The New Sudoku Players' Forum

[SpAce]

(6=7)r4c4 - (7=5)r6c356 - (5=76)r7c65 => -6r6c5,-7r4c6; stte
Better (6,7)r4c4,r6c56 = 5r6c356 - (5=76)r7c56 ?

I'd stick to the developed form: (6=7)r4c4 - *(7=685)r6c356 - (5=7)r7c6* - (7=6)r7c5 => -7 r4c6*, -6 r6c5

I actually like them both, but I'd prefer a couple of cosmetic changes (which you already applied). The subchain version is definitely easier to read if the last term is split, because it makes it easier to see the subchain. As an added bonus it makes the chain more symmetrical too. (Btw, I'd rather use something other than '*' to mark the subchain because most people use it as a memory marker -- it always confuses me for a second when it's used for something else in chains.) Both versions (especially the latter) would benefit from using the full complement of ALS digits. With those small changes I have no problem at all with either one. The second variant is clearly something I might write myself:

(6,7)b5p189 = (685)r6c356 - (5=76)r7c65 => -6 r6c5,r7c4, -7 r4c6

(It seems that the only problem I had with the original was the omitted bystanders. They do make a difference!)

Same side remark as to SpAce: this chain with embedded subchain is hard to put into a matrix (as well as my own, BTW)

Well, writing it is not hard but reading it requires similar extra effort as a subchained AIC because the end points are scattered. I guess it's against the spirit of matrices anyway, but it's certainly possible. I might do something like this with eleven's first chain:

Code: Select all

 6r4c4   7r4c4
        {7r6c56 685r6c356
                  5r7c6   7r7c6}
 6r7c5                    7r7c5
--------------------------------
-6r6c5 -{7r4c6}
-6r7c4


The same could be done with yours, but since it only needs one ANDed end point, I'd rather do it that way to avoid the extra hassle:

Code: Select all

 7r4c7       7r6c7
             5r6c7 5r6c6
                   5r7c6 7r7c6
                         7r7c5 6r7c5
 7r7c6&6r4c4       5r7c6       6r7c4
------------------------------------
-7r4c46

How would you do them?

[Cenoman]

Code: Select all

 7r4c7       7r6c7
             5r6c7 5r6c6
                   5r5c4 3r5c4
 7r7c6&7r3c4       5r7c6 3r3c4
------------------------------
-7r4c46

Would you accept that?

Code: Select all

 6r4c4   7r4c4
        {7r6c56 685r6c356
                  5r7c6   7r7c6}
 6r7c5                    7r7c5
--------------------------------
-6r6c5 -{7r4c6}
-6r7c4

The same could be done with yours, but since it only needs one ANDed end point, I'd rather do it that way to avoid the extra hassle:

Code: Select all

 7r4c7       7r6c7
             5r6c7 5r6c6
                   5r7c6 7r7c6
                         7r7c5 6r7c5
 7r7c6&6r4c4       5r7c6       6r7c4
------------------------------------
-7r4c46

How would you do them?

I have no better solution. For my own path, I'd rather a matrix similar to the second one.

Code: Select all

 {7r4c7       7r6c7
             5r6c7 5r6c6
                   5r7c6 7r7c6}
                         7r7c5 6r7c5
 6r4c4                         6r7c4
------------------------------------
-7r4c4
{-7r4c6}

If we consider that a matrix can contain such kind of combo (non native) strong links, as the last lines of the examples above, no concern. The only alternative is two matrices. The choice depends upon the audience one is addressing. The above suits sudoku experts; for less experimented players, I'd stick to the two matrices. I keep in mind that the puzzles proposed in this forum are not in the hardests, and could all be solved easily either in one lclste step, or in two "simple" steps. The "one step, ste finish" challenge added here leads to use complex strategies on puzzles not that hard. This is not a criticism at all, I personnally find it much funny.

[SpAce]

I have no better solution. For my own path, I'd rather a matrix similar to the second one. If we consider that a matrix can contain such kind of combo (non native) strong links, as the last lines of the examples above, no concern. The only alternative is two matrices.

Glad to hear! You've been a good teacher Once more I'd like to thank you for helping me use matrices. I think they're a valuable tool!

The choice depends upon the audience one is addressing. The above suits sudoku experts; for less experimented players, I'd stick to the two matrices.

I agree, of course.

I keep in mind that the puzzles proposed in this forum are not in the hardests, and could all be solved easily either in one lclste step, or in two "simple" steps. The "one step, ste finish" challenge added here leads to use complex strategies on puzzles not that hard. This is not a criticism at all, I personnally find it much funny.

Yes, I think it's a great format! Even though the puzzles aren't hard, the challenge makes it worthwhile to practice advanced techniques with them. All of the tricks learned here are directly transferable to harder puzzles which actually require them, and they also help understand even more advanced techniques that are never used here. They're just more fun to practice like this. It's like the old military slogan says: "train hard, fight easy." Even ridiculously simple scenarios can yield learning experiences. A recent example -- the puzzle wasn't even valid, and the desired effect was achievable with a single, yet the following discussion eventually yielded valuable insights for both myself and StrmCkr.

That said, it would be interesting and educational to occasionally have similar challenges with actually hard puzzles too. To avoid the drag of multi-steppers, especially extremely hard ones, the challenge could be restricted to (for example) finding any elimination in a particular puzzle state. When the puzzle state is > SE 9.0, finding (and documenting) that single elimination can be hard enough.

[SpAce]

Hi Dan,

Code: Select all

 *--------------------------------------------------*
 |  9  7  3    |  1     5     6       |  8    2  4  |
 |  5  6  1    |  2     48    48      |  3    9  7  |
 |  2  8  4    |  37    379   379     |  1    6  5  |
 |-------------+----------------------+-------------|
 |  4  5  26   | a67    1     29-7    |  79   8  3  |
 |  7  1  28   |  35    389   23589   |  59   4  6  |
 |  3  9 b68   |  4    b78-6 b578     |  57   1  2  |
 |-------------+----------------------+-------------|
 |  8  4  9    |  567  c67   c57      |  2    3  1  |
 |  6  3  5    |  8     2     1       |  4    7  9  |
 |  1  2  7    |  9     34    34      |  6    5  8  |
 *--------------------------------------------------*

(6=7)r3c4 - (7=5)r6c356 - (5=76)r7c65 => -6r6c5,-7r4c6; stte
Better (6,7)r4c4,r6c56 = 5r6c356 - (5=76)r7c56 ?

[(7|6=5)r4c4,r6c356-(5=76)r7c65]-7r4c6,-6r6c5; ste
??

Looks like no one ever answered your implied question. Being the resident killjoy, I'll take the bait

Your proposed AIC would be nice and elegant, but unfortunately it has a couple of problems. First of all, your start node (7|6) is ORed, so it can't eliminate both 7 and 6 (or either of them, actually), so it doesn't work as an end point for this AIC. It needs to be ANDed (76) like the last node (the other end point). Also, (7|6=5) is not exactly a meaningful strong link here anyway, because we rather have (7|6=DP). It's because (7|6) in those cells is a known truth (i.e. if neither 7 nor 6 is there we have a contradiction) and should only be followed by a weak link, like you did here (I liked that!).

However, those problems are easy to fix because the ANDed version of the start node is actually valid: (76=5)r4c4,r6c356. The only problem left is that it contains our elimination cell r6c5 with a 6 possibly there (so it can't eliminate 6r6c5). Therefore, that chain would only eliminate 7r4c6. To eliminate the 6r6c5 as well, the first end point must not include that cell or not have 6 in that cell. The (6,7)r4c4,r6c56 in eleven's second solution does the trick. Even though it includes the r6c5 cell, the comma notation indicates that 6 is not there (because it's in r4c4). It's basically a shorthand for 6r4c4&7r6c56.