The New Sudoku Players' Forum

[Sojourner9]

I have only looked upon the discussions of puzzles and minimum clues and puzzle symmetries, etc. from afar.

But my uninvolved opinion has been that I don't like using minLex for puzzles. I can see the attraction if all you want is a simple way of getting a puzzle.
The reason I dislike minLex puzzles is because they result in random reorderings of the minLex solution grid.
How anyone could look at minLex puzzles and try to do any serious analysis is not clear to me.
But that is my bias, I understand it, and I own it.

That being said, I have been analyzing the structure of Sudoku grids since 2018 and have never used minLex nor looked at puzzles.

So recently I decided to look at the 49158 17 clue puzzles, which are in puzzle minLex order.

To do this, I read them in and used JCZSolver() solver to get the solution grid in whatever random reordering it currently occupied.
JCZSolver() is likely the first software I've used that I had not written myself. I had written solvers before but didn't feel like writing another and I heard it was quick.

Then I wrote a minLex routine for grids which lead me to writing a minLex routine for bands and stacks to pick a winner for band 1.
With a little intelligence bands can be minLexed in 108 steps for M0 and 72 steps for non-M0, instead of the 7776 steps for a brute force band minLex.
The results for each band was a minLex 416 member. So the lowest 416 number would end up as band1.

I had not really understood what automorphisms were until after I had done my own Nmin enumeration (5472730538) and could confirm the numbers of automorphic grids.
So to me the automorphism count(AMC) is just the number of steps I do divided by the number of results I get back. So for bands 7776/3888 = AMC 2, or in my case 72/36 or 108/54.

The trick I have been using is to take the result and store it in a list and a hash. Then the AMC = list.count/hash.count. So I have been applying this trick on all the new code I write.
When I was doing the band minLex I was getting AMC numbers. I had always known that many of the 416 had reduced counts but did not understand why.
When I saw this I ask why this was happening and where the extra results were going and I had an epiphany. The hash was a coset, which implies a subgroup.
So each of the 416 has an associated subgroup as does all the automorphic grids.

So when I get the minLex for the 6 bands and stacks I can pick a winner unless there are duplicate minimum 416 values, in which case I let them duke it out.
Here's a Truth. The smaller the object the higher the odds that it will be automorphic. When we extend the object we reduce the odds of automorphism.
If I use a hash and store the reorders as the values I can get both reorders that causes a AMC of 2 and the distance between them is the same as the subgroup distance.
But this takes time and memory. If all I do is keep track of the minimum value from the 7776 gyrations then I can get both reorders by using the subgroup.

When I get a winner for band1 then all the columns have been set and symbols relabeled, so all I have to do is reorder rows in band 2 and 3 to get column 1 in minimum order.
But if band 1 is automorphic I have to try all the elements of the subgroup and then work on band 2 and 3 again, keeping track of the string for the grid and keep the minimum.

Then knowing the reorder map to get to grid minLex, I can use it to navigate the puzzle into its grid minLex version.
Then I can group the puzzles that have the same solution grid and sort them.

This allows me to look at the Strangely Familiar 29 or 20, etc.

Now to my point.

There are six of the 17 clue puzzles that lead to AMC 2 solution grids 12498, 18679, 23847, 34844, 35709, and 35716.
When we reorder the puzzles we end up with two solutions for each.

Code: Select all

12498 123456789456789231789132546268573914374961825915824673537618492691247358842395167 count 2
12498 .2.4.......6....3....1..5.........14....6....9.5.2..........4.2.........8..39....
12498 1......8......92..7...3....26.5.....................73.3.6.8....9.............1.7

18679 123456789456789123798231564261894357534627891987513246315968472642175938879342615 count 2
18679 ...4...8...6...1...........2..8.............1..7.1.2.6....6..7..4...5.3.8........
18679 ..34.........8.1................4...5.....8....75.3.4..1..6...2....7..3.8........

23847 123456789456789123789123465238574691671892354945361872392647518517938246864215937 count 2
23847 ..3.5.7............8.1..........4..1...8.2..49.5............5...1...8.........93.
23847 .2...67.............91...........6....1.9..........8.2....4..1....93..4.86.......

34844 123456789456789132789213456245698371638147925971532864364975218517824693892361547 count 2
34844 ...4....9..6.8....7..2...5........7..3.1...........86.....75....1......3....6....
34844 ...4....9..6.8...27......5........7..3.1...........86.....75....1......3....6....

35709 123456789456789132789213456248695371635147928971832564364978215592361847817524693 count 4
35709 ...4...89.56......7..2............7..3.1...........56.....78.......6.....1......3
35716 ...4...89.56.....27...............7..3.1...........56.....78.......6.....1......3
35716 ...45...9..6......78.2............7..3.1...........56.....78.......6.....1......3
35709 ...45...9..6.....278..............7..3.1...........56.....78.......6.....1......3


The last two had the same solution grid.

So each AMC 2 grid has two puzzles that reduce to the same minLex puzzle and the same solution grid.

If I just look at the input puzzles and count the number of resultant grids I get this distribution.

Code: Select all

Total Grids = 46301
[1: 44,134, 2: 1,778, 3: 252, 4: 83, 5: 17, 6: 21, 7: 6, 8: 4, 9: 1, 11: 1, 12: 1, 14: 1, 20: 1, 29: 1] = 49,158


But if I look at the solution grids and count the 17 clue puzzles I get this distribution.

Code: Select all

Total Grids = 46301
[1: 44,130, 2: 1,781, 3: 252, 4: 84, 5: 17, 6: 21, 7: 6, 8: 4, 9: 1, 11: 1, 12: 1, 14: 1, 20: 1, 29: 1] = 49,164

Do I deny that these 5 grids their full count of 17 clue puzzles?
That their grids are automorphic should be irrelevant.
I think the second distribution is more accurate, with maybe an asterisk saying 4 grids are AMC2 with 1 solution and 1 is AMC2 with 2 solutions.
Not sure if anyone has posted this distribution before.

Wade

[Serg]

Hi, Sojourner9!
Your observation that some Sudoku solution grids, containing 17-clue puzzles, are automorphic is new to me. Well observed! I checked those 5 solution grids, all of them are really automorphic (each grid has 2 automorphisms).

I think if one must answer - how many 17-clue puzzles does given solution grid contain, one should count essentially different puzzles only. Puzzles mutating to each other after applying grid automorphisms should be treated as essentially the same.

Serg

[coloin]

Indeed I assumed too that none of the grids containing a 17C puzzle were automorphic..

And the first pair of puzzles

Code: Select all

12498 123456789456789231789132546268573914374961825915824673537618492691247358842395167 count 2
12498 .2.4.......6....3....1..5.........14....6....9.5.2..........4.2.........8..39....
12498 1......8......92..7...3....26.5.....................73.3.6.8....9.............1.7

are mutually disjoint !!
and probably making the quadriga challenge somewhat easy.
the inverse 47 clues

Code: Select all

..3.567.945.78...1.89..2.46..8.739..3749.1825.1.8.46..5.7.1..9.6.1247358.42..5.6.


are all individually superfluos, so it would probably be possible to find another disjoint pair of puzzles in these clues.

[Leren]

Hi Serg,

I recall that you said that none of the solution grids was automorhic here and that consequently the number of absolutely different 17 clue puzzles was 59,923,509,000,929,280.

Does this new information affect that result a bit ?

Leren

[blue]

I recall that you said that none of the solution grids was automorhic here and that consequently the number of absolutely different 17 clue puzzles was 59,923,509,000,929,280.

Hi Leren,

It was the puzzles, not the solution grids, that didn't have automorphisms.

[blue]

Code: Select all

12498 123456789456789231789132546268573914374961825915824673537618492691247358842395167 count 2
12498 .2.4.......6....3....1..5.........14....6....9.5.2..........4.2.........8..39....
12498 1......8......92..7...3....26.5.....................73.3.6.8....9.............1.7

18679 123456789456789123798231564261894357534627891987513246315968472642175938879342615 count 2
18679 ...4...8...6...1...........2..8.............1..7.1.2.6....6..7..4...5.3.8........
18679 ..34.........8.1................4...5.....8....75.3.4..1..6...2....7..3.8........

23847 123456789456789123789123465238574691671892354945361872392647518517938246864215937 count 2
23847 ..3.5.7............8.1..........4..1...8.2..49.5............5...1...8.........93.
23847 .2...67.............91...........6....1.9..........8.2....4..1....93..4.86.......

34844 123456789456789132789213456245698371638147925971532864364975218517824693892361547 count 2
34844 ...4....9..6.8....7..2...5........7..3.1...........86.....75....1......3....6....
34844 ...4....9..6.8...27......5........7..3.1...........86.....75....1......3....6....

35709 123456789456789132789213456248695371635147928971832564364978215592361847817524693 count 4
35709 ...4...89.56......7..2............7..3.1...........56.....78.......6.....1......3
35716 ...4...89.56.....27...............7..3.1...........56.....78.......6.....1......3
35716 ...45...9..6......78.2............7..3.1...........56.....78.......6.....1......3
35709 ...45...9..6.....278..............7..3.1...........56.....78.......6.....1......3


Hi Sojourner9,

I'm curious: Why is the 2nd grid not in minlex form, but the others are ?
Not a bug I hope. It seems unlikely, if you came up with the correct ED grid count.

With a little intelligence bands can be minLexed in 108 steps for M0 and 72 steps for non-M0

P.S.: What does it mean ... "M0"-vs-"non-M0" ?

[Sojourner9]

Blue,

I have a lot of new code and I rechecked puzzle 18679 and the result looks good to me. Could you give me the correct minLex for that puzzle.
The puzzle minLex puzzle and solution are:

Code: Select all

..............1..2..3....4.......2....4.5.....2...61.5....4.....1.....6.7..3...8.
285964317946731852173528946651873294394152678827496135538649721419287563762315489

P.S.: What does it mean ... "M0"-vs-"non-M0" ?

What I call M0 is a band that does not have any displaced symbols. This is all the minLex band 1s that have 456 in r2c123.
Non-M0 are all the minLex bands that have 457 in r2c123.

Stack 1 in puzzle 1 is an M0:

Code: Select all

        984723156
        561489372
        327615849


We can see that r1c123, r2c456 and r3c789 contain the same symbols. As does r2c123, r3c456, and r1c789, and r3c123, r1c456, and r2c789.

Stack 2 in puzzle 1 is a non-M0:

Code: Select all

        179862534
        635947281
        842351967


We can see that for the same places we only have two symbols that are the same.
There are three displaced symbols, 1, 4, and 6, that are displaced in each box.

[blue]

Hi Sojourner9,

The for #18679, the minlex solution and the two puzzles are:

Code: Select all

123456789456789123798132465219867354567394812834521697381975246675248931942613578
..3..........8.1..7......6..19..7.........8.2....2.6...8....2....5...........3.7.
.2............91..7.......5..9...3...6...........2...7.81.7..........93......35..

Stack2 in the solution to the minlex puzzle, makes the best band1.

Thanks for the explanation on "M0"-vs-"non-M0".
I guessed that M0 meant "repeating" mini-rows, but I didn't understand why you said 72, instead of 36 operations for non-M0 bands.
It seems like it should be 36, since only 2 of the 6 ways to permute columns in stack 1, will have the r2c12 numbers appearing in the same stack in row 1.
[ Added: Equivalently, only 2 of the 6 ways to permute columns in stack 2, will have the r1c45 numbers appearing in the same mini-row in box 1. ]

It's all a neat idea, and one that I haven't seen exposed before. Thank you.

Cheers,
Blue.

P.S.: I don't know if you know, and maybe it isn't important, but the list of minlex 17's that you're using, has the wrong minlex form for some of the puzzles.
See this thread: 17 Clue Puzzles Apparently Not in Row-Based Minlex Form.

[Sojourner9]

Hi blue,
Sorry for the delay. I had to go back through my program to find the bug and now I am in alignment with you answer.
Problem sorting the old bad gangOf416 after finding the proper minLex for the bands.

I guessed that M0 meant "repeating" mini-rows, but I didn't understand why you said 72, instead of 36 operations for non-M0 bands.
It seems like it should be 36, since only 2 of the 6 ways to permute columns in stack 1, will have the r2c12 numbers appearing in the same stack in row 1.
[ Added: Equivalently, only 2 of the 6 ways to permute columns in stack 2, will have the r1c45 numbers appearing in the same mini-row in box 1. ]

I call them boxLines because they are the intersection of a box and a line(row or column).

I am not sure I understand what you are saying. Permuting the columns in stack 1 will always keep all the symbols in the same rows in stack 1.
Since you are talking about r2c12 and r1c45 I can assume you are presenting this as a band, which only make sense.
Since you don't care which symbol is first there are 6 pick 2 ways to set r2c12.
But having picked one there is only 6 pick 1 ways of filling r1c45.
The same could be said if you did stack 2 first.
So unless you have any correction in my interpretation of what your saying, I think I understand.

As far as the accounting for the number of operations, here is my method.

For both types of bands I start by setting the parity which makes sure r2c123 and r1c456 have the most in common.
This is done by swapping row 2 and 3, if necessary. I don't consider this as an operation as it is done once and save time throughout.

Then for both types of bands I permuting the boxes, while maintaining parity, and rotating the rows, (6 x 3) operations.
Permuting the boxes involves swapping two boxes six times and for each swap, I swap row 2 and 3 again to maintain parity.
This gives each row boxLine an opportunity to be in the r1c123 position and to see B2/B3 in both positions.

I think Red Ed rotated the boxes but had to compare and swap B2 and B3 at the inner most level.
This is why he had 36288 candidates, I have 72576 candidates and may explain the difference.

Now for the intelligent steps.
For non-M0:
I swap c123 to get the displaced symbol into r2c3 and I swap c789 to get the same displaced symbol into r1c7.
I swap c456 to set the non-displaced symbols into r1c45 and in the same order as r2c12.
After relabeling this will always result in minLex for c367.
What you said, I think.

The rest of the dumb operations involve permuting c12 and c45 in tandem and permuting col89, so (2 x 2) operations.
Multiply the numbers in brackets together and you get 6 x 3 x 2 x 2 = 72

For M0:
I swap c456 to get r1c456 into the same order as r2c123. After relabeling this will always result in minLex.
I swap c789 to get r1c789 in the same order as r2c456, and for the same reason.

The rest of the dumb operations involves permuting c123, c456, and c789 in tandem, so (6) operations.
Multiply the numbers in brackets together and you get 6 x 3 x 6 = 108.

And of course at the bottom level there is relabeling and comparing to find the minimum representative.

P.S.: I don't know if you know, and maybe it isn't important, but the list of minlex 17's that you're using, has the wrong minlex form for some of the puzzles.
See this thread: 17 Clue Puzzles Apparently Not in Row-Based Minlex Form.

If this has to do with puzzle minLex, I does not effect me as long as each puzzle is unique. Is the a corrected list?

I should have realized that you and others would not be using brute force to find band minLex. You are using index416.c I assume.
My code isn't optimized for speed as of yet.
I am curious how you get minLex on the grid, once you pick a winning band?

[Sojourner9]

Here is the solution grid and the 29 17 clue puzzles in band minLex order

Code: Select all

14913 123456789456789123798231564234675918815943276967812435379164852582397641641528397 count 29
45609 ..3..........8.1.....2..........5...81..4...........35.7....8....23.7......5...9.
30785 ..3..........8.1.....2.........7....81..4...........35.7....8....23........5...97
45526 ..3..........8.1.....2.........7....81..4...........35.7....8....23........5.8.9.
31638 ..3..........8.1.....2.........7....81..4...........35.7....8....23.....6..5....7
30781 ..3..........8.1.....2.........7....81..4...........35.7....8....23..6.....5....7
45528 ..3..........8.1.....2.........7....81..4...........35.7....8....23.7......5...9.
32006 ..3..........8.1.....2.........7....81..4...........35.7....8....239.......5....7
30790 ..3..........8.1.....2.........7....81..4...........35.7....8...823........5...9.
32035 ..3..........8.1.....2.........7....81..4...........35.7..6.8....23........5....7
30845 ..3..........8.1.....2.........7....81..4...........35.79...8....23........5....7
30787 ..3..........8.1.....2.........7....81..4.....6.....35.7....8....23........5....7
31637 ..3..........8.1.....2.........7....81..4....9......35.7....8....23........5....7
30782 ..3..........8.1.....2.........7....81..4...6.......35.7....8....23........5....7
38859 ..3..........8.1.....2.........7....81..4...6.......35.7....8....23........5.8...
38858 ..3..........8.1.....2.........7....81..4...6.......35.7....8....23.7......5.....
30791 ..3..........8.1.....2.........7....81..4...6.......35.7....8...823........5.....
31969 ..3..........8.1.....2.........7....81.94...........35.7....8....23........5....7
30780 ..3..........8.1.....2.........7.9..81..4...........35.7....8....23........5....7
31974 ..3..........8.1.....2........67....81..4...........35.7....8....23........5....7
30784 ..3..........8.1.....2...6.....7....81..4...........35.7....8....23........5....7
30789 ..3..........8.1...9.2.........7....81..4...........35.7....8....23........5....7
46749 ..3..........891.....2.........7....81..4...........35.7....8....23........5....7
30800 ..3........6.8.1.....2.........7....81..4...........35.7....8....23........5....7
30783 ..3.....9....8.1.....2.........7....81..4...........35.7....8....23........5....7
14913 ..3...7......8.1.....2..............81..4...........35.7....8....23.7......5...9.
14921 ..3...7......8.1.....2..............81..4...........35.7....8....239.......5....7
16399 ..3...7......8.1.....2..............81..4...6.......35.7....8....23.7......5.....
48690 ..3..6.......8.1.....2.........7....81..4...........35.7....8....23........5....7
14914 ..3.5........8.1.....2..............81..4...........35.7....8....23.7......5...9.


Has this been posted before?
There are 13 fixed symbols and 4 symbols that move between puzzles.
I'll not comment on what the fixed symbols are doing until I have a chance to analyze the puzzle.
Unless someone else wants to give it a crack?

[JPF]

Has this been posted before?

Yes. here

Edit:
actually there are 14 common digits:

Code: Select all

:
+---+---+---+
|..3|...|...|
|...|.8.|1..|
|...|2..|...|
+---+---+---+
|...|...|...|
|81.|.4.|...|
|...|...|.35|
+---+---+---+
|.7.|...|8..|
|..2|3..|...|
|...|5..|...|
+---+---+---+


JPF

[champagne]

Here is the solution grid and the 29 17 clue puzzles in band minLex order

I'll not comment on what the fixed symbols are doing until I have a chance to analyze the puzzle.
Unless someone else wants to give it a crack?

IMO, the best way to answer to this question is to think in terms of unavoidable sets.

The simpler case is a solution grid with 2 ED 17 clues differing by one digit in the same cell.
Usually, this means that all unavoidable sets (with no subset) are hit by the 16 clues except one unavoidable set of 2 digits (usually of size 4).

[blue]

Hi Sojourner9,

For non-M0:
I swap c123 to get the displaced symbol into r2c3 and I swap c789 to get the same displaced symbol into r1c7.
I swap c456 to set the non-displaced symbols into r1c45 and in the same order as r2c12.
After relabeling this will always result in minLex for c367.
What you said, I think.

The rest of the dumb operations involve permuting c12 and c45 in tandem and permuting col89, so (2 x 2) operations.
Multiply the numbers in brackets together and you get 6 x 3 x 2 x 2 = 72

After the c456 order is set ("c12 and c45 in tandem"), you don't actually need to try both "col89" options.
Only the one that has the r1c89 numbers in the same relative order as they appear in r2c456, can produce a minlex band.

P.S.: I don't know if you know, and maybe it isn't important, but the list of minlex 17's that you're using, has the wrong minlex form for some of the puzzles.
See this thread: 17 Clue Puzzles Apparently Not in Row-Based Minlex Form.

If this has to do with puzzle minLex, I does not effect me as long as each puzzle is unique. Is the a corrected list?

I would have thought so, but I can't seem to find a link to one.
Probably I shouldn't have mentioned it.

I should have realized that you and others would not be using brute force to find band minLex. You are using index416.c I assume.
My code isn't optimized for speed as of yet.
I am curious how you get minLex on the grid, once you pick a winning band?

In effect:

Preliminaries ...
1) set up a "best case" (result) grid that r4c1 set to 10, the best band result in band 1, and garbage everywhere else.
2) look up the "one of 416" index for the best case band 1, and use it to look up a list of band 1 automorphisms
The automorphism list should include the identity transformation.

Then the main work is ...
1) for each band/stack having the best type, transform the grid to put it in the band 1 position and in minlex form.
2) loop over the band 1 automorphisms, and extend each one to include the ordering for r456789 that puts (r456789)c1 in minlex order
3) starting with r4c1, going cell by cell: calculate the cell value in the transformed grid

if it's < the value in a "best case" grid, apply the entire transformation and store the result as the new "best case".
if it's > the value in a "best case" grid, move on to the next automorphism.
if it matches, move on to the next cell.

In the cell loop, you can skip column 9 and row 9, since if c1-8 match, c9 must as well, and if r1-8 match, r9 must match.

If you like, you can count automorphisms for the full grid, by resetting a counter to 1, each time the best case grid changes, and incrementing it each time the cell loop completes with all cells matching the best case.

[champagne]

Here is the solution grid and the 29 17 clue puzzles in band minLex order

I'll not comment on what the fixed symbols are doing until I have a chance to analyze the puzzle.
Unless someone else wants to give it a crack?

IMO, the best way to answer to this question is to think in terms of unavoidable sets.

The simpler case is a solution grid with 2 ED 17 clues differing by one digit in the same cell.
Usually, this means that all unavoidable sets (with no subset) are hit by the 16 clues except one unavoidable set of 2 digits (usually of size 4).

one partial example of this

take this subset of your list

Code: Select all

14913 123456789456789123798231564234675918815943276967812435379164852582397641641528397 count 29
30785 ..3..........8.1.....2.........7....81..4...........35.7....8....23........5...97
45528 ..3..........8.1.....2.........7....81..4...........35.7....8....23.7......5...9.
32006 ..3..........8.1.....2.........7....81..4...........35.7....8....239.......5....7
30845 ..3..........8.1.....2.........7....81..4...........35.79...8....23........5....7
31637 ..3..........8.1.....2.........7....81..4....9......35.7....8....23........5....7
31969 ..3..........8.1.....2.........7....81.94...........35.7....8....23........5....7
30780 ..3..........8.1.....2.........7.9..81..4...........35.7....8....23........5....7
30789 ..3..........8.1...9.2.........7....81..4...........35.7....8....23........5....7
46749 ..3..........891.....2.........7....81..4...........35.7....8....23........5....7
30783 ..3.....9....8.1.....2.........7....81..4...........35.7....8....23........5....7
14913 ..3...7......8.1.....2..............81..4...........35.7....8....23.7......5...9.
14921 ..3...7......8.1.....2..............81..4...........35.7....8....239.......5....7

all these 17s differ by one of the digits 7 or 9

the location of 7,9 in the grid is

Code: Select all

... ... 7.9
... 7.9 ...
79. ... ...

... .7. 9..
... 9.. .7.
9.7 ... ...

.79... ...
....97 ...
...... .97

in fact 3 disjoint unavoidable sets of 6 cells, one per stack.

If you apply the brute force to one of these 17 reduced to 16, erasing one of the digit 7/9, you should find 2 solutions differing by the corresponding unavoidable set.

[Sojourner9]

Hi Blue,
Sorry again for the long delay, I had a huge git commit needing done.

Only the one that has the r1c89 numbers in the same relative order as they appear in r2c456, can produce a minlex band.

This was subtle and I did not see that. Thanks.
Can we make as a general statement, without having to go off and confirm this, that all non-M0 minLex bands have 7189 starting in r2c3?

I am used to thinking in terms of the displaced and non-displaced symbols.
I know after some intelligent reordering of columns a displaced symbols is in r2c3, r1c6 and r1c7 as previously described.
That leaves two cases, one of r1c89 is in r2c6 or one of r1c89 is not in r2c6.
Assuming the general statement is true and one of r1c89 is not in r2c6,
then this r1c123 candidate (1 of 18) will not contribute to the minLex and can be guarded against this before we do the loop.
With this guard in place, we know c9 and thus c8. We also know c4 and c5 and then c1 and c2.
So no need for loops at all. This becomes deterministic. So instead of (36) it is (18). This seem right?
I checked and other rowRotates will contribute so this check is done for each of the 18.

Probably I shouldn't have mentioned it.

No I am glad you did. It is hard to find all the interesting topics without following a link. So the more links the better.
I did find time to look at this. Is there an improved puzzle minLex code now?
Also from the link in another post, all of Nmin has been searched and the once and for all total is 49158, and you were a major contributor? If so nice work.
That was one of the main reasons I was going to look at 17 clue puzzles was to see if I could give hints as to where to look. No need now.

Preliminaries ...
1) set up a "best case" (result) grid that r4c1 set to 10, the best band result in band 1, and garbage everywhere else.
2) look up the "one of 416" index for the best case band 1, and use it to look up a list of band 1 automorphisms
The automorphism list should include the identity transformation.

Then the main work is ...
1) for each band/stack having the best type, transform the grid to put it in the band 1 position and in minlex form.
2) loop over the band 1 automorphisms, and extend each one to include the ordering for r456789 that puts (r456789)c1 in minlex order
3) starting with r4c1, going cell by cell: calculate the cell value in the transformed grid

if it's < the value in a "best case" grid, apply the entire transformation and store the result as the new "best case".
if it's > the value in a "best case" grid, move on to the next automorphism.
if it matches, move on to the next cell.

In the cell loop, you can skip column 9 and row 9, since if c1-8 match, c9 must as well, and if r1-8 match, r9 must match.

If you like, you can count automorphisms for the full grid, by resetting a counter to 1, each time the best case grid changes, and incrementing it each time the cell loop completes with all cells matching the best case.

I am not sure what setting r4c1 to 10 does but I assume it is specific to you code and methodology.
I like that you mentioned the automorphisms and including the identity. I would be interested in seeing what a list of automorphisms looks like, even if it is just in a PM.
Starting with 3, for each best type and full grid and automorphism and minLex order which includes relabeling the whole grid, you know the order of the columns and the relabeling of bands 2 and 3.
I just sort the rows in band 2 using r456c1 and sort the rows in band 3 r789c1 and then sort band 2 and 3 using r3c1 and r7c1.

I don't think I would know if the solution grid was automorphic by rotation for example.
All I know is when swapping the columns and relabeling, results in the same band2 and band3.