I have noticed the existence of this topic for long but have never read through the text. Today is my first time browsing them, and as a competitive sudoku player, I have my things to share. There is so much thing so I would just tell a little bit now.

In competition any marking(or try&error, even try-without-error) is allowed, and the only criterium is the solving time. Personally my act is similar to RW's, but I would make some Pencilmarks, but limited to Double candidates, or cells with "unordinary" pencilmark deletions. I learned from grandmasters to mark intersections/pairs between cells, and summarized what deleting a candidate can make. Marking bivalue cells could help discovering (Naked Pairs), (Naked Singles after Intersection/..), (XY-Wing-like chains), or (ab-bc-ca triplets)..

Also I would keep the pencilmarks in bivalue cells ordered. If we have

12 23 13

in a minirow, I would mark

12 23 31,

such that two possibility of all these cells have two possibilities-- all left or all right. If any of their columns has the same property, then ordered bivalue can spread and eliminate some candidates.

This is like branching, and I think it suits competing.

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And I wonder what this topic should be about.When I first see this topic, instead of "How to solve without marking on the grid", I thought this topic is about "

How to discover(and use) informations other than pencilmarks" as pencilmark is not the only form of information. I think this is more open than what we discussed in the beginning.

In 999_Spring's

post, the 67-UR becomes the bottleneck of some of the puzzles, while I immediately see that UR is useless. There is a Sue De Coq, which is also what I saw.

SE:UR(with hidden triplet, 4.7, eliminate 3@r5c5)

SE sees the UR as a non-candidate information"r56c6 cannot both be 67", while r3c6's 67 directly shows the same. But it can only utilize such kind of information directly in a UR. And if we change the 67 pair to multivalue "half-ur"(like a 67-75-56) SE uses forcing chains. This confuses me as two problems, one is "

why only two-value URs count", and the other is "

Why only UR-form non-candidate information can be utilized".

Below is the puzzle (Simplified the UR part).

- Code: Select all
`+----------------+----------------+----------------+`

| 6 2 9 | 3 1 5 | 7 4 8 |

| 5 7 1 | 9 4 8 | 6 3 2 |

| 3 4 8 | 27 26 #67 | 1 5 9 |

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

| 8 5 4 | 12 23 139 | 39 6 7 |

| 2 9 *67 | 4 356 367 | 35 8 1 |

| 1 3 *67 | 57 8 679 | 59 2 4 |

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

| 9 8 2 | 15 35 13 | 4 7 6 |

| 4 1 5 | 6 7 2 | 8 9 3 |

| 7 6 3 | 8 9 4 | 2 1 5 |

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

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During years of my sudoku career I have used non-candidate methods uncountable number of times, while the first time I discover it was when drawing 6*6 jigsaw sudoku empty grids. I found that some shapes can directly determine equalities.

- Code: Select all
`o o `

o o q...--> o a . . o o --> a b

p o o o a o o o o o o o a b o o o o

These equalities can be used to construct jigsaw sudokus that JSudoku even fail to solve logically.

- Code: Select all
`a a a a b b`

c a a b b d

c c b b e d

c c e e e d

c e e f f d

f f f f d d with column 2=1~6

And the first time I used equation in an actual puzzle is like this(when I was 10 and haven't heard of this forum):

Original Puzzle:...........3.....1.8..9..7......6....7....8.....1.3.9.........679..5........4...3 (

edit:I cannot find the source anyway.. Bought a book and didn't find the puzzle)

- Code: Select all
`+-------+-------+-------+`

| . . 7 | . 3 . | 6 . 9 |

| 9 . 3 | . 6 . | . . 1 |

| 6 8 . | . 9 1 | 3 7 . |

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

| . . 9 | . . 6 | 1 3 . |

| 3 7 1 | . 2 . | 8 6 . |

| . 6 . | 1 . 3 | . 9 . |

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

| . 3 . | . 1 . | . . 6 |

| 7 9 6 | 3 5 2 | 4 1 8 |

| . . . | 6 4 . | . . 3 |

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

Above is the bottleneck of a 17-cell puzzle, by a UR we can get an 8 in r7c6 but it doesn't break the puzzle. Its bottleneck is a chain on 4(SE7.7).

And here's what I solved it:

According to r5 and c5, (suppose)a=45=r4c4=r5c9, then r3c3=a(hidden single in row), and r9c2=a(hidden single in column), a≠4, a=5.

Such ways can be used in multiple ways, it is hard to tell with a fixed pattern, as another example in a PK is:

Original puzzle: 689.2..3.2..7.9.......3...8.61..4...............8..39.8...4.......5.6..3.7..1.542 (source:oubk.com)

After singles

- Code: Select all
`+-------+-------+-------+`

| 6 8 9 | 4 2 . | . 3 . |

| 2 . . | 7 8 9 | . . . |

| . . . | 6 3 . | 9 2 8 |

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

| . 6 1 | . . 4 | . . . |

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

| . . . | 8 6 . |*3*9 1 |

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

| 8 . . | . 4 . | . . 9 |

| . . . | 5 . 6 | . . 3 |

| . 7 6 | . 1 8 | 5 4 2 |

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

r9c1=39, hidden single in b4 we have r5c2=r9c1=39, then we get an intersection on 2, so r6c6=7.

It also works on some 999999111/ 999991110-like puzzles. Partial grids solvable by this have various SE ratings up to 8.9 ...

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Still a bit too long.. For things not about specific grids I would post to

Coffee Bar: What I See Here before the new year.