The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |...|...|...|
 |.32|...|45.|
 |.5.|9.7|.2.|
 |---+---+---|
 |..5|.9.|6..|
 |...|1.2|...|
 |..4|.6.|9..|
 |---+---+---|
 |.8.|5.3|.7.|
 |.27|...|38.|
 |...|...|...|
 *-----------*


Play/Print this puzzle online

[Leren]

Code: Select all

*--------------------------------------*
| 18  9   18 | 2    5  4   | 7 6    3  |
| 7   3   2  | 68   18 16  | 4 5    9  |
| 4   5   6  | 9    3  7   | 8 2    1  |
|------------+-------------+-----------|
| 23  17  5  | 347  9  8   | 6 134  27 |
| 9   6  d38 | 1   b47 2   | 5 34  c78 |
| 28  17  4  | 37   6  5   | 9 13   28 |
|------------+-------------+-----------|
| 6   8   9  | 5    2  3   | 1 7    4  |
| 5-1 2   7  | 46  a14 9   | 3 8    56 |
| 135 4  e13 | 78   78 6-1 | 2 9    56 |
*--------------------------------------*

XY Chain : (1=4) r8c5 - (4=7) r6c5 - (7=8) r6c9 - (8=3) r6c3 - (3=1) r9c3 => - 1 r8c1, r9c6; stte

Alternatively here is a simple colouring move: Simple coloring on 1. Parity Group details: | r1c1 r2c5 r8c1 r9c3 r9c6 | r1c3 r2c6 r8c5 | Eliminate 1 from cells seeing both Parity Values => - 1 r1c1, r8c1, r9c3, r9c6; stte

Code: Select all

*---------------------------------------*
|a8-1 9  b18  | 2    5   4   | 7 6   3  |
| 7   3   2   | 68  a18 b16  | 4 5   9  |
| 4   5   6   | 9    3   7   | 8 2   1  |
|-------------+--------------+----------|
| 23  17  5   | 347  9   8   | 6 134 27 |
| 9   6   38  | 1    47  2   | 5 34  78 |
| 28  17  4   | 37   6   5   | 9 13  28 |
|-------------+--------------+----------|
| 6   8   9   | 5    2   3   | 1 7   4  |
|a5-1 2   7   | 46  b14  9   | 3 8   56 |
| 135 4  a3-1 | 78   78 a6-1 | 2 9   56 |
*---------------------------------------*

Does anybody actually understand any of that ?

Leren

[SpAce]

Code: Select all

.------------------.-----------------.--------------.
| a(1)8     9   18 |  2      5    4  | 7   6     3  |
|   7       3   2  |  68     18   16 | 4   5     9  |
|   4       5   6  |  9      3    7  | 8   2     1  |
:------------------+-----------------+--------------:
|  a23      17  5  | b34+7   9    8  | 6  b14+3  27 |
|   9       6   38 |  1     c47   2  | 5   34    78 |
|  a28      17  4  |  37     6    5  | 9   13    28 |
:------------------+-----------------+--------------:
|   6       8   9  |  5      2    3  | 1   7     4  |
|   5-1     2   7  |  46   c(1)4  9  | 3   8     56 |
|  A35+(1)  4   13 |  78     78   16 | 2   9     56 |
'------------------'-----------------'--------------'

BUG+3

(1)r9c1 == [(1=823)r164c1 - (3==7)r4c84 - (7=41)r58c5] => -1 r8c1; stte

[SpAce]

Alternatively here is a simple colouring move: Simple coloring on 1. Parity Group details: | r1c1 r2c5 r8c1 r9c3 r9c6 | r1c3 r2c6 r8c5 | Eliminate 1 from cells seeing both Parity Values => - 1 r1c1, r8c1, r9c3, r9c6; stte
...
Does anybody actually understand any of that ?

To be honest, no I don't see any available Simple Coloring moves here, and neither does Hodoku nor SudokuWiki. On 1s I can see three conjugate clusters: {r1c13 r9c3} & {r46c28} & {r2c56 r8c15 r9c6}, but none of them produce any eliminations either by themselves or together (via multi-coloring or X-Chains).

Furthermore, I don't think single digit eliminations are even theoretically possible on any digit here, based on keith's simple rules (which I use in p&p solving to quickly determine each digit's X-Chain potential). 1s are the closest possibility, since boxes 1,2,7,8 form a rectangle, but the candidates in b12 are aligned (i.e. known braid), so no cigar. If you run your template code, I think it would agree.

[SteveG48]

Code: Select all

 *--------------------------------------------------*
 | 18   9    18   | 2    5    4    | 7    6    3    |
 | 7    3    2    | 68  c18  b16   | 4    5    9    |
 | 4    5    6    | 9    3    7    | 8    2    1    |
 *----------------+----------------+----------------|
 | 23   17   5    | 347  9    8    | 6    134  27   |
 | 9    6    8-3  | 1   c47   2    | 5   d34   78   |
 | 28   17   4    | 37   6    5    | 9    13   28   |
 *----------------+----------------+----------------|
 | 6    8    9    | 5    2    3    | 1    7    4    |
 | 15   2    7    | 46   14   9    | 3    8    56   |
 | 135  4   a13   | 78  c78  a16   | 2    9    56   |
 *--------------------------------------------------*


(3=16)r9c36 - (6=1)r2c6 - (1=478)r259c5 - (4=3)r5c8 => -3 r5c3 ; stte

[SteveG48]

Alternatively here is a simple colouring move: Simple coloring on 1. Parity Group details: | r1c1 r2c5 r8c1 r9c3 r9c6 | r1c3 r2c6 r8c5 | Eliminate 1 from cells seeing both Parity Values => - 1 r1c1, r8c1, r9c3, r9c6; stte

Code: Select all

*---------------------------------------*
|a8-1 9  b18  | 2    5   4   | 7 6   3  |
| 7   3   2   | 68  a18 b16  | 4 5   9  |
| 4   5   6   | 9    3   7   | 8 2   1  |
|-------------+--------------+----------|
| 23  17  5   | 347  9   8   | 6 134 27 |
| 9   6   38  | 1    47  2   | 5 34  78 |
| 28  17  4   | 37   6   5   | 9 13  28 |
|-------------+--------------+----------|
| 6   8   9   | 5    2   3   | 1 7   4  |
|a5-1 2   7   | 46  b14  9   | 3 8   56 |
| 135 4  a3-1 | 78   78 a6-1 | 2 9   56 |
*---------------------------------------*

Does anybody actually understand any of that ?

Leren

I'm afraid I don't. You have r1c1 and r8c1 in the same parity group. Since they're in the same column, that eliminates 1 in both those cells, solving the puzzle immediately, but I don't see how you establish that coloring. The 1 in r9c1 seems to block things.

[SpAce]

You have r1c1 and r8c1 in the same parity group. Since they're in the same column, that eliminates 1 in both those cells, solving the puzzle immediately, but I don't see how you establish that coloring. The 1 in r9c1 seems to block things.

The 1r9c1 makes no difference, actually. Removing it would connect the two clusters {r1c13 r9c3} & {r2c56 r8c15 r9c6}, but it still wouldn't produce any eliminations. It would be a dead loop just like the X-Wing in the middle band. With the braid in the top band and the X-Wing in the middle, there's no simple coloring possibility here no matter what's in the bottom band.

[Cenoman]

Code: Select all

 +------------------+------------------+------------------+
 |  18    9    18   |  2     5    4    |  7    6     3    |
 |  7     3    2    |  68    18   16   |  4    5     9    |
 |  4     5    6    |  9     3    7    |  8    2     1    |
 +------------------+------------------+------------------+
 |  23    17*  5    | d347*  9    8    |  6    134*  27   |
 |  9     6   b38   |  1     e47  2    |  5   c34    78   |
 |  28    17*  4    |  37*   6    5    |  9    13*   28   |
 +------------------+------------------+------------------+
 |  6     8    9    |  5     2    3    |  1    7     4    |
 |  15    2    7    |  46   f14   9    |  3    8     56   |
 |  135   4   a3-1  |  78    78  g16   |  2    9     56   |
 +------------------+------------------+------------------+

DP(137)r46c248 (BUG lite) using mixed external-internal
(3)r9c3 = r5c3 - (3)r5c8==(4)r4c4 - r5c5 = (4-1)r8c5 = (1)r9c6 => -1 r9c3; ste

Alternatively here is a simple colouring move: Simple coloring on 1.
Parity Group details: | r1c1 r2c5 r8c1 r9c3 r9c6 | r1c3 r2c6 r8c5 | Eliminate 1 from cells seeing both Parity Values => - 1 r1c1, r8c1, r9c3, r9c6; stte

Does anybody actually understand any of that ?

Yes. Such a contradiction is the first inference learnt in coloring methods, aka "full tagging". Tagging this puzzle (easy to do manually...) I get the mentionned parity groups of the 1s, but I can't get this with a simple coloring alone (coloring only the 1s) I need to use other conjugates of 1r1c1 (or of any other known member of its parity group), e.g. 8r1c1

Proof of the contradiction considered by Steve:
(1=8)r1c1 - r1c3 - (8=3)r5c3 - (3=4)r5c8 - r5c5 = (4-1)r8c5 = (1)r8c1

I'd just replace "simple coloring" in your statement by "coloring"

[SpAce]

Yes. Such a contradiction is the first inference learnt in coloring methods, aka "full tagging".

Sorry about nitpicking again. There are quite a few coloring methods, "full tagging" being one of them (and the one I know the least about). Therefore I wouldn't say they're synonyms (which "aka" implies), but I'm sure you didn't mean that either. I solve practically all non-trivial puzzles with coloring but I've never used full tagging, at least knowingly. From what I understand, its full implementation is very powerful but more suitable for software than manual solvers. Personally I use my adaptation of David's GEM, which is about as powerful as manually practicable coloring can get (but of course I'd be happy to learn about even more powerful methods, if possible). Here's a recent example of how I actually do it using Hodoku colors (all dark eliminations being due to "traps", i.e. pincers, instead of a "wrap", i.e. a contradiction). This puzzle is much simpler, and only two colors would be needed for a wrap solution because almost all candidates are paired (due to the near BUG situation). Still, it's not "Simple Coloring" because multiple digits are needed (I'd call it 3D Medusa).

Tagging this puzzle (easy to do manually...)

Can you show an example of how you actually do "tagging" manually? (Not with this example, though, because it's too simple for a useful demo.) I've never looked into that technique more deeply because I've been quite happy with GEM, and full tagging has seemed more complex. In this case the full power of neither technique is required (or even close), because conjugate pairs can be used all the way and no weak links or groups or ALS etc are needed. When that's the case, I call it 3D Medusa, which is the multi-digit version of Simple Coloring.

3D Medusa works here very well and produces Leren's eliminations and much more. In fact a full conjugate coloring produces 25 immediate wrap eliminations, which practically solves the puzzle without further ado. In GEM, the initial seed cluster is (usually) a 3D Medusa, unless more complex nodes are used as seeds, but it's rare that it alone produces a solution (or even eliminations) like here. Here's one example where I used one, though. Almost always weak links and more complex nodes need to be used, and then it's no longer a 3D Medusa (nor presentable as a pattern).

I'd just replace "simple coloring" in your statement by "coloring"

Yes. Or "3D Medusa", which would be more specific, because it's the simplest kind of coloring that works here. Even SudokuWiki can do that:

3DMedusa.png (99.02 KiB) Viewed 693 times

[Leren]

I checked Medusa Colouring with my own solver and got the same 25 eliminations, but could i really say I understood it all ? I would say not really, but I trust the computer .

At least with complex moves like Exocets, Multifish and the like, when you stare at the board for a while you can eventually see how it all works, whereas a human friendly documented colouring move, seems to me to be difficult to achieve.

Still, colouring solved the puzzle, which is quite rare for Dan's dailies, so it was fun to post it. Provoked plenty of discussion. Surprise, surprise .

Leren

[SpAce]

I checked Medusa Colouring with my own solver and got the same 25 eliminations, but could i really say I understood it all ? I would say not really, but I trust the computer .

In that case, I hope you don't mind if I try to explain. With Simple Coloring and 3D Medusa we're coloring only conjugate (XOR) pairs (bilocation or bivalue candidates) with opposite colors, never using strong-only or weak-only links (unlike more advanced coloring methods), which produces two groups of candidates with a XOR relationship. In other words, one of those groups must be true and the other false. If we find a contradiction for one of the colors, then all candidates of that color must be false and all candidates of the other color must be true. (Even if we don't find a contradiction, we can find pincer eliminations when non-colored candidates see both colors -- but they don't have a similar global effect. I personally prefer those finds because they have the least "assumptive" taste and are simpler to translate into chains.)

In our example (using the SudokuWiki image) it's easy to see that the yellow group can't be true because we have two yellow 1s in c1 and b7 as well as two yellow candidates in r9c1. They're normal contradiction conditions which would also be produced if we assumed any of the yellow candidates to be true and tried to solve the puzzle (thus proving the assumed candidate false). The conjugate coloring simply tells us that any of the yellow candidates would produce those same contradictions, so they can all be eliminated at once. It also proves directly that all the blue candidates must be true, so they can be placed immediately. Of course it's all just gravy because eliminating any one of the yellow candidates would solve the rest of the coloring cluster (and in this case the puzzle) with singles, so a coloring solution is not more powerful than a single chain.

At least with complex moves like Exocets, Multifish and the like, when you stare at the board for a while you can eventually see how it all works, whereas a human friendly documented colouring move, seems to me to be difficult to achieve.

That's an interesting pov! I learned coloring before any other non-basic techniques, so for me it's the most natural way to solve. It's the patterns that can't be easily explained with coloring, such as those you mentioned, that are more difficult for me. (In practice I use coloring with them too, but with different semantics).

As for the documentation, there's no need to document the full coloring solution (which would be very cumbersome in most cases). If a conjugate coloring (i.e. Simple Coloring or 3D Medusa) produces a contradiction, like here, it's enough to build a simple chain that proves any one of the false color eliminations. Singles will take care of the rest. The results of more complex coloring methods, such as GEM, can always be explained with an AIC, a kraken or a net, as well. However, if the coloring produces several pincer eliminations, then it may take more than one chain/net to explain them all (because they're not similarly related like the contradiction eliminations).

[eleven]

Agreed. Looking for a single elimination directly in most cases will be more effective than this mechanical way of that coloring method - with the same progress in the end.
And you can have many starting points in hard puzzles ...

I also wouldn't want to assure mechanically, that i have not missed any singles, before i look for pairs.