The New Sudoku Players' Forum

[Pat]

I don't have the means to sweep through a sudoku and identify the redundant clues before solving it

and its not a skill I'm trying to develop

manually?
nobody would even try.

i do hope gsf shows us how to do it with his software

[champagne]

Hi,

When I said "easy", I didn't mean "straight forward easy".
But it's still much much more easier than the hellish forcing chains/nets without considering the symmetry.

The key to solving with this symmetry is to focus on the "windmills" (groups of 4 cells forming a cycle).

For example, look at the windmill r2c7+r7c8+r8c3+r3c2. Then look at the cell r2c3. What eliminations can you make on this windmill?

...

Each of us has it's own way to crack puzzles. In that case, I considered several possibilities:

What does my solver: nothing exiting, a long way, mainly due to a lot of parallel roads,
Allan Barker Model: I was waiting for something quick and easy in such a situation, but nothing came.

Finally, I try a mix between the way my solver works in such a situation and what symmetry tells us. It's not exactly the wind mills of udosuk, but I guess not so far.

Here the map of candidates with a limited tagging of main strong links including effect of symmetry.

Code: Select all

6      4b78   1a78   |378     1A3f8  2      |3e4B78   5      9     
5      2      789    |3e789   4      679    |3678     1      3a67   
1A489  4789   3      |5       16p8   1679   |2        678o   4b67   
-------------------------------------------------------------------
3      67s8   26o78  |1       9      4      |5        2678N  267   
2B49   1      279    |6       5      8      |479      3      24B7 
489    46N89  5      |2       7      3      |468o9    689s   1     
-------------------------------------------------------------------
2b89   6o89   4      |3789    368w   5      |1        2679   23A67
1a89   3      1689   |789     2      1e679  |679      4      5     
7      5      1e269  |4       13A6   169    |3a69     2b69   8     

We will work on r2c79. r2c79 is an AC2/AAHS with six possible super candidates. Let's eliminate all supercandidates not having digit 3.

Code: Select all

(7&8)r2c79 not valid.
  (7&8)r2c79 => 'A', e , P (6r79c5)
  this leads to (6&9)in AC2/AAHS r7c45r8c4 then cell r8c6 should have both 1e7t.

(6&7)r2c79 <=> (8&9)r8c13 not valid   
  (6&7) => A,b,o,e => 8r4c2=>8r8c1;9r8c3 
                         =>7r45c3->8r1c3 - 8r1c4
                  => 1r8c6 - 7r7c6 = 7r78c4 - 7rc4
                  - 3r1c4    No more candidate in r1c4
(6&8)r2c79 <=> (6&8) r8c13 not valid  8r2c7 6r2c8 8r8c1 6r8c3 => A,b,e,O,
           =>8r4c1;6r8c6=>7r3c1;9r9c7=>7r4c2;9r6c8
           ==============>49r56c1;27r45c9=>2r5c3;4r5c7
           and now a conflict 6;8 in the cells r4c3,r4c8,r6c2

3r2c79 is now compulsory
we can eliminate 'e' 3r1c6,3r2c4,1r8c6,1r9c3


The 2 first eliminations of supercandidates do not require an intensive use of symmetry. The solver had a slightly longer path.
The third one is very hard to establish without symmetry effects.

I'll edit that post to give eliminations in an equivalent AIC's net form.

From what I see in the solver print, similar things can be done in other places.


champagne

EDIT AICs nets


<> X==(7&8)r2c79

Code: Select all

3r1c4 - 3r1c7 = 3r2c79 - X
  |
 7r1c4 - 7r78c4 = 7r8c6 - (1r8c6.e;3r1c7.e) = 3r2c79 - X
  |
 8r1c4 - 8r7c4 = 8r7c5 - 6r7c5
                          |
                         6r3c5 - 6r3c8 = 78r3c8 - X
                          |
                         6r9c5 - 3r9c5.A = 3r2c9.a - X   

<> X==(6&7)r2c79

Code: Select all

3r1c4 - 3r1c7 = 3r2c79 - X
  |
 7r1c4 - 7r78c4 = 7r8c6 - (1r8c6.e;3r1c7.e) = 3r2c79 - X
  |
 8r1c4 - 8r1c3
          |
        ( 1rc3a;3r1c3.a) - X
          |
         7r1c3 - 7r45c3 = 7r4c2 - 8r4c2
                                   |
                                  8r4c3 - (6r4c3.o;8r3c8.o) = 67r3c8 - X
                                   |
                                  8r4c8 - 8r3c8 = 67r3c8 - X

Last step
<>X=(6&8)r2c79 <=> (6&8) r8c13

from above, we have still valid

Code: Select all

3r1c4 - 3r1c7 = 3r2c79 - X
  |
 7r1c4 - 7r78c4 = 7r8c6 - (1r8c6.e;3r1c7.e) = 3r2c79 - X
  |
 (8r1c4;6r9c6) - 8r1c3
                   |
                (1rc3a;3r1c3.a) - X
                   |
                (7r1c3;9r9c7)

That I usually summarize as X =>(7r1c3;9r9c7)

Second piece we need

Code: Select all

                         (X)6r3c9 - 6r4c9
                                       |
X - 3r2c79 = 3r1c7 - 4r1c7 = 4r3c9 - 4r5c9
                                       |
                                    27r45c9;49r56c1

summarized as X => 27r45c9;49r56c1

And the final

Code: Select all

8r4c3 - 68r4c8r6c2 <= X 
 |
7r4c3 - 7r1c3 <= X
 |
6r4c3.o - X
 |
2r4c3 - 2r5c3 = 79r5c3 - X

[udosuk]

Looks very complicated. My current energy level doesn't allow me to spend the effort to fully learn the "full-tagging" or AIC notations, apologies! (Not sure my notations are much more understandable to the general public though?)

Anyway, will post my symmetry-based solution soon. But I'll probably do it on the following min-min () version:

Code: Select all

.....1..2
..1.3..4.
.3.4..56.
..4..71..
.7.3....8
5...2....
..8..3.1.
.4.8....7
9...6.8..

[champagne]

Anyway, will post my symmetry-based solution soon. But I'll probably do it on the following min-min () version:

Code: Select all

.....1..2
..1.3..4.
.3.4..56.
..4..71..
.7.3....8
5...2....
..8..3.1.
.4.8....7
9...6.8..

as far as I am concerned, additionnal comments would we welcome

champagne

[udosuk]

But I'll probably do it on the following min-min () version:

In case anyone is wondering, "min-min" here actually stands for "minimal and min-lex".

"Minimal" means that all given clues are essential for the validity of the puzzle - remove any one of them, the puzzle will no longer have a unique solution.

"Min-lex" means "minimum lexicographically" - i.e. the puzzle string of 81 chars (with "0" representing empty cells) for this isomorph of the puzzle is smaller than such string of any other isomorph of this very puzzle.

gsf's program can probably help you verify both of these are true for my puzzle, but I find I enjoy more using a solver (JSudoku) and do all the checking/shuffling manually.

So, the next step is, we have this puzzle. Without seeing any symmetry explicitly, how would a Sudoku Master go about tackling it? Stay tuned for the next episode.