The New Sudoku Players' Forum

by **SpAce** » Sat May 11, 2019 1:48 am

Cenoman wrote:Steve K's BTM example:
...
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.

I feel more assured if you think that way. However, I'd still keep the question open because the same interpretation doesn't seem to work for the Mixed Block Matrix. I finally dared to look at that example, and its logic seems different. Based on the similar names, I would have imagined that the BTM is just a special case of MBM, where both (or all?) blocks are triangular while the latter allows other types of blocks as well (such as PM) -- but having the same definition and usage of the blocks. I'm not sure if that's the case. Either I've misunderstood something (quite possible), or they're somewhat different concepts.

Anyway, here's how I see SteveK's MBM example:

Mixed Block Matrix: Show

Code: Select all: Easter Monster after SK-Loop: .------------------.-----------------.------------------. | 1 478 3458 | 3567 389 5678 | 3489 369 2 | | 238 9 378 | 4 126 1267 | 138 5 368 | | 3458 248 6 | 1235 389 1258 | 7 139 3489 | :------------------+-----------------+------------------: | 2468 5 1478 | 9 126 3 | 128 1267 678 | | 389 126 389 | 126 7 4 | 3589 126 3589 | | 2369 1267 1379 | 8 5 126 | 1239 4 3679 | :------------------+-----------------+------------------: | 7 148 4589 | 1235 348 1258 | 6 239 3459 | | 456 3 145 | 1267 126 9 | 245 8 457 | | 4589 468 2 | 3567 348 5678 | 3459 379 1 | '------------------'-----------------'------------------'

The full 9x9 MBM:

Code: Select all: | 1c2,1b7 6c2,6b7 2r5 4n8 2b9 7b9 8n4 8n5 ---+--------------------------------------------------------------- 1B7| 1r8c3 1r7c2 6B7| 6r8c1 6r9c2 5N2| 1r5c2 6r5c2 2r5c2 2C8| 2r5c8 2r4c8 2r7c8 7C8| 7r4c8 7r9c8 7R8| 7r8c9 7r8c4 2R8| 2r8c7 2r8c4 2r8c5 1R8| 1r8c3 1r8c4 1r8c5 6R8| 6r8c1 6r8c4 6r8c5 ---+--------------------------------------------------------------- | (1==6)r8c13 | (1--6)r79c2

In other words, the matrix doesn't prove any direct eliminations but two derived inferences (or actually one, the other being a result of the first), but that's not relevant. The real question is how the blocks are defined in this case. The only piece of instruction we have is this:

SteveK wrote:The bold items represent a 6x6 pigeonhole matrix. The result column of this 6x6 pigeonhole matrix is :[(1)r8c3=(6)r8c1]=(2)r5c8. The other items are part of a general Triangular matrix.

I can't mark the "bold items" in the [code block] but they're rows 4..9, columns 5..9 -- quite confusingly since that would be a 6x5 matrix. Based on the other information I guess we should add the other cells in rows 4, 8, and 9 to get the actual 6x6 PM:

Code: Select all: | 4n8 2b9 7b9 8n4 8n5 ---+------------------------------------- 2C8| 2r5c8 2r4c8 2r7c8 7C8| 7r4c8 7r9c8 7R8| 7r8c9 7r8c4 2R8| 2r8c7 2r8c4 2r8c5 1R8| 1r8c3 1r8c4 1r8c5 6R8| 6r8c1 6r8c4 6r8c5 ---+------------------------------------- <=> (2)r5c8 == (1|6)r8c13

Once that part is removed from the full matrix, what's left is this 3x3 TM (isn't it a PM also?):

Code: Select all: | 1c2 6c2 ---+--------------------- 1B7| 1r8c3 1r7c2 6B7| 6r8c1 6r9c2 5N2| 2r5c2 1r5c2 6r5c2 ---+--------------------- <=> (1|6)r8c13 == (2)r5c2

When those two are combined we get an AIC:

(1|6)r8c13 =TM= (2)r5c2 - (2)r5c8 =PM= (1|6)r8c13

...which does correspond with SteveK's description of the MBM: "An AIC of matrices".

Do you agree with that workout and conclusion? On the other hand, as an AIC the BTM example would seem more like this to me: TMa == TMb. Maybe I'm missing something, but to me the two concepts seem different, at least based on this very small sample size. Conceptually the MBM example seems actually easier to understand because of the clearly defined relationship and boundaries between the blocks.

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).

Ok! I guess we can at least assume so for now.

Puzzle of May 6, 2019
...
So, my prefered is the second one. It is the matrix of a simple kraken column

Nice! Why didn't I see that in the first place?

Doesn't it mean that we can (and should) drop the ugly '&'s as well? I think they make it incorrect or at least very confusing, because it looks like a 2x3 matrix (but is really a disguised 2x2). Without the '&'s it would be an honest 3x3, right? I would prefer that too. Similarly, I now think my first attempt is wrong as written, and should be more clearly a 2x2 by adding the '|':

Code: Select all: | 6r1 6r2,6r5 -------+--------------------- 6C7,6B3| 6r1c7 6r2c8&6r5c7 6C6 | 6r1c6 6r2c6|6r5c6 -------+--------------------- | -6 r1c35

Do you agree? (In the fourth one I might accept the way it's written because it's not that hard to see as a 3x3 despite the weird shape, and the double row improves readability. It's otherwise so ugly that the question is moot, though.)

The matrix (TM and PM, but not symmetric)

I guess what breaks the symmetry is that we have two of the same entries in the result column which can't be weakly linked with each other?

by **Cenoman** » Mon May 13, 2019 8:43 pm

SpAce wrote:I now think my first attempt is wrong as written, and should be more clearly a 2x2 by adding the '|':
Hidden Text: Show
Code: Select all
| 6r1 6r2,6r5 -------+--------------------- 6C7,6B3| 6r1c7 6r2c8&6r5c7 6C6 | 6r1c6 6r2c6|6r5c6 -------+--------------------- | -6 r1c35

Do you agree?

It is the way I read it (to me the OR '|' was implicit) Otherwise I'd have commented... So, of course, I agree

SpAce wrote:I guess what breaks the symmetry is that we have two of the same entries in the result column which can't be weakly linked with each other?

Exactly.

I say nothing at the moment about Steve K's Mixed Block Matrices. I need time to study this combo that I never used. Topic to re-open later.

I propose to close this thread, that has been fruitful for both of us (at least it has been to me).

EDIT: Added, May 27, 2019

SpAce wrote:Anyway, here's how I see SteveK's MBM example:

Do you agree with that workout and conclusion? On the other hand, as an AIC the BTM example would seem more like this to me: TMa == TMb. Maybe I'm missing something, but to me the two concepts seem different, at least based on this very small sample size. Conceptually the MBM example seems actually easier to understand because of the clearly defined relationship and boundaries between the blocks.

Having found a little time to study again Steve K's MTM example, here are my comments on:
- your MBM interpretation,
- your thoughts about linking sub-matrices.

1) SteveK's MBM example:
Using your re-writing, Steve K's original matrix:

Hidden Text: Show

I agree with you, that the 6x6 PM considered by Steve K in his short explanation is the one you wrote:

Hidden Text: Show

... and I agree with your "chain": (1|6)r8c13 =TM= (2)r5c2 - (2)r5c8 =PM= (1|6)r8c13

This makes me think that the original matrix could also have been written with the leftest items in rows 8, 9 placed in the first column

Hidden Text: Show

2) As regards your comparison of the 'OR' relation between matrices in both examples (BTM and MBM), I also agree with you. The overall chain for the BTM is rather of the type:
(1)r6c6 => TM a
||
(6)r6c6 => TM b
Written as a chain: TM a = NOT(1r6c6) - (1=6)r6c6 - NOT(6r6c6) = TM b
For such chains, I'd personally use a notation TM a = !(1)r6c6 - (1=6)r6c6 - !(6)r6c6 = TM b
in which the NOT symbol '!' is borrowed to Java programming language, but I do not use it on the forum ... I'd rather use the implication symbol '=>" (for once, I'm the one who likes the shortest

)

Now, further this formal point of view, if you consider the items in column 3, below 1r6c6, you can see the 3x4 resulting matrix as an almost-almost TM:

Hidden Text: Show

and similarly, if you consider the items in column 5, below 6r6c6, you can see the 9x10 resulting matrix as an almost-almost TM:

Hidden Text: Show

You then get the resulting chain TM a = [(1)r7c6|(1)r5c4] - (1=6)r6c6 - [(6)r5c4|(1)r6c6] = TM b

As a summary, for the BTM example, the rationale is:
WHETHER (1)r6c6 is True, then TM a, OR (6)r6c6 is True, then TM b
For the MBM example, the rationale is:
WHETHER (2)r5c2 is False, then TM a, OR (2)r5c2 is True, then PM b
The question: 'are these slightly different processes typical of each type of complex matrices ?', remain open to me. Both are valid logics, and could be used independantly of the matrix type. Let's keep it open, until examples or counter-examples are found.

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