HP03 Solving “hardest puzzles” continued

 

P4 EXOCET cousins

 

Sometimes a puzzle has nearly an EXOCET pattern, but fails to get it. A typical one is Platinum Blonde 

 

000000012000000003002300400001800005060070800000009000008500000900040500470006000

 

1||35678 34589 34679 |4679__ 5689__ 4578__ |679___ 1_____ 2____

2||15678 14589 4679_ |124679 125689 124578 |679___ 56789_ 3____

3||15678 1589_ 2____ |3_____ 15689_ 1578__ |4_____ 56789_ 6789_

 

4||237__ 2349_ 1____ |8_____ 236___ 234___ |23679_ 234679 5____

5||235__ 6____ 349__ |124___ 7_____ 12345_ |8_____ 2349__ 149__

6||23578 23458 347__ |1246__ 12356_ 9_____ |12367_ 23467_ 1467_

 

7||1236_ 123__ 8____ |5_____ 1239__ 1237__ |123679 234679 14679

8||9____ 123__ 36___ |127___ 4_____ 12378_ |5_____ 23678_ 1678_

9||4____ 7____ 5____ |129___ 12389_ 6_____ |1239__ 2389__ 189__

 

That puzzle has a promising pattern with a base r12c7 and a potential target r4c8 r6c9, but it does not work.

 

“ttt” has shown that nevertheless,  it works nearly as an EXOCET, which means that combining super candidates in r12c7 and AURs in band 1;  boxes 1 and 2, the puzzle collapses.

 

In fact, not seen at that time as far as I remember, that puzzle has a rank 0 “square” logic (p3 Property) based on rows 4;7 and columns 3;4.

 

That rank0 logic eliminates 17 candidates which means that combining both, we are very close to an elegant solution.

 

One should note that even after the rank0 has been applied, the EXOCET is not there.

Another point is that we have a “square rank 0 pattern” without the mini row. The potential base for the missing EXOCET plays the same role.

 

To come back to the main point,  when the EXOCET is not there, we still have highly favorable AUR’s potential patterns in the band/stack perpendicular to the “base”. This can be enough to give a nice way to crack the puzzle.

 

I am far from having found how to use in a proper way all AUR’s effects.

 

If we go back to the puzzle cola 199 studied in the rank 0 logic,

The potential for eliminations without AUR’s is

 

1: r2c5 r3c1 r4c7 r5c1 r9c5

2: r1c5 r4c9 r7c15

4: r1c357 r9c35

7: r1c35 r4c39 r67c5

And the cells where digits out floors are eliminated : r2c4 r4c28 r5c28 r8c4

 

If we eliminate permutations having AUR’s, the potential is increased by

 

2 :  r2c4 r4c5 r5c5 r6c19 r7c6 r8c4

7:  r1c4 r2c5 r4c2 r5c39 r6c6 r7c4 r8c5

 

Both rank 0 logic and “eliminations of super candidates” should show increased efficiency using AUR’s.

 

 

P5 SYMMETRY

 

The symmetry of given has been long discussed  in several places. For anybody not aware of that topic, I suggest to go first thru these 2 threads

 

About Red Ed's Sudoku symmetry group

Down Under Upside Down - a Sudoku puzzle

 

The symmetry of given is a tiny group but some puzzles are in the family of hardest.

 

Moreover, a symmetry of given is easy to detect if the puzzle is in the proper form, (but nearly impossible to find if the puzzles has been scrambled, except for the solver).

 

My solver handles the main symmetry patterns:

 

_ diagonal (and anti diagonal)

_ rotational 90°

_stick

_central

 

The patterns easing significantly the path are the first 3 ones. The central symmetry alone brings less simplification in the path.

 

Examples of each family will come later. Some are discussed in the threads pointed above.

My interest for symmetry came after Mauricio published that one here

 

How hard is really this sudoku?

 

+-------+-------+-------+

| . . . | . . 1 | . . 2 |

| . . 3 | . . . | . 4 . |

| . 5 . | . 6 . | 7 . . |

+-------+-------+-------+

| . . . | 8 . . | . 7 . |

| . . 7 | . . 3 | 8 . . |

| 9 . . | . 5 . | . . 1 |

+-------+-------+-------+

| . . 6 | . 8 . | 2 . . |

| . 4 . | 6 . . | . . 7 |

| 2 . . | . . 9 | . 6 . |

+-------+-------+-------+ ED=11.3, Shining Mirror  (unknown ER)

 

With the following comments

To solve it, I only need to suppose uniqueness and to use not more than hidden pair.

 

That puzzle is not solved by my solver not using the symmetry property. It has no SK loop, no EXOCET pattern and no rank 0 logic up to five floors.

 

Mauricio published in the next page of the same thread  that one.

 

 

+-------+-------+-------+

| . . . | . . 1 | . . 2 |

| . 1 . | . 3 . | . 4 . |

| . . 5 | 6 . . | . . . |

+-------+-------+-------+

| . . 2 | 7 . . | 8 . . |

| . 8 . | . . 4 | . 3 . |

| 1 . . | . 9 . | . . . |

+-------+-------+-------+

| . . . | 3 . . | 5 . . |

| . 9 . | . 8 . | . 1 . |

| 6 . . | . . . | . . 7 |

+-------+-------+-------+ ER=11.3/11.3/10.5

 

 Again, it has none of the properties shown earlier, but the solver can crack it not using the symmetry property. That puzzle is still “relatively” difficult after the symmetry property has been applied.

Some files including all puzzles having a symmetry property will be available for downloading later;