The New Sudoku Players' Forum

[SpAce]

Applying the process that you have explored in the other thread (here), I guess this matrix can be identified as a BTM:

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3r5c4&5r7c5 8r5c4  3r7c5               
            8r6c5 35r67c5               
                   3r3c5   8r3c5       
                           8r3c9 8r5c9 
   8r9c4                   8r1c4 8r5c4 
=> -35 r9c4

TM a
   3r5c4    8r5c4                 
            8r6c5 35r67c5               
                   3r3c5   8r3c5       
                           8r3c9 8r5c9 
   8r9c4                   8r1c4 8r5c4
=> -3 r9c4

TM b
   5r7c5     3r7c5               
             3r3c5   8r3c5       
                     8r3c9 8r5c9 
   8r9c4             8r1c4 8r5c4
=> -5r9c4

Of course, I am not sure whether the first term in the first row can be split that way.

That was my exact problem why I didn't think it was a valid BTM. Now that you did it anyway, I think you're right:

It seems to make sense with the subsequent OR of the two matrices, which maintains an "OR" operator between 8r5c4 and 3r7c5 and thus requires the AND operator to link 3r5c4 and 5r7c5.

Yes, why not! Furthermore, I don't think it would be actually wrong to write the sub-matrices with the full ANDed first term either, since we're assuming the other option to be false (i.e. [[a&b = c|d] & ~d] <=> [a&b = c]). Right?

[Edit: On further thought, I don't think it's right, actually. I guess we can only assume the other option false when trying each case if the options are also weakly linked, i.e. have a XOR relationship (as in SteveK's example). Here 8r5c4 and 3r7c5 aren't weakly linked. Thus I don't think my suggestion would work, but your sub-matrices still would. However, I'm not sure if the correct relationship between the two sub-matrices is OR in this case. Isn't it AND, because they're independent and produce different results? As such, I would count them as two moves. It makes me doubt if this is a BTM after all, as it seems to me that the sub-matrices of a BTM should have an OR relationship and prove the same result. What do you think?]

[Edit2: I feel like I'm over-thinking this. I'm not sure about any of my arguments above. However, there seems to be one point in SteveK's BTM definition that would prevent the BTM status here anyway:

Each column, except the first column, has the following quality:

• The top non-empty entry is in conflict with each entry below it
• Note there is a no difference here from the triangular matrix.

I don't think that rule holds for this matrix:

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3r5c4&5r7c5 8r5c4  3r7c5               
            8r6c5 35r67c5               
                   3r3c5   8r3c5       
                           8r3c9 8r5c9 
   8r9c4                   8r1c4 8r5c4 
=> -35 r9c4

...as 3r7c5 is not in conflict with 35r67c5. However, I don't really see how it affects the logic as we don't really need a weak link between them.

Note another way to escape:

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Following Steve's chain closely:

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   8r9c4    SS 8c49
             8r3c5   3r3c5
                    35r67c5 8r6c5
3r5c4&5r7c5          3r7c5  8r6c4
=> -35r9c4

I guess that including a pattern such as a wing, a fish or other, is valid. This is similar to what can be done with ALS's

Yes, of course. I've also understood that any boolean nodes can be used as cells, just like with AICs. However, in this case using the full Skyscraper as a cell kind of defeats the purpose of the matrix to me, as the same can be written more concisely with an AIC. To me the real benefit of the matrix view is the ability to show the actual internal workings of a complex pattern, which is why I like to keep them quite uncompressed (as opposed to my AICs). It also makes it possible to see the relative complexity of patterns by exposing the sets needed, which you did very nicely for the solutions in this thread. Combined with the "alien fish" view it can also help find simplifications by removing redundant sets (which is what happened when I combined eleven's and Steve's solutions). That being said, I'm sure it's useful to include larger pattern elements as cells in very complex patterns to keep the matrix size reasonable.

[SteveG48]

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 *------------------------------------------------------------*
 | 57    6     57    | b348   1     348   | 39    389   2     |
 | 4     3     8     |  9     7     2     | 6     1     5     |
 | 2     1     9     |  6    c38    5     | 347   378  b48    |
 *-------------------+--------------------+-------------------|
 | 8     579   1357  |  345   6     1347  | 2     379   19    |
 | 137   2     6     |be38    9     1378  | 5     4    b18    |
 | 1357  579   4     |  2   de358   1378  | 379   3789  6     |
 *-------------------+--------------------+-------------------|
 | 9     4     35    |  1   de35    6     | 8     2     7     |
 | 6     8     2     |  7     4     9     | 1     5     3     |
 | 1357  57    1357  | a8-35  2     38    | 49    6     49    |
 *------------------------------------------------------------*


8r9c4 = [Skyscraper 8c49] - (8=3)r3c5 - 3r67c5 = (5r7c5)&(8r6c5)&(3r5c4) => -35 r9c4 ; stte

Hi Steve! Related to the recent matrix discussion here I did some practicing with this interesting puzzle and the various solutions (partly because Cenoman had so nicely provided matrices to compare with). I like yours very much, especially since it's the most difficult one to turn into a matrix, but I can't quite agree with the last combo node of the chain. I understand that it's just taking a shortcut, but being a nitpicker, I wanted to try some options:

1. AB:(8)r9c4 = [Skyscraper] - (8=3)r3c5 - A:(3=58)r76c5 - B:(8=3)r5c4 => -35 r9c4
2. (8)r9c4 = [Skyscraper] - (8=35)r37c5 - (3|5=8)r6c5 - (8)r5c4|(5)r6c5 = (35)r5c4,r7c5 => -35 r9c4
3. (8)r9c4 = [Skyscraper] - r3c5 = r6c5 - (8)r5c4|(5)r6c5 = (35)r5c4,r7c5 => -35 r9c4
4. (8)r9c4 = [Skyscraper] - r3c5 = (83)b5p84&(85)r67c5 => -35 r9c4

Needless to say I like the last one best (and the first multi-header the least). What do you think?

I agree that my solution is a bit awkward. However, I don't understand your favorite. How does not 8 r3c5 get you to (85)r67c5? On the other hand, your favorite suggests this to me:

(8)r9c4 = [Skyscraper] - 8r3c5 = (83)b5p84&(35)r37c5 => -35 r9c4

[SpAce]

1. AB:(8)r9c4 = [Skyscraper] - (8=3)r3c5 - A:(3=58)r76c5 - B:(8=3)r5c4 => -35 r9c4
2. (8)r9c4 = [Skyscraper] - (8=35)r37c5 - (3|5=8)r6c5 - (8)r5c4|(5)r6c5 = (35)r5c4,r7c5 => -35 r9c4
3. (8)r9c4 = [Skyscraper] - r3c5 = r6c5 - (8)r5c4|(5)r6c5 = (35)r5c4,r7c5 => -35 r9c4
4. (8)r9c4 = [Skyscraper] - r3c5 = (83)b5p84&(85)r67c5 => -35 r9c4

Needless to say I like the last one best (and the first multi-header the least). What do you think?

I agree that my solution is a bit awkward. However, I don't understand your favorite. How does not 8 r3c5 get you to (85)r67c5?

Hi Steve! The reason is spelled out in the third option. The key to both ANDed terms is the 8r6c5 which forces out both 8r5c4 and 5r6c5, which gives us 3r4c4 as a naked single and 5r7c5 as a hidden single. So, the corresponding pairs (83) and (85) happen for slightly different reasons, but I treat them equally because a pair is a pair.

On the other hand, your favorite suggests this to me:

(8)r9c4 = [Skyscraper] - 8r3c5 = (83)b5p84&(35)r37c5 => -35 r9c4

Stylewise I would like that better because it doesn't repeat the same 8r6c5 in both terms. However, it has the familiar problem: now the (83) pair doesn't work as intended because the digits could go either way. I would fix that with a comma, but I know you don't like that. At first glance it looks like my favorite is equally flawed for the same reason, but technically it's not because the ANDed (85)r67c5 (which can only go one way) fixes the 8 in r6c5 and doesn't leave room for the 3 there in any case. (It would be clearer if the 85 pair were written first, however.) Your pairs don't have a similar relationship that would fix all of the digits into specific cells (except in the aftermath, but that doesn't count, afaic). Personally I might prefer this, though it's probably not very popular:

(8)r9c4 = [Skyscraper] - (8)r3c5 = (8,5,3)r67c5,r5c4 => -35 r9c4

The first comma isn't technically necessary but I think it looks better with it. I guess you'd rather write it like this:

(8)r9c4 = [Skyscraper] - 8r3c5 = (8r6c5)&(5r7c5)&(3r5c4) => -35 r9c4

I'd accept that too, but I'd kind of have the same problem as with your original. When written as separate ANDed terms, instead of a locked set, the reason for the 5r7c5 and 3r5c4 remains somewhat questionable. For me the locked set style (with or without commas) usually implies internal interactions between the involved cells and digits, but the ANDed style doesn't. It rather implies that all the ANDed terms are independent and direct results of the previous strong link (which is not true here). That's one reason why I've adopted the comma style because it preserves the locked set style when digits need to be fixed in certain cells. (Of course, our example is not really a traditional locked set because the cells are in two houses -- but they do interact via the common r6c5.)

Btw, if I were truly consistent with that rule both ways, I would write the last node of my second and third option as (3)r5c4&(5)r7c5 instead of (35)r5c4,r7c5 since they have no direct interaction. Perhaps I should. (I know Cenoman would like that.)

[Cenoman]

Back to this matrix:

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3r5c4&5r7c5 8r5c4  3r7c5               
            8r6c5 35r67c5               
                   3r3c5   8r3c5       
                           8r3c9 8r5c9 
   8r9c4                   8r1c4 8r5c4 
=> -35 r9c4

Its third column is exactly the same as in my original search of a matrix for Steve's solution. At the time, I didn't pay attention to meeting Steve K's BTM definition, as it couldn't be a BTM (6 rows x 5 columns)
You recall BTM's definition, notably this requirement:

"The top non-empty entry is in conflict with each entry below it"

I don't think that rule holds for this matrix.

I don't think either. ("I don't think" is maybe too weak. Actually, I am convinced)

In another re-thinking,

I'm not sure if the correct relationship between the two sub-matrices is OR in this case. Isn't it AND, because they're independent and produce different results?

I had the same analyse meanwhile. In Steve K's example, the two sub-matrices produce the same eliminations. Thus, the OR relationship between these
demonstrate the eliminations (or any other result, such as a derived SIS). In the present case, TM a eliminates 3r9c4 and TM b eliminates 5r9c4. A AND relationship would be needed, as you correctly state. So, even without the issue on the 3rd column, the matrix is not shown to be a BTM (I was too optimistic...)

Last point, I proposed a version with the combined skyscraper. I couldn't find an arrangement with the unfolded SS, because of the same issue as in third column. There is no way to keep 8r3c5 as "the top non-empty entry".

Corrected the incorrect statements in my previous post above.

[SpAce]

You recall BTM's definition, notably this requirement:

"The top non-empty entry is in conflict with each entry below it"

I don't think that rule holds for this matrix.

I don't think either. ("I don't think" is maybe too weak. Actually, I am convinced)

Thanks for the confirmation! A pity, but I guess the rule is there for a reason. At least it makes it easier to follow the matrix both ways. In this case it's easy to miss the problem because the logic is quite easy to follow anyway, but in more complex situations it might cause confusion if the weak links are scattered within the columns. Here it also doesn't help if we flip the matrix upside down because then we have the same situation with the 8s column (between 8r1c4 and 8r3c9).

I had the same analyse meanwhile. In Steve K's example, the two sub-matrices produce the same eliminations. Thus, the OR relationship between these
demonstrate the eliminations (or any other result, such as a derived SIS). In the present case, TM a eliminates 3r9c4 and TM b eliminates 5r9c4. A AND relationship would be needed, as you correctly state. So, even without the issue on the 3rd column, the matrix is not shown to be a BTM (I was too optimistic...)

Thanks for that confirmation, too! Perhaps we'll find some other examples where the BTM (or even MBM) would work. Too bad SteveK's sample size is so small and the explanations so short. It's quite hard to draw definite conclusions based on them. For example, both of his BTM and MBM examples start with a binary strong link, but it's not stated as a requirement (unlike with TMs). It leaves one wondering if it's an implicit requirement after all or not (I'm guessing not, or at least hoping so). Of course that problem should only arise when none of the result sets is binary, or otherwise we could just reorder them.

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.

Ok! I'm still learning the ropes with the TMs and PMs, and it looks like they're quite powerful already. It probably takes some pretty complex situations to really need MBMs or even BTMs. If one presents itself, I'm pretty sure we can figure it out together!

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

That's probably a good idea (it's getting difficult to switch between threads). This discussion has been extremely fruitful to me, and I'm glad if you feel the same way! I know that without your help I probably would have gotten nowhere with this stuff. It had been bugging me for a long time, because I could see the usefulness of matrices but didn't really figure out how to actually use them until now. SteveK is no doubt a genius solver but not exactly stellar at explaining things.

Btw, I added a couple of matrices for my May 13 solution. Do they look all right? I like the second one better (of course), but the first one is a direct translation of my multi-headed AIC and depicts one way to tackle the multi-result problem we had earlier. It's really ugly, but do you think it's correct? I noticed your matrix too, and it's nice (as is your solution)!