The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |...|.32|...|
 |45.|...|2.9|
 |...|.9.|68.|
 |---+---+---|
 |...|...|94.|
 |.2.|917|.6.|
 |.61|...|...|
 |---+---+---|
 |.35|.7.|...|
 |6.7|...|.58|
 |...|16.|...|
 *-----------*


Play/Print this puzzle online

[SpAce]

Code: Select all

.----------------------.---------------.------------------.
| 178   b1789  a8(9)-6 | 4567  3   2   | 57    17    457  |
| 4      5     c3(6)   | 67    8   16  | 2     137   9    |
| 1237   17     23     | 457   9   14  | 6     8     3457 |
:----------------------+---------------+------------------:
| 3578  b78    c38     | 2368  25  368 | 9     4     1    |
| 358    2      4      | 9     1   7   | 358   6     35   |
| 9      6      1      | 348   45  348 | 3578  237   2357 |
:----------------------+---------------+------------------:
| 12     3      5      | 48    7   489 | 14    29    6    |
| 6      19     7      | 234   24  349 | 14    5     8    |
| 28     4      289    | 1     6   5   | 37    2379  237  |
'----------------------'---------------'------------------'

(9)r1c3 = (98)r14c2 - (8=36)r42c3 => -6 r1c3; stte

[Cenoman]

Code: Select all

 +----------------------+--------------------+-----------------------+
 | b178    1789   689   |  4567   3    2     |  57     17     457    |
 |  4      5      36    |  67     8   y16    |  2      137    9      |
 | a1237  A1-7    23    |  457    9   x14    |  6      8      3457   |
 +----------------------+--------------------+-----------------------+
 |  3578  z78    z38    |  2368   25  z368   |  9      4      1      |
 |  358    2      4     |  9      1    7     |  358    6      35     |
 |  9      6      1     |  348    45   348   |  3578   237    2357   |
 +----------------------+--------------------+-----------------------+
 | b12     3      5     |  48     7    489   |  14     29     6      |
 |  6      19     7     |  234    24   349   |  14     5      8      |
 | b28     4      289   |  1      6    5     |  37     2379   237    |
 +----------------------+--------------------+-----------------------+

Kraken row (1)r3c126
(1)r3c1 - (128=7)r179c1
(1)r3c2
(1)r3c6-(1=6)r2c6-(638=7)r4c236
=>-7r3c2; ste

[SteveG48]

Code: Select all

 *-----------------------------------------------------------*
 |c178  c1789  689   | 4567  3     2     | 57    17    457   |
 | 4     5     6-3   | 67    8     16    | 2     137   9     |
 |c1237 b17   c23    | 457   9     14    | 6     8     3457  |
 *-------------------+-------------------+-------------------|
 | 3578 b78   a38    | 2368  25    368   | 9     4     1     |
 | 358   2     4     | 9     1     7     | 358   6     35    |
 | 9     6     1     | 348   45    348   | 3578  237   2357  |
 *-------------------+-------------------+-------------------|
 | 12    3     5     | 48    7     489   | 14    29    6     |
 | 6    b19    7     | 234   24    349   | 14    5     8     |
 | 28    4     289   | 1     6     5     | 37    2379  237   |
 *-----------------------------------------------------------*


(3=8)r4c3 - (8=179)r348c2 - (1|9=2378)b1p1279 => -3 r2c3 ; stte

[eleven]

Code: Select all

 *-------------------------------------------------------------------*
 | b178   b1789   89-6  |  4567   3    2     |  57     17     457    |
 |  4      5      36    |  67     8    16    |  2      137    9      |
 |  1237   17     23    |  457    9    14    |  6      8      3457   |
 |----------------------+--------------------+-----------------------|
 |  3578   78     38    |  2368   25   368   |  9      4      1      |
 |  358    2      4     |  9      1    7     |  358    6      35     |
 |  9      6      1     |  348    45   348   |  3578   237    2357   |
 |----------------------+--------------------+-----------------------|
 | a12     3      5     |  48     7    489   |  14     29     6      |
 |  6     a19     7     |  234    24   349   |  14     5      8      |
 | a28     4      289   |  1      6    5     |  37     2379   237    |
 *-------------------------------------------------------------------*

Impossible hidden pair:
(8=9)b7p157 => -89r1c12 => 89r1c3, stte

[ArkieTech]

Code: Select all

 *-------------------------------------------------------------------*
 | b178   b1789   89-6  |  4567   3    2     |  57     17     457    |
 |  4      5      36    |  67     8    16    |  2      137    9      |
 |  1237   17     23    |  457    9    14    |  6      8      3457   |
 |----------------------+--------------------+-----------------------|
 |  3578   78     38    |  2368   25   368   |  9      4      1      |
 |  358    2      4     |  9      1    7     |  358    6      35     |
 |  9      6      1     |  348    45   348   |  3578   237    2357   |
 |----------------------+--------------------+-----------------------|
 | a12     3      5     |  48     7    489   |  14     29     6      |
 |  6     a19     7     |  234    24   349   |  14     5      8      |
 | a28     4      289   |  1      6    5     |  37     2379   237    |
 *-------------------------------------------------------------------*

Impossible hidden pair:
(8=9)b7p157 => -89r1c12 => 89r1c3, stte

[Ngisa]

Code: Select all

+----------------------+-------------------+----------------------+
| 178     1789    d689 | 456     3     2   | 57      17      457  |
| 4       5       e6-3 | 67      8     16  | 2       137     9    |
| 1237   b17       23  | 457     9     14  | 6       8       3457 |
+----------------------+-------------------+----------------------+
| 3578   a78      a38  | 2368    25    368 | 9       4       1    |
| 358     2        4   | 9       1     7   | 358     6       35   |
| 9       6        1   | 348     45    348 | 3578    237     2357 |
+----------------------+-------------------+----------------------+
| 12      3        5   | 48      7     489 | 14      29      6    |
| 6      b19       7   | 234     24    349 | 14      5       8    |
| 28      4       c289 | 1       6     5   | 37      2379    237  |
+----------------------+-------------------+----------------------+


(3=87)r4c23 - (7=19)r38c2 - (9)r9c3 = (9-6)r1c3 = (6)r2c3 => - 3r2c3; stte

Clement

[SpAce]

Impossible hidden pair:
(8=9)b7p157 => -89r1c12 => 89r1c3, stte

Very clever! Yeah, this time I got your point all right, but I still like it. I'm not an expert in that notation, but personally I might make a couple of minor tweaks:

(8|9)b7p157 -> -(89)r1c12 -> (8|9)r1c3 => -6 r1c3; stte

(The '=' looked out of place since it's not an AIC -- but I'm not sure which is better. The last node is definitely clearer (to me) with the '|'. Brackets and the explicit conclusion are just my preferences for readability. Matters of taste.)

Here's an AIC using boringly possible HPs:

(98)r1c23 = (8)r[1-9]c1 = (89)r91c3 => -6 r1c3

(Btw, I think that's as close to an AHS equivalent of an ALS XZ as possible. Does it have a name?)

[eleven]

You know, that i'm not interested any more in any discussions about notation and namings. What i want to know is, how people find solutions (and i find nothing bad with - sometimes surprising - program solutions).
Unfortunately contradictions have been banned in all the sudoku forums from the beginning (we discussed that earlier). But at least half of my non trivial solutions i found by contradiction. Here too.
If i write it as AIC, it just looks academic and complicated.
And i have troubles to understand your chain for my move.

[SpAce]

You know, that i'm not interested any more in any discussions about notation and namings.

Yes, I do, and I'm perfectly fine with it. Yet you sometimes express opinions about them, which I would normally see as an invitation for discussion and possible counter-arguments. If that's not the case, then I must presume you accept that those opinions can be simply ignored. (This is more of a general observation than directed at you specifically.)

What i want to know is, how people find solutions

Me too. It would be really interesting to know how different people find their solutions. I've pretty much revealed my methods in various threads, but I don't really know how anyone else does it. I think such a thread would be very educational. It's one thing to know various patterns and techniques, another to notate them, and a third to actually locate and use them. There's surprisingly little sharing about the third part even though it's obviously pretty crucial.

(and i find nothing bad with - sometimes surprising - program solutions).

Me neither. However, it would be kind of nice to know what kind of software help has been used for each solution.

Unfortunately contradictions have been banned in all the sudoku forums from the beginning (we discussed that earlier).

Well, who's enforcing those artificial rules these days? Can't we just decide amongst ourselves that (at least interesting) contradiction solutions are accepted? I wouldn't mind at all, unless they become the norm. Personally I prefer AICs and other verity solutions, but wouldn't mind a bit of variety.

But at least half of my non trivial solutions i found by contradiction. Here too.

That's an interesting detail. Like I said, it would be great if everyone shared something about their actual solving methods. Mine is mostly based on coloring, more specifically on my implementation of David's GEM, which is pretty mechanical and boring but quite effective at finding both verities and contradictions. That's one reason why I like to play with notations, because writing and naming the solution is often more interesting than finding it.

If i write it as AIC, it just looks academic and complicated.

Maybe, but I bet it would be easier to understand for most of us. This would be my direct translation:

(9)r1c3 = r1c2 - (9=8)b7p517 - (8)r1c1 = (89)r1c23 => -6 r1c3

And i have troubles to understand your chain for my move.

Which parts? I like specifics. (I presume you were talking about my tweaking of your implication chain, and not the AIC which wasn't the same chain at all.)

[eleven]

Maybe, but I bet it would be easier to understand for most of us. This would be my direct translation:

(9)r1c3 = r1c2 - (9=8)b7p517 - (8)r1c1 = (89)r1c23 => -6 r1c3

Maybe read from right to left would be ok for all:
hp89 r1c23 = 8r1c1 - (8=9)b7p715 - r1c2 = r1c3

And i have troubles to understand your chain for my move.

Which parts? I like specifics. (I presume you were talking about my tweaking of your implication chain, and not the AIC which wasn't the same chain at all.)

(98)r1c23 = (8)r[1-9]c1 = (89)r91c3 => -6 r1c3
The first link uses a hidden pair, the second not, though written the same way. And (8)r[1-9]c1 is just ugly and i have to translate it in my mind.

[SteveG48]

I agree with the remarks about contradictions. I see no reason why valid logic should be rejected.

Practically all my solutions start as contradictions. Then I look for the best way to formulate them as AICs. The result, IMO, is prettier than just leaving it as a demonstration of a contradiction. Maybe that's why people frown on contradictions.

[SpAce]

Maybe, but I bet it would be easier to understand for most of us. This would be my direct translation:

(9)r1c3 = r1c2 - (9=8)b7p517 - r1c1 = (89)r1c23 => -6 r1c3 [edit: removed (8) from r1c1]
(9)r1c[3=2] - (9=128)b7p517 - (8)r1c1 = (89)r1c23 => -6 r1c3 [added]

Maybe read from right to left would be ok for all:
hp89 r1c23 = 8r1c1 - (8=9)b7p715 - r1c2 = r1c3

What's the significant difference? I think both chains do the job and are understandable for everyone here. Obviously we both prefer our own versions visually (I added the version that I'd probably actually write, and edited the original to how it was supposed to be), but I'm not about to debate matters of taste. I can only tell what my taste is, but I don't expect anyone to share it.

For example, personally I don't see a single good reason to use the 'hp' prefix, but last time I mentioned that preference, it didn't end really well. So, by all means, feel free to use it, but don't expect me to fall in line -- unless you have an excellent argument why it's absolutely necessary Personally I also think that not using brackets with n-digits is less readable (especially with several of them) and so is leaving out ALS bystanders (I only did it the first time because it was meant as a direct translation of your logic) -- but I'm not absolute about them. Last but not least, leaving the end node without a digit hurts backwards readability and more importantly my sense of symmetry. (The 3D stuff is not something I'd yet even want to see in others' chains, as I'd still like to find a better way to implement it.)

So, we obviously have different tastes, but I think that's quite ok. I'm pretty sure we can still read each others' chains. It's also not unheard of that my tastes change over time, though who knows which way.

(98)r1c23 = (8)r[1-9]c1 = (89)r91c3 => -6 r1c3

The first link uses a hidden pair, the second not, though written the same way.

A pair is a pair, I don't care Poetry aside, I just can't see what difference it makes. If any link forces two digits into two cells then it's a pair, and as far as I know, it works exactly the same way whether it's a hidden pair, a naked pair, or some hybrid mongrel. So, what am I not seeing? Why should the pair type be specified? How would you categorize the second pair, anyway?

And (8)r[1-9]c1 is just ugly and i have to translate it in my mind.

No disagreement there. I wish I could find a better-looking way to do it, so think of it as a work in progress. Still, I'm probably not going back to plain old AIC way of writing bilocation links, because I think they're conceptually inferior (though easier on the eyes, I admit).

[eleven]

Why should the pair type be specified?

Because the logic of the link is different.
In a naked pair link you look at the other candidates of an ALS.
In a hidden pair link you look, where the candidates can be elsewhere in the unit.
And the link 89r19c3 = 8r9c1 depends on the strong links for 9 and 8.
I want to understand the move, not to solve the puzzle of the notation.

[SpAce]

Why should the pair type be specified?

Because the logic of the link is different.
In a naked pair link you look at the other candidates of an ALS.
In a hidden pair link you look, where the candidates can be elsewhere in the unit.
And the link 89r19c3 = 8r9c1 depends on the strong links for 9 and 8.
I want to understand the move, not to solve the puzzle of the notation.

eleven, I want to first thank you for taking the time to explain your reasons. However, I still don't get it. To me it seems that the only way to avoid any controversy would be to break the node and write it:

(9)r1c3 = (9-8)r9c3 = r9c1 - r1c1 = (89)r1c23

However, I do that automatically mentally if I see a multi-digit node with a strong link to something in a different house, as normally only one of the digits is actually linked. That's why this is equivalent to me:

(98)r19c3 = (8)r9c1...

Note that I always write the digits in that order to ease the translation (highlighting that 8 is the linking digit), though it makes no difference to its correctness. I also think it's easier the other way around, which is why I wrote it:

(8)r9c1 = (89)r91c3

Also, is it not a kind of a hidden pair indeed? The way I see it is that we originally have an AHP (89)r19c3, with one extra 8 in r4c3 as a spoiler (the 9s are locked). Removing that spoiler would be the natural way to make the HP true, but not the only way. The HP is true also if one of the 8s in the HP cells is made true. So, if there's no 8 in r9c1 then there must be one in r9c3 -- and the HP in c3 is (not so) magically true. So, the 8r9c1 can be seen as an "external" spoiler. I don't think it's anything more difficult than seeing the equivalence of internal and external guardians in DPs.

If that logic is accepted and understood, then even this should become clearer:

Code: Select all

.-----------.-----------------.--------------------.
| 1  8   6  | 245   235    45 |   23      9    7   |
| 5  4   23 | 8     9      7  |   123     123  6   |
| 9  23  7  | 1     23     6  |   8       5    4   |
:-----------+-----------------+--------------------:
| 7  5   23 | 6     8      1  |   9       4    23  |
| 6  23  8  | 57    4      9  |   1357    137  123 |
| 4  9   1  | 3   a(57)    2  | a(57)-6   67   8   |
:-----------+-----------------+--------------------:
| 3  7   59 | 245   256    8  |  146      16   19  |
| 8  6   4  | 9     1      3  |  27       27   5   |
| 2  1   59 | 457  b5(67)  45 | b34(6)    8    39  |
'-----------'-----------------'--------------------'

(57)r6c75 = (76)r9c57 => -6 r6c7; stte

The way I see it, we have two AHPs, with a shared and strongly linked spoiler digit (7), which makes the contained HPs strongly linked:

AHP (57)r6c57 (5s locked, internal spoiler: 7r6c8, external spoiler: 7r9c5)
AHP (67)r9c57 (6s locked, internal spoiler: 7r9c4, external spoiler: 7r6c5)

(Due to lack of imagination, I just did something I hate: overloaded well-known terms to mean something different. By "internal" and "external" I mean candidates that are either within the same house as the AHP (internal) or not (external). These terms aren't to be used outside of this discussion. In fact, they're really bad because if we had naked pairs in the discussion, then their spoilers should actually be called "internal" and it would be in line with the DP-use. HP-spoilers are in fact all external, but some are more external than others.)

Since the external spoilers are strongly linked, at least one of the HPs must be true. (Not really different from an ALS-XZ with weakly linked RCs -- in that case at most one ANS can contain a true RC, so at least one of the ANSs must be locked with the remaining digits. In this case the equivalents to RCs are strongly linked, forcing one or the other HP to be true).

Now, what complicates things slightly is that the second pair is not used like a normal HP, i.e. eliminating within its own cells. In fact, no elimination would occur if it had a full complement of its digits, because it would have no link to r6c7. So, we have to see that its digits are locked into their own cells (still doesn't make it non-HP), which makes the -6c7- link possible. However, that situation is no different from ANS nodes which have individual digits locked into certain houses or cells and used for weak links that aren't available to the full ANS node. (The naked/hidden/mongrel division is only meaningful in explaining how the pair/triple/etc is revealed; i.e. what kind of spoilers are neutralized. After the fact they're just generic locked sets and have the same weak links available in all dimensions. That's why to me "a pair is a pair.")

Btw, one way to make any link clearer, especially if one prefers to keep the digits in natural order, is to inline the reason:

(57)r6c57 =7c5= (67)r9c57 => -6 r6c7; stte

That's a perfectly standard mechanism, though usually seen with more obscure derived links. Theoretically, I don't see why it couldn't be used with any (or even every) link, if someone wants to be extremely explicit.