretty straight-forward mapping.

Added: I still haven't worked through your matrix transformations, but just one quick question about the bottom line:

Code: Select all
-358r3c9 -1r1c3 -9r6c5 -2r6c7 -5r5c9

Are you saying that -9r6c5 is a proven elimination too? That wasn't in my list, and I still don't see how you'd get that (and if you do, then why not -9r8c5 as well?).

by **Cenoman** » Fri May 03, 2019 11:00 pm

Are you saying that -9r6c5 is a proven elimination too? That wasn't in my list, and I still don't see how you'd get that (and if you do, then why not -9r8c5 as well?).

I have to check that. I'm not at home and I have neither the computer where I stored the matrices, nor the draft papers. I'm afraid I have put 9r1c46 as upper diagonal element, which would not be correct. I check as soon as I'm back home (next Monday). I'll post then the complete set of matrices.

Edit:Check done. 9r6c5 is NOT a proven elimination. Previous post edited. Matrices issued inside.

by **SpAce** » Fri May 03, 2019 11:55 pm

Cenoman wrote:I have to check that. I'm not at home and I have neither the computer where I stored the matrices, nor the draft papers. I'm afraid I have put 9r1c46 as upper diagonal element, which would not be correct. I check as soon as I'm back home (next Monday). I'll post then the complete set of matrices.

Excellent! In the mean time I'll try to get a better understanding of this stuff. About the BTM I was wondering if it simply means that the matrix can be split into two TMs, like this:

My original AIC matrix with the two cases marked as 'a' and 'b':

Code: Select all: 1r3c9 1r1c8 1r1c2 9r1c2 1r1c2 15r46c2 9r1c46 9r2c5 9r5c5 2r5c5:a 5r5c5:b 5r5c7 2r5c7 2r4c2 2r4c456 2r4c7 2r3c9 2r56c9

TM a:

Code: Select all: 1r3c9 1r1c8 1r1c2 9r1c2 1r1c2 15r46c2 9r1c46 9r2c5 9r5c5 2r5c5:a 2r4c2 2r4c456 2r4c7 2r3c9 2r56c9

TM b:

Code: Select all: 1r3c9 1r1c8 1r1c2 9r1c2 9r1c46 9r2c5 9r5c5 5r5c5:b 5r5c7 2r5c7 2r3c9 2r56c9

Is that what BTM means? As you neatly demonstrated, with this example it's simpler to just transform it into a single TM, but I guess it's not possible in more complex cases.

A bit related question. Is this a valid TM alternative:

Code: Select all: 1r3c9 1r1c8 1r1c2 9r1c2 1r1c2 15r46c2 2r3c9 2r56c9 2r5c7 5r5c7 2r4c2 2r4c7 2r4c456 9r1c46 9r2c5 5r5c5 2r5c5 9r5c5

Or is it significant that the two 3-SIS are at the bottom like in yours? (I think yours is more readable anyway. I'm just wondering about the correctness.)

Don't sweat on answering until you have time! I'm just thinking aloud while I try to process these things.

---

Added. I remembered you provided some neat matrices here. At the time I tried to reproduce them but ended up with more lines and probably not valid TMs. Now it seems easier, so I guess you've managed to teach me something! To practice some more I combined eleven's and Steve's solutions:

Code: Select all: .-----------------.----------------------.------------------. | 57 6 57 | ac34(8) 1 348 | 39 389 2 | | 4 3 8 | 9 7 2 | 6 1 5 | | 2 1 9 | 6 3-8 5 | 347 378 a4(8) | :-----------------+----------------------+------------------: | 8 579 1357 | c345 6 1347 | 2 379 19 | | 137 2 6 | ab38 9 1378 | 5 4 a18 | | 1357 579 4 | 2 c35(8) 1378 | 379 3789 6 | :-----------------+----------------------+------------------: | 9 4 35 | 1 35 6 | 8 2 7 | | 6 8 2 | 7 4 9 | 1 5 3 | | 1357 57 1357 | b35#8 2 38 | 49 6 49 | '-----------------'----------------------'------------------'

[Skyscraper 8c49] = (8,3)r95c4 - (3=[XY-Wing 84-45-58])r14c4,r6c5 => -8 r3c5; stte

Interestingly, it reduced to a pretty small "alien fish" (originally 8c4 being both a set and a link, thus canceling out):

Alien 5-Fish (Mixed Rank 3/2 (*)) {8C9 145N4 6N5} \ {8r35 8c5 8b2 35b5 34c4} => -8 r3c5 (Rank 2)

(*) link triplet (3c4b5) => Rank 2 on sets 45N4 extending to link 8r3

...with a corresponding matrix:

Code: Select all: | 8r3c5b2 8r5 3c4b5 4c4 5b5 ---+-------------------------------- 8C9| 8r3c9 8r5c9 5N4| 8r5c4 3r5c4 1N4| 8r1c4 3r1c4 4r1c4 4N4| 3r4c4 4r4c4 5r4c4 6N5| 8r6c5 3r6c5 5r6c5

After I got that on my own, I noticed it was the same as your matrix for eleven's solution minus the first line. Didn't really expect that result, as I thought it would rather be more complex. Thus it seems that matrices are a good way to see the essence of patterns.

by **SpAce** » Sun May 05, 2019 12:49 am

Hi Leren, while trying my hand with the matrix representations I noticed something interesting with your Sue de Coq.

Leren wrote:
Code: Select all
*---------------------------------------------------------------------* | 5 19 1789 | 2789 3 26789 |a268 c1-68 4 | | 3489 349 2 | 1 4689 5 | 7 c68 c368 | | 13478 6 13478 | 2478 248 278 |a2358 9 25-138 | |----------------------+---------------------+------------------------| | 6 12345 1347 | 23578 258 12378 | 48-25 14578 9 | | 1279 8 179 | 6 259 4 |b25 3 b1257 | | 123479 123459 13479 | 235789 2589 123789 | 468-25 145678 b125678 | |----------------------+---------------------+------------------------| | 23489 2349 5 | 23489 7 23689 | 1 468 368 | | 13489 7 134689 | 34589 45689 3689 | 34689-5 2 3568 | | 23489 2349 34689 | 234589 1 23689 | 34689-5 45678 35678 | *---------------------------------------------------------------------*

Sue de Coq: Base Cells r1c7 {268} r3c7 {2358} Pincer Cells r5c7 {25} + r1c8 {168} r2c8 {68} r2c9 {368} => - 68 r1c8 - 138 r3c9, - 25 r46c7, - 5 r89c7; lclste

At first I thought the r1c8 eliminations were wrong (as a direct result) because they seemed kind of cannibalistic or a combined result from a different SDC (without the r1c8 in the pattern). I almost missed the fact that 1r1c8 gets directly locked as a single, so I guess they're ok after all. It might be clearer (and stronger) to express it as +1r1c8, though (because otherwise you could also list 1r1 and 1c8 eliminations). There's also an option to use a simpler SDC without that cell, which I would probably choose.
--
Cenoman, here's my attempt to write the two available SDCs as matrices. They look pretty weird to me, but are those shapes ok as symmetric PMs (obviously not TMs, but we don't even want that since we're dealing with Rank 0, right)? The latter (corresponding with Leren's SDC) has another problem, though, related to the observation above. How do we deal with the locked 1r1c8?

First, the simpler variant (without r1c8); as a loop, as truths and links, and as a matrix:

(5=2)r5c7 - (2=368'5)b3p1567 - loop => -25 r46c7, -5 r89c7, -68 r1c8, -38 r3c9

Alien 5-Fish (Rank 0): {135N7 2N89} \ {25c7 368b3} => the same

Code: Select all: | 5c7 2c7 6b3 8b3 3b3 ---+------------------------------ 5N7| 5r5c7 2r5c7 1N7| 2r1c7 6r1c7 8r1c7 2N8| 6r2c8 8r2c8 2N9| 6r2c9 8r2c9 3r2c9 3N7| 5r3c7 2r3c7 8r3c7 3r3c7 ---+------------------------------ | -5c7 -2c7 -6b3 -8b3 -3b3

Is that ok?

Then the other one (Leren's SDC, including r1c8):

(5=2)r5c7 - (2=1368'5)b3p12567 - loop => -25 r46c7, -5 r89c7, +1r1c8, -38 r3c9

Alien 6-Fish (Rank 0): {135N7 1N8 2N89} \ {25c7 1368b3} => -25 r46c7, -5 r89c7, -138 r3c9

Code: Select all: | 5c7 2c7 6b3 8b3 3b3 1b3 ---+-------------------------------------- 5N7| 5r5c7 2r5c7 1N7| 2r1c7 6r1c7 8r1c7 2N8| 6r2c8 8r2c8 2N9| 6r2c9 8r2c9 3r2c9 1N8| 6r1c8 8r1c8 1r1c8 3N7| 5r3c7 2r3c7 8r3c7 3r3c7 (1r3c7) ---+-------------------------------------- | -5c7 -2c7 -6b3 -8b3 -3b3 -1b3

As you can see, I didn't know what to do with the 1r1c8 so I introduced a ghost candidate in r3c7. That doesn't seem right at all. How would you solve the problem? I guess the simplest (and best) way is to revert to the simpler matrix without that extra cell. Using the Alien Fish view it's easy to cancel the extra set (1N8) by using 1n8 also as a link (instead of 1b3):

{135N7 1N8 2N89} \ {25c7 1n8 368b3} <=> {135N7 2N89} \ {25c7 368b3}

Added. After a bit of rethinking, I guess it's not that complicated after all. Why not simply accept it as a single value column proving a single? Since it's a symmetric PM we know that every column is a SIS -- so why not a SIS of size one? Then we can apply all four cover sets in parallel. Thus simply:

Siamese Alien 6-Fish (Rank 0): {135N7 1N8 2N89} \ {25c7 368b3 1n8|1r1|1c8|1b3} => -25 r46c7, -5 r89c7, -138 r3c9, -68 r1c8, -1 r1c2,r46c8

Code: Select all: | 1n8| | 1r1| | 1c8| | 5c7 2c7 6b3 8b3 3b3 1b3 ---+-------------------------------------- 5N7| 5r5c7 2r5c7 1N7| 2r1c7 6r1c7 8r1c7 2N8| 6r2c8 8r2c8 2N9| 6r2c9 8r2c9 3r2c9 1N8| 6r1c8 8r1c8 1r1c8 3N7| 5r3c7 2r3c7 8r3c7 3r3c7 ---+-------------------------------------- | -5c7 -2c7 -6b3 -8b3 -3b3 +1r1c8

(Still I think the simpler variant is better.)
--
Added. Here's another unrelated exercise I did for this puzzle.

Code: Select all: .-------------------.------------------.--------------. | 6 58 1 | 58 9 3 | 4 7 2 | | 457 457 9 | 6 2 (47)5 | 1 8 3 | | 247 3 278 | 1 (8)-4 (47) | 69 69 5 | :-------------------+------------------+--------------: | 2459 4589 258 | 7 6 [245] | 3 1 48 | | 457 45678 5678 | 9 3 1 | 2 45 468 | | 3 1 256 | 458 458 245 | 7 459 469 | :-------------------+------------------+--------------: | 8 2 3 | 45 7 9 | 56 46 1 | | 157 567 567 | 3 1(4)5 8 | 59 2 49 | | 159 59 4 | 2 15 6 | 8 3 7 | '-------------------'------------------'--------------'

Kraken Cell (245)r4c6

(2)r4c6 - r6c6 = (26-9)r6c39 = (94)r8c95
||
(48)r4c69 - (4|8)r4c2 = (59,8)r491c2 - r3c3 = (8)r3c5
||
(5)r4c6 - (5=74)r23c6

=> -4 r3c5; stte

Alien 12-Fish (Rank 2):
{8R3 26R6 4R8 9C9 19N2 23N6 4N269} \ {48r4 59c2 4c5 25c6 8b1 47b2 3n5 6n3 68n9} => -4 r3c5

Code: Select all: | 3n5 | 47b2 | 4c5 8b1 5c2 9c2 48r4 5c6 2c6 6n3 6n9 8n9 -----+---------------------------------------------------------------- 8R3 | 8r3c5 8r3c3 1N2 | 8r1c2 5r1c2 9N2 | 5r9c2 9r9c2 4N29| 5r4c2 9r4c2 48r4c29 23N6 | 47r23c6 5r2c6 4N6 | 4r4c6 5r4c6 2r4c6 2R6 | 2r6c6 2r6c3 6R6 | 6r6c3 6r6c9 9C9 | 9r6c9 9r8c9 4R8 | 4r8c5 4r8c9 -----+---------------------------------------------------------------- | -4r3c5

Is that an ok TM? Presuming so, I'd next like to break down the combo sets. I (think I) can do it for 47r23c6:

Code: Select all: | 3n5 | 4b2 | 4c5 8b1 5c2 9c2 48r4 7b2 5c6 2c6 6n3 6n9 8n9 -----+---------------------------------------------------------------------- 8R3 | 8r3c5 8r3c3 1N2 | 8r1c2 5r1c2 9N2 | 5r9c2 9r9c2 4N29| 5r4c2 9r4c2 48r4c29 3N6 | 4r3c6 7r3c6 2N6 | 4r2c6 7r2c6 5r2c6 4N6 | 4r4c6 5r4c6 2r4c6 2R6 | 2r6c6 2r6c3 6R6 | 6r6c3 6r6c9 9C9 | 9r6c9 9r8c9 4R8 | 4r8c5 4r8c9 -----+---------------------------------------------------------------------- | -4r3c5

...but I can see no way to maintain the TM shape if I try to do it for the other:

Code: Select all: | 3n5 | 4b2 | 4c5 8b1 5c2 9c2 8r4 4r4 7b2 5c6 2c6 6n3 6n9 8n9 -----+-------------------------------------------------------------------------- 8R3 | 8r3c5 8r3c3 1N2 | 8r1c2 5r1c2 9N2 | 5r9c2 9r9c2 4N2 | 5r4c2 9r4c2 8r4c2 4r4c2 4N9 | 8r4c9 4r4c9 3N6 | 4r3c6 7r3c6 2N6 | 4r2c6 7r2c6 5r2c6 4N6 | 4r4c6 5r4c6 2r4c6 2R6 | 2r6c6 2r6c3 6R6 | 6r6c3 6r6c9 9C9 | 9r6c9 9r8c9 4R8 | 4r8c5 4r8c9 -----+-------------------------------------------------------------------------- | -4r3c5

Is that thus a BTM? Can you see any other way to do it?
--
Yet another exercise relating to this.

Code: Select all: .-------------------.-----------------.------------------. | 1 9 4 | 8 3 5 | 2 6 7 | | 58 7 58 | 2 9 6 | 1 4 3 | | 3 6 2 | 1 4 7 | 59 8 59 | :-------------------+-----------------+------------------: | 4 1235 *3(9) | 7 6 *23(9) | 8 135-9 5-9 | | 2689 1235 68 | 345 25 23489 | 7 1359 46 | | 689 35 7 | 345 1 3489 | 46 359 2 | :-------------------+-----------------+------------------: | 56 4 56 | 9 7 1 | 3 2 8 | | 7 *23 1 | 345 8 *234 | 4569 59 46 | | 29 8 *39 | 6 25 234 | 45 7 1 | '-------------------'-----------------'------------------'

First, using Clement's original sets:

Alien 4-Fish (Mixed Rank 1/0 (*)) {2R8 3B7 4N36} \ {39r4 3c3 2c6 8n2} => -9 r4c89 (Rank 0)
(*) Link triplet 3r4c3 => Rank 0 in set 4N3 extending to link 9r4

Code: Select all: | 3c3 | 9r4 3r4 8n2 2c6 ---+-------------------------- 4N3| 9r4c3 3r4c3 3B7| 3r9c3 3r8c2 2R8| 2r8c2 2r8c6 4N6| 9r4c6 3r4c6 2r4c6 ---+-------------------------- | -9r4c89

I guess that's a valid TM. What about this slightly different view of the same:

Alien 5-Fish (Mixed Rank 1/0 (*)) {2R8 39C3 3B7 4N6} \ {39r4 2c6 8n2 9n3} => -9 r4c89 (Rank 0)
(*) Set triplet 3r9c3 => Rank 0 along link 9n3 extending to 9r4

Code: Select all: | 9r4 9n3 8n2 3r4 2c6 -------+-------------------------------- 9C3 | 9r4c3 9r9c3 3C3,3B7| 3r9c3 3r8c2&3r4c3 2R8 | 2r8c2 2r8c6 4N6 | 9r4c6 3r4c6 2r4c6 -------+-------------------------------- | -9r4c89

Would that work, or any cleaner way to do it (with those sets)? As written it looks like a 4x5 matrix (which shouldn't happen) but columns 3 and 4 are just separated for clarity and count as one, so it's still 4x4.

by **Cenoman** » Mon May 06, 2019 11:11 pm

Hi SpAce,
I was preparing to answer your question

SpAce wrote:Are you saying that -9r6c5 is a proven elimination too?

... and I find a lot of further questions !

To the above question, the answer is definitely No. I have edited my previous posts and confirmed my first conclusion (same eliminations as the alien-fish).

Now, you propose to split your original AIC matrix into two TMs, in order to demonstrate that it is a BTM

Hidden Text: Show

SpAce wrote:Is that what BTM means? As you neatly demonstrated, with this example it's simpler to just transform it into a single TM, but I guess it's not possible in more complex cases.

I have no answer to this question. The lack of experience in BTMs makes me very careful. I have just a concern with your TM a and TM b.
They are triangular, for sure, and row 5 in TM a is in a OR with row 4 in TM b, but I am troubled by those rows:
9r5c5 2r5c5 (TM a) is not a SIS in the puzzle pencilmarks, nor 9r5c5 5r5c5 (TM b). Then, in Steve K's example it seems that every line in the split matrices is a full puzzle SIS. Such a requirement is not written explicitly though. So, your question remain opened.

SpAce wrote:A bit related question. Is this a valid TM alternative:
Hidden Text: Show
Code: Select all
1r3c9 1r1c8 1r1c2 9r1c2 1r1c2 15r46c2 2r3c9 2r56c9 2r5c7 5r5c7 2r4c2 2r4c7 2r4c456 9r1c46 9r2c5 5r5c5 2r5c5 9r5c5

Or is it significant that the two 3-SIS are at the bottom like in yours? (I think yours is more readable anyway. I'm just wondering about the correctness.)

Yes, this a valid TM. The 3-SIS are not required to be at the bottom (and you could imagine other permutations of rows, but keep always 1r1c8 above 1r1c2).

As to your proposal for December 11,2018 , nice 5*5 matrix !

Hidden Text: Show

Why not try something more compact ?

Code: Select all: | 8r3c5b2 8r5 34c4b5 5b5 ----+-------------------------------- 8C9 | 8r3c9 8r5c9 5N4 | 8r5c4 3r5c4 14N4| 8r1c4 34r14c4 5r4c4 6N5 | 8r6c5 3r6c5 5r6c5

Puzzle of the day.

SpAce wrote:here's my attempt to write the two available SDCs as matrices. Is that ok?
Hidden Text: Show
Code: Select all
| 5c7 2c7 6b3 8b3 3b3 ---+------------------------------ 5N7| 5r5c7 2r5c7 1N7| 2r1c7 6r1c7 8r1c7 2N8| 6r2c8 8r2c8 2N9| 6r2c9 8r2c9 3r2c9 3N7| 5r3c7 2r3c7 8r3c7 3r3c7 ---+------------------------------ | -5c7 -2c7 -6b3 -8b3 -3b3

OK for this PM

I find it difficult to accept to derogate from the rule of filling matrices with SIS in rows and WIS in columns. So, including r1c8 in the SDC needs some contortions such as the ghost candidate or the trivial SIS size one. If there were a clear benefit, maybe. With this example I am not convinced.

Next exercise

SpAce wrote:
Hidden Text: Show
Code: Select all
Kraken Cell (245)r4c6 (2)r4c6 - r6c6 = (26-9)r6c39 = (94)r8c95 || (48)r4c69 - (4|8)r4c2 = (59,8)r491c2 - r3c3 = (8)r3c5 || (5)r4c6 - (5=74)r23c6 => -4 r3c5; stte | 3n5 | 47b2 | 4c5 8b1 5c2 9c2 48r4 5c6 2c6 6n3 6n9 8n9 -----+---------------------------------------------------------------- 8R3 | 8r3c5 8r3c3 1N2 | 8r1c2 5r1c2 9N2 | 5r9c2 9r9c2 4N29| 5r4c2 9r4c2 48r4c29 23N6 | 47r23c6 5r2c6 4N6 | 4r4c6 5r4c6 2r4c6 2R6 | 2r6c6 2r6c3 6R6 | 6r6c3 6r6c9 9C9 | 9r6c9 9r8c9 4R8 | 4r8c5 4r8c9 -----+---------------------------------------------------------------- | -4r3c5

Is that an ok TM?

Yes, valid TM

SpAce wrote:Presuming so, I'd next like to break down the combo sets. I (think I) can do it for 47r23c6:
...but I can see no way to maintain the TM shape if I try to do it for the other:
Hidden Text: Show
Code: Select all
| 3n5 | 4b2 | 4c5 8b1 5c2 9c2 8r4 4r4 7b2 5c6 2c6 6n3 6n9 8n9 -----+-------------------------------------------------------------------------- 8R3 | 8r3c5 8r3c3 1N2 | 8r1c2 5r1c2 9N2 | 5r9c2 9r9c2 4N2 | 5r4c2 9r4c2 8r4c2 4r4c2 4N9 | 8r4c9 4r4c9 3N6 | 4r3c6 7r3c6 2N6 | 4r2c6 7r2c6 5r2c6 4N6 | 4r4c6 5r4c6 2r4c6 2R6 | 2r6c6 2r6c3 6R6 | 6r6c3 6r6c9 9C9 | 9r6c9 9r8c9 4R8 | 4r8c5 4r8c9 -----+-------------------------------------------------------------------------- | -4r3c5

Is that thus a BTM? Can you see any other way to do it?

I don't know... Didn't try another way, since this matrix is a PM (as well as your first version)
To check if a matrix is a PM, just check the weak links in columns containing > 2 candidates (here column 3 and 6). Column 1 is out of the check.

Puzzle May 5, 2019

Hidden Text: Show

SpAce wrote:I guess that's a valid TM.

Agreed

SpAce wrote:What about this slightly different view of the same:
Hidden Text: Show
Code: Select all
| 9r4 9n3 8n2 3r4 2c6 -------+-------------------------------- 9C3 | 9r4c3 9r9c3 3C3,3B7| 3r9c3 3r8c2&3r4c3 2R8 | 2r8c2 2r8c6 4N6 | 9r4c6 3r4c6 2r4c6 -------+-------------------------------- | -9r4c89

Would that work, or any cleaner way to do it (with those sets)?

I tried something similar in the first example (May 1,2019) without benefit. Note here that the second column in your first version prevent the matrix from being a PM. It is the same with your modified version (third column, consider that 2r8c2 and 3r4c6 are in the same column). It is valid, but I can't see any benefit: same size of matrix, same eliminations and clarity not improved.

Now, you can take your own way to write matrices. They are a nice tool to present complex patterns, but they have quite no heuristic capacity.

by **SpAce** » Tue May 07, 2019 4:46 pm

Cenoman wrote:I was preparing to answer your question
SpAce wrote:Are you saying that -9r6c5 is a proven elimination too?

... and I find a lot of further questions !

Yes, sorry about that, and thanks a lot for taking the time to answer them anyway! I tend to get carried away when I'm trying to learn a new skill, especially if there's a good teacher around! This is kind of like learning to write AICs from scratch -- the basic idea is simple, but anything deviating from trivial examples spawns new questions.

To the above question, the answer is definitely No. I have edited my previous posts and confirmed my first conclusion (same eliminations as the alien-fish).

Thanks for confirming that! The opposite result would have been quite confusing.

Now, you propose to split your original AIC matrix into two TMs, in order to demonstrate that it is a BTM
...
I have no answer to this question. The lack of experience in BTMs makes me very careful. I have just a concern with your TM a and TM b.
They are triangular, for sure, and row 5 in TM a is in a OR with row 4 in TM b, but I am troubled by those rows:
9r5c5 2r5c5 (TM a) is not a SIS in the puzzle pencilmarks, nor 9r5c5 5r5c5 (TM b). Then, in Steve K's example it seems that every line in the split matrices is a full puzzle SIS. Such a requirement is not written explicitly though. So, your question remain opened.

Ok, thanks for that too! Let's keep it open then. Too bad SteveK's BTM example is of limited help. Perhaps there are other examples in some of his solutions, but I haven't digged into those. Here's how I see the provided one (now that I translated it into something I think I understand):

Code: Select all: SteveK's BTM example: | 2r7 1r6 6b5 1r3 *d* | 7n4 5n8 1b5 1c2 6n6 6c2 1r7 3n6 4n8 4n9 7c3 2n6 ---+--------------------------------------------------------------------------- 2C8| 2r7c8 2r5c8 6N6| 1r6c6 6r6c6 1R7| 1r7c4 1r7c6 1r7c2 1R5| 1r5c8 1r5c4 1r5c2 6R5| 6r5c8 6r5c4 6r5c2 *D*| 6r9c2 1r7c2 1C6| 1r6c6 1r7c6 1r3c6 1C8| 1r5c8 1r3c8 1r4c8 6B6| 6r45c8 6r4c9 7R4| 7r4c8 7r4c9 7r4c3 7R2| 7r2c3 7r2c6 2C6| 2r7c6 2r3c6 2r2c6 ---+--------------------------------------------------------------------------- | -2r7c4 *D* and *d* refer to the derived strong and weak sets used in Steve's logic.

As far as I see, the only real instruction for interpreting that is this:

SteveK wrote:Note that (1)r6c6 reverts cleanly back to matrix column 1 with a 3x3 Triangular matrix, whilst (6) r6c6 leaves in its wake a 9x9 Triangular matrix.

I take it means that we try both cases of row 2 (6N6) and end up with these two strongly linked TMs:

Code: Select all: Case 1r6c6 (3x3 TM): | 2r7 5n8 1c2 ---+------------------- 2C8| 2r7c8 2r5c8 1R5| 1r5c8 1r5c2 1R7| 1r7c4 1r7c2 ---+------------------- | -2r7c4

Code: Select all: Case 6r6c6 (9x9 TM): | 2r7 1r3 *d* | 7c4 5n8 6c2 1r7 3n6 4n8 4n9 7c3 2n6 ---+--------------------------------------------------------- 2C8| 2r7c8 2r5c8 6R5| 6r5c8 6r5c2 *D*| 6r9c2 1r7c2 1C6| 1r7c6 1r3c6 1C8| 1r5c8 1r3c8 1r4c8 6B6| 6r45c8 6r4c9 7R4| 7r4c8 7r4c9 7r4c3 7R2| 7r2c3 7r2c6 2C6| 2r7c6 2r3c6 2r2c6 ---+--------------------------------------------------------- | -2r7c4

Do you see it the same way? What's different from my earlier attempt is that the two cases are strongly linked to begin with, so they can be assumed as singles in the sub-matrices (with corresponding eliminations), which simplifies the cases. In my earlier example they were part of a 3-SIS turning into two cases of 2-SIS. I would think it's a logical extension to the same principle, but I can't be sure without seeing more examples. What do you think of that interpretation?

Yes, this a valid TM. The 3-SIS are not required to be at the bottom (and you could imagine other permutations of rows, but keep always 1r1c8 above 1r1c2).

Ok, thanks for that confirmation too!

As to your proposal for December 11,2018 , nice 5*5 matrix !

Thanks! Eleven's original solution was so cool that it was fun to work with. Combining it with Steve's also cool one made it even more interesting -- and surprisingly the end result (as a matrix) was simpler than the original.

Why not try something more compact ?
Code: Select all
| 8r3c5b2 8r5 34c4b5 5b5 ----+-------------------------------- 8C9 | 8r3c9 8r5c9 5N4 | 8r5c4 3r5c4 14N4| 8r1c4 34r14c4 5r4c4 6N5 | 8r6c5 3r6c5 5r6c5

[Added 5r4c4 into row 3.] Yes, in this case it's both shorter and probably closer to the way most would see it on the grid. I think I considered writing it like that, but at this time I like to keep my matrices as uncompressed as possible just like when I was learning AICs. It makes them simpler to understand for a matrix novice like myself. I also think that one benefit of matrices is that they can be more easily broken into smaller pieces than AICs (which often depend on group and ALS nodes to avoid nets), so one can see what actually happens on the cell level. With more experience it probably gets more natural to use larger nodes in matrices as well, but first I want to learn how far they can be uncompressed.

SpAce wrote:here's my attempt to write the two available SDCs as matrices. Is that ok?

I find it difficult to accept to derogate from the rule of filling matrices with SIS in rows and WIS in columns. So, including r1c8 in the SDC needs some contortions such as the ghost candidate or the trivial SIS size one. If there were a clear benefit, maybe. With this example I am not convinced.

I agree. I just wanted to find out how it should be done if need be. My initial feeling was the same as yours about the column WIS rule being broken, but maybe it's not really broken in a symmetric PM? I presume it wouldn't work in a TM or a non-symmetric PM.

Next exercise
SpAce wrote:Kraken Cell (245)r4c6
...
Is that thus a BTM? Can you see any other way to do it?

I don't know... Didn't try another way, since this matrix is a PM (as well as your first version)
The check if a matrix is a PM, just check the weak links in columns containing > 2 candidates (here column 3 and 6). Column 1 is out of the check.

Wow! That was a major revelation! Thanks! I didn't even think it that way, but you're right of course. I think I finally picked up a missing piece of the TM vs PM mystery. In set logic terms, it seems that we need a TM when there's at least one set or link triplet, which in the matrix translates into an ANDed row node or a double entry in a a column or otherwise a column being not fully weakly-linked, all preventing a PM. And vice versa, if a matrix has any of those features its set logic form has at least one triplet, so it can't be a PM. Here it doesn't, so it can -- which is nice. Would you agree with that?

Puzzle May 5, 2019
SpAce wrote:What about this slightly different view of the same:
...
Would that work, or any cleaner way to do it (with those sets)?

I tried something similar in the first example (May 1,2019) without benefit. Note here that the second column in your first version prevent the matrix from being a PM. It is the same with your modified version (third column, consider that 2r8c2 and 3r4c6 are in the same column). It is valid, but I can't see any benefit: same size of matrix, same eliminations and clarity not improved.

Yes, I definitely agree that the first version is simpler and easier to read. I just thought it was an interesting comparison because the first matrix corresponds with a link triplet and the second one with a set triplet. Both kinds prevent it from being a PM. The link triplet version seems cleaner, but sometimes the set triplet might give a simpler solution, like here. Does that seem ok? If I try to do it the other way, I end up with something like this:

Alien 3-Fish (Mixed Rank 1/0) 6:R3C67\r15b12 => -6 r1c35; stte
(link triplet 6r12c6 => Rank 0 in set C6 extending to 6r1)

...which I think is harder to explain as an AIC or a matrix:

Code: Select all: .-----------------------.-------------------------.-----------------------. | 2 457 458-6 | 1 8-6 d5(6)78 | a(6)7 9 3 | | 13579 13579 3569 | 25679 2369 d25(6)7 | 4 267 8 | | 379 379 f3(6)89 | e26789 e23689 4 | 1 5 27 | :-----------------------+-------------------------+-----------------------: | 3457 3457 1 | 568 468 9 | 2 34678 47 | | 3459 8 23459 | 256 7 c1256 | b369 1346 149 | | 6 2479 249 | 3 1248 128 | 789 1478 5 | :-----------------------+-------------------------+-----------------------: | 1459 12459 7 | 2689 12689 3 | 589 1248 1249 | | 1349 6 2349 | 2789 5 1278 | 3789 123478 12479 | | 8 12359 2359 | 4 129 127 | 3579 1237 6 | '-----------------------'-------------------------'-----------------------'

As a multi-headed AIC and my matrix translation of that:

AB:(6)r1c7 = r5c7 - r5c6 = A:(6)r12c6 - r3c45 = B:(6)r3c3 => -6 r1c35; stte

Code: Select all: | 6r1b1 6r5 6b2 ---+------------------- 6C7| 6r1c7 6r5c7 6C6| 6r5c6 6r12c6 6R3| 6r3c3 6r3c45 ---+------------------- | -6r1c3 -6r1c5

Seems simple, except for the second elimination. How are we to see the sub-chain 6r1c7==6r12c6 proving it?

...or as a split-node AIC and the corresponding matrix (the way I see it):

(6)r1c7 = r1c5 - r5c6 = r12c6 - r3c45|r5c6 = (6)r3c3&r12c6 => -6 r1c35; stte

Code: Select all: | 6r1b12 6r5 6b2c6 ----+--------------------- 6C7| 6r1c7 6r5c7 6C6| 6r5c6 6r12c6 [6R3| &6r3c3 6r3c45 [6C6| &6r12c6 6r5c6 ----+--------------------- | -6r1c3 | -6r1c5

Now both eliminations are perhaps easier to see, but it's otherwise complicated and redundant. How would you write that?

by **Cenoman** » Fri May 10, 2019 3:50 pm

Steve K's BTM example:

SpAce wrote:As far as I see, the only real instruction for interpreting that is this:

SteveK wrote:Note that (1)r6c6 reverts cleanly back to matrix column 1 with a 3x3 Triangular matrix, whilst (6) r6c6 leaves in its wake a 9x9 Triangular matrix.

I take it means that we try both cases of row 2 (6N6) and end up with these two strongly linked TMs:
[....]
Do you see it the same way? What's different from my earlier attempt is that the two cases are strongly linked to begin with, so they can be assumed as singles in the sub-matrices (with corresponding eliminations), which simplifies the cases. In my earlier example they were part of a 3-SIS turning into two cases of 2-SIS. I would think it's a logical extension to the same principle, but I can't be sure without seeing more examples. What do you think of that interpretation?

Without any certainty, I'd say your interpretation is the right one. I long wondered what Steve K meant in his quoted sentence. I had in mind a process where the questionable entries had to be erased from the matrices with the cascade of linked terms. Your interpretation of a trial of different possible cases is convincing. So thank you for opening my eyes on that.

Then, back to your original matrix for May 1, 2019 (this thread), your matrices TM a & TM b are in line with this process and you can claim that your original matrix is a BTM. It would be strange to find that a re-ordered valid TM is not a BTM (but this is not an argument).

Puzzle of May 6, 2019

You have proposed 4 different matrices

Hidden Text: Show

SpAce wrote: How would you write that?

I'dsay the first one is a symmetic PM (2x2 with first column=WIS is always a PM), not easy to read. Symmetry is of no use, as the second column eliminates nothing.
The fourth one is, as you state yourself "complicated and redundant"

SpAce wrote:How are we to see the sub-chain 6r1c7==6r12c6 proving it?

in the third one; nice question...

So, my prefered is the second one. It is the matrix of a simple kraken column:
(6)r1c6
(6)r2c6 - r2c8 = r1c7
(6)r5c6 - r5c7 = r1c7
=> -6 r1c35
The matrix (TM and PM, but not symmetric), as well as the kraken show clearly the three needed SIS: 6C6, 6C7, 6B3 and the eliminations.
Note these three SIS are present in all other trials, and they are sufficient, as shown by this version.

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