## Gurth's Puzzles

Everything about Sudoku that doesn't fit in one of the other sections
Four computer made samples of ravel's symmetry (or is this gurth's "tenfold symmetry"?), the first three are minimal. ST and singles.
Code: Select all
`1 . .|. . .|. 9 8. . .|. . .|. . .. 5 6|. . .|4 . .-----+-----+-----. . 2|. 7 .|. 5 .. 7 .|. 4 .|2 . 9. 6 .|5 . 1|. . 4-----+-----+-----. 3 .|. . 7|. . 6. . 7|. . 4|3 8 .. . 5|6 1 .|. 4 . ER 8.4`

Code: Select all
`7 . .|4 . .|. 9 8. . .|. . .|. . .. 3 2|. . .|4 . .-----+-----+-----. . 5|. 4 .|. 7 .2 . .|. 1 .|5 . .. 8 .|5 . 2|. . 3-----+-----+-----. 6 .|. . 4|. . 73 . .|. . 1|6 . .. . 9|6 3 .|. 2 . ER 9.2`

Code: Select all
`7 . .|1 . .|. 8 9. . .|. . .|. . .. 2 3|. . .|7 . .-----+-----+-----. . 9|. 3 .|. 6 .6 . .|. 1 .|5 . .. 4 .|. . 6|. . 2-----+-----+-----. 8 .|. . 2|. . 55 . .|. . 1|6 . .. . 4|. 5 .|. 3 . ER 9.2`

This is not minimal and hard without ST:
Code: Select all
`1 . .|. . .|. 3 2. . .|1 . .|4 . .. 9 8|. . .|. . .-----+-----+-----. . 1|. 7 .|. 6 .9 . .|. 1 .|2 . .. 2 .|8 . 5|. . 4-----+-----+-----. 1 .|. . 7|. . 58 . .|. . 1|3 . .. . 3|9 6 .|. 4 . ER 9.8`
Mauricio

Posts: 1174
Joined: 22 March 2006

Oh, i hoped so, that you would find some good ones, thanks
ravel

Posts: 998
Joined: 21 February 2006

Mauricio wrote:
gurth wrote:But that is not at present enough to prove the solution has DDS. Not to my understanding, until someone proves it rigorously. I wish to claim AS A CONJECTURE that if every clue has at least ONE DDS partner (ie relative to either one of the two diagonals), then that is enough to ensure that any unique solution has DDS.

I propose that this CONJECTURE be allowed as a valid solution technique, provided it DOES lead to a DDS solution, until such time as my CONJECTURE be disproved by counter-example.
.

...That conjecture must now be discarded, having been disproved. Nice counter-examples, Mauricio!

re "ravel's symmetry", I doubt whether that is my tenfold symmetry. Try as I may, I can't find any symmetry in it at all. Maybe one day someone will enlighten me.
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

To summarize, what i learned from Red Ed's classes for symmetry techniques:
Only the 26 transformations (or equivalents to them) with >0 invariant grids are possible candidates for grid automorphisms (but there are grids with more than one of these automorphisms).

The most interesting for ST are those, where one or more cells are fixed by the transformation/automorphism. When you know that a cell is fixed, it only can hold a number, that maps to itself, therefore all other candidates (with a cyclic mapping) can be eliminated.

The ones with fixed cells we already know now are (i added the class number):
37 diagonal symmetry (9 cells fixed on the diagonal)
86 90 degree symmetry (1 fixed in the center)
79 180 degree rotational symmetry (1 fixed in the center)
The "new" one is represented by class 134. Fixed are 3 minicolums in the 3 boxes of a band (or 3 minirows in the 3 boxes of a stack). The transformation can be described as: swap 2 bands and in each stack 2 columns. Let me call it "sticks" symmetry here.

E.g. exchanging bands 2 and 3 and columns 2 and 3, 5 and 6, 8 and 9 is a transformation, that maps
r123c1, r123c4 and r123c7 to themselves
r123c2 to r123c3 and vice versa, same for r123c4/r123c5 and r123c8/r123c9
r456c1 to r789c1 and vice versa, same for r456c4/r789c4 and r456c7/r789c7
r456c2 to r789c3 and vice versa, same for r456c3/r789c2, r456c5/r789c6, r456c6/r789c5, r456c8/r789c9, r456c9/r789c8

There are no other transformations under the 26 with a fixed cell.

7 chunk symmetry (e.g. one stack maps to the next)
28 box symmetry (box1->b6->b8, b2->b9->b4 and b3->b5->b7)
AFAIK there are no really ST's for them, but each time, when a "normal" technique allows you to assign/eliminate 1 number, you can do the same for 2 more (or generally for as much as the cycle length is for this cell, e.g 4 for the non center cells with 90 degree symmetry). But i am not sure, if not some Emerald techniques (symm. UR, turbot etc.) can be applied also here. Any samples?

Not for all of the 26 classes there can be found sudoku grids with such an automorphism. I only looked at class 40 closer, where it was rather easy to see, that there cant be such a grid.
ravel

Posts: 998
Joined: 21 February 2006

ravel wrote:The "new" one is represented by class 134. Fixed are 3 minicolums in the 3 boxes of a band (or 3 minirows in the 3 boxes of a stack). The transformation can be described as: swap 2 bands and in each stack 2 columns. Let me call it "sticks" symmetry here.

Graphically:
Code: Select all
`abcabcabcabcabcabcabcabcabcdefdefdefdefdefdefdefdefdefghighighighighighighighighi`

is transformed to (r[123789456]c[132465798]):
Code: Select all
`acbacbacbacbacbacbacbacbacbgihgihgihgihgihgihgihgihgihdfedfedfedfedfedfedfedfedfe`

But to make it nicely symmetrical I propose the following isomorphic way:
Code: Select all
`ihgihgihgihgihgihgihgihgihgfedfedfedfedfedfedfedfedfedcbacbacbacbacbacbacbacbacba`

i.e. r[789456123]c[321654987]...
Or: swap bands [13], columns [13|46|79]...
udosuk

Posts: 2698
Joined: 17 July 2005

### Sticks Symmetry

ravel,
Thanks for the explanation of the "Sticks Symetry" and transformation/automorphism symmetries in general. Are there any other interesting "new" classes, apart from class 134?

Sticks Symmetry is a really great symmetry! And Mauricio's 4 puzzles on it a delight to solve. I did them all: irresistible fun. Yours too.

The ER 9.8 puzzle is demolished in minutes (or seconds for 90-second aces), with no computer. The puzzle raters wll have to look sharp! It makes NO sense to ignore symmetry techniques, on the argument that examples are unlikely to occur. With Mauricio around, I wouldn't bank on that!
______________________________________________________________________________
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

### Re: Sticks Symmetry

gurth wrote:Are there any other interesting "new" classes, apart from class 134?
As i said, there is no other class under the possible 26 with invariant grids, that has a transformation with fixed cells, so we already know all 4 existing.
I suppose, that the transformations with 3-cycles - where grids exist with them - all provide similar properties as the chunk or box symmetries, but with different patterns.
Those with 9-cycles i think are of no worth for symmetry techniques. I have tried one, but i could not drop more than 4 of the cycles in a grid without getting multiple solutions.
And i assume, that it is not much better with the transformations, that contain 6-cycles, but i did not try.
ravel

Posts: 998
Joined: 21 February 2006

Four more computer generated minimal sudokus.

Code: Select all
`. . .|. . .|. 8 9. . .|. 6 5|7 . .. . .|1 . .|. . .-----+-----+-----8 . 4|5 . .|3 . .. 5 .|. . 1|. . 2. 7 .|. 3 .|. 4 .-----+-----+-----9 4 .|6 . .|2 . .. . 6|. 1 .|. 3 .. . 7|. . 2|. . 4 ER 9.7`

Code: Select all
`1 . .|. . .|. . .. . .|4 . .|. 6 5. . .|. 9 8|. . .-----+-----+-----2 . 7|. 3 .|9 . .. 5 .|2 . .|. . 3. 1 .|. . 7|. 4 .-----+-----+-----3 7 .|. . 2|8 . .. . 6|3 . .|. 2 .. . 1|. 7 .|. . 4 ER 10.0`

This is not so simple after ST:
Code: Select all
`. . .|7 . .|. 2 3. . .|. . .|4 . .. . .|. 8 9|. . .-----+-----+-----5 . 8|2 . .|. 1 .. 3 .|. . 5|. . 6. 4 .|. 7 .|. . .-----+-----+-----6 9 .|3 . .|. . 1. . 2|. 6 .|. 5 .. . 4|. . 7|. . . ER 9.9, 6.6 after ST`

This does not have symmetry right away, after singles it does:
Code: Select all
`7 . .|4 . .|. 5 6. . .|. . .|. . .. . .|. 8 9|7 . .-----+-----+-----2 . 5|9 . .|. . .. 8 .|. . 7|5 . .. 1 .|. 4 .|. . 8-----+-----+-----3 6 .|8 . .|. . 7. . 9|. 7 .|6 . .. . 1|. . 4|. 9 . ER 9.8, singles then ST then singles`
Mauricio

Posts: 1174
Joined: 22 March 2006

Mauricio,

thanks for the new puzzles (is it a kind of Maurice Ravel production ? )

In the 3rd sample i think you also only need singles.
Code: Select all
` *--------------------------------------------------------------------* | 1489   1568   1569   | 7      145    146    | 15689  2      3      | | 1789   15678  15679  | 156    23     23     | 4      6789   5789   | | 147    2      3      | 1456   8      9      | 1567   67     57     | |----------------------+----------------------+----------------------| | 5      67     8      | 2      349    346    | 379    1      479    | | 1279   3      179    | 1489   149    5      | 2789   4789   6      | | 129    4      169    | 1689   7      1368   | 23589  389    2589   | |----------------------+----------------------+----------------------| | 6      9      57     | 3      245    248    | 278    478    1      | | 1378   178    2      | 1489   6      148    | 3789   5      4789   | | 138    158    4      | 1589   1259   7      | 23689  3689   289    | *--------------------------------------------------------------------*`
You can see from the symmetry, that 23, 56 and 89 are the pairs of exchanging numbers, where 147 are fixed. Then r3c4 has to be 4, r2c4=1, r1c7=1, r3c7=7, r3c1=1, r1c1=4 and r2c1=7. The rest is singles.
ravel

Posts: 998
Joined: 21 February 2006

ravel wrote:In the 3rd sample i think you also only need singles.
Code: Select all
` *--------------------------------------------------------------------* | 1489   1568   1569   | 7      145    146    | 15689  2      3      | | 1789   15678  15679  | 156    23     23     | 4      6789   5789   | | 147    2      3      | 1456   8      9      | 1567   67     57     | |----------------------+----------------------+----------------------| | 5      67     8      | 2      349    346    | 379    1      479    | | 1279   3      179    | 1489   149    5      | 2789   4789   6      | | 129    4      169    | 1689   7      1368   | 23589  389    2589   | |----------------------+----------------------+----------------------| | 6      9      57     | 3      245    248    | 278    478    1      | | 1378   178    2      | 1489   6      148    | 3789   5      4789   | | 138    158    4      | 1589   1259   7      | 23689  3689   289    | *--------------------------------------------------------------------*`
You can see from the symmetry, that 23, 56 and 89 are the pairs of exchanging numbers, where 147 are fixed. Then r3c4 has to be 4, r2c4=1, r1c7=1, r3c7=7, r3c1=1, r1c1=4 and r2c1=7. The rest is singles.

... Until you reach the following state:
Code: Select all
` *--------------------------------------------------* | 4    8    9    | 7    5    6    | 1    2    3    | | 7    5    6    | 1    23   23   | 4    89   89   | | 1    2    3    | 4    8    9    | 7    6    5    | |----------------+----------------+----------------| | 5    6    8    | 2    349  34   | 39   1    7    | | 29   3    7    | 89   1    5    | 289  4    6    | | 29   4    1    | 6    7    38   | 5    389  289  | |----------------+----------------+----------------| | 6    9    5    | 3    24   248  | 28   7    1    | | 38   7    2    | 89   6    1    | 389  5    4    | | 38   1    4    | 5    29   7    | 6    389  289  | *--------------------------------------------------*`

Then you need something more powerful than singles...
udosuk

Posts: 2698
Joined: 17 July 2005

udosuk wrote:Then you need something more powerful than singles...
Ah yes, e.g. the unique loop in r247c56, which gives r4c5=9 and r7c6=8.
ravel

Posts: 998
Joined: 21 February 2006

ravel wrote:Ah yes, e.g. the unique loop in r247c56, which gives r4c5=9 and r7c6=8.

... Or the turbot fish/skyscraper on 8s, using the strong links r5c47 & r7c67 to eliminate the 8 in r8c4...
udosuk

Posts: 2698
Joined: 17 July 2005

udosuk wrote:

"... Or the turbot fish/skyscraper on 8s, using the strong links r5c47 & r7c67 to eliminate the 8 in r8c4... "

Or the turbot fish/skyscraper on 9s, using the strong links d57 and h47 to eliminate 9k5 and 9e4.
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

ST is buried in this thread and in Gurth's "Isomorphs and normalization thread", but the "isomorphs and normaliztion" thread is very obscure and vague.

I suggest that someone with some spare time (I am very busy now, I will have some time in April) that fully understands ST creates a new thread about ST.

That thread must contain a formal proof to ST, or at least a link to the "proof" I gave (it is a proof for a particular case). It must contain links to the most representative sudokus for each given symmetry.

Mauricio.
Mauricio

Posts: 1174
Joined: 22 March 2006

### ai12 : Amethyst Class 1

ai12 : SE 9.4, AC1, ns, Pearl 9.4, Turquoise 6.7

Code: Select all
` . 5 . . . . . 3 . . . . . . 1 . . 5 . . 3 . 5 . 8 . . . . 8 . 9 . . . 6 . 7 . . . 2 . . . 1 . . 4 . . . . . . . 9 . 8 . . 5 . . 2 . . . . 6 . . 4 8 . 7 . . . . 1`

- I started with tarek's World Champion Gold Nugget, fiddled around a bit, and managed to knock 2.5 points off its SE rating! Resulting in a new high of 9.4 for my own compositions.
I don't use a computer to generate my puzzles. I let tarek do it, then degrade his puzzle. So I pull myself up.
I get the rating from my Sudoku Explainer version 1.1. I think I read somewhere that the new version 1.2.1 gives higher ratings. Is that true?

This puzzle is not a diamond, as it has a backdoor cell. But one only: 5h3: this makes it an Amethyst Class 1.

A puzzle of level SE 9.4 is more often than not a diamond, but even a SE 9.9 puzzle can have a backdoor cell and thus fail to be a diamond.

ai12 easily yields to EBC under singles, since it is not a Sapphire.
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

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