The New Sudoku Players' Forum

[eleven]

Thanks Denis, it works fine now for me with reading from and writing to a file.

[mith]

I was able to construct a 9.9 (only depth 2) using the 15-cell pattern without too much effort. Here's a singles-expanded minimal of it:

Code: Select all

..7.9.45..9....6.74.567..89...45.76.7.49.68.5....87.94...8...7.3..7.....27......8  skfr_ER=10.1

The particular form I used is such that it's also one cell away from a trivalue oddagon, but even providing the digit in the guardian cell doesn't collapse the 15-cell pattern. The 15-cell has two guardians (the original 9.9 only has one), but one can be eliminated pretty easily. Making the elimination reduces the puzzle to 8.3.

I may do a bit of iterating on this to see if I can find a depth 3, but probably it would be more fruitful to just search the expanded database for this pattern vs. the 12-cell trivalue oddagon. (I'm not currently generating any puzzles, and the expanded database sits at 375103 puzzles. Still need to find time to get the solution-minlex scripts set up, and I'm letting SE rate the remaining puzzles - about 100k left - in the meantime.)

[denis_berthier]

.
Great example, mith: the two known T&E(3) patterns in the same puzzle!

[denis_berthier]

.
eleven,
I'm curious about CLIPS running on other versions of Unix: in your system, are you able to copy/paste the config file into CLIPS or do you have to batch it?
.

[eleven]

I only used the command line version. My notebook does not have enough memory to check harder techniques, so i stopped trying it.
But this one also looks quite hard.

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 .  .  . |  .  .  X |  .  .  X
 .  .  . |  .  X  . |  .  X  .
 .  .  X |  .  .  X |  .  X  .
-------------------------------
 .  .  . |  .  X  . |  X  .  X
 .  .  X |  .  .  . |  .  X  .
 .  X  X |  .  X  . |  .  .  .   


[ryokousha]

I
But this one also looks quite hard.

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 .  .  . |  .  .  X |  .  .  X
 .  .  . |  .  X  . |  .  X  .
 .  .  X |  .  .  X |  .  X  .
-------------------------------
 .  .  . |  .  X  . |  X  .  X
 .  .  X |  .  .  . |  .  X  .
 .  X  X |  .  X  . |  .  .  .   


(assuming an empty band on top for numbering)
r7c5 = r8c8 = r4c9 = red
as guardian of oddagon (r6c6, r4c6, r5c5, r5c8, r6c8), r6c6 must also be red
r9c5 = r8c3 = green -> r6c3 = r5c5 = yellow, which excludes yellow from c8

[eleven]

Yes, as soon as you have 2 cells in a minirow/col the pattern is simplified - different to TH and #37 (a morph here):

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------------------------------
  X  .  . | X  .  . | X  .  . 
  .  X  . | .  X  . | .  X  . 
  .  .  X | .  .  X | .  .  . 
------------------------------
  X  .  . | .  .  X | .  .  . 
  .  .  . | .  X  . | X  .  . 
  .  .  X | X  .  . | .  X  . 
------------------------------


[eleven]

This 15 cell pattern is not as easy to identify as the TH (exactly one rectangle).
Starting with a TH, a cell of the rectangle is missing (#). Two of the outside cells then must be on the rectangle lines (a, b in different columns).

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 .  .  . |  .  .  X |  .  .  A
 .  .  b |  #  .  . |  .  X  .   ----
 .  c  . |  .  X  . |  X  .  .
-------------------------------
 .  .  - |  .  X  . |  .  .  B
 .  .  x |  .  .  X |  X  .  .
 .  a  . |  X  .  . |  .  X  .   ----

But the other two are somewhat tricky.
If there is one in c, in the box with 3 cells A can be both b and c. Then the other digit must not be in the line with B.

Same, if c is in the other cell:

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 .  c  . |  .  .  X |  .  .  X
 .  .  b |  #  .  . |  .  X  .   ----
 .  .  . |  .  X  . |  A  .  .
-------------------------------
 .  .  x |  .  X  . |  .  .  X
 .  .  - |  .  .  X |  B  .  .
 .  a  . |  X  .  . |  .  X  .   --

--

[marek stefanik]

Haven't seen it mentioned anywhere, but TH with one guardian cell produces three remote triples:
Edit: Only if the guardian cell is part of the rectangle.

Code: Select all

.--------------------.---------------------.------------------.
| 168   12368  7     |#123   9      1238   | 4      5   #123  |
| 18    9      1238  | 1235  1234   123458 | 6     #123  7    |
| 4     123    5     | 6     7     #123    |#123    8    9    |
:--------------------+---------------------+------------------:
| 189   1238   12389 | 4     5     #123    | 7      6   #123  |
| 7     123    4     | 9    #123    6      | 8     #123  5    |
| 156   12356  1236  |#123   8      7      |#123    9    4    |
:--------------------+---------------------+------------------:
| 1569  1456   169   | 8     12346  123459 | 12359  7    1236 |
| 3     14568  1689  | 7     1246   12459  | 1259   124  126  |
| 2     7      169   | 135   1346   13459  | 1359   134  8    |
'--------------------'---------------------'------------------'

123b356 all have the same permutation parity.
Obviously b36 and b56 have different permutations. The same is true for b35, otherwise the remaining TH cells in b2 would have to contain the same digit.
Therefore each of b356 has a different permutation of the same parity, so each position can contain each digit only once, and must contain all three digits.
remote triples b356p3, b356p5, b356p7
Edit: With the missing cell being part of the rectangle, all such patterns are isomorphic.

using the second one:

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.--------------------.---------------------.------------------.
|c1–68  12368  7     | 123   9      1238   | 4      5    123  |
| 18    9      1238  | 1235  1234   123458 | 6    a#1–23 7    |
| 4     123    5     | 6     7      123    | 123    8    9    |
:--------------------+---------------------+------------------:
| 189   1238   12389 | 4     5      123    | 7      6    123  |
| 7    b1–23   4     | 9    #123    6      | 8     #123  5    |
| 156   12356  1236  | 123   8      7      | 123    9    4    |
:--------------------+---------------------+------------------:
| 1569  1456   169   | 8     12346  123459 | 12359  7    1236 |
| 3     14568  1689  | 7     1246   12459  | 1259   124  126  |
| 2     7      169   | 135   1346   13459  | 1359   134  8    |
'--------------------'---------------------'------------------'

whichever digit appears in r2c8 is eliminated from r5c58 by the RT, hence forced into r5c2, and in b1 forced into r1c1 => –23r2c8, –23r5c2, –68r1c1; 6.6 skfr

It makes for a simple way to prove the pattern:

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------------------------------
  A  .  . | A  .  . | a  .  .
  .  B  . | .  B  . | .  b  .
  .  .  X | .  .  X | .  .  .
------------------------------
  X  .  . | .  .  X | .  .  .
  .  .  . | . bB  . | X  .  .
  .  .  X |aA  .  . | .  X  .
------------------------------

[Edit]missing cell r5c2 is part of the rectangle (r25c25) => [/Edit] RTs A[r1c14, r6c4], B[r2c25, r5c5]
Let ab be the digits in b3p15, in that order. HSs aA, bB.
b6p48 must both contain the remaining digit, ie. contra.

Marek

[ryokousha]

TH with one guardian cell produces three remote triples

This does not have the expected remote triples? Maybe I misunderstood something regarding the morphs.

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------------------------------
  .  .  A | .  .  . | .  .  .
  .  B  . | .  A  . | .  .  .
  C  .  . | .  .  B | .  .  .
------------------------------
  A  .  . | C  .  . | .  .  .
  .  C  . | .  B  . | .  .  .
  .  .  B | .  .  A | .  .  .
------------------------------

But the following is true (TH-1 Lemma):
Any single missing cell in TH sees one digit by row, a different digit by column and the third digit by box.

(proof omitted, it's not hard but interesting to think about)

If found that helpful for certain things, for example here

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---------------------------------
|           123 |      123      |               
|      123      |           123 |               
| 123           | 123           |               
---------------------------------
| 123           | 12            |               
|      12       |               |               
|           123 |           12  |               
---------------------------------

follows 3r1c5

or here

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---------------------------------
|           123 |           12  |             
|      123      |      123      |             
| 123           | 123           |             
---------------------------------
|           123 | 124           |             
|      123      |      124      |             
| 12            |           124 |             
---------------------------------


follows -4r6c6

[marek stefanik]

Yes, I made a mistake. It only works if the guardian cell is part of the rectangle.
Thank you for catching it. I'll edit my previous post.

Marek

[eleven]

Nice observation with the remote triples.

I have generated now 680 non equivalent 3-digit patterns in 6 boxes (10 cells: 31, 12: 38, 13: 290, 14: 159, 15: 102, 16: 10). This should be all, if i made no mistake.
I didn't see one more, which i would suspect to be as hard as the TH or the mentioned 15 cell pattern.

Added: i have copied a zip file to my old google drive: here (list). As patterns here (wait a bit, until it is shown). [Edit: updated link to patterns (10 cells were missing)]
Added: the latter link did not work anymore (?). I made a new copy 3digit6boxesAsPatterns.zip.

[ryokousha]

I'm still exploring the topic in various directions, maybe there will be a bigger update in the future.
Until then, some interesting things that came up on the way:

The trivalue oddagon / Thor's Hammer can, with help of extra constraints, be realized in rows, columns or even disjoint groups instead of boxes.
Here's a row version using both diagonals (Sudoku X):

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  .  .  X | X  .  . | X  .  .
  .  X  . | .  X  . | .  X  .
  .  .  . | .  .  . | .  .  .
------------------------------
  .  .  . | .  .  . | .  .  .
  .  .  . | .  .  . | .  .  .
  .  .  . | .  .  . | .  .  .
------------------------------
  .  .  . | .  .  . | .  .  .
  .  .  X | X  .  . | X  .  .
  X  .  . | .  X  . | .  .  X
------------------------------


This is an anti-knight version using 5 columns instead of the 4 boxes:

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------------------------------
  X  .  . | .  .  . | .  X  .
  .  .  . | X  .  X | .  .  .
  .  .  X | .  .  . | .  .  .
------------------------------
  .  .  X | .  .  . | .  .  .
  .  .  . | X  .  X | .  .  .
  X  .  . | .  .  . | .  X  .
------------------------------
  X  .  . | .  .  . | .  X  .
  .  .  . | .  .  X | .  .  .
  .  .  X | X  .  . | .  .  .
------------------------------


With diagonals or anti-knight, 3 row or column versions can be made. But these have an extra triangle and are therefore less complex to prove not 3-colorable.

This is a version with 4 triangles in disjoint groups, marked A,B,C,D for better visibility:

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------------------------------
  A  .  . | .  B  . | .  .  .
  .  .  D | .  .  . | .  C  .
  .  .  . | .  .  . | .  .  .
------------------------------
  .  B  . | .  .  . | A  .  .
  .  C  . | .  .  . | .  .  D
  .  .  . | .  .  . | .  .  .
------------------------------
  A  .  . | .  .  . | .  B  .
  .  .  D | .  C  . | .  .  .
  .  .  . | .  .  . | .  .  .
------------------------------


Aside from Thor's Hammer here is a - maybe - promising pattern that just came up yesterday:
(classic)

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------------------------------
  .  X  . | X  .  . | X  .  .
  X  .  . | .  .  . | .  .  .
  .  .  . | .  .  . | .  .  .
------------------------------
  X  .  . | .  X  . | X  .  .
  .  .  . | X  .  . | .  .  .
  .  .  . | .  .  . | .  .  .
------------------------------
  X  .  . | X  .  . | .  X  .
  .  .  . | .  .  . | X  .  .
  .  .  . | .  .  . | .  .  .
------------------------------


It is not completely trivial to prove, needing the use of graph symmetry or a bifurcation. Ultimately it relies on the combinatorial fact that any 3-coloring of a 6-cycle produces at least one opposing pair of edges with the same two colors. The 6-cycle in question being the single cells in b236478.

[ryokousha]

Two interesting new patterns appeared while toying around:

The first one I initially thought being in T&E(3), but marek showed on the CTC discord that it is likely not. The proof is still quite symmetric and interesting:

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------------------------------
  .  .  . | .  .  . | .  .  .
  .  .  . | .  .  . | .  .  .
  .  .  X | .  .  X | .  .  X
------------------------------
  .  .  X | .  .  . | X  .  .
  .  .  . | .  .  . | .  X  .
  .  .  . | .  .  . | .  .  X
------------------------------
  .  .  . | X  .  . | .  .  X
  .  .  X | .  X  . | .  X  .
  .  .  . | .  .  X | X  .  .
------------------------------

Hidden Text: Show

r8c3 has two possible arrangements in boxes 8 and 9 - either r9c6/r7c9 or r7c4/r9c7. The first one does not work as the value would be excluded from r3. The same argument can be made for r3c9, b6, b9 and c3 to show that r3c9 also goes in r9c7, so r8c3=r3c9. This quickly reveals a 9-loop bivalue oddagon.

(explanation thanks to marek and Youri on the discord)

The second pattern is interesting because it is - as far as I can tell - the first nontrivial example having a 3-chromatic subgraph induced by the occupied cells, but is 4-chromatic considering the whole sudoku graph.


"concerningly short sausage roll" [Edit: typo corrected]

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  .  X  . | X  .  . | X  .  .
  X  .  . | .  X  . | .  X  .
  .  .  . | .  .  . | .  .  .
------------------------------
  X  .  . | .  .  . | X  .  .
  .  X  . | .  .  . | .  X  .
  .  .  . | .  .  . | .  .  .
------------------------------
  X  .  . | X  .  . | .  X  .
  .  X  . | .  X  . | X  .  .
  .  .  . | .  .  . | .  .  .
------------------------------


Hidden Text: Show

We color the digit pairs in the boxes with three colors - think of assigning each missing digit to a color. B2 and b8 have opposite permutation parities which means they must be the same color. The same holds for b4 and b6. Considering the possible colorings of b1379, the b2/b8 color and the b4/b6 color must be the same. This assigns two different values to r6c6.

Marek gave an elegant explanation in more conventional terms on discord:

Hidden Text: Show

16 cells with 3 digits => at least one digit must appear in all six columns
Either that digit appears in r1c4, r8c5, r4c7, and r5c2 or in r7c4, r2c5, r4c1, and r5c8.
WLOG (diagonal symmetry) let it be 1 in r1c4, r8c5, r4c7, and r5c2.
x-wing 1r27 \ c18
x-wing 23r18 \ c27
1 takes one diagonal of its x-wing, the other is forced to contain twice the same digit, eliminating it from r7c4, r2c5, r4c1, and r5c8
The remaining digit must therefore take all four of them.
Both 1 and that digit are then forced into r6c6 in b5, ie. contra.

[denis_berthier]

.
The first pattern can be proven contradictory in T&E(3, BRT) but not in T&E(2, BRT).
The second pattern can't be proven contradictory in T&E(3, BRT).
Marek's proofs may prove that the patterns are contradictory; but they don't allow any conclusion about the T&E-depth necessary for the proof.

Do you have any real puzzle with any of these patterns?